Holonomy of Lie groups

Question

Simple compact Lie groups have unique bi-invariant metrics. Hence, they are Riemannian manifolds in a unique way, so we can ask what is their holonomy group. Is there a relationship between the group $G$ and its holonomy group? For example, is the holonomy group $G$ itself?

Answer 1

Let $(G,b)$ be a compact simple group equipped with the biinvariant metric. Let $Hol_e^0$ denote the identity component of the holonomy group of $G$ at $e\in G$.

Claim. $Hol_0$ is the identity component of the image of $G$ under the adjoint representation.

Proof. Let $H$ denote the isometry group of $(G,b)$. Then $H=L(G)R(G)$ meaning that every isometry of $(G,b)$ has the form $$ I_{g_1,g_2}: g\mapsto g_1 g g_2, g\in G $$ for some fixed $g_1, g_2$. (There might be a finite kernel of the map $I: G\times G\to H$.) A proof can be found for instance in Helgason's book "Differential Geometry and Symmetric spaces".

The stabilizer $H_e$ of $e$ in $H$ consists of isometries $I_{g,g^{-1}}$, which, therefore, acts on $T_eG$ via the adjoint representation of $G$. Next, $(G,b)$ is a symmetric space; it can be identified with $H/H_e$. Cartan proved that for each symmetric space without flat factors $X=H/K$, where $H$ is the full isometry group of $X$ and $K$ is the stabilizer of a point $p$ in $X$, $Hol_p^0=K^0$, the identity component of $K$. (This should be again in Helgason's book.) Putting it all together, we obtain that in our setting $Hol_e^0=Ad(G)^0$. qed

One can extend this proof to the case of nonsimple compact groups. The minor difference is that holonomy equals the holonomy of the semisimple factor of $G$; at the same time, the abelian part of $G$ does not contribute to the adjoint representation.

I am not sure what happens when you consider not only identity components but the entire $G$ and $Hol_e$.

Edit. To my dismay, Helgason's book does not contain Cartan's theorem on holonomy. A proof can be found in

J.-H. Eschenburg, Lecture Notes on Symmetric Spaces, Theorem 7.2.