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). This is, because the quadratic bi-invariant functions form a 1-dim. vector space in this case.

So it is already isometric up to a conformal factor. Multiply it away. 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.

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.

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). So it is already isometric up to a conformal factor. You do this for each simple part of the Lie algebra, stacking matrices, and get it for semisimple. Then you play with the center.

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). This is, because the quadratic bi-invariant functions form a 1-dim. vector space in this case.

So it is already isometric up to a conformal factor. Multiply it away. 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.