---
title: "Problem Sets: Lie Algebras"
subtitle: "Problem Set 5"
author:
- name: D. Zack Garza
  affiliation: University of Georgia 
  email: dzackgarza@gmail.com 
date: Fall 2022
order: 5
---
 
# Problem Set 5

## Section 9

:::{.proposition title="Humphreys 9.2"}
Prove that $\Phi\dual$ is a root system in $E$, whose Weyl group is naturally isomorphic to $\mathcal{W}$; show also that $\left\langle\alpha\dual, \beta\dual\right\rangle=\langle\beta, \alpha\rangle$, and draw a picture of $\Phi\dual$ in the cases $A_1, A_2, B_2, G_2$.
:::

:::{.solution}

We recall and introduce some notation:
\[
\norm{ \alpha}^2 &\da (\alpha, \alpha) \\ \\
\inp{ \beta}{ \alpha} &\da {2 (\beta, \alpha)\over \norm{ \alpha}^2} \\
&= {2 (\beta, \alpha) \over ( \alpha, \alpha)} \\ \\
s_\alpha( \beta) 
&= \beta - {2 (\beta, \alpha) \over \norm{\alpha}^2 } \alpha \\
&= \beta - {2 (\beta, \alpha) \over (\alpha, \alpha) } \alpha \\ \\
\alpha\dual &\da {2 \over \norm{\alpha}^2} \alpha = {2\over (\alpha, \alpha)} \alpha
.\]

:::{.claim}
\[
\inp{\alpha\dual}{\beta\dual} = \inp \beta \alpha
.\]
:::

:::{.proof title="?"}
This is a computation:
\[
\inp{ \alpha\dual} { \beta\dual} 
&= {2 (\alpha\dual, \beta\dual) \over \norm{\beta\dual}^2 } \\
&= {2 (\alpha\dual, \beta\dual) \over (\beta\dual, \beta\dual) } \\
&= {2\qty{ {2\alpha\over \norm{\alpha}^2}, {2 \beta\over \norm{\beta}^2} } \over \qty{{2 \beta\over \norm{\beta}^2}, {2 \beta\over \norm{\beta}^2} } } \\
&= {2^3 \norm{\beta}^4 (\alpha, \beta)\over 2^2 \norm{\alpha}^2 \norm{\beta}^2 (\beta, \beta)} \\
&= {2^3 \norm{\beta}^4 (\alpha, \beta)\over 2^2 (\alpha, \alpha) \norm{\beta}^2 \norm{\beta}^2} \\
&= {2( \alpha, \beta) \over (\alpha, \alpha)} \\
&= \inp{\beta}{ \alpha}
.\]


:::

:::{.claim}
$\Phi\dual$ is a root system.
:::

:::{.proof title="?"}
The axioms can be checked individually:

- $R1$: there is a bijection of sets
\[
\Phi &\iso \Phi\dual \\
\alpha &\mapsto \alpha\dual
,\]
  thus $\size \Phi\dual = \size \Phi < \infty$.
  To see that $\RR \Phi\dual = \EE$, for $\vector v\in \EE$, use the fact that $\RR \Phi = \EE$ to write $\vector v = \sum_{ \alpha\in \Phi} c_\alpha \alpha$, then
  \[
  \vector v 
  &= \sum_{ \alpha\in \Phi} c_ \alpha \alpha \\
  &= \sum_{ \alpha\in \Phi} c_{\alpha} {\norm{ \alpha}^2\over 2}\cdot {2\over \norm{ \alpha}^2} \alpha \\
  &\da \sum_{ \alpha\in \Phi} c_{\alpha} {\norm{ \alpha}^2\over 2} \alpha\dual \\
  &= \sum_{\alpha\dual \in \Phi\dual} d_{\alpha\dual} \alpha\dual, \qquad d_{ \alpha\dual} \da {1\over 2}c_ \alpha \norm{ \alpha^2}
  ,\]
  so $\vector v\in \RR \Phi\dual$.
  Finally, $\vector 0\not\in \Phi\dual$ since ${2\over (\alpha, \alpha)} \alpha\neq \vector 0$ since $\alpha \in \Phi\implies \alpha\neq \vector 0$, and $2/(\alpha, \alpha)$ is never zero.

- $R2$:
  It suffices to show that if $\lambda \alpha\dual = \beta\dual \in \Phi\dual$ then $\lambda = \pm 1$ and $\beta\dual = \alpha\dual$.
  So suppose \( \lambda \alpha\dual = \beta\dual \), then 
  \[
  \lambda {2\over \norm{ \alpha}^2} \alpha = {2\over \norm{ \beta}^2} \beta \implies \beta = \lambda{\norm{\beta}^2 \over \norm{ \alpha}^2} \alpha \da \lambda' \alpha
  ,\]
  and since $\Phi$ satisfies $R2$, we have \( \lambda' = \pm 1 \) and $\beta = \alpha$.
  But then 
  \[
\pm 1 = \lambda {\norm{ \beta}^2 \over \norm{ \alpha}^2} = \lambda{\norm{ \alpha}^2 \over \norm{ \alpha}^2} = \lambda
  .\]
  Finally, if \( \alpha = \beta \) then \( \alpha\dual = \beta\dual \) since $\Phi, \Phi\dual$ are in bijection.

:::


:::{.proof title="of R3 and R4"}
Continuing:

- $R3$:
  It suffices to show that if \( \alpha\dual, \beta\dual \in \Phi\dual \) then $s_{\alpha\dual}(\beta\dual) = \gamma\dual$ for some \( \gamma\dual\in \Phi \dual \).
  This follows from a computation:
  \[
  s_{ \alpha\dual}(\beta\dual) 
  &= \beta\dual - \inp{\beta\dual }{ \alpha\dual} \alpha\dual \\
  &= \beta\dual - \inp{ \alpha}{ \beta} \alpha\dual \\
  &= {2\beta\over \norm{ \beta}^2 }- \inp{ \alpha}{ \beta} {2 \alpha\over \norm{ \alpha}^2 } \\
  &= {2\beta\over \norm{ \beta}^2 }- {2 (\alpha, \beta) \over (\beta, \beta) } {2 \alpha\over \norm{ \alpha}^2 } \\
  &= {2\beta\over \norm{ \beta}^2 }- {2 (\alpha, \beta) \over \norm{\beta}^2 } {2 \alpha\over \norm{ \alpha}^2 } \\
  &= {2\over \norm{\beta}^2} \qty{ \beta -  {2 (\alpha, \beta) \over \norm{\alpha}^2 } \alpha } \\
  &= {2\over ( \beta, \beta)} \qty{ \beta -  {2 (\beta, \alpha) \over \norm{\alpha}^2 } \alpha } \\
  &= {2\over ( \beta, \beta)} \sigma_{ \alpha}(\beta) \\
  &= {2\over ( \sigma_{ \alpha}(\beta), \sigma_{ \alpha}(\beta) )} \sigma_{ \alpha}(\beta) \\
  &\da (\sigma_ \alpha( \beta))\dual
  ,\]
  where we've used that \( \sigma_{\alpha} \) is an isometry with respect to the symmetric bilinear form $(\wait, \wait)$.

- $R4$: 
  This follows directly from the formula proved in the claim at the beginning:
  \[
  \inp{ \alpha\dual}{\beta\dual} = \inp{ \beta}{ \alpha}\in \ZZ
  ,\]
  since \( \alpha, \beta\in \Phi \) and $\Phi$ satisfies $R4$.

:::


:::{.claim}
There is an isomorphism of groups $\mcw(\Phi) \iso \mcw(\Phi\dual)$.
:::

:::{.proof title="?"}
There is a map of Weyl groups
\[
\tilde \psi: \mcw(\Phi) &\iso \mcw(\Phi\dual) \\
s_ \alpha &\mapsto s_{\alpha\dual}
,\]
which is clearly a bijection of sets with inverse $s_{\alpha\dual} \mapsto s_{ \alpha}$.
Since it is also a group morphism, this yields an isomorphism of groups.
:::

:::{.remark}
The following are pictures of $\Phi\dual$ in the stated special cases:

- $A_1$: Writing $\Phi(A_1) = \ts{\alpha = e_1 - e_2, - \alpha = e_2 -e_1} \subseteq \RR^2$, we have $(\alpha, \alpha) = \sqrt 2$ and thus $\Phi(A_1)\dual = \ts{\sqrt{2} \alpha, - \sqrt{2} \alpha}$:

![](figures/2022-10-31_22-18-46.png)

- $A_2$: Writing $\Phi(A_2) = \ts{e_1 - e_2, e_1 - e_3, e_2 - e_1, e_2 - e_3, e_3 - e_1, e_3 - e_2} \subseteq \RR^3$, noting that every root has length $\sqrt{2}$, the dual results in a scaled version of $A_2$:

![](figures/2022-10-31_22-58-28.png)

- $B_2$: Let $\Phi(B_2) = \ts{e_1, -e_1, e_2, -e_2, e_1 + e_2, e_1 - e_2, -e_1 + e_2, -e_1 -e_2}$ with $\alpha = e_1$ the short root at $\beta = -e_1 + e_2$ the long root, taking the dual fixes the short roots $\pm e_1$ and $\pm e_2$, and normalizes the lengths of the long roots $\pm e_1 \pm e_2$ to 1:

![](figures/2022-10-31_23-18-28.png)

- $G_2$: $\Phi(G_2)$ is shown here in gray, with $\Phi(G_2)\dual$ in green:

![](figures/2022-10-31_23-45-39.png)


:::

:::

:::{.proposition title="Humphreys 9.3"}
In Table 1, show that the order of $\sigma_\alpha \sigma_\beta$ in $\mathcal{W}$ is (respectively) $2,3,4,6$ when $\theta=\pi / 2, \pi / 3$ (or $2 \pi / 3$ ), $\pi / 4$ (or $3 \pi / 4$ ), $\pi / 6$ (or $5 \pi / 6$ ). 

> Note that $\sigma_\alpha \sigma_\beta=$ rotation through $2 \theta$.

:::

:::{.solution}
Given the hint, this is immediate: if $s_\alpha s_\beta = R_{2\theta}$ is a rigid rotation through an angle of $2\theta$, then it's clear that
\[
R_{2 \cdot {\pi \over 2}}^2 = R_{2 \cdot {\pi \over 3}}^3 = R_{2\cdot {\pi \over 4}}^4 = R_{2\cdot {\pi \over 6}}^6 = \id
,\]
since these are all rotations through an angle of $2\pi$.

To prove the hint, note that in any basis, a reflection has determinant $-1$ since it fixes an $n-1\dash$dimensional subspace (the hyperplane $H_\alpha$ of reflection) and negates its 1-dimensional complement (generated by the normal to $H_\alpha$). On the other hand, $\det(s_\alpha s_\beta) = (-1)^2 = 1$ and is an isometry that only fixes the intersection $H_\alpha = H_\beta = \ts{\vector 0}$, so it must be a rotation.

To see that this is a rotation through an angle of exactly $2\theta$, consider applying $s_\beta\circ s_\alpha$ to a point $P$. Letting $H_\alpha, H_\beta$ by the corresponding hyperplanes. We then have the following geometric situation:

![](figures/2022-11-01_00-54-53.png)

We then have $\theta_1 + \theta_2 = \theta$, noting that the angle between $\alpha$ and $\beta$ is equal to the angle between the hyperplanes $H_\alpha, H_\beta$. The total angle measure between $P$ and $s_\beta(s_\alpha(P))$ is then $2\theta_1 + 2\theta_2 = 2\theta$.
:::

:::{.proposition title="Humphreys 9.4"}
Prove that the respective Weyl groups of $A_1 \times A_1, A_2, B_2, G_2$ are dihedral of order $4,6,8,12$. If $\Phi$ is any root system of rank 2 , prove that its Weyl group must be one of these.
:::

:::{.solution}
In light of the fact that 
\[
D_{2n} = \gens{s, r \st r^n = s^2 = 1, srs\inv = r^{-1}}
\]
where $r$ is a rotation and $s$ is a reflection, for the remainder of this problem, let $s \da s_\alpha$ and $r \da s_\alpha s_\beta$ after choosing roots $\alpha$ and $\beta$.

- $A_1\times A_1$: we have $\Phi(A_2) = \pm e_1, \pm e_2$, and setting $\alpha = e_1, \beta = e_2$ yields $\theta = \pi/2$.
  We have 
  \[
  \mcw(A_2) = \ts{\id, s_{\alpha}, s_{\beta}, s_\alpha s_\beta}
  \]
  where $s_{\alpha}^2 = s_{\beta}^2 = 1$, $s_\alpha s_\beta$ is rotation through $2\theta = \pi$ radians, and $(s_\alpha s_\beta)^2 = \id$.
  Setting $r= s_\alpha, s= s_\alpha s_\beta$ yields $r^2 = s^2 = \id$ and $srs = s$, which are the defining relations for $D_4$.

- $A_2$: there is an inscribed triangle in the regular hexagon formed by the convex hull of the roots (see the dotted triangle below), and the reflections $s_\alpha$ about the hyperplanes $H_\alpha$ restrict to precisely the symmetries of this triangle, yielding $D_{6}$:
![](figures/2022-11-01_01-35-46.png)
  Alternatively, choose a simple system $\Delta = \ts{\alpha= e_1, \beta= -e_1 - e_2}$, then $\mcw(A_2) = \gens{s_\alpha, s_\alpha s_\beta}$ is enough to generate the Weyl group. Since we have $s \da s_\alpha \implies s^2 = 1$ and $r\da s_\alpha s_\beta \implies r_3 = 1$ (since $\theta = \pi/3$), these satisfy the relations of $D_6$.

- $B_2$: there is similarly a square on which the hyperplane reflections act on, highlighted with dotted lines here:
![](figures/2022-11-01_01-53-11.png)
  Since the $s_\alpha$ act faithfully as the symmetries of a square, we have $\mcw(B_2)\cong D_{8}$. Alternatively, take $\alpha = e_1$ and $\beta = -e_1 + e_2$ and set $s = s_\alpha, r = s_\alpha s_\beta$.
  Then $\mcw(B_2) = \gens{s, r}$ and since $s^2 = r^4 = e$ (since here $\theta = \pi/4$) and they satisfy the proper commutation relation, this yields precisely the relations for $D_{2n}, n=4$.
  
- $G_2$: In this case, the convex hull of the short roots form a hexagon, on which the hyperplane reflections precisely restrict to symmetries:
![](figures/2022-11-01_02-01-32.png)

  This yields $\mcw(G_2)\cong D_{12}$.
  Alternatively, take $\alpha = e_1$ and $\beta$ the long root in quadrant II, set $s = s_\alpha, r= s_\alpha s_\beta$, then $s^2 = r^6 = 1$ since $\theta = \pi/6$ and again the commutation relations for $D_{2n}, n=6$ are satisfied.

Finally, for any root system $\Phi$ of rank 2, we will have $\mce(\Phi) = \gens{ s \da s_ \alpha, r\da s_ \alpha s _{\beta} }$. Because $\theta$ is restricted to one of the angles in Table 1 in Humphreys $\S 9.4$, i.e. the angles discussed in problem 9.3 above, the order of $s$ is always 2 and the order of $r$ is one of $4,6,8,12$. Since $srs\inv =srs = r\inv$ in all cases, this always yields a dihedral group.
:::

## Section 10

:::{.proposition title="Humphreys 10.1"}
Let $\Phi\dual$ be the dual system of $\Phi, \Delta\dual=\left\{\alpha\dual \mid \alpha \in \Delta\right\}$. 
Prove that $\Delta\dual$ is a base of $\Phi\dual$. 

> Compare Weyl chambers of $\Phi$ and $\Phi\dual$.

:::

:::{.solution}
Suppose that $\Delta$ is a base of $\Phi$.
We can use the fact that bases are in bijective correspondence with Weyl chambers via the correspondence
\[
\Delta \to \mathrm{WC}(\Delta) \da \ts{v\in \EE\st (v, \delta) > 0 \,\, \forall \delta\in \Delta}
,\]
sending $\Delta$ to all of the vectors making an acute angle with all simple vectors $\delta \in \Delta$, or equivalently the intersection of the positive half-spaces formed by the hyperplanes $H_{\delta}$ for $\delta\in \Delta$.

The claim is that $\mathrm{WC}(\Delta\dual) = \mathrm{WC}(\Delta)$, i.e. the Weyl chamber is preserved under taking duals.
This follows the fact that if $v\in \mathrm{WC}(\Delta)$, then $(v,\delta)> 0$ for all $\delta\in \Delta$.
Letting $\delta\dual \in \Delta\dual$, we have
\[
(v,\delta\dual) > \qty{v, {2 \over (\delta, \delta)}\delta } = {2 (v, \delta) \over (\delta,\delta)} > 0
\]
using that every term in the last step is non-negative.
Since this works for every $\delta\dual \in \Delta\dual$, this yields $v\in \mathrm{WC}(\Delta\dual)$, and a similar argument shows the reverse containment.
So $\Delta\dual$ corresponds to a fundamental Weyl chamber and thus a base.
:::

:::{.proposition title="Humphreys 10.9"}
Prove that there is a unique element $\sigma$ in $\mathcal{W}$ sending $\Phi^{+}$to $\Phi^{-}$(relative to $\Delta$ ). Prove that any reduced expression for $\sigma$ must involve all $\sigma_\alpha(\alpha \in \Delta)$. Discuss $\ell(\sigma)$.
:::

:::{.solution}
The existence and uniqueness of such an element follows directly from the fact that $W$ acts simply transitively on the set of bases, and since $\Delta$ and $-\Delta$ are both bases, there is some $w_0\in W$ such that $w_0(\Delta) = - \Delta$ and consequently $w_0(\Phi^+) = \Phi^-$.
Since $\ell(\alpha) = n(\alpha) \leq \size \Phi^+$ for any root $\alpha$ and $n(w_0) = \size \Phi^+$ by definition, $w_0$ must be the longest element in $W$, i.e. $\ell(w_0)$ is maximal.

Any reduced expression for $w_0$ must involve all $s_\alpha$ -- if not, and say $s_\alpha$ doesn't occur in any reduced expression for $w_0$, then $w_0$ does not change the sign of $\alpha$ since every $s_\beta$ for $\beta\neq \alpha \in \Delta$ changes the sign of $\beta$ and acts by permutations on $\Phi^+\sm\ts{\beta}$.
However, in this case, $w_0' \da w_0s_\alpha$ satisfies $n(w_0') = n(w_0) + 1$ since $w_0'$ necessarily changes the sign of $\alpha$, contradicting maximality of $w_0$.

Finally, we have $\ell(w_0) = n(w_0) = \size \Phi^+$.
:::

## Section 11

:::{.proposition title="Humphreys 11.3"}
Use the algorithm of (11.1) to write down all roots for $G_2$. 

Do the same for $C_3$:
\[
\left(\begin{array}{rrr}2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -2 & 2\end{array}\right)
\]

:::

:::{.solution}
Note that it suffices to find all positive roots, since $\Phi = \Phi^+ \disjoint \Phi^-$ once a simple system $\Delta$ is chosen.
Since $\size \Phi(G_2) = 12$, it thus suffices to find 6 positive roots.
For $G_2$, the Dynkin diagram indicates one long and one short root, so let $\alpha$ be short and $\beta$ be long.
In this system we have
\[
\inp \alpha \alpha = \inp \beta \beta &= 2 \\
\inp \alpha \beta &= -1 \\
\inp \beta \alpha &= -3
.\]


- The $\beta$ root string through $\alpha$: since $\height(\alpha) = 1$ and $\beta-\alpha\not\in \Phi$, we have $r=0$.
  Since $q = -\inp \alpha \beta = -(-1) = 1$, we obtain the string $\alpha, \alpha + \beta$.

- The $\alpha$ root string through $\beta$: since $\height( \beta) = 1$ and $\alpha -\beta \not\in \Phi$ we have $r=0$ again.
  Here $q = - \inp \alpha \beta = - (-3) = 3$, we obtain \( \beta, \beta+ \alpha, \beta + 2 \alpha, \beta+ 3 \alpha \) 

- We know that the $\alpha$ root strings through any of the above roots will yield nothing new.

- The $\beta$ root strings through $\alpha + \beta, \beta + 2\alpha$ turn out to yield no new roots.

- The $\beta$ root string through $\beta + 3 \alpha$: since $(\beta + 3\alpha) - \beta = 3 \alpha\not\in\Phi$, using that only $\pm \alpha\in \Phi$, we have $r=0$.
  We also have 
  \[
  r-q = \inp {\beta + 3 \alpha}{ \beta} = \inp \beta \beta + 3 \inp \alpha \beta = 2 + 3(-1) =-1
  ,\]
  we have $q=1$ and obtain $\beta+ 3 \alpha, 2\beta + 3 \alpha$.

Combining these yields 6 positive roots: 
\[
\Phi^+(G_2) = \ts{ \alpha, \alpha+ \beta, \beta, \beta+ 2 \alpha, \beta+ 3 \alpha, 2 \beta +3 \alpha} 
.\]

---

For $C_3$, there are $2\cdot 3^2 = 18$ total roots and thus 9 positive roots to find. 
Let $\alpha, \beta, \gamma$ be the three ordered simple roots, then the Cartan matrix specifies
\[
\inp \alpha \alpha = \inp \beta \beta = \inp \gamma \gamma &= 2 \\
\inp \beta \alpha = \inp \alpha \beta &= -1 \\
\inp \alpha \gamma = \inp \gamma \alpha &= 0 \\
\inp \beta \gamma &= -1 \\
\inp \gamma \beta &= -2
.\]

- The $\alpha$ root string through $\beta$: $r=0,\, q = - \inp \alpha \beta = 1\leadsto \alpha, \alpha + \beta$.
- The $\beta$ root string through $\gamma$: $r=0,\, q = - \inp \gamma \beta = 2 \leadsto \gamma, \gamma+ \beta, \gamma + 2 \beta$.
- The $\gamma$ root string through $\alpha$: here $r=q=0$ since $\inp \gamma \alpha = 0$.
- The $\alpha$ root string through \( \alpha + \beta \): $r=0$ since \( \alpha + \beta - \gamma\not\in \Phi \), and $r-1 = \inp{\alpha+ \beta}{ \gamma}= -1\implies q=1 \leadsto \alpha + \beta+ \gamma$.
- The $\beta$ root string through \( \alpha + \beta + \gamma \): here $r=1$ since $(\alpha + \beta + \gamma) - 2 \gamma \not\in \Phi$, and $r-q = \inp{\alpha+ \beta + \gamma}{ \beta} = -1 + 2 + 2 = -1 \implies q=2 \leadsto \alpha + 2 \beta + \gamma, \alpha + 3 \beta+ \gamma$.

This yields 9 positive roots:
\[
\Phi^+(C_3) = \ts{\alpha, \beta, \gamma, \alpha + \beta, \gamma + \beta, \gamma+ 2 \beta, \alpha+ \beta+ \gamma, \alpha+ 2 \beta + \gamma, \alpha + 3 \alpha + \gamma}
.\]
:::