Hodge Star Operator on a Riemannian Manifold

Chris L T521

Well-known member
Staff member
Hi all,

This is a question that wasn't overly difficult, but there is one part that I'm still having issues justifying.

On any Riemannian manifold we have the duality or 'index lowering' map $\varphi:F\rightarrow F^{\flat}$ from vector fields to one-forms. If $M$ is oriented, then we also have the Hodge star map $*:\bigwedge^k T^*M\rightarrow \bigwedge^{n-k}T^*M$. In the case $M=\mathbb{R}^3$ with the standard Euclidean metric, show that $\psi:F\rightarrow *(F^{\flat})$ sends the vector field $(F_1,F_2,F_3)$ to the two-form $F_1\,dx^2\wedge dx^3 + F_2\,dx^3\wedge dx^1+F_3\,dx^1\wedge dx^2$.
Here are my thoughts: Let $F\rightarrow F^{\flat}$, where $(F_1,F_2,F_3)\mapsto F_1\,dx^1+F_2\,dx^2+F_3\,dx^3$. Since $M=\mathbb{R}^3$, the Hodge star operator will map one-forms to two-forms $*:\bigwedge^1 T^*M\rightarrow \bigwedge^2 T^*M$. Since $\dim\bigwedge^1 T^*M = \dim\bigwedge^2 T^*M=3$, each has three basis elements: $dx^1,\,dx^2,\,dx^3\in\bigwedge^1 T^*M$ and $dx^1\wedge dx^2,\,dx^2\wedge dx^3\,dx^3\wedge dx^1\in \bigwedge^2 T^*M$.

However, I want to say that under $*$,

\begin{aligned}dx^1 &\mapsto dx^2\wedge dx^3\\ dx^2 &\mapsto dx^3\wedge dx^1\\ dx^3 &\mapsto dx^1\wedge dx^2\end{aligned}

but I can't justify that in my mind. If I were to believe this is true, then I'd have that $\psi:F\rightarrow *(F^{\flat})$ by having

\begin{aligned}\psi(F_1,F_2,F_3) &= *(F_1\,dx^1 + F_2\,dx^2 + F_3\,dx^3)\\ &= F_1 *(dx^1) + F_2 *(dx^2) + F_3 *(dx^3)\\ &= F_1\,dx^2\wedge dx^3 + F_2\,dx^3\wedge dx^1 + F_3\,dx^1\wedge dx^2\end{aligned}

and then this would complete the problem. I have a feeling that the justification for the mapping of the $dx^i$ under $*$ is hidden in the fact that our metric on $M$ is the standard Euclidean metric $ds^2=(dx^1)^2+(dx^2)^2+(dx^3)^2$, but I'm not sure how to prove it.

Any suggestions or ideas would be much appreciated!

Euge

MHB Global Moderator
Staff member
This is an excellent question. The Hodge star operator is determined up to sign, so that's why you need $M$ to be oriented (your signs would've been negative if you chose a negatively oriented basis).

Let's consider an arbitrary $n$-dimensional, real inner product space $V$. The inner product $(\cdot|\cdot)$ on $V$ satisfies bilinearity, symmetry, and nondegeneracy (i.e., $(v|w) = 0$ for all $w\in V$ implies $v = 0$). Note that positive definiteness is not assumed, so $(\cdot|\cdot)$ may take negative values. Choose an orientation $\gamma$ for $V$ (an equivalence class of orthonormal bases of $V$), and let $\{v_1,\ldots, v_n\}\in \gamma$. Let $\nu = v_1\wedge \cdots \wedge v_n$, the basis vector of $\Lambda^n V$ corresponding to $\{v_1,\ldots, v_n\}$. If $p\in \{0,1,\ldots, n\}$, then given $\alpha \in \Lambda^pV$, there is a linear map $\beta \mapsto \alpha\wedge \beta$ from $\Lambda^{n-p}V$ to $\Lambda^nV$. Hence, there is a unique linear functional $f_\alpha : \Lambda^{n-p}V \to \Bbb R$ for which $\alpha\wedge \beta = f_\alpha(\beta)\nu$. This, in turn, produces a unique element of $\Lambda^{n-p}V$, called $*\alpha$, such that

$$\displaystyle (1) \qquad (*\alpha|\beta) = f_\alpha(\beta).$$

Thus

$$\displaystyle (2) \qquad \alpha\wedge\beta = (*\alpha|\beta)\nu\quad \text{for all}\quad \beta\in \Lambda^{n-p}V.$$

The Hodge star operator is the map $\alpha \mapsto *\alpha$ from $\Lambda^pV$ to $\Lambda^{n-p}V$.

In the case of your example, $V = T^*\Bbb R^3$ and $(dx_i|dx_j) = \delta_{ij}$. Let $\nu = dx_1\wedge dx_2 \wedge dx_3$. Consider $*dx_1$, a two form on $\Bbb R^3$. It can be expressed as

$$\displaystyle (3) \qquad a\, dx_2\wedge dx_3 + b\, dx_3\wedge dx_1 + c\, dx_1 \wedge dx_2,$$

where $a$, $b$, and $c$ are real numbers to be determined. By the equation in $(2)$,

$$\displaystyle (4) \qquad (*dx_1|dx_2\wedge dx_3)\nu = dx_1\wedge dx_2\wedge dx_3 = \nu.$$

Using the expression $(3)$,

$$\displaystyle (5) \qquad (*dx_1|dx_2\wedge dx_3) = a(dx_2\wedge dx_3|dx_2\wedge dx_3) + b(dx_3\wedge dx_1|dx_2\wedge dx_3) + c(dx_1\wedge dx_2|dx_2\wedge dx_3).$$

Now

$$\displaystyle (dx_2\wedge dx_3|dx_2\wedge dx_3) = \text{det}\left(\begin{bmatrix}(dx_2|dx_2) & (dx_2|dx_3)\\(dx_3|dx_2) & (dx_3|dx_3)\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\right) = 1,$$

$$\displaystyle (dx_3\wedge dx_1|dx_2\wedge dx_3) = \text{det}\left(\begin{bmatrix}(dx_3|dx_2) & (dx_3|dx_3)\\(dx_1|dx_2) & (dx_1|dx_3)\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}\right) = 0,$$

$$\displaystyle (dx_1 \wedge dx_2|dx_2\wedge dx_3) = \text{det}\left(\begin{bmatrix}(dx_1|dx_2) & (dx_1|dx_3)\\(dx_2|dx_2) & (dx_2|dx_3)\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}\right) = 0.$$

Combining these results with $(4)$ and $(5)$, we deduce $a = 1$. By a similar analysis, taking inner products of $*dx_1$ with $dx_3\wedge dx_1$ and $dx_1 \wedge dx_2$ yields $b = 0$ and $c = 0$ respectively. Therefore $*dx_1 = dx_2\wedge dx_3$. Similarly, $*dx_2 = dx_3 \wedge dx_1$ and $*dx_3 = dx_1\wedge dx_2$.

If you chose $\nu = dx_2\wedge dx_1\wedge dx_3$ instead, then

$$\displaystyle *dx_1 = -dx_2\wedge dx_3$$

$$\displaystyle *dx_2 = -dx_3\wedge dx_1$$

$$\displaystyle *dx_3 = -dx_1\wedge dx_2$$

Last edited:

Chris L T521

Well-known member
Staff member
This is an excellent question. The Hodge star operator is determined up to sign, so that's why you need $M$ to be oriented (your signs would've been negative if you chose a negatively oriented basis).

Let's consider an arbitrary $n$-dimensional, real inner product space $V$. The inner product $(\cdot|\cdot)$ on $V$ satisfies bilinearity, symmetry, and nondegeneracy (i.e., $(v|w) = 0$ for all $w\in V$ implies $v = 0$). Note that positive definiteness is not assumed, so $(\cdot|\cdot)$ may take negative values. Choose an orientation $\gamma$ for $V$ (an equivalence class of orthonormal bases of $V$), and let $\{v_1,\ldots, v_n\}\in \gamma$. Let $\nu = v_1\wedge \cdots \wedge v_n$, the basis vector of $\Lambda^n V$ corresponding to $\{v_1,\ldots, v_n\}$. If $p\in \{0,1,\ldots, n\}$, then given $\alpha \in \Lambda^pV$, there is a linear map $\beta \mapsto \alpha\wedge \beta$ from $\Lambda^{n-p}V$ to $\Lambda^nV$. Hence, there is a unique linear functional $f_\alpha : \Lambda^{n-p}V \to \Bbb R$ for which $\alpha\wedge \beta = f_\alpha(\beta)\nu$. This, in turn, produces a unique element of $\Lambda^{n-p}V$, called $*\alpha$, such that

$$\displaystyle (1) \qquad (*\alpha|\beta) = f_\alpha(\beta).$$

Thus

$$\displaystyle (2) \qquad \alpha\wedge\beta = (*\alpha|\beta)\nu\quad \text{for all}\quad \beta\in \Lambda^{n-p}V.$$

The Hodge star operator is the map $\alpha \mapsto *\alpha$ from $\Lambda^pV$ from $\Lambda^{n-p}V$.

In the case of your example, $V = T^*\Bbb R^3$ and $(dx_i|dx_j) = \delta_{ij}$. Let $\nu = dx_1\wedge dx_2 \wedge dx_3$. Consider $*dx_1$, a two form on $\Bbb R^3$. It can be expressed as

$$\displaystyle (3) \qquad a\, dx_2\wedge dx_3 + b\, dx_3\wedge dx_1 + c\, dx_1 \wedge dx_2,$$

where $a$, $b$, and $c$ are real numbers to be determined. By the equation in $(2)$,

$$\displaystyle (4) \qquad (*dx_1|dx_2\wedge dx_3)\nu = dx_1\wedge dx_2\wedge dx_3 = \nu.$$

Using the expression $(3)$,

$$\displaystyle (5) \qquad (*dx_1|dx_2\wedge dx_3) = a(dx_2\wedge dx_3|dx_2\wedge dx_3) + b(dx_3\wedge dx_1|dx_2\wedge dx_3) + c(dx_1\wedge dx_2|dx_2\wedge dx_3).$$

Now

$$\displaystyle (dx_2\wedge dx_3|dx_2\wedge dx_3) = \text{det}\left(\begin{bmatrix}(dx_2|dx_2) & (dx_2|dx_3)\\(dx_3|dx_2) & (dx_3|dx_3)\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\right) = 1,$$

$$\displaystyle (dx_3\wedge dx_1|dx_2\wedge dx_3) = \text{det}\left(\begin{bmatrix}(dx_3|dx_2) & (dx_3|dx_3)\\(dx_1|dx_2) & (dx_1|dx_3)\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}\right) = 0,$$

$$\displaystyle (dx_1 \wedge dx_2|dx_2\wedge dx_3) = \text{det}\left(\begin{bmatrix}(dx_1|dx_2) & (dx_1|dx_3)\\(dx_2|dx_2) & (dx_2|dx_3)\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}\right) = 0.$$

Combining these results with $(4)$ and $(5)$, we deduce $a = 1$. By a similar analysis, taking inner products of $*dx_1$ with $dx_3\wedge dx_1$ and $dx_1 \wedge dx_2$ yields $b = 0$ and $c = 0$ respectively. Therefore $*dx_1 = dx_2\wedge dx_3$. Similarly, $*dx_2 = dx_3 \wedge dx_1$ and $*dx_3 = dx_1\wedge dx_2$.

If you chose $\nu = dx_2\wedge dx_1\wedge dx_3$ instead, then

$$\displaystyle *dx_1 = -dx_2\wedge dx_3$$

$$\displaystyle *dx_2 = -dx_3\wedge dx_1$$

$$\displaystyle *dx_3 = -dx_1\wedge dx_2$$
Someone finally took the plunge after 2.5 years! XD

This problem was still subconsciously aggravating me, so I appreciate the clear and thorough explanation! It makes a lot more sense now.