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!

  • $\begingroup$ 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

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.


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