III Algebras (Full)

Part III — Algebras

Based on lectures by C. J. B. Brookes

Notes taken by Dexter Chua

Lent 2017

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

The aim of the course is to give an introduction to algebras. The emphasis will be

on non-commutative examples that arise in representation theory (of groups and Lie

algebras) and the theory of algebraic D-modules, though you will learn something

ab out commutative algebras in passing.

Topics we discuss include:

–

Artinian algebras. Examples, group algebras of finite groups, crossed products.

Structure theory. Artin–Wedderburn theorem. Projective modules. Blocks.

–

No etherian algebras. Examples, quantum plane and quantum torus, differen-

tial operator algebras, enveloping algebras of finite dimensional Lie algebras.

Structure theory. Injective hulls, uniform dimension and Goldie’s theorem.

–

Hochschild chain and cochain complexes. Hochschild homology and cohomology.

Gerstenhab er algebras.

– Deformation of algebras.

– Coalgebras, bialgebras and Hopf algebras.

Pre-requisites

It will be assumed that you have attended a first course on ring theory, eg IB Groups,

Rings and Modules. Experience of other algebraic courses such as II Representation

Theory, Galois Theory or Number Fields, or III Lie algebras will be helpful but not

necessary.

Contents

0 Introduction

1 Artinian algebras

1.1 Artinian algebras

1.2 Artin–Wedderburn theorem

1.3 Crossed products

1.4 Projectives and blocks

1.5 K

2 Noetherian algebras

2.1 Noetherian algebras

2.2 More on A

(k) and U(g)

2.3 Injective modules and Goldie’s theorem

3 Hochschild homology and cohomology

3.1 Introduction

3.2 Cohomology

3.3 Star products

3.4 Gerstenhaber algebra

3.5 Hochschild homology

4 Coalgebras, bialgebras and Hopf algebras

0 Introduction

We start with the definition of an algebra. Throughout the course,

will be a

field.

Definition

(

-algebra)

A (unital) associative

-algebra is a

-vector space

together with a linear map

A ⊗ A → A

, called the product map, and linear

map u : k → A, called the unit map, such that

– The product induced by m is associative.

– u(1) is the identity of the multiplication.

In particular, we don’t require the product to be commutative. We usually

write m(x ⊗ y) as xy.

Example.

Let

K/k

be a finite field extension. Then

is a (commutative)

k-algebra.

Example.

The

n × n

matrices

(

) over

form a non-commutative

-algebra.

Example.

The quaternions

is an

-algebra, with an

-basis 1

, i, j, k

, and

multiplication given by

= j

= k

= 1, ij = k, ji = −k.

This is in fact a division algebra (or skew fields), i.e. the non-zero elements have

multiplicative inverse.

Example. Let G be a finite group. Then the group algebra

kG =

g : g ∈ G, λ

∈ k

with the obvious multiplication induced by the group operation is a k-algebra.

These are the associative algebras underlying the representation theory of

finite groups.

Most of the time, we will just care about algebras that are finite-dimensional

-vector spaces. However, often what we need for the proofs to work is not

that the algebra is finite-dimensional, but just that it is Artinian. These algebras

are defined by some finiteness condition on the ideals.

Definition

(Ideal)

A left ideal of

is a

-subspace of

such that if

x ∈ A

and

y ∈ I

, then

xy ∈ I

. A right ideal is one where we require

yx ∈ I

instead.

An ideal is something that is both a left ideal and a right ideal.

Since the multiplication is not necessarily commutative, we have to make the

distinction between left and right things. Most of the time, we just talk about

the left case, as the other case is entirely analogous.

The definition we want is the following:

Definition

(Artinian algebra)

An algebra

is left Artinian if it satisfies the

descending chain condition (DCC ) on left ideals, i.e. if we have a descending

chain of left ideals

≥ I

≥ · · · ,

then there is some N such that I

N+m

= I

for all m ≥ 0.

We say an algebra is Artinian if it is both left and right Artinian.

Example. Any finite-dimensional algebra is Artinian.

The main classification theorem for Artinian algebras we will prove is the

following result:

Theorem

(Artin–Wedderburn theorem)

Let

be a left-Artinian algebra such

that the intersection of the maximal left ideals is zero. Then

is the direct sum

of finitely many matrix algebras over division algebras.

When we actually get to the theorem, we will rewrite this in a way that

seems a bit more natural.

One familiar application of this theorem is in representation theory. The group

algebra of a finite group is finite-dimensional, and in particular Artinian. We

will later see that Maschke’s theorem is equivalent to saying that the hypothesis

of the theorem holds. So this puts a very strong constraint on how the group

algebra looks like.

After studying Artinian rings, we’ll talk about Noetherian algebras.

Definition

(Noetherian algebra)

An algebra is left Noetherian if it satisfies

the ascending chain condition (ACC ) on left ideals, i.e. if

≤ I

≤ · · ·

is an ascending chain of left ideals, then there is some

such that

N+m

for all m ≥ 0.

Similarly, we can define right Noetherian algebras, and say an algebra is

Noetherian if it is both left and right Noetherian.

We can again look at some examples.

Example. Again all finite-dimensional algebras are Noetherian.

Example.

In the commutative case, Hilbert’s basis theorem tells us a polynomial

algebra

[

, · · · , X

] in finitely many variables is Noetherian. Similarly, the

power series rings k[[X

, · · · , X

]] are Noetherian.

Example.

The universal enveloping algebra of a finite-dimensional Lie algebra

are the (associative!) algebras that underpin the representation theory of these

Lie algebras.

Example.

Some differential operator algebras are Noetherian. We assume

char k

= 0. Consider the polynomial ring

[

]. We have operators “multipli-

cation by

” and “differentiate with respect to

” on

[

]. We can form the

algebra

[

∂

∂x

] of differential operators on

[

], with multiplication given by

the composition of operators. This is called the Weyl algebra

. We will show

that this is a non-commutative Noetherian algebra.

Example.

Some group algebras are Noetherian. Clearly all group algebras of

finite groups are Noetherian, but the group algebras of certain infinite groups

are Noetherian. For example, we can take

G =











1 λ µ

0 1 ν

0 0 0





: λ, µ, ν ∈ Z







and this is Noetherian. However, we shall not prove this.

We will see that all left Artinian algebras are left Noetherian. While there is

a general theory of non-commutative Noetherian algebras, it is not as useful as

the theory of commutative Noetherian algebras.

In the commutative case, we often look at

Spec A

, the set of prime ideals of

. However, sometimes in the non-commutative there are few prime ideals, and

so Spec is not going to keep us busy.

Example. In the Weyl algebra A

, the only prime ideals are 0 and A

We will prove a theorem of Goldie:

Theorem

(Goldie’s theorem)

Let

be a right Noetherian algebra with no

non-zero ideals all of whose elements are nilpotent. Then

embeds in a finite

direct sum of matrix algebras over division algebras.

Some types of Noetherian algebras can be thought of as non-commutative

polynomial algebras and non-commutative power series, i.e. they are deformations

of the analogous commutative algebra. For example, we say

is a deformation

of the polynomial algebra

[

X, Y

], where instead of having

XY − Y X

= 0, we

have

XY − Y X

= 1. This also applies to enveloping algebras and Iwasawa

algebras. We will study when one can deform the multiplication so that it remains

associative, and this is bound up with the cohomology theory of associative

algebras — Hochschild cohomology. The Hochschild complex has rich algebraic

structure, and this will allow us to understand how we can deform the algebra.

At the end, we shall quickly talk about bialgebras and Hopf algebras. In a

bialgebra, one also has a comultiplication map

A → A⊗A

, which in representation

theory is crucial in saying how to regard a tensor product of two representations

as a representation.

1 Artinian algebras

1.1 Artinian algebras

We continue writing down some definitions. We already defined left and right

Artinian algebras in the introduction. Most examples we’ll meet are in fact

finite-dimensional vector spaces over

. However, there exists some more perverse

examples:

Example. Let

A =



r s

0 t



: r ∈ Q, s, t ∈ R



Then this is right Artinian but not left Artinian over

. To see it is not left

Artinian, note that there is an ideal

I =



0 s

0 0



: s ∈ R



∼

, and a matrix



r s

0 t



acts on this on the left by sending



0 s

0 0



. Since

is an infinite-dimensional

-vector space, one sees that there is an

infinite strictly descending chain of ideals contained in I.

The fact that it is right Artinian is a direct verification. Indeed, it is not

difficult to enumerate all the right ideals, which is left as an exercise for the

reader.

As in the case of commutative algebra, we can study the modules of an

algebra.

Definition

(Module)

Let

be an algebra. A left

-module is a

-vector space

M and a bilinear map

A ⊗ M M

a ⊗ m xm

such that (

)

(

) for all

a, b ∈ A

and

m ∈ M

. Right

-modules are

defined similarly.

-bimodule is a vector space

that is both a left

-module and a

right

-module, such that the two actions commute — for

a, b ∈ A

and

x ∈ M

we have

a(xb) = (ax)b.

Example.

The algebra

itself is a left

-module. We write this as

, and

call this the left regular representation. Similarly, the right action is denoted

These two actions are compatible by associativity, so A is an A-A-bimodule.

If we write

End

(

) for the

-linear maps

A → A

, then

End

is naturally a

algebra by composition, and we have a

-algebra homomorphism

A → End

(

)

that sends

a ∈ A

to multiplication by

on the left. However, if we want

to multiply on the right instead, it is no longer a

-algebra homomorphism

A → End

(A). Instead, it is a map A → End

(A)

, where

Definition

(Opposite algebra)

Let

be a

-algebra. We define the opposite

algebra

to be the algebra with the same underlying vector space, but with

multiplication given by

x · y = yx.

Here on the left we have the multiplication in

and on the right we have the

multiplication in A.

In general, a left A-module is a right A

-module.

As in the case of ring theory, we can talk about prime ideals. However, we

will adopt a slightly different definition:

Definition

(Prime ideal)

An ideal

is prime if it is a proper ideal, and if

and J are ideals with IJ ⊆ P , then either I ⊆ P or J ⊆ P .

It is an exercise to check that this coincides in the commutative case with

the definition using elements.

Definition

(Annihilator)

Let

be a left

-module and

m ∈ M

. We define

the annihilators to be

Ann(m) = {a ∈ A : am = 0}

Ann(M) = {a ∈ A : am = 0 for all m ∈ M} =

m∈M

Ann(m).

Note that

Ann

(

) is a left ideal of

, and is in fact the kernel of the

-module

homomorphism

A → M

given by

x 7→ xm

. We’ll denote the image of this map

by Am, a left submodule of M, and we have

Ann(m)

∼

Am.

On the other hand, it is easy to see that

Ann

(

) is an fact a (two-sided) ideal.

Definition

(Simple module)

A non-zero module

is simple or irreducible if

the only submodules of M are 0 and M.

It is easy to see that

Proposition.

Let

be an algebra and

a left ideal. Then

is a maximal left

ideal iff A/I is simple.

Example. Ann(m) is a maximal left ideal iff Am is irreducible.

Proposition.

Let

be an algebra and

a simple module. Then

∼

A/I

for some (maximal) left ideal I of A.

Proof.

Pick an arbitrary element

m ∈ M

, and define the

-module homomor-

phism

A → M

(

) =

. Then the image is a non-trivial submodule,

and hence must be

. Then by the first isomorphism theorem, we have

∼

A/ ker ϕ.

Before we start doing anything, we note the following convenient lemma:

Lemma.

Let

be a finitely-generated

module. Then

has a maximal

proper submodule M

Proof.

Let

, · · · , m

∈ M

be a minimal generating set. Then in particular

, · · · , m

k−1

is a proper submodule of

. Moreover, a submodule

containing

is proper iff it does not contain

, and this property is

preserved under increasing unions. So by Zorn’s lemma, there is a maximal

proper submodule.

Definition

(Jacobson radical)

The

(

) of

is the intersection of all maximal

left ideals.

This is in fact an ideal, and not just a left one, because

J(A) =

{maximal left ideals} =

m∈M,M simple

Ann(m) =

M simple

Ann(M),

which we have established is an ideal. Yet, it is still not clear that this is

independent of us saying “left” instead of “right”. It, in fact, does not, and this

follows from the Nakayama lemma:

Lemma

(Nakayama lemma)

The following are equivalent for a left ideal

(i) I ≤ J(A).

(ii)

For any finitely-generated left

-module

, if

, then

= 0,

where

is the module generated by elements of the form

, with

a ∈ I

and m ∈ M.

(iii) G = {1 + a : a ∈ I} = 1 + I is a subgroup of the unit group of A.

In particular, this shows that the Jacobson radical is the largest ideal satisfying

(iii), which is something that does not depend on handedness.

Proof.

–

(i)

⇒

(ii): Suppose

I ≤ J

(

) and

M 6

= 0 is a finitely-generated

-module,

and we’ll see that IM  M.

Let

be a maximal submodule of

. Then

M/N

is a simple module,

so for any

¯m ∈ M/N

, we know

Ann

(

¯m

) is a maximal left ideal. So

J(A) ≤ Ann(M/N ). So IM ≤ J(A)M ≤ N  M.

–

(ii)

⇒

(iii): Assume (ii). We let

x ∈ I

and set

= 1 +

. Hence

1 =

y − x ∈ Ay

. Since

is a left ideal, we know

. In

other words, we know





Now using (ii) on the finitely-generated module

A/Ay

(it is in fact generated

by 1), we know that

A/Ay

= 0. So

. So there exists

z ∈ A

such

that 1 =

(1 +

). So (1 +

) has a left inverse, and this left inverse

lies in

, since we can write

= 1

− zx

. So

is a subgroup of the unit

group of A.

–

(iii)

⇒

(i): Suppose

is a maximal left ideal of

. Let

x ∈ I

. If

x 6∈ I

then

by maximality of

. So 1 =

for some

y ∈ I

and

z ∈ A

. So

= 1

− zx ∈ G

. So

is invertible. But

y ∈ I

. So

This is a contradiction. So we found that

I < I

, and this is true for all

maximal left ideals I

. Hence I ≤ J(A).

We now come to the important definition:

Definition (Semisimple algebra). An algebra is semisimple if J(A) = 0.

We will very soon see that for Artinian algebras, being semi-simple is equiva-

lent to a few other very nice properties, such as being completely reducible. For

now, we shall content ourselves with some examples.

Example. For any A, we know A/J(A) is always semisimple.

We can also define

Definition (Simple algebra). An algebra is simple if the only ideals are 0 and

It is trivially true that any simple algebra is semi-simple — the Jacobson

radical is an ideal, and it is not

. A particularly important example is the

following:

Example.

Consider

(

). We let

be the matrix with 1 in the (

i, i

)th

entry and zero everywhere else. This is idempotent, i.e.

. It is also

straightforward to check that

















0 · · · 0 a

0 · · · 0

0 · · · 0 a

0 · · · 0

0 · · · 0 a

0 · · · 0

















The non-zero column is, of course, the

th column. Similarly,

is the matrices

that are zero apart from in the

th row. These are in fact all left and all right

ideals respectively. So the only ideals are 0 and A.

As a left A-module, we can decompose A as

A =

i=1

which is a decomposition into simple modules.

Definition

(Completely reducible)

A module

is completely reducible

iff it is a sum of simple modules.

Here in the definition, we said “sum” instead of “direct sum”, but the

following proposition shows it doesn’t matter:

Proposition. Let M be an A-module. Then the following are equivalent:

(i) M is completely reducible.

(ii) M is the direct sum of simple modules.

(iii)

Every submodule of

has a complement, i.e. for any submodule

there is a complement N

such that M = N ⊕ N

Often, the last condition is the most useful in practice.

Proof.

– (i) ⇒ (ii): Let M be completely reducible, and consider the set

≤ M} : S

are simple,

is a direct sum

Notice this set is closed under increasing unions, since the property of

being a direct sum is only checked on finitely many elements. So by Zorn’s

lemma, it has a maximal element, and let N be the sum of the elements.

Suppose this were not all of

. Then there is some

S ≤ M

such that

S 6⊆ N

. Then

S ∩ N ( S

. By simplicity, they intersect trivially. So

is a direct sum, which is a contradiction. So we must have

, and

M is the direct sum of simple modules.

– (ii) ⇒ (i) is trivial.

– (i) ⇒ (iii): Let N ≤ M be a submodule, and consider

≤ M} : S

are simple, N +

is a direct sum

Again this set has a maximal element, and let

be the direct sum of those

. Again if

P ⊕ N

is not all of

, then pick an

S ≤ M

simple such that

is not contained in

P ⊕ N

. Then again

S ⊕ P ⊕ N

is a direct sum, which

is a contradiction.

–

(iii)

⇒

(i): It suffices to show that if

N < M

is a proper submodule,

then there exists a simple module that intersects

trivially. Indeed, we

can take

to be the sum of all simple submodules of

, and this forces

N = M .

To do so, pick an

x 6∈ N

, and let

be submodule of

maximal among

those satisfying

P ∩ N

= 0 and

x 6∈ N ⊕ P

. Then

N ⊕ P

is a proper

submodule of M. Let S be a complement. We claim S is simple.

If not, we can find a proper submodule

. Let

be a complement of

N ⊕ P ⊕ S

. Then we can write

M = N ⊕ P ⊕ S

⊕ Q

x = n + p + s + q

By assumption,

and

are not both zero. We wlog assume

is non-

zero. Then

P ⊕ Q

is a larger submodule satisfying (

P ⊕ Q

)

∩ N

= 0 and

x 6∈ N ⊕

(

P ⊕ Q

). This is a contradiction. So

is simple, and we are

done.

Using these different characterizations, we can prove that completely reducible

modules are closed under the familiar ex operations.

Proposition.

Sums, submodules and quotients of completely reducible modules

are completely reducible.

Proof.

It is clear by definition that sums of completely reducible modules are

completely reducible.

To see that submodules of completely reducible modules are completely

reducible, let

be completely reducible, and

N ≤ M

. Then for each

x ∈ N

there is some simple submodule

S ≤ M

containing

. Since

S ∩ N ≤ S

and

contains x, it must be S, i.e. S ⊆ N. So N is the sum of simple modules.

Finally, to see quotients are completely reducible, if

is completely reducible

and N is a submodule, then we can write

M = N ⊕ P

for some P . Then M/N

∼

P , and P is completely reducible.

We will show that every left Artinian algebra is completely reducible over

itself iff it is semi-simple. We can in fact prove a more general fact for

-modules.

To do so, we need a generalization of the Jacobson radical.

Definition

(Radical)

For a module

, we write

Rad

(

) for the intersection

of maximal submodules of M, and call it the radical of M.

Thus, we have Rad(

A) = J(A) = Rad(A

Proposition.

Let

be an

-module satisfying the descending chain condition

on submodules. Then M is completely reducible iff Rad(M) = 0.

Proof.

It is clear that if

is completely reducible, then

Rad

(

) = 0. Indeed,

we can write

M =

α∈A

where each S

is simple. Then

J(A) ≤

α∈A





β∈A\{α}





= {0}.

Conversely, if

Rad

(

) = 0, we note that since

satisfies the descending

chain condition on submodules, there must be a finite collection

, · · · , M

maximal submodules whose intersection vanish. Then consider the map

i=1

x (x + M

, x + M

, · · · , x + M

)

The kernel of this map is the intersection of the

, which is trivial. So this

embeds

as a submodule of

. But each

is simple, so

is a submodule

of a completely reducible module, hence completely reducible.

Corollary.

is a semi-simple left Artinian algebra, then

is completely

reducible.

Corollary.

is a semi-simple left Artinian algebra, then every left

-module

is completely reducible.

Proof.

Every

-module

is a quotient of sums of

. Explicitly, we have a

map

m∈M

A M

)

Then this map is clearly surjective, and thus M is a quotient of

is not semi-simple, then it turns out it is rather easy to figure out radical

of M , at least if M is finitely-generated.

Lemma.

Let

be left Artinian, and

a finitely generated left

-module, then

J(A)M = Rad(M ).

Proof.

Let

be a maximal submodule of

. Then

M/M

is simple, and is in

fact A/I for some maximal left ideal I. Then we have

J(A)





= 0,

since J(A) < I. Therefore J(A)M ≤ M

. So J(A)M ≤ Rad(M).

Conversely, we know

J(A)M

is an

A/J

(

)-module, and is hence completely

reducible as

A/J

(

) is semi-simple (and left Artinian). Since an

-submodule

J(A)M

is the same as an

A/J

(

)-submodule, it follows that it is completely

reducible as an A-module as well. So

Rad



J(A)M



= 0,

and hence Rad(M ) ≤ J(A)M.

Proposition. Let A be left Artinian. Then

(i) J(A) is nilpotent, i.e. there exists some r such that J(A)

= 0.

(ii)

is a finitely-generated left

-module, then it is both left Artinian and

left Noetherian.

(iii) A is left Noetherian.

Proof.

(i)

Since

is left-Artinian, and

(

)

r ∈ N}

is a descending chain of

ideals, it must eventually be constant. So

(

)

(

)

r+1

for some

. If

this is non-zero, then again using the descending chain condition, we see

there is a left ideal

with

(

)

I 6

= 0 that is minimal with this property

(one such ideal exists, say J(A) itself).

Now pick

x ∈ I

with

(

)

x 6

= 0. Since

(

)

(

)

, it follows

that

(

)

(

)

= 0. So by minimality,

(

)

x ≥ I

. But the other

inclusion clearly holds. So they are equal. So there exists some

a ∈ J

(

)

with x = ax. So

(1 − a)x = 0.

But 1 − a is a unit. So x = 0. This is a contradiction. So J(A)

= 0.

(ii)

Let

(

)

. Then

i+1

is annihilated by

(

), and hence

completely reducible (it is a module over semi-simple

A/J

(

)). Since

is a finitely generated left

-module for a left Artinian algebra, it satisfies

the descending chain condition for submodules (exercise), and hence so

does M

i+1

So we know

i+1

is a finite sum of simple modules, and therefore

satisfies the ascending chain condition. So

i+1

is left Noetherian,

and hence M is (exercise).

(iii) Follows from (ii) since A is a finitely-generated left A-module.

1.2 Artin–Wedderburn theorem

We are going to state the Artin–Wedderburn theorem for right (as opposed to

left) things, because this makes the notation easier for us.

Theorem

(Artin–Wedderburn theorem)

Let

be a semisimple right Artinian

algebra. Then

A =

i=1

for some division algebra D

, and these factors are uniquely determined.

A has exactly r isomorphism classes of simple (right) modules S

, and

End

) = {A-module homomorphisms S

→ S

}

∼

and

dim

) = n

If A is simple, then r = 1.

If we had the left version instead, then we need to insert op’s somewhere.

Artin–Wedderburn is an easy consequence of two trivial lemma. The key idea

that leads to Artin–Wedderburn is the observation that the map

→ End

(

)

sending

to left-multiplication by

is an isomorphism of algebras. So we need

to first understand endomorphism algebras, starting with Schur’s lemma.

Lemma

(Schur’s lemma)

Let

, M

be simple right

-modules. Then either

∼

, or

Hom

(

, M

) = 0. If

is a simple

-module, then

End

(

)

is a division algebra.

Proof.

A non-zero

-module homomorphism

→ M

must be injective, as

the kernel is submodule. Similarly, the image has to be the whole thing since

the image is a submodule. So this must be an isomorphism, and in particular

has an inverse. So the last part follows as well.

As mentioned, we are going to exploit the isomorphism

∼

End

(

This is easy to see directly, but we can prove a slightly more general result, for

the sake of it:

Lemma.

(i)

is a right

-module and

is an idempotent in

, i.e.

, then

∼

Hom

(eA, M ).

(ii) We have

eAe

∼

End

(eA).

In particular, we can take e = 1, and recover End

)

∼

Proof.

(i) We define maps

me (ex 7→ mex)

Me Hom(eA, M )

α(e) α

We note that

(

) =

(

) =

(

)

e ∈ Me

. So this is well-defined. By

inspection, these maps are inverse to each other. So we are done.

Note that we might worry that we have to pick representatives

and

for the map

, but in fact we can also write it as

(

)(

) =

, since

idempotent. So we are safe.

(ii) Immediate from above by putting M = eA.

Lemma. Let M be a completely reducible right A-module. We write

M =

where

}

are distinct simple

-modules. Write

End

(

), which we

already know is a division algebra. Then

End

)

∼

and

End

(M) =

)

Proof.

The result for

End

(

) is just the familiar fact that a homomorphism

→ S

is given by an

m × n

matrix of maps

S → S

(in the case of vector

spaces over a field

, we have

End

(

)

∼

, so they are matrices with entries in

k). Then by Schur’s lemma, we have

End

(M) =

End

)

∼

We now prove Artin–Wedderburn.

Proof of Artin–Wedderburn.

is semi-simple, then it is completely reducible

as a right A-module. So we have

∼

End(A

)

∼

We now decompose each

(

) into a sum of simple modules. We know each

(

) is a non-trivial

(

) module in the usual way, and the action of

the other summands is trivial. We can simply decompose each

(

) as the

sum of submodules of the form

















0 0 · · · 0 0

· · · a

−1

0 0 · · · 0 0

















and there are

components. We immediately see that if we write

for this

submodule, then we have

dim

) = n

Finally, we have to show that every simple module S of A is one of the S

. We

simply have to note that if

is a simple

-module, then there is a non-trivial

map

A → S

(say by picking

x ∈ S

and defining

(

) =

). Then in the

decomposition of

into a direct sum of simple submodules, there must be one

factor

such that

is non-trivial. Then by Schur’s lemma, this is in fact an

isomorphism S

∼

This was for semi-simple algebras. For a general right Artinian algebra, we

know that

A/J

(

) is semi-simple and inherits the Artinian property. Then

Artin–Wedderburn applies to A/J(A).

Some of us might be scared of division algebras. Sometimes, we can get away

with not talking about them. If

is not just Artinian, but finite-dimensional,

then so are the D

Now pick an arbitrary

x ∈ D

. Then the sub-algebra of

generated by

be commutative. So it is in fact a subfield, and finite dimensionality means it is

algebraic over

. Now if we assume that

is algebraically closed, then

must

live in k. So we’ve shown that these D

must be k itself. Thus we get

Corollary.

is algebraically closed and

is a finite-dimensional semi-simple

k-algebra, then

∼

(k).

This is true, for example, when k = C.

We shall end this section by applying our results to group algebras. Recall

the following definition:

Definition

(Group algebra)

Let

be a group and

a field. The group algebra

of G over k is

kG =

g : g ∈ G, λ

∈ k

This has a bilinear multiplication given by the obvious formula

(λ

g)(µ

h) = λ

(gh).

The first thing to note is that group algebras are almost always semi-simple.

Theorem

(Maschke’s theorem)

Let

be a finite group and

p - |G|

, where

p = char k, so that |G| is invertible in k, then kG is semi-simple.

Proof.

We show that any submodule

of a

-module

has a complement.

Let π : U → V be any k-vector space projection, and define a new map

|G|

g∈G

gπg

−1

: U → V.

It is easy to see that this is a

-module homomorphism

U → V

, and is a

projection. So we have

U = V ⊕ ker π

and this gives a kG-module complement.

There is a converse to Maschke’s theorem:

Theorem. Let G be finite and kG semi-simple. Then char k - |G|.

Proof.

We note that there is a simple

-module

, given by the trivial module.

This is a one-dimensional k vector space. We have

D = End

(S) = k.

Now suppose

is semi-simple. Then by Artin–Wedderburn, there must be

only one summand of S in kG.

Consider the following two ideals of kG: we let

g ∈ kG :

= 0

This is in fact a two-sided ideal of

. We also have the center of the algebra,

given by

g ∈ kG : λ ∈ k

Now if char k | |G|, then I

⊆ I

. So we can write

kG =

⊕ I

⊕ · · · .

But we know

acts trivially on

and

, and they both have dimension 1.

This gives a contradiction. So we must have char k - |G|.

We can do a bit more of representation theory. Recall that when

algebraically closed and has characteristic zero, then the number of simple

modules is the number of conjugacy classes of

. There is a more general result

for a general characteristic p field:

Theorem.

Let

be algebraically closed of characteristic

, and

be finite.

Then the number of simple

modules (up to isomorphism) is equal to the

number of conjugacy classes of elements of order not divisible by

. These are

known as the p-regular elements.

We immediately deduce that

Corollary.

|G|

for some

and

is prime, then the trivial module is the

only simple kG module, when char k = p.

Note that we can prove this directly rather than using the theorem, by

showing that

ker

(

kG → k

) is a nilpotent ideal, and annihilates all simple

modules.

Proof sketch of theorem.

The number of simple

modules is just the number

of simple

kG/J

(

) module, as

(

) acts trivially on every simple module.

There is a useful trick to figure out the number of simple

-modules for a given

semi-simple A. Suppose we have a decomposition

∼

i=1

(k).

Then we know

is the number of simple

-modules. We now consider [

A, A

the k-subspace generated by elements of the form xy − yx. Then we see that

[A, A]

∼

i=1

(k)

(k), M

(k)]

Now by linear algebra, we know [

(

)

, M

(

)] is the trace zero matrices, and

so we know

dim

(k)

(k), M

(k)]

= 1.

Hence we know

dim

[A, A]

= r.

Thus we need to compute

dim

kG/J(kG)

[kG/J(kG), kG/J(kG)]

We then note the following facts:

(i) For a general algebra A, we have

A/J(A)

[A/J(A), A/J(A)]

∼

[A, A] + J(A)

(ii) Let g

, · · · , g

be conjugacy class representatives of G. Then

+ [kG, kG]}

forms a k-vector space basis of kG/[kG, kG].

(iii)

, · · · , g

is a set of representatives of

-regular conjugacy classes, then



[kG, kG] + J(kG)

o

form a basis of kG/([kG, kG] + J(kG)).

Hence the result follows.

One may find it useful to note that [

kG, kG

] +

(

) consists of the elements

in kG such that x

∈ [kG, kG] for some s.

In this proof, we look at

[

A, A

]. However, in the usual proof of the result

in the characteristic zero looks at the center

(

). The relation between these

two objects is that the first is the 0th Hochschild homology group of

, while

the second is the 0th Hochschild cohomology group of A.

1.3 Crossed products

Number theorists are often interested in representations of Galois groups and

-modules where

is an algebraic number field, e.g.

. In this case, the

’s

appearing in Artin–Wedderburn may be non-commutative.

We have already met one case of a non-commutative division ring, namely

the quaternions H. This is in fact an example of a general construction.

Definition

(Crossed product)

The crossed product of a

-algebra

and a

group G is specified by the following data:

– A group homomorphism φ : G → Aut

(B), written

(λ) = λ

;

– A function

Ψ(g, h) : G × G → B.

The crossed product algebra has underlying set

g : λ

∈ B.

with operation defined by

λg · µh = λµ

Ψ(g, h)(gh).

The function Ψ is required to be such that the resulting product is associative.

We should think of the

as specifying what happens when we conjugate

pass µ, and then Ψ(g, h)(gh) is the product of g and h in the crossed product.

Usually, we take

, a Galois extension of

, and

Gal

(

K/k

). Then

the action

is the natural action of

on the elements of

, and we restrict to

maps Ψ : G × G → K

only.

Example.

Consider

, and

. Then

Gal

(

C/R

)

∼

{e, g}, where g is complex conjugation. The elements of H are of the form

e + λ

where λ

, λ

∈ C, and we will write

1 · g = g, i · g = k, 1 · e = 1, i · e = i.

Now we want to impose

−1 = j

= 1g · 1g = ψ(g, g)e.

So we set Ψ(

g, g

) =

−

1. We can similarly work out what we want the other

values of Ψ to be.

Note that in general, crossed products need not be division algebras.

The crossed product is not just a

-algebra. It has a natural structure of a

G-graded algebra, in the sense that we can write it as a direct sum

BG =

g∈G

Bg,

and we have Bg

· Bg

⊆ Bg

Focusing on the case where

K/k

is a Galois extension, we use the notation

(

K, G,

Ψ), where Ψ :

G × G → K

. Associativity of these crossed products is

equivalent to a 2-cocycle condition Ψ, which you will be asked to make precise

on the first example sheet.

Two crossed products (

K, G,

) and (

K, G,

) are isomorphic iff the map

G × G K

(g, h) Ψ

(g, h)(Ψ

(g, h))

−1

satisfies a 2-coboundary condition, which is again left for the first example sheet.

Therefore the second (group) cohomology

{2-cocycles : G × G → K

}

{2-coboundaries : G × G → K

}

determines the isomorphism classes of (associative) crossed products (K, G, Ψ).

Definition

(Central simple algebra)

A central simple

-algebra is a finite-

dimensional k-algebra which is a simple algebra, and with a center Z(A) = k.

Note that any simple algebra is a division algebra, say by Schur’s lemma.

So the center must be a field. Hence any simple

-algebra can be made into a

central simple algebra simply by enlarging the base field.

Example. M

(k) is a central simple algebra.

The point of talking about these is the following result:

Fact.

Any central simple

-algebra is of the form

(

) for some division

algebra

which is also a central simple

-algebra, and is a crossed product

(K, G, Ψ).

Note that when

and

, then the second cohomology group has

2 elements, and we get that the only central simple

-algebras are

(

) or

(H).

For amusement, we also note the following theorem:

Fact (Wedderburn). Every finite division algebra is a field.

1.4 Projectives and blocks

In general, if

is not semi-simple, then it is not possible to decompose

as a

direct sum of simple modules. However, what we can do is to decompose it as a

direct sum of indecomposable projectives.

We begin with the definition of a projective module.

Definition

(Projective module)

-module is projective

if given modules

M and M

and maps

M 0

then there exists a map

P → M

such that the following diagram commutes:

M 0

Equivalently, if we have a short exact sequence

0 N M

M 0,

then the sequence

0 Hom(P, N) Hom(P, M

) Hom(P, M) 0

is exact.

Note that we get exactness at

Hom

(

P, N

) and

Hom

(

P, M

) for any

at all.

Projective means it is also exact at Hom(P, M ).

Example. Free modules are always projective.

In general, projective modules are “like” free modules. We all know that free

modules are nice, and most of the time, when we want to prove things about

free modules, we are just using the property that they are projective. It is also

possible to understand projective modules in an algebro-geometric way — they

are “locally free” modules.

It is convenient to characterize projective modules as follows:

Lemma. The following are equivalent:

(i) P is projective.

(ii) Every surjective map φ : M → P splits, i.e.

∼

ker φ ⊕ N

where N

∼

P .

(iii) P is a direct summand of a free module.

Proof.

– (i) ⇒ (ii): Consider the following lifting problem:

M P 0

The lifting gives an embedding of

into

that complements

ker φ

(by

the splitting lemma, or by checking it directly).

–

(ii)

⇒

(iii): Every module admits a surjection from a free module (e.g. the

free module generated by the elements of P )

–

(iii)

⇒

(i): It suffices to show that direct summands of projectives are

projective. Suppose P is is projective, and

∼

A ⊕ B.

Then any diagram

M 0

can be extended to a diagram

A ⊕ B

M 0

˜α

by sending

to 0. Then since

A ⊕ B

∼

is projective, we obtain a lifting

A ⊕ B → M

, and restricting to A gives the desired lifting.

Our objective is to understand the direct summands of a general Artinian

-algebra

, not necessarily semi-simple. Since

is itself a free

-module, we

know these direct summands are always projective.

Since

is not necessarily semi-simple, it is in general impossible to decompose

it as a direct sum of simples. What we can do, though, is to decompose it as a

direct sum of indecomposable modules.

Definition

(Indecomposable)

A non-zero module

is indecomposable if

cannot be expressed as the direct sum of two non-zero submodules.

Note that since

is (left) Artinian, it can always be decomposed as a finite

sum of indecomposable (left) submodules. Sometimes, we are also interested in

decomposing A as a sum of (two-sided) ideals. These are called blocks.

Definition

(Block)

The blocks are the direct summands of

that are inde-

composable as ideals.

Example. If A is semi-simple Artinian, then Artin-Wedderburn tells us

A =

and the M

) are the blocks.

We already know that every Artinian module can be decomposed as a direct

sum of indecomposables. The first question to ask is whether this is unique. We

note the following definitions:

Definition

(Local algebra)

An algebra is local if it has a unique maximal left

ideal, which is J(A), which is the unique maximal right ideal.

If so, then

A/J

(

) is a division algebra. This name, of course, comes from

algebraic geometry (cf. local rings).

Definition

(Unique decomposition property)

A module

has the unique

decomposition property if

is a finite direct sum of indecomposable modules,

and if

M =

i=1

then n = m, and, after reordering, M

= M

We want to prove that

as an

-module always has the unique decomposi-

tion property. The first step is the following criterion for determining unique

decomposition property.

Theorem

(Krull–Schmidt theorem)

Suppose

is a finite sum of indecom-

posable

-modules

, with each

End

(

) local. Then

has the unique

decomposition property.

Proof. Let

M =

i=1

We prove by induction on

. If

= 1, then

is indecomposable. Then we

must have n = 1 as well, and the result follows.

For m > 1, we consider the maps

: M

M M

: M

M M

We observe that

i=1

◦ β

: M

→ M

Since

End

(

) is local, we know some

◦ β

must be invertible, i.e. a unit, as

they cannot all lie in the Jacobson radical. We may wlog assume

◦ β

is a

unit. If this is the case, then both

and

have to be invertible. So

∼

Consider the map id −θ = φ, where

θ : M M

i=2

−1

Then φ(M

) = M

. So φ|

looks like α

. Also

i=2

φ|

i=2

looks like the identity map. So in particular, we see that

surjective. However, if φ(x) = 0, this says x = θ(x), So

x ∈

i=2

But then

(

) = 0. Thus

= 0. Thus

is an automorphism of

with

φ(M

) = φ(M

). So this gives an isomorphism between

i=2

∼

i=2

and so we are done by induction.

Now it remains to prove that the endomorphism rings are local. Recall the

following result from linear algebra.

Lemma

(Fitting)

Suppose

is a module with both the ACC and DCC on

submodules, and let f ∈ End

(M). Then for large enough n, we have

M = im f

⊕ ker f

Proof. By ACC and DCC, we may choose n large enough so that

: f

(M) → f

(M)

is an isomorphism, as if we keep iterating

, the image is a descending chain and

the kernel is an ascending chain, and these have to terminate.

If m ∈ M, then we can write

(m) = f

)

for some m

. Then

m = f

) + (m − f

)) ∈ im f

+ ker f

and also

im f

∩ ker f

= ker(f

: f

(M) → f

(M)) = 0.

So done.

Lemma.

Suppose

is an indecomposable module satisfying ACC and DCC

on submodules. Then B = End

(M) is local.

Proof.

Choose a maximal left ideal of

, say

. It’s enough to show that if

x 6∈ I

then x is left invertible. By maximality of I, we know B = Bx + I. We write

1 = λx + y,

for some

λ ∈ B

and

y ∈ I

. Since

y ∈ I

, it has no left inverse. So it is not an

isomorphism. By Fitting’s lemma and the indecomposability of

, we see that

= 0 for some m. Thus

(1 + y + y

+ · · · + y

m−1

)λx = (1 + y + · · · + y

m−1

)(1 − y) = 1.

So x is left invertible.

Corollary.

Let

be a left Artinian algebra. Then

has the unique decompo-

sition property.

Proof.

We know

satisfies the ACC and DCC condition. So

is a finite

direct sum of indecomposables.

So if

is an Artinian algebra, we know

can be uniquely decomposed as a

direct sum of indecomposable projectives,

A =

For convenience, we will work with right Artinian algebras and right modules

instead of left ones. It turns out that instead of studying projectives in

, we

can study idempotent elements instead.

Recall that

End

(

)

∼

. The projection onto

is achieved by left

multiplication by an idempotent e

= e

The fact that the

decomposes as a direct sum of the

translates to the

condition

= 1, e

= 0

for i 6= j.

Definition

(Orthogonal idempotent)

A collection of idempotents

}

is or-

thogonal if e

= 0 for i 6= j.

The indecomposability of P

is equivalent to e

being primitive:

Definition

(Primitive idempotent)

An idempotent is primitive if it cannot be

expressed as a sum

e = e

+ e

where e

, e

are orthogonal idempotents, both non-zero.

We see that giving a direct sum decomposition of

is equivalent to finding

an orthogonal collection of primitive idempotents that sum to 1. This is rather

useful, because idempotents are easier to move around that projectives.

Our current plan is as follows — given an Artinian algebra

, we can quotient

out by

(

), and obtain a semi-simple algebra

A/J

(

). By Artin–Wedderburn,

we know how we can decompose

A/J

(

), and we hope to be able to lift this

decomposition to one of

. The point of talking about idempotents instead is

that we know what it means to lift elements.

Proposition.

Let

be a nilpotent ideal in

, and let

be an idempotent of

A/N ≡

A. Then there is an idempotent e ∈ A with f = ¯e.

In particular, we know

(

) is nilpotent, and this proposition applies. The

proof involves a bit of magic.

Proof.

We consider the quotients

A/N

for

i ≥

1. We will lift the idempotents

successively as we increase

, and since

is nilpotent, repeating this process

will eventually land us in A.

Suppose we have found an idempotent

i−1

∈ A/N

i−1

with

i−1

. We

want to find f

∈ A/N

such that

= f.

For

i >

1, we let

be an element of

A/N

with image

i−1

A/N

i−1

Then since

− x

vansishes in

A/N

i−1

, we know

− x ∈ N

i−1

. Then in

particular,

− x)

= 0 ∈ A/N

. (†)

We let

= 3x

− 2x

Then by a direct computation using (

†

), we find

, and

has image

i−1

−

i−1

A/N

i−1

(alternatively, in characteristic

, we can use

= x

). Since N

= 0 for some k, this process gives us what we want.

Just being able to lift idempotents is not good enough. We want to lift

decompositions as projective indecomposables. So we need to do better.

Corollary. Let N be a nilpotent ideal of A. Let

1 = f

+ · · · + f

with {f

} orthogonal primitive idempotents in A/N. Then we can write

1 = e

+ · · · + e

with {e

} orthogonal primitive idempotents in A, and ¯e

= f

Proof. We define a sequence e

∈ A inductively. We set

= 1.

Then for each

i >

1, we pick

a lift of

· · ·

∈ e

i−1

, since by

inductive hypothesis we know that f

+ · · · + f

∈ e

i−1

/N . Then

i+1

= e

i+1

= e

i+1

We let

= e

− e

i+1

Then

¯e

= f

Also, if j > i, then

= e

i+1

and so

= (e

− e

i+1

= 0.

Similarly e

= 0.

We now apply this lifting of idempotents to

(

), which we know is

nilpotent. We know

A/N

is the direct sum of simple modules, and thus the

decomposition corresponds to

1 = f

+ · · · + f

∈ A/J(A),

and these

are orthogonal primitive idempotents. Idempotent lifting then gives

1 = e

+ · · · + e

∈ A,

and these are orthogonal primitive idempotents. So we can write

A =

where

are indecomposable projectives, and

(

) =

is simple.

By Krull–Schmidt, any indecomposable projective isomorphic to one of these

The final piece of the picture is to figure out when two indecomposable

projectives lie in the same block. Recall that if

is a right

-module and

idempotent, then

∼

Hom

(eA, M ).

In particular, if M = fA for some idempotent f, then

Hom(eA, f A)

∼

fAe.

However, if e and f are in different blocks, say B

and B

, then

fAe ∈ B

∩ B

= 0,

since B

and B

are (two-sided!) ideals. So we know

Hom(eA, f A) = 0.

So if

Hom

(

eA, f A

)

= 0, then they are in the same block. The existence of a

homomorphism can alternatively be expressed in terms of composition factors.

We have seen that each indecomposable projective P has a simple “top”

P/P J(A)

∼

Definition

(Composition factor)

A simple module

is a composition factor

of a module M if there are submodules M

≤ M

with

∼

Suppose

is a composition factor of a module

. Then we have a diagram

S 0

So by definition of projectivity, we obtain a non-zero diagonal map

P → M

≤ M

as shown.

Lemma.

Let

be an indecomposable projective, and

-module. Then

Hom(P, M) 6= 0 iff P/P J(A) is a composition factor of M .

Proof.

We have proven

⇒

. Conversely, suppose there is a non-zero map

P →

M. Then it factors as

S =

P J(A)

→

im f

(im f )J(A)

Now we cannot have

im f

= (

im f

)

(

), or else we have

im f

= (

im f

)

(

)

= 0

for sufficiently large

since

(

) is nilpotent. So this map must be injective,

hence an isomorphism. So this exhibits S as a composition factor of M .

We define a (directed) graph whose vertices are labelled by indecomposable

projectives, and there is an edge

→ P

if the top

is a composition

factor of P

Theorem.

Indecomposable projectives

and

are in the same block if and

only if they lie in the same connected component of the graph.

Proof.

It is clear that

and

are in the same connected component, then

they are in the same block.

Conversely, consider a connected component X, and consider

I =

P ∈X

We show that this is in fact a left ideal, hence an ideal. Consider any

x ∈ A

. Then

for each P ∈ X, left-multiplication gives a map P → A, and if we decompose

A =

then this can be expressed as a sum of maps

P → P

. Now such a map can

be non-zero only if

is a composition factor of

. So if

= 0, then

∈ X

So left-multiplication by

maps

to itself, and it follows that

is an ideal.

1.5 K

We now briefly talk about the notion of K

Definition

(

)

For any associative

-algebra

, consider the free abelian

group with basis labelled by the isomorphism classes [

] of finitely-generated

projective A-modules. Then introduce relations

] + [P

] = [P

⊕ P

This yields an abelian group which is the quotient of the free abelian group by

the subgroup generated by

] + [P

] − [P

⊕ P

The abelian group is K

(A).

Example.

is an Artinian algebra, then we know that any finitely-generated

projective is a direct sum of indecomposable projectives, and this decomposition

is unique by Krull-Schmidt. So

(A) =



abelian group generated by the isomorphism

classes of indecomposable projectives



(

)

∼

, where

is the number of isomorphism classes of indecomposable

projectives, which is the number of isomorphism classes of simple modules.

Here we’re using the fact that two indecomposable projectives are isomorphic

iff their simple tops are isomorphic.

It turns out there is a canonical map

(

)

→ A/

[

A, A

]. Recall we have

met

[

A, A

] when we were talking about the number of simple modules. We

remarked that it was the 0th Hochschild homology group, and when

there is a

-basis of

[

A, A

] given by

+ [

A, A

], where

are conjugacy class

representatives.

To construct this canonical map, we first look at the trace map

(A) → A/[A, A].

This is a

-linear map, invariant under conjugation. We also note that the

canonical inclusion

(A) → M

n+1

(A)

X 7→



X 0

0 0



is compatible with the trace map. We observe that the trace induces an isomor-

phism

(A)

(A), M

(A)]

→

[A, A]

by linear algebra.

Now if

is finitely generated projective. It is a direct summand of some

Thus we can write

= P ⊕ Q,

for

P, Q

projective. Moreover, projection onto

corresponds to an idempotent

e in M

(A) = End

), and that

P = e(A

and we have

End

(P ) = eM

(A)e.

Any other choice of idempotent yields an idempotent

conjugate to

(

Therefore the trace of an endomorphism of

is well-defined in

[

A, A

independent of the choice of e. Thus we have a trace map

End

(P ) → A/[A, A].

In particular, the trace of the identity map on

is the trace of

. We call this

the trace of P .

Note that if we have finitely generated projectives

and

, then we have

⊕ Q

= A

⊕ Q

= A

Then we have

⊕ P

) ⊕ (Q

⊕ Q

) = A

m+n

So we deduce that

tr(P

⊕ P

) = tr P

+ tr P

Definition

(Hattori-Stallings trace map)

The map

(

)

→ A/

[

A, A

] induced

by the trace is the Hattori–Stallings trace map.

Example.

Let

, and

be finite. Then

[

A, A

] is a

-vector space

with basis labelled by a set of conjugacy class representatives

}

. Then we

know, for a finitely generated projective P , we can write

tr P =

where the

(

) may be regarded as class functions. However,

may be regarded

as a k-vector space.

So the there is a trace map

End

(P ) → k,

and also the “character”

G → k

, where

(

) =

tr g

. Hattori proved that if

(g) is the centralizer of g ∈ G, then

(g) = |C

(G)|r

−1

). (∗)

char k

= 0 and

is is algebraically closed, then we know

is semi-simple.

So every finitely generated projective is a direct sum of simples, and

(kG)

∼

with r the number of simples, and (∗) implies that the trace map

∼

(kG) →

[kG, kG]

∼

is the natural inclusion.

This is the start of the theory of algebraic

-theory, which is a homology

theory telling us about the endomorphisms of free

-modules. We can define

(A) to be the abelianization of

GL(A) = lim

n→∞

(A).

(

) tells us something about the relations required if you express

(

) in

terms of generators and relations. We’re being deliberately vague. These groups

are very hard to compute.

Just as we saw in the i = 0 case, there are canonical maps

(A) → HH

(A),

where

∗

is the Hochschild homology. The

= 1 case is called the Dennis

trace map. These are analogous to the Chern maps in topology.

2 Noetherian algebras

2.1 Noetherian algebras

In the introduction, we met the definition of Noetherian algebras.

Definition

(Noetherian algebra)

An algebra is left Noetherian if it satisfies

the ascending chain condition (ACC ) on left ideals, i.e. if

≤ I

≤ · · ·

is an ascending chain of left ideals, then there is some

such that

N+m

for all m ≥ 0.

Similarly, we say an algebra is Noetherian if it is both left and right Noethe-

rian.

We’ve also met a few examples. Here we are going to meet lots more. In

fact, most of this first section is about establishing tools to show that certain

algebras are Noetherian.

One source of Noetherian algebras is via constructing polynomial and power

series rings. Recall that in IB Groups, Rings and Modules, we proved the Hilbert

basis theorem:

Theorem

(Hilbert basis theorem)

is Noetherian, then

[

] is Noetherian.

Note that our proof did not depend on

being commutative. The same

proof works for non-commutative rings. In particular, this tells us

[

, · · · , X

]

is Noetherian.

It is also true that power series rings of Noetherian algebras are also Noethe-

rian. The proof is very similar, but for completeness, we will spell it out

completely.

Theorem. Let A be left Noetherian. Then A[[X]] is Noetherian.

Proof.

Let

be a left ideal of

[[

]]. We’ll show that if

is left Noetherian,

then I is finitely generated. Let

= {a : there exists an element of I of the form aX

+ higher degree terms}.

We note that J

is a left ideal of A, and also note that

≤ J

≤ · · · ,

as we can always multiply by

. Since

is left Noetherian, this chain terminates

for some

. Also,

, J

, · · · , J

are all finitely generated left ideals.

We suppose

, · · · , a

generates

for

= 1

, · · · , N

. These correspond to

elements

(X) = a

+ higher odder terms ∈ I.

We show that this finite collection generates

as a left ideal. Take

(

)

∈ I

and suppose it looks like

+ higher terms,

with b

6= 0.

Suppose n < N. Then b

∈ J

, and so we can write

f(X) −

(X) ∈ I

has zero coefficient for X

, and all other terms are of higher degree.

Repeating the process, we may thus wlog

n ≥ N

. We get

(

) of the form

+ higher degree terms. The same process gives

f(X) −

(X)

with terms of degree

+ 1 or higher. We can repeat this yet again, using the

fact J

= J

N+1

, so we obtain

f(X) −

(x) −

N+1,j

(X) + · · · .

So we find

f(X) =

(X)f

(X)

for some

(

). So

is in the left ideal generated by our list, and hence so is

Example.

It is straightforward to see that quotients of Noetherian algebras are

Noetherian. Thus, algebra images of the algebras

[

] and

[[

]] would also be

Noetherian.

For example, finitely-generated commutative

-algebras are always Noethe-

rian. Indeed, if we have a generating set

as a

-algebra, then there is an

algebra homomorphism

k[X

, · · · , X

] A

We also saw previously that

Example. Any Artinian algebra is Noetherian.

The next two examples we are going to see are less obviously Noetherian,

and proving that they are Noetherian takes some work.

Definition

(

th Weyl algebra)

The

th Weyl algebra

(

) is the algebra

generated by X

, · · · , X

, Y

, · · · , Y

with relations

− X

= 1,

for all i, and everything else commutes.

This algebra acts on the polynomial algebra

[

, · · · , X

] with

acting

by left multiplication and

∂

∂X

. Thus

[

, · · · , X

] is a left

(

) module.

This is the prototype for thinking about differential algebras, and

-modules in

general (which we will not talk about).

The other example we have is the universal enveloping algebra of a Lie

algebra.

Definition

(Universal enveloping algebra)

Let

be a Lie algebra over

, and

take a

-vector space basis

, · · · , x

. We form an associative algebra with

generators x

, · · · , x

with relations

− x

= [x

, x

and this is the universal enveloping algebra U(g).

Example.

is abelian, i.e. [

, x

] = 0 in

, then the enveloping algebra is

the polynomial algebra in x

, · · · , x

Example. If g = sl

(k), then we have a basis



0 1

0 0





0 0

1 0





1 0

0 −1



They satisfy

[e, f ] = h, [h, e] = 2e, [h, f ] = −2f,

To prove that

(

) and

(

) are Noetherian, we need some machinery, that

involves some “deformation theory”. The main strategy is to make use of a

natural filtration of the algebra.

Definition

(Filtered algebra)

A (

-)filtered algebra

is a collection of

-vector

spaces

· · · ≤ A

−1

≤ A

≤ · · ·

such that A

· A

⊆ A

i+j

for all i, j ∈ Z, and 1 ∈ A

For example a polynomial ring is naturally filtered by the degree of the

polynomial.

The definition above was rather general, and often, we prefer to talk about

more well-behaved filtrations.

Definition (Exhaustive filtration). A filtration is exhaustive if

= A.

Definition (Separated filtration). A filtration is separated if

= {0}.

Unless otherwise specified, our filtrations are exhaustive and separated.

For the moment, we will mostly be interested in positive filtrations.

Definition (Positive filtration). A filtration is positive if A

= 0 for i < 0.

Our canonical source of filtrations is the following construction:

Example. If A is an algebra generated by x

, · · · , x

, say, we can set

– A

is the k-span of 1

– A

is the k-span of 1, x

, · · · , x

– A

is the k-span of 1, x

, · · · , x

, x

for i, j ∈ {1, · · · , n}.

In general,

is elements that are of the form of a (non-commutative) polynomial

expression of degree ≤ r.

Of course, the filtration depends on the choice of the generating set.

Often, to understand a filtered algebra, we consider a nicer object, known as

the associated graded algebra.

Definition

(Associated graded algebra)

Given a filtration of

, the associated

graded algebra is the vector space direct sum

gr A =

i−1

This is given the structure of an algebra by defining multiplication by

(a + A

i−1

)(b + A

j−1

) = ab + A

i+j−1

∈

i+j

i+j−1

In our example of a finitely-generated algebra, the graded algebra is generated

by x

+ A

, · · · , x

+ A

∈ A

The associated graded algebra has the natural structure of a graded algebra:

Definition

(Graded algebra)

A (

-)graded algebra is an algebra

that is of

the form

B =

i∈Z

where

are

-subspaces, and

⊆ B

i+j

. The

’s are called the homogeneous

components.

A graded ideal is an ideal of the form

where J

is a subspace of B

, and similarly for left and right ideals.

There is an intermediate object between a filtered algebra and its associated

graded algebra, known as the Rees algebra.

Definition

(Rees algebra)

Let

be a filtered algebra with filtration

}

. Then

the Rees algebra

Rees

(

) is the subalgebra

of the Laurent polynomial

algebra A[T, T

−1

] (where T commutes with A).

Since 1 ∈ A

⊆ A

, we know T ∈ Rees(A). The key observation is that

– Rees(A)/(T )

∼

gr A.

– Rees(A)/(1 − T )

∼

Since

(

) and

(

) are finitely-generated algebras, they come with a

natural filtering induced by the generating set. It turns out, in both cases, the

associated graded algebras are pretty simple.

Example.

Let

(

), with generating set

, · · · , X

and

, · · · , Y

. We

take the filtration as for a finitely-generated algebra. Now observe that if

∈ A

and a

∈ A

, then

− a

∈ A

i+j−2

So we see that gr A is commutative, and in fact

gr A

(k)

∼

, · · · ,

where

are the images of

and

respectively. This is not

hard to prove, but is rather messy. It requires a careful induction.

Example.

Let

be a Lie algebra, and consider

(

). This has generating

set

, · · · , x

, which is a vector space basis for

. Again using the filtration for

finitely-generated algebras, we get that if a

∈ A

and a

∈ A

, then

− a

∈ A

i+j−1

So again gr A is commutative. In fact, we have

gr A

∼

k[¯x

, · · · , ¯x

The fact that this is a polynomial algebra amounts to the same as the Poincar´e-

Birkhoff-Witt theorem. This gives a k-vector space basis for U(g).

In both cases, we find that

gr A

are finitely-generated and commutative, and

therefore Noetherian. We want to use this fact to deduce something about

itself.

Lemma.

Let

be a positively filtered algebra. If

gr A

is Noetherian, then

left Noetherian.

By duality, we know that A is also right Noetherian.

Proof. Given a left ideal I of A, we can form

gr I =

I ∩ A

i−1

where I is filtered by {I ∩ A

}. By the isomorphism theorem, we know

I ∩ A

i−1

∼

I ∩ A

+ A

i−1

⊆

i−1

Then gr I is a left graded ideal of gr A.

Now suppose we have a strictly ascending chain

< I

< · · ·

of left ideals. Since we have a positive filtration, for some

, we have

∩ A

(

∩ A

and I

∩ A

i−1

= I

∩ A

i−1

. Thus

gr I

( gr I

( · · · .

This is a contradiction since

gr A

is Noetherian. So

must be Noetherian.

Where we need positivity is the existence of that transition from equality to

non-equality. If we have a

-filtered algebra instead, then we need to impose

some completeness assumption, but we will not go into that.

Corollary. A

(k) and U(g) are left/right Noetherian.

Proof. gr A

(

) and

gr U

(

) are commutative and finitely generated algebras.

Note that there is an alternative filtration for

(

) yielding a commutative

associated graded algebra, by setting A

= k[X

, · · · , X

] and

= k[X

, · · · , X

] +

j=1

k[X

, · · · , X

i.e. linear terms in the

, and then keep on going. Essentially, we are filtering on

the degrees of the

only. This also gives a polynomial algebra as an associative

graded algebra. The main difference is that when we take the commutator,

we don’t go down by two degrees, but one only. Later, we will see this is

advantageous when we want to get a Poisson bracket on the associated graded

algebra.

We can look at further examples of Noetherian algebras.

Example.

The quantum plane

[

X, Y

] has generators

and

, with relation

XY = qY X

for some

q ∈ k

. This thing behaves differently depending on whether

is a

root of unity or not.

This quantum plane first appeared in mathematical physics.

Example.

The quantum torus

[

X, X

−1

, Y, Y

−1

] has generators

−1

with relations

−1

= Y Y

−1

= 1, XY = qY X.

The word “quantum” in this context is usually thrown around a lot, and

doesn’t really mean much apart from non-commutativity, and there is very little

connection with actual physics.

These algebras are both left and right Noetherian. We cannot prove these

by filtering, as we just did. We will need a version of Hilbert’s basis theorem

which allows twisting of the coefficients. This is left as an exercise on the second

example sheet.

In the examples of

(

) and

(

), the associated graded algebras are

commutative. However, it turns out we can still capture the non-commutativity

of the original algebra by some extra structure on the associated graded algebra.

So suppose

is a (positively) filtered algebra whose associated graded algebra

gr A

is commutative. Recall that the filtration has a corresponding Rees algebra,

and we saw that Rees A/(T )

∼

gr A. Since gr A is commutative, this means

[Rees A, Rees A] ⊆ (T ).

This induces a map

Rees

(

)

× Rees

(

)

→

(

)

(

), sending (

r, s

)

7→ T

+ [

r, s

Quotienting out by (T ) on the left, this gives a map

gr A × gr A →

(T )

)

We can in fact identify the right hand side with gr A as well. Indeed, the map

gr A

∼

Rees(A)

(T )

)

mult. by T

is an isomorphism of gr A

∼

Rees A/(T )-modules. We then have a bracket

{ · , · } : gr A × gr A gr A

(¯r, ¯s) {r, s}

Note that in our original filtration of the Weyl algebra

(

), since the commu-

tator brings us down by two degrees, this bracket vanishes identically, but using

the alternative filtration does not give a non-zero { · , · }.

This { · , · } is an example of a Poisson bracket.

Definition

(Poisson algebra)

An associative algebra

is a Poisson algebra if

there is a k-bilinear bracket { · , · } : B × B → B such that

– B is a Lie algebra under { · , · }, i.e.

{r, s} = −{s, r}

and

{{r, s}, t} + {{s, t}, r} + {{t, r}, s} = 0.

– We have the Leibnitz rule

{r, st} = s{r, t} + {r, s}t.

The second condition says {r, · } : B → B is a derivation.

2.2 More on A

(k) and U(g)

Our goal now is to study modules of

(

). The first result tells us we must

focus on infinite dimensional modules.

Lemma.

Suppose

char k

= 0. Then

(

) has no non-zero modules that are

finite-dimensional k-vector spaces.

Proof.

Suppose

is a finite-dimensional module. Then we’ve got an algebra

homomorphism θ : A

(k) → End

(M)

∼

(k), where m = dim

In A

(k), we have

− X

= 1.

Applying the trace map, we know

tr(θ(Y

)θ(X

) − θ(X

)θ(Y

)) = tr I = m.

But since the trace is cyclic, the left hand side vanishes. So

= 0. So

trivial.

A similar argument works for the quantum torus, but using determinants

instead.

We’re going to make use of our associated graded algebras from last time,

which are isomorphic to polynomial algebras. Given a filtration

}

, we

may filter a module with generating set S by setting

= A

Note that

⊆ M

i+j

which allows us to form an associated graded module

gr M =

i+1

This is a graded

gr A

-module, which is finitely-generated if

is. So we’ve got a

finitely-generated graded module over a graded commutative algebra.

To understand this further, we prove some results about graded modules

over commutative algebras, which is going to apply to our gr A and gr M.

Definition

(Poincar´e series)

Let

be a graded module over a graded algebra

S, say

V =

∞

i=0

Then the Poincar´e series is

P (V, t) =

∞

i=0

(dim V

Theorem

(Hilbert-Serre theorem)

The Poincar´e series

(

V, t

) of a finitely-

generated graded module

V =

∞

i=0

over a finitely-generated generated commutative algebra

S =

∞

i=0

with homogeneous generating set x

, · · · , x

is a rational function of the form

f(t)

(1 − t

)

where f (t) ∈ Z[t] and k

is the degree of the generator x

Proof.

We induct on the number

of generators. If

= 0, then

and

is therefore a finite-dimensional

-vector space. So

(

V, t

) is a polynomial.

Now suppose

m >

0. We assume the theorem is true for

< m

generators.

Consider multiplication by x

. This gives a map

i+k

and we have an exact sequence

0 K

i+k

→ 0,

(∗)

where

K =

= ker(x

: V → V )

and

L =

i+k

= coker(x

: V → V ).

Then

is a graded submodule of

and hence is a finitely-generated

-module,

using the fact that

is Noetherian. Also,

V/x

is a quotient of

, and it

is thus also finitely-generated.

Now both

and

are annihilated by

. So they may be regarded as

, · · · , x

m−1

]-modules. Applying dim

to (∗), we know

dim

) − dim

) + dim(V

i+k

) − dim(L

i+k

) = 0.

We multiply by t

i+k

, and sum over i to get

P (K, t) − t

P (V, t) + P (V, t) − P (L, t) = g(t),

where

(

) is a polynomial with integral coefficients arising from consideration

of the first few terms.

We now apply the induction hypothesis to K and L, and we are done.

Corollary.

If each

, · · · , k

= 1, i.e.

is generated by

and homoge-

neous elements

, · · · , x

of degree 1, then for large enough

, then

dim V

(

)

for some polynomial

(

)

∈ Q

[

] of

d −

1, where

is the order of the pole of

P (V, t) at t = 1. Moreover,

j=0

dim V

= χ(i),

where χ(t) ∈ Q[t] of degree d.

Proof. From the theorem, we know that

P (V, t) =

f(t)

(1 − t)

for some d with f (1) 6= 0, f ∈ Z[t]. But

(1 − t)

−1

= 1 + t + t

+ · · ·

By differentiating, we get an expression

(1 − t)

−d



d + i − 1

d − 1



f(t) = a

+ a

t + · · · + a

then we get

dim V

= a



d + i − 1

d − 1



+ a



d + i − 2

d − 1



+ · · · + a



d + i − s − 1

d − 1



where we set



d−1



= 0 if

r < d −

1, and this expression can be rearranged to

give φ(i) for a polynomial φ(t) ∈ Q[t], valid for i − s > 0. In fact, we have

φ(t) =

f(1)

(d − 1)!

d−1

+ lower degree term.

Since f (1) 6= 0, this has degree d − 1.

This implies that

j=0

dim V

is a polynomial in Q[t] of degree d.

This

(

) is the Hilbert polynomial, and

(

) the Samuel polynomial. Some

people call χ(t) the Hilbert polynomial instead, though.

We now want to apply this to our cases of

gr A

, where

(

) or

(

filtered as before. Then we deduce that

dim

j−1

= χ(i),

for a polynomial χ(t) ∈ Q[t]. But we also know

j=0

dim

j−1

= dim M

We are now in a position to make a definition.

Definition

(Gelfand-Kirillov dimension)

Let

(

) or

(

) and

finitely-generated

-module, filtered as before. Then the Gelfand-Kirillov di-

mension

(

) of

is the degree of the Samuel polynomial of

gr M

as a

gr A-module.

This makes sense because

gr A

is a commutative algebra in this case. A

priori, it seems like this depends on our choice of filtering on

, but actually, it

doesn’t. For a more general algebra, we can define the dimension as below:

Definition

(Gelfand-Kirillov dimension)

Let

be a finitely-generated

algebra, which is filtered as before, and a finitely-generated

-module

, filtered

as before. Then the GK-dimension of M is

d(M) = lim sup

n→∞

log(dim M

)

log n

In the case of

(

) or

(

), this matches the previous definition. Again,

this does not actually depend on the choice of generating sets.

Recall we showed that no non-zero

(

)-module

can have finite dimension

as a

-vector space. So we know

(

)

0. Also, we know that

(

) is an

integer for cases

(

), since it is the degree of a polynomial. However,

for general

, we can get non-integral values. In fact, the values we can

get are 0

2, and then any real number

≥

2. We can also have

∞

if the

lim sup

doesn’t exist.

Example.

, then we have

dim

(

)

< ∞

iff

has a subgroup

of finite index with

embedding into the strictly upper triangular integral

matrices, i.e. matrices of the form







1 ∗ · · · ∗

0 1 · · · ∗

0 0 · · · 1







This is a theorem of Gromoll, and is quite hard to prove.

Example.

We have

dim

(

) = 0 iff

is finite-dimensional as a

-vector

space.

We have

GK-dim(k[X]) = 1,

and in general

GK-dim(k[X

, · · · , X

]) = n.

Indeed, we have

dim

(mth homogeneous component) =



m + n



So we have

χ(t) =



t + n



This is of degree n, with leading coefficient

We can make the following definition, which we will not use again:

Definition

(Multiplicity)

Let

be a commutative algebra, and

-module.

The multiplicity of M with d(M) = d is

d! × leading coefficient of χ(t).

On the second example sheet, we will see that the multiplicity is integral.

We continue looking at more examples.

Example.

We have

(

)) = 2

, and

(

)) =

dim

. Here we are using

the fact that the associated graded algebras are polynomial algebras.

Example.

We met

[

, · · · , X

] as the “canonical”

(

)-module. The filtra-

tion of the module matches the one we used when thinking about the polynomial

algebra as a module over itself. So we get

d(k[X

, · · · , X

]) = n.

Lemma. Let M be a finitely-generated A

-module. Then d(M ) ≤ 2n.

Proof.

Take generators

, · · · , m

. Then there is a surjective filtered

module homomorphism

⊕ · · · ⊕ A

, · · · , a

)

It is easy to see that quotients can only reduce dimension, so

GK-dim(M) ≤ d(A

⊕ · · · ⊕ A

But

⊕···⊕A

= sχ

has degree 2n.

More interestingly, we have the following result:

Theorem

(Bernstein’s inequality)

Let

be a non-zero finitely-generated

(k)-module, and char k = 0. Then

d(M) ≥ n.

Definition

(Holonomic module)

(

) module

is holonomic iff

(

) =

If we have a holonomic module, then we can quotient by a maximal submodule,

and get a simple holonomic module. For a long time, people thought all simple

modules are holonomic, until someone discovered a simple module that is not

holonomic. In fact, most simple modules are not holonomic, but we something

managed to believe otherwise.

Proof.

Take a generating set and form the canonical filtrations

}

(

)

and

}

. We let

(

) be the Samuel polynomial. Then for large enough

i, we have

χ(i) = dim M

We claim that

dim A

≤ dim Hom

, M

) = dim M

× dim M

Assuming this, for large enough i, we have

dim A

≤ χ(i)χ(2i).

But we know

dim A



i + 2



which is a polynomial of degree 2

. But

(

)

) is a polynomial of degree

2d(M). So we get that

n ≤ d(M ).

So it remains to prove the claim. It suffices to prove that the natural map

→ Hom

, M

given by multiplication is injective.

So we want to show that if

a ∈ A

= 0, then

= 0. We prove this by

induction on

. When

= 0, then

, and

is a finite-dimensional

-vector

space. Then the result is obvious.

i >

0, we suppose the result is true for smaller

. We let

a ∈ A

is non-zero.

If aM

= 0, then certainly a 6∈ k. We express

a =

αβ

· · · X

· · · Y

where α = (α

, · · · , α

), β = (β

, · · · , β

), and c

α,β

∈ k.

If possible, pick a

such that

α,α

= 0 for some

with

= 0 (this happens

when there is an X involved). Then

, a] =

α,β

· · · X

−1

· · · X

· · · Y

and this is non-zero, and lives in A

i−1

If aM

= 0, then certainly aM

i−1

= 0. Hence

, a]M

i−1

= (Y

a − aY

i−1

= 0,

using the fact that Y

i−1

⊆ M

. This is a contradiction.

If a only has Y ’s involved, then we do something similar using [X

, a].

There is also a geometric way of doing this.

We take k = C. We know gr A

is a polynomial algebra

gr A

= k[

, · · · ,

which may be viewed as the coordinate algebra of the cotangent bundle on affine

-space

. The points of this correspond to the maximal ideals of

gr A

. If

is a left ideal of

(

), then we can form

gr I

and we can consider the set of

maximal ideals containing it. This gives us the characteristic variety

(

We saw that there was a Poisson bracket on

gr A

, and this may be used to

define a skew-symmetric form on the tangent space at any point of the cotangent

bundle. In this case, this is non-degenerate skew-symmetric form.

We can consider the tangent space

(

) at a non-singular point,

and there’s a theorem of Gabber (1981) which says that

U ⊇ U

⊥

, where

⊥

with respect to the skew-symmetric form. By non-degeneracy, we must have

dim U ≥ n, and we also know that

dim Ch(A

/I) = d(A

/I).

So we find that d(A

/I) ≥ n.

In the case of

(

), we can think of

gr A

as the coordinate algebra on

∗

, the vector space dual of

. The Poisson bracket leads to a skew-symmetric

form on tangent spaces at points of

∗

. In this case, we don’t necessarily get

non-degeneracy. However, on

, we have the adjoint action of the corresponding

Lie group

, and this induces a co-adjoint action on

∗

. Thus

∗

is a disjoint

union of orbits. If we consider the induced skew-symmetric form on tangent

spaces of orbits (at non-singular points), then it is non-degenerate.

2.3 Injective modules and Goldie’s theorem

The goal of this section is to prove Goldie’s theorem.

Theorem

(Goldie’s theorem)

Let

be a right Noetherian algebra with no

non-zero ideals all of whose elements are nilpotent. Then

embeds in a finite

direct sum of matrix algebras over division algebras.

The outline of the proof is as follows — given any

, we embed

in an

“injective hull”

(

). We will then find that similar to what we did in Artin–

Wedderburn, we can decompose

End

(

)) into a direct sum of matrix algebras

over division algebras. But we actually cannot. We will have to first quotient

End(E(A)) by some ideal I.

On the other hand, we do not actually have an embedding of

∼

End

(

)

into

End

(

)). Instead, what we have is only a homomorphism

End

(

)

→

End

(

))

, where we quotient out by the same ideal

. So actually the two

of our problems happen to cancel each other out.

We will then prove that the kernel of this map contains only nilpotent

elements, and then our hypothesis implies this homomorphism is indeed an

embedding.

We begin by first constructing the injective hull. This is going to involve

talking about injective modules, which are dual to the notion of projective

modules.

Definition

(Injective module)

-module

is injective if for every diagram

of A-module maps

0 M N

such that

is injective, there exists a map

that makes the diagram commute.

Equivalently, Hom( · , E) is an exact functor.

Example. Take A = k. Then all k-vector spaces are injective k-modules.

Example. Take A = k[X]. Then k(X) is an injective k[X]-module.

Lemma. Every direct summand of an injective module is injective, and direct

products of injectives is injective.

Proof. Same as proof for projective modules.

Lemma. Every A-module may be embedded in an injective module.

We say the category of

-modules has enough injectives. The dual result for

projectives was immediate, as free modules are projective.

Proof.

Let

be a right

-module. Then

Hom

(

A, M

) is a right

-module via

(fa)(x) = f(ax).

We claim that Hom

(A, M ) is an injective module. Suppose we have

0 M

Hom

(A, M )

We consider the k-module diagram

0 M

where

(

) =

(

)(1). Since

is injective as a

-module, we can find the

such that α = βθ. We define ψ : N

→ Hom

(A, M ) by

ψ(n

)(x) = β(n

x).

It is straightforward to check that this does the trick. Also, we have an embedding

M → Hom

(A, M ) by m 7→ (φ

: x 7→ mx).

The category theorist will write the proof in a line as

Hom

( · , Hom

(A, M ))

∼

Hom

( · ⊗

A, M )

∼

Hom

( · , M ),

which is exact since M is injective as a k-module.

Note that neither the construction of

Hom

(

A, M

), nor the proof that it is

injective requires the right

-modules structure of

. All we need is that

an injective k-module.

Lemma.

-module is injective iff it is a direct summand of every extension

of itself.

Proof.

Suppose

is injective and

is an extension of

. Then we can form

the diagram

0 E E

and then by injectivity, we can find ψ. So

= E ⊕ ker ψ.

Conversely, suppose

is a direct summand of every extension. But by the

previous lemma, we can embed

in an injective

. This implies that

is a

direct summand of E

, and hence injective.

There is some sort of “smallest” injective a module embeds into, and this

is called the injective hull, or injective envelope. This is why our injectives are

called

. The “smallness” will be captured by the fact that it is an essential

extension.

Definition

(Essential submodule)

An essential submodule

of an

-module

is one where

M ∩ V 6

{

}

for every non-zero submodule

. We say

is an essential extension of M.

Lemma. An essential extension of an essential extension is essential.

Proof.

Suppose

M < E < F

are essential extensions. Then given

N ≤ F

, we

know

N ∩ E 6

{

}

, and this is a submodule of

. So (

N ∩ E

)

∩ M

N ∩ M 6

= 0.

So F is an essential extension of M.

Lemma. A maximal essential extension is an injective module.

Such maximal things exist by Zorn’s lemma.

Proof.

Let

be a maximal essential extension of

, and consider any embedding

E → F

. We shall show that

is a direct summand of

. Let

be the set of

all non-zero submodules

with

V ∩ E

{

}

. We apply Zorn’s lemma to

get a maximal such module, say V

Then

embeds into

F/V

as an essential submodule. By transitivity of

essential extensions,

F/V

is an essential extension of

, but

is maximal. So

∼

F/V

. In other words,

F = E ⊕ V

We can now make the following definition:

Definition

(Injective hull)

A maximal essential extension of

is the injective

hull (or injective envelope) of M, written E(M ).

Proposition.

Let

be an

-module, with an inclusion

M → I

into an injective

module. Then this extends to an inclusion E(M ) → I.

Proof. By injectivity, we can fill in the diagram

0 M E(M)

We know

restricts to the identity on

. So

ker ψ ∩ M

{

}

. By Since

(

)

is essential, we must have ker ψ = 0. So E(M ) embeds into I.

Proposition.

Suppose

is an injective essential extension of

. Then

∼

E(M). In particular, any two injective hulls are isomorphic.

Proof.

By the previous lemma,

(

) embeds into

. But

(

) is a maximal

essential extension. So this forces E = E(M).

Using what we have got, it is not hard to see that

Proposition.

E(M

⊕ M

) = E(M

) ⊕ E(M

Proof.

We know that

(

)

⊕ E

(

) is also injective (since finite direct sums

are the same as direct products), and also

⊕ M

embeds in

(

)

⊕ E

(

So it suffices to prove this extension is essential.

Let V ≤ E(M

) ⊕ E(M

). Then either V/E(M

) 6= 0 or V /E(M

) 6= 0.

We wlog it is the latter. Note that we can naturally view

E(M

)

≤

E(M

) ⊕ E(M

)

E(M

)

∼

E(M

Since M

⊆ E(M

) is essential, we know

∩ (V/E(M

)) 6= 0.

So there is some

∈ V

such that

∈ E

(

) and

∈ M

. Now

consider

{m ∈ E(M

) : am

+ m ∈ V for some a ∈ A}.

This is a non-empty submodule of

(

), and so contains an element of

say n. Then we know am

+ n ∈ V ∩ (M

⊕ M

), and we are done.

The next two examples of injective hulls will be stated without proof:

Example. Take A = k[X], and M = k[X]. Then E(M) = k(X).

Example.

Let

[

] and

be the trivial module, where

acts by 0.

Then

E(V ) =

k[X, X

−1

]

Xk[X]

which is a quotient of A-modules. We note V embeds in this as

∼

k[X]

Xk[X]

→

k[X, X

−1

]

Xk[X]

Definition

(Uniform module)

A non-zero module

is uniform if given non-zero

submodules V

, V

, then V

∩ V

6= {0}.

Lemma. V is uniform iff E(V ) is indecomposable.

Proof.

Suppose

(

) =

A ⊕ B

, with

A, B

non-zero. Then

V ∩ A 6

{

}

and

V ∩B 6

{

}

since the extension is essential. So we have two non-zero submodules

of V that intersect trivially.

Conversely, suppose

is not uniform, and let

, V

be non-zero submod-

ules that intersect trivially. By Zorn’s lemma, we suppose these are maximal

submodules that intersect trivially. We claim

E(V

) ⊕ E(V

) = E(V

⊕ V

) = E(V )

To prove this, it suffices to show that

is an essential extension of

⊕ V

, so

that E(V ) is an injective hull of V

⊕ V

Let

W ≤ V

be non-zero. If

W ∩

(

⊕ V

) = 0, then

⊕

(

⊕ W

) is a larger

pair of submodules with trivial intersection, which is not possible. So we are

done.

Definition

(Domain)

An algebra is a domain if

= 0 implies

= 0 or

= 0.

This is just the definition of an integral domain, but when we have non-

commutative algebras, we usually leave out the word “integral”.

To show that the algebras we know and love are indeed domains, we again

do some deformation.

Lemma.

Let

be a filtered algebra, which is exhaustive and separated. Then

if gr A is a domain, then so is A.

Proof.

Let

x ∈ A

\ A

i−1

, and

y ∈ A

\ A

j−1

. We can find such

i, j

for any

elements

x, y ∈ A

because the filtration is exhaustive and separated. Then we

have

¯x = x + A

i−1

6= 0 ∈ A

i−1

¯y = y + A

j−1

6= 0 ∈ A

j−1

gr A

is a domain, then we deduce

¯x¯y 6

= 0. So we deduce that

xy 6∈ A

i+j−1

. In

particular, xy 6= 0.

Corollary. A

(k) and U(g) are domains.

Lemma.

Let

be a right Noetherian domain. Then

is uniform, i.e.

(

)

is indecomposable.

Proof.

Suppose not, and so there are

and

non-zero such that

xA ∩ yA

{0}. So xA ⊕ yA is a direct sum.

But

is a domain and so

∼

as a right

-module. Thus

yxA ⊕ yyA

a direct sum living inside yA. Further decomposing yyA, we find that

xA ⊕ yxA ⊕ y

xA ⊕ · · · ⊕ y

is a direct sum of non-zero submodules. But this is an infinite strictly ascending

chain as n → ∞, which is a contradiction.

Recall that when we proved Artin–Wedderburn, we needed to use Krull–

Schmidt, which told us the decomposition is unique up to re-ordering. That

relied on the endomorphism algebra being local. We need something similar

here.

Lemma.

Let

be an indecomposable injective right module. Then

End

(

)

is a local algebra, with the unique maximal ideal given by

I = {f ∈ End(E) : ker f is essential}.

Note that since

is indecomposable injective, given any non-zero

V ≤ E

, we

know

(

) embeds into, and hence is a direct summand of

. Hence

(

) =

ker f

being essential is the same as saying

ker f

being non-zero. However,

this description of the ideal will be useful later on.

Proof.

Let

E → E

and

ker f

{

}

. Then

(

) is an injective module, and

so is a direct summand of

. But

is indecomposable. So

is surjective. So it

is an isomorphism, and hence invertible. So it remains to show that

I = {f ∈ End(E) : ker f is essential}

is an ideal.

ker f

and

ker g

are essential, then

ker

(

)

≥ ker f ∩ ker g

, and the

intersection of essential submodules is essential. So ker(f + g) is also essential.

Also, if

ker g

is essential, and

is arbitrary, then

ker

(

f ◦ g

)

≥ ker g

, and is

hence also essential. So I is a maximal left ideal.

The point of this lemma is to allow us to use Krull–Schmidt.

Lemma.

Let

be a non-zero Noetherian module. Then

is an essential

extension of a direct sum of uniform submodules N

, · · · , N

. Thus

E(M)

∼

E(N

) ⊕ · · · E(N

)

is a direct sum of finitely many indecomposables.

This decomposition is unique up to re-ordering (and isomorphism).

Proof.

We first show any non-zero Noetherian module contains a uniform one.

Suppose not, and

is in particular not uniform. So it contains non-zero

, V

with

∩ V

= 0. But

is not uniform by assumption. So it contains non-zero

and V

with zero intersection. We keep on repeating. Then we get

⊕ V

⊕ · · · ⊕ V

is a strictly ascending chain of submodules of M , which is a contradiction.

Now for non-zero Noetherian

, pick

uniform in

. Either

is essential

, and we’re done, or there is some

non-zero with

∩ N

= 0. We pick

uniform in N

. Then either N

⊕ N

is essential, or. . .

And we are done since M is Noetherian. Taking injective hulls, we get

E(M) = E(N

) ⊕ · · · ⊕ E(N

and we are done by Krull–Schmidt and the previous lemma.

This is the crucial lemma, which isn’t really hard. This allows us to define

yet another dimension for Noetherian algebras.

Definition

(Uniform dimension)

The uniform dimension, or Goldie rank of

M is the number of indecomposable direct summands of E(M).

This is analogous to vector space dimensions in some ways.

Example.

The Goldie rank of domains is 1, as we showed

is uniform. This

is true for A

(k) and U(g).

Lemma.

Let

, · · · , E

be indecomposable injectives. Put

⊕ · · · ⊕ E

Let I = {f ∈ End

(E) : ker f is essential}. This is an ideal, and then

End

(E)/I

∼

) ⊕ · · · ⊕ M

)

for some division algebras D

Proof. We write the decomposition instead as

E = E

⊕ · · · ⊕ E

Then as in basic linear algebra, we know elements of

End

(

) can be written as

an r × r matrix whose (i, j)th entry is an element of Hom(E

, E

Now note that if

∼

, then the kernel of a map

→ E

is essential in

. So quotienting out by I kills all of these “off-diagonal” entries.

Also

Hom

(

, E

) =

(

End

(

)), and so quotienting out by

gives

(End(E

)/{essential kernel})

∼

), where

∼

End(E

)

essential kernel

which we know is a division algebra since I is a maximal ideal.

The final piece to proving Goldie’s theorem is the following piece

Lemma.

is a right Noetherian algebra, then any

→ A

with

ker f

essential in A

is nilpotent.

Proof. Consider

0 < ker f ≤ ker f

≤ · · · .

Suppose

is not nilpotent. We claim that this is a strictly increasing chain.

Indeed, for all n, we have f

) 6= 0. Since ker f is essential, we know

) ∩ ker f 6= {0}.

This forces ker f

n+1

> ker f

, which is a contradiction.

We can now prove Goldie’s theorem.

Theorem

(Goldie’s theorem)

Let

be a right Noetherian algebra with no

non-zero ideals all of whose elements are nilpotent. Then

embeds in a finite

direct sum of matrix algebras over division algebras.

Proof. As usual, we have a map

A End

)

x left multiplication by x

For a map A

→ A

, it lifts to a map E(A

) → E(A

) by injectivity:

0 A

E(A

)

E(A

)

We can complete the diagram to give a map

(

)

→ E

(

), which

restricts to

. This is not necessarily unique. However, if we have two

lifts

and

, then the difference

− f

has

in the kernel, and hence has

an essential kernel. So it lies in I. Thus, if we compose maps

End

) End(E(A

))/I .

The kernel of this consists of

which when multiplying on the left has essential

kernel. This is an ideal all of whose elements is nilpotent. By assumption, any

such ideal vanishes. So we have an embedding of

End

(

))

, which we

know to be a direct sum of matrix algebras over division rings.

Goldie didn’t present it like this. This work in injective modules is due to

Matlis.

We saw that (right Noetherian) domains had Goldie rank 1. So we get that

End

(

))

∼

for some division algebra

. So by Goldie’s theorem, a right

Noetherian algebra embeds in a division algebra. In particular, this is true for

(k) and U(g).

3 Hochschild homology and cohomology

3.1 Introduction

We now move on to talk about Hochschild (co)homology. We will mostly talk

about Hochschild cohomology, as that is the one that is interesting. Roughly

speaking, given a

-algebra

and an

-bimodule

, Hochschild cohomology is

an infinite sequence of

-vector spaces

(

A, M

) indexed by

n ∈ N

associated

to the data. While there is in theory an infinite number of such vector spaces,

we are mostly going to focus on the cases of

= 0

2, and we will see that

these groups can be interpreted as things we are already familiar with.

The construction of these Hochschild cohomology groups might seem a bit

arbitrary. It is possible to justify these a priori using the general theory of

homological algebra and/or model categories. On the other hand, Hochschild

cohomology is sometimes used as motivation for the general theory of homological

algebra and/or model categories. Either way, we are not going to develop these

general frameworks, but are going to justify Hochschild cohomology in terms of

its practical utility.

Unsurprisingly, Hochschild (co)homology was first developed by Hochschild

in 1945, albeit only working with algebras of finite (vector space) dimension. It

was introduced to give a cohomological interpretation and generalization of some

results of Wedderburn. Later in 1962/1963, Gerstenhaber saw how Hochschild

cohomology was relevant to the deformations of algebras. More recently, it’s

been realized that that the Hochschild cochain complex has additional algebraic

structure, which allows yet more information about deformation.

As mentioned, we will work with

-bimodules over an algebra

. If our

algebra has an augmentation, i.e. a ring map to

, then we can have a decent

theory that works with left or right modules. However, for the sake of simplicity,

we shall just work with bimodules to make our lives easier.

Recall that a

-bimodule is an algebra with both left and right

actions

that are compatible. For example,

is an

-bimodule, and we sometimes write

it as

to emphasize this. More generally, we can view

⊗(n+2)

A⊗

· · ·⊗

as an A-A-bimodule by

x(a

⊗ a

⊗ · · · ⊗ a

n+1

)y = (xa

) ⊗ a

⊗ · · · ⊗ (a

n+1

y).

The crucial property of this is that for any

n ≥

0, the bimodule

⊗(n+2)

is a

free

-bimodule. For example,

A ⊗

is free on a single generator 1

⊗

whereas if {x

} is a k-basis of A, then A ⊗

A ⊗

A is free on {1 ⊗

⊗

1}.

The general theory of homological algebra says we should be interested in

such free things.

Definition

(Free resolution)

Let

be an algebra and

-bimodule. A

free resolution of M is an exact sequence of the form

· · · F

where each F

is a free A-A-bimodule.

More generally, we can consider a projective resolution instead, where we

allow the bimodules to be projective. In this course, we are only interested in

one particular free resolution:

Definition

(Hochschild chain complex)

Let

be a

-algebra with multiplication

map µ : A ⊗ A. The Hochschild chain complex is

· · · A ⊗

A ⊗

A A ⊗

A A 0.

We refer to

⊗

(n+2)

as the degree

term. The differential

⊗

(n+3)

→

⊗

(n+2)

is given by

d(a

⊗

· · · ⊗

n+1

) =

n+1

i=0

(−1)

⊗

· · · ⊗

i+1

⊗

· · · ⊗

n+2

This is a free resolution of

(the exactness is merely a computation, and

we shall leave that as an exercise to the reader). In a nutshell, given an

bimodule

, its Hochschild homology and cohomology is obtained by applying

· ⊗

A-A

and

Hom

A-A

(

· , M

) to the Hochschild chain complex, and then taking

the homology and cohomology of the resulting chain complex. We shall explore

in more detail what this means.

It is a general theorem that we could have applied the functors

· ⊗

A-A

and

Hom

A-A

(

· , M

) to any projective resolution of

and take the (co)homology,

and the resulting vector spaces will be the same. However, we will not prove

that, and will just always stick to this standard free resolution all the time.

3.2 Cohomology

As mentioned, the construction of Hochschild cohomology involves applying

Hom

A-A

(

· , M

) to the Hochschild chain complex, and looking at the terms

Hom

A-A

(

⊗(n+2)

, M

). This is usually not very convenient to manipulate, as

it involves talking about bimodule homomorphisms. However, we note that

⊗(n+2)

is a free

-bimodule generated by a basis of

⊗n

. Thus, there is a

canonical isomorphism

Hom

A-A

⊗(n+2)

, M )

∼

Hom

⊗n

, M ),

and k-linear maps are much simpler to work with.

Definition

(Hochschild cochain complex)

The Hochschild cochain complex

of an

-bimodule

is what we obtain by applying

Hom

A-A

(

· , M

) to the

Hochschild chain complex of A. Explicitly, we can write it as

Hom

(k, M) Hom

(A, M ) Hom

(A ⊗ A, M ) · · · ,

where

(δ

f)(a) = af(1) − f(1)a

(δ

f)(a

⊗ a

) = a

f(a

) − f(a

) + f(a

(δ

f)(a

⊗ a

) = a

f(a

⊗ a

) − f(a

⊗ a

)

+ f(a

⊗ a

) − f(a

⊗ a

(δ

n−1

f)(a

⊗ · · · ⊗ a

) = a

f(a

⊗ · · · ⊗ a

)

i=1

(−1)

f(a

⊗ · · · ⊗ a

i+1

⊗ · · · ⊗ a

)

+ (−1)

n+1

f(a

⊗ · · · ⊗ a

n−1

The reason the end ones look different is that we replaced

Hom

A-A

(

⊗(n+2)

, M

)

with Hom

⊗n

, M ).

The crucial observation is that the exactness of the Hochschild chain complex,

and in particular the fact that d

= 0, implies im δ

n−1

⊆ ker δ

Definition (Cocycles). The cocycles are the elements in ker δ

Definition (Coboundaries). The coboundaries are the elements in im δ

These names came from algebraic topology.

Definition (Hochschild cohomology groups). We define

(A, M ) = ker δ

(A, N ) =

ker δ

im δ

n−1

These are k-vector spaces.

If we do not want to single out

, we can extend the Hochschild cochain

complex to the left with 0, and setting

= 0 for

n <

0 (or equivalently extending

the Hochschild chain complex to the right similarly), Then

(A, M ) =

ker δ

im δ

−1

= ker δ

The first thing we should ask ourselves is when the cohomology groups van-

ish. There are two scenarios where we can immediately tell that the (higher)

cohomology groups vanish.

Lemma.

Let

be an injective bimodule. Then

(

A, M

) = 0 for all

n ≥

Proof. Hom

A-A

( · , M ) is exact.

Lemma.

is a projective bimodule, then

(

A, M

) = 0 for all

and

all n ≥ 1.

If we believed the previous remark that we could compute Hochschild coho-

mology with any projective resolution, then this result is immediate — indeed,

we can use

· · · →

→

→ A → A →

0 as the projective resolution. However,

since we don’t want to prove such general results, we shall provide an explicit

computation.

Proof.

is projective, then all

⊗n

are projective. At each degree

, we

can split up the Hochschild chain complex as the short exact sequence

⊗(n+3)

ker d

⊗(n+2)

im d

n−1

The im d is a submodule of A

⊗(n+1)

, and is hence projective. So we have

⊗(n+2)

∼

⊗(n+3)

ker d

⊕ im d

n−1

and we can write the Hochschild chain complex at n as

ker d

⊕

⊗(n+3)

ker d

⊗(n+3)

ker d

⊕ im d

n−1

⊗(n+1)

im d

n−1

⊕ im d

n−1

(a, b) (b, 0)

(c, d) (0, d)

n−1

Now

Hom

(

· , M

) certainly preserves the exactness of this, and so the Hochschild

cochain complex is also exact. So we have HH

(A, M ) = 0 for n ≥ 1.

This is a rather strong result. By knowing something about

itself, we

deduce that the Hochschild cohomology of any bimodule whatsoever must vanish.

Of course, it is not true that

(

A, M

) vanishes in general for

n ≥

1, or

else we would have a rather boring theory. In general, we define

Definition (Dimension). The dimension of an algebra A is

Dim A = sup{n : HH

(A, M ) 6= 0 for some A-A-bimodule M }.

This can be infinite if such a sup does not exist.

Thus, we showed that

embeds as a direct summand in

A ⊗ A

, then

Dim A = 0.

Definition

(

-separable)

An algebra

-separable if

embeds as a

direct summand of A ⊗ A.

Since A ⊗ A is a free A-A-bimodule, this condition is equivalent to A being

projective. However, there is some point in writing the definition like this.

Note that an

-bimodule is equivalently a left

A ⊗ A

-module. Then

is a direct summand of

A ⊗ A

if and only if there is a separating idempotent

e ∈ A ⊗ A

so that

viewed as A ⊗ A

-bimodule is (A ⊗ A

)e.

This is technically convenient, because it is often easier to write down a

separating idempotent than to prove directly A is projective.

Note that whether we write

A⊗A

is merely a matter of convention.

They have the same underlying set. The notation

A ⊗ A

is more convenient

when we take higher powers, but we can think of

A ⊗ A

as taking

as a

left-

right-

module and

as a left-

right-

, and tensoring them gives a

A-A-bimodule.

We just proved that separable algebras have dimension 0. Conversely, we

have

Lemma. If Dim A = 0, then A is separable.

Proof. Note that there is a short exact sequence

0 ker µ A ⊗ A A 0

If we can show this splits, then

is a direct summand of

A ⊗ A

. To do so, we

need to find a map A ⊗ A → ker µ that restricts to the identity on ker µ.

To do so, we look at the first few terms of the Hochschild chain complex

· · · im d ⊕ ker µ A ⊗ A A 0

By assumption, for any

, applying

Hom

A-A

(

· , M

) to the chain complex gives

an exact sequence. Omitting the annoying

A-A

subscript, this sequence looks like

0 −→ Hom(A, M)

∗

−→ Hom(A ⊗ A, M)

(∗)

−→ Hom(ker µ, M) ⊕ Hom(im d, M)

∗

−→ · · ·

Now

∗

sends

Hom

(

ker µ, M

) to zero. So

Hom

(

ker µ, M

) must be in the image

of (∗). So the map

Hom(A ⊗ A, M ) −→ Hom(ker µ, M)

must be surjective. This is true for any

. In particular, we can pick

ker µ

Then the identity map

ker µ

lifts to a map

A ⊗ A → ker µ

whose restriction to

ker µ is the identity. So we are done.

Example. M

(

) is separable. It suffices to write down the separating idem-

potent. We let

be the elementary matrix with 1 in the (

i, j

)th slot and 0

otherwise. We fix j, and then

⊗ E

∈ A ⊗ A

is a separating idempotent.

Example.

Let

with

char k - |G|

. Then

A ⊗ A

kG ⊗

(

)

∼

kG ⊗ kG. But this is just isomorphic to k(G × G), which is again semi-simple.

Thus, as a bimodule,

A ⊗ A

is completely reducible. So the quotient of

is a direct summand (of bimodules). So we know that whenever

char k - |G|

then kG is k-separable.

The obvious question is — is this notion actually a generalization of separable

field extensions? This is indeed the case.

Fact.

Let

L/K

be a finite field extension. Then

is separable as a

-algebra

iff L/K is a separable field extension.

However

-separable algebras must be finite-dimensional

-vector spaces. So

this doesn’t pass on to the infinite case.

In the remaining of the chapter, we will study what Hochschild cohomology

in the low dimensions mean. We start with

. The next two propositions

follow from just unwrapping the definitions:

Proposition. We have

(A, M ) = {m ∈ M : am − ma = 0 for all a ∈ A}.

In particular, HH

(A, A) is the center of A.

Proposition.

ker δ

= {f ∈ Hom

(A, M ) : f (a

) = a

f(a

) + f(a

These are the derivations from A to M . We write this as Der(A, M).

On the other hand,

im δ

= {f ∈ Hom

(A, M ) : f (a) = am − ma for some m ∈ M}.

These are called the inner derivations from A to M. So

(A, M ) =

Der(A, M )

InnDer(A, M )

Setting A = M , we get the derivations and inner derivations of A.

Example. If A = k[x], and char k = 0, then

DerA =



p(X)

: p(x) ∈ k[X]



and there are no (non-trivial) inner derivations because

is commutative. So

we find

(k[X], k[X])

∼

k[X].

In general, Der(A) form a Lie algebra, since

− D

∈ End

(A)

is in fact a derivation if D

and D

are.

There is another way we can think about derivations, which is via semi-direct

products.

Definition

(Semi-direct product)

Let

be an

-bimodule. We can form

the semi-direct product of

and

, written

A n M

, which is an algebra with

elements (a, m) ∈ A × M , and multiplication given by

, m

) · (a

, m

) = (a

, a

+ m

Addition is given by the obvious one.

Alternatively, we can write

A n M

∼

A + Mε,

where

commutes with everything and

= 0. Then

Mε

forms an ideal with

(Mε)

= 0.

In particular, we can look at

A n A

∼

Aε

. This is often written as

[

Previously, we saw first cohomology can be understood in terms of derivations.

We can formulate derivations in terms of this semi-direct product.

Lemma. We have

Der

(A, M )

∼

{algebra complements to M in A n M isomorphic to A}.

Proof. A complement to M is an embedded copy of A in A n M,

A A n M

a (a, D

)

The function

A → M

given by

a 7→ D

is a derivation, since under the embedding,

we have

ab 7→ (ab, aD

+ D

b).

Conversely, a derivation

A → M

gives an embedding of

A n M

given by

a 7→ (a, f(a)).

We can further rewrite this characterization in terms of automorphisms of

the semi-direct product. This allows us to identify inner derivations as well.

Lemma. We have

Der(A, M )

∼



automorphisms of A n M of the form

a 7→ a + f (a)ε, mε 7→ mε



where we view A n M

∼

A + Mε.

Moreover, the inner derivations correspond to automorphisms achieved by

conjugation by 1 + mε, which is a unit with inverse 1 − mε.

The proof is a direct check.

This applies in particular when we pick

, and the Lie algebra of

derivations of A may be thought of as the set of infinitesimal automorphisms.

Let’s now consider

(

A, M

). This is to be understood in terms of exten-

sions, of which the “trivial” example is the semi-direct product.

Definition

(Extension)

Let

be an algebra and

and

-bimodule. An

extension of

. An extension of

is a

-algebra

containing a

2-sided ideal I such that

– I

= 0;

– B/I

∼

A; and

– M

∼

I as an A-A-bimodule.

Note that since

= 0, the left and right multiplication in

induces an

A-A-bimodule structure on I, rather than just a B-B-bimodule.

We let

B → A

be the canonical quotient map. Then we have a short

exact sequence

0 I B A 0 .

Then two extensions

and

are isomorphic if there is a

-algebra isomorphism

θ : B

→ B

such that the following diagram commutes:

0 I A 0

Note that the semi-direct product is such an extension, called the split extension.

Proposition.

There is a bijection between

(

A, M

) with the isomorphism

classes of extensions of A by M.

This is something that should be familiar if we know about group cohomology.

Proof.

Let

be an extension with, as usual,

B → A

ker π

= 0.

We now try to produce a cocycle from this.

Let

be any

-linear map

A → B

such that

(

)) =

. This is possible

since

is surjective. Equivalently,

(

)) =

b mod I

. We define a

-linear map

: A ⊗ A → I

∼

⊗ a

7→ ρ(a

)ρ(a

) − ρ(a

Note that the image lies in I since

ρ(a

)ρ(a

) ≡ ρ(a

) (mod I).

It is a routine check that f

is a 2-cocycle, i.e. it lies in ker δ

If we replace ρ by any other ρ

, we get f

, and we have

⊗ a

) − f

⊗ a

)

= ρ(a

)(ρ(a

) − ρ

)) − (ρ(a

) − ρ

)) + (ρ(a

) − ρ

))ρ

)

= a

· (ρ(a

) − ρ

)) − (ρ(a

) − ρ

)) + (ρ(a

) − ρ

)) · a

where · denotes the A-A-bimodule action in I. Thus, we find

− f

= δ

(ρ − ρ

noting that ρ − ρ

actually maps to I.

So we obtain a map from the isomorphism classes of extensions to the second

cohomology group.

Conversely, given an

-bimodule

and a 2-cocycle

A ⊗ A → M

, we

let

= A ⊕ M

as k-vector spaces. We define the multiplication map

, m

)(a

, m

) = (a

, a

+ m

+ f(a

⊗ a

)).

This is associative precisely because of the 2-cocycle condition. The map (

a, m

)

→

yields a homomorphism

B → A

, with kernel

being a two-sided ideal of

which has

= 0. Moreover,

∼

as an

-bimodule. Taking

A → B

ρ(a) = (a, 0) yields the 2-cocycle we started with.

Finally, let

be another 2 co-cycle cohomologous to

. Then there is a

linear map τ : A → M with

f − f

= δ

τ.

That is,

f(a

⊗ A

) = f

⊗ a

) + a

· τ(a

) − τ(a

) + τ(a

) · a

Then consider the map B

→ B

given by

(a, m) 7→ (a, m + τ(a)).

One then checks this is an isomorphism of extensions. And then we are done.

In the proof, we see 0 corresponds to the semi-direct product.

Corollary. If HH

(A, M ) = 0, then all extensions are split.

We now prove some theorems that appear to be trivialities. However, they

are trivial only because we now have the machinery of Hochschild cohomology.

When they were first proved, such machinery did not exist, and they were written

in a form that seemed less trivial.

Theorem (Wedderburn, Malcev). Let B be a k-algebra satisfying

– Dim(B/J(B)) ≤ 1.

– J(B)

= 0

Then there is an subalgebra A

∼

B/J(B) of B such that

B = A n J(B).

Furthermore, if

Dim

(

B/J

(

)) = 0, then any two such subalgebras

A, A

are

conjugate, i.e. there is some x ∈ J(B) such that

= (1 + x)A(1 + x)

−1

Notice that 1 + x is a unit in B.

In fact, the same result holds if we only require

(

) to be nilpotent. This

follows from an induction argument using this as a base case, which is messy

and not really interesting.

Proof. We have J(B)

= 0. Since we know Dim(B/J(B)) ≤ 1, we must have

(A, J(B)) = 0

where

∼

J(B)

Note that we regard

(

) as an

-bimodule here. So we know that all

extension of A by J(B) are semi-direct, as required.

Furthermore, if

Dim

(

B/J

(

)) = 0, then we know

(

A, J

(

)) = 0. So by

our older lemmas, we see that complements are all conjugate, as required.

Corollary.

is algebraically closed and

dim

B < ∞

, then there is a subal-

gebra A of B such that

∼

J(B)

and

B = A n J(B).

Moreover,

is unique up to conjugation by units of the form 1+

with

x ∈ J

(

Proof.

We need to show that

Dim

(

) = 0. But we know

B/J

(

) is a semi-

simple

-algebra of finite dimension, and in particular is Artinian. So by

Artin–Wedderburn, we know

B/J

(

) is a direct sum of matrix algebras over

(since k is algebraically closed and dim

(B/J(B))).

We have previously observed that

(

) is

-separable. Since

-separability

behaves well under direct sums, we know

B/J

(

) is

-separable, hence has

dimension zero.

It is a general fact that J(B) is nilpotent.

3.3 Star products

We are now going to do study some deformation theory. Suppose

is a

-algebra.

We write

for the underlying vector space of

. Then there is a natural algebra

structure on

V ⊗

[[

]] =

[[

]], which we may write as

[[

]]. However, we

might we to consider more interesting algebra structures on this vector space.

Of course, we don’t want to completely forget the algebra structure on

. So we

make the following definition:

Definition

(Star product)

Let

be a

-algebra, and let

be the underlying

vector space. A star product is an associative

[[

]]-bilinear product on

[[

]]

that reduces to the multiplication on A when we set t = 0.

Can we produce non-trivial star products? It seems difficult, because when

we write down an attempt, we need to make sure it is in fact associative, and

that might take quite a bit of work. One example we have already seen is the

following:

Example.

Given a filtered

-algebra

, we formed the Rees algebra associated

with the filtration, and it embeds as a vector space in (

gr A

)[[

]]. Thus we get a

product on (gr A

)[[t]].

There are two cases where we are most interested in — when

(

) or

(

). We saw that

gr A

was actually a (commutative) polynomial algebra.

However, the product on the Rees algebra is non-commutative. So the

∗

-product

will be non-commutative.

In general, the availability of star products is largely controlled by the

Hochschild cohomology of

. To understand this, let’s see what we actually

need to specify to get a star product. Since we required the product to be a

k[[t]]-bilinear map

f : V [[t]] × V [[t]] → V [[t]],

all we need to do is to specify what elements of

are sent to. Let

a, b ∈ V = A. We write

f(a, b) = ab + tF

(a, b) + t

(a, b) + · · · .

Because of bilinearity, we know

are

-bilinear maps, and so correspond to

k-linear maps V ⊗ V → V . For convenience, we will write

(a, b) = ab.

The only non-trivial requirement f has to satisfy is associativity:

f(f(a, b), c) = f(a, f (b, c)).

What condition does this force on our

? By looking at coefficients of

, this

implies that for all λ = 0, 1, 2, · · · , we have

m+n=λ

m,n≥0



(a, b), c) − F

(a, F

(b, c))



= 0. (∗)

For

= 0, we are just getting the associativity of the original multiplication on

A. When λ = 1, then this says

(b, c) − F

(ab, c) + F

(a, bc) − F

(a, b)c = 0.

All this says is that

is a 2-cocycle! This is not surprising. Indeed, we’ve

just seen (a while ago) that working mod

, the extensions of

are

governed by

. Thus, we will refer to 2-cocycles as infinitesimal deformations

in this context.

Note that given an arbitrary 2 co-cycle

A ⊗ A → A

, it may not be possible

to produce a star product with the given 2-cocycle as F

Definition

(Integrable 2-cocycle)

Let

A ⊗ A → A

be a 2-cocycle. Then it

is integrable if it is the F

of a star product.

We would like to know when a 2-cocycle is integrable. Let’s rewrite (

∗

) as

(†

m+n=λ

m,n>0



(a, b), c) − F

(a, F

(b, c))



= (δ

)(a, b, c). (†

)

Here we are identifying F

with the corresponding k-linear map A ⊗ A → A.

For λ = 2, this says

(ab), c) − F

(a, F

(b, c)) = (δ

)(a, b, c).

is a 2-cocycle, then one can check that the LHS gives a 3-cocycle. If

is integrable, then the LHS has to be equal to the RHS, and so must be a

coboundary, and thus has cohomology class zero in HH

(A, A).

In fact, if

, · · · , F

λ−1

satisfy (

†

)

, · · · ,

(

†

λ−1

), then the LHS of (

†

) is also

a 3-cocycle. If it is a coboundary, and we have defined

, · · · , F

, then we can

define

such that (

†

) holds. However, if it is not a coboundary, then we get

stuck, and we see that our choice of

, · · · , F

λ−1

does not lead to a

∗

-product.

The 3-cocycle appearing on the LHS of (

†

) is an obstruction to integrability.

If, however, they are always coboundaries, then we can inductively define

, F

, · · · to give a ∗-product. Thus we have proved

Theorem

(Gerstenhaber)

(

A, A

) = 0, then all infinitesimal deformations

are integrable.

Of course, even if

(

A, A

)

= 0, we can still get

∗

-products, but we need

to pick our F

more carefully.

Now after producing star products, we want to know if they are equivalent.

Definition

(Equivalence of star proeducts)

Two star products

and

are

equivalent on

V ⊗ k

[[

]] if there is a

[[

]]-linear automorphism Φ of

[[

]] of the

form

Φ(a) = a + tφ

(a) + t

(a) + · · ·

sch that

f(a, b) = Φ

−1

g(Φ(a), Φ(b)).

Equivalently, the following diagram has to commute:

V [[t]] ⊗ V [[t]] V [[t]]

Φ⊗Φ

Star products equivalent to the usual product on A ⊗ k[[t]] are called trivial.

Theorem

(Gerstenhaber)

Any non-trivial star product

is equivalent to one

of the form

g(a, b) = ab + t

(a, b) + t

n+1

(a, b) + · · · ,

where

is a 2-cocycle and not a coboundary. In particular, if

(

A, A

) = 0,

then any star product is trivial.

Proof. Suppose as usual

f(a, b) = ab + tF

(a, b) + t

(a, b) + · · · ,

and suppose F

, · · · , F

n−1

= 0. Then it follows from (†) that

= 0.

If F

is a coboundary, then we can write

= −δφ

for some φ

: A → A. We set

(a) = a + t

(a).

Then we can compute that

−1

(f(Φ

(a), Φ

(b)))

is of the form

ab + t

n+1

(a, b) + · · · .

So we have managed to get rid of a further term, and we can keep going until

we get the first non-zero term not a coboundary.

Suppose this never stops. Then

is trivial — we are using that

· · · ◦

n+2

◦

n+1

◦

converges in the automorphism ring, since we are adding terms of

higher and higher degree.

We saw that derivations can be thought of as infinitesimal automorphisms.

One can similarly consider k[[t]]-linear maps of the form

Φ(a) = a + tφ

(a) + t

(a) + · · ·

and consider whether they define automorphisms of

A ⊗ k

[[

]]. Working modulo

, we have already done this problem — we are just considering automorphisms

of A[ε], and we saw that these automorphisms correspond to derivations.

Definition

(Integrable derivation)

We say a derivation is integrable if there is

an automorphism of A ⊗ k[[t]] that gives the derivation when we mod t

In this case, the obstructions are 2-cocycles which are not coboundaries.

Theorem

(Gerstenhaber)

Suppose

(

A, A

) = 0. Then all derivations are

integrable.

The proof is an exercise on the third example sheet.

We haven’t had many examples so far, because Hochschild cohomology is

difficult to compute. But we can indeed do some examples.

Example. Let A = k[x]. Since A is commutative, we have

(A, A) = A.

Since A is commutative, A has no inner derivations. So we have

(A, A) = DerA =



f(x)

: f(x) ∈ k[x]



For any i > 1, we have

(A, A) = 0.

So we have

Dim(A) = 1.

We can do this by explicit calculation. If we look at our Hochschild chain

complex, we had a short exact sequence

0 ker µ A ⊗ A A 0 (∗)

and thus we have a map

A ⊗ A ⊗ A A ⊗ A

whose image is ker µ .

The point is that

ker µ

is a projective

-bimodule. This will mean that

(

A, M

) = 0 for

i ≥

2 in the same way we used to show that when

is a

projective A-A-bimodule for i ≥ 1. In particular, HH

(A, A) = 0 for i ≥ 2.

To show that ker µ is projective, we notice that

A ⊗ A = k[X] ⊗

k[X]

∼

k[X, Y ].

So the short exact sequence (∗) becomes

0 (X − Y )k[X, Y ] k[X, Y ] k[X] 0 .

So (X − Y )k[X, Y ] is a free k[X, Y ] module, and hence projective.

We can therefore use our theorems to see that any extension of

[

] by

[

] is split, and any

∗

-product is trivial. We also get that any derivation is

integrable.

Example. If we take A = k[X

, X

], then again this is commutative, and

(A, A) = A

(A, A) = DerA.

We will talk about HH

later, and similarly

(A, A) = 0

for i ≥ 3.

From this, we see that we may have star products other than the trivial ones,

and in fact we know we have, because we have one arising from the Rees algebra

(

). But we know that any infinitesimal deformation yields a star product.

So there are much more.

3.4 Gerstenhaber algebra

We now want to understand the equations (

†

) better. To do so, we consider the

graded vector space

(A, A) =

∞

A=0

(A, A),

as a whole. It turns out this has the structure of a Gerstenhaber algebra

The first structure to introduce is the cup product. They are a standard tool

in cohomology theories. We will write

(A, A) = Hom

⊗n

, A) = Hom

A-A

⊗(n+2)

The Hochschild chain complex is then the graded chain complex S

(A, A).

Definition (Cup product). The cup product

: S

(A, A) ⊗ S

(A, A) → S

m+n

(A, A)

is defined by

(f  g)(a

⊗ · · · ⊗ a

⊗ b

⊗ · · · ⊗ b

) = f(a

⊗ · · · ⊗ a

) · g(b

⊗ · · · ⊗ b

where a

, b

∈ A.

Under this product, S

(A, A) becomes an associative graded algebra.

Observe that

δ(f  g) = δf  g + (−1)

f  δg.

So we say

is a (left) graded derivation of the graded algebra

(

A, A

). In

homological (graded) algebra, we often use the same terminology but with

suitable sign changes which depends on the degree.

Note that the cocycles are closed under



. So cup product induces a product

(

A, A

). If

f ∈ S

(

A, A

) and

g ∈ S

(

A, A

), and both are cocycles, then

(−1)

(g  f − (−1)

(f  g)) = δ(f ◦ g),

where f ◦ g is defined as follows: we set

f ◦

g(a

⊗ · · · ⊗ a

i−1

⊗ b

⊗ · · · ⊗ b

⊗ a

i+1

· · · ⊗ a

)

= f(a

⊗ · · · ⊗ a

i−1

⊗ g(b

⊗ · · · ⊗ b

) ⊗ a

i+1

⊗ · · · ⊗ a

Then we define

f ◦ g =

i=1

(−1)

(n−1)(i−1)

f ◦

This product

◦

is not an associative product, but is giving a pre-Lie structure to

(A, A).

Definition (Gerstenhaber bracket). The Gerstenhaber bracket is

[f, g] = f ◦ g − (−1)

(n+1)(m+1)

g ◦ f

This defines a graded Lie algebra structure on the Hochschild chain complex,

but notice that we have a degree shift by 1. It is a grade Lie algebra on

(A, A).

Of course, we really should define what a graded Lie algebra is.

Definition (Graded Lie algebra). A graded Lie algebra is a vector space

L =

with a bilinear bracket [ · , · ] : L × L → L such that

– [L

, L

] ⊆ L

i+j

;

– [f, g] − (−1)

[g, f]; and

– The graded Jacobi identity holds:

(−1)

[[f, g], h] + (−1)

[[g, h], f] + (−1)

[[h, f ], g] = 0

where f ∈ L

, g ∈ L

, h ∈ L

In fact,

(

A, A

) is a differential graded Lie algebra under the Gerstenhaber

bracket.

Lemma. The cup product on HH

(A, A) is graded commutative, i.e.

f  g = (−1)

(g  f).

when f ∈ HH

(A, A) and g ∈ HH

(A, A).

Proof. We previously “noticed” that

(−1)

(g  f − (−1)

(f  g)) = δ(f ◦ g),

Definition

(Gerstenhaber algebra)

A Gerstenhaber algebra is a graded vector

space

H =

with

a graded Lie algebra with respect to a bracket [

· , ·

] :

× H

→

m+n−1

, and an associative product



× H

→ H

m+n

which is graded

commutative, such that if

f ∈ H

, then [

f, ·

] acts as a degree

m −

1 graded

derivation of :

[f, g  h] = [f, g]  h + (−1)

(m−1)

ng  [f, h]

if g ∈ H

This is analogous to the definition of a Poisson algebra. We’ve seen that

(A, A) is an example of a Gerstenhaber algebra.

We can look at what happens in low degrees. We know that

is a

commutative k-algebra, and : H

× H

→ H

is a module action.

Also,

is a Lie algebra, and [

· , ·

] :

× H

→ H

is a Lie module

action, i.e.

gives us a Lie algebra representation of

. In other words,

the corresponding map [

· , ·

] :

→ End

(

) gives us a map of Lie algebras

→ Der(H

The prototype Gerstenhaber algebra is the exterior algebra

DerA

for a

commutative algebra A (with A in degree 0).

Explicitly, to define the exterior product over

, we first consider the tensor

product over A of two A-modules V and W , defined by

V ⊗

W =

V ⊗

hav ⊗ w − v ⊗ awi

The exterior product is then

V ∧

V =

V ⊗

hv ⊗ v : v ∈ V i

The product is given by the wedge, and the Schouten bracket is given by

[λ

∧ · · · ∧ λ

, λ

∧ · · · ∧ λ

]

= (−1)

(m−1)(n−1)

i,j

(−1)

i+j

[λ

, λ

] λ

∧ · · · λ

| {z }

ith missing

∧ λ

∧ · · · ∧ λ

| {z }

jth missing

For any Gerstenhaber algebra

, there is a canonical homomorphism

of Gerstenhaber algebras

→ H.

Theorem

(Hochschild–Kostant–Ronsenberg (HKR) theorem)

is a “smooth”

commutative k-algebra, and char k = 0, then the canonical map

(DerA) → HH

∗

(A, A)

is an isomorphism of Gerstenhaber algebras.

We will not say what “smooth” means, but this applies if

[

, · · · , X

or if

and

is appropriate functions on smooth manifolds or algebraic

varieties.

In the 1960’s, this was stated just for the algebra structure, and didn’t think

about the Lie algebra.

Example. Let A = k[X, Y ], with char k = 0. Then HH

(A, A) = A and

(A, A) = DerA

∼



p(X, Y )

∂

∂y

+ q(X, Y )

∂

∂Y

: p, q ∈ A



So we have

(A, A) = DerA ∧

DerA,

which is generated as an A-modules by

∂

∂X

∧

∂

∂Y

. Then

(A, A) = 0 for all i ≥ 3

We can now go back to talk about star products. Recall when we considered

possible star products on

V ⊗

[[

]], where

is the underlying vector space of

the algebra

. We found that associativity of the star product was encapsulated

by some equations (†

). Collectively, these are equivalent to the statement

Definition (Maurer–Cartan equation). The Maurer–Cartan equation is

δf +

[f, f]

Gerst

= 0

for the element

f =

where F

(a, b) = ab.

When we write [

· , ·

]

Gerst

, we really mean the

[[

]]-linear extension of the

Gerstenhaber bracket.

If we want to think of things in cohomology instead, then we are looking

at things modulo coboundaries. For the graded Lie algebra

(

DerA

), the

Maurer–Cartan elements, i.e. solutions of the Maurer–Cartan equation, are the

formal Poisson structures. They are formal power series of the form

Π =

for π

∈ DerA ∧ DerA, satisfying

[Π, Π] = 0.

There is a deep theorem of Kontsevich from the early 2000’s which implies

Theorem (Kontsevich). There is a bijection



equivalence classes

of star products



←→

classes of formal

Poisson structures

This applies for smooth algebras in

char

0, and in particular for polynomial

algebras A = k[X

, · · · , X

This is a difficult theorem, and the first proof appeared in 2002.

An unnamed lecturer once tried to give a Part III course with this theorem as

the punchline, but the course ended up lasting 2 terms, and they never reached

the punchline.

3.5 Hochschild homology

We don’t really have much to say about Hochschild homology, but we are morally

obliged to at least write down the definition.

To do Hochschild homology, we apply

· ⊗

A-A

for an

-bimodule

the Hochschild chain complex.

· · · A ⊗

A ⊗

A A ⊗

A A 0.

d d

We will ignore the

→ A →

0 bit. We need to consider what

· ⊗

A-A

means. If

we have bimodules

and

, we can regard

as a right

A ⊗ A

-module. We

can also think of W as a left A ⊗ A

module. We let

B = A ⊗ A

and then we just consider

V ⊗

W =

V ⊗

hvx ⊗ w − v ⊗ xw : w ∈ Bi

V ⊗

hava

⊗ w − v ⊗ a

wai

Thus we have

· · · (A ⊗

A ⊗

A) ⊗

A-A

M (A ⊗

A) ⊗

A-A

∼

Definition (Hochschild homology). The Hochschild homology groups are

(A, M ) =

im b

(A, M ) =

ker b

i−1

im b

for i > 0.

A long time ago, we counted the number of simple

-modules for

alge-

braically closed of characteristic

when

is finite. In the proof, we used

[A,A]

and we pointed out this is HH

(A, A).

Lemma.

(A, M ) =

hxm − mx : m ∈ M, x ∈ Ai

In particular,

(A, A) =

[A, A]

Proof. Exercise.

4 Coalgebras, bialgebras and Hopf algebras

We are almost at the end of the course. So let’s define an algebra.

Definition (Algebra). A k-algebra is a k-vector space A and k-linear maps

µ : A ⊗ A → A u : k → A

x ⊗ y 7→ xy λ 7→ λI

called the multiplication/product and unit such that the following two diagrams

commute:

A ⊗ A ⊗ A A ⊗ A

A ⊗ A A

µ⊗id

id ⊗µ

k ⊗ A A ⊗ A A ⊗ k

∼

u⊗id

id ⊗u

∼

These encode associativity and identity respectively.

Of course, the point wasn’t to actually define an algebra. The point is to

define a coalgebra, whose definition is entirely dual.

Definition (Coalgebra). A coalgebra is a k-vector space C and k-linear maps

∆ : C → C ⊗ C ε : C → k

called comultiplication/coproduct and counit respectively, such that the following

diagrams commute:

C ⊗ C ⊗ C C ⊗ C

C ⊗ C C

id ⊗∆

∆⊗id

∆

k ⊗ C C ⊗ C C ⊗ k

ε⊗id id ⊗ε

∼

These encode coassociativity and coidentity

A morphism of coalgebras

C → D

is a

-linear map such that the following

diagrams commute:

C D

C ⊗ C D ⊗ D

∆

f⊗f

C D

k k

A subspace

is a co-ideal if ∆(

)

≤ C ⊗ I

I ⊗ C

, and

(

) = 0. In this

case, C/I inherits a coproduct and counit.

A cocommutative coalgebra is one for which

τ ◦

∆ = ∆, where

V ⊗ W →

W ⊗ V given by the v ⊗ w 7→ w ⊗ v is the “twist map”.

It might be slightly difficult to get one’s head around what a coalgebra actually

is. It, of course, helps to look at some examples, and we will shortly do so. It

also helps to know that for our purposes, we don’t really care about coalgebras

per se, but things that are both algebras and coalgebras, in a compatible way.

There is a very natural reason to be interested in such things. Recall that

when doing representation theory of groups, we can take the tensor product of

two representations and get a new representation. Similarly, we can take the

dual of a representation and get a new representation.

If we try to do this for representations (ie. modules) of general algebras, we

see that this is not possible. What is missing is that in fact, the algebras

and

(

) also have the structure of coalgebras. In fact, they are Hopf algebras,

which we will define soon.

We shall now write down some coalgebra structures on kG and U(g).

Example. If G is a group, then kG is a co-algebra, with

∆(g) = g ⊗ g



(g)



We should think of the specification ∆(

) =

g ⊗ g

as saying that our groups act

diagonally on the tensor products of representations. More precisely, if

V, W

are

representations and v ∈ V, w ∈ W , then g acts on v ⊗ w by

∆(g) · (v ⊗ w) = (g ⊗ g) · (v ⊗ w) = (gv) ⊗ (gw).

Example.

For a Lie algebra

over

, the universal enveloping algebra

(

) is

a co-algebra with

∆(x) = x ⊗ 1 + 1 ⊗ x

for x ∈ g, and we extend this by making it an algebra homomorphism.

To define ε, we note that elements of U(g) are uniquely of the form

λ +

,...,i

· · · x

where {x

} is a basis of g (the PBW theorem). Then we define



λ +

,...,i

· · · x



= λ.

This time, the specification of ∆ is telling us that if

X ∈ g

and

v, w

are elements

of a representation of g, then X acts on the tensor product by

∆(X) · (v ⊗ w) = Xv ⊗ w + v ⊗ Xw.

Example. Consider

O(M

(k)) = k[X

: 1 ≤ i, j ≤ n],

the polynomial functions on

n × n

matrices, where

denotes the

th entry.

Then we define

∆(X

) =

i=1

⊗ X

and

ε(X

) = δ

These are again algebra maps.

We can also talk about

(

)) and

(

)). The formula of the

determinant gives an element

D ∈ O

(

)). Then

(

)) is given by

adding a formal inverse to

(

)), and

(

)) is obtained by

quotienting out O(GL

(k)) by the bi-ideal hD − 1i.

From an algebraic geometry point of view, these are the coordinate algebra

of the varieties M

(k), GL

(k) and SL

(k).

This is dual to matrix multiplication.

We have seen that we like things that are both algebras and coalgebras,

compatibly. These are known as bialgebras.

Definition

(Bialgebra)

A bialgebra is a

-vector space

and maps

µ, υ,

∆

, ε

such that

(i) (B, µ, u) is an algebra.

(ii) (B, ∆, ε) is a coalgebra.

(iii) ∆ and ε are algebra morphisms.

(iv) µ and u are coalgebra morphisms.

Being a bialgebra means we can take tensor products of modules and still

get modules. If we want to take duals as well, then it turns out the right notion

is that of a Hopf algebra:

Definition

(Hopf algebra)

A bialgebra (

H, µ, u,

∆

, ε

) is a Hopf algebra if there

is an antipode S : H → H that is a k-linear map such that

µ ◦ (S ⊗ id) ◦ ∆ = µ ◦ (id ⊗S) ◦ ∆ = u ◦ ε.

Example. kG is a Hopf algebra with S(g) = g

−1

Example. U(g) is a Hopf algebra with S(x) = −x for x ∈ U(g).

Note that our examples are all commutative or co-commutative. The term

quantum groups usually refers to a non-commutative non-co-commutative Hopf

algebras. These are neither quantum nor groups.

As usual, we write

∗

for

Hom

(

V, k

), and we note that if we have

V → W

then this induces a dual map α

∗

: W

∗

→ V

∗

Lemma.

is a coalgebra, then

∗

is an algebra with multiplication ∆

∗

(that

is, ∆

∗

⊗C

∗

) and unit ε

∗

. If C is co-commutative, then C

∗

is commutative.

However, if an algebra

is infinite dimensional as a

-vector space, then

∗

may not be a coalgebra. The problem is that (

∗

⊗ A

∗

) is a proper subspace of

(

A ⊗ A

)

∗

, and

∗

of an infinite dimensional

need not take values in

∗

⊗ A

∗

However, all is fine for finite dimensional

, or if

is graded with finite

dimensional components, where we can form a graded dual.

In general, for a Hopf algebra H, one can define the Hopf dual,

= {f ∈ H

∗

: ker f contains an ideal of finite codimension}.

Example.

Let

be a finite group. Then (

)

∗

is a commutative non-co-

commutative Hopf algebra if G is non-abelian.

Let

{g}

be the canonical basis for

, and

{φ

}

be the dual basis of (

)

∗

Then

∆(φ

) =

⊗ φ

There is an easy way of producing non-commutative non-co-commutative Hopf

algebras — we take a non-commutative Hopf algebra and a non-co-commutative

Hopf algebra, and take the tensor product of them, but this is silly.

The easiest non-trivial example of a non-commutative non-co-commutative

Hopf algebra is the Drinfeld double, or quantum double, which is a general

construction from a finite dimensional hopf algebra.

Definition (Drinfeld double). Let G be a finite group. We define

D(G) = (kG)

∗

⊗

as a vector space, and the algebra structure is given by the crossed product

(kG)

∗

o G, where G acts on (kG)

∗

(x) = f(gxg

−1

Then the product is given by

⊗ g

)(f

⊗ g

) = f

−1

⊗ g

The coalgebra structure is the tensor of the two coalgebras (

)

∗

and

, with

∆(φ

⊗ h) =

⊗ h ⊗ φ

⊗ h.

(

) is quasitriangular, i.e. there is an invertible element

(

)

⊗ D

(

)

such that

R∆(x)R

−1

= τ(∆(x)),

where τ is the twist map. This is given by

R =

(φ

⊗ 1) ⊗ (1 ⊗ g)

(φ

⊗ 1) ⊗ (1 ⊗ g

−1

The equation

∆

−1

∆ results in an isomorphism between

U ⊗V

and

V ⊗U

for D(G)-bimodules U and V , given by flip follows by the action of R.

is non-abelian, then this is non-commutative and non-co-commutative.

The point of defining this is that the representations of

(

) correspond to the

G-equivariant k-vector bundles on G.

As we said, this is a general construction.

Theorem

(Mastnak, Witherspoon (2008))

The bialgebra cohomology

(

H, H

) for a finite-dimensional Hopf algebra is equal to

(

)

, k

where k is the trivial module, and D(H) is the Drinfeld double.

In 1990, Gerstenhaber and Schack defined bialgebra cohomology, and proved

results about deformations of bialgebras analogous to our results from the

previous chapter for algebras. In particular, one can consider infinitesimal

deformations, and up to equivalence, these correspond to elements of the 2nd

cohomology group.

There is also the question as to whether an infinitesimal deformation is

integrable to give a bialgebra structure on

V ⊗ k

[[

]], where

is the underlying

vector space of the bialgebra.

Theorem

(Gerstenhaber–Schack)

Every deformation is equivalent to one where

the unit and counit are unchnaged. Also, deformation preserves the existence of

an antipode, though it might change.

Theorem

(Gerstenhaber–Schack)

All deformations of

(

)) or

(

))

are equivalent to one in which the comultiplication is unchanged.

We nwo try to deform

(

)). By the previous theorems, we only have

to change the multiplication. Consider O

(k)) defined by

= qX

= X

− X

= (q

−1

− q)X

We define the quantum determinant

det

= X

− q

−1

= X

− qX

Then

∆(det

) = det

⊗ det

, ε(det

) = 1.

Then we define

O(SL

(k)) =

O(M

(k))

(det

−1)

where we are quotienting by the 2-sided ideal. It is possible to define an antipode,

given by

S(X

) = X

S(X

) = −qX

S(X

) = −q

−1

S(X

) = X

and this gives a non-commutative and non-co-commutative Hopf algebra. This

is an example that we pulled out of a hat. But there is a general construction

due to Faddeev–Reshetikhin–Takhtajan (1988) via

-matrices, which are a way

of producing a k-linear map

V ⊗ V → V ⊗ V,

where V is a fintie-dimesnional vector space.

We take a basis

, · · · , e

, and thus a basis

⊗ e

V ⊗ V

. We write

for the matrix of R, defined by

R(e

⊗ e

) =

`,m

⊗ e

The rows are indexed by pairs (

, m

), and the columns by pairs (

i, j

), which are

put in lexicographic order.

The action of

V ⊗ V

induces 3 different actions on

V ⊗ V ⊗ V

. For

s, t ∈ {

}

, we let

be the invertible map

V ⊗ V ⊗ V → V ⊗ V ⊗ V

which

acts like

on the

th and

th components, and identity on the other. So for

example,

⊗ e

⊗ v) =

i,j

⊗ e

⊗ v.

Definition

(Yang–Baxter equation)

. R

satisfies the quantum Yang–Baxter

equation (QYBE ) if

= R

and the braided form of QYBE (braid equation) if

= R

Note that

satisfies QYBE iff

Rτ

satisfies the braid equation. Solutions to

either case are R-matrices.

Example. The identity map and the twist map τ satisfies both.

Take V to be 2-dimensional, and R to be the map







q 0 0 0

0 1 0 0

0 q − q

−1

1 0

0 0 0 q







where q 6= 0 ∈ K. Thus, we have

R(e

⊗ e

) = qe

⊗ e

R(e

⊗ e

) = e

⊗ e

R(e

⊗ e

) = e

⊗ e

+ (q − q

−1

⊗ e

R(e

⊗ e

) = qe

⊗ e

and this satisfies QYBE. Similarly,

(Rτ)







q 0 0 0

0 0 1 0

0 1 q − q

−1

0 0 0 q







satisfies the braid equation.

We now define the general construction.

Definition (R-symmetric algebra). Given the tensor algebra

T (V ) =

∞

n=0

⊗n

we form the R-symmetric algebra

(V ) =

T (V )

hz − R(z) : z ∈ V ⊗ V i

Example. If R is the identity, then S

(V ) = T (V ).

Example. If R = τ, then S

(V ) is the usual symmetric algebra.

Example. The quantum plane O

) can be written as S

(V ) with

R(e

⊗ e

) = qe

⊗ e

R(e

⊗ e

) = e

⊗ e

R(e

⊗ e

) = q

−1

⊗ e

R(e

⊗ e

) = e

⊗ e

Generally, given a

which is finite-dimensional as a vector space, we can

identify (V ⊗ V )

∗

with V

∗

⊗ V

∗

We set

V ⊗ V

∗

∼

End

(

)

∼

(

). We define

and

∗

E ⊗ E →

E ⊗ E

, where

acts like

on terms 1 and 3 in

V ⊗ V

∗

⊗ V ⊗ V

∗

, and

identity on the rest; R

∗

acts like R

∗

on terms 2 and 4.

Definition

(Coordinate algebra of quantum matrices)

The coordinate algebra

of quantum matrices associated with R is

T (E)

(z) − R

∗

(z) : z ∈ E ⊗ Ei

= S

(E),

where

T = R

∗

−1

The coalgebra structure remains the same as

(

)), and for the antipode,

we write

for the image of

(

), and similarly

for

. Then we map

7→

j=1

⊗ E

7→

i=1

⊗ X

This is the general construction we are up for.

Example. We have

(k)) = A

Rτ

(V )

for







q 0 0 0

0 1 0 0

0 q − q

−1

1 0

0 0 0 q





