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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?

Myshkin
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2 Answers2

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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.

Lukas Geyer
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$SL_{n+1}(C)$ 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 $n\ge 2$ and if $n=1$ the fundamental group is Z (the fundamental group of the circle).

i. m. soloveichik
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  • I think you mean $n>2$ and $n=2$ instead of $n>1$ and $n=1$. – Lukas Geyer Oct 27 '12 at 16:59
  • @LukasGeyer thanks – i. m. soloveichik Oct 27 '12 at 17:01
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    No. The unitary group $U(n)$ is never simply-connected (well, except for $U(0)$). It has fundamental group $\mathbb{Z}$ as shown for instance in http://math.stackexchange.com/questions/222038/fundamental-group-of-un – Bruno Le Floch Nov 05 '16 at 14:42
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    Not only this, but $U(n)$ is not a subgroup of $SL_n(\mathbb{C})$! The group $SL_n(\mathbb{C})$ has $SU(n)$ as a deformation retract, which is simply connected as seen by considering the fiber bundles $SU(n-1) \to SU(n) \to S^{2n-1}$ and that $SU(1)$ is trivial. Hence, $SL_n(\mathbb{C})$ is simply connected for all $n \geq 1$. – Joshua Mundinger Mar 14 '19 at 04:25