\classheader{2012-10-24}

Neat fact: given closed $A_1,\ldots,A_{n+1}\subset\S^n$ such that $\bigcup A_i=\S^n$, there exists a $j$ such that $A_j$ contains a pair of antipodal points.


\subsection*{The Lefschetz Fixed Point Theorem}

Last time, we talked about the Hopf trace formula, where we had a complex $\cal{C}=\{C_n,\partial_n\}$ and a chain map $\varphi:\cal{C}\to\cal{C}$, which induces maps $(\varphi_n)_*:H_n(\cal{C})/T_n(\cal{C})\rcirclearrowleft$, and deduced that
\[\sum (-1)^n\tr(\varphi_n)=\sum(-1)^n\tr((\varphi_n)_*).\]


Let's do some setup first. Let $X$ be a finite $\Delta$-complex, and let $f:X\to X$ be a continuous map. Consider $(f_n)_*:H_n(X)/T_n(X)\rcirclearrowleft$.

\begin{definition}
The Lefschetz number of $f$ is
\[\Lambda(f)=\sum_{n\geq 0}(-1)^n\tr((f_n)_*).\]
\end{definition}

\begin{theorem}[LFPT]
If $\Lambda(f)\neq0$, then $f$ has a fixed point.
\end{theorem}
\begin{corollary}
Let $X$ be a contractible $\Delta$-complex, e.g. $X=\D^n$. Then any $f:X\to X$ has a fixed point.
\end{corollary}
\begin{proof}[Proof of Corollary]
By hypothesis, we have that $X$ is ``aspherical'', i.e.
\[H_n(X)=\begin{cases}
\Z & \text{ if }n=0,\\
0 & \text{ if }n>0
\end{cases}\]

This implies that $(f_n)_*=0$ for all $n>0$, and hence $\tr((f_n)_*)=0$ for all $n>0$.  We also have (from your current homework)
\[\tr((f_0)_*)=\tr(\id)=\tr([1])=1.\]
Thus,
\[\Lambda(f)=1-0+0-\cdots=1\neq 0.\qedhere\]
\end{proof}
\begin{proof}[Proof of Lefschetz]
Suppose that $f(x)\neq x$ for all $x\in X$.

Put a metric $d$ on $X$. There is some $\delta>0$ such that $d(f(x),x)>\delta$ for all $x\in X$ because $f$ is continuous, $f(x)\neq x$ for all $x$, and because $X$ is compact. Now, by barycentrically subdividing, we can put a $\Delta$-complex structure on $X$ such that $\diam(\sigma)<\frac{d}{100}$ for all simplices $\sigma$. 

By simplicial approximation (which we haven't covered in detail, but you should know the statement at least), we have that, possibly after further subdividing, there is a simplicial map $h:X\to X$ such that $h\sim f$ and $d(h(x),f(x))\leq\frac{\delta}{2}$ for all $x\in X$.

Thus, for all $x\in X$, we have $d(h(x),x)>\frac{\delta}{10}>\frac{\delta}{100}$, and because $h$ is simplicial, we have $h(\sigma)\cap\sigma=\varnothing$ for all simplices $\sigma$.

We claim that this implies $\tr((h_n)_\#)=0$ for all $n\geq 0$. The matrix for $(h_n)_\#$ acting on $C_n(X)\cong\Z^d$ looks like
\[\bordermatrix{ ~ & \sigma_1 & \sigma_2 & \cdots & \sigma_d\cr
\sigma_1 & 0 & & & \cr
\sigma_2 &  & 0 & & \cr
\vdots & & & & \cr
\sigma_d & & & & 0}\]
so, applying the Hopf trace formula,
\[\Lambda(h)=\sum(-1)^n\tr((h_n)_*)=\sum(-1)^n\tr((h_n)_\#)=0.\]
But $f\sim h$, so $f_*=h_*$, and so $\Lambda(f)=\Lambda(h)=0$.
\end{proof}
\begin{corollary}[of Brouwer]
Let $A$ be an $n\times n$ matrix with real entries $a_{ij}>0$. Then there exists a real eigenvalue $\lambda$ with a real eigenvector $(v_1,\ldots,v_n)$ with all $v_i>0$.
\end{corollary}
\begin{proof}
This has applications to the adjacency matrix of a graph, and to probability.

Let $X=\{\text{rays through 0 in a positive orthant}\}$, which is homeomorphic to $\Delta^{n-1}$. We have that $A$ maps $X$ to $X$ by hypothesis. Then Brouwer implies that there exists an $x\in X$ such that $Ax=x$, i.e. there is a ray $\R\vec{x}$ such that $A\vec{v}=\lambda\vec{v}$  Then we must have $\lambda>0$ and $v_i>0$ for all coordinates of $\vec{v}=(v_1,\ldots,v_n)$.
\end{proof}

\begin{remark}[Weil's Observation]
Suppose that $f:X\to X$ is a homeomorphism of compact manifolds.

\textit{Assumption:} Suppose that $\Lambda(f)=\#$ of fixed points of $f$, and that $\Lambda(f^n)=\#$ of periodic points of $f$ of period $n$, i.e. points with $f^n(x)=x$.

\textit{Claim:} There exist algebraic integers $\alpha_1,\ldots,\alpha_r,\beta_1,\ldots,\beta_s$ such that
\[\Lambda(f^n)=\#\text{ of periodic points of $f$ of period $n$} =\sum \alpha_i^n-\sum \beta_j^n.\]
The amazing thing is that the left side is obviously an integer, but the right side rarely is if you choose arbitrary $\alpha_i$ and $\beta_i$.

The proof is just that
\[\Lambda(f^n)=\sum_{i}(-1)^i\tr((f_i^n)_*)\]
where $(f_i^n)_*:H_n(X)/T_n(X)\rcirclearrowleft$, and $H_n(X)/T_n(X)\cong \Z^d$, so that $(f_i)_*\in\M_{d\times d}(\Z)$ and hence its eigenvalues $\lambda_1,\ldots,\lambda_s$ are algebraic integers, so that $\tr((f_i^n)_*)=\sum\lambda_i^n$.

Moreover, he observed that, if $X$ is a projective smooth variety defined over $\Z$, then
\[X(\F_{p^n})=\sum\alpha_i^n-\sum\beta_j^n\]
and he conjectured there was a relationship. Indeed, $X(\overline{\F_p})\rcirclearrowleft\operatorname{Frob}_p$, and $\operatorname{fix}(\operatorname{Frob}_p^n)=X(\F_{p^n})$.
\[\sum(-1)^i\operatorname{Frob}_p^n\lcirclearrowright\underbrace{H^i_\text{etal}(X)}_{\cong\,H^i(X(\C))}\]
\end{remark}