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Kunneth formula

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How does taking products of complexes affect the homology?

In other words, given two cell complexes $K$ and $L$, express $H(K \times L)$ in terms of $H(K)$ and $H(L)$.

It would be so easy if the answer was "it's the product of the homology groups". There is some evidence for and against that idea...

Let's assume that our homology is over field $F$; in that case the homology groups are vector spaces and the products are understood in the usual way: $${\bf R}^n \times {\bf R}^m = {\bf R}^{n+m}.$$

Let's consider a few examples:

  • $H_0(point \times point)=H_0(point)=F$ vs $H_0(point) \times H_0(point)= F^2$. NO. Same for all path-connected spaces!
  • $H_k(point \times point)=H_k(point)=0$ vs $H_k(point) \times H_k(point)= 0 \times 0 = 0,k>0$. YES. Same for all complexes of dimension $>k$!
  • $H_1(Torus)=H_1({\bf S}^1 \times {\bf S}^1) = F^2$ vs $H_1({\bf S}^1) \times H_1({\bf S}^1) = F^2$. YES.
  • $H_2(Torus)=H_2({\bf S}^1 \times {\bf S}^1) = F$ vs $H_2({\bf S}^1) \times H_2({\bf S}^1) = 0 \times 0 = 0$. NO.

The answer seems to be No here, but, clearly, there is a relation.

Especially, the last two examples suggest this reasoning.

  • First, the torus, a $2$-manifold, is the product of the two circles, $1$-manifolds.
  • Second, maybe the $2$-homology class of the whole torus is made, as some kind of product, of the two $1$-classes of the circles.
Torus.JPG

This reasoning applies to the other homology classes of the torus:

  • First, the longitude (latitude) of the torus, a $1$-manifold, is the product of a circle, $1$-manifold, and a point, $0$-manifold.
  • Second, maybe the $1$-homology class represented by the longitude (latitude) is made, as some kind of product, of the $1$-class of the circle and the $0$-class of the point in the other circle.

This product is called the cross product. It is an operation by which an $i$-cycle in $K$ and a $j$-cycle in $L$ are combined to create an $(i+j)$-cycle in $K × L$.

Of course, we start the construction with the cells; after all they are the basis elements of the chain complex. So the goal is construct an $(i+j)$-cell from an $i$-cell in $K$ and a $j$-cell. It's tricky for general cell complexes, but for cubical complexes it's fully implemented already.

The product $K×L$ of two cubical complexes is the set of all pairwise products $a \times b$ of cells in $K$ and $L$: $$K \times L = \{a \times b: a \in K, b \in L \}.$$ The reason this works is that the product of an $i$-cube and a $j$-cube is an $(i+j)$-cube: $$I^i \times I^j = I^{i+j}.$$

For example in a $2$-cell as the product of two $1$-cells we have the full list of cells:

  • $0$-cells: $n \times m, n \times (m+1), (n+1) \times m, (n+1) \times (m+1)$;
  • $1$-cells: $n \times (m,m+1), (n+1) \times (m,m+1), (n,n+1) \times m, (n,n+1) \times (m+1)$;
  • $2$-cells: $(n,n+1) \times (m,m+1)$.

To complete the picture, define the boundary operator $$\partial \colon C(K \times L) \rightarrow C(K \times L)$$ by$$\partial(a \times b) = \partial a \times b + a \times \partial b.$$ Note: the computation is via the distributive property.

Now, what happens to the homology classes? You can see that in this example:

Product of circle and segment complexes.jpg

Here, for example, $(a+b+c+d) \times e$ is a $1$-cycle.

More generally, the multiplication of chains works as follows. Suppose we are given two chains: $$a_0+a_1+...+a_n \in C_*(K), b_0+b_1+...+b_n \in C_*(L), $$ where $a_i \in C_i(K), b_i \in C_i(L)$. Then $$(a_0+a_1+...+a_n) \times (b_0+b_1+...+b_n) $$ $$= (a_0 \times b_0)$$ $$+(a_0 \times b_1 + a_1 \times b_0)$$ $$+(a_0 \times b_2 + a_1 \times b_1+a_2 \times b_0)$$ $$...$$ $$+(a_0 \times b_i + a_1 \times b_{1-1}+a_2 \times b_{i-2}+...+a_i \times b_0)$$ $$...$$ $$+ (a_n \times b_n).$$ Here $i$-th row gives you the component of the product chain in $C_i(K \times L)$.

This isn't the end of the story because homology as an equivalence relation on chains complicates things.

We need an algebraic construction called the tensor product.