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I want to find the Lie algebra of $O(n)$. I know $Lie(O(n))=\{X \in \mathfrak{gl}_n(\mathbb{R}) : \exp(tX)\in O(n) \forall t\in \mathbb{R}\}$

So $\exp(tX)^T \exp(tX)=I_n$ i.e. $\exp(tX^T)\exp(tX)=I_n$ then it seems that $tX^T$ and $tX$ don't commute in general so I don't know (this is the point where I need some help) if I can say $\exp(tX^T+tX)=\exp(0)$ then take $t$ small enough such that $tX\in U$ such that $\exp|_U$ is a diffeomorphism so in particular injective. Then $tX^T+tX=0$ so $X^T+X=0$.

glS
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roi_saumon
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  • Can you share your source, please? – zkutch Jul 19 '20 at 16:47
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    https://math.stackexchange.com/q/1823418/96384, https://math.stackexchange.com/q/356499/96384, https://math.stackexchange.com/q/123288/96384. – Torsten Schoeneberg Jul 19 '20 at 17:07
  • @Torsten Schoeneberg. Thanks. I'll be grateful if you can add some books, monographs or something such kind also. – zkutch Jul 19 '20 at 17:58
  • @zkutch: Misunderstanding. I'm not the post author and my comment was not a reply to yours, rather a collection of posts on this site strongly related to the question, so that OP might find information in there. – Torsten Schoeneberg Jul 19 '20 at 18:30
  • @TorstenSchoeneberg in your link, it is suggested by someone "exponentiate $X+X^T=0$ to get $e^Xe^{X^T}=1$". But how do you justify this. It seems $[X,X^T] \neq 0$ in general no? – roi_saumon Jul 19 '20 at 20:51
  • Indeed they don't necessarily commute, but in the answer to the third link it's explained how to get the result from an explicit power series computation (the answer is more general, here we just can take $K=Id$). – Torsten Schoeneberg Jul 20 '20 at 01:25
  • Since Lie algebra is a local structure, isn't Lie algebra of () is that same as Lie algebra of $SO()$? – wonderich Jul 20 '20 at 17:02

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