## Weight lattices

Last update: 19 September 2012

## $G{L}_{n}\left(ℂ\right)$

$T= { ( x1 0 ⋱ 0 xn ) ∣xi∈ ℂ× }$

Define

$Xεi: T⟶ℂ× byXεi ( x1 0 ⋱ 0 xn ) =Xi, hεi∨: ℂ×⟶T byhεi∨ (t)= ( ( 1 0 ⋱ 1 t 1 ⋱ 0 1 ) ) .$

Then

$𝔥ℤ* = { (Xε1)λ1 … (Xεn)λn ∣λ1,…, λn∈ℤ } 𝔥ℤ = { (hε1∨)λ1 … (hε1∨)λn ∣λ1,…, λn∈ℤ } N(T) = { n×nmatrices with exactly one nonzero entry on each row and each column } W0 = N(T)T= {wT∣w∈Sn} where Sn = {permutation matrices}.$

## $S{L}_{n}\left(ℂ\right)$

$T= { ( x1 0 ⋱ 0 xn ) ∣x1…xn=1 }$

With ${X}^{{\epsilon }_{i}}:\phantom{\rule{0.2em}{0ex}}T\to {ℂ}^{×}$ and ${h}_{{\epsilon }_{i}^{\vee }}:\phantom{\rule{0.2em}{0ex}}{ℂ}^{×}\to {T}_{G{L}_{n}}$ as in ???

$𝔥ℤ* = ⟨ Xε1,…, Xεn∣ Xε1…Xεn =1 ⟩ 𝔥ℤ = ⟨ hε1∨-ε2∨ ,…, hεn-1∨-εn∨ ⟩ = { h λ1ε1∨ +…+ λnεn∨ ∣ λ1+…+λn=0 } N(T) = { n×nmatrices with (a) exactly one nonzero entry in each row and each column (b) product of the nonzero entries is 1$

## $PG{L}_{n}\left(ℂ\right)$

$T= { [ x1 0 ⋱ 0 xn ] ∣ [ x1 0 ⋱ 0 xn ] = [ λx1 0 ⋱ 0 λxn ] forλ∈ℂ× }$

Then with ${X}^{{\epsilon }_{i}}:\phantom{\rule{0.2em}{0ex}}{T}_{G{L}_{n}}\to {ℂ}^{×}$ and ${h}_{{\epsilon }_{i}^{\vee }}:\phantom{\rule{0.2em}{0ex}}{ℂ}^{×}\to {T}_{G{L}_{n}}\to {T}_{PGL}$

$𝔥ℤ* = { xλ= x λ1ε1+…+ λnεn ∣λi∈ℤ ,λ1+…+ λn=0 } 𝔥ℤ = ⟨ hε1∨ ,…, hεn∨ ∣ hε1∨ ,…, hεn∨ =1 ⟩ .$

## For $S{L}_{2}$

$𝔥ℤ* {\epsilon }_{2} {\epsilon }_{1}$

Identify points on diagonal lines.

The 2-fold cover of $ℤ\text{–span}\phantom{\rule{0.2em}{0ex}}\left\{{\epsilon }_{1}-{\epsilon }_{2}\right\}$ is very visible.

$𝔥ℤ {\epsilon }_{2}^{\vee } {\epsilon }_{1}^{\vee }$ $𝔥ℤ=ℤ–span {ε1∨-ε2∨} .$

## $S{L}_{3}$

$ε2+ε3 ε1+ε3 ε1+ε2 ε1+ε2+ε3 ε2 ε3 ε1$

The weight lattice of $S{L}_{3}$ as a 3–fold cover of

$span { ε1-ε2, ε2-ε3 } = { λ1ε1+ λ2ε2+ λ3ε3 ∣ λ1+λ2+ λ3=0 } .$