When computing homology groups, I have seen some people use simplicial complexes, delta complexes or CW-complexes.
Does it matter which one I use in general? Do they always yield the same homology groups?
Thank you!
algebraic-topologycw-complexeshomology-cohomologysimplicial-complex
When computing homology groups, I have seen some people use simplicial complexes, delta complexes or CW-complexes.
Does it matter which one I use in general? Do they always yield the same homology groups?
Thank you!
Yes, this is true. Here's one way to prove it. For $0\leq j<n$, write $s_j^{n-1}:\Delta^n\to\Delta^{n-1}$ for the map induced by the order-preserving surjection $\{0,\dots,n\}\to\{0,\dots,n-1\}$ that maps both $j$ and $j+1$ to $j$. Say that a singular simplex $\Delta^n\to X$ is degenerate if it factors through $s_j^{n-1}$ for some $j$. Note that the boundary of a degenerate simplex is a linear combination of degenerate simplices: all but possibly two of its faces are degenerate (the two exceptions being the faces corresponding to omitting the vertices $j$ and $j+1$), and those two faces cancel out. So if we write $D_n(X)\subseteq C_n(X)$ for the span of the degenerate simplices, $D_\bullet(X)$ is a subcomplex of $C_\bullet(X)$. Write $N_\bullet(X)=C_\bullet(X)/D_\bullet(X)$.
Now note that given an order-preserving simplicial map $f:X\to Y$, the induced maps $C^\Delta_\bullet(X)\to C^\Delta_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)$ and $C^\Delta_\bullet(X)\to C_\bullet(X)\to C_\bullet(Y)\to N_\bullet(Y)$ are equal, since the $n$-simplices $\sigma$ that you're worried about have the property that $f\circ c_\sigma$ is degenerate. So to show that $f^\Delta_{\bullet_*}=f_{\bullet_*}$, it suffices to show that the map $C_\bullet(Y)\to N_\bullet(Y)$ induces isomorphisms on homology. By the long exact sequence in homology associated to the short exact sequence $0\to D_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)\to 0$, it suffices to show that $D_\bullet(Y)$ has trivial homology.
We can show this by constructing a chain homotopy. Given a degenerate $n$-simplex $\sigma:\Delta^n\to Y$, let $j(\sigma)\in\{0,\dots,n-1\}$ be the least $j$ such that $\sigma$ factors through $s_j^{n-1}$. Now define $H:D_n(Y)\to D_{n+1}(Y)$ by $H(\sigma)=(-1)^{j(\sigma)}\sigma\circ s^n_{j(\sigma)}$. An elementary computation then shows that $H\partial+\partial H:D_n(Y)\to D_n(Y)$ is the identity map. It follows that $D_\bullet(Y)$ has no homology.
To give a sense of the computation $H\partial+\partial H=1$, let's show what happens when you take a $3$-simplex $\sigma$ with $j(\sigma)=1$; the general case works very similarly. Let's write $\sigma=[a,b,b,c]$; all the simplices built from $\sigma$ will be written using similar expressions with the obvious meaning (here $a$, $b$, and $c$ are the vertices of $\sigma$, with $b$ repeated since it factors through $s^2_1$). We then have $H(\sigma)=-[a,b,b,b,c]$, so $$\begin{align}\partial H(\sigma)=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,c]+[a,b,b,c]-[a,b,b,b]\\=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,b].\end{align}$$ On the other hand, $\partial\sigma=[b,b,c]-[a,b,c]+[a,b,c]-[a,b,b]=[b,b,c]-[a,b,b]$. We have $H([b,b,c])=[b,b,b,c]$ and $H([a,b,b])=-[a,b,b,b]$. Thus we get $$H(\partial\sigma)=[b,b,b,c]+[a,b,b,b].$$ When we add together $\partial H(\sigma)$ and $H(\partial\sigma)$, all the terms cancel except $[a,b,b,c]$, which is just $\sigma$.
There are actually a few things that can mean the "cohomology of a topological group $G$" in different contexts. It could mean
consider $G$ as a space (forgetting group structure), and take its singular cohomology $H^*(G)$.
consider $G$ as a group (forgetting the topology) and take its group cohomology $H^*_{grp}(G)$
take the singular cohomology of its classifying space $H^*(BG)$.
If $G$ is discrete it's a theorem that the 2) and 3) agree, but if $G$ has a non-trivial topology then usually all three are different. I'm assuming you really mean case 1).
By polar decomposition there are homotopy equivalences $GL_n(\mathbb{R}) \simeq O(n)$ and $SL_n(\mathbb{R})\simeq SO(n)$, so if you're interested in homotopy invariants you can consider these spaces instead. One thing to notice is that, topologically, $O(n)$ is just a disjoint union of two copies of $SO(n)$, so $H_*(O(n)) \cong H_*(SO(n)) \oplus H_*(SO(n))$ (of course as a group $O(n)$ is a non-trivial extension of $SO(n)$ by $\mathbb{Z}/2$). The interesting question is then whether $H_*(SO(n))\cong H_*(SO(m))$ for $n\neq m$, and this is answered in section 7 of "The integral homology and cohomology rings of $SO(n)$ and $Spin(n)$" by Pittie, though I haven't deciphered the result far enough to tell whether they really are all different but it seems so.
If you want an answer to 1) or 3) in terms of cohomology then I recommend Mimura and Toda's "Topology of Lie Groups" as a reference, as they compute the cohomology of $O(n)$ and $SO(n)$ as well as $BO(n)$ and $BSO(n)$, and I forget but they might actually compute the homology as well (most sources I know only compute the cohomology of the classifying spaces).
The cohomology of the classifying spaces is a very different story than just the groups because now the functorial map $B\iota\colon BSO(n) \to BO(n)$ is not an inclusion of a component but a double-cover. The $\mathbb{Z}/2$ cohomologies of $BO(n)$ and $BSO(n)$ are distinguished by the fact that the first universal Steifel-Whitney class vanishes in $H^1(BSO(n);\mathbb{Z}/2)$, and the integral cohomolgies can be distinguished by the fact that there is an integral Steifel-Whitney class $W_2 = \beta(w_1)\in H^2(BO(n);\mathbb{Z})$ but not in the cohomology of $SO(n)$, as well as the existence of an Euler class $e \in H^{n}(BSO(n);\mathbb{Z})$. Moreover $H^*(BO(n)) \ncong H^*(BO(m))$ and $H^*(BSO(n)) \ncong H^*(BSO(m))$ when $n\neq m$. I believe the (singular) cohomologies of $O(n)$ and $SO(n)$ have similar distinctions but I don't have access to Mimura and Toda at the moment to make sure.
As for option 2), I don't know the computations of the group cohomologies $H^*_{grp}(GL_n(\mathbb{R}))$ and $H^*_{grp}(SL_n(\mathbb{R})$, it may be similar to the link that E. KOW posted in the comments about $GL_n(\mathbb{Z})$.
Best Answer
To be precise, we have a way to define homology for each different kinds of cell complexes:
So, how are these different homologies related? The answer is that they are the same in a very strong sense: not only are the homology groups isomorphic, but the chain complexes themselves are the same.
To make this precise, note first that a simplicial complex is a special case of a $\Delta$-complex and a $\Delta$-complex is a special case of a CW-complex (see my answer at Simplicial Complex vs Delta Complex vs CW Complex for more on this). Given a simplicial complex, there is a canonical isomorphism between its simplicial chain complex (whose homology is simplicial homology) and its $\Delta$-chain complex (whose homology is $\Delta$-homology). Indeed, this is completely trivial, since the definitions of the two chain complexes are pretty much exactly the same (whether they are literally the same depends on some technical details of how you choose to define them, but in any case it is immediate that they are canonically isomorphic). Similarly, given a $\Delta$-complex, its $\Delta$-chain complex is canonically isomorphic to its cellular chain complex. This is not quite as obvious since the definition of the cellular chain complex is a bit complicated but again is not difficult to verify (see my answer at If X is a simplicial complex, are the complex of simplicial chains and the one of cellular chains of X identical?).
Now, a separate and very natural question would be, what if I have a space $X$ and I (say) give it the structure of a simplicial complex in one way and the structure of a CW-complex in a different way (not the canonically way in which every simplicial complex is a CW-complex). Then will the simplicial homology be the same as the cellular homology? The answer is again yes, but this is a much deeper result. The typical way to prove this is to define a fourth homology theory called singular homology which is defined on a mere topological space with no extra structure whatsoever. You can then prove that for any CW-complex structure on a space, the cellular homology is canonically isomorphic to the singular homology (and the corresponding statements for simplicial complex or $\Delta$-complex structures then follow since they are special cases of CW-complex structures).
So, what is the point of having all these different homologies, if they're actually the same? The point is that sometimes some of them are easier to compute than others. Simplicial complexes and simplicial homology are very combinatorial which is sometimes nice for computations (and for pedagogy when you are first learning homology). On the other hand, finding a usable simplicial complex structure on a space is often much harder than finding a usable CW-complex structure, since CW-complex structures are much more flexible. $\Delta$-complexes are somewhere in the middle. Singular homology is totally horrendous to compute directly from the definition, but tends to be very useful for proving general theorems since its definition depends on no extra structure. There are also methods of computation that don't involve looking directly at any particular definition of homology but instead use high-level theorems about how the homology of different spaces are related (so, you compute the homology of a complicated space by relating it to simpler spaces).