Homology of product of spheres: Difference between revisions

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<math>b_q(A) = {r \choose q}</math>
<math>b_q(A) = {r \choose q}</math>


An alternative interpretation of the above result is that <math>b_q(A)</math> is the coefficient of <math>x^q</math> in the product:
An alternative interpretation of the above result is that <math>b_q(A)</math> is the coefficient of <math>x^q</math> in the following:


<math>\prod_{i=1}^r (1 + x^{m_i})</math>
<math>\prod_{i=1}^r (1 + x^{m_i})</math>
In other words, the [[Poincare polynomial]] of <math>A</math> is the product of the Poincare polynomials of the individual spheres (note that the Poincare polynomial of a product of topological spaces is not in general the product of the Poincare polynomials.


==Related invariants==
==Related invariants==

Revision as of 21:34, 3 November 2007

Statement

Let (m1,m2,,mr) be a tuple of nonnegative integers. Let A be the space Sm1×Sm2×Sm3××Smr. Then the homologies of A are free Abelian, and the qth Betti number is given by the following formula:

bq(A)=|{T{1,2,3,,r}|iTmi=q}|

In other words bq(A) is the number of ways q can be obtained by summing up subsets of (m1,m2,,mr).

A particular case of this is when all the mis are 1, viz the torus. In this case:

bq(A)=(rq)

An alternative interpretation of the above result is that bq(A) is the coefficient of xq in the following:

i=1r(1+xmi)

In other words, the Poincare polynomial of A is the product of the Poincare polynomials of the individual spheres (note that the Poincare polynomial of a product of topological spaces is not in general the product of the Poincare polynomials.

Related invariants

Euler characteristic

The Euler characteristic of the product of spheres can be obtained by plugging (1) in the above polynomial. From this it turns out that the Euler characteristic is 0 if any of the spheres has odd dimension, and is 2r if all the spheres have even dimension.

Proof

Using exact sequence for join and product

Further information: exact sequence for join and product

The above claim can be easily proved using induction, and the exact sequence for join and product.

Using a CW-decomposition