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Let $G$ be a matrix Lie group. Its Lie algebra $\mathfrak{g}$ comes equipped with a symmetric (2,0) tensor known as its Killing from, denoted $K$ and defined by $$ K(X,Y) = -\text{Tr}(\text{ad}_X \circ \text{ad}_Y)$$

where $\text{ad}_X \in \text{GL}(\mathfrak{g})$ defined by $\text{ad}_X(Y)= [X,Y]$. Show that $K$ is Ad-invariant, in the sense that $$ K(\text{Ad}_A(X), \text{Ad}_A(Y)) = K(X,Y) \quad \forall X,Y \in \mathfrak{g}, A \in G $$.

Here $\text{Ad}_A(X) = AXA^{-1} \in \mathfrak{g}$.

I suppose that it suffices to prove it on a basis, but it isn't a big deal to let $X$ and $Y$ be arbitrary. Choosing a basis $E_j$ for $\mathfrak{g}$, I write $$ [E_j,E_k] = {c_{jk}}^l E_l $$ $$ X = a^kE_k, Y = b^kE_k $$ and $$ AE_kA^{-1} = {d_k}^lE_l. $$ Then $$ [X,[Y,E_m]] = [a^kE_k,[b^lE_l,E_m]] = a^kb^l {c_{kn}}^j {c_{lm}}^n E_j$$ and so the trace of this operator is $a^kb^l{c_{kn}}^j{c_{lj}}^n$. The left hand side is $$ [AXA^{-1},[AYA^{-1},E_m]] = [a^k{d_k}^lE_l,[b^n {d_n}^j E_j,E_m]]= a^kb^n{d_k}^l{d_n}^j {c_{jm}}^r {c_{lr}}^k E_k$$ so the trace of this operator is $$ a^kb^n{d_n}^j{d_k}^l{c_{jk}}^r{c_{lr}}^k $$ assuming the indices are correct. Now these two have to be the same somehow. Using that $\text{Ad}_A{[X,Y]}=[\text{Ad}_A{X}, \text{Ad}_A{Y}]$ may be promising, but I haven't found anything useful with it. A hint would be appreciated.

user110503
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1 Answers1

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Hint: Start by verifying (best without expanding in terms of a basis) that $ad_{Ad_A(X)}=Ad_A\circ ad_X\circ Ad_{A^{-1}}$, then use what you know about compatibility of the trace with similarity of linear maps.

Andreas Cap
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  • Thank you. I was able to show this, and therefore that $ad_{Ad_A(X)} \circ ad_{Ad_A(Y)} = Ad_A \circ ad_X \circ ad_Y \circ Ad_{A^{-1}}$. The problem is I don't view $X \to AXA^{-1}$ as a similarity transform but as a linear transformation on $\mathfrak{g}$, if that is what you mean. I'm not sure what else I can glean from it. – user110503 Aug 29 '17 at 18:42
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    The similarity argument is not applied to the linear maps $X$ and $AXA^{-1}$ but to linear maps $\mathfrak g\to\mathfrak g$. Just take into account that $Ad_{A^{-1}}=(Ad_A)^{-1}$. – Andreas Cap Aug 30 '17 at 07:08