Our professor gave us exercise to show that $G=SO(n,\mathbb R)$ is path connected. He gave some hints, using them I have come upto this far:

  1. I have shown that $SO(n)$ acts on $S^{n-1}$ transitively by matrix multiplication: $A.x=Ax $ $\forall A\in SO(n), x\in \mathbb R^n$.

  2. I have also shown that Stabilizer of $e_1=(1,0,\cdots0)$ : $H=Stab(e_1)\cong SO(n-1)$. By induction, I assume that $SO(n-1)$ is path connected.

  3. Then, if $G/H$ is given the quotient topology from $G$, then $G/H$ is homeomorphic to $S^{n-1}$ via the map $gH \rightarrow ge_1$.

So now, the situation is that: I have $SO(n-1)$ path connected, and $SO(n)/SO(n-1)$ is path connected. But from these two facts, I can't prove that $SO(n)$ is path connected.

Thanks in advance.

  • 2
    $\begingroup$ Alternatively, you can show that it is connected and locally path connected. $\endgroup$ – Prahlad Vaidyanathan Sep 19 '13 at 9:04
  • $\begingroup$ @PrahladVaidyanathan: Thanks! I am able to show that $SO(n)$ is connected using induction, and then by locally connectedness it also implies that $SO(n)$ is path connected. $\endgroup$ – yojusmath Sep 19 '13 at 12:56

There are some details here that needs ironing out, but this approach should work with the results you already have:

Take any two points in $SO(n)$, map them via quotient map $q$ to $SO(n)/SO(n-1)$. Connect them via a path in that space. If the path doesn't go through the point $q(SO(n-1))$, lift the path via $q^{-1}$. Rejoice.

However, if it does, remove that point from the path, lift the remaining parts of the path to $SO(n)$. The closure of the image of the path has two endpoints contained in $SO(n-1)$. Connect them. Rejoice.

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  • $\begingroup$ what do you mean by 'lifting' the path? $\endgroup$ – yojusmath Sep 19 '13 at 9:34
  • $\begingroup$ @mathmansujo I just mean take the set of points that make up the path in $SO(n)/SO(n-1)$, and apply the inverse quotient map, $q^{-1}$, to that set. Since $q$ is a bijection outside of $SO(n-1)$ this presents no problems. A bit stricter, say $l:[0, 1]\to SO(n)/SO(n-1)$ is a path. Then the lifting of $l$ would be the path $q^{-1}\circ l:[0, 1]\to SO(n)$. $\endgroup$ – Arthur Sep 19 '13 at 10:25

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