# $\pi$-cohomology class — a variant of cohomology class

Let $X$ be a topological space with a triangulation. The triangulation defines a chain complex in $X$. Let $\mu_d$ be a cochain and $M^d$ be a chain. We use $< \mu_d, M^d > \in M$ to denote the evaluation of the cochain $\mu_d$ on the chain $M^d$. Here $M$ is a module and $M=R/Z$.

A $\pi$-cocycle $\mu_d$ is defined as a cochain that satisfy $$< \mu_d, M^d > = < \mu_d, N^d >$$ for any pair of cycles $M^d$ and $N^d$ that can "deform" into each other continuously.

In contrast, a cocycle $\nu_d$ is a cochain that satisfy $$< \nu_d, M^d > = < \nu_d, N^d >$$ for any $M^d$ and $N^d$ such that $M^d-N^d$ is a boundary.

Let $W_d$ be the collection of $\pi$-cocycles. Let $Z_d$ be the collection of cocycles. Let $B_d$ be the collection of coboundaries. We have $B_d \subset Z_d \subset W_d$.

$H^d(X,M) = Z_d/B_d$ is the usual cohomology class. ${\cal H}^d(X,M) = W_d/B_d$ is the new $\pi$-cohomology class.

Is such $\pi$-cohomology class a well defined concept? Was it already studied under a different name?

Thanks!

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In order for this to be well-defined, you'd need to say precisely what it means for two cycles to '"deform" into each other continuously.' If both cycles are subcomplexes of $X$, you could ask for them to be simplicially homotopic, but then you'd want your simplicial set to be Kan, i. e. probably the singular complex of the space $X$ rather than a triangulation. Then you'd need to make sure that your definition extends well to the abelian group of cycles. – Paul VanKoughnett Mar 4 '13 at 3:32
@Paul: Thanks for the clarification. As a physicist, I cannot distinguish the subtle difference between "triangulation" and "singular complex". If it is helpful to make the definition valid, we can replace "triangulation" by "singular complex" and replace "deform continuously" by "simplicially homotopic" in the above definition. I will be happy with it. – Xiao-Gang Wen Mar 4 '13 at 4:30
@Xiao: You should also decide of what are the "$\pi$-coboundaries". – Misha Mar 4 '13 at 15:07
@Misha: as described in the above definition, "$\pi$-coboundaries" = usual coboundaries. – Xiao-Gang Wen Mar 4 '13 at 17:15

This answer is actually more like something between a comment and a new question but as it has some bearing on

Is such π-cohomology class a well defined concept?

I will post it here.

Following Paul VanKoughnett's comment, what you need is to make the definition of the $\pi$-cocycle'' more precise. In particular, you are defining $\pi$-cocycles to be the set of ordinary cocycles which agree on all cycles satisfying some kind of equivalence relation that you have not really specified here.

I believe that you want the cycles to be considered equivalent here if they are somehow homotopic.''

However, a cycle is just a formal sum of simplices in $X$ and it is not clear without saying more what a homotopy of a cycle is. In the singular chain complex, I can at least say that two simplices $S_1,S_2$ are homotopic if there exists a (continuous) map $h:T\times[0,1]\rightarrow X$ such that $h_0=S_1$ and $h_1=S_2$.

Your question, however, is posed in terms of the simplicial chain complex which is a rather rigid object. I am not sure what two homotopic simplices in a triangulation would be, perhaps something involving the notion of collapses and expansions from Whitehead's simple homotopy theory?

In any case it would help the most if you described an example - perhaps tell us in detail how you compute say $\mathcal{H}^1(S^1\times S^1)$ or something like that.

Here is a zeroth order try in the context of singular chains. Let us say that two cycles $M^d$ and $N^d$ are $\pi$-equivalent if there exist sums $S_{M^d}$, $S_{N^d}$ of simplices representing $M^d$ and $N^d$ such that

1. there is a bijection $b$ between the terms in the two sums
2. the coefficients of $s$ and $b(s)$ agree for all terms $s$ of the sums
3. the simplices $s$ and $b(s)$ are homotopic as maps into $X$ for all $s$.

Now there is a problem here with 3. This is because the homotopies of these simplices may very well break the cycle condition. Thus I am guessing that this will lead to something you don't want. It might be better to require the existence of a homotopy deforming all of the simplices at once such that everything in between is also a cycle. We then should strengthen 3 to 3':

3'. Let $T_{M^d}$ be the set of simplices which appear as terms of the sum $S_{M^d}$ and likewise for $T_{N^d}$. Suppose the cardinality of that set is $k$. There is a continuous map $h:(d-\text{simplex})^k\times[0,1]\rightarrow X$ such that $h_0$ is (the product of) $T_{M^d}$ and $h_1$ is $T_{N^d}$ and $h_t$ for all $t\in[0,1]$ is a cycle when each of the component simplices is summed with the coefficients from 2.

(Apologies for being a bit informal - if this is confusing, I can expand later, but I didn't want to introduce a host of projection maps and more notation here when the idea is pretty straightforward).

I am not sure what you get out of this definition, I suspect it will be rather complicated though as computing this could well involve understanding the sets of homotopy classes of all $d$-dimensional simplicial complexes into $X$.

One last comment / question. Suppose you are able to compute the $\pi$-cohomology'' that arises out of the above or some variant. Is it clear why this is truly a cohomology theory? In other words, what connects the $\mathcal{H}^d(X,M)$ with $\mathcal{H}^{d+1}(X,M)$? To me it seems the nice structure of ordinary cohomology is gone because we have lost the connection with the coboundary operator.

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A serious issue with both 3 and 3' is that the dependence on the set of simplices is quite strong. For instance, two 1-cycles which are both polygons (triangulations of $S^1$) in $X$ would be considered distinct already if they had different numbers of sides, whereas you might like them to be equivalent if they induce the same path in the set of homotopy classes $[S^1,X]$ (not the group $\pi_1(X)$ which requires a basepoint). One way to fix this might be to incorporate the collapses and expansions I mentioned but I haven't thought this through. The last two paragraphs above still apply though. – j.c. Jun 8 '13 at 10:29