# Isometric embedding of a compact Lie Group in $M(n,\mathbb{C})$

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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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) ? – Amin Dec 2 '12 at 9:25
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 ? – Dmitri Dec 2 '12 at 9:36
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. – Robert Bryant Dec 2 '12 at 13:34

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.