Let $K$ be a simplicial complex and denote by $K_1$ and $K_2$ the complexes obteined from $K$ with two different orientations. I want to prove that the simplicial homology groups of $K_1$ and $K_2$ are isomorphic, that is, for each $n$ we have $$ H_n(K_1) \cong H_n(K_2)$$

I have thought about an explicit isomorphism given by sending a simplex to itself but changing the orientation. But I do not know how to explicitly prove that this function induces a ismorphisms in simplicial homology.

  • $\begingroup$ What does "orientation" mean in this context? An ordering of the vertices of $K$? $\endgroup$ – Eric Wofsey Dec 8 '15 at 2:28
  • $\begingroup$ Yes, thats the definition of orientation that I meant. $\endgroup$ – Marco Armenta Dec 8 '15 at 2:31

Let $<_1$ and $<_2$ be the two orderings of the vertices of $K$. Given a simplex $x\in K$, $<_1$ and $<_2$ induce two different orderings on the set of vertices of $x$. These orderings differ by a permutation of that set; let $\sigma(x)$ denote the sign of that permutation. Define a map $f:C_*(K_1)\to C_*(K_2)$ by $f(x)=\sigma(x)x$ for each simplex $x$. This is clearly an isomorphism on the individual chain groups $C_n(K)$; I claim it is in fact compatible with the boundary maps as well, so it is an isomorphism of chain complexes and thus induces an isomorphism $H_*(K_1)\to H_*(K_2)$.

To see that $f$ is compatible with the boundary map, let us fix an $n$-simplex $x$ of $K$, and say the vertices of $x$ are the set $S=\{0,1,\dots,n\}$. Let us write $<$ for the usual ordering of $S$ and assume that ${<_1}={<}$. Let $p:S\to S$ be the permutation that turns $<$ into $<_2$, so $a<_2b$ iff $p(a)<p(b)$ and $\sigma(x)$ is the sign of $p$. Let $y_i$ be the face of $x$ with vertex set $S\setminus \{i\}$. Note that $y_i$ appears in $\partial_{K_1}x$ with sign $(-1)^i$ and in $\partial_{K_2}x$ with sign $(-1)^{p(i)}$. Since $f(x)=\sigma(x)x$ and $f(y_i)=\sigma(y_i)y_i$, the condition that $f(\partial_{K_1}x)=\partial_{K_2}f(x)$ is equivalent to the assertion that $$\sigma(y_i)=\sigma(x)(-1)^i(-1)^{p(i)}$$ for each $i$. Note that $\sigma(y_i)$ is the sign of the restriction of $p$ to a bijection $S\setminus\{i\}\to S\setminus\{p(i)\}$, where you identify both the domain and codomain with $\{0,1,\dots,n-1\}$ using the standard ordering.

The identity $\sigma(y_i)=\sigma(x)(-1)^i(-1)^{p(i)}$ can be proven in a variety of ways, and I encourage you to try to prove it yourself before reading on. One way to prove it is by using the following characterization of the sign of a permutation: the sign of $p$ is equal to the parity of the number of pairs $(a,b)$ such that $a<b$ but $p(a)>p(b)$. From this description, we see that $\sigma(y_i)$ differs from $\sigma(x)$ by the number of such pairs $(a,b)$ for which either $a$ or $b$ is equal to $i$. Define the following sets: $$A=\{j:j<i\text{ and }p(j)<p(i)\}$$ $$B=\{j:j<i\text{ and }p(j)>p(i)\}$$ $$C=\{j:j>i\text{ and }p(j)<p(i)\}$$ We are interested in the parity of the number $|B|+|C|$. Note that $|A|+|B|=i$ and $|A|+|C|=p(i)$, so $2|A|+|B|+|C|=i+p(i)$. Thus the parity of $|B|+|C|$ is the same as the parity of $i+p(i)$. This corresponds to the factors $(-1)^i(-1)^{p(i)}$ in the identity we wanted, and thus proves the identity.

If you find all of this wizardry with signs of permutations baffling (or even if you don't), it may be enlightening to think concretely about what you're doing for small values of $n$. You start by defining $f$ to take each vertex to itself. You then want to have $f$ take each edge to itself, but you can see that for this to be compatible with the boundary maps, you need $f$ to put a minus sign on edges if the order of their vertices is different in $<_1$ and $<_2$. Next, you look at triangles. Again, you will need to put appropriate signs on the triangles to be compatible with your signs on the edges, and with some trial and error you can come up with the rule that you should have a plus sign if the ordering of the vertices is unchanged or is cyclically permuted, and should have a minus sign if two of the vertices have been swapped. You can then try 3-simplices, but it starts to get complicated to keep track of all the cases. Still, from the case of the low-dimensional simplices, you can reasonably guess that the correct sign to give is the sign of the permutation involved.

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  • $\begingroup$ How does simplicial homology depend on an orientation of the simplicial complex? The way I always understood it, given a simplicial complex $K$, you would start with a free abelian group generated by all oriented $p$-simplices in $K$, and quotient this by a relation which says that each oriented simplex is equal to minus the same simplex with reversed orientation -- this way you'd get the group $C_p(K)$. And then you define the boundary maps as usual. None of this is depending on having some God given orientation of $K$ -- or am I wrong? $\endgroup$ – Pedro Dec 8 '15 at 5:36
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    $\begingroup$ That's one way you can define it. Another way is to fix a total ordering on all the vertices in your complex and use the orientation it provides on everything, and that is presumably the definition being used here. This way is perhaps less elegant if all you care about is simplicial complexes, but is better-suited to generalizations (if you try using your definition to define singular homology, for instance, you run into some annoying technicalities because your chain groups will not be free). $\endgroup$ – Eric Wofsey Dec 8 '15 at 5:51
  • $\begingroup$ @EricWofsey "Let $ < 1$ and $ < 2$ be the two orderings of the vertices of $K$. Given a simplex $x \in K$, $<1$ and $<2$ induce two different orderings on the set of vertices of $x$." Could you explain how to "induce" in this context, here: math.stackexchange.com/q/2563941/22872 $\endgroup$ – kηives Jan 2 '18 at 21:45

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