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\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<a_0,
      \ldots, a_n> & \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<v_0,
    \ldots, v_n> \\
    - \sgn(\sigma) \sgn[v_{\sigma(0)}, \ldots,
    v_{\sigma(n)}] \simplex<v_{\sigma(0)}, \ldots,
    v_{\sigma(n)}> \\
    &= 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<a_0,
    \ldots, a_n>) \\
    &= \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}
