Composition of boundary homomorphism from Hatcher

We have $\partial_n ( \sigma) = \sum_{i} (-1)^{i} \sigma [ v_1, \dots, \overset { \wedge} v_i, \dots, v_n]$ and we want to show that the composition $$\delta_n(X) \overset{\partial_n} \to \delta_{n-1}(X) \overset{\partial_{n-1}} \to \delta_{n-2}(X)$$ is zero.

The book goes on to simply write, $$\partial_{n-1}\partial_n(\sigma)= \sum_{j < i} (-1)^{i} (-1)^{j} \sigma [ v_0, \dots, \overset{\wedge}v_j, \dots, \overset{\wedge}v_i, \dots v_n ]+$$ $$\sum_{i < j} (-1)^{i} (-1)^{j-1} \sigma [ v_0, \dots, \overset{\wedge}v_i, \dots, \overset{\wedge}v_j, \dots v_n ]$$

I am not quite able to produce this summation on my own using the definition of the boundary homomorphism. I would like to see the above result more fleshed out if possible.

• what summation do you get then when you try to do this problem yourself? – Dan Rust Nov 6 '14 at 21:38
• @asdvn: Have you had any luck with this problem? – Michael Albanese May 23 '15 at 20:46
• +1, I face the same problem. I have two questions, how do we get $j-1$ power for the second summation, and what happens to the $i=j$ term? – yoyostein Nov 6 '15 at 14:38

A lot of confusion arises in this problem due to an index shift performed during the calculation. I'm going to change the notation used slightly, but I'll hopefully define everything below (let me know if I miss something). First, I'll write the i-th face map $d^i:\Delta^{n-1}\rightarrow\Delta^n$:

$$d^i(t_0,\ldots,t_{n-1})=(t_0,\ldots,t_{i-1},0,t_i,\ldots,t_{n-1}).$$

Next, define the boundary map in terms of the face maps:

$$\partial_n(\sigma)=\sum\limits_{i=0}^n(-1)^i(\sigma\circ d^i).$$

Note that whenever $i\leq j$ we have $d^i\circ d^j=d^{j+1}\circ d^i$ because if we put a 0 in before the $t_i$ co-ordinate first, then to put a 0 in before the $t_j$ co-ordinate it needs to go in at the $(j+1)$-th place (due to everything including and after the $t_i$ co-ordinate being shifted up an index by the placement of the first 0). This will allow us to simplify the algebra a little in the following.

Now, we get: \begin{align*} \partial_{n-1}\circ\partial_n(\sigma)&=\partial_{n-1}\sum\limits_{i=0}^n(-1)^i(\sigma\circ d^i)\\ &=\sum\limits_{i=0}^n(-1)^i\sum\limits_{j=0}^{n-1}(-1)^j(\sigma\circ d^i\circ d^j)\\ &=\sum\limits_{i=0}^n\Big\lbrack\sum\limits_{j=0}^{i-1}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)+\sum\limits_{j=i}^{n-1}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)\Big\rbrack\\ &=\sum_{\substack{0\leq i\leq n\\j<i}}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)+\sum_{\substack{0\leq i\leq n\\j\geq i}}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j) \end{align*} Now, using the fact that $i\leq j$ implies $d^i\circ d^j=d^{j+1}\circ d^i$ in the second sum, we have: \begin{align*} \partial_{n-1}\circ\partial_n(\sigma)&=\sum_{\substack{0\leq i\leq n\\j<i}}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)+\sum_{\substack{0\leq i\leq n\\j\geq i}}(-1)^i(-1)^j(\sigma\circ d^{j+1}\circ d^i) \end{align*} Here's where the index shift happens. We change the index in the second sum to get rid of the $j+1$. In particular note that the limit of the sum changes to a strict inequality because we run j from $i+1$ to $n$ instead of from $i$ to $n-1$: \begin{align*} \partial_{n-1}\circ\partial_n(\sigma)&=\sum_{\substack{0\leq i\leq n\\j<i}}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)+\sum_{\substack{0\leq i\leq n\\j> i}}(-1)^i(-1)^{j-1}(\sigma\circ d^j\circ d^i) \end{align*} and this is the sum you asked about. To see why this is zero, notice that we can take the alternate view in the second sum of first choosing $j$, then exchange the roles of the indices: \begin{align*} \partial_{n-1}\circ\partial_n(\sigma)&=\sum_{\substack{0\leq i\leq n\\j<i}}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)+\sum_{\substack{0\leq j\leq n\\j> i}}(-1)^i(-1)^{j-1}(\sigma\circ d^j\circ d^i)\\ &=\sum_{\substack{0\leq i\leq n\\j<i}}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)+\sum_{\substack{0\leq i\leq n\\i> j}}(-1)^j(-1)^{i-1}(\sigma\circ d^i\circ d^j)\\ &=\sum_{\substack{0\leq i\leq n\\j<i}}(-1)^i(-1)^j(\sigma\circ d^i\circ d^j)-\sum_{\substack{0\leq i\leq n\\i> j}}(-1)^j(-1)^i(\sigma\circ d^i\circ d^j) \end{align*} and now the second sum is just the negative of the first.