The group $\mathrm{SU}(2)$ and the manifold $S^3$

Question

The group manifold of $\mathrm{SU}(2)$ is $S^3$. $\mathrm{SU}(2)$ has three symmetry generators, while $S^3$ is a maximally symmetric space with 6 symmetry generators (Killing vectors). How can these two facts be reconciled?

Answer 1

There's not really anything to reconcile. One the one hand you're talking about a certain Lie group structure on $S^3$, and on the other hand you're talking about a certain Riemannian structure on $S^3$. It happens that the group structure is compatible with the Riemannian structure, so translation by any group element gives an isometry. But there's no reason there can't be additional symmetries of the metric that don't come from translation with respect to the group structure. That happens in this case, as there are 3 extra "dimensions" of symmetries of the metric.

Answer 2

In fact on any compact Lie group, left translations and right translations are isometries for the Killing metric . This gives a map from $G\times G$ to $Isom(G)$. One checks that the kernel of this morphism $(g,h)\to (x\to gxh^{-1}$ is the center $Z(G)$, and finite in your case. So for every compact Lie group with finite center, the dimension of isometry group is (at least) twice the dimension of the group.