lowest weight representation of loop groups - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T18:01:41Zhttp://mathoverflow.net/feeds/question/63978http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/63978/lowest-weight-representation-of-loop-groupslowest weight representation of loop groupssolbap2011-05-05T08:36:21Z2011-05-06T20:21:34Z
<p>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.</p>
<p>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. </p>
<p>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 $(B_{11} = B_{22} = 2, B_{12} = B_{21} = -1)$.</p>
<p>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 <em>Loop Groups</em> (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 $a,b$.</p>
<p>According to (9.3.7) on pg 180 of <em>Loop Groups</em> the $\tilde\mu =(m, \mu,3)$ which satisfy</p>
<p>$3\langle \mu,\mu\rangle - 6m = 6 = 3\langle -\alpha_3,-\alpha_3\rangle$</p>
<p>appear among the weight of $V_{\tilde \lambda}$. This says </p>
<p>$m = {1 \over 2}\langle\mu,\mu\rangle-1 = {1\over 2}[a\ \ b]B[a\ \ b]^T - 1 = a^2 + b^2- ab -1$. </p>
<p>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 <em>sum</em> of positive roots. So what gives?</p>
http://mathoverflow.net/questions/63978/lowest-weight-representation-of-loop-groups/64153#64153Answer by charris for lowest weight representation of loop groupscharris2011-05-06T20:21:34Z2011-05-06T20:21:34Z<p>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).</p>