0
$\begingroup$

Given simplicial complexes $K,L$, a map $f:K\to L$ is simplicial if it sends each simplex of $K$ to a simplex of $L$ by a linear map taking vertices to vertices, which means it has form: $\sum_{i}t_i v_i \to \sum_{i} t_i f(v_i)$, where $v_i$ denotes vertex and each $t_i$ is a number.

Now, since $H_n(K)$ and $H_n(L)$ are free abelian groups, we can talk about the trace of induced map on the torsion free part of the groups $f_*: H_n(K)/Torsion \to H_n(L)/Torsion$.

In Hatcher’s, he says that if a simplicial map $g$ satisfies $g(\sigma)\cap \sigma =\emptyset$ for all simplex in $K$, then the matrix for $g_*$ has zeros down the diagonal, thus has trace zero.

I don’t know why this statement holds? If we see vertices as a base since $g$ is a linear map on vertices, it becomes a linear algebra question. But I can’t go further. Hope someone could help. Thanks!

$\endgroup$
  • $\begingroup$ How is the induced map on homology defined? Are you using simplicial or singular homology? $\endgroup$ – Connor Malin Aug 13 at 11:19
  • $\begingroup$ Could you provide a reference for where he says this? You can only take traces of a map from a space into itself, and I believe there must be more conditions because if you give the circle a simplicial structure by splitting it into 4 line segments, we may rotate the circle and obtain a map satisfying your condition that is homotopic to the identity. This map has trace 1. $\endgroup$ – Connor Malin Aug 13 at 11:36
  • $\begingroup$ If you have a finite simplicial complex on $K$, then the chain complex $C_n(K)$ is given by the span of all $n$-simplices as abstract generators. If a simplicial map $g:K\to K$ satisfies $g(\sigma)\cap\sigma=\emptyset$ for all $\sigma$ then the map induced by $g$ on $C_n(K)$ must send the basis vector $\sigma$ to a linear combination of the other basis vectors. In particular $\widetilde{\sigma}( g_*(\sigma))=0$ where $\widetilde\sigma$ is the "dual" of $\sigma$. Since the trace of $g_*$ is $\sum_{\sigma \text{ $n$ simplex}}\widetilde\sigma(g_*(\sigma))$ you get that the trace is $0$. $\endgroup$ – s.harp Aug 13 at 11:38
  • $\begingroup$ @s.harp This is on the chain complex level though. You may get nonzero trace in the homology. For example, the map swapping $(0,1)$ and $(1,0)$ has trace 0, but when restricting to the diagonal and making the trivial quotient, it has trace $1$. $\endgroup$ – Connor Malin Aug 13 at 11:50
  • $\begingroup$ @Connor thanks, looks like I got a bit confused. But as you remark the statement is false for homology. Rotating the circle $S^1$ is an example where the induced map on homology is the identity but you can choose a simplicial complex so that the condition $g(\sigma)\cap\sigma=\emptyset$ is satisfied. $\endgroup$ – s.harp Aug 13 at 11:52
1
$\begingroup$

Some context: the question refers to Hatcher's proof of the Lefschetz fixed point theorem. The proof goes like this: Given a finite simplicial complex $X$, he wants to show that if some map $f:X \to X$ has no fixed points, then $\tau(f_*) = 0$. He does this by constructing a map $g$, homotopic to $f$, which is simplicial when considered as a map from some subdivision $K$ of $X$ to itself, such that $g(\sigma) \cap \sigma = \emptyset$ for all simplices $\sigma$ in this subdivision. He further shows that the Lefschetz number $\tau(f)$ can be computed by $$ \tau(f) = \tau(g) = \sum (-1)^n \text{tr} \left( g_*: H_n(K_n, K_{n-1}) \to H_n(K_n, K_{n-1})\right) $$ Then he claims that each summand on the right side is zero. This is what the question is asking about. The group $H_n(K_n, K_{n-1})$ is free abelian, with basis the $n$-simplices of $K$, and the map $g_*: H_n(K_n, K_{n-1}) \to H_n(K_n, K_{n-1})$ will take the generator for an $n$-simplex $\sigma$ to (up to a sign) the generator for the simplex $g(\sigma)$, or to zero if $g(\sigma)$ is not $n$-dimensional. So since $g(\sigma) \cap \sigma$ is empty, the matrix for $g_*$ has zeroes down the diagonal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.