3Hochschild homology and cohomology
III Algebras
3.3 Star products
We are now going to do study some deformation theory. Suppose
A
is a
k
-algebra.
We write
V
for the underlying vector space of
A
. Then there is a natural algebra
structure on
V ⊗
k
k
[[
t
]] =
V
[[
t
]], which we may write as
A
[[
t
]]. 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
A
. So we
make the following definition:
Definition
(Star product)
.
Let
A
be a
k
-algebra, and let
V
be the underlying
vector space. A star product is an associative
k
[[
t
]]-bilinear product on
V
[[
t
]]
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
k
-algebra
A
0
, we formed the Rees algebra associated
with the filtration, and it embeds as a vector space in (
gr A
0
)[[
t
]]. Thus we get a
product on (gr A
0
)[[t]].
There are two cases where we are most interested in — when
A
0
=
A
n
(
k
) or
A
0
=
U
(
g
). We saw that
gr A
0
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
A
. 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
V
=
A
are sent to. Let
a, b ∈ V = A. We write
f(a, b) = ab + tF
1
(a, b) + t
2
F
2
(a, b) + · · · .
Because of bilinearity, we know
F
i
are
k
-bilinear maps, and so correspond to
k-linear maps V ⊗ V → V . For convenience, we will write
F
0
(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
F
i
? By looking at coefficients of
t
, this
implies that for all λ = 0, 1, 2, · · · , we have
X
m+n=λ
m,n≥0
F
m
(F
n
(a, b), c) − F
m
(a, F
n
(b, c))
= 0. (∗)
For
λ
= 0, we are just getting the associativity of the original multiplication on
A. When λ = 1, then this says
aF
1
(b, c) − F
1
(ab, c) + F
1
(a, bc) − F
1
(a, b)c = 0.
All this says is that
F
1
is a 2-cocycle! This is not surprising. Indeed, we’ve
just seen (a while ago) that working mod
t
2
, the extensions of
A
by
A
A
A
are
governed by
HH
2
. 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
1
.
Definition
(Integrable 2-cocycle)
.
Let
f
:
A ⊗ A → A
be a 2-cocycle. Then it
is integrable if it is the F
1
of a star product.
We would like to know when a 2-cocycle is integrable. Let’s rewrite (
∗
) as
(†
λ
):
X
m+n=λ
m,n>0
F
m
(F
n
(a, b), c) − F
m
(a, F
n
(b, c))
= (δ
2
F
λ
)(a, b, c). (†
λ
)
Here we are identifying F
λ
with the corresponding k-linear map A ⊗ A → A.
For λ = 2, this says
F
1
(F
1
(ab), c) − F
1
(a, F
1
(b, c)) = (δ
2
F
2
)(a, b, c).
If
F
1
is a 2-cocycle, then one can check that the LHS gives a 3-cocycle. If
F
1
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
3
(A, A).
In fact, if
F
1
, · · · , F
λ−1
satisfy (
†
1
)
, · · · ,
(
†
λ−1
), then the LHS of (
†
λ
) is also
a 3-cocycle. If it is a coboundary, and we have defined
F
1
, · · · , F
λ
1
, then we can
define
F
λ
such that (
†
λ
) holds. However, if it is not a coboundary, then we get
stuck, and we see that our choice of
F
1
, · · · , F
λ−1
does not lead to a
∗
-product.
The 3-cocycle appearing on the LHS of (
†
λ
) is an obstruction to integrability.
If, however, they are always coboundaries, then we can inductively define
F
1
, F
2
, · · · to give a ∗-product. Thus we have proved
Theorem
(Gerstenhaber)
.
If
HH
3
(
A, A
) = 0, then all infinitesimal deformations
are integrable.
Of course, even if
HH
3
(
A, A
)
6
= 0, we can still get
∗
-products, but we need
to pick our F
1
more carefully.
Now after producing star products, we want to know if they are equivalent.
Definition
(Equivalence of star proeducts)
.
Two star products
f
and
g
are
equivalent on
V ⊗ k
[[
t
]] if there is a
k
[[
t
]]-linear automorphism Φ of
V
[[
t
]] of the
form
Φ(a) = a + tφ
1
(a) + t
2
φ
2
(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]]
V [[t]] ⊗ V [[t]] V [[t]]
f
Φ⊗Φ
Φ
g
Star products equivalent to the usual product on A ⊗ k[[t]] are called trivial.
Theorem
(Gerstenhaber)
.
Any non-trivial star product
f
is equivalent to one
of the form
g(a, b) = ab + t
n
G
n
(a, b) + t
n+1
G
n+1
(a, b) + · · · ,
where
G
n
is a 2-cocycle and not a coboundary. In particular, if
HH
2
(
A, A
) = 0,
then any star product is trivial.
Proof. Suppose as usual
f(a, b) = ab + tF
1
(a, b) + t
2
F
2
(a, b) + · · · ,
and suppose F
1
, · · · , F
n−1
= 0. Then it follows from (†) that
δ
2
F
n
= 0.
If F
n
is a coboundary, then we can write
F
n
= −δφ
n
for some φ
n
: A → A. We set
Φ
n
(a) = a + t
n
φ
n
(a).
Then we can compute that
Φ
−1
n
(f(Φ
n
(a), Φ
n
(b)))
is of the form
ab + t
n+1
G
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
f
is trivial — we are using that
· · · ◦
Φ
n+2
◦
Φ
n+1
◦
Φ
n
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
1
(a) + t
2
φ
2
(a) + · · ·
and consider whether they define automorphisms of
A ⊗ k
[[
t
]]. Working modulo
t
2
, 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
2
.
In this case, the obstructions are 2-cocycles which are not coboundaries.
Theorem
(Gerstenhaber)
.
Suppose
HH
2
(
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
HH
0
(A, A) = A.
Since A is commutative, A has no inner derivations. So we have
HH
1
(A, A) = DerA =
f(x)
d
dx
: f(x) ∈ k[x]
.
For any i > 1, we have
HH
i
(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
d
whose image is ker µ .
The point is that
ker µ
is a projective
A
-
A
-bimodule. This will mean that
HH
i
(
A, M
) = 0 for
i ≥
2 in the same way we used to show that when
A
A
A
is a
projective A-A-bimodule for i ≥ 1. In particular, HH
i
(A, A) = 0 for i ≥ 2.
To show that ker µ is projective, we notice that
A ⊗ A = k[X] ⊗
k
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
k
[
X
] by
k
[
X
] is split, and any
∗
-product is trivial. We also get that any derivation is
integrable.
Example. If we take A = k[X
1
, X
2
], then again this is commutative, and
HH
0
(A, A) = A
HH
1
(A, A) = DerA.
We will talk about HH
2
later, and similarly
HH
i
(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
of
A
1
(
k
). But we know that any infinitesimal deformation yields a star product.
So there are much more.