Weight lattices

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

Last update: 19 September 2012

GLn(ℂ)

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}.

SLn(ℂ)

T= { ( x1 0 ⋱ 0 xn ) ∣x1…xn=1 }

With Xεi:T→ℂ× and hεi∨:ℂ×→TGLn 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

PGLn(ℂ)

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

Then with Xεi: TGLn→ℂ× and hεi∨: ℂ×→TGLn →TPGL

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

For SL2

𝔥ℤ* ε2 ε1

Identify points on diagonal lines.

The 2-fold cover of ℤ–span{ε1-ε2} is very visible.

𝔥ℤ ε2∨ ε1∨ 𝔥ℤ=ℤ–span {ε1∨-ε2∨} .

SL3

ε2+ε3 ε1+ε3 ε1+ε2 ε1+ε2+ε3 ε2 ε3 ε1

The weight lattice of SL3 as a 3–fold cover of

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

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