Lefschetz number equal to Euler Characteristic of Fixed Points

Question

If $X$ is a finite simplicial complex and $f:X\rightarrow X$ is a simplicial homeomorphism, show that the Lefschetz number $\tau(f)$ equals the Euler characteristic of the set of fixed points of $f$.

My progress: We know the set of fixed points $X_f$ of $f$ can be represented as a simplicial complex by barycentric subdivision on $X$. Then, the faces of $X_f$ are generated by the fixed vertices of each maximal face and the barycentric point of such a face. I've created some examples to verify the problem statement, but this creates no intuition on how to proceed. I realize that there are symmetries that determine what faces could be mapped into what other faces. Any hints on how to proceed? Especially how to relate the trace of $f_*:H_n(X)\rightarrow H_n(X)$ to the rank of $H_n(X_f)$ or the number of cells of $X_f$?

EDIT: I've tried to work on this for a bit, but the terminology and conclusions that user8268 makes are too complicated for what I know, so I'll list what I'm working with.

We have the short exact sequences $0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0$ and $0\rightarrow B_n\rightarrow Z_n\rightarrow H_n\rightarrow 0$.

Then, we can write $C_n=B_{n-1}\oplus Z_n=H_n\oplus B_n\oplus B_{n-1}$, so I'm assuming this is what the graded vector space. Then, through some manipulation, the trace of the induced endomorphism on $H_n$ is equal to the trace of the map $\frac{ker\partial_n}{im\partial_{n+1}}f_*:H_n\oplus B_n\oplus B_{n-1}\rightarrow H_n$, but how does this matrix look like (this doesn't even look like a square matrix)?

Can someone offer a simpler explanation or an alternative approach?

Answer 1

After barycentric subdivision, $f$ gives you a permutation of the $k$-dim simplexes, the simplexes which are fixed by this permutation are those which are in the fixes point set of $f$. Hence $f_*:C_k(X)\to C_k(X)$ is a permutation matrix, so its trace is the number of the fixed simplexes, i.e. $\dim C_k(X_f)$.

Answer 2

I want to (hopefully) improve upon the answer here, which I believe glosses over some important points which contribute to this result. For the rest of this post let $K$ be a finite simplicial complex with a maximal dimension of $m$ and $f: K \rightarrow K$ is a simplicial homeomorphism with the set of fixed points under $f$ denoted by $K_f$. I will denote the barycentric subdivision of a simplicial complex $L$ by $BL$. The following results will be vital in establishing our argument.

1. The fixed points of a simplicial homeomorphism of a standard $n$-simplex $\Delta^n$ form a simplicial subcomplex of $B\Delta^n$

2. The fixed points of a simplicial homeomorphism of a simplicial complex $K$ forms a simplicial subcomplex of $BK$.

3. $\tau(f) \equiv \sum_{i=0}^m (-1)^i \text{tr}(f_i: C_i \rightarrow C_i)= \sum_{i=0}^m (-1)^i \text{tr}(H_i(f))$ where $f_i$ is induced map by $f$ on the free abelian group $C_i$ (one generator for each $i$-simplex in $K$) and $H_i(f)$ is the $i$th induced map on homology by $f$.

4. Let $C'_i$ be the free abelian group generated by the $i$-simplices of $BK$. Then, the induced map $f'_i: C'_i \rightarrow C'_i$ (via $f$) satisfies $\tau(f)= \sum_{i=0}^m (-1)^i \text{tr}(f'_i: C'_i \rightarrow C'_i)$.

5. $\text{tr}(f'_i: C'_i \rightarrow C'_i) = \text{rank}(D_i)$ where $D_i$ is the free abelian group generated by the $i$-simplices in the simplicial subcomplex $K_f$ (regarded as a subcomplex of $BK$).

6. $\chi(K_f) = \sum_{i=0}^m (-1)^i \text{rank}(D_i)$.

The result is immediate if one applies $(1) -(6)$.

To show (1) notice that every simplicial homeomorphism can be represented as a permutation of vertices. Every permutation is a disjoint union of cycles. We can write any simplicial homeomorphism as a permutation of vertices as $(a_1, ..., a_j)...(b_1, ..., b_k)$ where the cycles are disjoint. Consider $B\Delta^n$. For every disjoint cycle we can find a vertex within $B\Delta^n$ which remains fixed under the action of $f$, namely the vertices $\{1/j (a_1 + ... + a_j), ..., 1/k (b_1 + ... + b_k)\}$. The set of vertices we obtain via this procedure under the typical simplicial linear combinations will span $\Delta^n_f$. To find all the individual simplices which lie in this span, take linear combinations of $c -1$ of the vertices above where $c$ is the cardinality of that set, with the central vertex, under the restriction that the vertices you pick must lie in the span of one of the faces of $\Delta^n$. (1) $\implies$ (2) since for every simplex in $K$ the set of fixed points will be a subcomplex of their image ($K_f$) in $BK$. (3),(5) follow from the other answer on this post. (3) $\implies$ (4) immediately because the induced map $f'_i$ is also an induced map under $f$ meaning that $\sum_{i=0}^m (-1)^i \text{tr}(f'_i: C'_i \rightarrow C'_i) = \sum_{i=0}^m (-1)^i \text{tr}(H_i(f))$. (6) is a commonly known fact relating the Euler characteristic to the alternating sum of numbers of $i$-simplices in a simplicial complex.

Note: One of the most crucial aspects of this proof is noticing that we do not necessarily have a simplicial subcomplex until we take the barycentric subdivision. So, the main idea of this argument is to prove the result in terms of subcomplexes, which we have nice results for, then translate this back into the original situation where $K_f$ is not a subcomplex of $K$.