0
$\begingroup$

I have a field with zero characteristic, like $K=\mathbb{C},\mathbb{R}$ and I want to show that the homology groups and cohomology groups with coefficients in these fields satisfy:

$$H_n(X,K) \approx H_n(X)\otimes K, \ H^n(X,K) \approx {\rm Hom}\ (H_n(X),K)$$

for $n\geq 0$.

Do you have reference for this proof?

Thanks in advance.

$\endgroup$
  • 3
    $\begingroup$ Open any book on algebraic topology and look for "universal coefficient theorem". $\endgroup$ – Martin Brandenburg Jan 10 '14 at 10:59
3
$\begingroup$

In the case of coefficients in the field of charactertistic 0, this is more simple than the universal coefficient theorem. (Although anyone seriously interested in algebraic topology should learn $UCT$ in the general case by heart.)

Let $K$ be a field of charactersitic $0$ and notice that as an abelian group it is a rational vector space and is in particulat flat over $\mathbb{Z}$, ie. the functor $- \otimes _{\mathbb{Z}} K$ is exact.

The singular homology of a space $X$ with coefficients in $\mathbb{R}$ is defined to be the homology of the singular chain complex $C_{n}(X, R) = \bigoplus _{\sigma \in Sing_{n}(X)} R$, where $Sing_{n}(X)$ is the set of $n$-singular simplices of $X$. In particular, we have $C_{n}(X, R) \simeq C_{n}(X, \mathbb{Z}) \otimes R$.

Now if $R$ is flat (for example, $R = K$, $char(K) = 0$), then tensoring with it preserves homology and so we have

$H_{n}(X, R) = H_{n}(C_{\bullet}(X, R)) \simeq H_{n}(C_{\bullet}(X, \mathbb{Z}) \otimes R) \simeq H_{n}(C_{\bullet}(X, \mathbb{Z})) \otimes R = H_{n}(X,\mathbb{Z}) \otimes R$,

where we used flatness in the second-to-last isomorphism.

The version for homology is analogous, since by inspection of the definitions we obtain the cochain complex of $X$ over $K$ by applying $Hom _{Ab}(-, K)$ to $C_{\bullet}(X, \mathbb{Z})$. This functor is again exact, since

$Hom _{Ab}(-, K) \simeq Hom _{K}(- \otimes K, K)$

and so it is a composition of two exact functors (tensoring with $K$ and taking a dual vector space).

$\endgroup$
1
$\begingroup$

For your first question look at Universal coefficient theorem (http://en.wikipedia.org/wiki/Universal_coefficient_theorem )

The second part, if I am not mistaken is just a question in linear algebra. The chain complexes are dual (including the duality of morphisms) and then isomorpism follows from isomorphism of linear spaces $V \subset W, (W/V)^*=W^*/V^*$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.