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How to show that $SU(n)$ is simply connected for $n>2$ if I don't know about fibrations yet? For $SU(2) \cong S^3$ the fact is said to be known. For any matrix $A \in SU(n)$ there is a matrix $S \in U(n)$ such that $$ A = S^{-1}DS, $$ where D has a form $$ D = \begin{bmatrix} e^{i \varphi_1} &0&0 & \cdots &0 \\ 0 & e^{i\varphi_2} & 0 & \cdots & 0 \\ 0 & 0 & e^{i\varphi_3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \\ 0 & 0 & 0 & \cdots & e^{i\varphi_n} \end{bmatrix} $$ and since $\det D =1$ we have $\varphi_1+\ldots+\varphi_n = 0 \mod 2\pi$. Maybe one can find some deformation retract of $SU(n)$ with easy computable fundamental group (which will be trivial)?

Appliqué
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    You can reduce this to the case $n = 2$ using the long exact sequence associated to a suitable fibration. – Qiaochu Yuan Apr 23 '13 at 23:52
  • @QiaochuYuan But I don't know about fibrations yet :( It there a way using homotopy\homology theories? – Appliqué Apr 24 '13 at 10:28
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    I'm guessing you won't find a deformation retract of $SU(n)$ because of this :http://math.stackexchange.com/questions/258367/is-it-possible-for-a-closed-manifold-to-deformation-retract-onto-a-proper-subset/258409#258409 – Olivier Bégassat Apr 24 '13 at 11:01
  • @OlivierBégassat Thank you, I see – Appliqué Apr 24 '13 at 11:12
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    There is a cell structure on $SU(n)$ with a single zero cell and no one cells: http://projecteuclid.org/euclid.pja/1195525543. This implies that $SU(n)$ is simply connected, but it takes a bit of work. – Justin Young May 08 '13 at 12:14
  • It seems to me that the simplest approach is to learn about fibrations. It's not THAT hard. – Igor Rivin Dec 30 '13 at 20:26

2 Answers2

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Using Morse theory (I will let you fill in the details):

Consider the Morse function $f:SU(n)\to\mathbb{R}$ given by $[z_{ij}]\mapsto\text{Re}(\sum_ic_iz_{ii})$ for fixed constants $1<c_1<<c_2<\ldots<c_n\in\mathbb{R}$. It has a critical point of index 0, and the next smallest index is 3 (figure out why we want $c_1<<c_2$). Thus $SU(n)$ has no 1-handles, meaning it is simply-connected.

For $n=2$, it has only two critical points $\lbrace\pm\text{id}\rbrace$. In particular, $SU(2)$ is built up by two 3-disks and hence is diffeomorphic to $S^3$.

Chris Gerig
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Perhaps you might try thinking about it starting with $SU(2)$. What is this homeomorphic to?

Ted Shifrin
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  • I said that $n>2$ because for $n=2$ $SU(2) \simeq S^3$ – Appliqué Apr 23 '13 at 22:00
  • Hm, I can use homotopy $t\varphi_k$ that transforms $D$ into identity matrix, right? But what is a problem to use it in the case of $U_n$ instead of $SU_n$ (because $U_n$ is not simply connected)? – Appliqué Apr 23 '13 at 22:05
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    So do you know what manifold $SU(3)/SU(2)$ is? You really want to be thinking about fiber bundles here. – Ted Shifrin Apr 23 '13 at 23:48
  • But I know nothing about fibrations! – Appliqué Apr 24 '13 at 09:20
  • But note that your approach gives us a circle in $SU(2)$. Since the circle has a nontrivial fundamental group, you'd better hope you don't have a retraction onto that circle! – Ted Shifrin Apr 24 '13 at 11:08