0.0 Exercise

Week 2

Q1. For $A,B\in \mathrm{End}(V)$, show

$$\begin{array}{rl}& A{B}^n=B^{n}A+\sum _{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)B^{n-i}[\cdots [[A,B],B],\cdots ,B]\text{ (i’s brackets)}\\ & A^{n}B=B{A}^n+\sum _{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(\mathrm{ad}A{)}^i(B)A^{n-i}\end{array}$$

Proof. By induction. And $\left(\genfrac{}{}{0ex}{}{n}{i}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{i-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{i-1}\right)$ is used.

Week 3

Q1. Suppose $V$ is a finite dimensional linear space and $L_1,L_2\in \mathrm{End}(V)$. The generalized eigenspaces of $L_1$ are $L_2$-invariant iff $(\mathrm{ad}{L}_1{)}^m{L}_2=0$ for some $m$.

Proof. i) Suppose ${\tilde{V}}_{\lambda }$ is a generalized eigenspace of $L_1$ and $(\mathrm{ad}{L}_1{)}^m{L}_2=0$ for some $m$. Then it suffices to show for any $v\in {\tilde{V}}_{\lambda }$ there is a $k$ such that $(L_1-\lambda I{)}^k{L}_{2}v=0$. By Week1-Q1, we have

$$\begin{array}{rl}& (L_1-\lambda I{)}^k{L}_2\\ =& L_2(L_1-\lambda I{)}^k+\sum _{i=1}^k(\genfrac{}{}{0ex}{}{k}{i})(\mathrm{ad}(L_1-\lambda I){)}^i{L}_2(L_1-\lambda I{)}^{k-i}\\ =& L_2(L_1-\lambda I{)}^k+\sum _{i=1}^k(\genfrac{}{}{0ex}{}{k}{i})(\mathrm{ad}{L}_1{)}^i{L}_2(L_1-\lambda I{)}^{k-i}\end{array}$$

as $[L_1-\lambda I,L_2]=[L_1,L_2]$. For any $v\in {\tilde{V}}_{\lambda }$, suppose $(L_1-\lambda I{)}^{n}v=0$. Then $(L_1-\lambda I{)}^k{L}_{2}v=0$ if $k\ge m+n$.

ii) Conversely, suppose ${\tilde{V}}_{\lambda }$ is a generalized eigenspace of $L_1$ and it is $L_2$-invariant. By induction, we can prove

$$(\mathrm{ad}{L}_1{)}^n{L}_2=\sum _{k=1}^n(-1{)}^{n-k}\left(\genfrac{}{}{0ex}{}{n}{k}\right)L_1^k{L}_2{L}_1^{n-k}.\text{\hspace{1em}(*)}$$

Also note that $[L_1,L_2]=[L_1-\lambda I,L_2]$, so $(\mathrm{ad}{L}_1{)}^n{L}_2=(\mathrm{ad}(L_1-\lambda I){)}^n{L}_2$. For any ${\tilde{V}}_{\lambda }$, there is a $n_{\lambda }$ s.t. $(L_1-\lambda I{)}^{n_{\lambda }}v=0$ for all $v\in {\tilde{V}}_{\lambda }$. By $(\ast )$, we have $(\mathrm{ad}{L}_1{)}^{2n_{\lambda }}{L}_2({\tilde{V}}_{\lambda })=0$. If ${\tilde{V}}_{\lambda }$ runs over all generalized eigenspaces of $L_1$ and take

$$m=\max \{2n_{\lambda }:\lambda \text{ is a eigenvalue }\}.$$

Then $(\mathrm{ad}{L}_1{)}^m{L}_{2}v=0$ for all $v\in V$ and so $(\mathrm{ad}{L}_1{)}^m{L}_2=0$. $■$

Q2. Let $g$ be a finite dimensional complex Lie algebra and $X_1,X_2\in g$.

• If the generalized eigenspaces of $\mathrm{ad}{X}_1$ are $\mathrm{ad}{X}_2$-invariant, then $(\mathrm{ad}{X}_1{)}^m(X_2)=0$ for some $m$.
• $(\mathrm{ad}{X}_1{)}^m(X_2)=0$ for some $m$ iff for any finite dimensional complex representation $(\rho ,V)$ of $g$, the generalized eigenspaces of $\rho (X_1)$ is $\rho (X_2)$-invariant.

Proof. i) Since $\mathrm{ad}{X}_1(X_1)=0$, $X_1$ is in the $0$-generalized eigenspace of $\mathrm{ad}{X}_1$. Then $\mathrm{ad}{X}_2(X_1)$ is still in the $0$-generalized eigenspace of $\mathrm{ad}{X}_1$ and so

$$(\mathrm{ad}{X}_1{)}^m([X_2,X_1])=(\mathrm{ad}{X}_1{)}^{m+1}{X}_2=0.$$

ii) Assume that $(\mathrm{ad}{X}_1{)}^m{X}_2=0$ for some $m$. Since $\rho$ is a homomorphism, $(\mathrm{ad}\rho (X_1){)}^m\rho (X_2)=0$. Use the same argument as Week3 Q1 i) to finish the proof. Now suppose that for any finite dimensional complex representation $(\rho ,V)$ of $g$, the generalized eigenspaces of $\rho (X_1)$ is $\rho (X_2)$-invariant. Take $\rho$ as the adjoint representation then we get back to i).

Q3. Suppose $h\subseteq g$ is an ideal, then $h^i\subseteq g$ is also an ideal.

Proof. By induction. For any $X\in g$,

$$\left[h^i,X\right]=[[h,h^{i-1}],X]=-[X,[h,h^{i-1}]]=[h,[h^{i-1},X]]+[h^{i-1},[X,h]]\subseteq h^i.$$

\end{proof}

Q4. For a complex finite dimensional Lie algebra, if its each finite dimensional complex representation can be decomposed as a direct product sum of generalized weight spaces, it is nilpotent.

Proof. By Q2, we only need to show each generalized weight space of adjoint representation is $g$-invariant. If it is true, then for any $X_1,X_2\in g$, generalized eigenspaces of $\mathrm{ad}{X}_1$ are $\mathrm{ad}{X}_2$-invariant. Then $(\mathrm{ad}{X}_1{)}^m([X_1,X_2])=0$ for some $m$. By the arbitrary of $X_2$, $\mathrm{ad}{X}_1$ is nilpotent. Now it remains to show generalized weight spaces are $g$-invariant.

Since the adjoint representation can be decomposed as a direct product sum of generalized weight space, each generalized weight space is a $\mathrm{ad}g$-submodule and so is $g$-invariant. Now we finish the proof. $■$

Week 4

Q1. For a Lie algebra $g$ and $y\in g$, if $\mathrm{ad}y=s+n$ is the Jordan-Chevalley decomposition of $\mathrm{ad}y$, then the Jordan-Chevalley decomposition of $\mathrm{adad}y$ is

$$\mathrm{adad}y=\mathrm{ad}s+\mathrm{ad}n.$$

\begin{proof}For any $h\in \mathrm{End}(g)$, $\mathrm{adady}(h)=[\mathrm{ad}y,h]=[s+n,h]=\mathrm{ad}s(h)+\mathrm{ad}n(h)$. So $\mathrm{adad}y=\mathrm{ad}s+\mathrm{ad}n$. Since $\mathrm{ad}$ is a homomorphism, $[\mathrm{ad}s,\mathrm{ad}n]=\mathrm{ad}([s,n])=0$. It remains to verify $\mathrm{ad}s$ is diagonal and $\mathrm{ad}n$ is nilpotent.

Note that $\mathrm{ad}y\in \mathrm{End}(g)$ and $\mathrm{adad}y\in \mathrm{End}(\mathrm{End}(g))$. Suppose $\dim g=n$, then $\{(e_{ij}):i,j\in n\}$ is a set of basis of $\mathrm{End}(g)$. For any $(e_{ij})$, $\mathrm{ad}s(e_{ij})=[s,e_{ij}]=[\mathrm{diag}\{s_{11},\cdots ,s_{nn}\},e_{ij}]=(s_{ii}-s_{jj})e_{ij}$. So $\mathrm{ad}s$ is diagonal. Similarly, $\mathrm{ad}n(e_{ij})=E_{i-1,j}-E_{i,j+1}$. Reorder the basis $\{e_{ij}\}$ like $e_{n,1},e_{n,2},e_{n-1,1},e_{n-2,1},e_{n-1,2},\cdots ,e_{1,n}$ and then $\mathrm{ad}n$ is nilpotent. By the uniqueness of Jordan-Chevalley decomposition we complete the proof. \end{proof}

Q2. For a complex diagonal matrix $A$, there is a polynomial $f$ with $0$ constant term such that $f(A)=\overline{A}$.

Proof. Take a Lagrange polynomial, which maps $a_{ii}$ to $\overline{a_{ii}}$. $■$

Week 7

Q1. Show that $g:=s{l}_{n,\mathbb{C}}$ has a Cartan subalgebra

$$h:=\{\mathrm{diag}({\mathrm{a}}_1,\cdots ,{\mathrm{a}}_{\mathrm{n}}):{\mathrm{a}}_{\mathrm{i}}\in \mathbb{C},\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}=0\}.$$

Proof. Since $h$ is abelian, it is nilpotent. And it is easy to verify $N(h)=h$. $■$

Week 8

Q1. Let $(V,R)$ be a root system according to Definition 1’, then there exists an inner product which is unique up to scalars so that $(V\text{ with }(\cdot ,\cdot ),R)$ is a root system according to Definition 1.

Proof. Firstly, Prof. Qiu gives us a lemma. I write it down here.

Lemma. Suppose finite dimensional $R\subseteq V\backslash \{0\}$ and $R$ spans $V$. Then for any $\alpha \in R$. a symmetry with vector $\alpha$ is unique if it exists. In the other word, a hyperplane fixed by a symmetry is unique.

Proof. Suppose $S_1,S_2$ are two symmetries with vector $\alpha$, then $S_1(\alpha )=S_2(\alpha )=-\alpha$ and induce identity maps on $V/\mathbb{R}\alpha$. Note that $S_1{S}_2(\alpha )=\alpha$ and $S_1{S}_2$ is identity on $V/\mathbb{R}\alpha$. (Now the corresponding matrix of $S_1{S}_2$ is upper triangular, we hope it is diagonal. That’s why we consider the order of $S_1{S}_2$ as a permutation. )

Since $S_1{S}_2$ is a permutation of $R$, there is an integer $w$ s.t. $(S_1{S}_2{)}^w=1$. Thus $t^w-1$ is a polynomial killing $S_1{S}_2$ and hence the characteristic polynomial of $S_1{S}_2$ has no multiple roots. Also all eigenvalues of $S_1{S}_2$ are $1$. So $S_1{S}_2=1$ and hence $S_1=S_2$. $■$

By lemma, $S_{\alpha }$ is unique. So we can denote the hyperplane fixed by $S_{\alpha }$ to $H_{\alpha }$. Since $S_{\alpha }$ is a linear isomorphism, with a set of basis $\alpha ,\{h_1,\cdots ,h_{n-1}\}$ where $\{h_i:i\in \{1,\cdots ,n-1\}\}$ spans $H_{\alpha }$, $S_{\alpha }$ can be written as

$$\left[\begin{array}{cc}-\alpha & 0\\ 0& I_{n-1}\end{array}\right]=I_n-\left[\begin{array}{cc}2\alpha & 0\\ 0& 0\end{array}\right].$$

So $S_{\alpha }=\mathrm{id}-{\alpha }^{\ast }\otimes \alpha$ where $\alpha$ is a linear functional such that ${\alpha }^{\ast }(\alpha )=2$ and ${\alpha }^{\ast }(H_{\alpha })=0$. For any inner product $(\cdot ,\cdot )$ of $V$, there is a $v_0\in V$ such that ${\alpha }^{\ast }=(v_0,\cdot )$. Then $(v_0,H_{\alpha })=0$ and $v_0\in H_{\alpha }^{\perp }=\mathbb{R}\alpha$.

Now suppose $(\cdot ,\cdot {)}_1$ and $(\cdot ,\cdot {)}_2$ are two inner products which make $(V,R)$ a root system according to Definition 1, then ${\alpha }^{\ast }=(k_1\alpha ,\cdot {)}_1=(k_2\alpha ,\cdot {)}_2$. For all $\alpha \in R$, we may assume

$$(\alpha ,\cdot {)}_1=(k_{\alpha }\alpha ,\cdot {)}_2.$$

Then for any $\alpha ,\beta \in R$, there is

$$k_{\beta }(\alpha ,\beta {)}_2=(\alpha ,\beta {)}_1=k_{\alpha }(\alpha ,\beta {)}_2$$

and so $k_{\alpha }\equiv k$ for some $k\in \mathbb{R}$. Therefore, $(\cdot ,\cdot {)}_1=k(\cdot ,\cdot {)}_2$ for this $k$, which is what we want. $■$

Q2. Let $(V,R)$ be a root system according to Definition 1’ for each $\alpha \in R$, write the symmetry with vector $\alpha$ as $S_{\alpha }=\mathrm{id}-{\alpha }^{\ast }\otimes \alpha$, where ${\alpha }^{\ast }\in V^{\ast }$. Then $(V^{\ast },R^{\ast })$ is a root system according to Definition 1’.

Proof. In 4.3 Root system, we have proved that $(V^{\ast },R^{\ast })$ is a root system with “dual inner product”, i.e. it is a root system according to Definition 1. Since Definition 1 and Definition 1’ are equivalent, $(V^{\ast },R^{\ast })$ is also a root system according to Definition 1’ and it is done. However, I want to prove it with inner product omitted.

Note that ${\alpha }^{\ast }$ is defined in the symmetry with $\alpha$. Now we define $S_{{\alpha }^{\ast }}=\mathrm{id}-\alpha \otimes {\alpha }^{\ast }$, where $\alpha (f):=f(\alpha )$. Since $R^{\ast }$ spans $V^{\ast }$, $\alpha \in \mathrm{End}(V^{\ast })$ is well-defined. Also define $H_{{\alpha }^{\ast }}=\mathrm{span}\{v^{\ast }:v\in H_{\alpha }\}$. By Q1, for any inner product $(\cdot ,\cdot )$, ${\alpha }^{\ast }=(k\alpha ,\cdot )$ for some real number $k$. So there is

$$\begin{array}{rl}{S}_{{\alpha }^{\ast }}({\alpha }^{\ast })& ={\alpha }^{\ast }-\alpha ({\alpha }^{\ast }){\alpha }^{\ast }=-{\alpha }^{\ast },\\ S_{{\alpha }^{\ast }}(v^{\ast })& =v^{\ast }-\alpha (v^{\ast }){\alpha }^{\ast }=v^{\ast }-v^{\ast }(\alpha ){\alpha }^{\ast }=v^{\ast }-c_{\alpha }{\alpha }^{\ast }(v){\alpha }^{\ast }=v^{\ast },\forall v^{\ast }\in H_{{\alpha }^{\ast }},\\ S_{{\alpha }^{\ast }}({\beta }^{\ast })& ={\beta }^{\ast }-\alpha ({\beta }^{\ast }){\alpha }^{\ast }={\beta }^{\ast }-{\beta }^{\ast }(\alpha ){\alpha }^{\ast },{\beta }^{\ast }(\alpha )\in \mathbb{Z}.\end{array}$$

Therefore, $(V^{\ast },R^{\ast })$ is a root system. $■$

Week 12

Q1. Let $G$ be a Lie group. If $H\subseteq G$ is a subgroup and a regular sub-manifold, then $H$ is an embedding Lie group.

\begin{proof}Firstly, it is easy to verify $H$ is a Lie subgroup. Then it is enough to show $i:H\to G$ is an embedding. Since $H$ is a regular sub-manifold, $i$ is automatically an embedding and we finish the proof. \end{proof}

Q2. Show that $\mathrm{Lie}{\mathrm{GL}}_n(\mathbb{R})\cong M_{n\times n}(\mathbb{R})$ and $\exp :M_{n\times n}(\mathbb{R})\to {\mathrm{GL}}_n(\mathbb{R})$ is given by $\exp (X)={\sum }_{n=0}^{\infty }{X}^n/n!$, where $\mathrm{Lie}G$ denotes the Lie algebra of a Lie group $G$.

\begin{proof}Since ${\mathrm{GL}}_n(\mathbb{R})$ is an open set of $M_{n\times n}(\mathbb{R})$, the tangent space at $I\in {\mathrm{GL}}_n(\mathbb{R})$ is $M_{n\times n}(\mathbb{R})$. Thus $\mathrm{Lie}{\mathrm{GL}}_n(\mathbb{R})\cong M_{n\times n}(\mathbb{R})$. Another proof from gz: for a small enough $t$, $tM+I$ is invertible for any $M\in M_{n\times n}(\mathbb{R})$. Therefore $\mathrm{Lie}{\mathrm{GL}}_n(\mathbb{R})\cong M_{n\times n}(\mathbb{R})$.

For any $X\in M_{n\times n}(\mathbb{R})$, define

$$\begin{array}{rl}{\phi }_X:\mathbb{R}& \to M_{n\times n}(\mathbb{R}),\\ t& \mapsto \sum _{n=0}^{\infty }(tX{)}^n/n!.\end{array}$$

Since ${\phi }_X(t){\phi }_X(-t)=\mathrm{id}$, ${\phi }_X$ maps $\mathbb{R}$ to ${\mathrm{GL}}_n(\mathbb{R})$. Note

$$d{\phi }_X(t)/dt{|}_{t=0}=\underset{t\to 0}{\lim }(\exp (tX)-I)/t=X.$$

By ^3c468f, ${\phi }_X=h_X$ and so $\exp (1)={\phi }_X(1)={\sum }_{i=0}^{\infty }{X}^n/n!$.

\end{proof}

Week 15

Q1. Figure out the Lie algebra of $\mathrm{Sp}(n)$.

\begin{proof}Note that

$$\mathrm{Sp}(n)=\{M:M\in {\mathrm{GL}}_{2n}(\mathbb{R}),M^T\Omega M=\Omega \}$$

where $\Omega =\left[\begin{array}{cc}0& I\\ -I& 0\end{array}\right]$. Suppose $\mathrm{Lie}\mathrm{Sp}(n)=sp(n)$. Then for any $X\in sp(n)$ and any $t\in \mathbb{R}$, there is

$$\exp (tX{)}^T\Omega \exp (tX)=\Omega .$$

Then

$$d/dt(\exp (tX{)}^T\Omega \exp (tX)){|}_{t=0}=X^T\Omega +\Omega X=0.$$

Conversely, for any $X\in M_{2n\times 2n}(\mathbb{R})$ with $X^T\Omega +\Omega X=0$,

$$\begin{array}{rl}\exp (tX{)}^T\Omega \exp (tX)& =(\exp (tX{)}^T\Omega )\exp (tX)\\ & =(\sum _{n=0}^{\infty }(-1{)}^n\Omega \frac{(tX{)}^n}{n!})\exp (tX)\\ & =\Omega \exp (-tX)\exp (tX)\\ & =\Omega .\end{array}$$

Therefore, $sp(n)=\{X\in M_{2n\times 2n}:X^T\Omega +\Omega X=0\}$.

\end{proof} Q2. Figure out the Lie algebra of $\mathrm{U}(n)$.

\begin{proof}Note that

$$U(n)=\{M:M\in {\mathrm{GL}}_n(\mathbb{C}),M{M}^{\ast }=I_n\}$$

where $M^{\ast }=\overline{M^T}$. Suppose $\mathrm{Lie}U(n)=u(n)$. For any $X\in u(n)$, we have

$$\exp (tX)\exp (tX{)}^{\ast }=\exp (tX)\exp (t{X}^{\ast })=I_n.$$

Then

$$d/dt\exp (tX)\exp (tX{)}^{\ast }{|}_{t=0}=X+X^{\ast }=0.$$

Conversely, if $X+X^{\ast }=0$, then

$$\exp (tX)\exp (tX{)}^{\ast }=\exp (tX)\exp (-tX)=I_n.$$

Therefore, $u(n)=\{X\in M_{n\times n}(\mathbb{C}):X+X^{\ast }=0\}$. \end{proof}

Q3. Figure out the Lie algebra of ${\mathrm{SL}}_n(\mathbb{R})$.

\begin{proof}Suppose $sl(n)=\mathrm{Lie}{\mathrm{SL}}_n(\mathbb{R})$. Then for any $X\in sl(n)$, $\det \exp (tX)=1$. Note that

$$d\det (X)=d/dt\det (I+tX){|}_{t=0}=\mathrm{Tr}(X).$$

Define

$$\gamma :\mathbb{R}\to {\mathrm{GL}}_1(\mathbb{R}),t\mapsto \det \exp (tX),$$

then the derivative of $\gamma$ satisfies ${\gamma }^{\prime }(t)=\gamma (t)\mathrm{Tr}(X)$. Since $\alpha :t\mapsto e^{t\mathrm{Tr}(X)}$ also allows ${\alpha }^{\prime }(t)=\alpha \mathrm{Tr}(X)$, we have $\det \exp (tX)=e^{t\mathrm{Tr}X}$. Thus $\exp (tX)\in {\mathrm{SL}}_n(\mathbb{R})$ yields $t\mathrm{Tr}(X)=0$ and $\mathrm{Tr}(X)=0$.

Conversely, if $X$ satisfies $\mathrm{Tr}(X)=0$, then $\phi (t):=\det \exp (tX)$ is a constant map and $\phi (t)\equiv \phi (0)=1$. Therefore,

$$sl(n)=\{X\in M_{n\times n}(\mathbb{R}):\mathrm{Tr}(X)=0\}.$$

\end{proof} Q4. Figure out the Lie algebra of $\mathrm{SU}(n)$.

\begin{proof}Note that

$$SU(n)=\{M:M\in {\mathrm{SL}}_n(\mathbb{C}),M{M}^{\ast }=I_n\}$$

where $M^{\ast }=\overline{M^T}$. Suppose $\mathrm{Lie}\mathrm{SU}(n)=su(n)$. For any $X\in u(n)$, we have

$$\exp (tX)\exp (tX{)}^{\ast }=\exp (tX)\exp (t{X}^{\ast })=I_n,\det \exp (tX)=1.$$

Then

$$\begin{array}{rl}& d/dt\exp (tX)\exp (tX{)}^{\ast }{|}_{t=0}=X+X^{\ast }=0,\\ & d/dt\det \exp (tX)=\mathrm{Tr}(X)=0.\end{array}$$

Conversely, if $X+X^{\ast }=0$ and $\mathrm{Tr}(X)=0$, then

$$\exp (tX)\exp (tX{)}^{\ast }=\exp (tX)\exp (-tX)=I_n,\det \exp (tX)=1.$$

Therefore, $u(n)=\{X\in M_{n\times n}(\mathbb{C}):X+X^{\ast }=0,\mathrm{Tr}(X)=0\}$. \end{proof}

Week 16

Q1. A connected Lie group is generated by an arbitrary neighborhood of the identity element.

\begin{proof}see here. In fact, this conclusion is true for topological group.\end{proof}

midterm

Q1. an exercise in week 3.

Q2. see LieAlgebrasOfFiniteAndAffineType_carter, 4.4.

Q3. Recall that a root system is a finite set $R$ in a finite-dimensional real vector space $V$ satisfying the following three conditions.

• $R$ spans $V$ and $0\notin R$;
• For each $\alpha \in R$, there exists a symmetry $S_{\alpha }$ in $V$ with vector $\alpha$ leaving $R$ invariant; (a symmetry in $V$ with vector $\alpha$ is a linear isomorphism $V\to V$ that sends $\alpha$ to $-\alpha$ and whose fixed point set is a hyperplane.)
• For $\alpha ,\beta \in R$, $S_{\alpha }(\beta )-\beta$ is an integer multiple of $\alpha$.

Let $R$ be a root system in $V$. For each $\alpha \in R$, let ${\mathcal{H}}_{\alpha }$ be the hyperplane fixed by the symmetry $S_{\alpha }$ which vector $\alpha$, then $S_{\alpha }$ is given by the formula

$$S_{\alpha }(v)=v-{\alpha }^{\ast }(v)\alpha$$

where ${\alpha }^{\ast }$ is the linear functional vanishing on ${\mathcal{H}}_{\alpha }$ and taking value $2$ on $\alpha$.

• Show that the subgroup $W\subseteq \mathrm{GL}(V)$ generated by all $S_{\alpha }(\alpha \in R)$ is a finite group.
• Show that there exists an inner product $(\cdot ,\cdot )$ on $V$ that is $W$-invariant. Show also that if $(\cdot ,\cdot {)}_1$ and $(\cdot ,\cdot {)}_2$ are two such inner products on $V$, then there exists $c\in {\mathbb{R}}_{>0}$ such that $(\cdot ,\cdot {)}_2=c(\cdot ,\cdot {)}_1$.
• Let $V^{\ast }=\mathrm{Hom}(V,\mathbb{R})$ be the dual space of $V$ and put $R^{\ast }=\{{\alpha }^{\ast }:\alpha \in R\}$. For $\alpha \in R$ let $S_{\alpha }^{\ast }:V^{\ast }\to V^{\ast }$ be the map naturally induced by $S_{\alpha }$, namely, $S_{\alpha }^{\ast }(l)=l\circ S_{\alpha }$ for $l\in V^{\ast }$. Show that $S_{\alpha }^{\ast }$ is a symmetry in $V^{\ast }$ with vector ${\alpha }^{\ast }$ and it leaves $R^{\ast }$ invariant.
• Show that $R^{\ast }$ is a root system in $V^{\ast }$.

\begin{proof}i) Define $W=⟨S_{\alpha }:\alpha \in R⟩$. Since $S_{\alpha }(R)=R$ for any $\alpha \in R$, each $S_{\alpha }$ is a permutation of $R$. Since $R$ spans $V$, $W\subseteq \mathrm{GL}(V)$. And $W\le \mathrm{Sym}(R)$ yields $W$ is a finite group.

ii) Define an inner product $(\cdot ,\cdot )$ by $(v,w):={\sum }_{f\in W}(fv,fw)$. It is easy to verify it is an inner product and it is $W$-invariant. Then show that it is unique up to scalar.

Firstly, show that a symmetry with vector $\alpha$ is unique. Suppose $S_1,S_2$ are symmetries with vector $\alpha$. Then $S_1{S}_2(\alpha )=\alpha$. And $S_1{S}_2$ acting on $V/\mathbb{C}\alpha$ is trivial. Hence $S_1{S}_2$ has a representation

$$M:=\left[\begin{array}{ccc}1& & \ast \\ & \ddots & \\ & & 1\end{array}\right].$$

Since $S_1{S}_2\in \mathrm{Sym}(R)$, there is $m\in {\mathbb{Z}}_{+}$ such that $(S_1{S}_2{)}^m=1$. By Cayley-Hamilton’s theorem, $\det (M-tI)|t^m-1$. Note that $t^m-1$ has no multiple roots, $S_1{S}_2=1$. Therefore the symmetry is unique.

Since the symmetry with vector $\alpha$ is unique, $H_{\alpha }$ is unique. Suppose $(\cdot ,\cdot {)}_1$ and $(\cdot ,\cdot {)}_2$ are two inner products on $V$. Then there exist $v_1,v_2\in V$ such that ${\alpha }^{\ast }=(v_1,\cdot {)}_1=(v_2,\cdot {)}_2$. Since $(v_1,H_{\alpha }{)}_1=(v_2,H_{\alpha }{)}_2=0$, $v_i\in H_{\alpha }^{\perp }$. Thus $v_i=c_i\alpha$. For each $\alpha \in R$, there is $c_{\alpha }$ such that $(\alpha ,\cdot {)}_1=c_{\alpha }(\alpha ,\cdot {)}_2$. Since ${\alpha }^{\ast }(\alpha )=2$, $c_{\alpha }>0$. Note that $c_{\alpha }(\alpha ,\beta {)}_2=(\alpha ,\beta )=c_{\beta }(\alpha ,\beta {)}_2$, $c_{\alpha }=c_{\beta }$ for all $\alpha ,\beta \in R$. Then $R$ spans $V$ yields $(\cdot ,\cdot {)}_2=c(\cdot ,\cdot {)}_1$, where $c=c_{\alpha }>0$.

iii) Since $S_{\alpha }^{\ast }({\alpha }^{\ast })(v)=-{\alpha }^{\ast }(v)$ and $S_{\alpha }^{\ast }(w^{\ast })=w^{\ast }$ for any $w^{\ast }:w\in H_{\alpha }$, $S_{\alpha }^{\ast }$ is a symmetry. Since $S_{\alpha }^{\ast }({\beta }^{\ast })=(S_{\alpha }(\beta ){)}^{\ast }$, $S_{\alpha }^{\ast }$ leaves $R^{\ast }$ invariant. The key point is, ${\alpha }^{\ast }(v)=2(\alpha ,v)/(\alpha ,\alpha )$.

iv) It remains to show ${\beta }^{\ast }(\alpha )\in \mathbb{Z}$, which can be yielded by $S_{\alpha }(\beta )-\beta =-{\alpha }^{\ast }(\beta )\alpha \in \mathbb{Z}\alpha$. \end{proof}

Q4. Let $g$ be a finite dimensional semi-simple Lie algebra over $\mathbb{C}$ and $(V,\Phi )$ the according root system. Choose a set of simple roots $\Delta =\{{\alpha }_1,\cdots {\alpha }_l\}$. Show that any positive root $\alpha \in {\Phi }^{+}$ can be written in the form

$$\alpha ={\alpha }_{i_1}+\cdots +{\alpha }_{i_k}$$

with each ${\alpha }_{i_j}\in \Delta$ and with each partial summand ${\sum }_{j=1}^m{\alpha }_{i_j}$ from the left equal to a positive root.

\begin{proof}The following lemma is used: For any $\alpha \in {\Phi }^{+}\backslash \Delta$, there is a $\beta \in \Delta$ such that $(\alpha ,\beta )>0$. Then $\alpha -\beta \in {\Phi }^{+}$ and by induction we finish the proof.

Proof of lemma: otherwise, $(\alpha ,\beta )<0$ for all $\beta \in \Delta$. Then $\Delta \cup \{\alpha \}$ is linear independent, contradiction.

\end{proof}