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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.
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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
\begin{aligned} \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 \end{aligned}
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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