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Hodge Star Operator on a Riemannian Manifold

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Chris L T521

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Jan 26, 2012
995
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! (Nod)
 

Euge

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Jun 20, 2014
1,902
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

\(\displaystyle *dx_1 = -dx_2\wedge dx_3\)

\(\displaystyle *dx_2 = -dx_3\wedge dx_1\)

\(\displaystyle *dx_3 = -dx_1\wedge dx_2\)
 
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Chris L T521

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Jan 26, 2012
995
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. (Bigsmile)