2

I'm reading the Fulton and Harris Representation Theory book, trying to learn about Lie Algebras.

On pg. 213, they compute the killing form on $\mathfrak{h}^*$ for $\mathfrak{sl}_n(\mathbb{C})$. I understand the computation for $\mathfrak{h}$, and I know that for $h_1,h_2 \in \mathfrak{h}^*$, $B(h_1,h_2) = B(X,Y)$ where $h_1 = B(X,\cdot)$ and $h_2=B(Y,\cdot)$, but for the life of me I can't figure out how to compute the $\mathfrak{h}^*$ killing form?

Also, on pg. 162 when they discuss the representations of $\mathfrak{sl}_3(\mathbb{C})$, they say that since we know diagonalizable commuting matrices are simultaneously diagonalizable, using the Jordan Decomposition Theorem we know that $\rho(H)$ admits a direct sum decomposition of $V$ into eigenspaces. I know that by the Jordan Decomposition theorem, $H$ diagonalizable implies $\rho(H)$ is diagonalizable, but if $H_1$ and $H_2$ commute in $\mathfrak{h}$, how do I know $\rho(H_1)$ and $\rho(H_2)$ commute?

Bill Cook
  • 29,244
Lauren
  • 21
  • 2
    when you have two separate questions you'll usually get better feedback if you ask them separately, but your questions are interesting so, I may be proved wrong here shortly. In any event, welcome to the MSE. – James S. Cook Oct 05 '13 at 19:32

1 Answers1

0

I don't have Fulton and Harris around right now so I'm not sure of exactly what you're looking at. However, for your first question, Killing forms are forms on Lie algebras, so there is a Killing form for $\mathfrak{h}$ (namely $\kappa(x,y) = \mathrm{tr}(\mathrm{ad}_x \circ \mathrm{ad}_y)$). By the "Killing form on $\mathfrak{h}^*$" the authors must be referring to the bilinear form induced by the Killing form on $\mathfrak{h}$ (which is induced in the way you indicated in your question: If $h_1=\kappa(x,\cdot)$ and $h_2=\kappa(y,\cdot)$, then $\kappa(h_1,h_2)=\kappa(x,y)=\mathrm{tr}(\mathrm{ad}_x \circ \mathrm{ad}_y)$). So to compute the "killing form on $\mathfrak{h}^*$" you merely need to compute the killing form on $\mathfrak{h}$ and use your isomorphism from $\mathfrak{h}$ to $\mathfrak{h}^*$: $h \mapsto \kappa(h,\cdot)$ to translate.

Your second question is a bit simpler to answer. Let $\rho:\mathfrak{g} \to \mathfrak{gl}(V)$ be a representation of some Lie algebra $\mathfrak{g}$ on some vector space $V$. Then for all $x,y \in \mathfrak{g}$, $\rho([x,y]) = [\rho(x),\rho(y)] = \rho(x) \circ \rho(y) - \rho(y) \circ \rho(x)$ (this is part of the definition of a representation). So if $[x,y]=0$ ($x$ and $y$ commute) you have $\rho(x) \circ \rho(y) - \rho(y) \circ \rho(x) = [\rho(x),\rho(y)] = \rho([x,y]) = \rho(0)=0$ so that $\rho(x) \circ \rho(y) = \rho(y) \circ \rho(x)$.

So since Cartan subalgebras like $\mathfrak{h}$ are Abelian Lie algebras, $\rho(\mathfrak{h})$ is a collection of commuting endomorphisms.

Bill Cook
  • 29,244
  • Ohh okay, that makes sense about why Cartan subalgebras are abelian. Thanks! I think what is messing me up about the Killing Form, is that we want to find $B(\sum a_iL_i,\sum b_i L_i)$ where $L_1+\ldots+L_n=0$. So under my isomorphism, $a_1L_1+...+a_nL_n=B(\sum\frac{1}{2n}a_iH_i,-)$, but I also have the equivalence class part to account for... and I'm not sure how the formula changes when taking that part into consideration. – Lauren Oct 06 '13 at 19:31
  • Ohh I figured it out! What I said above is wrong, because we may have $a_1+\ldots +a_n\neq 0$ in which case $\sum\frac{1}{2n}a_iH_i$ is not in $h$. Instead, given an element in $h^$ $\sum a_iL_i$ it equals $\sum a_iL_i - (a_1+\ldots+a_n)\sum L_i$, since $\sum L_i=0$ which under our isomorphism corresponds to the element $B(\sum\frac{1}{2n}(a_i-\frac{a_1+\ldots +a_n}{n}H_i,-)$. Using this correspondence, I do get the correct formula for the Killing Form on $h^$ given on pg. 99 of Fulton and Harris. Phew. – Lauren Oct 07 '13 at 16:26