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D_\gamma for LSL_3 Part one: Lifting the Weyl group

By trdunlap2

Let me recap D_\gamma a bit. Let \gamma=w\cdot\Lambda_i where w is an element of the Weyl group and \Lambda_i is a fundamental weight. Before calculating D_\gamma we’ll need to choose weight vectors v_\gamma\in V_\gamma such that v_{w\cdot\gamma}=\bar w\cdot v_\gamma where \bar w indicates the lift of w.

For SL_3 W is basically the set of permutation matrices only I feel like there is a trouble with signs. Ignoring signs for now think of it as generated by

\bar{s_1}=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right)

and

\bar{s_2}=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right).

In that case we take

v_{\Lambda_1}=\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right) v_{\Lambda_2}=\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)\wedge\left(\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right).

(Note/check that \bar {s_i} v_{\Lambda_j}=\pm v_{\Lambda_j} when j\neq i.)

This information is more completely presented in a set of diagram I have in my notes — The fastest way to get it up will probably be to scan it on Monday.

I’ve made a similary diagram for W_{\text{aff}}. The lifts are much the same as for the finite case only there will be t’s in places. What I don’t have nailed down yet is the v_{w\cdot \Lambda_3} vectors: I don’t even know where they live.

UPDATE: Two of the Scans I promised — (a) the Weyl diagram for SL_3 and (b) the Diagram for the affine Weyl group with my guess at appropriate matrix representations (once again ignoring sign issues).

(a)sl3w.png

(b)lsl3wm.png

This entry was posted on February 17, 2008 at 11:24 pm and is filed under Current, Examples/exercises, Open. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “D_\gamma for LSL_3 Part one: Lifting the Weyl group”

  1. trdunlap2 Says:
    February 19, 2008 at 10:52 am | Reply

    Some notes on the image (b):
    There is a central hexagon consisting of the permutation matrices. If you picture the plane tesselated by these hexagons you will see that the form of the corresponding cells of each hexagon have the same configuration of their non-zero entries. Whereas within each hexagon the the corressponding rows of each matrix have the same degree. For example in the hexagon just below the center one each first row has degree 0, each second row has degree -1 and each third row has degree 1.

  2. trdunlap2 Says:
    February 19, 2008 at 11:02 am | Reply

    A guess at the v_{\Lambda_i} – I don’t think its right, but I’ll try to make my case anyway – v_{\Lambda_3} should be fixed by all permutation matrices (up to sign perhaps). v_{\Lambda_1} should be fixed by the s_2 above and by \left(\begin{array}{ccc} 0 & 0 & z \\ 0 & 1 & 0 \\ z^{-1} & 0 & 0 \end{array}\right). The idea suggests itself to me that we let v_ {\Lambda_3}=\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right) and v_{\Lambda_1}=\left(\begin{array}{c} 1 \\ z^{-1} \\ z^{-1} \end{array}\right).

  3. trdunlap2 Says:
    March 3, 2008 at 9:59 pm | Reply

    The suggestion of the previous comment leaves me with a bad taste because e.g. we may as well take (z,z,z) as take (1,1,1) for v_{\Lambda_3} or (z+1,z+1,z+1) for that matter. And indeed if v_{\Lambda_2}=(z,1,1) and v_{-\Lambda_2}=(1,z,z) then their sum is (z+1,z+1,z+1) which could be v_{\Lambda_3} — that’s crazy. Um, not to mention these should be vectors in different representations as they are on different parts of the fundamental triangle.

    Ideally all of the fundamental representations together should be \wedge (H^+\oplus \bar{H^-}) if I understand chapter 10 of Loop groups correctly. Taking these v’s would only represent some subset of H^+ when taken altogether. (By H^+ I mean \mathbb{C}[z]^3 taken as a vector space over \mathbb{C}. Or its closure or something like that.)

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