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!

  • $\begingroup$ 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

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 $g \mapsto g^{-1}$.

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.