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I am trying to understand lowest representations of loop groups as developed in Pressley and Segal's book. Specifically I want to be able to compute the weight spaces that appear in a lowest weight representation. I realize there is a formula for this -my question is along the lines of how to apply the formula correctly.

I tried to do a small example with $LSL_3$ (actually $\mathbb{C}^\times_{\theta} \ltimes \tilde LSL_3$) and something fishy happened so I was hoping someone could point out my mistake. The maximal torus is $\mathbb{C}^\times_\theta \times T \times \mathbb{C}^\times$ where $\mathbb{C}^\times_\theta$ is the loop rotations and the other $\mathbb{C}^\times$ is central.

The fundamental weights are $w_0 = (0,0,1)$, $w_1 = (0,-\omega_1,1)$, $w_2 = (0, -\omega_2,1)$. The positive roots are $(0,\alpha_1,0), (0,\alpha_2,0),(1,-\alpha_3,0)$. Where $\omega_i,\alpha_i$ are fundamental weights and positive roots of $SL_3$. Pressley and Segal normalize the Killing form so $\langle H_{\alpha_i},H_{\alpha_i}\rangle = 2$. Choosing coordinates $H_{\alpha_1} = [1\ \ 0]^T$, $H_{\alpha_2} = [0\ \ 1]^T$, $\alpha_1 = [2 \ \ -1]$, $\alpha_2 = [-1 \ \ 2]$, $\omega_1 = [1\ \ 0]$, $\omega_2 = [0\ \ 1]$ and the restriction of the Killing form to the torus is just the Cartan matrix $(B_{11} = B_{22} = 2, B_{12} = B_{21} = -1)$.

I'm interested in the representation $V_{\tilde \lambda}$ of lowest weight $\tilde\lambda = (0, - \alpha_3,3) = w_0 + w_1 + w_2$. Let $\tilde \mu = (m,\mu, 3)$ be a weight of $V_{\tilde \lambda}$. According to Loop Groups (11.1.1) it is the case that $\tilde \mu - \tilde \lambda = (m,\mu +\alpha_3, 0)$ is a sum of positive roots. Viewing $B$ as a map from co-characters to characters and noting that $\alpha_i = BH_{\alpha_i}$ it follows that we can write $\mu = B[a\ \ b]^T$ for some $a,b$.

According to (9.3.7) on pg 180 of Loop Groups the $\tilde\mu =(m, \mu,3)$ which satisfy

$3\langle \mu,\mu\rangle - 6m = 6 = 3\langle -\alpha_3,-\alpha_3\rangle$

appear among the weight of $V_{\tilde \lambda}$. This says

$m = {1 \over 2}\langle\mu,\mu\rangle-1 = {1\over 2}[a\ \ b]B[a\ \ b]^T - 1 = a^2 + b^2- ab -1$.

Taking $a,b = 0$ produces the weight $\tilde \mu = (-1, 0, 3)$ but then $\tilde \mu - \tilde \lambda = -(1,-\alpha_3,0)$. Which is certainly not a sum of positive roots. So what gives?

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I'm tempted in a situation like this to suggest that you try contacting one or both authors by email. That doesn't invariably get a helpful response (or any response), but on the other hand the authors of books have a vested interest in clarifying things for their readers. –  Jim Humphreys May 5 '11 at 14:32
    
Why is the equation (9.3.7) as you have it? I would have thought $||\mu||^2-2mh=||\widetilde \lambda||^2$ should be $\langle \mu, \mu \rangle - 6m = \langle -\alpha_3, -\alpha_3 \rangle=2$ –  charris May 6 '11 at 0:06
    
@charris My original thinking was that for a level $h$ representation you have to take $h$ times the standard pairing. But perhaps its as you say; that would certainly prevent negative values of $m$. –  solbap May 6 '11 at 19:51
    
@charris actually that business of taking $h$ times the standard pairing is something I confused with representation of loop groups of tori (9.5.10) so I think you are absolutely right. –  solbap May 6 '11 at 20:31
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1 Answer

up vote 4 down vote accepted

The formula for the invariant bilinear form is given in $(4.9.3)$ on page 64 $$\langle (x_1,\xi_1, y_1),(x_2,\xi_2,y_2) \rangle=\langle \xi_1, \xi_2 \rangle - x_1 y_2-y_1x_2$$ As I mentioned in the comments, $(9.3.7)$ becomes then $||\mu||^2-6m=2$. So your last equation would be $m=\frac{1}{3}(a^2-ab+b^2)-\frac{1}{3}$. As you said, there's no more worries about negative $m$ and as a consistency check, for $m=0$, the solutions for $[a \ \ b]$ are $[\pm 1 \ \ 0]$, $[0 \ \ \pm 1]$, $[1 \ \ 1 ]$, and $[-1 \ \ -1]$. Applying, $B$ gives you the six weights in the Weyl orbit of $-\alpha_3$ (the roots).

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