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Prove that $SL_n(C)$ is path connected. We need a continuous path joining any two points in $SL_n(C).$ Let $A,B\in SL_n(C)$ be any two points. Then can anyone give a path define $\gamma:[0,1]\to SL_n(C)$ such that $\gamma(0)=A, \gamma(1)=B.$

Mathlover
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  • forget about path for some time. Given two matrices from special linear group can you produce another matrix from special linear group? Do you know more such? –  May 15 '18 at 17:43
  • $\textrm{SL}_2(\Bbb C)$ is generated by elementary matrices. – Angina Seng May 15 '18 at 17:55
  • I recommend first finding a path $\gamma : [0, 1]\to \text{SL}_n(\mathbb{C})$ between $I_n$ and arbitrary $A\in \text{SL}_n(\mathbb{C})$, and then taking the path $\gamma_1$ from $I_n$ to $A$ and the path $\gamma_2$ from $I_n$ to $B$, and consider $\gamma(t) = \gamma_1(1-t)\gamma_2(t)$. – Michael L. May 15 '18 at 17:55
  • @MichaelLee I doubt that this hint would make the problem easier.. –  May 15 '18 at 17:59
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    On the contrary... – Michael L. May 15 '18 at 18:01
  • Alternatively, there is an explanation here of why the special linear group is even simply connected (which of course implies that it is path connected). – Michael L. May 15 '18 at 18:20
  • I tried as your suggestion, but not succeeded to find such path. Please help with the proper path. – Mathlover May 16 '18 at 03:24

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