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Suppose you have a compact connected Lie group $G$, with an inner product on its Lie algebra $\mathfrak{g}$ defined by the Killing form. I've seen it written that the adjoint representation $Ad: G\rightarrow GL(\mathfrak{g})$ in fact takes values in $SO(\mathfrak{g})$.

Why is this? Many thanks.

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This is because $Ad(g)$ acts by change of basis in any representation of $\mathfrak{g}$. Since the Killing form is the trace form on the adjoint representation and trace is independent of the choice of basis, we get that $Ad(g)$ is an element of $SO(\mathfrak{g})$.

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  • $\begingroup$ Thanks! So does it land in $SO(\mathfrak{g})$ because $G$ is connected and the representation is continuous? If $G$ is not connected, could it land in $O(\mathfrak{g})$ but not $SO(\mathfrak{g})$? $\endgroup$ – ougoah Mar 25 '17 at 2:42
  • $\begingroup$ I don't think so. The reasoning above is independent of $G$ being connected. $\endgroup$ – David Hill Mar 25 '17 at 3:16
  • $\begingroup$ Does $Ad(g)$ always act by an orientation-preserving change of basis though? That it preserves the Killing form only gives you $O(\mathfrak{g})$, right? $\endgroup$ – ougoah Mar 25 '17 at 5:33
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    $\begingroup$ You are probably right. For some reason I thought conjugation would be a determinant 1 operation. $\endgroup$ – David Hill Mar 26 '17 at 20:22

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