I have another question about $SU(n)$, again I hope it's not too basic. For $n=2$, the action of $SU(2)$ on $C^2$ is free since it's equal to the group of rotations. In general, the group of rotations is properly contained in $SU(n)$, does this mean that its action on $C^n$ is no longer free.
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asked Feb 5, 2010 at 17:45
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2 Answers
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$SU(n)$ acts freely on $\Bbb C^n\setminus 0$ (not on $\Bbb C^n$) for $n=1,2$, but not for higher $n$ (the stabilizer of a point is $SU(n-1)$). If this is what you are asking about, the answer is "yes". (This IS a basic question, though.)
answered Feb 5, 2010 at 18:00
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1$\begingroup$ I meant all nonzero vectors, not only unit ones. The action on these of $SU(2)$ is free, although not transitive. $\endgroup$ Commented Feb 6, 2010 at 21:49
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Perhaps the generalization you want of the case $n=2$ is that $SU(n)$ acts freely and transitively on itself, and thus also on orthonormal $(n-1)$-tuples of vectors since the last column of an element of $SU(n)$ is determined by the first $n-1$. You can relax the condition to independence if you only want a free action.
answered Feb 6, 2010 at 12:39