Prove that the induced metric on SO(n) is bi-invariant. The inner product is given by the Frobenius inner product on matrices.
SO(n) bi-invariant metric
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differential-geometry
lie-groups
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1 Answers
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Hint 1:$\quad$ Note that for invertible matrices $A$ and $B$, which way you multiply two matrices does not affect the trace, i.e. $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ (do you see why?), hence the trace is similarity-invariant: $\mathrm{tr}(A^{-1}BA)=\mathrm{tr}((BA)A^{-1})=\mathrm{tr}(B(AA^{-1}))=\mathrm{tr}(B).$
Hint 2:$\quad$ In $SO$ (or $O$ for that matter), what relationship do $A^T$ and $A^{-1}$ have to each other?
Now you should be able to check both left and right invariance with direct computations.