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Greetings,

Let $G$ be a compact Lie group with a bi-invariant inner product $h$ on it. Can one embedd $G$ in $M(n,\mathbb{C})$ isometrically for some $n \in \mathbb{N}$. By isometrically I mean that the restriction of the standard metric in $M(n,\mathbb{C}) = \mathbb{C}^{n^{2}}$ restricts to $h$ on $G$. Is this possible? If no, why it isnt possible? If yes, how could one prove this?

Dmitri

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  • $\begingroup$ Maybe I missed a point, but why does that not trivially follow from Nash ? At what point do you use seriously the group structure here (except for picking a particular metric) ? $\endgroup$
    – Amin
    Commented Dec 2, 2012 at 9:25
  • $\begingroup$ I mean if this embedding is also a group homomorphism. By Nash it is clear that such an embedding exists. But is it also a grouphomomorphism ? $\endgroup$
    – Dmitri
    Commented Dec 2, 2012 at 9:36
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    $\begingroup$ The answer to your exact question is 'no'. It's not even possible for most invariant metrics on $G = S^1$, as is easy to see since you know all of its homomorphisms into $M(n,\mathbb{C})$ up to conjugacy. As Peter points out below, in the case $G$ is simple or $S^1$, you can easily get a homomorphism into $M(n,\mathbb{C})$ that is isometric up to a constant multiple; in fact, any homomorphism will do. However, as soon as the group is a nontrivial product, its bi-invariant metrics are not unique up to an overall constant multiple, and so most of them will not admit a homothetic homomorphism. $\endgroup$
    – Robert Bryant
    Commented Dec 2, 2012 at 13:34

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For a simple Lie group $G$ and representation $\rho:G\to GL(n,\mathbb C)$ with infinitesimal representation $\rho':\mathfrak g \to \mathfrak g\mathfrak l(n,\mathbb C)$ we have $Trace(\rho'(X).\rho'(Y)) = j_{\rho} B(X,Y)$ where $B$ is the Cartan-Killing form, for a constant $j_{\rho}$, which is called the Dynkin index (up to a possible factor due to normalization; if done right, $j_\rho$ is always an integer). This is, because the quadratic invariant functions form a 1-dim. vector space in this case.

So it is already isometric up to a conformal factor.

Edit: Multiplying it away is not possible (Thanks, Robert).

You do this for each simple part of the Lie algebra, stacking matrices, and get it for semisimple compact Lie groups. Then you play with the center.

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  • $\begingroup$ is there any reference on this where there is a precise proof? $\endgroup$
    – Dmitri
    Commented Dec 2, 2012 at 10:05
  • $\begingroup$ @Dmitri, the Dynkin index is discussed in Onishchik and Vinberg (books.google.com/…). The rest seems to be essentially a matter of definition. $\endgroup$
    – LSpice
    Commented Dec 29, 2014 at 3:00

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