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I am trying to use Algebra; Linear and otherwise to find the co-differential ,i.e., the differential

operator d in a Cohomology Theory starting with homology. For now, I just wanted to start with a

specific example I am familiar with, that of Simplicial Homology, where the differential is given

by the restriction map from an $n$-simplex S_n to all the $(n-1)-$ simplices $\{S_{n-1}\}$ incident

with $S_n$

I have some ideas, but I am far from a solution.

I know the p-th Homology group $H^{p}(M; R)$ ; M is a Topological space and $R$ is a coeficient

ring is given ( up to isomorphism) by (edit *) $$ H^P(M;R)=Hom(C_p(M;R), R)$$ , where $H_p$ is the p-th Homology group.

I am trying to make use of the dual map in linear algebra. Given (finite-dimensional, I believe)

vector spaces V,W over the same field, and a linear map L: $V \rightarrow W$ , there is a contravariant functorial

dual map between the associated duals $V^{*}, W^{*}$ given by sending $w^{*} \in W^{*}$ to $$w*L(v)$$
trying to use this dual map to define the co-differential in Simplicial Cohomology, starting

with the differential d betwen chain groups $C_n$ , given by:

restricting : an n-chain $a*c_n$ is sent to the formal sum $a*ci_{n-1}$ , where $c_{n-1}$ is

incident with $c_n$. The dual differential, aka , the co-differential would then map

from $C_{n-1}\rightarrow C_n$ through means I don't have clear. Clearly I am kind of rambling here

and would appreciate hints. Thank you.

*Subbed $C_p$ for $H_p$

MSIS
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1 Answers1

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Couple of things first

$H^p(M;R) \cong \text{Hom} (H_p(M;R),R)$

is not necessarily true, see the Universal Coefficients Theorem. Perhaps you meant to replace $H$ with $C$?

As for the intuition for $\delta$ (the co-differential as you call it), linear algebra will work, but let's think about it a little clearly.

Given $L:V \to W$, you're right that you get the dual map $L^*:W^* \to V^*$ given by $L^*(w^*) (v) = w^*(L(v))$. Essentially, you "precompose" with $L$ to give you something that an element $w^*$ can act on. The same idea works with the cochain groups.

If you have a chain complex $\rightarrow C_n \rightarrow C_{n-1} \rightarrow$, and $\partial$ is the differential (boundary operator), then note that dualizing gives you $\rightarrow C^{n-1} \to C^n \to$ where $C^\circ = \text{Hom}(C_\circ, R)$. Now note that we can define $\delta$ the same way, as $\delta(c^{n-1})(c_n)= c^{n-1}(\partial(c_n))$

In other words, we use the boundary operator to push an $n$-cycle to an $(n-1)$-cycle, and then you can use the $(n-1)$ cocycle. So you "precompose" with the boundary.

Osama Ghani
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  • Thank you. Does your definition of the coboundary generalize to all homology/cohomology theories? – MSIS May 03 '21 at 01:07
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    Yes, for any theory where cohomology is defined in terms of dualizing homology. That may seem a weird statement but for example, I'm not aware of a homology theory for which de Rham cohomology is the corresponding dualized theory, and so de Rham has a more concrete definition of what the differential is. – Osama Ghani May 03 '21 at 01:09
  • Yes, this is part of what motivated me. I kept hearing about DeRham, Cech Cohomoloy theories but not their corresponding dualizations and was trying to figure out what they were. – MSIS May 03 '21 at 01:10
  • I am also interested in seeing if I can dualize persistent homology. – MSIS May 03 '21 at 01:15
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    de Rham cohomology uses an existing cochain complex and so doesn't need a dualization. I'm not sure how one would define a de Rham homology, but it's probably not of much interest. As for Cech, the naive idea of Cech homology fails some important properties, but it does work on a decent class of spaces, see https://math.stackexchange.com/questions/33117/what-is-cech-homology#:~:text=1%20Answer&text=There%20is%20a%20notion%20of,like%20with%20a%20homology%20theory. I'm not fully sure what you would mean or accomplish by dualizing persistent homology. All persistent homology diagrams give is... – Osama Ghani May 03 '21 at 01:17
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    ...a calculation of Betti numbers of spaces given different resolution sizes. Simplicial homology would be the easiest to calculate these Betti numbers. Simplicial cohomology for example won't give you anything different. – Osama Ghani May 03 '21 at 01:18
  • Sorry for the necropost. If I may: So, for an infinite (Co)chain complex, I can obtain a Chain(Co) -Complex by:1) Subbing a boundary operator $\delta$ with its dual map $\beta$ as above, and sub in each appearing $H_p(M,R)$ with a $H_p(Hom(C_p(M;R),R)$ ( And) a similar conversion i the opposite direccion ? Are the latter computed in practice? – MSIS Aug 08 '22 at 23:21
  • And if we represented $\delta$ as a matrix $M$ with respect to a basis $B$, the codifferential $\beta$ is given by $M^T$? – MSIS Aug 08 '22 at 23:26
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    So there are two steps to clear up what is going on. First, the objects in a chain complex are chain groups, not homology groups themselves! So yes, given a chain complex, you can dualize by replacing each $C_p(M;R)$ with $\text{Hom} (C_p(M;R), R)$ which we usually denote $C^p(M;R)$. Now, yes if you had fixed a basis for the group (if that makes sense), and you represented the boundary by a matrix, then the coboundary would be a transpose (again, if it makes sense!) – Osama Ghani Aug 09 '22 at 01:28
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    The second step now is to understand what the coboundary on the cohomology groups are! We understand it on the cochain complex. This is the exact same as understanding boundary on chains vs boundary on homology. For any (co)homology class, we take a representative (co)chain and apply the (co)boundary and take its equivalence class. How do we know this works? You need to check that the (co)boundary is the same regardless of which representative we chose, and so the (co)boundary descends to a map in (co)homology. This is just basic quotient groups from algebra! – Osama Ghani Aug 09 '22 at 01:32
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    As an exercise, just forget about manifolds or anything and consider an abstract chain complex with each $C_n = \mathbb{Z} \oplus \mathbb{Z}$ and each $\delta | (x,y) \mapsto (0, x)$. See if you can write the boundary in a matrix. Now dualize, what is the coboundary matrix? – Osama Ghani Aug 09 '22 at 01:34
  • Thanks. Just havent looked at the material for a while. Yes, I now remember: Chain Complex is for chain groups, not Homology groups. I get we need to do the quotient by the Boundary subgroups to get the n-th Homology. – MSIS Aug 10 '22 at 21:51