On the meaning of formal sums of $k$-cubes, i.e. $k$-chains (in integration on manifolds)

Solution 1:

Let's see what happens when we interpret these sums pointwise. For example, $\partial I^2$ is an 1-chain given by $$\partial I^2 = -I^2_{(1,0)} + I^2_{(1,1)} + I^2_{(2,0)} - I^2_{2,1}$$ Now evaluating this "sum" at a point $x \in [0,1]$ gives $$\partial I^2 = -(0,x) + (1, x) + (x, 0) - (x, 1) = (1, -1) $$ Try the same calculation for $\partial I^3$ and you'll get $(1,-1,1)$. Clearly this doesn't make sense.

To see how Spivak "summed" the $(n-2)$-chain in his proof of $\partial^2=0$, let's take $\partial^2I^2$ as an example. Now we must sum the boundaries of each $I^2_{(i, \alpha)}$ with the correct orientation. So, for instance, \begin{align} \partial (-I^2_{(1,0)})(0) &= -[-(I^2_{(1,0)})_{(1,0)}(0) + (I^2_{(1,0)})_{(1,1)}(0)] \\ &= -[-I^2_{(1,0)}(I^1_{(1,0)}(0)) + I^2_{(1,0)}(I^1_{(1,1)}(0))] \\ &= -[-I^2_{(1,0)}(0) + I^2_{(1,0)}(1)] \\ &= -[-(0,0) + (0,1)]\\ &= (0,0) + - (0,1) \end{align} We don't add these: they are just "oriented" points in $\mathbb{R}^2$.

Similarly, we get \begin{align} \partial(I^2_{(1,1)})(0) &= -(1,0) + (1,1)\\ \partial(I^2_{(2,0)})(0) &= -(0,0) + (1,0)\\ \partial(-I^2_{(2,1)})(0) &= (0,1) + -(1,1) \end{align}

Each of the four corners of the square are appear twice with opposite signs. For instance, $(0,1) = (I^2_{(1,0)})_{(1,1)} = (I^2_{(2,1)})_{(1,0)}$ (this is the statement $(I^n_{(i,\alpha)})_{(j,\beta)} = (I^n_{(j+1,\beta)})_{(i,\alpha)}$ in Spivak's proof). But it's orientation is negative in the former and positive in the latter. So they cancel each other out - they are not "added".

To summarize, we don't add two different n-cubes in an n-chain. We add the coefficients of the same n-cube if it appears more than once.