Representation Theory Lecture 12

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

Last update: 20 May 2013

Representation Theory Lecture 12

$$\begin{array}{c}{SO}_{10}=\{g\in {GL}_{10}\hspace{0.17em}|\hspace{0.17em}\text{det}\hspace{0.17em}g=1,\hspace{0.17em}g{g}^t=1\}\\ {𝔰𝔬}_{10}=\{x\in {𝔤𝔩}_{10}\hspace{0.17em}|\hspace{0.17em}\text{tr}\hspace{0.17em}x=0,\hspace{0.17em}x+x^t=0\}\end{array}$$

since

$$1=\text{det}\left(e^{tx}\right)=\text{det}\left(\begin{array}{c}{e}^{t{h}_1}\\ & \ddots \\ & & e^{t{h}_n}\end{array}\right)=e^{t(h_1+\dots +h_n)}=e^{t·\text{tr}\hspace{0.17em}\left(x\right)}$$

and

$$1=e^{tx}{\left(e^{tx}\right)}^t=e^{tx}{e}^{t{x}^t}=e^{t(x+x^t)}$$

Then

$${𝔰𝔬}_5=\left\{\left(\begin{array}{ccccc}0& a_{12}& a_{13}& a_{14}& a_{15}\\ -a_{12}& 0& a_{23}& a_{24}& a_{25}\\ -a_{13}& -a_{23}& 0& a_{34}& a_{35}\\ -a_{14}& -a_{24}& -a_{34}& 0& a_{45}\\ -a_{15}& -a_{25}& -a_{35}& -a_{45}& 0\end{array}\right)\right\},\text{dim}\hspace{0.17em}\left({𝔰𝔬}_5\right)=\frac{5·4}{2}=10$$

Another choice is

$$\begin{array}{c}{SO}_{10}=\{g\in {GL}_{10}\hspace{0.17em}|\hspace{0.17em}\text{det}\hspace{0.17em}g=1,\hspace{0.17em}gJ{g}^t=J\}\\ {𝔰𝔬}_{10}=\{x\in {𝔤𝔩}_{10}\hspace{0.17em}|\hspace{0.17em}\text{tr}\hspace{0.17em}x=0,\hspace{0.17em}xJ+J{x}^t=0\}\end{array}$$

where

$$J=\left(\begin{array}{ccccc}0& & & & 1\\ & & & 1& \\ & & ⋰\\ & 1\\ 1& & & & 0\end{array}\right)\text{or}J=\left(\begin{array}{cc}0& \begin{array}{ccc}1& & 0\\ & \ddots \\ 0& & 1\end{array}\\ \begin{array}{ccc}1& & 0\\ & \ddots \\ 0& & 1\end{array}& 0\end{array}\right)\text{.}$$

Since

$$\left(\begin{array}{cccccc}{a}_{11}& a_{12}& a_{13}& a_{1-3}& a_{1-2}& a_{1-1}\\ a_{21}& a_{22}& a_{23}& a_{2-3}& a_{2-2}& a_{2-1}\\ a_{31}& a_{32}& a_{33}& a_{3-3}& a_{3-2}& a_{3-1}\\ a_{-31}& a_{-32}& a_{-33}& a_{-3-3}& a_{-3-2}& a_{-3-1}\\ a_{-21}& a_{-22}& a_{-23}& a_{-2-3}& a_{-2-2}& a_{-2-1}\\ a_{-11}& a_{-12}& a_{-13}& a_{-1-3}& a_{-1-2}& a_{-1-1}\end{array}\right)\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0\end{array}\right)=\left(\begin{array}{cccccc}{a}_{1-1}& a_{1-2}& a_{1-3}& a_{13}& a_{12}& a_{11}\\ a_{2-1}& a_{2-2}& a_{2-3}& a_{23}& a_{22}& a_{21}\\ a_{3-1}& a_{3-2}& a_{3-3}& a_{33}& a_{32}& a_{31}\\ a_{-3-1}& a_{-3-2}& a_{-3-3}& a_{-33}& a_{-32}& a_{-31}\\ a_{-2-1}& a_{-2-2}& a_{-2-3}& a_{-23}& a_{-22}& a_{-21}\\ a_{-11}& a_{-12}& a_{-13}& a_{-1-3}& a_{-1-2}& a_{-1-1}\end{array}\right),$$ $${𝔰𝔬}_4=\left(\begin{array}{cccccc}{a}_{11}& a_{12}& a_{13}& a_{1-3}& a_{1-2}& 0\\ a_{21}& a_{22}& a_{23}& a_{2-3}& 0& -a_{1-2}\\ a_{31}& a_{32}& a_{33}& 0& -a_{2-3}& -a_{1-2}\\ a_{31}& a_{32}& 0& -a_{33}& -a_{23}& -a_{13}\\ a_{21}& 0& -a_{-32}& -a_{32}& -a_{22}& -a_{12}\\ 0& -a_{-21}& -a_{-31}& -a_{31}& -a_{21}& -a_{11}\end{array}\right)$$

with

$$𝔥=\left\{\left(\begin{array}{cccccc}{\lambda }_1& & & & & 0\\ & {\lambda }_2\\ & & {\lambda }_3\\ & & & -{\lambda }_3\\ & & & & -{\lambda }_2\\ 0& & & & & -{\lambda }_1\end{array}\right)\right\}$$

and if $𝔤={𝔰𝔬}_6$ then

$$𝔤=𝔥+a_{12}{X}_{{\epsilon }_1-{\epsilon }_2}+a_{13}{X}_{{\epsilon }_1-{\epsilon }_3}+a_{23}{X}_{{\epsilon }_2-{\epsilon }_3}+a_{21}{X}_{{\epsilon }_2-{\epsilon }_1}+a_{31}{X}_{{\epsilon }_3-{\epsilon }_1}+a_{32}{X}_{{\epsilon }_3-{\epsilon }_2}+a_{1-2}{X}_{{\epsilon }_1+{\epsilon }_2}+a_{1-3}{X}_{{\epsilon }_1+{\epsilon }_3}+a_{2-3}{X}_{{\epsilon }_2+{\epsilon }_3}+a_{-21}{X}_{-({\epsilon }_1+{\epsilon }_2)}+a_{-31}{X}_{-({\epsilon }_1+{\epsilon }_3)}+a_{-32}{X}_{-({\epsilon }_2+{\epsilon }_3)}$$

where, for $i<j,$

$$\begin{array}{cc}{X}_{{\epsilon }_i-{\epsilon }_j}=E_{ij}-E_{-j-i},& X_{{\epsilon }_i+{\epsilon }_j}=E_{i-j}-E_{j-i}\\ X_{{\epsilon }_j-{\epsilon }_i}=E_{ji}-E_{-i-j},& X_{-{\epsilon }_i-{\epsilon }_j}=E_{-ji}-E_{-ij}\end{array}$$

and

$$\begin{array}{ccc}[h,X_{{\epsilon }_i-{\epsilon }_j}]& =& [h,E_{ij}-E_{-j-i}]=({\lambda }_i-{\lambda }_j)E_{ij}-(-{\lambda }_j+{\lambda }_i)E_{-j-i}\\ & =& ({\lambda }_i-{\lambda }_j)(E_{ij}-E_{-j-i})=({\lambda }_i-{\lambda }_j)X_{{\epsilon }_i-{\epsilon }_j},\\ [h,X_{{\epsilon }_i+{\epsilon }_j}]& =& [h,E_{i-j}-E_{j-i}]=({\lambda }_i+{\lambda }_j)E_i-({\lambda }_j+{\lambda }_i)E_{j-i}\\ & =& ({\lambda }_i+{\lambda }_j)(E_{i-j}-E_{j-i})=({\lambda }_i+{\lambda }_j)X_{{\epsilon }_i+{\epsilon }_j},\end{array}$$

if

$$h=\left(\begin{array}{c}{\lambda }_1\\ & {\lambda }_2\\ & & {\lambda }_3\\ & & & -{\lambda }_3\\ & & & & -{\lambda }_2\\ & & & & & -{\lambda }_1\end{array}\right)$$

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

The root system

$$\begin{array}{ccc}R& =& \{\pm {\epsilon }_i\pm e_j\hspace{0.17em}|\hspace{0.17em}1\le i\ne j\le r\}\text{and}\\ R^{+}& =& \{{\epsilon }_i\pm {\epsilon }_j\hspace{0.17em}|\hspace{0.17em}1\le i<j\le r\}\text{.}\end{array}$$

The Dynkin diagram

The fundamental chamber is

$$\begin{array}{ccc}{\left({𝔥}_{ℝ}^{*}\right)}^{+}& =& \{\lambda \in {𝔥}_{ℝ}^{*}\hspace{0.17em}|\hspace{0.17em}⟨\lambda ,{\alpha }^{\vee }⟩\ge 0\hspace{0.17em}\text{for}\hspace{0.17em}{\alpha }^{\vee }\in {\left(R^{\vee }\right)}^{+}\}\\ & =& \{{\lambda }_1{\epsilon }_1+\dots +{\lambda }_r{\epsilon }_r\hspace{0.17em}|\hspace{0.17em}{\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_{r-1}\ge \left|{\lambda }_r\right|\}\end{array}$$

This chamber is on the positive side of the hyperplanes

$${𝔥}^{{\alpha }^{\vee }}=\{\lambda \in {𝔥}_{ℝ}^{*}\hspace{0.17em}|\hspace{0.17em}⟨\lambda ,{\alpha }^{\vee }⟩=0\}\text{for}{\alpha }^{\vee }\in {\left(R^{\vee }\right)}^{+}$$

where

$${\left(R^{\vee }\right)}^{+}\{{\epsilon }_i^{\vee }\pm {\epsilon }_j^{\vee }\hspace{0.17em}|\hspace{0.17em}1\le i<j\le e\}$$

and

$$⟨{\epsilon }_i,{\epsilon }_j^{\vee }⟩={\delta }_{ij}\text{.}$$

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

$${𝔥}^{{\epsilon }_1^{\vee }-{\epsilon }_2^{\vee }},{𝔥}^{{\epsilon }_2^{\vee }-{\epsilon }_3^{\vee }},\dots ,{𝔥}^{{\epsilon }_{r-1}^{\vee }-{\epsilon }_r^{\vee }},{𝔥}^{{\epsilon }_{r-1}^{\vee }+{\epsilon }_r^{\vee }}$$

and the Dynkin diagram is

$$\begin{array}{cc} ε1∨-ε2∨ ε2∨-ε3∨ ε3∨-ε4∨ ε4∨-ε5∨ ε4∨+ε5∨ & \text{for}{𝔰𝔬}_{10}\end{array}$$

The Weyl group $W_0$

$W_0$ is generated by the reflection in the hyperplanes ${𝔥}^{{\epsilon }_i^{\vee }\pm {\epsilon }_j^{\vee }}\text{.}$ Using the basis ${\epsilon }_1,\dots ,{\epsilon }_r$ for ${𝔥}^{*},$ the group $W_0$ is generated by

$$s_{ij}=\left(\begin{array}{cc}\begin{array}{c}1\\ & \ddots \\ & & 0& 1\\ & & 1& 0\\ & & & & & 1\\ & & & & & & \ddots \\ & & & & & & & 1\end{array}& \\ & \begin{array}{c}1\\ & \ddots \\ & & 0& 1\\ & & 1& 0\\ & & & & & 1\\ & & & & & & \ddots \\ & & & & & & & 1\end{array}\end{array}\right)$$

and

$$s_{{\epsilon }_i+{\epsilon }_j}=\begin{array}{cc}\begin{array}{c}\phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ i\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ j\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ -j\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ -i\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\end{array}& \left(\begin{array}{cc}\begin{array}{c}1\\ & \ddots \\ & & 1\\ & & & 0\\ & & & & 1\\ & & & & & \ddots \\ & & & & & & 1\\ & & & & & & & 0\\ & & & & & & & & 1\\ & & & & & & & & & \ddots \\ & & & & & & & & & & 1\end{array}& \begin{array}{c}\phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ & & & & & & & 1\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ & & & 1\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\end{array}\\ \begin{array}{c}\phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ & & & & & & & 1\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\\ & & & 1\\ \phantom{1}\\ \phantom{\ddots }\\ \phantom{1}\end{array}& \begin{array}{c}1\\ & \ddots \\ & & 1\\ & & & 0\\ & & & & 1\\ & & & & & \ddots \\ & & & & & & 1\\ & & & & & & & 0\\ & & & & & & & & 1\\ & & & & & & & & & \ddots \\ & & & & & & & & & & 1\end{array}\end{array}\right)\end{array}$$ $$W_0\cong \left\{\begin{array}{c}r\times r\hspace{0.17em}\text{matrices with}\\ \text{(a) exactly one nonzero entry in each row and column}\\ \text{(b) nonzero entries are}\hspace{0.17em}\pm 1\text{.}\\ \text{(c)}\hspace{0.17em}\prod _{\text{nonzero entries}}{a}_{ij}=1\end{array}\right\}$$

The character of the adjoint representation $𝔤$

Let $M$ be a $𝔤\text{-module.}$

The character of $M$ is

$$\begin{array}{c}\text{char}\left(M\right)=\sum _{\mu \in {𝔥}^{*}}\left(\text{dim}\hspace{0.17em}{M}_{\mu }\right)e^{\mu },\text{where}\\ M_{\mu }=\{m\in M\hspace{0.17em}|\hspace{0.17em}hm=\mu \left(h\right)m\hspace{0.17em}\text{for each}\hspace{0.17em}h\in 𝔥\}\end{array}$$

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

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

$$\begin{array}{cccccccc}{\epsilon }_1-{\epsilon }_2& {\epsilon }_1-{\epsilon }_3& {\epsilon }_1-{\epsilon }_4& {\epsilon }_1-{\epsilon }_5& {\epsilon }_1+{\epsilon }_5& {\epsilon }_1+{\epsilon }_4& {\epsilon }_1+{\epsilon }_3& {\epsilon }_1+{\epsilon }_2\\ & {\epsilon }_2-{\epsilon }_3& {\epsilon }_2-{\epsilon }_4& {\epsilon }_2-{\epsilon }_5& {\epsilon }_2+{\epsilon }_5& {\epsilon }_2+{\epsilon }_4& {\epsilon }_2+{\epsilon }_3& \\ & & {\epsilon }_3-{\epsilon }_4& {\epsilon }_3-{\epsilon }_5& {\epsilon }_3+{\epsilon }_5& {\epsilon }_3+{\epsilon }_4& & \\ & & & {\epsilon }_4-{\epsilon }_5& {\epsilon }_4+{\epsilon }_5& & & \end{array}$$

their negatives and the weight 0:

$${𝔤}_0=𝔥\text{and}\text{dim}\left({𝔤}_0\right)=\text{dim}\left(𝔥\right)=5\text{.}$$

The character of $𝔤$ is

$$\begin{array}{ccc}{s}_{{\epsilon }_1+{\epsilon }_2}& =& x_1{x}_2^{-1}+x_1{x}_3^{-1}+x_1{x}_4^{-1}+x_1{x}_5^{-1}+x_1{x}_5+x_1{x}_4+x_1{x}_3+x_1{x}_2\\ & & x_2{x}_3^{-1}+x_2{x}_4^{-1}+x_2{x}_5^{-1}+x_2{x}_5+x_2{x}_4+x_2{x}_3\\ & & x_3{x}_4^{-1}+x_3{x}_5^{-1}+x_3{x}_5+x_3{x}_4\\ & & x_4{x}_5^{-1}+x_4{x}_5\\ & & +5\\ & & +x_1^{-1}{x}_2+x_1^{-1}{x}_3+x_1^{-1}{x}_4+x_1^{-1}{x}_5+x_1^{-1}{x}_5^{-1}+x_1^{-1}{x}_4^{-1}+x_1^{-1}{x}_3^{-1}+x_1^{-1}{x}_2^{-1}\\ & & +x_2^{-1}{x}_3+x_2^{-1}{x}_4+x_2^{-1}{x}_5+x_2^{-1}{x}_5^{-1}+x_2^{-1}{x}_4^{-1}+x_2^{-1}{x}_3^{-1}\\ & & +x_3^{-1}{x}_4+x_3^{-1}{x}_5+x_3^{-1}{x}_5^{-1}+x_3^{-1}{x}_4^{-1}\\ & & +x_4^{-1}{x}_5+x_4^{-1}{x}_5^{-1}\end{array}$$

where

$$x_i=e^{{\epsilon }_i},\text{for}i=1,2,\dots ,5\text{.}$$

In this example

$$p=\frac{1}{2}\sum _{\alpha \in R^{+}}\alpha =\frac{1}{2}(8{\epsilon }_1+6{\epsilon }_2+4{\epsilon }_3+2{\epsilon }_4)=4{\epsilon }_1+3{\epsilon }_2+2{\epsilon }_3+{\epsilon }_4\text{.}$$

The Weyl denominator formula says

$$\begin{array}{ccc}{a}_p& =& \sum _{w\in W_0}\text{det}\left(w\right)w\left(x_1^4{x}_2^3{x}_3^2{x}_4\right)=x_1^4{x}_2^3{x}_3^2{x}_4+x_1^{-4}{x}_2^3{x}_3^2{x}_4+\dots +x_4^{-4}{x}_3^{-3}{x}_2^{-2}{x}_1^{-1}\\ & =& e^p\prod _{\alpha \in R^{+}}(1-e^{-\alpha })=x_1^4{x}_2^3{x}_3^2{x}_4\prod _{i<j}(1-x_i^{-1}{x}_j)(1-x_i^{-1}{x}_j^{-1})\end{array}$$

and the Weyl character formula says

$$s_{{\epsilon }_1+{\epsilon }_2}=\frac{a_{{\epsilon }_1+{\epsilon }_2+p}}{a_p}=\frac{\sum _{w\in W_0}\text{det}\left(w\right)w\left(x_1^5{x}_2^4{x}_3^2{x}_4\right)}{x_1^4{x}_2^3{x}_3^2{x}_4\prod _{1\le i<j\le 5}(1-x_i^{-1}{x}_j)(1-x_i^{-1}{x}_j^{-1})}$$

The crystal $B({\epsilon }_1+{\epsilon }_2)$

${𝔥}_{ℝ}^{*}$ 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

$${\stackrel{\sim }{e}}_1,{\stackrel{\sim }{e}}_2,{\stackrel{\sim }{e}}_3,{\stackrel{\sim }{e}}_4,{\stackrel{\sim }{e}}_5,{\stackrel{\sim }{f}}_1,{\stackrel{\sim }{f}}_2,{\stackrel{\sim }{f}}_3,{\stackrel{\sim }{f}}_4,{\stackrel{\sim }{f}}_5$$

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

$$\begin{array}{c}{𝔥}^{{\epsilon }_1^{\vee }-{\epsilon }_2^{\vee }},{𝔥}^{{\epsilon }_2^{\vee }-{\epsilon }_3^{\vee }},{𝔥}^{{\epsilon }_3^{\vee }-{\epsilon }_4^{\vee }},{𝔥}^{{\epsilon }_4^{\vee }-{\epsilon }_5^{\vee }},{𝔥}^{{\epsilon }_4^{\vee }+{\epsilon }_6^{\vee }}\\ ε1∨-ε2∨ ε2∨-ε3∨ ε3∨-ε4∨ ε4∨-ε5∨ ε4∨+ε5∨ \end{array}$$

The weights of $𝔤$ are $\pm ({\epsilon }_i\pm {\epsilon }_j),$ $1\le i<j\le r$ and these are some of the vertices of the 5 dimensional cube.

The highest weight path in $B({\epsilon }_1+{\epsilon }_2)$ can be taken to be the straight line path from 0 to ${\epsilon }_1+{\epsilon }_2\text{.}$

Most of the time (in $B({\epsilon }_1+{\epsilon }_2)\text{)}$ the root operators are taking a straight line path to a straight line path. The only exceptions are

$$\begin{array}{ccccccc}{p}_{{\epsilon }_4+{\epsilon }_5}& \stackrel{{\stackrel{\sim }{f}}_5}{\to }& \left(\frac{1}{2}{p}_{-{\epsilon }_4-{\epsilon }_5}\right)& \otimes & \left(\frac{1}{2}{p}_{{\epsilon }_4+{\epsilon }_5}\right)& \stackrel{{\stackrel{\sim }{f}}_5}{\to }& p_{-{\epsilon }_4-{\epsilon }_5}\\ p_{{\epsilon }_4-{\epsilon }_5}& \stackrel{{\stackrel{\sim }{f}}_4}{\to }& \left(\frac{1}{2}{p}_{-{\epsilon }_4+{\epsilon }_5}\right)& \otimes & \left(\frac{1}{2}{p}_{{\epsilon }_4-{\epsilon }_5}\right)& \stackrel{{\stackrel{\sim }{f}}_4}{\to }& p_{-{\epsilon }_4+{\epsilon }_5}\\ p_{{\epsilon }_3-{\epsilon }_4}& \underset{{\stackrel{\sim }{e}}_3}{\overset{{\stackrel{\sim }{f}}_3}{\rightleftarrows }}& \left(\frac{1}{2}{p}_{-{\epsilon }_3+{\epsilon }_4}\right)& \otimes & \left(\frac{1}{2}{p}_{{\epsilon }_3-{\epsilon }_4}\right)& \underset{{\stackrel{\sim }{e}}_3}{\overset{{\stackrel{\sim }{f}}_3}{\rightleftarrows }}& p_{-{\epsilon }_3+{\epsilon }_4}\\ p_{{\epsilon }_2-{\epsilon }_3}& \underset{{\stackrel{\sim }{e}}_2}{\overset{{\stackrel{\sim }{f}}_2}{\rightleftarrows }}& \left(\frac{1}{2}{p}_{-{\epsilon }_2+{\epsilon }_3}\right)& \otimes & \left(\frac{1}{2}{p}_{{\epsilon }_2-{\epsilon }_3}\right)& \underset{{\stackrel{\sim }{e}}_2}{\overset{{\stackrel{\sim }{f}}_2}{\rightleftarrows }}& p_{-{\epsilon }_2+{\epsilon }_3}\\ p_{{\epsilon }_1-{\epsilon }_2}& \underset{{\stackrel{\sim }{e}}_1}{\overset{{\stackrel{\sim }{f}}_1}{\rightleftarrows }}& \left(\frac{1}{2}{p}_{-{\epsilon }_1+{\epsilon }_2}\right)& \otimes & \left(\frac{1}{2}{p}_{{\epsilon }_1-{\epsilon }_2}\right)& \underset{{\stackrel{\sim }{e}}_1}{\overset{{\stackrel{\sim }{f}}_1}{\rightleftarrows }}& p_{-{\epsilon }_1+{\epsilon }_2}\end{array}$$

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

$$\begin{array}{cc}{SO}_{10}& \\ ⊆ & ⊆ \\ {SU}_5& \\ ⊆ & ⊆ \\ {SU}_3\times U_1\times {SU}_2& \end{array}$$

These restrictions are obtained by ignoring the operators

$${\stackrel{\sim }{f}}_6,\text{for}{SO}_{10}\supseteq {SU}_5,$$

and

$${\stackrel{\sim }{f}}_3,\text{for}{SU}_5\supseteq {SU}_3\times U_1\times {SU}_2\text{.}$$

The crystal graph $B({\epsilon }_1+{\epsilon }_2)$ is (all paths are straight line paths except the 5 exceptional ones listed above):

The crystal graph of $B({\epsilon }_1+{\epsilon }_2)$

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

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

Notes and References

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

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