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- Jan 26, 2012

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

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 $dx^1\wedge dx^2,\,dx^2\wedge dx^3\,dx^3\wedge dx^1\in \bigwedge^2 T^*M$.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 $F_1\,dx^2\wedge dx^3 + F_2\,dx^3\wedge dx^1+F_3\,dx^1\wedge dx^2$.

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!