In the proof of Lefschetz Fixed point Theorem (Rotman's Algebraic topology)

Question

I'm reading the Rotman's An introduction to Algebraic Topology, p. 250~251, Theorem 9.19 (Lefschetz) and trying to understand some statement.

Theorem 9.19 (Lefschetz) Let $X$ be a compact polyhedron and let $f:X\to X$ be continuous. If $\lambda(f) \neq 0$ , the $f$ has a fixed point.

Here $\lambda(f)$ is the Lefschetz Number (defined below)

Proof of Theorem 9.19 is as follows :

Here I arrange some definitions and theorems :

Def.1. : If $K$ is a simplicial complex, then $\operatorname{mesh}K := \operatorname{sup}_{s\in K}\{\operatorname{diam}(s)\}$

Def.2. : For a simplicial complex $K$, $\operatorname{Sd}K$ is the barycentric subdivision.

Def.3.(Lefschetz Number) Let $G_0, G_1, \cdots G_m$ be a sequence of f.g. abelian groups and let $h=(h_0, \cdots h_m)$, where $h_i :G_i \to G_i$ is a homomorphism for every i. The Lefschetz number of $h$, denoted by $\lambda(h)$, is $$ \lambda(h) := \Sigma_{i=0}^{m} (-1)^m \operatorname{tr}(h_i)$$.

Example 9.6. Let $K$ be an $m$-dimensional (finite) simplicial complex, and let $f:K \to K$ be a simplicial map. Let $f_{\#}:=(f_{0\#},\cdots , f_{m\#})$, where $f_{i\#} : C_{i}(K) \to C_{i}(K)$ is the $i$th term of the chain map $f_{\#}$. Then

$$ \lambda(f_{\#}) = \Sigma_{i=0}^{m}(-1)^{i}\operatorname{tr}(f_{i\#})$$

Lemma 9.17 means : If $K$ is a finite simplicial complex and $\varphi : \operatorname{Sd}K \to K$ is a simplicial approximation to the identity $|\operatorname{Sd}K| \to |K|$, then $$\varphi_{*} = (\operatorname{Sd}_{*})^{-1} : \tilde{H}_{*}(\operatorname{Sd}K) \cong \tilde{H}_{*}(K)$$

Lemma 9.18 means : Let $C$ be a chain complex of the form $0 \to C_m \to \cdots \to C_0 \to 0$ in which each $C_i$ is f.g., and let $f:C\to C$ be a chain map. Then $$\lambda(f) = \lambda(f_{*})$$ , where $f_{i*} : H_i(C) \to H_i(C)$.

Exercise 7.10 is, "Prove that a simplicial map $\varphi : K \to L$ is a simplicial approximation to $f:|K| \to |L| $ if and only if, whenever $x\in |K|$ and $f(x) \in s^{\circ}$ (where s is a simplex of $L$), then $|\varphi|(x) \in s$".

If needed, I will upload more associtated definitions and theorems.

I can't understand the underlined statement.

Can anyone helps?

Answer 1

Maybe we can track the Rotman's proof with some concrete example. Let $X$ be the circle, as a realization of the simplicial complex mentioned in the question Definition of map to barycentric subdivision? and let $f: X \to X$ be the function that rotate it $\frac{4}{5}\pi$ counterclockwise. It's obvious that $\mathrm{tr}f_{0*}=\mathrm{tr}f_{1*}=1$. Let's try to use this to check the proof.

Take the metric $\|\cdot\|$ as the difference between angles. Since $f$ has no fixed points, so that the compactness of $X$ provides $\delta > 0$ with $\|x-f(x)\| \geq \delta$ for all $x \in X$. Here $\delta$ we can choose $\frac{4}{5}\pi$. Let $X=|K|$ for some finite simplicial complex $K$, here $K$ can be chosen as $\{[a, b], [b, c], [a, c], [a], [b], [c]\}$, and choose $n$ so that mesh $\mathrm{Sd}^n K < \frac{1}{2}\delta$. here $n$ is choosen as $1$, since $\frac{1}{3}\pi < \frac{2}{5}\pi$.

Choose $t$ so that there is a simplicial approximation $g: \mathrm{Sd}^{n+t} K \to \mathrm{Sd}^n K$ to $f$. Here $t$ can be chosen as $1$, since $g: \mathrm{Sd}^{2} K \to \mathrm{Sd}^1 K$ defined as shown in the figure. It's easy to check that $f(\mathrm{st} ~v) \subseteq \mathrm{st} ~ g(v)$ holds for all vertices in $\mathrm{Sd}^{2} K$.

If $h: \mathrm{Sd}^2 K \to \mathrm{Sd}^1 K$ is a simplicial approximation to the identity $|\mathrm{Sd}^2 K| \to |\mathrm{Sd}^1 K|$, then $|g| \cong f \cong f|h|$ and $|g|_* = f_* |h|_*$. (Here we don't need the explicit formula for $h$). And Lemma 9.17 givs $h_* = (\mathrm{Sd}_*)^{-1}$(We dont need to iterate, siince here t=1). Hence

$$ g_\# \mathrm{Sd}_\#: C_*(\mathrm{Sd}^1 K) \to C_*(\mathrm{Sd}^1 K) $$

is a chain map inducing $f_*$. Hence

$$\lambda(f) = \lambda(g_* \mathrm{Sd}_*) = \lambda(g_\# \mathrm{Sd}_\#) = tr(g_{0\#} \mathrm{Sd}_{0\#}) - tr(g_{1\#} \mathrm{Sd}_{1\#}) $$

Here for this concrete case we have that(the notation in the screenshot changed):

$$ g_{0\#} \mathrm{Sd}_{0\#}: C_0(\mathrm{Sd}^1 K) \to C_0(\mathrm{Sd}^1 K) \\ <p0> \mapsto <p1> \\ <p1> \mapsto <p2> \\ <p2> \mapsto <p0> \\ <q0> \mapsto <q1> \\ <q1> \mapsto <q2> \\ <p3> \mapsto <q0> $$

Show that matrix's diagnal elements are all zero. It's also easy to check for the case $p=1$ too:

$$ g_{1\#} \mathrm{Sd}_{1\#}: C_1(\mathrm{Sd}^1 K) \to C_1(\mathrm{Sd}^1 K) \\ <p0, q0> \mapsto g_{1, \#} (<p0, r0>) + g_{1, \#}(<r0, q0>) = <p1, q1> + <q1, q1> = <p1, q1> $$

So since all elements of both the two matrixs' diagnal are $0$, $tr(g_{0\#} \mathrm{Sd}_{0\#}) = tr(g_{1\#} \mathrm{Sd}_{1\#}) = 0$, hence $\lambda(f) = 0$.

Back to the original proof in the screenshot for the general case, the keypoint of the final paragraph is to show that the diagnal elements in the basis transformation matrix of $g_{p\#} \mathrm{Sd}_{p\#}$ are all zeros. Since the basis of $C_p(\mathrm{Sd}^n K)$ is all the ordered $p$-simplexes. Proof via contracondition, suppose for some $p$, some diagnal elements in the basis transformation matrix of

$$ g_{p\#} \mathrm{Sd}_{p\#}^t: C_p(\mathrm{Sd}^n K) \to C_p(\mathrm{Sd}^n K) $$ is not zero. Then equivalently it means that for some $p$-simplex $\sigma$, $(g_{p\#} \mathrm{Sd}_{p\#}^t)(\sigma) = \cdots + m \cdot \sigma + \cdots$ where integer $m \neq 0 $. Then it follows the underlined sentence: Since $\mathrm{Sd}_p^t$ is just the barycentric subdivision intutively(it can be defined rigorous), for example:

$$ \mathrm{Sd}_2^1(<p_0, p_1, p_2>) = <p_0, p_{01}, p_{012}> + <p_{012}, p_{01}, p_1> + \cdots $$

We can write(here the notation $\mathrm{Sd}_p^t \sigma$ used for two purpose, one is algebraic and the other is geometric)

$$ \mathrm{Sd}_p^t \sigma = \sum_{\rho \in \{p \text{ simplexes in } \mathrm{Sd}_p^t \sigma \}} a_\rho \rho $$ where all $a_\rho$s are non-zero integers and as the underlined sentence said, the $\rho$s are simplexes in $\sigma$. Then it comes the second half part: the proof has assumed that some diagnal elements in the basis transformation matrix of $g_{p\#} \mathrm{Sd}_{p\#}$ is not zero, or equivalently $(g_{p\#} \mathrm{Sd}_{p\#})(\sigma) = \cdots + m \cdot \sigma + \cdots$ where integer $m \neq 0 $. Since $$ \begin{aligned} (g_{p\#} \mathrm{Sd}_{p\#}^t)(\sigma) &= g_{p\#}(\mathrm{Sd}_{p\#}^t(\sigma)) \\ &= g_{p\#}\big(\sum_{\rho \in \{p \text{ simplexes in } \mathrm{Sd}_p^t \sigma \}} a_\rho \rho\big) \\ &= \sum_{\rho \in \{p \text{ simplexes in } \mathrm{Sd}_p^t \sigma \}} a_\rho g_{p\#}(\rho) \end{aligned} $$

Here recall that $g_{p\#}$ is a chain map from $C_p(\mathrm{Sd}^{n+t} K)$ to $C_p(\mathrm{Sd}^{n} K)$. Now then there has to be some $\rho_{i_0}$ such taht $g_{p\#}(\rho_{i_0}) = \sigma$(As Plantation said in the chat, it's weird that the screenshot said $g_{p\#}(\rho_{i_0}) \subset \sigma$, I think it should be $g_{p\#}(\rho_{i_0}) = \sigma$ too.) Otherwise there will be no such part $m \cdot \sigma$ in $(g_{p\#} \mathrm{Sd}_{p\#})(\sigma)$. Then the rest proof follows the same as the screenshot.