III Algebraic Topology

8Cup product

8 Cup product

So far, homology and cohomology are somewhat similar. We computed them,

saw they are not the same, but they seem to contain the same information

nevertheless. However, cohomology is 10 times better, because we can define a

ring structure on them, and rings are better than groups.

Just like the case of homology and cohomology, we will be able to write down

the definition easily, but will struggle to compute it.

Definition

(Cup product)

Let

be a commutative ring, and

φ ∈ C

(

;

ψ ∈ C

(X; R). Then φ ^ ψ ∈ C

k+`

(X; R) is given by

(φ ^ ψ)(σ : ∆

k+`

→ X) = φ(σ|

,...,v

]

) · ψ(σ|

,...,v

k+`

]

Here the multiplication is multiplication in

, and

, · · · , v

are the vertices of

∆

k+`

⊆ R

k+`+1

, and the restriction is given by

σ|

,...,x

]

, . . . , t

) = σ





This is a bilinear map.

Notation. We write

∗

(X; R) =

n≥0

(X; R).

This is the definition. We can try to establish some of its basic properties.

We want to know how this interacts with the differential

with the cochains.

The obvious answer

(

φ ^ ψ

) = (d

)

) doesn’t work, because the degrees

are wrong. What we have is:

Lemma. If φ ∈ C

(X; R) and ψ ∈ C

(X; R), then

d(φ ^ ψ) = (dφ) ^ ψ + (−1)

φ ^ (dψ).

This is like the product rule with a sign.

Proof. This is a straightforward computation.

Let σ : ∆

k+`+1

→ X be a simplex. Then we have

((dφ) ^ ψ)(σ) = (dφ)(σ|

,...,v

k+1

]

) · ψ(σ|

k+1

,...,v

k+`+1

]

)

= φ

k+1

i=0

(−1)

σ|

,...,ˆv

,...,v

k+1

]

· ψ(σ|

k+1

,...,v

k+`+1

]

)

(φ ^ (dψ))(σ) = φ(σ|

,...,v

]

) · (dψ)(σ|

,...,v

k+`+1

]

)

= φ(σ|

,...,v

]

) · ψ

k+`+1

i=k

(−1)

i−k

σ|

,...,ˆv

,...,v

k+`+1

]

= (−1)

φ(σ|

,...,v

]

) · ψ

k+`+1

i=k

(−1)

σ|

,...,ˆv

,...,v

k+`+1

]

We notice that the last term of the first expression, and the first term of the

second expression are exactly the same, except the signs differ by

−

1. Then the

remaining terms overlap in exactly 1 vertex, so we have

((dφ) ^ ψ)(σ) + (−1)

φ ^ (dψ)(σ) = (φ ^ ψ)(dσ) = (d(φ ^ ψ))(σ)

as required.

This is the most interesting thing about these things, because it tells us this

gives a well-defined map on cohomology.

Corollary. The cup product induces a well-defined map

^: H

(X; R) × H

(X; R) H

k+`

(X; R)

([φ], [ψ]) [φ ^ ψ]

Proof. To see this is defined at all, as dφ = 0 = dψ, we have

d(φ ^ ψ) = (dφ) ^ ψ ± φ ^ (dψ) = 0.

φ ^ ψ

is a cocycle, and represents the cohomology class. To see this is

well-defined, if φ

= φ + dτ, then

^ ψ = φ ^ ψ + dτ ^ ψ = φ ^ ψ + d(τ ^ ψ) ± τ ^ (dψ).

Using the fact that

dψ

= 0, we know that

^ ψ

and

φ ^ ψ

differ by a

boundary, so [φ

^ ψ] = [φ ^ ψ]. The case where we change ψ is similar.

Note that the operation

is associative on cochains, so associative on

∗

too.

Also, there is a map 1 : C

(X) → R sending σ 7→ 1 for all σ. Then we have

[1] ^ [φ] = [φ].

So we have

Proposition. (H

∗

(X; R), ^, [1]) is a unital ring.

Note that this is not necessarily commutative! Instead, we have the following

graded commutative condition.

Proposition.

Let

be a commutative ring. If

α ∈ H

(

;

) and

β ∈ H

(

;

then we have

α ^ β = (−1)

β ^ α

Note that this is only true for the cohomology classes. It is not true in general

for the cochains. So we would expect that this is rather annoying to prove.

The proof relies on the following observation:

Proposition.

The cup product is natural, i.e. if

X → Y

is a map, and

α, β ∈ H

∗

(Y ; R), then

∗

(α ^ β) = f

∗

(α) ^ f

∗

(β).

So f

∗

is a homomorphism of unital rings.

Proof of previous proposition. Let ρ

: C

(X) → C

(x) be given by

σ 7→ (−1)

n(n+1)/2

σ|

n−1

,...,v

]

The

σ|

n−1

,...,v

]

tells us that we reverse the order of the vertices, and the

factor of (

−

n(n+1)/2

is the sign of the permutation that reverses 0

, · · · , n

. For

convenience, we write

= (−1)

n(n+1)/2

Claim.

We claim that

is a chain map, and is chain homotopic to the identity.

We will prove this later.

Suppose the claim holds. We let

φ ∈ C

(

;

) represent

and

ψ ∈ C

(

;

)

represent β. Then we have

(ρ

∗

φ ^ ρ

∗

ψ)(σ) = (ρ

∗

φ)(σ|

,...,v

]

(ρ

∗

ψ)(σ|

,...,v

k+`

]

)

= φ(ε

· σ|

,...,v

]

)ψ(ε

σ|

k+`

,...,v

]

Thus, we can compute

∗

(ψ ^ φ)(σ) = (ψ ^ φ)(ε

k+`

σ|

k+`

,...,v

]

)

= ε

k+`

ψ(σ|

k+`

,...,v

])φ(σ|

,...,v

]

)

= ε

k+`

(ρ

∗

φ ^ ρ

∗

ψ)(σ).

By checking it directly, we can see that ε

n+`

= (−1)

. So we have

α ^ β = [φ ^ ψ]

= [ρ

∗

φ ^ ρ

∗

ψ]

= (−1)

[ρ

∗

(ψ ^ φ)]

= (−1)

[ψ ^ φ]

= (−1)

β ^ α.

Now it remains to prove the claim. We have

dρ(σ) = ε

i=0

(−1)

σ|

,...,ˆv

n−i

,...,v

]

ρ(dσ) = ρ

i=0

(−1)

σ|

,...,ˆv

,....,v

]

= ε

n−1

j=0

(−1)

σ|

,...,ˆv

]

We now notice that ε

n−1

(−1)

n−i

= ε

(−1)

. So this is a chain map!

We now define a chain homotopy. This time, we need a “twisted prism”. We

let

(−1)

n−i

, · · · , v

, w

, · · · , w

] ∈ C

n+1

([0, 1] × ∆

where

, · · · , v

are the vertices of

{

} ×

∆

and

, · · · , w

are the vertices of

{1} × ∆

We let

: [0

∆

→

∆

be the projection, and let

(

)

→

n+1

(X) be given by

σ 7→ (σ ◦ π)

We calculate

(σ) = (σ ◦ π)

(dP

)

= (σ ◦ π

)









j≤i

(−1)

n−i

, · · · , ˆv

, · · · , v

, w

, · · · , w

]









j≥i

(−1)

n+i+1−j

(−1)

n−i

, · · · , v

, w

, · · · , ˆw

, · · · , v

]









The terms with j = i give

(σ ◦ π)

n−i

, · · · , v

i−1

, w

, · · · , w

]

(−1)

n+1

(−1)

n−i

, · · · , v

, w

, · · · , w

i+1

]

= (σ ◦ π)

(ε

, · · · , w

] − [v

, · · · , v

])

= ρ(σ) − σ

The terms with

j 6

are precisely

−F

n−1

) as required. It is easy to see that

the terms are indeed the right terms, and we just have to check that the signs

are right. I’m not doing that.

There are some other products we can define. One example is the cross

product:

Definition

(Cross product)

Let

X × Y → X

X × Y → Y

be the

projection maps. Then we have a cross product

× : H

(X; R) ⊗

(Y ; R) H

k+`

(X × Y ; R)

a ⊗ b (π

∗

a) ^ (π

∗

Note that the diagonal map ∆ :

X → X × X

given by ∆(

) = (

x, x

) satisfies

∆

∗

(a × b) = a ^ b

for all a, b ∈ H

∗

(X; R). So these two products determine each other.

There is also a relative cup product

^: H

(X, A; R) ⊗ H

(X; R) → H

k+`

(X, A; R)

given by the same formula. Indeed, to see this is properly defined, note that if

φ ∈ C

(X, A; R), then φ is a map

φ : C

(X, A) =

(X)

(A)

→ R.

In other words, it is a map

(

)

→ R

that vanishes on

(

). Then if

σ ∈ C

k+`

(A) and ψ ∈ C

(X; R), then

(φ ^ ψ)(σ) = φ(σ|

,...,v

]

) · ψ(σ|

,...,v

k+`

]

We now notice that [

, · · · , v

]

∈ C

(

). So

kills it, and this vanishes. So

this is a term in H

k+`

(X, A; R).

You might find it weird that the two factors of the cup product are in different

things, but note that a relative cohomology class is in particular a cohomology

class. So this restricts to a map

^: H

(X, A; R) ⊗ H

(X, A; R) → H

k+`

(X, A; R),

but the result we gave is more general.

Example.

Suppose

is a space such that the cohomology classes are given by

k 0 1 2 3 4 5 6

(X, Z) Z 0 0 Z = hxi 0 0 Z = hyi

What can x ^ x be? By the graded commutativity property, we have

x ^ x = −x ^ x.

So we know 2(x ^ x) = 0 ∈ H

(X, Z)

∼

Z. So we must have x ^ x = 0.