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?

  • $\begingroup$ 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... $\endgroup$
    – user99914
    Commented Sep 8, 2016 at 9:54
  • $\begingroup$ @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.) $\endgroup$ Commented Sep 8, 2016 at 10:11
  • $\begingroup$ I missed the word simple.. anyway $SU(2)$ is isometric to $\mathbb S^3$. $\endgroup$
    – user99914
    Commented Sep 8, 2016 at 10:15
  • 2
    $\begingroup$ @JohnMa $\mathrm{Hol}(G)=\mathrm{Ad}(G)$ would generalize both examples... $\endgroup$ Commented Sep 8, 2016 at 15:35

1 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.

  • $\begingroup$ 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. $\endgroup$ Commented Sep 10, 2016 at 16:01
  • $\begingroup$ @SimonParker: Berger only listed holonomies not coming from symmetric spaces and your question is about a compact symmetric space. $\endgroup$ Commented Sep 10, 2016 at 16:14
  • $\begingroup$ Right ! Thanks ! +1 $\endgroup$ Commented Sep 10, 2016 at 17:10

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