Proposition

Let $c$ be a singular $k$-cube and $p:[0, 1]^k \rightarrow [0, 1]^k$ a 1-1 function such that $p([0, 1]^k) = [0, 1]^k$ and $\det(p’(x)) \geq 0$ for $x \in [0, 1]^k$. If $\omega$ is a $k$-form, show that

\[\begin{align*} \int_{c} \omega = \int_{c \circ p} \omega. \end{align*}\]

Solution

Let $\omega_{i_1, \cdots, i_k}$ be chosen such that $c^*\omega = \sum_{i_1 < \cdots < i_k} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k}$.

\[\begin{align*} \int_{c \circ p} \omega &= \int_{[0, 1]^k} (c \circ p)^*\omega \\ &= \int_{[0, 1]^k} p^*(c^*\omega) \\ &= \int_{[0, 1]^k} p^*(\sum_{i_1 < \cdots < i_k} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k}) \\ &= \int_{[0, 1]^k} p^*(\sum_{i_1 < \cdots < i_k} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k}) \\ &= \sum_{i_1 < \cdots < i_k} \int_{[0, 1]^k} p^*(\omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k}) & \text{(Theorem 4.8(2)[Spivak])} \\ &= \sum_{i_1 < \cdots < i_k} \int_{[0, 1]^k} (\omega_{i_1, \cdots, i_k} \circ p) (\det p') dx^{i_1} \wedge \cdots \wedge dx^{i_k} & \text{(Theorem 4.9[Spivak])} \\ &= \sum_{i_1 < \cdots < i_k} \int_{[0, 1]^k} (\omega_{i_1, \cdots, i_k} \circ p) \abs{\det p'} dx^{i_1} \wedge \cdots \wedge dx^{i_k} & (\det p' > 0) \\ &= \sum_{i_1 < \cdots < i_k} \int_{[0, 1]^k} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k} & \text{(Theorem 3-13)} \\ &= \int_{[0, 1]^k} c^*\omega \\ &= \int_{c} \omega. \end{align*}\]