A folding example

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 10 April 2012

A folding example

For the group $G=S{L}_3\left(ℂ\left(\right(t\left)\right)\right),$ $$\begin{array}{rcrrcrrcr}{x}_{{\alpha }_1}\left(c\right)& =& \left(\begin{array}{ccc}1& c& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),& h_{{\alpha }_1^{\vee }}\left(c\right)& =& \left(\begin{array}{ccc}c& 0& 0\\ 0& c^{-1}& 0\\ 0& 0& 1\end{array}\right),& n_1& =& \left(\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right),\\ x_{{\alpha }_2}\left(c\right)& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& c\\ 0& 0& 1\end{array}\right),& h_{{\alpha }_2^{\vee }}\left(c\right)& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& c& 0\\ 0& 0& c^{-1}\end{array}\right),& n_2& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right),\\ x_{{\alpha }_0}\left(c\right)& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& c\\ ct& 0& 1\end{array}\right),& h_{{\alpha }_0^{\vee }}\left(c\right)& =& \left(\begin{array}{ccc}{c}^{-1}& 0& 0\\ 0& 1& 0\\ 0& 0& c\end{array}\right),& n_0& =& \left(\begin{array}{ccc}0& 0& -t^{-1}\\ 0& 1& 0\\ t& 0& 0\end{array}\right).\end{array}$$

Let $w=s_2{s}_1{s}_0{s}_2{s}_0{s}_1{s}_0{s}_2{s}_0$ and $v=s_2{s}_1{s}_0{s}_2{s}_1{s}_2{s}_0$ so that $$w=\left(\begin{array}{ccc}{t}^2& 0& 0\\ 0& 0& 1\\ 0& -t^{-2}& 0\end{array}\right)andv=\left(\begin{array}{ccc}0& -1& 0\\ t^2& 0& 0\\ 0& 0& t^{-2}\end{array}\right).$$ We shall use Theorem 7.1 to show that the points of $IwI\cap U^{-}vI$ are $$x_2\left(c_1\right)n_2^{-1}{x}_1\left(c_2\right)n_1^{-1}{x}_0\left(c_3\right)n_0^{-1}{x}_2\left(c_4\right)n_2^{-1}{x}_0\left(c_5\right)n_0^{-1}{x}_1\left(c_6\right)n_1^{-1}{x}_0\left(c_7\right)n_0^{-1}{x}_2\left(c_8\right)n_2^{-1}{x}_0\left(c_9\right)n_0^{-1}I,$$ with $c_1,...,c_9\in ℂ$ such that $$\begin{array}{ll}{c}_1=0,c_2=0,c_3=0,c_4=0,c_5\ne 0,c_6=0,c_7\ne 0,c_9=c_7^{-1}{c}_8.& (FEg\; 1)\end{array}$$ Precisely, $$x_2\left(0\right)n_2^{-1}{x}_1\left(0\right)n_1^{-1}{x}_0\left(0\right)n_0^{-1}{x}_2\left(0\right)n_2^{-1}{x}_0\left(c_5\right)n_0^{-1}{x}_1\left(0\right)n_1^{-1}{x}_0\left(c_7\right)n_0^{-1}{x}_2\left(c_8\right)n_2^{-1}{x}_0\left(c_7^{-1}{c}_8\right)n_0^{-1}$$ is equal to $u_9{v}_9{b}_9,$ with $u_9\in U^{-}, v_9\in N, b_9\in I$ given by $$\begin{array}{ll}{u}_9=\left(\begin{array}{ccc}1& 0& 0\\ c_5^{-1}-c_5^{-2}{c}_7^{-1}{c}_{8}t& 1& 0\\ c_5^{-1}{c}_7^{-1}{t}^{-2}& 0& 1\end{array}\right),v_9=\left(\begin{array}{ccc}0& 1& 0\\ -t^2& 0& 0\\ 0& 0& t^{-2}\end{array}\right),b_9=\left(\begin{array}{ccc}{c}_5^{-1}-c_5^{-2}{c}_7^{-1}{c}_{8}t& -c_5^{-2}{c}_7^{-1}{c}_8^2& c_5^{-2}{c}_7^{-2}{c}_8^2\\ -t^2& c_5{c}_7+c_{8}t& -c_5-c_7^{-1}{c}_{8}t\\ -c_5^{-1}{c}_7^{-1}{t}^2& -c_5^{-1}{c}_7^{-1}{c}_{8}t& c_7^{-1}+c_5^{-1}{c}_7^{-2}{c}_{8}t\end{array}\right),& (FEg\; 2)\end{array}$$ so that $u_9=x_{-{\alpha }_2}\left(d_1\right)x_{-\phi }\left(d_2\right)x_{-{\alpha }_2-\delta }\left(d_4\right)x_{-{\alpha }_1}\left(d_5\right)x_{-{\alpha }_2-2\delta }\left(d_6\right)x_{-\phi -3\delta }\left(d_7\right)x_{-{\alpha }_1+\delta }\left(d_8\right)x_{-{\alpha }_2-3\delta }\left(d_9\right)$ with $$d_1=d_2=d_3=d_4=0,d_5=c_5^{-1},d_6=0,d_7=c_5^{-1}{c}_7^{-1},d_8=-c_5^{-2}{c}_7^{-1}{c}_8,d_9=0.$$ Pictorially, the walk with labels $c_1,...,c_9$ $$\begin{array}{c}\end{array}becomes\begin{array}{c}\end{array},$$ the labeled folded path with labels $d_1,...,d_9.$

The step by step computation is as follows:

Step 1

If $c_1=0$ then $$x_2\left(c_1\right)n_2^{-1}=x_{-{\alpha }_2}\left(0\right)n_2^{-1}=u_1{v}_1{b}_1,with$$ $$u_1=x_{-{\alpha }_2}\left(0\right),v_1=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right),and{b}_1=1.$$

Step 2

If $c_2=0$ then, since $v_1{x}_1\left(c_2\right)v_1^{-1}=x_{\phi }\left(c_2\right),$ $$u_1{v}_1{b}_1{x}_1\left(c_2\right)n_1^{-1}=u_1{x}_{\phi }\left(c_2\right)v_1{n}_1^{-1}{b}_1=u_1{x}_{-\phi }\left(0\right)v_1{n}_1^{-1}{b}_1=u_2{v}_2{b}_2,with$$ $$u_2=u_1{x}_{-\phi }\left(0\right),v_2=v_1{n}_1^{-1}=\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right)and{b}_2=1.$$

Step 3

If $c_3=0$ then, since $v_2{x}_0\left(c_3\right)v_2^{-1}=x_{{\alpha }_2+\delta }(-c_3),$ $$u_2{v}_2{b}_2{x}_0\left(c_3\right)n_0^{-1}=u_2{x}_{{\alpha }_2+\delta }(-c_3)v_2{n}_0^{-1}{b}_2=u_2{x}_{-{\alpha }_2-\delta }\left(0\right)v_2{n}_0^{-1}{b}_2=u_3{v}_3{b}_3,with$$ $$u_3=u_2{x}_{-{\alpha }_2-\delta }\left(0\right),v_3=v_2{n}_0^{-1}=\left(\begin{array}{ccc}0& -1& 0\\ t& 0& 0\\ 0& 0& t^{-1}\end{array}\right),and{b}_3=1.$$

Step 4

If $c_4=0$ then, since $v_3{x}_2\left(c_4\right)v_3^{-1}=x_{\phi +\delta }(-c_4),$ $$u_3{v}_3{b}_3{x}_2\left(c_4\right)n_2^{-1}=u_3{x}_{\phi +\delta }(-c_4)v_3{n}_2^{-1}{b}_3=u_3{x}_{-\phi -\delta }\left(0\right)v_3{n}_2^{-1}{b}_3=u_4{v}_4{b}_4,with$$ $$u_4=u_3{x}_{-\phi -\delta }\left(0\right),v_4=v_3{n}_2^{-1}=\left(\begin{array}{ccc}0& 0& 1\\ t& 0& 0\\ 0& t^{-1}& 0\end{array}\right)and{b}_4=1.$$

Step 5

If $c_5\ne 0$ then by the folding law and the fact that $v_4{x}_{-{\alpha }_0}\left(c_5^{-1}\right)v_4^{-1}=x_{-{\alpha }_1}\left(c_5^{-1}\right),$ $$u_4{v}_4{b}_4{x}_0\left(c_5\right)n_0^{-1}=u_4{v}_4{x}_{-{\alpha }_0}\left(c_5^{-1}\right)x_{{\alpha }_0}(-c_5)h_{{\alpha }_0^{\vee }}\left(c_5\right)b_4=u_4{x}_{-{\alpha }_1}\left(c_5^{-1}\right)v_4{b}_5=u_5{v}_5{b}_5,$$ where $$u_5=u_4{x}_{-{\alpha }_1}\left(c_5^{-1}\right),v_5=v_4,and{b}_5=x_{{\alpha }_0}(-c_5)h_{{\alpha }_0^{\vee }}\left(c_5\right)b_4=\left(\begin{array}{ccc}{c}_5^{-1}& 0& 0\\ 0& 1& 0\\ -t& 0& c_5\end{array}\right).$$

Step 6

If $c_5^{-1}{c}_6=0$ (so $c_6=0$) then $$u_5{v}_5{b}_5{x}_1\left(c_6\right)n_1^{-1}=u_5{v}_5{x}_1\left(c_5^{-1}{c}_6\right)n_1^{-1}{b\prime }_5=u_5{x}_{-{\alpha }_2-2\delta }\left(0\right)v_5{n}_1^{-1}{b\prime }_5=u_6{v}_6{b}_6,$$ with $$u_6=u_5{x}_{-{\alpha }_2-2\delta }\left(0\right),v_6=v_5{n}_1^{-1}=\left(\begin{array}{ccc}0& 0& 1\\ 0& -t& 0\\ t^{-1}& 0& 0\end{array}\right)and{b}_6={b\prime }_5=\left(\begin{array}{ccc}1& 0& 0\\ 0& c_5^{-1}& 0\\ -c_{6}t& t& c_5\end{array}\right)$$ so that $b_5{x}_1\left(c_6\right)n_1^{-1}=x_1\left(c_5^{-1}{c}_6\right)n_1^{-1}{b\prime }_5.$

Step 7

If $c_5{c}_7\ne 0$ then, since $v_6{x}_{-{\alpha }_0}\left(c\right)v_6^{-1}=x_{-\phi -2\delta }\left(c\right),$ $$\begin{array}{rcl}{u}_6{v}_6{b}_6{x}_0\left(c_7\right)n_0^{-1}& =& u_6{v}_6{x}_0\left(c_5{c}_7\right)n_0^{-1}{b\prime }_6\\ & =& u_6{v}_6{x}_{-{\alpha }_0}\left(c_5^{-1}{c}_7^{-1}\right)x_{{\alpha }_0}(-c_5{c}_7)h_{{\alpha }_0^{\vee }}\left(c_5{c}_7\right){b\prime }_6\\ & =& u_6{x}_{-\phi -2\delta }\left(c_5^{-1}{c}_7^{-1}\right)v_6{b}_7\\ & =& u_7{v}_7{b}_7,\end{array}$$ where $$u_7=u_6{x}_{-\phi -2\delta }\left(c_5^{-1}{c}_7^{-1}\right),v_7=v_6,and$$ $${b\prime }_6=\left(\begin{array}{ccc}{c}_5& -1& 0\\ 0& c_5^{-1}& 0\\ 0& 0& 1\end{array}\right)and{b}_7=x_{{\alpha }_0}(-c_5{c}_7)h_{{\alpha }_0^{\vee }}\left(c_5{c}_7\right){b\prime }_6=\left(\begin{array}{ccc}{c}_7^{-1}& -c_5^{-1}{c}_7^{-1}& 0\\ 0& c_5^{-1}& 0\\ -c_{5}t& t& c_5{c}_7\end{array}\right),$$ so that $b_6{x}_0\left(c_7\right)n_0^{-1}=x_0\left(c_5{c}_7\right)n_0^{-1}{b\prime }_6.$

Step 8

No restrictions on $c_5^{-2}{c}_7^{-1}{c}_8.$ Since $v_7{x}_{{\alpha }_2}\left(c\right)v_7^{-1}=x_{-{\alpha }_1+\delta }(-c),$ $$u_7{v}_7{b}_7{x}_2\left(c_8\right)n_2^{-1}=u_7{v}_7{x}_2\left(c_5^{-2}{c}_7^{-1}{c}_8\right)n_2^{-1}{b\prime }_7=u_7{x}_{-{\alpha }_1+\delta }(-c_5^{-2}{c}_7^{-1}{c}_8)v_7{n}_2^{-1}{b\prime }_7=u_8{v}_8{b}_8,$$ with $$u_8=u_7{x}_{-{\alpha }_1+\delta }(-c_5^{-2}{c}_7^{-1}{c}_8),v_8=v_7{n}_2^{-1}=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& t\\ t^{-1}& 0& 0\end{array}\right),and$$ $$b_8={b\prime }_7=\left(\begin{array}{ccc}{c}_7^{-1}& -c_5^{-1}{c}_7^{-1}{c}_8& c_5^{-1}{c}_7^{-1}\\ -c_{5}t& c_5{c}_7+c_{8}t& -t\\ -c_5^{-1}{c}_7^{-1}{c}_{8}t& c_5^{-2}{c}_7^{-1}{c}_8^{2}t& c_5^{-1}-c_5^{-2}{c}_7^{-1}{c}_{8}t\end{array}\right),$$ so that $b_7{x}_2\left(c_8\right)n_2^{-1}=x_2\left(c_5^{-2}{c}_7^{-1}{c}_8\right)n_2^{-1}{b\prime }_7.$

Step 9

If $c_5^{-1}{c}_7{c}_9-c_5^{-1}{c}_8=0$ (so $c_9=c_7^{-1}{c}_8$) then $$u_8{v}_8{b}_8{x}_0\left(c_9\right)n_0^{-1}=u_8{v}_8{x}_0(c_5^{-1}{c}_7{c}_9-c_5^{-1}{c}_8)n_0^{-1}{b\prime }_8=u_8{x}_{-{\alpha }_2-3\delta }\left(0\right)v_8{n}_0^{-1}{b\prime }_8=u_9{v}_9{b}_9$$ with $u_9,v_9$ and $b_9$ as in (FEg 2).

Notes and References

These notes are taken from section 8 of the paper

[PRS] J. Parkinson, A. Ram, and C. Schwer, Combinatorics in affine flag varieties, J. Algebra, 321 (2009), 3469-3493.

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