Proposition

If $\omega \ne 0$, show that there is a chain $c$ such that $\int_{c} \omega \ne 0$. Use this fact, Stokes’ theorem and $\partial^2 = 0$ to prove $d^2 = 0$.

Solution

Let $\omega$ be a $k$-form on an open set $A \subset \mathbb{R}^n$. Then there exist $\omega_{i_1, \cdots, i_k}: A \rightarrow \mathbb{R}$ such that $\omega(p) = \sum_{i_1 < \cdots < i_k} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k}$. Suppose that $\omega \ne 0$. Then there exist $a = (a^1, \cdots, a^n) \in A$ and $i_1 < \cdots < i_k$ such that $\omega_{i_1, \cdots, i_k}(a) \ne 0$. For simplicity, we will assume $i_1 = 1, i_2 = 2, \cdots, i_k = k$. Moreover, we will assume $\omega_{1, \cdots, k}(a) > 0$. Since $\omega_{1, \cdots, k}$ is smooth, there exists an $\epsilon > 0$ such that $\omega_{1, \cdots, k}(x) > 0$ for all $x \in B(a, \epsilon)$.

Let $h = \epsilon / 2$. Define a $k$-chain $c:I^k \rightarrow A$ such that

\[\begin{align*} c(t^1, \cdots, t^k) = (f(a^1, t^1), \cdots, f(a^k, t^k), a^{k + 1}, \cdots, a^n) \end{align*}\]

where $f(a^i, t^i) = (a^i - h) + 2ht^i$ represents the straight line between $a^i - h$ and $a^i + h$.

  • Claim 1: $\int_{c} \omega_{1, \cdots, k} dx^1 \wedge \cdots \wedge dx^k \ne 0$.
\[\begin{align*} \int_{c} \omega_{1, \cdots, k} dx^1 \wedge \cdots \wedge dx^k &= \int_{[0, 1]^k} c^*(\omega_{1, \cdots, k}) dx^1 \wedge \cdots \wedge dx^k \\ &= \int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) (c^*dx^1) \wedge \cdots \wedge (c^*dx^k) \\ &= \int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) (\sum \frac{\partial c^1}{\partial x^j}dx^j) \wedge \cdots \wedge (\sum \frac{\partial c^k}{\partial x^j}dx^j) \\ &= \int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) (\frac{\partial c^1}{\partial x^1}dx^1) \wedge \cdots \wedge (\frac{\partial c^k}{\partial x^k}dx^k) \\ &= \int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) 2hdx^1 \wedge \cdots \wedge 2hdx^k \\ &= (2h)^k\int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) dx^1 \wedge \cdots \wedge dx^k \\ &= (2h)^k\int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c)(x^1, \cdots, x^k) dx^1 \cdots dx^k. \end{align*}\]

Since $(\omega^{1, \cdots, k} \circ c)(x) > 0$ for any $x \in I^k$, so $\int_{c} \omega_{1, \cdots, k} dx^1 \wedge \cdots \wedge dx^k > 0$.

  • Claim 2: $\int_{c} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k} = 0$ if $i_1 < \cdots < i_k$ and $(i_1, \cdots, i_k) \ne (1, \cdots, k)$.

Let $i_1 < \cdots < i_k$ such that $(i_1, \cdots, i_k) \ne (1, \cdots, k)$. Then $i_k > k$. Therefore, $\frac{\partial c^{i_k}}{\partial x^j}$ for any $j$.

\[\begin{align*} \int_{c} \omega_{1, \cdots, k} dx^{i_1} \wedge \cdots \wedge dx^{i_k} &= \int_{[0, 1]^k} c^*(\omega_{1, \cdots, k}) dx^{i_1} \wedge \cdots \wedge dx^{i_k} \\ &= \int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) (c^*dx^{i_1}) \wedge \cdots \wedge (c^*dx^{i_k}) \\ &= \int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) (\sum \frac{\partial c^{i_1}}{\partial x^j}dx^j) \wedge \cdots \wedge (\sum \frac{\partial c^{i_k}}{\partial x^j}dx^j) \\ &= \int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c) (\sum \frac{\partial c^{i_1}}{\partial x^j}dx^j) \wedge \cdots \wedge 0 \\ &= 0. \end{align*}\]

By combining Claim 1 and 2, we obtain

\[\begin{align*} \int_{c} \omega &= \int_{c} \sum_{i_1 < \cdots < i_k} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k} \\ &= \sum_{i_1 < \cdots < i_k} \int_{c} \omega_{i_1, \cdots, i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k} \\ &= (2h)^k\int_{[0, 1]^k} (\omega_{1, \cdots, k} \circ c)(x^1, \cdots, x^k) dx^1 \cdots dx^k \\ &\ne 0. \end{align*}\]

Let $\omega$ be a $k$-form on an open set $A \subset \mathbb{R}^n$. Suppose $d^2\omega \ne 0$. By the argument above, we obtain a $k + 2$ chain $c$ such that $\int_{c} d^2\omega \ne 0$. By applying Stokes’ Theorem [Theorem 4-13, Spivak] twice, we obtain $\int_{\partial^2 c} \omega \ne 0$. However, $\partial^2 c = 0$ by Theorem 4-12 [Spivak], so this is a contradiction. Therefore, $d^2\omega = 0$.