%! TEX root = AlgT.tex % vim: tw=50 % 23/02/2024 11AM \setcounter{section}{7} \setcounter{subsection}{0} \subsection{Simplicial homology} \begin{flashcard}[On-K-Z-module-defn] \begin{definition*} \glssymboldefn{Ogp}{$O_n(K)$}{$O_n(K)$} Let $K$ be a \gls{simpcomp}. We define $\mathcal{O}_n(K)$ to be \cloze{the free abelian group ($\ZZ$-module) with basis \[ \{[v_0, v_1, \ldots, v_n] \st \text{the $v_i$ are \glspl{vert} of $K$ which span a \gls{simp}}\} .\] The $v_i$ are considered to be \emph{ordered}, and could span a \gls{simp} of $\dim < n$, i.e. could have repeats.} \end{definition*} \end{flashcard} \begin{flashcard}[Tn-K-defn] \begin{definition*} \glssymboldefn{Tgp}{$T_n(K)$}{$T_n(K)$} \glssymboldefn{Cgp}{$C_n(K)$}{$C_n(K)$} Define $T_n(K) \cloze{\le \Ogp_n(K)}$ \cloze{to be the subgroup spanned by: \begin{enumerate}[(i)] \item $[v_0, v_1, \ldots, v_n]$ containing a repeat. \item $[v_0, v_1, \ldots, v_n] - \sgn(\sigma)[v_{\sigma(0)}, \ldots, v_{\sigma(n)}]$ for a permutation $\sigma$ of $\{0, 1, \ldots\}$. \end{enumerate} Define $C_n(K) = \Ogp_n(K) / T_n(K)$, the quotient group. } \end{definition*} \end{flashcard} \begin{flashcard}[C-n-K-non-canonical-iso-lemma] \begin{lemma*} There is a non-canonical isomorphism $C_n(K) \cong \ZZ \{\text{\glsref[simp]{$n$-simplices} of $K$}\}$. \end{lemma*} \begin{proof} \cloze{Choose a total order $\prec$ of $\vset K$. Then each \nsimp{n} $\sigma$ of $K$ determines a canonical ordered \gls{simp} $[\sigma] \in \Ogp_n(K)$ by ordering its \glspl{vert} such that $a_0 \prec a_1 \prec \cdots \prec a_n$. This gives: \begin{align*} \phi : \ZZ\{\text{\glsref[simp]{$n$-simplices of $K$}}\} &\to \Ogp_n(K) \\ \sigma &\mapsto [\sigma] \end{align*} ans so gives \begin{align*} \phi' : \ZZ\{\text{\glsref[simp]{$n$-simplices of $K$}}\} &\to \Cgp_n(K) \\ \end{align*} For each $[a_0, a_1, \ldots, a_n] \in \Ogp_n(K)$, there is a \emph{unique} permutation $\tau$ of $\{0, 1, \ldots, n\}$ such that $a_{\tau(0)} \prec a_{\tau(1)} \prec \cdots \prec a_{\tau(n)}$. Let \[ \sgn[a_0, \ldots, a_n] \defeq \sgn(\tau) \in \{\pm 1\} .\] Define \begin{align*} \rho : \Ogp_n(K) &\to \ZZ\{\text{\glsref[simp]{$n$-simplices} of $K$}\} \\ [a_0, \ldots, a_n] &\mapsto \begin{cases} (\sgn[a_0, \ldots, a_n]) \simplex & \text{no repeats} \\ 0 & \text{repeats} \end{cases} \end{align*} For this to descend to $\Cgp_n(K)$, need $\Tgp_n(K)$ to be in $\Ker(\rho)$. Certainly $[a_0, \ldots, a_n]$'s with repeats are in $\Ker(\rho)$. \begin{align*} \rho([v_0, \ldots, v_n] - \sgn(\sigma)[v_{\sigma(0)}, \ldots, v_{\sigma(n)}]) &= \sgn[v_0, \ldots, v_n] \simplex \\ - \sgn(\sigma) \sgn[v_{\sigma(0)}, \ldots, v_{\sigma(n)}] \simplex \\ &= 0 \end{align*} as required. So we get $\rho' : \Cgp_n(K) \to \ZZ\{\text{\glsref[simp]{$n$-simplices} of $K$}\}$. Not $\rho' \circ \phi'(\sigma) = \sigma$. If $[a_0, \ldots, a_n]$ has no repeats, then \begin{align*} \phi' \circ \rho'([a_0, \ldots, a_n]) &= \phi'(\sgn[a_0, \ldots, a_n]\simplex) \\ &= \sgn[a_), \ldots, a_n] [a_{\tau(0)}, \ldots, a_{\tau(n)}] \\ &= [a_0, \ldots, a_n] \mod T_n(K) \end{align*} So $\phi'$ and $\rho'$ are inverse.} \end{proof} \end{flashcard} \begin{flashcard}[dn-hom-defn] \begin{definition*}[$d_n$] \glssymboldefn{dhom}{$d_n$}{$d_n$} \glssymboldefn{skip}{widehat $v_i$}{widehat $v_i$} \cloze{Define a homomorphism \begin{align*} d_n : \Ogp_n(K) &\to \Ogp_{n - 1}(K) \\ [v_0, \ldots, v_n] &\mapsto \sum_{i = 0}^n (-1)^i [v_0, \ldots, \widehat{v_i}, \ldots, v_n] \end{align*} where the $\widehat{v_i}$ means ``skipel this element''.} \end{definition*} \end{flashcard} \begin{flashcard}[dn-Tn-to-Tn-lemma] \begin{lemma*} $\dhom_n$ sends $\Tgp_n(K)$ into $\Tgp_{n - 1}(K)$. \end{lemma*} \begin{proof} \cloze{Note \begin{align*} &\!\!\!\!\!\dhom_n([v_0, \ldots, v_n] - \sgn(\sigma)[v_{\sigma(0)}, \ldots, v_{\sigma(n)}]) \\ &= \sum_{i = 0}^n (-1)^i [v_0, \ldots, \skipel{v_i}, \ldots, v_n] - \sum_{i = 0}^n (-1)^i \sgn(\sigma) [v_{\sigma(0)}, \ldots, \skipel{v_{\sigma(i)}}, \ldots, v_n] \end{align*} Need to show that this is trivial in $\Ogp_{n - 1}(K) / \Tgp_{n - 1}(K)$. Suppose first $\sigma = (j, j + 1)$, a transposition, so $\sgn(\sigma) = -1$. Then \begin{align*} &\!\!\!\!\!\sum_{i = 0}^n (-1)^i \sgn(\sigma) [v_{\sigma(0)}, \ldots, \skipel{v_{\sigma(i)}}, \ldots, v_{\sigma(n)}] \\ &= \sum_{i = 0}^{j - 1} (-1)^{i + 1} [v_0, \ldots, \skipel{v_i}, \ldots, v_{j + 1}, v_j, v_{j + 2}, \ldots, v_n] \\ &~~~~ + (-1)^{j + 1} [v_0, \ldots, v_{j - 1}, v_j, v_{j + 2}, \ldots, v_n] \\ &~~~~ + (-1)^{j + 2} [v_0, \ldots, v_{j - 1}, v_{j + 1}, \ldots, v_n] \\ &~~~~ + \sum_{i = j + 2}^n (-1)^{i + 1} [v_0, \ldots, v_{j - 1}, v_{j + 1}, v_j, \ldots, v_{j + 2}, \ldots, \skipel{v_i}, \ldots, v_n] \end{align*} In the first sum \[ [v_0, \ldots, \skipel{v_i}, \ldots, v_{j - 1}, v_{j + 1}, v_{j + 2}, \ldots, v_n] \equiv -[v_0, \ldots, \skipel{v_i}, \ldots, v_n] \mod \Tgp_{n - 1}(K) .\] In the second sum, \[ [v_0, \ldots, v_{j - 1}, v_{j + 1}, v_j, v_{j + 2}, \ldots, \skipel{v_i}, \ldots, v_n] \equiv -[v_0, \ldots, \skipel{v_i}, \ldots, v_n] \mod \Tgp_{n - 1}(K) .\] Then \[ \RHS \equiv \sum_{i = 0}^n (-1)^i [v_0, \ldots, \skipel{v_i}, \ldots, v_n] \mod \Tgp_{n - 1}(K) \] as required. Any $\sigma$ is a product of $(j, j + 10$, so we get the same for any $\sigma$. Now suppose $[v_0, \ldots, v_n]$ with $v_j = v_{j + 1}$ (if it has repeats, we can assume this without loss of generality by using permutations since we showed this doesn't affect values $\mod \Tgp_{n - 1}(K)$). Then: \begin{align*} \dhom_n[v_0, \ldots, v_n] &= \sum_{i = 0}^{j - 1} (-1)^i [v_0, \ldots, \skipel{v_i}, \ldots, v_j, v_{j + 1}, \ldots, v_n] \\ &~~~~ + (-1)^j [v_0, \ldots, v_{j - 1}, v_{j + 1}, \ldots, v_n] \\ &~~~~ + (-1)^{j + 1} [v_0, \ldots, v_j, v_{j + 2}, \ldots, v_n] \\ &~~~~ + \sum_{i = j + 2}^n (-1)^j [v_0, \ldots, v_j v_{j + 1}, \ldots, \skipel{v_i}, \ldots, v_n] \\ &\in \Tgp_{n - 1}(K) \end{align*} This is because the sums are both in $\Tgp_{n - 1}(K)$ since each term has a repeat, and the middle alone terms cancel each other since $v_j = v_{j + 1}$.} \end{proof} \end{flashcard} So $\dhom_n$ induces a homomorphism $\dhom_n : \Cgp_n(K) \to \Cgp_{n - 1}(K)$ given by the same formula. \begin{flashcard}[dn-composition-is-zero-lemma] \begin{lemma*} The composition $d_{n - 1} \circ d_n : \Cgp_n(K) \to \Cgp_{n - 1}(K)$ is zero. \end{lemma*} \begin{proof} \cloze{At the level of $\Ogp_n(K)$, compute \begin{align*} &\!\!\!\! \dhom_{n - 1} \circ \dhom_n [v_0, \ldots, v_n] \\ &= \dhom_{n - 1} \left( \sum_{i = 0}^n (-1)^i [v_0, \ldots, \skipel{v_i}, \ldots, v_n]\right) \\ &= \sum_{i = 0}^n (-1)^i \left[ \sum_{k = 0}^{i - 1} (-1)^k [v_0, \ldots, \skipel{v_i}, \ldots, v_n] + \sum_{k = i}^{n - 1} (-1)^k [v_0, \ldots, \skipel{v_i}, \ldots, v_n] \right] \end{align*} Coefficient of $[v_0, \ldots, \skipel{v_a}, \ldots, \skipel{v_b}, \ldots, v_n]$ is $(-1)^a(-1)^b + (-1)^a (-1)^{b - 1} = 0$. As the $[v_0, \ldots, v_n]$ generate, we get $\dhom_{n - 1} \circ \dhom_n = 0$.} \end{proof} \end{flashcard}