## Representation Theory Lecture 12

Last update: 20 May 2013

## Representation Theory Lecture 12

$SO10= { g∈GL10 | det g=1, ggt=1 } 𝔰𝔬10= { x∈𝔤𝔩10 | tr x=0, x+xt=0 }$

since

$1=det(etx)= det ( eth1 ⋱ ethn ) =et(h1+…+hn) =et·tr (x)$

and

$1=etx (etx)t= etx etxt= et(x+xt)$

Then

$𝔰𝔬5= { ( 0 a12 a13 a14 a15 -a12 0 a23 a24 a25 -a13 -a23 0 a34 a35 -a14 -a24 -a34 0 a45 -a15 -a25 -a35 -a45 0 ) } ,dim (𝔰𝔬5) =5·42=10$

Another choice is

$SO10= { g∈GL10 | det g=1, gJgt=J } 𝔰𝔬10= { x∈𝔤𝔩10 | tr x=0, xJ+Jxt=0 }$

where

$J= ( 01 1 ⋰ 1 10 ) orJ= ( 0 10 ⋱ 01 10 ⋱ 01 0 ) .$

Since

$( a11 a12 a13 a1-3 a1-2 a1-1 a21 a22 a23 a2-3 a2-2 a2-1 a31 a32 a33 a3-3 a3-2 a3-1 a-31 a-32 a-33 a-3-3 a-3-2 a-3-1 a-21 a-22 a-23 a-2-3 a-2-2 a-2-1 a-11 a-12 a-13 a-1-3 a-1-2 a-1-1 ) ( 000001 000010 000100 001000 010000 100000 ) = ( a1-1 a1-2 a1-3 a13 a12 a11 a2-1 a2-2 a2-3 a23 a22 a21 a3-1 a3-2 a3-3 a33 a32 a31 a-3-1 a-3-2 a-3-3 a-33 a-32 a-31 a-2-1 a-2-2 a-2-3 a-23 a-22 a-21 a-11 a-12 a-13 a-1-3 a-1-2 a-1-1 ) ,$ $𝔰𝔬4= ( a11 a12 a13 a1-3 a1-2 0 a21 a22 a23 a2-3 0 -a1-2 a31 a32 a33 0 -a2-3 -a1-2 a31 a32 0 -a33 -a23 -a13 a21 0 -a-32 -a32 -a22 -a12 0 -a-21 -a-31 -a31 -a21 -a11 )$

with

$𝔥= { ( λ10 λ2 λ3 -λ3 -λ2 0-λ1 ) }$

and if $𝔤={𝔰𝔬}_{6}$ then

$𝔤=𝔥+ a12Xε1-ε2 + a13Xε1-ε3 + a23Xε2-ε3 + a21Xε2-ε1 + a31Xε3-ε1 + a32Xε3-ε2 + a1-2Xε1+ε2 + a1-3Xε1+ε3 + a2-3Xε2+ε3 + a-21X-(ε1+ε2) + a-31X-(ε1+ε3) + a-32X-(ε2+ε3)$

where, for $i

$Xεi-εj= Eij-E-j-i, Xεi+εj= Ei-j- Ej-i Xεj-εi= Eji-E-i-j, X-εi-εj= E-ji-E-ij$

and

$[h,Xεi-εj] = [ h, Eij- E-j-i ] =(λi-λj) Eij- (-λj+λi) E-j-i = (λi-λj) ( Eij- E-j-i ) =(λi-λj) Xεi-εj, [h,Xεi+εj] = [ h,Ei-j- Ej-i ] =(λi+λj) Ei-(λj+λi) Ej-i = (λi+λj) ( Ei-j- Ej-i ) =(λi+λj) Xεi+εj,$

if

$h= ( λ1 λ2 λ3 -λ3 -λ2 -λ1 )$

and ${\epsilon }_{i}:𝔥\to ℂ$ is given by ${\epsilon }_{i}\left(h\right)={\lambda }_{i}\text{.}$

## The root system

$R = { ±εi±ej | 1≤i≠j ≤r } and R+ = { εi±εj | 1≤i

## The Dynkin diagram

The fundamental chamber is

$(𝔥ℝ*)+ = { λ∈𝔥ℝ* | ⟨λ,α∨⟩ ≥0 for α∨∈ (R∨)+ } = { λ1ε1+…+ λrεr | λ1≥λ2≥… ≥λr-1≥ |λr| }$

This chamber is on the positive side of the hyperplanes

$𝔥α∨= { λ∈𝔥ℝ* | ⟨λ,α∨⟩ =0 } forα∨∈ (R∨)+$

where

$(R∨)+ { εi∨±εj∨ | 1≤i

and

$⟨εi,εj∨⟩ =δij.$

The walls of ${\left({𝔥}_{ℝ}^{*}\right)}^{+}$ are

$𝔥ε1∨-ε2∨, 𝔥ε2∨-ε3∨,…, 𝔥εr-1∨-εr∨, 𝔥εr-1∨+εr∨$

and the Dynkin diagram is

$ε1∨-ε2∨ ε2∨-ε3∨ ε3∨-ε4∨ ε4∨-ε5∨ ε4∨+ε5∨ for𝔰𝔬10$

## The Weyl group ${W}_{0}$

${W}_{0}$ is generated by the reflection in the hyperplanes ${𝔥}^{{\epsilon }_{i}^{\vee }±{\epsilon }_{j}^{\vee }}\text{.}$ Using the basis ${\epsilon }_{1},\dots ,{\epsilon }_{r}$ for ${𝔥}^{*},$ the group ${W}_{0}$ is generated by

$sij= ( 1 ⋱ 01 10 1 ⋱ 1 1 ⋱ 01 10 1 ⋱ 1 )$

and

$sεi+εj= 1 ⋱ 1 i 1 ⋱ 1 j 1 ⋱ 1 1 ⋱ 1 -j 1 ⋱ 1 -i 1 ⋱ 1 ( 1 ⋱ 1 0 1 ⋱ 1 0 1 ⋱ 1 1 ⋱ 1 1 1 ⋱ 1 1 1 ⋱ 1 1 ⋱ 1 1 1 ⋱ 1 1 1 ⋱ 1 1 ⋱ 1 0 1 ⋱ 1 0 1 ⋱ 1 )$ $W0≅ { r×r matrices with (a) exactly one nonzero entry in each row and column (b) nonzero entries are ±1. (c) ∏nonzero entriesaij=1 }$

## The character of the adjoint representation $𝔤$

Let $M$ be a $𝔤\text{-module.}$

The character of $M$ is

$char(M)= ∑μ∈𝔥* (dim Mμ) eμ,where Mμ= { m∈M | hm=μ (h)m for each h∈𝔥 }$

is the $\mu \text{-weight}$ space of $M\text{.}$

The weights of the adjoint representation for $𝔤={𝔰𝔬}_{10}$ are

$ε1-ε2 ε1-ε3 ε1-ε4 ε1-ε5 ε1+ε5 ε1+ε4 ε1+ε3 ε1+ε2 ε2-ε3 ε2-ε4 ε2-ε5 ε2+ε5 ε2+ε4 ε2+ε3 ε3-ε4 ε3-ε5 ε3+ε5 ε3+ε4 ε4-ε5 ε4+ε5$

their negatives and the weight 0:

$𝔤0=𝔥and dim(𝔤0)= dim(𝔥)=5.$

The character of $𝔤$ is

$sε1+ε2 = x1x2-1+ x1x3-1+ x1x4-1+ x1x5-1+ x1x5+ x1x4+ x1x3+ x1x2 x2x3-1+ x2x4-1+ x2x5-1+ x2x5+ x2x4+ x2x3 x3x4-1+ x3x5-1+ x3x5+ x3x4 x4x5-1+ x4x5 +5 +x1-1x2 +x1-1x3 +x1-1x4 +x1-1x5 +x1-1x5-1 +x1-1x4-1 +x1-1x3-1 +x1-1x2-1 +x2-1x3 +x2-1x4 +x2-1x5 +x2-1x5-1 +x2-1x4-1 +x2-1x3-1 +x3-1x4 +x3-1x5 +x3-1x5-1 +x3-1x4-1 +x4-1x5 +x4-1x5-1$

where

$xi=eεi, fori=1,2,…,5.$

In this example

$p=12∑α∈R+ α=12 ( 8ε1+6ε2+ 4ε3+2ε4 ) =4ε1+3ε2+ 2ε3+ε4.$

The Weyl denominator formula says

$ap= ∑w∈W0 det(w)w ( x14 x23 x32x4 ) = x14x23 x32x4+ x1-4 x23x32 x4+…+ x4-4 x3-3 x2-2 x1-1 = ep∏α∈R+ (1-e-α)= x14x23x32 x4∏i

and the Weyl character formula says

$sε1+ε2= aε1+ε2+pap = ∑w∈W0 det(w)w ( x15x24 x32x4 ) x14x23x32 x4∏1≤i

## The crystal $B\left({\epsilon }_{1}+{\epsilon }_{2}\right)$

${𝔥}_{ℝ}^{*}$ has basis ${\epsilon }_{1},{\epsilon }_{2},{\epsilon }_{3},{\epsilon }_{4},{\epsilon }_{5}$ and crystals are sets of paths in ${𝔥}_{ℝ}^{*}\simeq {ℝ}^{5}$ which are closed under the action of the root operators

$e∼1, e∼2, e∼3, e∼4, e∼5, f∼1, f∼2, f∼3, f∼4, f∼5$

corresponding the the wall of ${\left({𝔥}_{ℝ}^{*}\right)}^{+}:$

$𝔥ε1∨-ε2∨, 𝔥ε2∨-ε3∨, 𝔥ε3∨-ε4∨, 𝔥ε4∨-ε5∨, 𝔥ε4∨+ε6∨ ε1∨-ε2∨ ε2∨-ε3∨ ε3∨-ε4∨ ε4∨-ε5∨ ε4∨+ε5∨$

The weights of $𝔤$ are $±\left({\epsilon }_{i}±{\epsilon }_{j}\right),$ $1\le i and these are some of the vertices of the 5 dimensional cube.

The highest weight path in $B\left({\epsilon }_{1}+{\epsilon }_{2}\right)$ can be taken to be the straight line path from 0 to ${\epsilon }_{1}+{\epsilon }_{2}\text{.}$

Most of the time (in $B\left({\epsilon }_{1}+{\epsilon }_{2}\right)\text{)}$ the root operators are taking a straight line path to a straight line path. The only exceptions are

$pε4+ε5 →f∼5 (12p-ε4-ε5) ⊗ (12pε4+ε5) →f∼5 p-ε4-ε5 pε4-ε5 →f∼4 (12p-ε4+ε5) ⊗ (12pε4-ε5) →f∼4 p-ε4+ε5 pε3-ε4 ⇄e∼3f∼3 (12p-ε3+ε4) ⊗ (12pε3-ε4) ⇄e∼3f∼3 p-ε3+ε4 pε2-ε3 ⇄e∼2f∼2 (12p-ε2+ε3) ⊗ (12pε2-ε3) ⇄e∼2f∼2 p-ε2+ε3 pε1-ε2 ⇄e∼1f∼1 (12p-ε1+ε2) ⊗ (12pε1-ε2) ⇄e∼1f∼1 p-ε1+ε2$

For the "standard model" in particle physics it is important to understand how this representation decomposes under the action of the subalgebras

$SO10 \subseteq \subseteq SU5 \subseteq \subseteq SU3× U1× SU2$

These restrictions are obtained by ignoring the operators

$f∼6,for SO10⊇ SU5,$

and

$f∼3,for SU5⊇ SU3×U1 ×SU2.$

The crystal graph $B\left({\epsilon }_{1}+{\epsilon }_{2}\right)$ is (all paths are straight line paths except the 5 exceptional ones listed above):

## The crystal graph of $B\left({\epsilon }_{1}+{\epsilon }_{2}\right)$

$ε1+ε2 f∼2 ε1+ε3 f∼1 f∼3 ε2+ε3 ε1+ε4 f∼3 f∼1 f∼5 f∼4 ε2+ε4 ε1-ε5 ε1+ε5 f∼1 f∼1 f∼4 f∼2 f∼5 f∼1 f∼4 f∼5 ε2+ε5 ε3+ε4 ε2-ε5 ε1-ε4 f∼2 f∼4 f∼5 f∼2 f∼4 f∼1 f∼3 ε3+ε5 ε3-ε5 ε2-ε4 ε1-ε3 f∼3 f∼3 f∼4 f∼2 f∼3 f∼1 f∼2 ε4+ε5 ε4-ε5 ε3-ε4 ε2-ε3 ε1-ε2 f∼5 f∼4 f∼3 f∼2 f∼1 12(-ε4-ε5) 12(-ε4+ε5) 12(-ε3+ε4) 12(-ε2+ε3) 12(-ε1+ε2) +12(ε4+ε5) +12(ε4-ε5) +12(ε3-ε4) +12(ε2-ε3) +12(ε1-ε2) f∼5 f∼4 f∼3 f∼2 f∼1 -ε4-ε5 -ε4+ε5 -ε3+ε4 -ε2+ε3 -ε1+ε2 f∼3 f∼3 f∼4 f∼2 f∼3 f∼1 f∼2 -ε3-ε5 -ε3+ε5 -ε2+ε4 -ε1+ε3 f∼2 f∼4 f∼5 f∼2 f∼4 f∼1 f∼3 -ε2-ε5 -ε3-ε4 -ε2+ε5 -ε1+ε4 f∼1 f∼1 f∼4 f∼2 f∼5 f∼1 f∼4 f∼5 -ε2-ε4 -ε1+ε5 -ε1-ε5 f∼3 f∼1 f∼5 f∼4 -ε2-ε3 -ε1-ε4 f∼1 f∼3 -ε1-ε3 f∼2 -ε1-ε2$

$ε1-ε3 f∼1 f∼2 ε4-ε5 ε2-ε3 ε1-ε2 f∼4 f∼2 f∼1 12(-ε4-ε5) 12(-ε3+ε4) 12(-ε2+ε3) 12(-ε1+ε2) +12(ε4-ε5) +12(ε3-ε4) +12(ε2-ε3) +12(ε1-ε2) f∼4 f∼2 f∼1 -ε4+ε5 -ε2+ε3 -ε1+ε2 f∼1 f∼2 -ε1+ε3 3 bosons 1 photon 8 gluons of QCD w+,z,w-$

## Notes and References

This is a typed copy of handwritten notes by Arun Ram on 28/10/2008.