First part Spivak's proof of Stokes' theorem

Question

This is a question regarding the first part of Spivak's proof of Stokes' theorem. Let $\omega$ be a $(k-1)$-form on $[0,1]^k$. Then $\omega$ is the sum of $(k-1)$-forms of the type $$ fdx^1\wedge\dots\wedge\hat{dx^i}\wedge\dots\wedge dx^k, $$ and it suffices to prove the theorem for each of these. Now Spivak claims the following (for the case $j=i$): $$ \int_{[0,1]^{k-1}}I^{k*}_{(j,\alpha)}(fdx^1\wedge\dots\wedge\hat{dx^i}\wedge\dots\wedge dx^k) =\int_{[0,1]^k}f(x^1,\dots,\alpha,\dots,x^k)dx^1\dots dx^k. $$ Now, I know that a couple of posts have already been made on this part of the proof, but those were not helpful to me, so I will proceed to clarify my own confusion.

Now, I understand that we have $f(x^1,\dots,\alpha,\dots,x^k)$, because $$ f^*\omega(p)(v_1,\dots,v_k)=\omega(f(p))(f_*(v_1),\dots,f_*(v_k)), $$ and in our case $f=I^k_{(j,\alpha)}$. So the part $\omega(f(p))$ is clear to me. I just don't understand why we get $dx^1\dots dx^k$.

And how do we go from $[0,1]^{k-1}$ to $[0,1]^k$? Initially, I wanted to equate the two integrands, but I don't believe that's possible because we don't integrate over the same set, so I'm kind of stuck right now.

Answer 1

You have already started out with the right formula ($f^*(w) = \dots$). Now ask yourself the following questions (and try to answer them):

What is $I^k_{(j,\alpha)*}$?

What is $\int_0^1 f(x^1, \dots, \alpha, \ldots, x^k) \, dx^i$ (with $\alpha$ at position $i$)? -- Spoiler: note that the interval over which you take the integral has length $=1$ and the integrand is constant wrt to the integration variable -- and

What do you get if you integrate the result of the integral in the previous question over $[0,1]^{k-1}$ viewed as $[0,1]^k$ with the $i$-th factor missing?

Answer 2

I know I'm super late to the party but I got stuck same as you and I just want a record of my solution in case I forget. First, to simplify the notation a little bit, define $I := I_{(j,\alpha)}^k$. Then you're trying to compute $$\int_{[0,1]^{k-1}} I^* (f \; dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^k).$$ Now let $p = (x_1,x_2, \ldots, x_{k-1})$.This implies $$I(p) = (x_1,\ldots, x_{j-1},\alpha,x_j,\ldots,x_{k-1})$$ Then we know $$I^* (f \; dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^k)=(f\circ I) \; I^*(dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^k)$$ Now $$f\circ I(p) = f(x_1,\ldots, x_{j-1},\alpha,x_j,\ldots,x_{k-1})$$ And $$I^*(dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^k)(p)(v_1,\ldots, v_{i-1},v_{i+1},\ldots,v_{k})\\ = (dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^k)(I(p))(I_*(v_1),\ldots, I_*(v_{i-1}),I_*(v_{i+1}),\ldots,I_*(v_{k}))$$ To make the notation cleaner, we'll drop the $I(p)$ keeping in mind that our $dx^i$ are duals to $T_{I(p)}\mathbb{R}^n$

$$I^*(dx^i)(v_{i}) = dx^i \left( \sum_{j=1}^{k-1}v_i^j D_j I^1(p),\ldots,\sum_{j=1}^{k-1}v_i^j D_j I^k(p)\right)$$ So that means $$I^*(dx^i)(p)(v_{i}) = \sum_{j=1}^{k-1}v_i^j D_j I^i(p) = \sum_{j=1}^{k-1} D_j I^i \;dx_j (p)(v_i) $$ And we have $$I^*(dx^i) =\sum_{j=1}^{k-1} \frac{\partial I^i}{\partial x_j} \;dx_j $$ Now if $i = j$, then $I^i = \alpha$ so $I^*(dx_i) = 0$ and if $i\neq j$ then either $i < j$ and so $I^*(dx^i) = dx^i$ or $i > j$ in which case $I^*(dx^i) = dx^{i-1}$.

Therefore if $i=j$ $$I^*(dx^1\wedge \cdots \wedge dx^{i-1}\wedge \widehat{dx^i} \wedge dx^{i+1} \wedge \cdots \wedge dx^{k})\\ = dx^1\wedge \cdots \wedge dx^{i-1}\wedge \widehat{\,0\,} \wedge dx^{i+1-1} \wedge \cdots \wedge dx^{k-1}\\ =dx^1\wedge \cdots \wedge dx^{k-1}$$ If however $i\neq j$, then the $0$ occurs at a non-disappearing term, which makes the entire expression disappear, so altogether we have: $$I^*(dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^k) = \begin{cases}0 & i \neq j\\ dx^1\wedge \cdots \wedge dx^{k-1} & i= j\end{cases}$$ and so our original expression becomes: $$\int_{[0,1]^{k-1}} I^* (f \; dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^k) \\ = \begin{cases}0 & i \neq j \\ \int_{[0,1]^{k-1}} f(x_1,\ldots, x_{j-1},\alpha,x_j,\ldots,x_{k-1}) dx^1\cdots dx^{k-1} & i=j \end{cases}$$ Last thing to note is just that $$\int_{[0,1]^{k-1}} f(x_1,\ldots, x_{j-1},\alpha,x_j,\ldots,x_{k-1}) dx^1\cdots dx^{k-1}\\ = \int_{[0,1]^{k-1}} f(x_1,\ldots, x_{j-1},\alpha,x_{j+1},\ldots,x_{k}) dx^1\cdots dx^{k}$$ with a change of variable for the last $k - j$ variables from $x^l \mapsto x^{l+1}$ and integrating along another singular 1-cube.

Answer 3

A simple example might clarify the previous post: suppose $n=2$ and $\alpha =1$. As $\int^1_0dy=1$,
$$\int^1_0f(x,1)dx= \int^1_0f(x,1)\int^1_0dy\,dx=\int^1_0\int^1_0 f(x,1)dy\,dx$$