What is the relationship between the dimension of the manifold of a lie group and the dimension of the manifold the lie group describes the symmetries of?
For example, the dimension of the sphere, S^2, is 2. And the dimension of the Lie group, SO(3), that describes the symmetries of the sphere, is 3.
Is there a general relationship between the dimension of the two manifolds, i.e. that of the manifold of a Lie group and that of the manifold which the Lie group 'acts on'? Thanks.
On
Another way to interpret your question is that you are refering to symmetries of a manifold as the Groups which act transitively on the manifold. There we have a theorem stating that if we have a Lie group $G$ which acts smooth and transitive on a smooth manifold $M$ than the quotient of $G$ with the stabilizer for some point $b\in M$ denoted $H=\operatorname{Stab}_b(G)$ induces a diffeomorphism
$G /H \to M$ via $gH\mapsto gb$
First, a manifold by itself doesn't have any natural finite dimensional set of symmetries. (If you allow infinite dimensional examples, then the group of diffeomorphisms, or orientation preserving diffeomorphisms work. There may be some other examples).
Now, if you equip your manifold with a Riemannian metric, then it has an isometry group. Based on your post, it seems as though this is what you mean: when we equip $S^2$ with its standard round metric, the full isometry group is $O(3)$ and the orientation preserving subgroup is $SO(3)$ (not $SU(3)$, by the way), which as dimension $3$.
But there are other metrics on the sphere. For example, if you deform it into an ellipsoid with circular cross section, then the symmetry group reduces to $O(2)$, which has dimension $1$. Further, a generic "bumpy" metric has no symmetries at all!
In general, the isometry group $G$ with respect to a Riemannian metric and the dimension of the manifold are only remotely related. For example, if $n = \dim M$, then one always has $0\leq \dim G\leq n + \frac{n(n-1)}{2}$, but you generally can't say more than this. (Roughly, the upper bound comes from the fact that an isometry is determined by where it sends one points ($n$ degrees of freedom) and what the differential does at one point ($n(n-1)/2$ degrees of freedom - the ability to move one orthonormal basis to any other.)
One thing to keep in mind is that a generic perturbation of any Riemannian metric has trivial isometry group. (I'd love a reference for this result - it's a "folklore" theorem to me.)
There are, however, a few cool facts: For example, if $\dim G = n + \frac{n(n-1)}{2}$, then $M$ must be isometric to a sphere or $\mathbb{R}P^n$ with their standard metrics. There is some othe result (which I don't remember exactly) which shows that you'll never get $\dim G = n + \frac{n(n-1)}{2} - 1$, except possibly for really small $n$. More details can be found the book Introduction to Compact Transformation Groups by Bredon.