## A folding example

Last update: 10 April 2012

## A folding example

For the group $G=S{L}_{3}\left(ℂ\left(\left(t\right)\right)\right),$ $xα1 (c) = 1 c 0 0 1 0 0 0 1 , hα1∨ (c) = c 0 0 0 c-1 0 0 0 1 , n1 = 0 1 0 -1 0 0 0 0 1 , xα2 (c) = 1 0 0 0 1 c 0 0 1 , hα2∨ (c) = 1 0 0 0 c 0 0 0 c-1 , n2 = 1 0 0 0 0 1 0 -1 0 , xα0 (c) = 1 0 0 0 1 c ct 0 1 , hα0∨ (c) = c-1 0 0 0 1 0 0 0 c , n0 = 0 0 -t-1 0 1 0 t 0 0 .$

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 $w = t2 0 0 0 0 1 0 -t-2 0 and v = 0 -1 0 t2 0 0 0 0 t-2 .$ We shall use Theorem 7.1 to show that the points of $IwI\cap {U}^{-}vI$ are $x2(c1) n2-1 x1(c2) n1-1 x0(c3) n0-1 x2(c4) n2-1 x0(c5) n0-1 x1(c6) n1-1 x0(c7) n0-1 x2(c8) n2-1 x0(c9) n0-1 I,$ with ${c}_{1},...,{c}_{9}\in ℂ$ such that $c1=0, c2=0, c3=0, c4=0, c5≠0, c6=0, c7≠0, c9 = c7-1c8. (FEg 1)$ Precisely, $x2(0) n2-1 x1(0) n1-1 x0(0) n0-1 x2(0) n2-1 x0(c5) n0-1 x1(0) n1-1 x0(c7) n0-1 x2(c8) n2-1 x0(c7-1c8) n0-1$ is equal to ${u}_{9}{v}_{9}{b}_{9},$ with given by $u9 = 1 0 0 c5-1-c5-2c7-1c8t 1 0 c5-1c7-1t-2 0 1 , v9 = 0 1 0 -t2 0 0 0 0 t-2 , b9 = c5-1-c5-2c7-1c8t -c5-2c7-1c82 c5-2c7-2c82 -t2 c5c7+c8t -c5-c7-1c8t -c5-1c7-1t2 -c5-1c7-1c8t c7-1+c5-1c7-2c8t , (FEg 2)$ 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 $d1 = d2 = d3 = d4 = 0, d5 = c5-1, d6 = 0, d7 = c5-1 c7-1, d8 = -c5-2 c7-1 c8, d9 = 0.$ Pictorially, the walk with labels ${c}_{1},...,{c}_{9}$ $becomes ,$ 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 $x2(c1) n2-1 = x-α2(0) n2-1 = u1v1b1, with$ $u1 = x-α2(0), v1 = 1 0 0 0 0 -1 0 1 0 , and b1 = 1.$

### Step 2

If ${c}_{2}=0$ then, since ${v}_{1}{x}_{1}\left({c}_{2}\right){v}_{1}^{-1}={x}_{\phi }\left({c}_{2}\right),$ $u1v1b1 x1(c2) n1-1 = u1xφ(c2)v1 n1-1 b1 = u1x-φ(0) v1 n1-1 b1 = u2v2b2, with$ $u2 = u1x-φ(0), v2 = v1n1-1 = 0 -1 0 0 0 -1 1 0 0 and b2 = 1.$

### Step 3

If ${c}_{3}=0$ then, since ${v}_{2}{x}_{0}\left({c}_{3}\right){v}_{2}^{-1}={x}_{{\alpha }_{2}+\delta }\left(-{c}_{3}\right),$ $u2v2b2 x0(c3) n0-1 = u2 xα2+δ(-c3) v2 n0-1 b2 = u2 x-α2-δ(0) v2 n0-1 b2 = u3v3b3, with$ $u3 = u2x-α2-δ(0), v3 = v2n0-1 = 0 -1 0 t 0 0 0 0 t-1 , and b3 = 1.$

### Step 4

If ${c}_{4}=0$ then, since ${v}_{3}{x}_{2}\left({c}_{4}\right){v}_{3}^{-1}={x}_{\phi +\delta }\left(-{c}_{4}\right),$ $u3v3b3 x2(c4) n2-1 = u3xφ+δ(-c4) v3 n2-1 b3 = u3x-φ-δ(0) v3 n2-1 b3 = u4v4b4, with$ $u4 = u3x-φ-δ(0), v4 = v3n2-1 = 0 0 1 t 0 0 0 t-1 0 and b4 = 1.$

### Step 5

If ${c}_{5}\ne 0$ then by the folding law and the fact that ${v}_{4}{x}_{-{\alpha }_{0}}\left({c}_{5}^{-1}\right){v}_{4}^{-1}={x}_{-{\alpha }_{1}}\left({c}_{5}^{-1}\right),$ $u4v4b4 x0(c5) n0-1 = u4v4 x-α0(c5-1) xα0(-c5) hα0∨(c5) b4 = u4 x-α1(c5-1) v4b5 = u5v5b5,$ where $u5 = u4 x-α1(c5-1), v5 = v4, and b5 = xα0(-c5) hα0∨(c5) b4 = c5-1 0 0 0 1 0 -t 0 c5 .$

### Step 6

If ${c}_{5}^{-1}{c}_{6}=0$ (so ${c}_{6}=0$) then $u5v5b5 x1(c6) n1-1 = u5v5 x1(c5-1c6) n1-1 b′5 = u5 x-α2-2δ(0) v5 n1-1 b′5 = u6v6b6,$ with $u6 = u5 x-α2-2δ(0), v6 = v5n1-1 = 0 0 1 0 -t 0 t-1 0 0 and b6 = b′5 = 1 0 0 0 c5-1 0 -c6t t c5$ 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 ${v}_{6}{x}_{-{\alpha }_{0}}\left(c\right){v}_{6}^{-1}={x}_{-\phi -2\delta }\left(c\right),$ $u6v6b6 x0(c7) n0-1 = u6v6 x0(c5c7) n0-1 b′6 = u6v6 x-α0(c5-1c7-1) xα0(-c5c7) hα0∨(c5c7) b′6 = u6 x-φ-2δ(c5-1c7-1) v6b7 = u7v7b7,$ where $u7 = u6 x-φ-2δ(c5-1c7-1), v7 = v6, and$ $b′6 = c5 -1 0 0 c5-1 0 0 0 1 and b7 = xα0(-c5c7) hα0∨(c5c7) b′6 = c7-1 -c5-1c7-1 0 0 c5-1 0 -c5t t c5c7 ,$ 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 ${v}_{7}{x}_{{\alpha }_{2}}\left(c\right){v}_{7}^{-1}={x}_{-{\alpha }_{1}+\delta }\left(-c\right),$ $u7v7b7 x2(c8) n2-1 = u7v7 x2(c5-2c7-1c8) n2-1 b′7 = u7 x-α1+δ(-c5-2c7-1c8) v7 n2-1 b′7 = u8v8b8,$ with $u8 = u7 x-α1+δ(-c5-2c7-1c8), v8 = v7n2-1 = 0 1 0 0 0 t t-1 0 0 , and$ $b8 = b′7 = c7-1 -c5-1c7-1c8 c5-1c7-1 -c5t c5c7+c8t -t -c5-1c7-1c8t c5-2c7-1c82t c5-1-c5-2c7-1c8t ,$ 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 $u8v8b8 x0(c9) n0-1 = u8v8 x0(c5-1c7c9 - c5-1c8) n0-1 b′8 = u8 x-α2-3δ(0) v8 n0-1 b′8 = u9v9b9$ 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.