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Let $\phi:\mathbb{R}^n\to\mathbb{R}^n$ be an orthogonal linear map. Prove that $\phi^*(*\alpha) = *\phi^*(\alpha)$ for all $k$-forms $\alpha$ on $\mathbb{R}^n$.

I tried to write out $\phi^*(*\alpha)$ and $*\phi^*(\alpha)$, but I don't see where linearity and orthogonality comes into the proof. Any ideas?

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HINT: Let's write $\phi\colon V\to W$, both $V$ and $W$ being $\Bbb R^n$. If $\alpha_i$ give an orthonormal basis for $W^*$, let $\beta_i = \phi^*\alpha_i$ and show that these give an orthonormal basis for $V^*$. It suffices to consider $\alpha = \alpha_{i_1}\wedge\dots\wedge\alpha_{i_k}$. What is $\star\alpha$? Now express $\phi^*(\alpha)$ and $\phi^*(\star\alpha)$ in terms of the $\beta_i$'s.

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  • $\begingroup$ Oh, sorry. What is $W^*$? $\endgroup$ – QD666 Feb 11 at 2:08
  • $\begingroup$ The dual space of $W$, i.e., the space of linear functionals on $W$. That's where $dx_1, \dots, dx_n$ live. $\endgroup$ – Ted Shifrin Feb 12 at 7:11
  • $\begingroup$ thank you so much $\endgroup$ – QD666 Feb 12 at 16:02

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