Diffeomorphism of $\mathcal O(n)$ into itself

Asked Dec 31 '22 at 18:41

Active Dec 31 '22 at 18:48

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I have a exercice where I need to proof that
$f:\mathcal O(n)\to\mathcal O(n)\;\;,\;\;f(A) =A^{-1}\;$ is a diffeomorphism.
$f\;$ is cleary bijective because $\,f\,$ maps a matrix $\,B\,$ to $\,B^{-1}=B^T\,$ since $\,B\,$ is an orthogonal matrix, so the inverse function of $\,f\,$ is just the identity map on $\,\mathcal O(n)$.
But I struggle to show that $\,f\,$ is smooth. Should I try to do it with charts ? Or maybe should I use the fact that $\,\mathcal O(n)\,$ is a vector space of dimension $\,n(n − 1)/2$ ?

Thanks you

edited Dec 31 '22 at 18:48

Angelo

12,328

asked Dec 31 '22 at 18:41

RAT

What is $O(n)$? – José Carlos Santos Dec 31 '22 at 18:42
The orthogonal group – RAT Dec 31 '22 at 18:44
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You've already observed that on $O(n)$ we have $A^{-1} = A^\top$. Although you can certainly prove, in general, that inversion is a smooth map on $GL(n)$ (Hint: Use the formula for $A^{-1}$ in terms of cofactors), it is much easier here. Why is the transpose map smooth on the vector space of all $n\times n$ matrices? ... $O(n)$ is a manifold of dimension $n(n-1)/2$; it is certainly not a vector space. Why? – Ted Shifrin Dec 31 '22 at 18:45
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Do you believe that the orthogonal group is a vector space? – José Carlos Santos Dec 31 '22 at 18:45
You are right, it is not a vector space. My bad I'm a bit tired sorry – RAT Dec 31 '22 at 18:48
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The inverse function of your $f$ is $f$ itself, not the identity map on $O(n)$. – Andreas Blass Dec 31 '22 at 20:57

Diffeomorphism of $\mathcal O(n)$ into itself

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