Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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. –  Bunuelian Trick Oct 27 '12 at 17:21
add comment

2 Answers

up vote 3 down vote accepted

$SL_n(C)$ 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).

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.