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I've scanned some of the literature on this, but couldn't find an answer to the following simple questions (probably because I'm not an expert):

Q1: Let G be a Lie group with a left-invariant metric. What are some simple criteria for G to be symmetric, namely, for G to admit, for any point and geodesic through that point, an isometry reversing that geodesic?

Q2: In three dimensions, in terms of the structure constants, one can easily work out essentially all simply-connected groups very concretely. Is there a criterion for going through the list of 3D Lie groups, looking at the structure constants, and deciding which ones are symmetric spaces?

Thank you for your time!

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  • $\begingroup$ For Q2, I've implicitly chosen a basis of the Lie algebra, declared it orthonormal, and then took the left-invariant metric corresponding to that. It is in terms of the structure constants arising that way that I ask the question. $\endgroup$ – Guest333 May 15 '15 at 19:21
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Suppose the metric is bi-invariant. (You can always find such a metric if $G$ is compact; see Amitesh Datta's comment below.) Then geodesics starting at the identity are one-parameter subgroups (and by invariance this determines what geodesics are everywhere else), and the isometry reversing those geodesics is $g \mapsto g^{-1}$.

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    $\begingroup$ Hi @Qiaochu, I'm sure you know this already, but you can always find a bi-invariant metric on a compact Lie group: simply pick an inner product at the identity which is Ad-invariant ("Ad" denoting the adjoint representation of the Lie group), and extend it in the obvious left-invariant manner to a Riemannian metric on all of $G$. $\endgroup$ – Amitesh Datta May 15 '15 at 19:58
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An answer to Q2 is: $G$ is a symmetric space if and only if the left-invariant metric is bi-invariant.

The answer of Qiaochu Yuan explains why this is a sufficient criterion. It remains to check that it is also a necessary criterion. To this end, assume $G$ is endowed with a left-invariant, but not a bi-invariant metric, and is a symmetric space. Then, $G$ must be one of the symmetric spaces from Cartan's classification https://en.wikipedia.org/wiki/Symmetric_space#Classification_result . None of those symmetric spaces have dimension 3, which is a contradiction.

An answer to Q1 that is useless for general proofs, but may be helpful in special cases, is: For $G$ to be a symmetric space, it has to be in one of two classes:

  1. $G$ is endowed with a bi-invariant metric.
  2. $G$ is endowed with a left-invariant, but not bi-invariant metric.

In case 1 it is automatically a symmetric space, as stated above.

In case 2 it is isometric to a symmetric space $M=B/K$. The Iwasawa decomposition $\mathfrak{b}=\mathfrak{k \oplus a \oplus n}$ gives rise to a subgroup $AN \subset B$. This subgroup carries a left-invariant metric and is isometric to $M$, and therefor to $G$. So, a necessary and sufficient criterion for $G$ to be a symmetric space is that it is isometric to the Iwasa subgroup $AN$ of some symmetric space. Of course, this in itself is not helpful at all. However, one could write down a list of all possible $AN$, and then check if a given $G$ appears in the list. If it does, it is a Riemannian space. If it does not, one has to check whether it is isometric to any entry in the list, which is admittedly difficult.

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