Left invariant Riemannian metrics which are symmetric and Einstein

Question

Is there a list of connected Lie groups which admit a left invariant Riemannian metric which is Einstein, locally symmetric and its infinitesimal holonomy is irreducible?

Answer 1

There is such a list, but I don't know where it might already be written down. It's not hard to compile it, though, using known facts about the classification of symmetric spaces.

First, one can easily reduce to the case that the Lie group $S$ is simply connected, so that the given left-invariant metric $g$ is actually symmetric. The assumption that the infinitesimal holonomy is irreducible, then implies that $S = G/K$ where $(G,K)$ is an irreducible symmetric pair. In particular, the Einstein constant cannot be zero and hence one can treat the two cases of positive or negative curvature separately.

If the Einstein constant is positive, then $S$ is compact, as is $G$. Thus, $\pi_3(S)$ contains an infinite cyclic group. Since $\pi_2(K)=1$, it follows that $\pi_3(K)\to \pi_3(G)$ has an infinite cokernel, which implies that $G/K$ cannot be of Type I. Hence $G = K\times K$ and $K$ is embedded diagonally in $G$. Thus, $S = K$ and the metric is bi-invariant on $K$. Irreducibility then implies that $S=K$ must be simple, endowed with its bi-invariant metric, which is Einstein. The finite quotients are of the form $S = K/Z$ where $Z$ is a subgroup of the (finite) center of $K$.

If the Einstein constant is negative, then $S$ is not compact, nor is $G$. Because left multiplication by $S$ is a $g$-isometry, $S$ is then embedded into $G$, the identity component of the group of isometries of $g$, and $S\cap K = \{e\}$ and $G = SK$. Since $S$ is a symmetric space of noncompact type, it follows that $S$ is contractible and hence solvable. Conversely, this always happens: Since $K$ is a maximal compact in $G$, one can write $G$ in the form $G = NAK$ where $N$ is a nilpotent subgroup and $A$ is abelian so that $S = NA$ is a solvable subgroup. Thus, $S = NA = G/K$ carries the irreducible symmetric, Einstein metric that it inherits through its identification with the irreducible symmetric space $G/K$. I think that, in this case, $S$ has no nontrivial discrete central subgroups, so there are no nontrivial quotients of these examples.