# hodge star and pull back

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?

## 1 Answer

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.

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