Why is the trace of such a simplicial map is zero?

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!

• How is the induced map on homology defined? Are you using simplicial or singular homology? – Connor Malin Aug 13 at 11:19
• 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. – Connor Malin Aug 13 at 11:36
• 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 $0$. – s.harp Aug 13 at 11:38
• @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 $1$. – Connor Malin Aug 13 at 11:50
• @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. – 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.