~mht/cra

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933b2735 — Martin Hafskjold Thoresen Fix warnings and remove some dead code 10 months ago
                                                                                
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\section{Words and things}

\def\cp{\textsf{c}_p}
\def\cpm1{\textsf{c}_{p-1}}
\def\boundary#1{\partial #1}
\def\xor{\texttt{XOR}}
\def\homgroup{\tilde{H}_{p}(K)}

\begin{definition}[p-chain]
$\cp \subseteq K^p$: some of the p-cells of dimension $p$.
The set of all p-chains is a complex $K$ is $C_p(K)$.
\end{definition}

\begin{definition}[boundary]
The boundary of a p-chain $\cp$ is a $(p-1)$-chain $\cpm1$ such that  
all cells in $\cpm1$ share an odd number of cells in $\cp$.
\end{definition}

\begin{definition}[p-boundary]
A p-boundary is the boundary of a p-chain.
The set of all p-boundaries is a complex $K$ is $B_p(K)$.
\end{definition}

\begin{definition}[p-cycle]
A p-cycle is a p-chain which boundary is $\emptyset$.
The set of all p-cycles is a complex $K$ is $Z_p(K)$.
\end{definition}

Remember that we have a $(-1)$-cell, which is the empty cell. This is a face
of all other cells, which means that the boundary of a single vertex 
is the empty cell, so a single vertex is not a 0-cycle.
Two vertices are a 0-cycle however: 
the empty cell is shared between both vertices, so the boundary is $\emptyset$,
making it a 0-cycle.


Apparently, $\boundary{(a + b)} = \boundary{a} + \boundary{b}$.
Then we can show why the definition of a boundary makes sense:
let $P$ be a $p$-chain, and look at the $p$-cells in $P$ (here denoted $p$):
The boundary of $P$ is 
\[
\boundary{P} = \boundary{\sum_{p \in P} p} = \sum_{p \in P} \boundary{p} = \sum_{p \in P}\sum_{f \in p} f 
\]
Since we're operating with \xor{} addition, the $(p-1)$ faces that survive this is exactly
the ones that are shared by an odd number of $p$-cells in $P$.

\begin{lemma}
The boundary of a boundary is $\emptyset$.
\begin{proof}

\end{proof}
\end{lemma}

The boundaries are the trivial cycles of a $p$-chain. From the conversation
with Ondra, this is analogous to a vector space in which we chose one direction
to be trivial, and have from then all all vectors that differ by this trivial
direction be the same.  In this context, the trivial cycles are the ``zeroes'',
and so adding a boundary to a cycle doesn't really ``change'' it: we put both
chains into the same class, and denote the set of all classes by $\homgroup{} =
Z_p(K)/B_p(K)$, and the \emph{Betti number} is the rank of $\homgroup{}$, that
is the number of non-trivial cycles.

A geometrical intuition for why we have chosen the boundaries to be the trivial
cycles is that these boundaries are filled, since there is a $p$-chain that
have this $(p-1)$-chain as a boundary, and we're interested in the cycles that
are not filled.