## Unipotent Hecke algebras: the structure, representation theory, and combinatorics

Last updated: 26 March 2015

## A basis with multiplication in the $G={GL}_{n}\left({𝔽}_{q}\right)$ case

### Unipotent Hecke algebras

#### The group ${GL}_{n}\left({𝔽}_{q}\right)$

Let $G={GL}_{n}\left({𝔽}_{q}\right)$ be the general linear group over the finite field ${𝔽}_{q}$ with $q$ elements. Define the subgroups $T={diagonal matrices}, N={monomial matrices}, W={permutation matrices},and U={(1*⋱01)}, (4.1)$ where a monomial matrix is a matrix with exactly one nonzero entry in each row and column $\text{(}N=WT\text{).}$ If necessary, specify the size of the matrices by a subscript such as ${G}_{n},{U}_{n},{W}_{n},$ etc. If $a\in {G}_{m}$ and $b\in {G}_{n}$ are matrices, then let $a\oplus b\in {G}_{m+n}$ be the block diagonal matrix $a⊕b= (a00b).$

Let ${x}_{ij}\left(t\right)\in U$ be the matrix with $t$ in position $\left(i,j\right),$ ones on the diagonal and zeroes elsewhere; write ${x}_{i}\left(t\right)={x}_{i,i+1}\left(t\right)\text{.}$ Let ${h}_{{\epsilon }_{i}}\left(t\right)\in T$ denote the diagonal matrix with $t$ in the $i\text{th}$ slot and ones elsewhere, and let ${s}_{i}\in W\subseteq N$ be the identity matrix with the $i\text{th}$ and $\left(i+1\right)\text{st}$ columns interchanged. That is, $xi(t) = Idi-1⊕ (1t01)⊕ Idn-i-1, hεi(t) = Idi-1⊕(t)⊕ Idn-i, si = Idi-1⊕ (0110) ⊕Idn-i-1, (4.2)$ where ${\text{Id}}_{k}$ is the $k×k$ identity matrix. Then $W=⟨s1,s2,…,sn-1⟩, T=⟨hεi(t) | 1≤i≤n,t∈𝔽q*⟩, N=WT, U=⟨xij(t) | 1≤i The Chevalley group relations for $G$ are (see also Section 3.3.1) $xij(a) xrs(b) = xij(b) xrs(a) xis (δjrab) xrj (-δisab), (i,j)≠(r,s), (U1) xij(a) xij(b) = xij(a+b), (U2) si2 = 1, (N1) sisi+1si = si+1sisi+1 andsisj= sjsi, ∣i-j∣>1, (N2) hεi(b) hεi(a) = hεi(ab), (N4) hεj(b) hεi(a) = hεj(b) hεi(a), (N5) srxij(t) = xsr(i)sr(j) (t)sr, (UN1) xij(a) hεr(t) = hεr(t) xij(t-δritδrja), (UN2) sixi(t)si = xi(t-1) sixi(-t) hεi(t) hεi+1 (-t-1), t≠0, (UN3)$ where ${\delta }_{ij}$ is the Kronecker delta.

#### A pictorial version of ${GL}_{n}\left({𝔽}_{q}\right)$

For the results that follow, it will be useful to view these elements of $ℂG$ as braid-like diagrams instead of matrices. Consider the following depictions of elements by diagrams with vertices, strands between the vertices, and various objects that slide around on the strands. View $si as \cdots \cdots i , (4.4) hεi(t) as \cdots \cdots 1 1 t 1 1 i , (4.5) xij(ab) as \cdots \cdots \cdots \underset{\circ }{\overset{\to }{a}} \underset{←}{\overset{\circ }{b}} i j , (4.6)$ where each diagram has two rows of $n$ vertices. Multiplication in $G$ corresponds to the concatenation of two diagrams; for example, ${s}_{2}{s}_{1}$ is $s2s1= {s}_{1} {s}_{2} = = .$ In the following Chevalley relations, curved strands indicate longer strands, so for example (UN1) indicates that $\underset{\circ }{\overset{\to }{a}}$ and $\underset{←}{\overset{\circ }{b}}$ slide along the strands they are on (no matter how long). The Chevalley relations are $\colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{1}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} = \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{1}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} (U1) \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{1}}} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{1}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} = \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{1}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{a+b}}} (U2) = (N1) = and = (N2) \colorbox[rgb]{1,1,1}{a} \colorbox[rgb]{1,1,1}{b} = \colorbox[rgb]{1,1,1}{ab} (N4) \colorbox[rgb]{1,1,1}{a} \colorbox[rgb]{1,1,1}{b} = \colorbox[rgb]{1,1,1}{a} \colorbox[rgb]{1,1,1}{b} (N5) \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} = \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} (UN1) \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} r s = \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{‾}{{ar}^{-1}}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{bs}}} r s (UN2) \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} = \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{b}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{-a}}} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{{b}^{-1}}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{{a}^{-1}}}} ab -{\left(ab\right)}^{-1} ab≠0. (UN3)$

#### The unipotent Hecke algebra ${ℋ}_{\mu }$

Fix a nontrivial group homomorphism $\psi :{𝔽}_{q}^{+}\to {ℂ}^{*},$ and a map $μ: {(i,j) | 1≤i Then $ψμ: U ⟶ ℂ* xij(t) ⟼ ψ(μijt) (4.8)$ is a group homomorphism. Since ${\mu }_{ij}=0$ for all $j\ne i+1,$ write $μ=(μ(1),μ(2),…,μ(n-1),μ(n)), where μ(i)= μi,i+1 and μ(n)=0. (4.7)$

The unipotent Hecke algebra ${ℋ}_{\mu }$ of the triple $\left(G,U,{\psi }_{\mu }\right)$ is $ℋμ=EndG (IndUG(ψμ)) ≅eμℂGeμ, whereeμ=1∣U∣ ∑u∈Uψμ (u-1)u. (4.9)$

The type of a linear character ${\psi }_{\mu }:U\to {ℂ}^{*}$ is the composition $\nu$ such that $μ(i)=0if and only if i∈ν≤(with ν≤ as in Section 2.3.2).$

Let $\nu$ be a composition. Suppose $\chi$ and $\chi \prime$ are two type $\nu$ linear characters of $U\text{.}$ Then $ℋ(G,U,χ)≅ ℋ(G,U,χ′).$

Proof.

Note that for $h=\text{diag}\left({h}_{1},{h}_{2},\dots ,{h}_{n}\right)\in T,$ $hxij(t)h-1= xij(hihj-1t) for 1≤i Thus, $T$ normalizes ${U}_{ij}=⟨{x}_{ij}\left(t\right) | t\in {𝔽}_{q}⟩$ and acts on linear characters of $U$ by $hχ(u)=χ (huh-1), for h∈T,u∈U.$ This action preserves the type of $\chi \text{.}$ The map $IndUG(χ) ≅ ℂGeχ ⟶∼ ℂGeχ′ ≅ IndUG(χ′) geχ ⟼ gheχ′ where χ′=hχ for h∈T,$ is a $G\text{-module}$ isomorphism, so $ℋ\left(G,U,\chi \right)\cong ℋ\left(G,U,{}^{h}\chi \right)\text{.}$ It therefore suffices to prove that $T$ acts transitively on the linear characters of type $\mu \text{.}$

By Proposition 3.2 (c), the number of $T\text{-orbits}$ of type $\mu$ linear characters is $∣Zμ∣(q-1)n-ℓ(μ) ∣T∣ = ∣{h∈T | hxi(t)h-1=xi(t),μ(i)≠0}∣(q-1)n-ℓ(μ) (q-1)n = (q-1)ℓ(μ)(q-1)n-ℓ(μ) (q-1)n = 1.$ Therefore $T$ acts transitively on the type $\mu$ linear characters of $U\text{.}$

$\square$

Proposition 4.1 implies that given any fixed character $\psi :{𝔽}_{q}^{+}\to {ℂ}^{*},$ it suffices to specify the type of the linear character ${\psi }_{\mu }\text{.}$ Note that the map (given by example) ${Compositionsμ=(μ1,μ2,…,μℓ)⊨n} ⟷ {μ=(μ(1),μ(2),…,μ(n))μ(i)∈{0,1} and μ(n)=0} μ= 1 0 1 1 1 1 0 1 1 0 1 1 1 0 ⟷ (1,0,1,1,1,1,0,1,1,0,1,1,1,0) (4.10)$ is a bijection. In the following sections identify the compositions $\mu =\left({\mu }_{1},\dots ,{\mu }_{\ell }\right)$ with the map $\mu =\left({\mu }_{\left(1\right)},\dots ,{\mu }_{\left(n\right)}\right)$ using this bijection.

The classical examples of unipotent Hecke algebras are the Yokonuma algebra ${ℋ}_{\left({1}^{n}\right)}$ [Yok1969-2] and the Gelfand-Graev Hecke algebra ${ℋ}_{\left(n\right)}$ [Ste1967].

### An indexing for the standard basis of ${ℋ}_{\mu }$

The group $G$ has a double-coset decomposition $G=\underset{v\in N}{⨆}UvU,$ so if $Nμ = {v∈N | eμveμ≠0} = {v∈N | u,vuv-1∈U implies ψμ(u)=ψμ(vuv-1)} (4.11)$ then the isomorphism in (4.9) implies that the set $\left\{{e}_{\mu }v{e}_{\mu } | v\in {N}_{\mu }\right\}$ is a basis for ${ℋ}_{\mu }$ [CRe1981, Prop. 11.30].

Suppose $a\in {M}_{\ell }\left({𝔽}_{q}\left[X\right]\right)$ is an $\ell ×\ell$ matrix with polynomial entries. Let $d\left({a}_{ij}\right)$ be the degree of the polynomial ${a}_{ij}\text{.}$ Define the degree row sums and the degree column sums of $a$ to be the compositions $d→(a)= ( d→(a)1, d→(a)2,…, d→(a)ℓ ) and d↓(a)= ( d↓(a)1, d↓(a)2,…, d↓(a)ℓ ) ,$ where $d→(a)i= ∑j=1ℓd (aij)and d↓(a)j =∑i=1ℓd (aij).$ Let $Mμ= { a∈Mℓ(μ) (𝔽q[X]) | d→(a) =d↓(a)=μ ,aij monic, aij(0) ≠0 } . (4.12)$ For example, $( X+1 11 X+2 ) (1+0+0+1=2) X+3 X3+2X+3 1 X+2 (1+3+0+1=5) 1 X2+4X+2 X+2 1 (0+2+1+0=3) 1 1 X2+3X+1 X2+2 (0+0+2+2=4) (1+1+0+0=2) (0+3+2+0=5) (0+0+1+2=3) (1+1+0+2=4) ∈M(2,5,3,4). (*)$

Suppose $\left(f\right)=\left({a}_{0}+{a}_{1}{X}^{{i}_{1}}+{a}_{2}{X}^{{i}_{2}}+\cdots +{a}_{r}{X}^{{i}_{r}}+{X}^{n}\right)\in {M}_{\left(n\right)}$ is a $1×1$ matrix with ${a}_{0},{a}_{1},\dots ,{a}_{r}\ne 0\text{.}$ Let $v(f)= \left(f\right) =w(n) ( w(i1)a0⊕ w(i2-i1) a1⊕⋯⊕w(n-ir) ar ) ∈N, (4.13)$ where $w(k)= \cdots \cdots ∈Wk,$ and by convention ${v}_{\left(1\right)}$ is the empty strand (not a strand). For example, ${v}_{\left(a+b{X}^{3}+c{X}^{4}+{X}^{6}\right)}$ is $\left(a+b{X}^{3}+c{X}^{4}+{X}^{6}\right) = a a a b c c = a a a b c c .$

For each $a\in {M}_{\mu }$ define a matrix ${v}_{a}\in N$ pictorially by $va= \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \left({a}_{\ell 1}\right) \left({a}_{21}\right) \left({a}_{11}\right) \left({a}_{\ell 2}\right) \left({a}_{22}\right) \left({a}_{12}\right) \left({a}_{\ell \ell }\right) \left({a}_{2\ell }\right) \left({a}_{1\ell }\right) , (4.14)$ where ${a}_{ij}$ is the $\left(i,j\right)\text{th}$ entry of $a,$ and the top vertex associated with ${a}_{ij}$ goes to the bottom vertex associated with ${a}_{ji}\text{.}$ Note that if ${a}_{ij}=1,$ then both its strand and corresponding vertices vanish. The fact that $a\in {M}_{\mu }$ guarantees that we are left with two rows of exactly $n$ vertices.

For example, if $a$ is as in $\left(*\right),$ then ${v}_{a}$ is $\left(3+X\right) \left(1+X\right) \left(2+4X+{X}^{2}\right) \left(3+2X+{X}^{3}\right) \left(1+3X+{X}^{2}\right) \left(2+X\right) \left(2+{X}^{2}\right) \left(2+X\right) \left(2+X\right) = \left[3\right] \left[1\right] \left[2 4\right] \left[3 2 2\right] \left[1 3\right] \left[2\right] \left[2 2\right] \left[2\right] \left[2\right] .$

Viewed as a matrix, $va= \stackrel{\phantom{\rule{15em}{0ex}}}{⏞} \stackrel{\phantom{\rule{15em}{0ex}}}{⏞} \stackrel{\phantom{\rule{15em}{0ex}}}{⏞} \cdots {\mu }_{1} \text{columns} {\mu }_{2} \text{columns} {\mu }_{3} \text{columns} \right\} \right\} \right\} ⋮ {\mu }_{1} \text{rows} {\mu }_{2} \text{rows} {\mu }_{3} \text{rows} ( v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) 0 v(a11) v(a11) ⋯ v(a11) v(a11) 0 v(a11) v(a11) 0 v(a11) v(a11) v(a13) v(a11) 0 v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a12) v(a11) 0 v(a11) 0 v(a11) 0 v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) 0 v(a11) 0 v(a11) 0 v(a11) 0 v(a11) 0 v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) 0 v(a11) 0 v(a11) 0 v(a11) ⋯ v(a11) v(a11) 0 v(a11) 0 v(a11) 0 v(a11) v(a11) v(a23) 0 v(a11) 0 v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a22) 0 0 0 0 v(a11) 0 v(a11) v(a11) v(a11) v(a21) 0 0 0 0 0 0 0 v(a11) 0 v(a11) v(a11) v(a11) v(a11) v(a11) 0 0 0 0 v(a11) 0 0 v(a11) ⋯ v(a11) 0 0 0 v(a11) v(a33) 0 0 v(a11) 0 v(a11) v(a11) v(a11) v(a11) v(a32) 0 0 0 0 0 0 v(a11) 0 v(a11) v(a11) v(a31) 0 0 0 0 0 0 0 0 0 0 v(a11) 0 v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) ⋮ 0 0 0 ⋮ 0 0 0 ⋮ 0 0 0 v(a11) ⋱ v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) v(a11) )$

Let ${N}_{\mu }$ be as in (4.11) and ${M}_{\mu }$ be as in (4.12). The map $Mμ ⟶ Nμ a ⟼ va,$ given by (4.14) is a bijection.

Remark. When $\mu =\left(n\right)$ this theorem says that the map $\left(f\right)↦{v}_{\left(f\right)}$ of (4.13) is a bijection between ${M}_{\left(n\right)}$ and ${N}_{\left(n\right)}\text{.}$

Proof.

Using the remark following the theorem, it is straightforward to reconstruct $a$ from ${v}_{a}\text{.}$ Therefore the map is invertible, and it suffices to show (a) the map is well-defined $\left({v}_{a}\in {N}_{\mu }\right),$ (b) the map is surjective. To show (a) and (b), we investigate the diagrams of elements in ${N}_{\mu }\text{.}$ Suppose $v\in {N}_{\mu }\text{.}$ Let $vi be the entry above the ith vertex of v, v(i) be the number of the bottom vertex connected to the ith top vertex,$ so that $\left\{{v}_{1},{v}_{2},\dots ,{v}_{n}\right\}$ are the labels above the vertices of $v$ and $\left(v\left(1\right),v\left(2\right),\dots ,v\left(n\right)\right)$ is the permutation determined the diagram (and ignoring the labels ${v}_{i}\text{).}$ By (4.8), $ψμ(xij(t))= { ψ(t), if j=i+1 and μ(i)=1, 1, otherwise. (A)$ Recall that $v\in {N}_{\mu }$ if and only if $u,vu{v}^{-1}\in U$ implies ${\psi }_{\mu }\left(u\right)={\psi }_{\mu }\left(vu{v}^{-1}\right)\text{.}$ That is, $v\in {N}_{\mu }$ if and only if for all $1\le i such that $v\left(i\right) $ψμ(xij(t)) = ψμ(vxij(t)v-1) = ψμ(xv(i)v(j)(vitvj-1)) = { ψ(vitvj-1), if v(j)=v(i) +1 and μ(v(i))=1, 1, otherwise. (B)$ Compare (A) and (B) to obtain that $b\in {N}_{\mu }$ if and only if for all $1\le i such that $v\left(i\right) (i) If ${\mu }_{\left(i\right)}=1$ and ${\mu }_{\left(v\left(i\right)\right)}=0,$ then $j\ne i+1,$ (ii) If ${\mu }_{\left(i\right)}=0$ and ${\mu }_{\left(v\left(i\right)\right)}=1,$ then $v\left(j\right)\ne v\left(i\right)+1,$ (iii) If ${\mu }_{\left(i\right)}={\mu }_{\left(v\left(i\right)\right)}=1,$ then $j=i+1$ if and only if $v\left(j\right)=v\left(i\right)+1,$ $\text{(iii)}\prime$ If ${\mu }_{\left(i\right)}={\mu }_{\left(v\left(i\right)\right)}=1$ and $v\left(j\right)=v\left(i\right)+1,$ then ${v}_{i}={v}_{i+1}\text{.}$ We can visualize the implications of the conditions $\text{(i)âˆ’(iii)}\prime$ in the following way. Partition the vertices of $v\in {N}_{\mu }$ by $\mu \text{.}$ For example, $\mu =\left(2,3,1\right)$ partitions $v$ according to ${v}_{1} {v}_{2} {v}_{3} {v}_{4} {v}_{5} {v}_{6} .$ Suppose the $i\text{th}$ top vertex is not next to a dotted line and the $v\left(i\right)\text{th}$ bottom vertex is immediately to the left of a dotted line. Then condition (i) implies that $v\left(i+1\right) so ${v}_{i} {v}_{i+1} (I)$ Similarly, condition (ii) implies ${v}_{i} (II)$ and conditions (iii) and $\text{(iii)}\prime$ imply ${v}_{i} {v}_{i+1} or {v}_{i} {v}_{i} (vi+1= vi in the second). (III)$ In the case $\mu =\left(n\right)$ condition (III) implies that every $v\in {N}_{\left(n\right)}$ is of the form ${a}_{1} {a}_{1} {a}_{1} {a}_{2} {a}_{2} {a}_{2} {a}_{r} {a}_{r} {a}_{r} \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots =w(n) ( a1w(i1)⊕ a2w(i2)⊕⋯⊕ arw(ir) ) ,$ where $\left({i}_{1},{i}_{2},\dots ,{i}_{r}\right)\models n,$ ${a}_{i}\in {𝔽}_{q}^{*},$ and ${w}_{\left(k\right)}\in {W}_{k}$ is as in (4.13). In fact, this observation proves that the map $\left(f\right)↦{v}_{\left(f\right)}$ is a bijection between ${M}_{\left(n\right)}$ and ${N}_{\left(n\right)}$ (mentioned in the remark).

Note that since the diagrams ${v}_{a}$ satisfy (I), (II) and (III), ${v}_{a}\in {N}_{\mu },$ proving (a). On the other hand, (I), (II) and (III) imply that each $v\in {N}_{\mu }$ must be of the form $v={v}_{a}$ for some $a\in {M}_{\mu },$ proving (b).

$\square$

Remark. This bijection will prove useful in developing a generalization of the RSK correspondence in Chapter 5.

Let $mμ= { a∈Mℓ(ℤ≥0) | row-sums and column-sums are μ }$ and for $a\in {m}_{\mu },$ let $ℓ(a)=Card {aij≠0 | 1≤i,j≤ℓ}.$

Let $\mu \models n\text{.}$ Then $dim(ℋμ)= ∑a∈mμ (q-1)ℓ(a) qn-ℓ(a)$

Proof.

Theorem 4.2 gives $dim(ℋμ)= ∣Mμ∣.$ We can obtain a matrix $\stackrel{‾}{a}\in {M}_{\mu }$ from a matrix $a\in {m}_{\mu }$ by selecting for each $1\le i,j\le \ell \left(\mu \right),$ a monic polynomial ${f}_{ij}$ with nonzero constant term and degree ${a}_{ij}\text{.}$ Conversely, every $\stackrel{‾}{a}\in {M}_{\mu }$ arises uniquely out of such a construction.

For a fixed $a\in {m}_{\mu },$ the total number of ways to choose appropriate polynomials is $∏1≤i,j≤ℓaij≠0 Card { f∈𝔽q[X] | f monic, f(0)=0 and d(f)=aij } =(q-1)ℓ(a) qn-ℓ(a).$ The result follows by summing over all $a\in {m}_{\mu }\text{.}$

$\square$

### Multiplication in ${ℋ}_{\mu }$

This section examines the implications of the unipotent Hecke algebra multiplication relations from Chapter 3 in the case $G={GL}_{n}\left({𝔽}_{q}\right)\text{.}$ The final goal is the algorithm given in Theorem 4.5.

#### Pictorial versions of ${e}_{\mu }v{e}_{\mu }$

Note that the map $π: N = WT ⟶ W wh ⟼ w, for w∈W,h∈T, (4.15)$ is a surjective group homomorphism. Since $W\subseteq N,$ adjust the decomposition of (3.4) as follows. Let $u\in N$ with $\pi \left(u\right)={s}_{{i}_{1}}\cdots {s}_{{i}_{r}}$ for $r$ minimal. Then $u$ has a unique decomposition $u=u1u2⋯uruT, where uk=sik, uT∈T. (4.16)$ For $t\in {𝔽}_{q},$ write ${u}_{k}\left(t\right)={s}_{{i}_{k}}{x}_{{i}_{k}}\left(t\right)\text{.}$

Fix a composition $\mu \text{.}$ The decomposition $U=∏1≤i implies $eμ=∏1≤i

Pictorially, let $uk as \cdots \cdots {i}_{k} \colorbox[rgb]{1,1,1}{k} , (4.18) uk(tk) as \cdots \cdots {i}_{k} \colorbox[rgb]{1,1,1}{\overline{)k}} , (4.19) eij(k) as \cdots \cdots \cdots \cdots i j \colorbox[rgb]{1,1,1}{\overline{)k}} , (4.20) eμ as \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} . (4.21)$

Examples: 1. If $u={u}_{1}{u}_{2}\cdots {u}_{8}{u}_{T}\in N$ decomposes according to ${s}_{3}{s}_{1}{s}_{2}{s}_{3}{s}_{1}{s}_{4}{s}_{2}{s}_{3}\in W$ with ${u}_{T}=\text{diag}\left(a,b,c,d,e\right),$ then $u= a b c d e \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{2} \colorbox[rgb]{1,1,1}{3} \colorbox[rgb]{1,1,1}{4} \colorbox[rgb]{1,1,1}{5} \colorbox[rgb]{1,1,1}{6} \colorbox[rgb]{1,1,1}{7} \colorbox[rgb]{1,1,1}{8} ,u1(t1) u2(t2)⋯ u8(t8)uT= a b c d e \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}}$ and $eμueμ= a b c d e \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(4\right)}} {\mu }_{\left(1\right)} {\mu }_{\left(2\right)} {\mu }_{\left(3\right)} {\mu }_{\left(4\right)} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{2} \colorbox[rgb]{1,1,1}{3} \colorbox[rgb]{1,1,1}{4} \colorbox[rgb]{1,1,1}{5} \colorbox[rgb]{1,1,1}{6} \colorbox[rgb]{1,1,1}{7} \colorbox[rgb]{1,1,1}{8} .$ 2. If $n=5,$ then (4.17) implies $\colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(4\right)}} = \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}}$ $\text{(}{e}_{\mu }={e}_{45}\left(1\right){e}_{35}\left(1\right){e}_{34}\left(1\right){e}_{25}\left(1\right){e}_{24}\left(1\right){e}_{23}\left(1\right){e}_{15}\left(1\right){e}_{14}\left(1\right){e}_{13}\left(1\right){e}_{12}\left(1\right)\text{).}$

The elements ${e}_{ij}\left(k\right)$ also interact with $U$ and $N$ as follows (see also Section 3.3.1) $sreij(k) sr = esr(i)sr(j) (k), (E1) eμveij(1) = eμv, v∈Nμ,(πv) (i)<(πv) (j), (E2) eij(k) hεr(t) = hεr(t) eij(tδrit-δrjk), (E3) eμxij(t) = ψ(μijt) eμ=xij(t) eμ, (E4)$ or pictorially, $\colorbox[rgb]{1,1,1}{\overline{)k}} = \colorbox[rgb]{1,1,1}{\overline{)k}} (E1) \colorbox[rgb]{1,1,1}{\overline{)k}} r s = \colorbox[rgb]{1,1,1}{\overline{)kr{s}^{-1}}} r s (E3) \colorbox[rgb]{1,1,1}{k} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} =ψ(kab) \colorbox[rgb]{1,1,1}{k} and \colorbox[rgb]{1,1,1}{k} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{b}}} =ψ(kab) \colorbox[rgb]{1,1,1}{k} (E4)$ and for $v\in {N}_{\mu },$ $\colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\overline{)1}} v \colorbox[rgb]{1,1,1}{{\mu }_{\left(i\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(j-1\right)}} = \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{\cdots } v \colorbox[rgb]{1,1,1}{{\mu }_{\left(i\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(j-1\right)}} (E2)$

Suppose $u={u}_{1}{u}_{2}\cdots {u}_{r}{u}_{T}\in {N}_{\mu }$ with ${u}_{T}=\text{diag}\left({h}_{1},{h}_{2},\dots ,{h}_{n}\right)\text{.}$ Then using (4.17), (E3), (E1) and (E2), we can rewrite ${e}_{\mu }u{e}_{\mu }$ as $\colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} \colorbox[rgb]{1,1,1}{{h}_{1}} \colorbox[rgb]{1,1,1}{{h}_{2}} \colorbox[rgb]{1,1,1}{{h}_{3}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{h}_{n}} {u}_{1}{u}_{2}\cdots {u}_{r} \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} =1qr∑t∈𝔽qr (ψ∘fu)(t) \colorbox[rgb]{1,1,1}{{h}_{1}} \colorbox[rgb]{1,1,1}{{h}_{2}} \colorbox[rgb]{1,1,1}{{h}_{3}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{h}_{n}} {u}_{1}\left({t}_{1}\right){u}_{2}\left({t}_{2}\right)\cdots {u}_{r}\left({t}_{r}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} (4.22)$ where ${f}_{u}\in {𝔽}_{q}\left[{y}_{1},{y}_{2},\dots ,{y}_{r}\right]$ is given by $fu(y1,y2,…,yr)=- μi1j1hi1 hj1-1y1- μi2j2hi2 hj2-1y2-⋯- μirjrhir hjr-1yr, (4.23)$ for $\left({i}_{k},{j}_{k}\right)=\left(a,b\right),$ if the $k\text{th}$ crossing in $u$ crosses the strands coming from the $a\text{th}$ and $b\text{th}$ top vertices.

Example (continued). Suppose $u={u}_{1}{u}_{2}\cdots {u}_{8}{u}_{T}\in N$ decomposes according to ${s}_{3}{s}_{1}{s}_{2}{s}_{3}{s}_{1}{s}_{4}{s}_{2}{s}_{3}\in W$ and ${u}_{T}=\text{diag}\left(a,b,c,d,e\right)\in T$ (as in Example 1 above). Consider the ordering $eμ= \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)1}} ,$ or ${e}_{\mu }={e}_{13}\left(1\right){e}_{23}\left(1\right){e}_{12}\left(1\right){e}_{45}\left(1\right){e}_{15}\left(1\right){e}_{25}\left(1\right){e}_{14}\left(1\right){e}_{35}\left(1\right){e}_{24}\left(1\right){e}_{34}\left(1\right)\text{.}$ Then $eμueμ = eμu e13(1) e23(1) ← e12(1) e45(1) e15(1) e25(1) e14(1) e35(1) e24(1) e34(1) = eμu e12(1) e45(1) e15(1) e25(1) e14(1) e35(1) e24(1) e34(1) (by (E2)) = eμu1u2⋯u8 uT e12(1) e45(1) e15(1) e25(1) e14(1) e35(1) e24(1) e34(1) ← = eμu1u2⋯u8 e12(ab) e45(de) e15(ae) e25(be) e14(ad) e35(ce) e24(bd) e34(cd) ← UT(by (E3)) = eμu1 e34(ab)u2 e12(de)u3 e23(ae)u4 e34(be)u5 e12(ad)u6 e45(ce)u7 e23(bd)u8 e34(cd)uT (by (E1)) = eμq8∑t∈𝔽q8 (ψ∘f)(t)u1 x34(abt1)u2 x12(det2)u3 x23(aet3)⋯ u8x34(cdt8) uT,$ where by (4.17), $f=-{y}_{1}-{y}_{2}-{y}_{8}\text{.}$ Therefore, by renormalizing $eμueμ = eμq8∑t′∈𝔽q8 (ψ∘fu)(t′) u1x34(t1) u2x12(t2) u3x23(t3)⋯ u8x34(t8)uT = eμq8∑t′∈𝔽q8 (ψ∘fu)(t′) u1(t1) u2(t2) u3(t3)⋯ u8(t8)uT,$ where ${f}_{u}=-b{a}^{-1}{y}_{1}-e{d}^{-1}{y}_{2}-d{c}^{-1}{y}_{8}\text{.}$ Pictorially, by sliding the ${e}_{ij}\left(1\right)$ down along the strands until they get stuck, this computation gives $eμueμ= a b c d e \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(4\right)}} {\mu }_{\left(1\right)} {\mu }_{\left(2\right)} {\mu }_{\left(3\right)} {\mu }_{\left(4\right)} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{2} \colorbox[rgb]{1,1,1}{3} \colorbox[rgb]{1,1,1}{4} \colorbox[rgb]{1,1,1}{5} \colorbox[rgb]{1,1,1}{6} \colorbox[rgb]{1,1,1}{7} \colorbox[rgb]{1,1,1}{8} =1q8 ∑t∈𝔽q8 (ψ∘fu)(t) a b c d e {\mu }_{\left(1\right)} {\mu }_{\left(2\right)} {\mu }_{\left(3\right)} {\mu }_{\left(4\right)} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} (4.24)$ where ${f}_{u}=-b{a}^{-1}{y}_{1}-e{d}^{-1}{y}_{2}-d{c}^{-1}{y}_{8}$ (as in (4.23)), since $\left({i}_{1},{j}_{1}\right)=\left(1,2\right),$ $\left({i}_{2},{j}_{2}\right)=\left(4,5\right),$ $\left({i}_{3},{j}_{3}\right)=\left(1,5\right),$ etc.

### Relations for multiplying basis elements.

Let $u={u}_{1}{u}_{2}\cdots {u}_{r}{u}_{T}\in {N}_{\mu }$ decompose according to a minimal expression in $W$ as in (4.16). Let $v\in {N}_{\mu }$ and use (N3) and (N4) to write ${u}_{T}v=w·\text{diag}\left({a}_{1},{a}_{2},\cdots ,{a}_{n}\right)$ for some $w=\pi \left(v\right)\in W$ (see (4.7)). Then use (4.22) to write $(eμueμ) (eμveμ)= 1qr∑t∈𝔽qr (𝓅∘fu)(t) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{2}} \colorbox[rgb]{1,1,1}{{a}_{3}} \colorbox[rgb]{1,1,1}{{a}_{n}} w {u}_{1}\left({t}_{1}\right){u}_{2}\left({t}_{2}\right)\cdots {u}_{r}\left({t}_{r}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} (4.25)$ (This form corresponds to ${\mathrm{\Xi }}^{\varnothing }\left(u,{u}_{T}v\right)$ of Corollary 3.9).

Example (continued). If $u$ is as in (4.24) and $v={s}_{2}{s}_{3}{s}_{2}{s}_{1}{s}_{2}·\text{diag}\left(f,g,h,i,j\right)\in N,$ then $(eμueμveμ)= 1q8∑t∈𝔽q8 (ψ∘fu)(t) fd gc ah bi ej \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(4\right)}} {\mu }_{\left(1\right)} {\mu }_{\left(2\right)} {\mu }_{\left(3\right)} {\mu }_{\left(4\right)} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} .$

Consider the crossing in (4.25) corresponding to ${u}_{r}\left({t}_{r}\right)\text{.}$ There are two possibilities.

Case 1 the strands that cross at $\overline{)r}$ do not cross again as they go up to the top of the diagram $\text{(}\ell \left({u}_{r}w\right)>\ell \left(w\right)\text{),}$

Case 2 the strands that cross at $\overline{)r}$ cross once on the way up to the top of the diagram $\text{(}\left({u}_{r}w\right)<\ell \left(w\right)\text{).}$

In the first case, $eμueμveμ= 1qr∑t∈𝔽qr (ψ∘fu)(t) \colorbox[rgb]{1,1,1}{\overline{)r}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(i\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(j-1\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{i}} \colorbox[rgb]{1,1,1}{{a}_{j}} \colorbox[rgb]{1,1,1}{{a}_{n}} w {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} ,$ so (UN1), (UN2) and (E4) imply $eμueμveμ= 1qr∑t∈𝔽qr (ψ∘f(+0)) (t) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{2}} \colorbox[rgb]{1,1,1}{{a}_{3}} \colorbox[rgb]{1,1,1}{{a}_{n}} {u}_{r}w {u}_{1}\left({t}_{1}\right)\cdots {u}_{r}\left({t}_{r}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} (4.26)$ where ${f}^{\left(+0\right)}={f}_{u}+{\mu }_{ij}{a}_{j}{a}_{i}^{-1}{y}_{r}\text{.}$

In the second case, $eμueμveμ= 1qr∑t∈𝔽qr (ψ∘fu)(t) \colorbox[rgb]{1,1,1}{\overline{)r}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(i\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(j-1\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{i}} \colorbox[rgb]{1,1,1}{{a}_{j}} \colorbox[rgb]{1,1,1}{{a}_{n}} w {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} .$ Use (UN3) and (N1) to split the sum into two parts corresponding to ${t}_{r}=0$ and ${t}_{r}\ne 0,$ $eμueμveμ = 1qr∑t∈𝔽qrtr=0 (ψ∘f(-0))(t) \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(i\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(j-1\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{i}} \colorbox[rgb]{1,1,1}{{a}_{j}} \colorbox[rgb]{1,1,1}{{a}_{n}} {u}_{r}w {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} +1qr ∑t∈𝔽qrtr∈𝔽q* (ψ∘fu)(t) \colorbox[rgb]{1,1,1}{r} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{1}}} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{-1}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{{t}_{r}}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{{t}_{r}^{-1}}}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(i\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(j-1\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{i}{t}_{r}} \colorbox[rgb]{1,1,1}{-{a}_{j}{t}_{r}^{-1}} \colorbox[rgb]{1,1,1}{{a}_{n}} w {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}}$ where ${f}^{\left(-0\right)}={f}_{u}\text{.}$ Now use (UN1), (UN2), (U2), (U1) and (E4) on the second sum to get $eμueμveμ = 1qr∑t∈𝔽qrtr=0 (ψ∘f(-0))(t) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{2}} \colorbox[rgb]{1,1,1}{{a}_{3}} \colorbox[rgb]{1,1,1}{{a}_{n}} {u}_{r}w {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} +1qr ∑t∈𝔽qrtr∈𝔽q* (ψ∘f(1))(t) \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(i\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(j-1\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{a}_{1}} \colorbox[rgb]{1,1,1}{{a}_{i}{t}_{r}} \colorbox[rgb]{1,1,1}{-{a}_{j}{t}_{r}^{-1}} \colorbox[rgb]{1,1,1}{{a}_{n}} w {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} (4.27)$ where ${f}^{\left(1\right)}={\phi }_{r}\left({f}_{u}\right)+{\mu }_{ij}{a}_{j}{a}_{i}^{-1}{y}_{r}^{-1},$ and ${\phi }_{r}\left(f\right)$ is defined by $∑t∈𝔽qrtr∈𝔽q* (ψ∘f)(t) \colorbox[rgb]{1,1,1}{r} \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{1}}} \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{{t}_{r}^{-1}}}} {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} =∑t∈𝔽qrtr∈𝔽q* (ψ∘φr(f))(t) \colorbox[rgb]{1,1,1}{r} {u}_{1}\left({t}_{1}\right)\cdots {u}_{r-1}\left({t}_{r-1}\right) \colorbox[rgb]{1,1,1}{{\mu }_{\left(1\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(2\right)}} \colorbox[rgb]{1,1,1}{{\mu }_{\left(3\right)}} \colorbox[rgb]{1,1,1}{\cdots } \colorbox[rgb]{1,1,1}{{\mu }_{\left(n-1\right)}} . (*)$

Remarks: (a) We could have applied these steps for any $u,$ $u,$ and $v,$ so we can iterate the process with each sum. (b) The most complex step in these computations is determining ${\phi }_{r}\text{.}$ The following section will develop an efficient algorithm for computing the right-hand side of $\left(*\right)\text{.}$

#### Computing ${\phi }_{k}$ via painting, paths and sinks.

Painting algorithm $\left({u}^{\overline{)k}}\right)\text{.}$ Suppose $u={u}_{1}{u}_{2}\cdots {u}_{r}\in N$ decomposes according to ${s}_{{i}_{1}}{s}_{{i}_{2}}\cdots {s}_{{i}_{k}}\in W$ (assume ${u}_{T}=1\text{).}$ Paint flows down strands (by gravity). Each step is illustrated with the example $u={u}_{1}{u}_{2}\cdots {u}_{8},$ decomposed according to ${s}_{3}{s}_{1}{s}_{2}{s}_{3}{s}_{1}{s}_{4}{s}_{2}{s}_{3}\in W\text{.}$ (1) Paint the left [respectively right] strand exiting $\overline{)k}$ red [blue] all the way to the bottom of the diagram. $\colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} where k is 8 .$ (2) For each crossing that the red [blue] strand passes through, paint the right [left] strand (if possible) red [blue] until that strand either reaches the bottom or crosses the blue [red] strand of (1). $\colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} (4.28)$ Let $uk u1(t1) u2(t2)⋯ uk(tk) painted according to the above algorithm (4.29)$

Sinks and paths. The diagram ${u}^{\overline{)k}}$ has a crossed sink at $\overline{)j}$ if $\overline{)j}$ is a crossing between a red strand and a blue one, or $\colorbox[rgb]{1,1,1}{\overline{)j}}$ Note that since $u$ is decomposed according to a minimal expression in $U,$ there will be no crossings of the form $\colorbox[rgb]{1,1,1}{\overline{)j}} (since \colorbox[rgb]{1,1,1}{\overline{)j}} would imply \colorbox[rgb]{1,1,1}{\overline{)j\prime }} \colorbox[rgb]{1,1,1}{\overline{)j}} for some j′≥j.)$ The diagram ${u}^{\overline{)k}}$ has a bottom sink at $j$ if a red strand enters $j\text{th}$ bottom vertex and a blue strand enters the $\left(j+1\right)\text{st}$ bottom vertex, or $\colorbox[rgb]{1,1,1}{j} \colorbox[rgb]{1,1,1}{j+1}$ A red [respectively blue] path $p$ from a sink $s$ (either crossed or bottom) in ${u}^{\overline{)k}}$ is an increasing sequence $j1 such that in ${u}^{\overline{)k}}$ (a) $\overline{){j}_{m}}$ is directly connected (no intervening crossings) to $\overline{){j}_{m+1}}$ by a red [blue] strand, (b) if $s$ is a crossed sink, then $\overline{){j}_{1}}=s,$ (b') if $s$ is a bottom sink, then in a red path, the $s\text{th}$ bottom vertex connects to the crossing $\overline{){j}_{1}}$ with a red strand. in a blue path, the $\left(s+1\right)\text{st}$ bottom vertex connects to the crossing $\overline{){j}_{1}}$ with a blue strand. Let $P=(uk,s)= {red paths from s in uk} andP:(uk,s) ={blue paths from s in uk} (4.30)$ The weight of a path $p$ is $wt(p)= { ∏p switchesstrands at i yi, if p∈P= (uk,s), ∏p switchesstrands at i (-yi), if p∈P: (uk,s). (4.31)$ Each sink $s$ in ${u}^{\overline{)k}}$ (either crossed $\overline{)j}$ or bottom $j\text{)}$ has an associated polynomial ${g}_{s}\in {𝔽}_{q}\left[{y}_{1},{y}_{2},\dots ,{y}_{k-1},{y}_{k}^{-1}\right]$ given by $gs=∑p∈P=(uk,s)p′∈P:(uk,s) wt(p)yk-1wt(p′). (4.32)$

Example (continued). If $u={u}_{1}{u}_{2}\cdots {u}_{8}$ decomposes according to ${s}_{3}{s}_{1}{s}_{2}{s}_{3}{s}_{1}{s}_{4}{s}_{2}{s}_{3}$ (as in (4.28)), then $P=(u8,4)= { \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{\overline{)5}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} , \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)4}} \colorbox[rgb]{1,1,1}{\overline{)7}} \colorbox[rgb]{1,1,1}{\overline{)8}} } P:(u8,4) = { \colorbox[rgb]{1,1,1}{\overline{)6}} \colorbox[rgb]{1,1,1}{\overline{)8}} }$ with $\text{wt}\left(1<3<5<7<8\right)={y}_{5},$ $\text{wt}\left(1<4<7<8\right)={y}_{1}{y}_{7},$ and $\text{wt}\left(6<8\right)=1\text{.}$ The corresponding polynomial is $g4=y5y8-1 +y1y7y8-1. (4.33)$

Let $u={u}_{1}{u}_{2}\cdots {u}_{r}$ and ${\phi }_{r}$ be as in (4.27) and $\text{(}*\text{);}$ suppose ${u}^{\overline{)r}}$ is painted as above. Then $φr(f)=f |{yj↦yj-gj | j a crossed sink} +∑j a bottomsink μ(j)gj.$

Proof.

In the painting, $marks a strand travelled by \colorbox[rgb]{1,1,1}{\underset{\circ }{\overset{\to }{a}}}$ and $marks a strand travelled by \colorbox[rgb]{1,1,1}{\underset{←}{\overset{\circ }{a}}} .$ Substitutions due to crossed sinks correspond to the normalizations in relation (U2), and the sum over bottom sinks comes from applications of relation (E4).

$\square$

Example (continued). Recall $u={s}_{3}{s}_{1}{s}_{2}{s}_{3}{s}_{1}{s}_{4}{s}_{2}{s}_{3}\text{.}$ Then ${u}^{\overline{)8}}$ at $\overline{)2},$ $\overline{)3},$ and $\overline{)4}\text{.}$ The only bottom sink is at $4\text{.}$ Therefore, $φ8(f)=f |y4↦y4-g4y3↦y3-g3y2↦y2-g2+ μ(4)g4=f | y4↦y4+y7y8-1y6 y3↦y3+y5y8-1y6 y2↦y2+y2y8-1y6 + μ(4) (y5y8-1+y1y7y8-1).$ (for example, ${g}_{4}$ was computed in (4.33)).

#### A multiplication algorithm

(The algorithm). Let $G={GL}_{n}\left({𝔽}_{q}\right)$ and $u,v\in {N}_{\mu }\text{.}$ An algorithm for multiplying ${e}_{\mu }u{e}_{\mu }$ and ${e}_{\mu }v{e}_{\mu }$ is (1) Decompose $u={u}_{1}{u}_{2}\cdots {u}_{r}{u}_{T}$ according to some minimal expression in $W$ (as in (4.16)). (2) Put ${e}_{\mu }u{e}_{\mu }v{e}_{\mu }$ into the form specified by (4.25), with ${u}_{T}v=w·\text{diag}\left({a}_{1},{a}_{2},\dots ,{a}_{n}\right)$ $\text{(}w=\pi \left(v\right)\in W\text{).}$ (3) Complete the following (a) If $\ell \left({u}_{r}w\right)>\ell \left(w\right),$ then apply relation (4.26). (b) If $\ell \left({u}_{r}w\right)<\ell \left(w\right),$ then apply relation (4.27), using ${\left({u}_{1}{u}_{2}\cdots {u}_{r}\right)}^{\overline{)r}}$ and Lemma 4.4 to compute ${\phi }_{r}\text{.}$ (4) If $r>1,$ then reapply (3) to each sum with $r≔r-1$ and with (a) $w≔{u}_{r}w,$ after using (3a) or using (3b), in the first sum, (b) $w≔w,$ after using (3b), in the second sum. (5) Set all diagrams not in ${N}_{\mu }$ to zero.

Sample computation. Suppose $n=3$ and ${\mu }_{\left(i\right)}=1$ for all $1\le i\le 3$ (i.e.. the Gelfand-Graev case). Then $Nμ= { a a a , a a b , a b b , a b c | a,b,c∈𝔽q* }$ Suppose $u= a b c andv= d e e .$

1. Theorem 4.5 (1): Let $u={u}_{1}{u}_{2}{u}_{3}{u}_{T}\in {N}_{\mu }$ decompose according to ${s}_{2}{s}_{1}{s}_{2}\in W,$ with ${u}_{T}=\text{diag}\left(a,b,c\right)\text{.}$

2. Theorem 4.5 (2): By (4.25) $( eμ a b c eμ ) ( eμ d e e eμ ) =1q3 ∑t∈𝔽q3 (ψ∘fu)(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd ae be$ with ${u}_{T}v={s}_{2}{s}_{1}·\text{diag}\left(cd,ae,be\right)$ (so $w={s}_{2}{s}_{1}\text{),}$ and ${f}_{u}=-\frac{b}{a}{y}_{1}-\frac{c}{b}{y}_{3}$ (as in (4.23)).

3. Theorem 4.5 (3b): Since $\ell \left({u}_{3}w\right)<\ell \left(w\right),$ paint ${u}_{1}\left({t}_{1}\right){u}_{2}\left({t}_{2}\right){u}_{3}\left({t}_{3}\right)$ to get ${\left({u}_{1}{u}_{2}{u}_{3}\right)}^{\overline{)3}}$ (as in (4.29)), $=1q3∑t∈𝔽q3 (ψ∘fu)(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{\overline{)3}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd ae be$ Now apply (4.27), $=1q3∑t∈𝔽q3t3=0 (ψ∘f(-0))(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd ae be +1q3∑t∈𝔽q3t3∈𝔽q* (ψ∘f(1))(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{3} ae -be{t}_{3}^{-1}$ where ${f}^{\left(-0\right)}=-\frac{b}{a}{y}_{1}-\frac{c}{b}{y}_{3}$ and by Lemma 4.4, $f(1)=φ3(fu) +μ13becdy3-1 =-bay1+bay2 y3-1-cdy3+ y3-1.$

4. Theorem 4.5 (4): Set $r≔2$ with $w≔{u}_{r}w={s}_{1}$ in the first sum and $w≔w$ in the second sum.

5. Theorem 4.5 (3a) (3b): In the first sum, $\ell \left({u}_{2}{s}_{1}\right)<\ell \left({s}_{1}\right),$ so paint ${u}_{1}\left({t}_{1}\right){u}_{2}\left({t}_{2}\right)$ to get ${\left({u}_{1}{u}_{2}\right)}^{\overline{)2}}\text{.}$ In the second sum, $\ell \left({u}_{2}{s}_{2}{s}_{1}\right)>\ell \left({s}_{2}{s}_{1}\right),$ so apply (4.26), $=1q3∑t∈𝔽q3t3=0 (ψ∘f(-0))(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{\overline{)2}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd ae be +1q3∑t∈𝔽q3t3∈𝔽q* (ψ∘f(+0,1))(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{3} ae -be{t}_{3}^{-1}$ where ${f}^{\left(+0,1\right)}=-\frac{b}{a}{y}_{1}+\frac{b}{a}{y}_{2}{y}_{3}^{-1}-\frac{c}{d}{y}_{3}+{y}_{3}^{-1}-{\mu }_{\left(3\right)}\frac{b}{a}{y}_{2}{y}_{3}^{-1}=-\frac{b}{a}{y}_{1}-\frac{c}{b}{y}_{3}+{y}_{3}^{-1}\text{.}$ Now apply (4.27) to the first sum, $= 1q3∑t∈𝔽q3t2=t3=0 (ψ∘f(-0,-0))(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd ae be +1q3∑t∈𝔽q3t2∈𝔽q*,t3=0 (ψ∘f(1,-0)) (t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{2} -ae{t}_{2}^{-1} be +1q3∑t∈𝔽q3t3∈𝔽q* (ψ∘f(+0,1))(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{3} ae -be{t}_{3}^{-1}$ where ${f}^{\left(-0,-0\right)}=-\frac{b}{a}{y}_{1}-\frac{c}{b}{y}_{3}$ and ${f}^{\left(1,-0\right)}=\phi \left({f}^{\left(-0\right)}\right)+{\mu }_{\left(1\right)}\frac{ae}{cd}{y}_{2}^{-1}=-\frac{b}{a}{y}_{1}-{y}_{2}^{-1}{y}_{1}+\frac{ae}{cd}{y}_{2}^{-1}\text{.}$

6. Theorem 4.5 (4): Set $r=1$ with $w≔{u}_{2}{s}_{1}=1$ in the first sum, $w≔{s}_{1}$ in the second sum, and $w≔{s}_{1}{s}_{2}{s}_{1}$ in the third sum.

7. Theorem 4.5 (3a) (3a) (3b): In the first sum $\ell \left({s}_{2}1\right)>\ell \left(1\right),$ so apply (4.26); in the second sum $\ell \left({s}_{2}{s}_{1}\right)>\ell \left({s}_{1}\right),$ so apply (4.26); in the third sum, $\ell \left({s}_{2}{s}_{1}{s}_{2}{s}_{1}\right)<\ell \left({s}_{1}{s}_{2}{s}_{1}\right),$ so paint ${u}_{1}\left({t}_{1}\right)$ to get ${u}_{1}^{\overline{)1}},$ $= 1q3∑t∈𝔽q3t2=t3=0 (ψ∘f(+0,-0,-0))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd ae be +1q3∑t∈𝔽q3t2∈𝔽q*,t3=0 (ψ∘f(+0,1,-0))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{2} -ae{t}_{2}^{-1} be +1q3∑t∈𝔽q3t3∈𝔽q* (ψ∘f(+0,1))(t) \colorbox[rgb]{1,1,1}{\overline{)1}} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{3} ae -be{t}_{3}^{-1}$ where ${f}^{\left(+0,-0,-0\right)}=-\frac{b}{a}{y}_{1}-\frac{c}{b}{y}_{3}+\frac{b}{a}{y}_{1}=-\frac{c}{b}{y}_{3}$ and ${f}^{\left(+0,1,-0\right)}=-\frac{b}{a}{y}_{1}+\frac{ae}{cd}{y}_{2}^{-1}-{y}_{2}^{-1}{y}_{1}\text{.}$ Now apply (4.27) to the third sum $= 1q3∑t∈𝔽q3t2=t3=0 (ψ∘f(+0,-0,-0))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd ae be +1q3∑t∈𝔽q3t2∈𝔽q*,t3=0 (ψ∘f(+0,1,-0))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{2} -ae{t}_{2}^{-1} be +1q3∑t∈𝔽q3t1=0,t3∈𝔽q* (ψ∘f(-0,+0,1))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{3} ae -be{t}_{3}^{-1} +1q3∑t∈𝔽q3t1,t3∈𝔽q* (ψ∘f(1,+0,1))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{1}{t}_{3} -ae{t}_{1}^{-1} -be{t}_{3}^{-1}$ where ${f}^{\left(-0,+0,1\right)}=-\frac{c}{b}{y}_{3}+{y}_{3}^{-1}$ and $f(1,+0,1)= φ1(f(+0,1)) +μ(1)aecdy1-1 y3-1=-bay1- cby3+y3-1+ y1-1+aecd y1-1y3-1.$

8. Theorem 4.5 (5): The first sum contains no elements of ${N}_{\mu },$ so set it to zero. The second sum contains elements of ${N}_{\mu }$ when $be=-ae{t}_{2}^{-1},$ so set ${t}_{2}=-\frac{a}{b}\text{.}$ The third sum contains elements of ${N}_{\mu }$ when $cd{t}_{3}=ae,$ so set ${t}_{3}=\frac{ae}{cd}\text{.}$ All the terms in the fourth sum are basis elements. $= 0+1q3∑t∈𝔽q3t2=-ab,t3=0 (ψ∘f(+0,1,-0))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} -a{b}^{-1}cd be be +1q3∑t∈𝔽q3t1=0,t3=aecd (ψ∘f(-0,+0,1))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} ae ae -{a}^{-1}bcd +1q3∑t∈𝔽q3t1,t3∈𝔽q* (ψ∘f(1,+0,1))(t) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{1}{t}_{3} -ae{t}_{1}^{-1} -be{t}_{3}^{-1} = 1q2ψ(-becd) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} -a{b}^{-1}cd be be +1q2ψ(-aebd+cdae) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} ae ae -{a}^{-1}bcd +1q2∑t1,t3∈𝔽q* ψ ( -bat1 -cbt3 +t3-1 +t1-1 +aecd t1-1 t3-1 ) \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} \colorbox[rgb]{1,1,1}{1} cd{t}_{1}{t}_{3} -ae{t}_{1}^{-1} -be{t}_{3}^{-1} .$

## Notes and References

This is an excerpt of the PhD thesis Unipotent Hecke algebras: the structure, representation theory, and combinatorics by F. Nathaniel Edgar Thiem.

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