# Holonomy of Lie groups

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?

• For some example, the holomony group of $U(1)$ is trivial, while that of $SU(2)$ is $SO(3)$. Not sure how they are related...
– user99914
Commented Sep 8, 2016 at 9:54
• @JohnMa Many thanks. Do you have a reference that $\operatorname{Hol}(\operatorname{SU}(2))=\operatorname{SO}(3)$? (Note that $\operatorname{U}(1)$ is not simple.) Commented Sep 8, 2016 at 10:11
• I missed the word simple.. anyway $SU(2)$ is isometric to $\mathbb S^3$.
– user99914
Commented Sep 8, 2016 at 10:15
• @JohnMa $\mathrm{Hol}(G)=\mathrm{Ad}(G)$ would generalize both examples... Commented Sep 8, 2016 at 15:35

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.

• I have a concern: Let $\mathfrak{g}$ be a compact real form of the exceptional complex simple Lie algebra of type $E_6$, and let $G$ be the unique connected, simply connected Lie group with Lie algebra $\mathfrak{g}$. Then, $G$ is compact and its image $\mathrm{Ad}(G)$ is a compact connected Lie group with Lie algebra $\mathrm{ad}\mathfrak{g}\cong\mathfrak{g}$. Here $G$ is simply connected, so you claim that $\mathrm{Hol}(G)=\mathrm{Ad}(G)$. But there is no Lie group of type $E_6$ in Berger's list of possible holonomy groups. Commented Sep 10, 2016 at 16:01
• @SimonParker: Berger only listed holonomies not coming from symmetric spaces and your question is about a compact symmetric space. Commented Sep 10, 2016 at 16:14
• Right ! Thanks ! +1 Commented Sep 10, 2016 at 17:10