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By considering $S^3$ as the group manifold of $SU(2)$, the ordinary action of $SO(4)$ on the three sphere can be written as the $SU(2)\times SU(2)/\mathbb{Z}_2$ given by the group action of multiplication on the left and right:

$(g,h).x=gxh^{-1}$

From this, there are two obvious subgroups acting freely and transitively on the sphere, by restricting to left and right multiplication.

My question is, are there any others?

Of less importance to me directly, but perhaps more general interest, is the generalisation: How many subgroups of $SO(n)$ act freely and transitively on $S^{n-1}$?

For n=2, there's 1: $SO(2)$ itself.

For n=3, there are none.

For n=4, perhaps 2?

Thanks!

Holographer
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    It is very rare for an action to be free and transitive - in fact, if a Lie group $G$ acts on $M$ freely and transitively, then $M$ and $G$ are diffeomorphic. Since the only spheres which are Lie groups are $S^0$, $S^1$, and $S^3$, you've essentially exhausted all the examples. If you relax your question to just transitive actions, then you get much more interesting examples... – Jason DeVito - on hiatus Nov 29 '13 at 15:41
  • Sure, thanks Jason. So it's clear that I'm looking for $SU(2)\subseteq SO(4)$. In that context, the question is whether there are any more than 2 ways to imbed $SU(2)$ inside $SO(4)$, and if not, how to prove it? – Holographer Dec 01 '13 at 18:48

1 Answers1

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There are precisely two ways, up to conjugacy, of embedding $SU(2)$ into $SO(4)$. The restriction of the natural action of $SO(4)$ to either of these $SU(2)$s gives a free, transitive action of $SU(2)$ on $S^3$.

To see this, first note that the images of the two $SU(2)$s in $SO(4)$ which you've found are not conjugate in $SO(4)$ (because they are both normal in $SO(4)$, being the images of normal subgroups in $SU(2)^2$). However, any other subgroup of $SO(4)$ which is isomorphic to $SU(2)$ is necessarily conjugate to one of these two. Here a sketch of why that's true.

Consider an injective homomorphism $\overline{f}:SU(2)\rightarrow SO(4)$. By the lifting criteria for covering maps, there is a unique lift $f:SU(2)\rightarrow SU(2)\times SU(2)$ of $\overline{f}$ with the property that $f$ maps the identity to the identity. One can easily show that $f$ must also be an injective homomorphism.

I claim that there are only two such $f$, up to conjugacy, for which $\pi \circ f:SU(2)\rightarrow SO(4)$ is an injective homomorphism. Establishing this, because $\pi\circ f = \overline{f}$ (where $\pi:SU(2)^2\rightarrow SO(4)$ is the double covering map), this implies there are at most two such $\overline{f}$ up to conjugacy.

First, it's clear that there are at least two such $f$ given by the inclusions into each $SU(2)$ factor of $SU(2)^2$. Since these factors are both normal, they are not conjugate to each other. Now we show that these are the only two.

For this, notice that a map $f:SU(2)\rightarrow SU(2)^2$ is really given as a pair of maps $(f_1,f_2):SU(2)\rightarrow SU(2)^2$, where $f(A) = (f_1(A), f_2(A))$, so this reduces the problem to understanding all maps $f_1:SU(2)\rightarrow SU(2)$. If both $f_1$ and $f_2$ are trivial, then $f$ is not injective, so we assume wlog that $f_1$ is nontrivial.

We interpret $f_1$ as a $2$-d representation of $SU(2)$. But the representations of $SU(2)$ have been completely classified - every representation breaks into a direct sum of so-called irreducible representations, and there is a unique irreducible representation of every (complex) dimension. Because $2=2$ and $2=1+1$ are the only parititions of $2$, $f_1$ is equivalent, as a representation, so either the trivial representation or the standard representation. Since it's not trivial, $f_1$ is equivalent to the standard representation.

Mal'cev has proven that for maps into $SU(n)$ (and in fact, maps into $U(n), Sp(n), SO(2n+1)$, but not $SO(2n))$, the induced representations are equivalent iff the images are conjugate. Thus, applying this to $f_1$, we know that $f_1(A) = BAB^{-1}$ for some $B\in SU(2)$.

Working back to $f$, we now know $f$ either has the form $f(A) = (BAB^{-1}, CAC^{-1})$ for some $B, C\in SU(2)$, or $f(A) = (BAB^{-1},I)$ for some $B\in SU(2)$. In the first case, we note that $f(-I) = (-I, -I) = -(I,I)$, and hence, $f(-I)\in \ker \pi$. This implies $\overline{f} = \pi \circ f$ has nontrivial kernel, a contradiction. Hence, $f(A) = (BAB^{-1},I)$ for some $B\in SU(2)$. This proves there are precisely three nontrivial homomorphisms from $SU(2)\rightarrow SU(2)^2$, but only 2 which give injective homomorphisms $SU(2)\rightarrow SO(4)$.

Jason DeVito - on hiatus
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  • Great answer, thanks! The reason I was interested in this was in looking for homogeneous vector fields on homogeneous spaces. I suspect that this reduces to finding free transitive actions, though I've not managed to convince myself. I've just posted another question on that if you're interested. – Holographer Dec 08 '13 at 17:12