# Spivak Calculus on Manifolds, Definition of Boundary

In Chapter 4's section on Geometric Preliminaries, I am confused by the definition of "boundary."

The standard $$n$$-cube $$I^n$$ is defined to be $$I^n(x^1,...,x^n) = (x^1,...,x^n)$$, and two associated $$n-1$$-cubes are defined as $$I^n_{(i,0)}(x^1,...,x^{n-1}) = I^n(x^1,...,x^{i-1},0,x^i,...,x^{n-1})$$ $$I^n_{(i,1)}(x^1,...,x^{n-1}) = I^n(x^1,...,x^{i-1},1,x^i,...,x^{n-1})$$

With these definitions, the boundary of the standard $$n$$-cube is defined as: $$\partial I^n = \sum_{i=1}^n \sum_{a=0,1} (-1)^{i+a} I^n_{(i,a)}$$

I'm having trouble interpreting these definitions in the case of $$n=2$$. The image of $$I^2$$ in $$R^2$$ is just the unit square with endpoints at $$(0,0)$$ and $$(1,1)$$. The image of $$\partial I^2$$ is thus the outline of the square (as depicted in the book). However, when I work out the definition of boundary given above, I get:

\begin{align*} \partial I^2(x) &= \sum_{i=1}^2 \sum_{a=0,1} (-1)^{i+a} I^2_{(i,a)}(x)\\ &= (-1)^1 I^2_{(1,0)}(x) + (-1)^2 I^2_{(1,1)}(x) + (-1)^2 I^2_{(2,0)}(x) + (-1)^3 I^2_{(2,1)}(x)\\ &= -(0,x) + (1,x) + (x,0) - (x,1)\\ &= (1,-1) \end{align*}

Spivak's provided equation suggests that the image of $$\partial I^2$$ is a single point, which doesn't make any sense.

Does this equation contain a typo, and if not, where am I going wrong in my interpretation of it?

• For that matter, what does the notation $I^n(x^1,...,x^n) = (x^1,...,x^n)$ mean? Surely the RHS doesn't simply mean a single point. Does this mean the collection of all such points where the coordinates vary with $0\leq x^i \leq 1$ for each $i$? And what does the $\sum$ notation mean? It can't mean coordinate wise addition as you suggest.
– MPW
Mar 6 '20 at 21:26
• You don't do the algebraic sum of the functions. You take the singular chain, whose image is the union of the respective images. Mar 6 '20 at 23:09

It's indeed not the best choice of notation.

Geometrically it's clear what should happen (modulo the alternating sum): The boundary of an $$n$$ dimensional box is the 'collection' of its faces, each of which is an ($$n$$-$$1$$)-cube.

Now, instead of taking their collection we take their (alternating) formal sum (which can be rigorously achieved by introducing a free Abelian group generated by the set of all $$n$$-cubes).

The notation $$I^n(x^1,\dots,x^n)=(x^1,\dots, x^n)$$ defines the embedding of the standard $$n$$-cube $$[0,1]^n$$ into $$\Bbb R^n$$.

Then $$I^n_{i,0}$$ is the embedding $$[0,1]^{n-1}\to\Bbb R^n$$ that chooses the face on the hyperplane $$x_i=0$$.

Note also that the alternating sum is in accordance with the orientation of the (embeddings of the) faces.

Specifically for $$n=2$$, if we denote the (signed) segments as $$\def\segment#1#2#3#4{\big[(#1,#2)\multimap(#3,#4)\big]} \segment{x_1}{y_1}{x_2}{y_2}$$ we get: $$\partial I^2\ =\ -I^2_{(1,0)}+I^2_{(1,1)}+I^2_{(2,0)}-I^2_{(2,1)}\ =\\ -\segment0001 + \segment1011 + \segment0010 -\segment0111 \\ =\ \segment0010 + \segment1011 + \segment1101 + \segment0100\,.$$