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

Question

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!

Answer 1

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.