0
$\begingroup$

I am trying to find a proof of the fact that the special unitary group is a Lie Group, but I can't come up with any good ideas and I couldn't find anything by searching at google.

Could you please help me?

Thank you in advance

$\endgroup$
  • $\begingroup$ You could show that it is a sub manifold of the collection of all $n \times n$ matrices by showing that 0 is the only critical value of the determinant. $\endgroup$ – Wintermute Apr 18 '17 at 18:00
  • $\begingroup$ @Wintermute: That argument would work for $\operatorname{SL}(n,\mathbb C)$, but the OP asked about $\operatorname{SU}(n)$. $\endgroup$ – Jack Lee Apr 18 '17 at 18:54
  • $\begingroup$ Any ideas, please? $\endgroup$ – perlman Apr 20 '17 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.