# How to show the identity $d(*df)=(\Delta f)\nu$ where $\nu=dx_1\wedge\ldots \wedge dx_n$?

My textbook defines the Hodge star operator as follows:

Given $\omega\in \Omega^k(\mathbb R^n)$ we define $*\omega\in \Omega^{n-1}(\mathbb R^n)$ setting $$*(dx_{i_1}\wedge\ldots\wedge dx_{i_k})=(-1)^\sigma(dx_{j_1}\wedge \ldots\wedge dx_{j_{n-k}}),$$ and extending it linearly, where $i_1<\ldots<i_k, j_1<\ldots<j_{n-1}$, $(i_1, \ldots, i_k, j_1, \ldots, j_{n-k})$ is a permutation of $(1, \ldots, n)$, and $\sigma=0$ or $1$ according to the permutation is even or odd, respectively.

How would I use this to show $d(*df)=(\Delta f)\nu$ where $f:\mathbb R^n\longrightarrow \mathbb R$ is a differentiable function, $\Delta f=\sum_{j=1}^n \frac{\partial^2 f}{\partial x_j^2}$ and $\nu=dx_1\wedge\ldots\wedge dx_n$?

My Atempt: I know $df\in \Omega^1(\mathbb R^n)$ hence $*df\in \Omega^{n-1}(\mathbb R^n)$ and $$*df=\sum_{i_1=1}^n \frac{\partial f}{\partial x_{i_1}}*dx_{i_1}.$$ I'm having trouble for writing $*dx_i$. I guess it would be something like $$*dx_{i_1}=(-1)^\sigma dx_{j_1}\wedge\ldots \wedge dx_{j_{n-1}}$$ where $(i, j_1, \ldots, j_{n-1})$ is a permutation of the first $n$-integers with $j_1<\ldots<j_{n-1}$. Hence $$d(*df)=(-1)^\sigma\sum_{i_1=1}^nd\left(\frac{\partial f}{\partial x_{i_1}}\right)\wedge dx_{j_1}\wedge \ldots\wedge dx_{j_{n-1}}\\ =(-1)^\sigma\sum_{i_1=1}^n \sum_{j_n=1}^n \frac{\partial^2 f}{\partial x_{i_1}\partial x_{j_n}}dx_{j_n}\wedge dx_{j_1}\wedge \ldots\wedge dx_{j_{n-1}}.$$ Now I don't know how to go on.

Note that

$$*dx^i = (-1)^{i-1} dx^1 \wedge \cdots \wedge dx^{i-1} \wedge dx^{i+1} \wedge \cdots \wedge dx^n$$

As $df = \sum_i f_i dx^i$,

$$d(*df) = d\bigg(\sum_{i=1}^n (-1)^{i-1} f_i dx^1 \wedge\cdots \wedge dx^{i-1} \wedge dx^{i+1} \wedge \cdots \wedge dx^n\bigg)$$

$$= \sum_{i=1}^n (-1)^{i-1} df_i\wedge dx^1 \wedge\cdots \wedge dx^{i-1} \wedge dx^{i+1} \wedge \cdots \wedge dx^n$$

$$= \sum_{i=1}^n (-1)^{i-1} f_{ii} dx^i \wedge dx^1 \wedge\cdots \wedge dx^{i-1} \wedge dx^{i+1} \wedge \cdots \wedge dx^n$$

$$= \sum_{i=1}^n f_{ii} dx^1 \wedge\cdots \wedge dx^{i-1} \wedge dx^i \wedge dx^{i+1} \wedge \cdots \wedge dx^n = \Delta f \nu$$

• could you explain a bit more how you got to $*dx_i=(-1)^{i-1} dx_1\wedge \ldots\wedge dx_{i-1}\wedge dx_{i+1}\wedge \ldots\wedge dx_n$? – PtF Apr 28 '14 at 12:00
• @PtF: For example, can you see why $*dx^1 = dx^2\wedge \cdots \wedge dx^n$? – user99914 Apr 28 '14 at 12:07
• Well, I guess it is because $(i_1, j_1, \ldots, j_n)$ is a permutation of $\{1, \ldots, n\}$, so in this case $i_1=1$ and since a permutation is 1-1 none of $j_k$ can be $1$, right? – PtF Apr 28 '14 at 12:17