# Boundary of a cube

I was reading the chapter "Integration on Chains" from Spivak, and I was trying to understand the intuition behind the definition of the boundary of a cube: Initially we define $$I^n\colon [0,1]^n\to\mathbb{R}^n$$ by $$I^n(x)=x$$ for $$x\in[0,1]^n$$, and then, for each $$i$$ with $$1\le i\le n$$, we define

$$I^n_{(i,0)}(x_1, \ldots, x_{n-1}) =I^n(x_1,\dots,x_{i-1},0,x_i,\dots,x_{n-1})=(x_1,\dots,x_{i-1},0,x_i,\dots,x_{n-1})$$,

and

$$I^n_{(i,1)} (x_1, \ldots, x_{n-1}) =I^n(x_1,\dots,x_{i-1},1,x_i,\dots,x_{n-1})=(x_1,\dots,x_{i-1},1,x_i,\dots,x_{n-1})$$.

Finally, the boundary of $$I^n$$ is defined by:

$$\partial I^n=\sum_{i=1}^n\sum_{\alpha=0}^1(-1)^{i+\alpha}I_{(i,\alpha)}^n$$.

But I just find it weird. Is there a clear motive behind this definition, or is it just made to work well?

• Ummm... what is that definition? Do you really expect all of us to remember it?!? Jul 30 '20 at 15:08
• In line 3 do you mean $x\in [0,1]^n$? Otherwise it's nonsense. Jul 30 '20 at 16:00
• What is it exactly that you find confusing? This is just a precise way of naming each of the hypercubes that bound the unit $n$-cube, and the sum is the oriented sum (try drawing a picture for $n=2$ or $n=3$). Jul 30 '20 at 16:12
• I don't understand why we choose this orientation. What is the oriented sum that you mentioned? Jul 30 '20 at 16:19
• This is one spot where the whole "tiny book" model so this series fails badly --- this one chapter is all formulas and no motivation to speak of. Jul 30 '20 at 16:32

Despite @DavidGStork's objection, I actually do remember that definition, even though I really worked through the details about 50 years ago. :(

So part of the trick is to look at what happens in the case $$n = 1$$. In this case, $$I^1 : [0, 1] \to \Bbb R : x \mapsto x$$ is a "cube" whose image is a line segment from $$0$$ to $$1$$. We're hoping, soon, to do things like say "let's add to that a line segment from $$1$$ to $$2$$" and have the boundary of the composite thing be the "sum" of the boundaries of the parts. If we said that the boundary of $$I^1$$ consisted of "a 0-cube at $$0$$" and "a 0-cube at $$1$$", then we'd end up, when we computed boundary of the sum, with a 0-cube at 0, two of them at 1, and one of them at 2. We REALLY (to make things "add" nicely, so we can use linear algebra techniques, etc.) want "cancellation", so instead we say that the boundary of $$I^1$$ is a $$0$$-cube at 1 minus a $$0$$-cube at $$0$$.

Let's look at what the definition says the boundary of $$I^1$$ actually is. Well, we have a sum for $$i = 1$$ up to $$1$$, and for $$\alpha = 0, 1$$, so we're looking at $$\partial I^1 = (-1)^{1+0} I^1_{(1, 0)} + (-1)^{1+1} I^1_{(1,1)}.$$ That's $$\partial I^1 = - I^1_{(1, 0)} + I^1_{(1,1)}.$$ Now $$I^1_{(1,0)}$$ is just $$I^1(0) = 0$$; $$I^1_{(1,1)}$$ is $$I^1(1)$$. So the result is a chain that sends a 0-cube to $$1$$, minus a chain that sends the $$0$$-cube to $$0$$, just as desired.

NB: You definition of the subscripted $$I^n$$s uses $$I_n$$ when it should use $$I^n$$; your third line should have $$x \in [0, 1]^n$$.

What about $$I^2$$? The image of $$I^2$$ is a filled-in square, and the boundary should consist (pictorially) of four line-segments, which, in terms of 1-cubes, means "four 1-cubes", i.e., four functions from $$[0, 1]$$ to the plane. And we'd like, if we placed two unit squares next to each other, to have the "adjacent" line-segments "cancel", so the coefficient for the right-hand edge should be the negative of the coefficient for the left-hand edge. Well...that's what the "$$\alpha$$" in the $$(-1)^{i+\alpha}$$ factor is providing -- an sign switch for "opposite" sides of each segment/square/cube/...

What about the $$i$$ in there? What's up with that? Answer: if you draw just about any nice shape (a disk, a line-segment, a square, a cube) and look at its boundary, and then take the boundary of that, you get the empty set. We'd like, by analogy, to have the boundary of the boundary of an $$n$$-cube (or $$n$$-chain) always be the empty chain. [If we do this for an $$n$$-cube, we get it automatically for any $$n$$-chain -- that's part of the "linearity" thing I was talking about.]

So try it for the unit square. We get the right-hand edge pointing up, the left edge pointing down, the top edge pointing left, the bottom edge pointing right. If the square is

D------C
|//////|
|//////|
A------B


then I'm saying that the boundary of that filled-in square consists of four 1-cubes, with coefficients:

1. $$AB$$, coefficient $$+1$$
2. $$BC$$, coefficient $$+1$$
3. $$DC$$, coefficient $$-1$$
4. $$AD$$, coefficient $$-1$$.

The boundaries of these individually are

1. $$(+1) (B - A) = B - A$$
2. $$(+1) (C - B) = C - B$$
3. $$(-1) (C - D) = D - C$$
4. $$(-1) (D - A) = A - D$$

And if you sum these up, you get a zero-chain all of whose coefficients are zeroes. So that means we got the sign-choices right!

So...that's the motivation: (1) $$\partial^2 = 0$$, and (2) adjacent $$n$$-cubes have the shared part of their boundaries "cancel".

It is difficult to provide an intuition in $$\mathbb R^n$$, but you can certainly convince yourself that the definition makes geometric sense in the cases $$n=2,3$$. Spivak covers the $$n=2$$ case himself, so I'll cover the $$n=3$$ in this answer.

Ok, you have the "standard cube" in $$\mathbb R^3$$, but by this we don't mean the set $$[0,1]^3$$ but the function which sends each $$x \in [0,1]^3$$ to itself (we put things this way because later in the chapter you will consider other sets "parametrized" by the unit cube in this way, but which don't look like a cube, but rather like the smooth$$^1$$ image of a cube). Ok, what's the boundary of that cube? It consists of six faces, each parametrized by the unit square $$[0,1]^2\subset \mathbb R^2$$. We want to say that the boundary is the "sum" of these faces (which are really parametrizations, i.e. functions from $$[0,1]^2$$ to $$\mathbb R^3$$), each corrected by an appropiate sign accounting for the orientation (see below).

The parametrization of the left face (which lies in the $$yz$$ plane, that is, the $$x=0$$ plane) is given by [unlike Spivak, I'll use $$u,v$$ instead of $$x_1, x_2$$ in order for the disinction between the $$\mathbb R^2$$ variables and the $$\mathbb R^3$$ variables to remain clear]:

$$I^3 _{(1,0)} : (u, v) \mapsto (0, u, v)$$.

The parametrization of the right face (which lies in the $$x=1$$ plane) is given by

$$I^3 _{(1,1)} : (u, v) \mapsto (1, u, v)$$.

But we want the "outward pointing normal" (which for $$n=3$$ is given by the classical vector calculus formula$$^2$$ $$\overrightarrow N (u,v)=\frac{\partial \overrightarrow f(u,v)} {\partial u} \times \frac{\partial \overrightarrow f(u,v)} {\partial v}$$) to point left (i.e. for its $$x$$-coordinate to be negative) for the left face and to point right (i.e. for its $$x$$-coordinate to be positive) for the right face. This is achieved exactly by inserting the signs $$(-1)=(-1)^{1+0}$$ and $$1=(-1)^{1+1}$$ respectively.

Likewise, you should write down (I really recommend you do this in order to convince yourself, at least in the $$n=3$$ case) the parametrizations for the bottom and top faces (which lie respectively in the $$xz$$-plane, that is, the $$y=0$$ plane, and in the $$y=1$$ plane), compute the $$z$$-signs of their outward-pointing normals and check that the corrections by $$(-1)^{2+0}$$ and $$(-1)^{2+1}$$ respectively guarantee that the outward pointing normals will point down and up respectively (that is, they'll have negative and positive $$y$$-coordinates respectively). Finally, proceed analogously for the front and back faces.

Of course, there's a deeper geometric argument (valid for any dimension) lurking in the background, but you'll have time to learn about that when studying some differential geometry (e.g. from Lee's Smooth Manifolds book or even Spivak's own Comprehensive Introduction to Differential Geometry Vol. 1).

$$^1$$ Spivak asks for continuity rather than smoothness, but this is a mistake. For a list of errata check:

https://www.jirka.org/spivak-errata.html

http://www.petraaxolotl.com/mathematics/calculus-on-manifolds/

as well as the Addenda at the end of the book, and this amazon review.

$$^2$$ Actually, we then divide by the norm in order for the vector to have length $$1$$. But this doesn't change the sign of the coordinates which is all we care about, so we can safely omit this. Of course, $$f$$ is whatever parametrization we're dealing with, whose normal we want to calculate at a given point $$f(u,v)$$.