0
$\begingroup$

How do I prove the statement that for a Riemannian symmetric space $M$, the isometry group $Iso(M)$ is a Lie group? What can we say about the dimension of $Iso(M)$? And is its action transitive on $M$? And what can we say about the isometry group of a Lie group endowed with a left/right (not bi) invariant Riemannian metric (as we know if the metric is bi invariant, then the Lie group is a symmetric space)?

P.S I not only look for answers but also for resource and literature to probe such related questions. (Note: In the definition of Riemannian symmetric space that I come across, the space is not assumed to be simply connected)

$\endgroup$
  • $\begingroup$ $Iso(M)$ is a Lie group is proved by Myers and Steenrod. $\endgroup$ – user10354138 Jun 7 at 6:58
  • $\begingroup$ @user10354138 The poster asked about symmetric spaces, on which the isometry group does always act transitively. $\endgroup$ – Danu Jun 7 at 9:02
  • 1
    $\begingroup$ Try Helgasson's book and see if you can handle it. $\endgroup$ – Moishe Kohan Jun 7 at 10:32
  • $\begingroup$ Thanks a lot. What about isometry groups of lie groups with left invariant (not bi invariant) metric? $\endgroup$ – Rama Seshan Chandrasekaran Jun 13 at 3:43

Your Answer

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

Browse other questions tagged or ask your own question.