# compactness and fundamental group of $SL(n,\mathbb{C})$

could any one help me to prove the heading?

$SL(n,\mathbb{C})$ is closed I can prove only, what are the other tools I need?

$SL(n,\mathbb{C})$ connected?simply connected?

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I voted this question down because after your questions here and here no new ideas are required. Try harder before you ask! – commenter Oct 27 '12 at 17:11
@commenter thank you very much. – Un Chien Andalou Oct 27 '12 at 17:21

$SL_n(C)$ has the unitary group $U_n$ as a deformation retract (Gram-Schmidt) so the fundamental groups are the same. The unitary group is simply connected for $n>2$ and if $n=2$ the fundamental group is Z (the fundamental group of the circle).
I think you mean $n>2$ and $n=2$ instead of $n>1$ and $n=1$. – Lukas Geyer Oct 27 '12 at 16:59
The case $n=1$ is trivial, so assume $n\ge 2$. It is easy to see that $SL(n,\mathbb{C})$ is not compact, just look at diagonal matrices with entries $\lambda, \lambda^{-1}, 1, \ldots,1$, and let $\lambda \to \infty$. It is equally easy to see that it is connected, just deform the Jordan normal form to the identity matrix. For the fundamental group you need a little more sophisticated tools, see the other answer.