Let $U \subset \mathbb{R}^n$ be open and consider the codifferential $$d^* :\Omega^p(U) \to \Omega^{n-p}(U)$$ given by $$d^*(\omega)=(-1)^{np+n-1} *\circ d \circ *(\omega).$$ Show that $$d^*(fdx_1 \wedge \dots \wedge dx_p) = \sum_{j=1}^p(-1)^j \frac{\partial f}{\partial x_j}dx_1 \wedge \dots \wedge\widehat{dx_j} \wedge \dots\wedge dx_p.$$
Trying to compute this I got the following: $$\begin{align*} d^*(fdx_1 \wedge \dots \wedge dx_p)&= (-1)^{np+n-1} *\circ d \circ *(fdx_1 \wedge \dots \wedge dx_p) \\ &= (-1)^{np+n-1} *\circ d(f dx_{p+1} \wedge \dots \wedge dx_{n}) \\
&= (-1)^{np+n-1} *\left( \left( \sum_{j=1}^n \frac{\partial f}{\partial x_j} dx_j \right) \wedge dx_{p+1} \wedge \dots \wedge dx_n\right) \end{align*}$$
here is where I'm stuck. Is it so that all the terms with $j=p+1, … n$ will vanish and I'm left with $$\left( \sum_{j=1}^n \frac{\partial f}{\partial x_j} dx_j \right) \wedge dx_{p+1} \wedge \dots \wedge dx_n = \sum_{j=1}^p \frac{\partial f}{\partial x_j} dx_j$$ and then $$*\left(\sum_{j=1}^p \frac{\partial f}{\partial x_j} dx_j\right) = \sum_{j=1}^p \frac{\partial f}{\partial x_j} *(dx_j) = \sum_{j=1}^p \frac{\partial f}{\partial x_j} dx_1 \wedge \dots \wedge\widehat{dx_j} \wedge \dots\wedge dx_p?$$ This would result in $$d^*(fdx_1 \wedge \dots \wedge dx_p) = (-1)^{np+n-1}\sum_{j=1}^p \frac{\partial f}{\partial x_j}dx_1 \wedge \dots \wedge\widehat{dx_j} \wedge \dots\wedge dx_p$$ which is not yet what we want. Could someone help me with where I'm going wrong here?
Best Answer
Your error occurs here:
$$\left( \sum_{j=1}^n \frac{\partial f}{\partial x_j} dx_j \right) \wedge dx_{p+1} \wedge \dots \wedge dx_n = \sum_{j=1}^p \frac{\partial f}{\partial x_j} dx_j.$$
Note that the left hand side is a $(n - p + 1)$-form, and the right hand side is a one-form. Instead, you have
$$\left( \sum_{j=1}^n \frac{\partial f}{\partial x_j} dx_j \right) \wedge dx_{p+1} \wedge \dots \wedge dx_n = \sum_{j=1}^p \frac{\partial f}{\partial x_j} dx_j\wedge dx_{p+1} \wedge \dots \wedge dx_n.$$
From the discussion below, it seems that the OP knows how to complete the problem. With this in mind, here is the rest of the computation.
We have
\begin{align*} d^*(fdx_1\wedge\dots\wedge dx_p) &= (-1)^{np+n-1}\ast\left(\sum_{j=1}^n\dfrac{\partial f}{\partial x_j}dx^j\wedge dx_{p+1}\wedge\dots\wedge dx_n\right)\\ &= (-1)^{np+n-1}\sum_{j=1}^n\frac{\partial f}{\partial x_j}\ast(dx_j\wedge dx_{p+1}\wedge\dots\wedge dx_n). \end{align*}
Since $\alpha\wedge\ast\beta = \langle\alpha,\beta\rangle dx_1\wedge\dots\wedge dx_n$, and $\{dx_{i_1}\wedge\dots\wedge dx_{i_p} \mid 1\leq i_1 < \dots < i_p \leq n\}$ is an orthonormal basis for $p$-forms, it follows that $\ast(dx_{i_1}\wedge\dots\wedge dx_{i_p}) = \operatorname{sign}(\sigma)dx_{j_1}\wedge\dots\wedge dx_{j_{n-p}}$ where $\{j_1, \dots, j_{n-p}\} = \{1, \dots, n\}\setminus\{i_1, \dots, i_p\}$ with $j_1 < \dots < j_{n-p}$ and $\sigma \in S_n$ is the permutation with
$$\sigma(k) = \begin{cases} i_k & 1 \leq k \leq p\\ j_k & p + 1 \leq k \leq n. \end{cases}$$
In particular, $\ast(dx_j\wedge dx_{p+1}\wedge\dots\wedge dx_n) = \operatorname{sign}(\sigma)dx_1\wedge\dots\wedge\widehat{dx^j}\wedge\dots\wedge dx_p$. Instead of computing $\operatorname{sign}(\sigma)$ directly, we can calculate it implicitly as follows. Since
\begin{align*} dx_1\wedge\dots\wedge dx_n &= dx_j\wedge dx_{p+1}\wedge\dots\wedge dx_n\wedge\ast(dx_j\wedge dx_{p+1}\wedge\dots\wedge dx_n)\\ &= dx_j\wedge dx_{p+1}\wedge\dots\wedge dx_n\wedge\operatorname{sign}(\sigma)dx_1\wedge\dots\wedge\widehat{dx^j}\wedge\dots\wedge dx_p\\ &= (-1)^{n-p+j-1}\operatorname{sign}(\sigma)dx_{p+1}\wedge\dots\wedge dx_n\wedge dx_1\wedge\dots\wedge dx_p\\ &= (-1)^{n-p+j-1}(-1)^{(n-p)p}\operatorname{sign}(\sigma)dx^1\wedge\dots\wedge dx^p\wedge dx^{p+1}\wedge\dots\wedge dx_n\\ &= (-1)^{n-p+j-1+(n-p)p}\operatorname{sign}(\sigma)dx_1\wedge\dots\wedge dx_n, \end{align*}
we have $\operatorname{sign}(\sigma) = (-1)^{n-p+j-1 + (n-p)p} = (-1)^{n - p + j - 1 + np - p^2} = (-1)^{np + n - 1 + j}$. Therefore
$$\ast(dx_j\wedge dx_{p+1}\wedge\dots\wedge dx_n) = (-1)^{np+n-1+j}dx_1\wedge\dots\wedge\widehat{dx_j}\wedge\dots\wedge dx_p.$$
So we finally obtain
\begin{align*} d^*(fdx_1\wedge\dots\wedge dx_n) &= (-1)^{np+n-1}\sum_{j=1}^n\frac{\partial f}{\partial x_j}\ast(dx_j\wedge dx_{p+1}\wedge\dots\wedge dx_n)\\ &= (-1)^{np+n-1}\sum_{j=1}^n\frac{\partial f}{\partial x_j}(-1)^{np+n-1+j}dx_1\wedge\dots\wedge\widehat{dx_j}\wedge\dots\wedge dx_p\\ &= \sum_{j=1}^n(-1)^j\frac{\partial f}{\partial x_j}dx_1\wedge\dots\wedge\widehat{dx^j}\wedge\dots\wedge dx_n. \end{align*}