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In this delightful question, the poster mentioned that the isometry group of a compact Lie group $G$, equipped with the metric from the Killing form, is $G\times G/Z(G)$, where $Z(G)$ is the center of $G$. I'm specifically interested in the Riemannian case, so I will add the extra condition that $G$ be semisimple.

Is there a concise way to show that $G\times G/Z(G)$ is the isometry group, other than using the algorithm from the answer?

I can clearly see that it is a subgroup, since the metric coming from the Killing form is bi-invariant, and that quotienting out the center from one of the two copies of $G$ is necessary, because otherwise the action wouldn't be faithful. However, I cannot see how all isometries are of that form. In fact, I'm almost positive that this would exclude something of finite order, like $g\mapsto g^{-1}$.

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    $\begingroup$ Note that @RobertBryant's answer gives the Lie algebra of the isometry group, rather than the full isometry group, and as such shows only that the group that you have described is the identity component of the isometry group. It is not surprising that finite-order isometries live in different components! $\endgroup$ – LSpice Dec 29 '14 at 1:05
  • $\begingroup$ Googling "isometry group of a Lie group" turned up the paper Ochiai and Takahashi - "The group of isometries of a left-invariant Riemannian metric on a Lie group" (link.springer.com/article/10.1007%2FBF01360280), which confirms that it is indeed the identity component which is being identified here. (In turn that article attributes it to p. 207 of Helgason, "Differential geometry and symmetric spaces", I think Theorem 4.1 (books.google.com/books?id=xf_YVHF51MwC&pg=PA207).) $\endgroup$ – LSpice Dec 29 '14 at 1:10
  • $\begingroup$ @LSpice -- I see. I guess my confusion in that respect stems from no assumptions of connectedness being stated, so I just assumed that it meant the whole group. Thank you. $\endgroup$ – Robin Goodfellow Dec 29 '14 at 1:21
  • $\begingroup$ Note that the first sentence of @RobertBryant's answer is "Here is an algorithm to compute the Lie algebra of the group of isometries of a homogeneous space $G/H$ endowed with a $G$-invariant (pseudo-)Riemannian metric $g$" (emphasis mine). That is, he does explicitly mention (though I missed it at first!) that he is only identifying the Lie algebra; and the Lie algebra of a Lie group, which explicitly 'zooms in' near the identity, only ever has a chance of identifying the identity component. $\endgroup$ – LSpice Dec 29 '14 at 1:31
  • $\begingroup$ @LSpice -- Yes, I saw this. My question regards the question's post, not the answer. $\endgroup$ – Robin Goodfellow Dec 29 '14 at 1:32

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