%! TEX root = AlgT.tex % vim: tw=50 % 11/03/2024 11AM \subsubsection*{Example (non-orientable surfaces)} \begin{center} \includegraphics[width=0.6\linewidth]{images/4e88f59494cf41c9.png} \end{center} \begin{center} \includegraphics[width=0.6\linewidth]{images/0ba89b82208c425b.png} \end{center} Homology of $K$: the subcomplex $\polyh|W| \subset \polyh|K|$ with \begin{center} \includegraphics[width=0.2\linewidth]{images/b5401080a4f94210.png} \end{center} is such that $\polyh|K|$ deformation retracts to it, so $\Hgp_0(K) = \ZZ\{\text{vertex}\}$, $\Hgp_1(K) = \ZZ\{u\}$, $u = [1, 2] + [2, 3] + [3, 1]$ and $\Hgp_i(K) = 0$ for $i > 1$. Note $r = [4, 5] + [5, 6] + [6, 4]$ is homologous to $2u$. Homology of $K_n$: $K_n = K_{n - 1} \cup K$, $K_{n - 1} \cap K = \ssimp^1$. So Mayer-Vietoris gives: \begin{center} \includegraphics[width=0.6\linewidth]{images/e35fd3c4a1e14a74.png} \end{center} So inductively $\Hgp_0(K_n) = \ZZ\{\text{a vertex}\}$, $\Hgp_1(K_n) = \ZZ^n = \ZZ\{u_1, \ldots, u_n\}$, $\Hgp_2(K_n) = 0$ for $i \ge 2$. The boundary $L$ of $K_n$ is the cycle $r_1 + r_2 + \cdots + r_n = 2(u_1 + \cdots + u_n)$. Homology of $K_n \cup CL$: Mayer-Vietoris again: \begin{center} \includegraphics[width=0.6\linewidth]{images/197b4ea7992a48c9.png} \end{center} Note ($*$) sends the generator to $r_1 + r_2 + \cdots + r_n = 2(u_1 + \cdots + u_n)$. So ($*$) is injective, which tells us that the previous map $\partial_*$ is $0$. So $\Hgp_0(K_n \cup CL) = \ZZ$, $\Hgp_1(K_n \cup CL) = \frac{\ZZ\{u_1, \ldots, u_n\}}{\langle 2(u_1 + \cdots + u_n) \rangle} \cong \ZZ / 2\ZZ \oplus \ZZ^{n - 1}$, and finally $\Hgp_i(K_n \cup CL) = 0$ for $i \ge 2$. This is $H_*(S_n)$. The surfaces: \begin{center} \begin{tabular}{c|c|c|c} & $S^2$ & $\Sigma_g$ & $S_n$ \\ \hline $H_1$ & $0$ & $\ZZ^{2g}$ & $\ZZ / 2\ZZ \oplus \ZZ^{n - 1}$ \\ $H_2$ & $\ZZ$ & $\ZZ$ & $0$ \end{tabular} \end{center} are all not homotopically equivalent to each other. \begin{theorem*} Every triangulable surface is homeomorphic to one of these. \end{theorem*} \subsection{Rational homology, Euler and Lefschetz numbers} \begin{flashcard}[On-with-Q] \begin{definition*}[$O_n(K; \QQ)$] \glssymboldefn{Hqq}{$H(K; Q)$}{$H(K; Q)$} \cloze{For a \gls{simpcomp} $K$, let $O_n(K; \QQ)$ be the $\QQ$-vector space with basis the ordered \glspl{simp} of $K$. Define $T_n(K; \QQ)$ as usual, and so get $C_n(K; \QQ)$. Define $d_n$ by the same formula, and \[ H_n(K; \QQ) = \frac{\Ker(d_n : C_n(K; \QQ) \to C_{n - 1}(K, \QQ))}{\Im(d_{n + 1} : C_{n + 1}(K; \QQ) \to C_n(K; \QQ)} .\] This is a $\QQ$-vector space.} \end{definition*} \end{flashcard} \begin{lemma*} If $\Hgp_n(K) \cong \ZZ^r \oplus \text{(finite abelian group)}$, then $\Hqq_n(K; \QQ) \cong \QQ^r$. \end{lemma*} \begin{proof} See \es[5]{4}. \end{proof} \begin{example*} \[ \Hqq_i(S6n; \QQ) = \begin{cases} \QQ & i = 0, n \\ 0 & \text{otherwise} \end{cases} \] \[ \Hqq_i(\Sigma_g; \QQ) = \begin{cases} \QQ & i = 0, 2 \\ \QQ^{2g} & i = 1 \\ 0 & \text{otherwise} \end{cases} \] \[ \Hqq_i(S_n; \QQ) = \begin{cases} \QQ & i = 0 \\ \QQ^{n - 1} & i = 1 \\ 0 & \text{otherwise} \end{cases} \] \end{example*} \begin{flashcard}[lefschetz-number-defn] \begin{definition*}[Lefschetz number] \glssymboldefn{lef}{$L(f)$}{$L(f)$} \glsnoundefn{lefnum}{Lefschetz number}{Lefschetz numbers} \cloze{Let $X$ be a polyhedron ($\cong \polyh|K|$) and $f : X \to X$ be a continuous map. The \emph{Lefschetz number} of $f$ is \[ L(f) = \sum_{i = 0}^\infty (-1)^i \Trace(f_* : \Hqq_i(X; \QQ) \to \Hqq_i(X; \QQ)) .\]} \end{definition*} \end{flashcard} \begin{flashcard}[euler-char-defn] \begin{definition*}[Euler characteristic] \glssymboldefn{eul}{$\chi$}{$\chi$} \glsnoundefn{eulnum}{Euler characteristic}{Euler characteristics} \cloze{The \emph{Euler characteristic} of $X$ is \[ \chi(X) = \lef(\id) = \sum_{i = 0}^\infty (1)^i \dim_{\QQ} \Hqq_i(X; \QQ) .\]} \end{definition*} \end{flashcard} \begin{example*} \[ \eul(S^n) = 1 + (-1)^n = \begin{cases} 2 & \text{$n$ even} \\ 0 & \text{$n$ odd} \end{cases} \] \[ \eul(\Sigma_g) = 2 - 2g, \qquad \eul(S_n) = 2 - n .\] \end{example*} \begin{example*} The antipodal map $a : S^n \to S^n$ induces the identity map on $\Hqq_0(S^n; \QQ)$, and multiplication by $(-1)^{n + 1}$ on $\Hqq_n(S^n; \QQ)$. So $\lef(a) = 1 + (-1)^n(-1)^{n + 1} = 0$. \end{example*} \begin{flashcard}[linalg-trace-fact-lemma] \begin{lemma*} Let $V$ be a finite-dimensional vector space, $W \le V$ a subspace, and $A : V \to V$ a linear map such that $A(W) \le W$. Let $B = A|_W : W \to W$ and $C : V / W \to V / W$ the map induced by $A$. Then \[ \Trace(A) = \Trace(B) + \Trace(C) .\] \end{lemma*} \begin{proof} \cloze{Let $e_1, \ldots, e_r$ be a basis for $W$, and extend it to a basis of $V$. In this basis, \[ A = \begin{pmatrix} B & * \\ 0 & C \end{pmatrix} \] so $\Trace(A) = \Trace(B) + \Trace(C)$.} \end{proof} \end{flashcard} \begin{flashcard}[lefschetz-equality-coro] \begin{corollary*} For a \gls{cmap} $f_\bullet : C_\bullet(K; \QQ) \to C_\bullet(K; \QQ)$, we have \[ \cloze{\sum_{i = 0}^\infty (-1)^i \Trace(f_* : \Hqq_i(K; \QQ) \to \Hqq_i(K; \QQ)) = \sum_{i = 0}^\infty (-1)^i \Trace(f_i : C_i(K; \QQ) \to C_i(K; \QQ))} \] \end{corollary*} \begin{proof} \cloze{Consider \[ 0 \to B_i(K; \QQ) \to Z_i(K; \QQ) \to H_i(K; \QQ) \to 0 \] \[ 0 \to Z_i(K; \QQ) \to C_i(K; \QQ) \stackrel{d_i}{\to} B_{i - 1}(K; \QQ) \to 0 \] Let $f^C, f^Z, f^B, f^H$ be the induced maps on chains / cycles / boundary / homology. \begin{align*} \sum (-1)^i \Trace(f_i^H) &= \sum (-1)^i (\Trace(f_i^Z) - \Trace(f_i^B)) \\ &= \sum (-1)^i (\Trace(f_i^C) - \Trace(f_{i - 1}^B) - \Trace(f_i^B)) \\ &= \sum (-1)^i \Trace(f_i^C) \end{align*}} \end{proof} \end{flashcard} \begin{corollary*} \[ \eul(\polyh|K|) = \sum_{i = 0}^\infty (-1)^i \#\{\text{$i$-simplices of $K$}\} \] \end{corollary*}