# Two questions about the Killing Form in Fulton-Harris

I have two questions about the Killing Form that Fulton-Harris seems to leave unclear:

• First, in their proof that the killing form is non-degenerate on semi-simple Lie Algebras, they assert

Since the Killing form of a direct sum is the sum of the Killing forms of the factors, it follows that the Killing form is non-degenerate on a semisimple Lie algebra $\mathfrak{g}$

As I understand this, they mean that if $\mathfrak{g} = \mathfrak{g_1} \oplus \mathfrak{g_2}$ (just two factors for simplicity) then we have:

$$B(X_1+Y_1, X_2 +Y_2) = B(X_1,X_2) + B(Y_1, Y_2)$$ where $X_i \in \mathfrak{g_1}$ and $Y_i \in \mathfrak{g_2}$. This would require $$B(X_1,Y_2) +B(Y_1,X_2) =0$$ But, I do not see exactly why this should be the case. Could someone enlighten me? I attempted a proof that works provided that each root space is one dimensional, which I think is true in general. But, they seem to imply you don't need this.

• Secondly, they go on to do prove that if $\alpha$ is a positive root, then $\Omega_\alpha = \{ \beta \mid \beta(H_\alpha)\}$ is perpendicular to the line spanned by $\alpha$. They assert

It suffices to prove the dual assertion that $H \perp H_\alpha$ for all $H$ in the annihilator of $\alpha$.

I do not see why this should be the dual statement. I think that if $\alpha^\dagger = H_\alpha$ then this should be true (where $\cdot^\dagger$ is the map sending $\alpha$ to its dual given by the Killing Form). If this is the case, then $B(\alpha, \beta) = B(\alpha^\dagger, \beta^\dagger) = B(H_\alpha, \beta^\dagger)$. But $0=\beta(H_\alpha) = B(\beta^\dagger, H_\alpha) = \alpha(\beta^\dagger)$ , and so we are in good shape.

But, I cannot see whey $\alpha^\dagger = H_\alpha$. Is this the right route to go down? Or am I on the wrong track?

For the first question, non degenerated means that $B(X,.)$ is not zero if $X$ is not zero. Remark that if $X_1\in g_1, X_2\in g_2, B(X_1,X_2)=tr(ad(X_1)\circ ad(X_2))=0$. Write $X=X_1+X_2, X_1\in g_1, X_2\in g_2$,suppose that $X_1\neq 0$, there exists $Y_1\in g_1$ such that $B(X_1,Y_1)\neq 0$, this implies that $B(X,Y_1)=B(X_1,Y_1)+B(X_2,Y_1)=B(X_1,Y_1)\neq 0$. If $X_1=0, X_2\neq 0$ the argument is similar.
• Right, so that last bit I am completely comfortable with. The crucial fact I do not quite understand is why should $tr(ad(X_1) \circ ad(X_2)) = 0$ if $X_1$ and $X_2$ are not in the same direct summand? – msm Feb 24 '18 at 22:14
• Let $X_1,Y_1\in g_1, X_2,Y_2\in g_2$, $ad(X_1)\circ ad(X_2)(Y_1)=[X_1,[X_2,Y_1]]=0$ since $[g_1,g_2]=0$, $ad(X_1)\circ ad(X_2)(Y_2)=[X_1,[X_2,Y_2]]=0$ since $X_1\in g_1, [X_2,Y_2]\in g_2$, you deduce that $ad(X_1)\circ ad(X_2)=0$. – Tsemo Aristide Feb 24 '18 at 22:18
• Sorry to be so obtuse, but why should $[g_1, g_2]=0$? – msm Feb 24 '18 at 22:32
I think what Fulton and Harris then go on to say makes the dual problem clear. Still, they seem to allude that we do not need it to see that this is the dual problem. In any event, we know by construction that $H_\alpha = [X_\alpha, Y_\alpha]$ with $X_\alpha \in \mathfrak{g_\alpha}$ and $Y_\alpha \in \mathfrak{g_{-\alpha}}$. Then, we find $B(H_\alpha,H) = B([X_\alpha, Y_\alpha],H) = B(X_\alpha, [Y_\alpha,H]) = \alpha(H) B(X_\alpha, Y_\alpha)$ which tells us that $H_\alpha^\dagger \neq \alpha$ but is proportional to it.