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If \(k \geq 2\) and \(\sigma=\left[\mathbf{p}_{0}, \mathbf{p}_{1}, \ldots, \mathbf{p}_{k}\right]\) is an oriented affine \(k\) simplex, prove that \(\partial^{2} \sigma=0\), directly from the definition of the boundary operator \(\partial\). Deduce from this that \(\partial^{2} \Psi=0\) for every chain \(\Psi\).

Hint: For orientation, do it first for \(k=2, k=3\). In general, if \(i<j\), let \(\sigma_{i j}\) be the \((k-2)\)-simplex obtained by deleting \(\mathbf{p}_{i}\) and \(\mathbf{p}_{j}\) from \(\sigma\). Show that each \(\sigma_{i j}\) occurs twice in \(\partial^{2} \sigma\) with opposite sign.
in Mathematics by Diamond (55.6k points) | 118 views

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For \(k=2\) we have
\partial \sigma=\left[\mathbf{p}_{1}, \mathbf{p}_{2}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{2}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}\right]
so that
\partial^{2} \sigma=\left(\mathbf{p}_{2}-\mathbf{p}_{1}\right)-\left(\mathbf{p}_{2}-\mathbf{p}_{0}\right)+\left(\mathbf{p}_{1}-\mathbf{p}_{0}\right)=0
For \(k=3\) we have
\partial \sigma=\left[\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{2}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{1}, \mathbf{p}_{2}\right]
so that
\partial^{2} \sigma=&\left(\left[\mathbf{p}_{2}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{1}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{1}, \mathbf{p}_{2}\right]\right) \\
&-\left(\left[\mathbf{p}_{2}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{2}\right]\right) \\
&+\left(\left[\mathbf{p}_{1}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}\right]\right) \\
&-\left(\left[\mathbf{p}_{1}, \mathbf{p}_{2}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{2}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}\right]\right) \\
=& 0
In general the order in which \(\mathbf{p}_{i}\) and \(\mathbf{p}_{j}\) are omitted from \(\sigma\) determines the sign that \(\sigma_{i j}\) will have. If \(\mathbf{p}_{j}\) is omitted first, the resulting \((k-1)\)-simplex \(\sigma_{j}=\) \(\left[\mathbf{p}_{0}, \ldots, \mathbf{p}_{j-1}, \mathbf{p}_{j+1}, \ldots, \mathbf{p}_{k}\right]\) will acquire the sign \((-1)^{j}\). If \(\mathbf{p}_{i}\) is then omitted, the resulting \((k-2)\)-simplex will acquire a factor of \((-1)^{i}\), resulting in \((-1)^{i+j} \sigma_{i j}\).
However, if \(\mathbf{p}_{i}\) is omitted first, \(\mathbf{p}_{j}\) will move forward one position in the resulting \((k-1)\)-simplex \(\sigma_{i}\), and when it is subsequently omitted, a factor of

\((-1)^{j-1}\) will be affixed, resulting in \((-1)^{i+j-1} \sigma_{i j}\). Hence the two occurrences of \(\sigma_{i j}\) in the second boundary will cancel each other.

The linearity of the boundary operator, operating on a base of simplexes, then shows that \(\partial^{2}\) is the zero operator on all chains.
by Diamond (55.6k points)

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