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Suppose I have a chain complex of chains $C_n$. Then one can obtain the homology groups of this complex. Now if I choose any abelian group $G$ and I consider the cochain group $C_n^*=Hom(C_n,G)$ then I can obtain the cohomology groups. Now the question is: If I form the cocohomology group by considering $C_n^{**}=Hom(Hom(C_n,G),G)$ and defining the cocoboundary maps in the obvious ways, will I obtain the Homology groups again?

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At the very least this depends on $G$ (if $G$ is trivial, then both the cocomplex and the cococomplex becomes trivial no matter what you started with). – Arthur Jan 17 at 22:21
    
Very important question IMO – Werner Germán Busch Jan 19 at 23:41
up vote 5 down vote accepted

Not in general.

Suppose that all of the $C_n = \oplus_{i = 0}^{\infty} \mathbb{k}$, with differentials $0$, and let $G = \mathbb{k}$. Then $Hom(C_n, \mathbb{k}) = \Pi_{i = 0}^{\infty} \mathbb{k}$. Taking the $\mathbb{k}$ dual again gives something containing as a subspace $\Pi_{i = 0}^{\infty} \mathbb{k}$, with zero as the differentials. The cohomology groups are different, as $(\Pi_{i = 0}^{\infty} \mathbb{k})^* \not = \oplus_{i = 0}^{\infty} \mathbb{k}$

If you have torsion in your chain groups: for example $\mathbb{Z}/2$ as some component of some $C_n$, this data will disappear when you double dual. (You can concoct an example again with zero differentials.)

I guess one situation in which you can say that they will be the same is when you are working in the category of finite dimensional vector spaces.

In the more common algebraic topology situation (for example cellular homology), then $C_*$ is a bounded complex of finitely generated free $R$ modules, where $R$ is a PID. In this case one can say something by repeated applications of the universal coefficients theorem and computational facts about exts - the formula you'll get for ${}_iH$ (the ith cocohomology) will be some messy combination of direct sums and compositions of Exts and Homs, involving the groups $H^{i}, H^{i-1}, H_i, H_{i-2}$: https://en.wikipedia.org/wiki/Universal_coefficient_theorem

(Maybe somebody with more sophistication than me can interpret this in terms of composition of the left derived functors of $Hom(\_,G)$?)

Maybe you can play around with that to get some conditions under which the cocohomology agrees.

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Re: Fourth paragraph. I don't really understand how de Rham is "working in the category of finite-dimensional vector spaces". – Mike Miller Jan 17 at 23:04
    
@MikeMiller You are absolutely right, the chain complexes aren't. I was thinking about the cohomology groups. – AreaMan Jan 17 at 23:05
    
@MikeMiller Have taken that parenthetical out. Though the chains are finitely generated modules over $C^{\infty}(M)$ right? At least I think so in the compact case. I know they are not in general free, for example because of the hairy ball theorem on the sphere. – AreaMan Jan 17 at 23:11
    
That's true, but I don't think very helpful, because the differential is not $C^\infty$-linear. – Mike Miller Jan 17 at 23:13
    
@MikeMiller Yeah, true, good point. – AreaMan Jan 17 at 23:17

If $G = \mathbb R$ (or any field), and $C$ is the simplical chain complex associated to a countably infinite discrete space (e.g., the integers), then $H_0 = Z_0$ will be an infinite direct sum of copies of the reals, but $Z^0$ will be the dual space of that, and the double dual will not be isomorphic to the original (because of the inifinite dimensions), hence I believe that the $0$th co-cohomology will not be the same as the $0$th homology.

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