I'm working through Ziller's Lie Groups. Representation Theory and Symmetric Spaces, and in Proposition 1.36, he shows the following identity:

Let $\mathfrak{g}$ be a real or complex [finite-dimensional] Lie algebra with Killing form $B$... If $L \in \mathfrak{Der}(\mathfrak{g})$, then $B(LX, Y) + B(X, LY) = 0$.

He's already demonstrated that if $A \in \operatorname{Aut}(\mathfrak{g})$, then $B(AX, AY) = B(X, Y)$. He says that if $L$ is a derivation, then $e^{tL}$ is an automorphism of $\mathfrak{g}$, so $B\left( e^{tL} X , e^{tL} Y \right) = B(X, Y)$. This is where he loses me.

Differentiating at $t = 0$ proves [the] claim.

I don't understand how I'd flesh this step out. I think he wants to do something like this.

\begin{align*} B(X, Y) & = B \left( e^{tL} X , e^{tL} Y \right) \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} B(X, Y) & = \frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} B \left( e^{tL} X , e^{tL} Y \right) \\ = 0 & = \frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} \operatorname{tr} \left( \operatorname{ad}_{e^{tL} X} \circ \operatorname{ad}_{e^{tL} Y} \right) , \end{align*}

and somehow get to $\frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} B \left( e^{tL} X , e^{tL} Y \right) = B(LX, Y) + B(X, LY)$ or something, but I just don't know what to do with this. I expect it's a fairly straightforward computation, but I just don't have an intuition for where to take this.

Thanks in advance!

  • 1
    $\begingroup$ Write $B(e^{tL}X,e^{tL}Y)=B((1+tL+o(t))X,(1+tL+o(t))Y)=B(X,Y)+t(B(LX,Y)+B(X,LY))+o(t)$. Here the Landau notation is meant $X,Y,L$ being fixed, and $t\to 0$. $\endgroup$ – YCor Mar 24 at 20:04
  • $\begingroup$ @YCor Thanks, this makes sense! Could you perhaps explain how I could finish the above argument? $\endgroup$ – AJY Mar 24 at 21:42
  • $\begingroup$ how? I guess, by trying. $\endgroup$ – YCor Mar 24 at 22:07
  • $\begingroup$ @YCor how could I phrase what you wrote in terms of differential maps and the like? $\endgroup$ – AJY Mar 24 at 22:08

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