What is the a connection between isometry groups and lie manifolds?

Question

A Lie group is a smooth manifold that is also a group. e.g. U(1) is the circle, SU(2) is the 3-sphere , SO(3) is the real-projective-3-space.

A manifold can also have an isometry group e.g. the isometry group of a 2-sphere is O(3). (But a 2-sphere can not have a lie group structure).

Therefor I would like to know if there is a general way to get the isometry group of a Lie manifold. And is there a connection between these two groups?

i.e. given a manifold M (which is a lie group G) what is it's isometry group? And is this related to G?

Also I'm interested in if there are any manifolds apart from the circle where the Lie group is the same as the isometry group. U(1).

(It strikes me also that one could recursively get isometry group of a manifold and then find the isometry group of this group as a manifold as so on to get a chain of groups.)

Answer 1

The isometry group of a metric space is the set of all bijective isometries (i.e. bijective, distance-preserving maps) from the metric space onto itself, with the function composition as group operation. Its identity element is the identity function. The elements of the isometry group are called motions of the space.

Lie groups are differential manifolds whose group operations are smooth. It is a continuous group whose elements are several real parameters and it lies on the concept of continuous symmetries. The connection between the two is isometry group is a Lie group in the case of Riemannian symmetric spaces.

Answer 2

When we have a Lie group $G$, there exists always a left invariant Riemannian metric $\langle\cdot\,,\cdot\rangle$ on $G$. This implies that any left translation $L_g\colon G\to G$ given by $L_g(h) = g\cdot h$ is a Riemannian isometry, and thus a metric isometry with respect to $d$, the induced distance on $G$ by $\langle\cdot\,,\cdot\rangle$.

This implies that $G$ is itself a subgroup of $\mathrm{Isom}(G,d)$. But it is not necesarilly the whole group.

In the example of identifying the $3$-sphere $\mathbb{S}^3$ with the Lie group $\mathrm{SU}(2)$, we can find a left invariant metric with constant sectional curvature equal to $1$, i.e. a round metric. This metric has as group of isometries the Lie group $\mathrm{O}(4)$. The group $\mathrm{O}(4)$ has topological dimension equal to $\frac{4(4-1)}{2} =6$; but $\mathrm{dim}(\mathrm{SU}(2)) = 3$. Thus in this example we have more isometries than those given by the Lie group $\mathrm{SU}(2)$ acting on itself. I.e. $\mathrm{SU}(2)\subsetneq \mathrm{O}(4)$.

Summarizing for a Lie group $G$ with a left-invariant Riemannian distance $d$ (i.e. induced by a left-invariant Riemannian metric $\langle\cdot\,\cdot\rangle$, we have $G\subseteq \mathrm{Iso}(G,d)$, but in general it is a proper subgroup.

A general method for computing the dimension of $\mathrm{Iso}(G,g)$ for a Lie group $G$ with a left-invariant Riemannian distance $d$ is given in the following answer by Robert Bryant. In this way, we can see how larger the group $\mathrm{Iso}(G,d)$ is with respect to $G$.