Representation Theory IV 5 Problems for Epiphany

6 Solutions

Solution to (1).

1.

We have

$\exp\begin{pmatrix}s&0\\ 0&t\end{pmatrix}=\sum_{n=0}^{\infty}\begin{pmatrix}s^{n}/n!&0\\ 0&t^{n}/n!\end{pmatrix}=\begin{pmatrix}\exp(s)&0\\ 0&\exp(t)\end{pmatrix}.$

Since

$\begin{pmatrix}0&1\\ -1&0\end{pmatrix}^{n}=\begin{cases}(-1)^{n/2}\begin{pmatrix}1&0\\ 0&1\end{pmatrix}&\text{$n$ even}\\ (-1)^{(n-1)/2}\begin{pmatrix}0&1\\ -1&0\end{pmatrix}&\text{$n$ odd}\end{cases}$

we have

$\displaystyle\exp\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&t\\ \hidden@noalign{}\hfil\textstyle-t&0\end{pmatrix}$ $\displaystyle=(1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\ldots)I+(t-\frac{t^{3}}{3!% }+\frac{t^{5}}{5!}-\ldots)\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&1% \\ \hidden@noalign{}\hfil\textstyle-1&0\end{pmatrix}$

$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle\cos(t)&\sin(t)\\ \hidden@noalign{}\hfil\textstyle-\sin(t)&\cos(t)\end{pmatrix}.$

A very similar calculation (just lose all the minus signs) shows

$\exp\begin{pmatrix}0&t\\ t&0\end{pmatrix}=\begin{pmatrix}\cosh(t)&\sinh(t)\\ \sinh(t)&\cosh(t)\end{pmatrix}.$
2.

If $a\neq b$ , then $E_{a,b}^{n}=0$ for $n\geq 2$ so $\exp(tE_{a,b})=I+tE_{a,b}$ . If $a=b$ , then $E_{a,b}^{n}=E_{a,b}$ for $n\geq 1$ so $\exp(tE_{a,a})$ is diagonal with $e^{t}$ in the $a$ -th entry and $1$ in the other entries.

Solution to (2).

Using the definition we expand the LHS and the RHS; then it is enough they agree up to terms involving $t^{3}$ or higher powers of $t$ . Left hand side:

	$\displaystyle LHS$	$\displaystyle=$	$\displaystyle\left(I+tX+\frac{t^{2}X^{2}}{2}+O(t^{3})\right)\left(I+tY+\frac{t% ^{2}Y^{2}}{2}+O(t^{3})\right)$
		$\displaystyle=$	$\displaystyle I+t(X+Y)+\frac{t^{2}}{2}(X^{2}+2XY+Y^{2})+O(t^{3}).$

Right hand side:

$\displaystyle RHS$	$\displaystyle=$	$\displaystyle I+t(X+Y)+\frac{t^{2}}{2}(XY-YX)+\frac{t^{2}}{2}(X+Y)^{2}+O(t^{3})$
	$\displaystyle=$	$\displaystyle I+t(X+Y)+\frac{t^{2}}{2}(XY-YX+X^{2}+XY+YX+Y^{2})+O(t^{3})$
	$\displaystyle=$	$\displaystyle I+t(X+Y)+\frac{t^{2}}{2}(X^{2}+2XY+Y^{2})+O(t^{3}).$

Solution to (3).

1. Let $V=\mathbb{C}^{n}$ and, for $0\leq i\leq n$ , let $V_{i}$ be the span of $e_{1},\ldots,e_{i}$ , so

V_{0}=\{0\}\subset V_{1}=\left\langle e_{1}\right\rangle\subset V_{2}=\left% \langle e_{1},e_{2}\right\rangle\subset\ldots\subset V_{n}=V.

Then $Xe_{i}\in V_{i-1}$ for all $i$ , so $XV=XV_{n}\subset V_{n-1}$ , $X^{2}V\subset XV_{n-1}\subset V_{n-2}$ etc. Thus $X^{n}V=0$ , so $X^{n}=0$ .

2. Since each $X^{i}\in\mathfrak{n}\subset N$ , and $I\in N$ , we have

\exp(X)=I+\sum_{i=1}^{\infty}\frac{X^{i}}{i!}\in N

as $N$ is a closed subset of $GL_{n}(\mathbb{K})$ (alternatively, note that by part (1) the sum is actually a finite sum).

3. The sum is finite since, for $g\in N$ , $(g-I)\in\mathfrak{n}$ and so $(g-I)^{k}=0$ for all $k\geq n$ .

4. We have, as power series in $t$ , $\exp(\log(1+t))=1+t$ since this holds for all $t$ sufficiently small. Similarly, $\log(\exp(t))=t$ as power series in $t$ . Since all powers of $X$ commute with each other and there are no convergence issues (by the previous part and the fact that $\exp$ converges absolutely everywhere) we may substitute in $X$ .

In fact, this shows that $\exp$ and $\log$ define diffeomorphisms between a neighbourhood of zero in $\mathfrak{gl}_{n,\mathbb{K}}$ and a neighbourhood of the identity in $\operatorname{GL}_{n}(\mathbb{K})$ .

Solution to (4).

2. I claim that $g=\begin{pmatrix}-1&1\\ 0&-1\end{pmatrix}\in\operatorname{SL}_{2}(\mathbb{C})$ is not in the image of the exponential map from $\mathfrak{sl}_{2,\mathbb{C}}$ . Indeed, suppose it were of the form $g=\exp(A)$ where $\operatorname{tr}(A)=0$ . Then the eigenvalues of $A$ must add to zero; since they are nonzero (as they exponentiate to $-1$ ), they must be distinct. But then $A=PDP^{-1}$ for some diagonal $D$ and some $P\in\operatorname{GL}_{2}(\mathbb{C})$ since every matrix with distinct eigenvalues is diagonalizable. Then $g=P\exp(D)P^{-1}$ would be diagonalizable, which it is not.

For the second part, we take the same $g=\begin{pmatrix}-1&1\\ 0&-1\end{pmatrix}\in\operatorname{GL}^{+}_{2}(\mathbb{R})$ and suppose it is of the form $\exp(A)$ for $A\in\mathfrak{gl}_{2,\mathbb{R}}$ . Then the eigenvalues of $A$ are imaginary, since they exponentiate to $-1$ , and since $A$ is real they must be a complex conjugate pair. But then we get a contradiction in the same way: $A$ is diagonalisable but $g$ is not.

Solution to (5).

The function $f(\theta)$ is clearly continuous, and it is a one-parameter subgroup as $f(\theta+\theta^{\prime})$ is rotation by $\theta+\theta^{\prime}$ about $v$ , which is the same as rotation by $f(\theta^{\prime})$ followed by $f(\theta)$ :

f(\theta+\theta^{\prime})=f(\theta)f(\theta^{\prime}).

We compute $\frac{d}{d\theta}f(\theta)(e_{1})$ . If $e_{1}$ is parallel to $v$ , this is zero.

If $y$ is the projection of $e_{1}$ on $v$ , and $x=e_{1}-y$ , then rotation about $v$ initially moves $e_{1}$ in the direction $v\times e_{1}$ perpendicular to $v$ and $x$ (note that we consider rotations as being anticlockwise when looking from the ‘nose’ of $v$ towards the origin).

We have

f(\theta)(e_{1})=y+\sin(\theta)|x|\frac{v\times e_{1}}{|v\times e_{1}|}+\cos(% \theta)x.

Thus

f^{\prime}(0)(e_{1})=|x|\frac{v\times e_{1}}{|v\times e_{1}|}=v\times e_{1}

since $|v\times e_{1}|=|x|=\sin(\alpha)$ where $\alpha$ is the angle between $v$ and $e_{1}$ . This formula also holds in the parallel case.

Thus the infinitesimal generator is

f^{\prime}(0)=(v\times e_{1}\,v\times e_{2}\,v\times e_{3})=\begin{pmatrix}0&-% z&y\\ z&0&-x\\ -y&x&0\\ \end{pmatrix}

where $v=\begin{pmatrix}x&y&z\end{pmatrix}$ .

Solution to (6).

If $X+X^{\dagger}=0$ then, for all $t\in\mathbb{R}$ ,

\exp(tX)=\exp(-tX^{\dagger})=(\exp(tX)^{\dagger})^{-1}

and so $\exp(tX)\in\operatorname{U}(n)$ .

Conversely, if $\exp(tX)\in\operatorname{U}(n)$ for all $t\in\mathbb{R}$ , we have

\exp(tX)\exp(tX^{\dagger})=\exp(tX)\exp(tX)^{\dagger}=I

for all $t\in\mathbb{R}$ . Taking the derivative at $t=0$ gives

X+X^{\dagger}=0.

Thus the Lie algebra of $\operatorname{U}(n)$ is

\{X\in\mathfrak{gl}_{n,\mathbb{C}}:X+X^{\dagger}=0\}.

We have that $X+X^{{\dagger}}=0$ if and only if $x_{ii}$ is imaginary for all $i$ and

x_{ji}=-\overline{x}_{ij}

for all $i<j$ . Thus $X$ is determined by its $n$ imaginary diagonal entries and its $n(n-1)/2$ complex entries above the diagonal. Its real dimension is thus

2\frac{n(n-1)}{2}+n=n^{2}.

If $X\in\mathfrak{u}_{n}$ is nonzero, then

(iX)^{\dagger}=-iX^{\dagger}=iX\neq-iX

so $iX\not\in\mathfrak{u}_{n}$ . Thus $\mathfrak{u}_{n}$ is not a complex subspace of $\mathfrak{gl}_{n,\mathbb{C}}$ .

For a challenge, try to show that there is no complex structure on $\mathfrak{u}_{n}$ : there is no linear map

J:\mathfrak{u}_{n}\to\mathfrak{u}_{n}

such that

[J(X),Y]=J([X,Y])=[X,J(Y)]

for all $X,Y\in\mathfrak{u}_{n}$ and

J^{2}(X)=-X

for all $X\in uf_{n}$ (for $n$ odd this is easy, but it is trickier for $n$ even).

Solution to (7).

If $XI_{p,q}+I_{p,q}X^{T}=0$ then, for all $t\in\mathbb{R}$ , $I_{p,q}^{-1}tXI_{p,q}=tX^{T}$ and so

I_{p,q}^{-1}\exp(tX)I_{p,q}=\exp(tX)^{T}

whence $\exp(tX)\in\operatorname{O}(p,q)$ . Conversely, if $\exp(tX)\in\operatorname{O}(p,q)$ for all $t\in\mathbb{R}$ then

\exp(tX)I_{p,q}\exp(tX^{T})=I_{p,q}

for all $t\in\mathbb{R}$ . Differentiating with respect to $t$ at $t=0$ gives

XI_{p,q}+I_{p,q}X^{T}=0

as required.

For the last part, we need only show that $X\in\mathfrak{o}_{p,q}$ implies $\operatorname{tr}X=0$ . But if $X\in\mathfrak{o}_{p,q}$ then

\operatorname{tr}(X)=\operatorname{tr}(I_{p,q}^{-1}XI_{p,q})=\operatorname{tr}% (-X^{T})=-\operatorname{tr}(X)

so $\operatorname{tr}(X)=0$ as required.

Solution to (8).

1. Problem 55 suggests that we consider the following basis $J_{x},J_{y},J_{z}$ of infinitesimal rotations around the axes:

	$\displaystyle J_{x}$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&0&0\\ \hidden@noalign{}\hfil\textstyle 0&0&-1\\ \hidden@noalign{}\hfil\textstyle 0&1&0\end{pmatrix}$
	$\displaystyle J_{y}$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&0&1\\ \hidden@noalign{}\hfil\textstyle 0&0&0\\ \hidden@noalign{}\hfil\textstyle-1&0&0\end{pmatrix}$
	$\displaystyle J_{z}$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&-1&0\\ \hidden@noalign{}\hfil\textstyle 1&0&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0\end{pmatrix}.$

In fact, these are the images of $e_{1},e_{2},e_{3}$ under an isomorphism $(\mathbb{R}^{3},\times)\to\mathfrak{so}_{3}$ . By calculation we have

[J_{x},J_{y}]=J_{z},[J_{y},J_{z}]=J_{x},[J_{z},J_{x}]=J_{y}.

These may remind you of the quaternion group, whose irreducible two-dimensional representation leads us to consider the following basis for $\mathfrak{su}_{2}$ :

	$\displaystyle\mathcal{I}$	$\displaystyle=\tfrac{1}{2}\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&1% \\ \hidden@noalign{}\hfil\textstyle-1&0\end{pmatrix}$
	$\displaystyle\mathcal{J}$	$\displaystyle=\frac{1}{2}\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&i\\ \hidden@noalign{}\hfil\textstyle i&0\end{pmatrix}$
	$\displaystyle\mathcal{K}$	$\displaystyle=\frac{1}{2}\begin{pmatrix}\hidden@noalign{}\hfil\textstyle i&0\\ \hidden@noalign{}\hfil\textstyle 0&-i\end{pmatrix}.$

We have

[\mathcal{I},\mathcal{J}]=\mathcal{K},[\mathcal{J},\mathcal{K}]=\mathcal{I},[% \mathcal{K},\mathcal{I}]=\mathcal{J}

— we need the factors of $\frac{1}{2}$ for this, otherwise the right hand sides would be doubled.

It follows that the linear map $\mathfrak{su}_{2}\to\mathfrak{so}_{3}$ taking $\mathcal{I}$ to $J_{x}$ , $\mathcal{J}$ to $J_{y}$ and $\mathcal{K}$ to $J_{z}$ is an isomorphism of Lie algebras.

2. By the problems class (or problem 7), we know that $\mathfrak{so}_{2,1}$ has a basis $A, B, C$ with

	$\displaystyle A$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&0&0\\ \hidden@noalign{}\hfil\textstyle 0&0&1\\ \hidden@noalign{}\hfil\textstyle 0&1&0\end{pmatrix}$
	$\displaystyle B$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&0&1\\ \hidden@noalign{}\hfil\textstyle 0&0&0\\ \hidden@noalign{}\hfil\textstyle 1&0&0\end{pmatrix}$
	$\displaystyle C$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&1&0\\ \hidden@noalign{}\hfil\textstyle-1&0&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0\end{pmatrix}.$

We calculate that $[A,B]=C$ , $[A,C]=-B$ and $[B,C]=A$ . It follows that $2A,B,C$ satisfy the same commutation relations as

H=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix},E=\begin{pmatrix}0&1\\ 0&0\end{pmatrix},F=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}

so that there is a Lie algebra isomorphism sending $2A\mapsto H$ , $B\mapsto E$ , $C\mapsto F$ .

How might you think of this? Well, the eigenvalues of the linear map $X\mapsto[H,X]$ are $2,0,-2$ , with $E$ and $F$ the eigenvectors. So you might look for an element $H^{\prime}$ of $\mathfrak{so}_{2,1}$ such that the eigenvalues of $X\mapsto[H^{\prime},X]$ are $2,0,-2$ , and $2A$ above works; $B$ and $C$ are then the eigenvectors!

For a more conceptual approach, let $\left\langle X,Y\right\rangle=\operatorname{tr}(XY)$ , a bilinear form on $\mathfrak{sl}_{2,\mathbb{R}}$ . For each $g\in\operatorname{SL}_{2}(\mathbb{R})$ , $\rho(g):X\mapsto gXg^{-1}$ is a linear map $\mathfrak{sl}_{2,\mathbb{R}}\to\mathfrak{sl}_{2,\mathbb{R}}$ preserving this bilinear form. But it is possible to write down a basis $e_{1},e_{2},e_{3}=H,E+F,E-F$ of $\mathfrak{sl}_{2,\mathbb{R}}$ such that

\left\langle e_{i},e_{j}\right\rangle=\begin{cases}2&i=j=1,2\\ -2&i=j=3\\ 0&i\neq j\end{cases}.

With respect to this basis, $\left\langle\cdot,\cdot\right\rangle$ is the bilinear form determined by $2I_{2,1}$ and so $\rho(g)\in\operatorname{O}(2,1)$ for all $g$ . The derived map $D\rho$ on Lie algebras is the desired isomorphism.

3. Here is a possible approach. Show that $\operatorname{SL}_{2}(\mathbb{C})$ acts on the four-dimensional real vector space $V$ of Hermitian $2\times 2$ matrices by $g\cdot X=gXg^{\dagger}$ for $g\in\operatorname{SL}_{2}(\mathbb{C})$ and $X$ a Hermitian matrix. The quadratic form $-\det$ on $V$ is preserved by this action. It has signature $(3,1)$ ; indeed, it is positive definite on the space of matrices

\begin{pmatrix}x&z\\ \overline{z}&-x\end{pmatrix}:x\in\mathbb{R},z\in\mathbb{C}

and negative definite on the subspace of matrices

\operatorname{diag}(x,x):x\in\mathbb{R}.

We obtain a map $\operatorname{SL}_{2,\mathbb{C}}\rightarrow\operatorname{O}(3,1)$ ; its derivative is the required isomorphism.

Solution to (9).

We prove the first part. Let $J$ be the matrix defining the standard symplectic form, so that

\mathfrak{sp}_{2n}=\{X\in\mathfrak{gl}_{2n,\mathbb{C}}:X^{T}J+JX=0\}.

If $X\in\mathfrak{sp}_{2n}$ , then $JX=-X^{T}J$ , and hence

X=-J^{-1}X^{T}J.

Taking traces and using that trace is invariant under conjugation gives

\operatorname{tr}(X)=-\operatorname{tr}(J^{-1}X^{T}J)=-\operatorname{tr}(X^{T}% )=-\operatorname{tr}(X).

Therefore $\operatorname{tr}(X)=0$ .

For the second part, try induction.

Solution to (10).

Suppose that $h\in G^{0}$ and $g\in G$ . We must show $ghg^{-1}\in G^{0}$ . By assumption, there is a path $\gamma:[0,1]\rightarrow G$ with $\gamma(0)=I$ , $\gamma(1)=h$ . Then $t\mapsto g\gamma(t)g^{-1}$ is a path from $I$ to $ghg^{-1}$ so $ghg^{-1}\in G^{0}$ as required.

Alternatively, we have that, for any $g\in G$ , $gG^{0}g^{-1}$ is open and closed in $G$ and contains the identity. It therefore contains $G^{0}$ . But by the same argument $G^{0}$ contains $gG^{0}g^{-1}$ . So $G^{0}=gG^{0}g^{-1}$ .

Solution to (11).

1. Let $R(v,\theta)$ be rotation by $\theta$ about $v$ , for $v\in\mathbb{R}^{3}$ — every element of $\mathrm{SO}(3)$ has this form. Then, for any $v$ and $\theta$ , the path

t\mapsto R(v,t\theta)

is a continuous path in $\mathrm{SO}(3)$ from the identity to $R(v,\theta)$ , showing that this group is connected.

2. Induction on $n$ . Base case: $n=2$ , clear from the explicit description of $\mathrm{SO}(2)$ .

Suppose true for $n-1$ . Let $v\in\mathbb{R}^{n}$ be a unit vector and let $R\in\mathrm{SO}(n)$ . Let $w=Rv$ , and let $V$ be a plane containing $v, w$ . Then there is some rotation $S\in\mathrm{SO}(V)$ such that $w=Sv$ . Extend $S$ to an element of $\mathrm{SO}(n)$ by letting it act as the identity on $V^{\perp}$ : explicitly, $S(v+v^{\prime})=S(v)+v^{\prime}$ for $v\in V,v^{\prime}\in V^{\perp}$ . Since $\mathrm{SO}(V)\cong\mathrm{SO}(2)$ is connected, there is a path $\gamma(t)$ from $I$ to $S$ . Then $\gamma(t)^{-1}R$ is a path from $R$ to $S^{-1}R$ , and it suffices to show that $T=S^{-1}R$ is connected to the identity. Note that $T$ fixes $v$ and so is determined by its action on $W=\left\langle v\right\rangle^{\perp}$ . By the induction hypothesis applied to $\mathrm{SO}(W)$ , we can connect $T$ to an element that fixes $v$ and $W$ , and therefore fixes $V$ . In other words, $T$ is connected to the identity.

Solution to (12).

Let $g=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in\mathrm{SU}(2)$ . Then $g^{-1}=\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}$ since $\det(g)=1$ . But also $g^{-1}=g^{\dagger}$ and so $d=\overline{a}$ and $c=-\overline{b}$ . Thus $g=\begin{pmatrix}a&b\\ -\overline{b}&\overline{a}\end{pmatrix}$ and the determinant condition gives $a\overline{a}+b\overline{b}=1$ .

We deduce that the map $\mathrm{SU}(2)\rightarrow\{(a,b)\in\mathbb{C}^{2}:|a|^{2}+|b|^{2}=1\}$ is a diffeomorphism (it and its inverse are clearly smooth). Moreover, writing $a=w+ix$ , $b=y+iz$ , the latter space is

\{(w,x,y,z)\in\mathbb{R}^{4}:w^{2}+x^{2}+y^{2}+z^{2}=1\}

which is the three-sphere.

Solution to (14).

Firstly we will show that the Lie algebra of $Z$ is contained in $\mathfrak{z}$ . Indeed, suppose that $X\in\mathfrak{g}$ with

\exp(tX)\in Z

for all $t\in\mathbb{R}$ . Then for all $Y\in\mathfrak{g}$ ,

\exp(tX)\exp(sY)\exp(-tX)=\exp(sY)

for all $s,t\in\mathbb{R}$ . Taking the derivative at $s=0$ gives

\exp(tX)Y\exp(-tX)=Y

for all $t$ and taking the derivative of this at $t=0$ gives $[X,Y]=0$ for all $Y\in\mathfrak{g}$ , whence $X\in\mathfrak{z}$ .

Conversely, if $G$ is connected and $X\in\mathfrak{z}$ , then I claim that $\exp(tX)\in Z$ for all $t\in\mathbb{R}$ . Indeed, for $Y\in\mathfrak{g}$ ,

\exp(tX)\exp(Y)=\exp(tX+Y)=\exp(Y)\exp(tX)

as $[tX,Y]=0$ . So $\exp(tX)$ commutes with all elements of $G$ of the form $\exp(Y)$ . Since these generate $G$ by the connectedness assumption, we see $\exp(tX)\in Z$ .

Solution to (16).

1. I claim that for $Z\in\mathfrak{z}$ , $v\mapsto\rho(Z)v$ is a $\mathfrak{g}$ -homomorphism. Indeed, if $X\in\mathfrak{g}$ then

	$\displaystyle\rho(Z)\rho(X)v$	$\displaystyle=[\rho(Z),\rho(X)]v+\rho(X)\rho(Z)v$
		$\displaystyle=\rho([Z,X])v+\rho(X)\rho(Z)v$
		$\displaystyle=0+\rho(X)\rho(Z)v$
		$\displaystyle=\rho(X)\rho(Z)v.$

By Schur’s lemma, as $V$ is irreducible, there is $\alpha(Z)$ such that $\rho(Z)v=\alpha(Z)v$ for all $v\in V$ . As $\rho$ is linear, so is $\alpha$ .

2. If $A=(a_{ij})\in\mathfrak{z}$ , then it commutes with every $n\times n$ matrix. Let $D_{i}$ be the diagonal matrix with ‘1’ in position $i$ and ‘0’ elsewhere. Then $D_{i}A-AD_{i}=0$ implies that $a_{ij}=0$ for $j\neq i$ . Hence $A$ is diagonal, with entries $a_{1},\ldots,a_{n}$ . But then $A$ commutes with the elementary matrix $E_{ij}$ , with ‘1’ in row $i$ and column $j$ and ‘0’ elsewhere, if and only if $a_{i}=a_{j}$ . So all the $a_{i}$ are equal, so $A$ is scalar. Conversely, all scalar matrices are in $\mathfrak{z}$ . So

\mathfrak{z}=\{\lambda I:\lambda\in\mathbb{C}\}.

If $V=\Lambda^{k}\mathbb{C}^{n}$ , then for all $v_{1}\wedge\cdots\wedge v_{k}$ ,

(\lambda I)(v_{1}\wedge\cdots\wedge v_{k})=\sum_{i=1}^{k}v_{1}\wedge\cdots% \wedge(\lambda v_{i})\wedge\cdots\wedge v_{k}=k\lambda(v_{1}\wedge\cdots\wedge v% _{k}),

so $\alpha$ is the linear map

\alpha(\lambda I)=k\lambda.

Solution to (17).

We use induction on $m$ . The case $m=1$ is clear — in that case, we just get

[X,Y]=XY-YX.

Suppose the result is true for $m$ . Then

	$\displaystyle\operatorname{ad}_{X}^{m+1}(Y)$	$\displaystyle=\operatorname{ad}_{X}(\operatorname{ad}_{X}^{m}(Y))$
		$\displaystyle=\sum_{k=0}^{m}\binom{m}{k}\left(X^{k+1}Y(-X)^{m-k}-X^{k}Y(-X)^{m% -k}X\right)$
		$\displaystyle=\sum_{k=0}^{m}\binom{m}{k}\left(X^{k+1}Y(-X)^{m-k}+X^{k}Y(-X)^{m% -k+1}\right)$
		$\displaystyle=\sum_{k=0}^{m+1}\left(\binom{m}{k-1}+\binom{m}{k}\right)X^{k}Y(-% X)^{m+1-k}$
		$\displaystyle=\sum_{k=0}^{m+1}\binom{m+1}{k}X^{k}Y(-X)^{m+1-k}$

as required.

For the second part, for all $X$ and $Y$ in $\mathfrak{gl}_{n,\mathbb{C}}$ we have

	$\displaystyle\exp(ad_{X})(Y)$	$\displaystyle=\sum_{m=0}^{\infty}\frac{\operatorname{ad}_{X}^{m}}{m!}(Y)$
		$\displaystyle=\sum_{m=0}^{\infty}\sum_{k=0}^{m}\frac{1}{m!}\binom{m}{k}X^{k}Y(% -X)^{m-k}$
		$\displaystyle=\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{(k+l)!}\binom{k+l% }{k}X^{k}Y(-X)^{l}$
		$\displaystyle=\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{k!l!}X^{k}Y(-X)^{l}$
		$\displaystyle=\left(\sum_{k=0}^{\infty}\frac{X^{k}}{k!}\right)Y\left(\sum_{l=0% }^{\infty}\frac{(-X)^{l}}{l!}\right)$
		$\displaystyle=\exp(X)Y\exp(-X)$
		$\displaystyle=\operatorname{Ad}_{\exp(X)}(Y)$

as required.

Solution to (18).

1. Let $h=e^{is}$ . Then

\int_{G}\varphi(hg)dg=\int_{0}^{2\pi}\varphi(e^{is}e^{it})dt.

Making the linear substitution $u=s+t$ , we obtain

\int_{0}^{2\pi}\varphi(e^{iu})du=\int_{G}\varphi(g)dg

as required. Similarly for $\int_{G}\varphi(gh)dg$ .

2. Fix $v,w\in V$ . Then the $G$ -invariance is simply the previous part applied to the function

\varphi(g)=(\rho(g)v,\rho(g)w).

It is clear that the form $(\,,\,)_{\rho}$ is sesquilinear as $(\,,\,)$ is. Finally, it is positive definite: if $v=w\neq 0$ then

(v,v)_{\rho}=\int_{G}(\rho(g)v,\rho(g)v)dg>0

since the integrand is positive for all $g$ .

3. This is the proof of Maschke’s theorem. Suppose $V$ is a finite dimensional representation and $W$ is a subrepresentation. By part (2), there is a $G$ -invariant Hermitian inner product on $V$ , and then $V=W\oplus W^{\perp}$ . By the $G$ -invariance, $W^{\perp}$ is also a subrepresentation. This implies that $V$ is completely reducible.

Solution to (19).

1. Every orthogonal matrix has determinant $\pm 1$ (which follows from taking $\det$ of $AA^{T}=I$ ). So $\det$ is a surjective homomorphism $\operatorname{O}(2)\rightarrow\{\pm 1\}$ — it is surjective because $\det(s)=-1$ — and its kernel is $\mathrm{SO}(2)$ . This implies that $\mathrm{SO}(2)$ has index 2 in $\operatorname{O}(2)$ , by the first isomorphism theorem. As $s$ is an element in the non-identity coset, we have $\operatorname{O}(2)=\mathrm{SO}(2)\cup\mathrm{SO}(2)s$ ; as every element of $\mathrm{SO}(2)$ has the form $r_{\theta}$ , we get the second claim.

Finally,

	$\displaystyle sr_{\theta}s^{-1}$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&1\\ \hidden@noalign{}\hfil\textstyle 1&0\end{pmatrix}\begin{pmatrix}% \hidden@noalign{}\hfil\textstyle\cos(\theta)&-\sin(\theta)\\ \hidden@noalign{}\hfil\textstyle\sin(\theta)&\cos(\theta)\end{pmatrix}\begin{% pmatrix}\hidden@noalign{}\hfil\textstyle 0&1\\ \hidden@noalign{}\hfil\textstyle 1&0\end{pmatrix}$
		$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle\cos(\theta)&\sin% (\theta)\\ \hidden@noalign{}\hfil\textstyle-\sin(\theta)&\cos(\theta)\end{pmatrix}$
		$\displaystyle=r_{-\theta}$

as required. Geometrically, $r_{\theta}s$ is reflection about a line through the origin making an angle of $\pi/4+\theta/2$ with the $x$ -axis (it is the reflection that maps $(\cos(\pi/4),\sin(\pi/4))$ to $(\cos(\pi/4+\theta),\sin(\pi/4+\theta))$ ).

2. Suppose that $V$ is an irreducible finite-dimensional representation of $\operatorname{O}(2)$ . As a representation of $\mathrm{SO}(2)$ it is completely reducible; every irrep of $\mathrm{SO}(2)$ is one-dimensional of the form $r_{\theta}\mapsto e^{in\theta}$ for some integer $n$ . So there is an integer $n$ and nonzero $v\in V$ such that

r_{\theta}(v)=e^{in\theta}v

for all $\theta$ . I claim that $v, s v$ span a subrepresentation; clearly their span is preserved by $s$ , and

r_{\theta}(sv)=sr_{-\theta}(v)=se^{-in\theta}v=e^{-in\theta}sv

so it is preserved by $r_{\theta}$ for all $\theta$ as well. As $V$ is irreducible, we have

V=\left\langle v,sv\right\rangle.

Now, if $n\neq 0$ then, since $r_{\theta}$ has different eigenvalues on $v$ and $s v$ (for general $\theta$ ), they are linearly independent and $\dim V=2$ . If $n<0$ then we switch $v$ and $s v$ , and now $n>0$ . Now $V$ is determined up to isomorphism by $n$ , as in the basis $v, s v$ we have

\rho(r_{\theta})=\begin{pmatrix}e^{in\theta}&0\\ 0&e^{-in\theta}\end{pmatrix}

and

\rho(s)=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}.

Different $n$ give different eigenvalues for $r_{\theta}$ and hence non-isomorphic two-dimensional representations.

If $n=0$ then $v$ , $s v$ are fixed by $\mathrm{SO}(2)$ . In this case, either $v+sv$ or $v-sv$ is nonzero and spans a one-dimensional subrepresentation which must then be all of $V$ . Thus $V$ is one-dimensional, and either isomorphic to the trivial representation or the determinant representation (according as $sv=v$ or $sv=-v$ ).

Solution to (20).

We start with the definition: if $A\in\mathfrak{gl}_{2,\mathbb{C}}$ , then

(A\phi)(v)=\frac{\mathrm{d}}{\mathrm{d}t}\phi(\exp(-At)v)|_{t=0}.

For typesetting reasons I’ll write $(x,y)^{T}$ for the column vector $\begin{pmatrix}x\\ y\end{pmatrix}$ .

Starting with $A=X$ :

	$\displaystyle(X\phi)\left((x,y)^{T}\right)$	$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\left.\phi\left(\exp(-tX)(x,y)^{T}% \right)\right\|_{t=0}$
		$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\left.\phi\left((x-ty,y)^{T}\right% )\right\|_{t=0}$
		$\displaystyle=-y\frac{\partial}{\partial x}(\phi)$

by the multivariable chain rule. Thus $X$ acts as $-y\frac{\partial}{\partial x}$ and a very similar calculation shows that $Y$ acts as $-x\frac{\partial}{\partial y}$ .

Finally,

	$\displaystyle(H\phi)\left((x,y)^{T}\right)$	$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\left.\phi\left(\exp(-tH)(x,y)^{T}% \right)\right\|_{t=0}$
		$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\left.\phi\left((e^{-t}x,e^{t}y)^{% T}\right)\right\|_{t=0}$
		$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}(e^{t}y)\|_{t=0}\frac{\partial}{% \partial y}(\phi)-\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}x)\|_{t=0}\frac{\partial% }{\partial x}(\phi)$
		$\displaystyle=y\frac{\partial}{\partial y}(\phi)-x\frac{\partial}{\partial x}(\phi)$

so that $H$ acts as $y\frac{\partial}{\partial y}-x\frac{\partial}{\partial x}$ .

An alternative solution would be to compute

\frac{\mathrm{d}}{\mathrm{d}t}\phi(\exp(-At)v)|_{t=0}

using the multivariate chain rule. The derivative of $v\mapsto\exp(-At)v$ at $t=0$ -s $-Av$ and the derivative of $\phi$ is $\nabla\phi=(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y})$ so we get that the required derivative is

-\nabla\phi Av

which one can check agrees with the answers from before.

Remark. Another possible convention is to use $g^{T}$ rather than $g^{-1}$ , which leads to slightly different formulas. This second convention is the same as if we considered elements of $\mathbb{C}^{2}$ as row vectors, with matrices acting on the right, and defined instead

(g\phi)(v)=\phi(vg).

Solution to (21).

1. Let $e_{1}\wedge e_{2}$ be the standard basis vector for $\Lambda^{2}(\mathbb{C})$ . Then

\begin{pmatrix}a&b\\ c&d\end{pmatrix}(e_{1}\wedge e_{2})=(ae_{1}+ce_{2})\wedge(be_{1}+de_{2}).

Multiplying out, and noting that $e_{1}\wedge e_{1}=e_{2}\wedge e_{2}=0$ while $e_{2}\wedge e_{1}=-e_{1}\wedge e_{2}$ , we get that

\begin{pmatrix}a&b\\ c&d\end{pmatrix}(e_{1}\wedge e_{2})=(ad-bc)e_{1}\wedge e_{2}

which is what we want (since $\det\begin{pmatrix}a&b\\ c&d\end{pmatrix}=ad-bc$ ).

2. This is similar. We get, with $X=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ ,

X(e_{1}\wedge e_{2})=(Xe_{1})\wedge e_{2}+e_{1}\wedge(Xe_{2}).

This simplifies to

(a+d)(e_{1}\wedge e_{2})=\operatorname{tr}(X)(e_{1}\wedge e_{2})

as required.

3. Compute the action of $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ on the basis vectors $e_{1}^{2},e_{1}e_{2},e_{2}^{2}$ . Skipping the working, the result is

\rho\left(\begin{pmatrix}a&b\\ c&d\end{pmatrix}\right)=\begin{pmatrix}a^{2}&ab&b^{2}\\ 2ac&ad+bc&2bd\\ c^{2}&cd&d^{2}\end{pmatrix}.

Solution to (22).

We have

H(v\otimes w)=Hv\otimes w+v\otimes Hw=(\alpha v)\otimes w+v\otimes\beta w=(% \alpha+\beta)v\otimes w

as required. If $v, w$ are highest weight vectors then

X(v\otimes w)=(Xv)\otimes w+v\otimes Xw=0\otimes w+v\otimes 0=0

so $v\otimes w$ is a highest weight vector.

Solution to (23).

1.

If $v_{1},\ldots,v_{n}$ is a basis of weight vectors of $V$ with weights $\lambda_{1},\ldots,\lambda_{n}$ , then the dual basis $v_{1}^{*},\ldots,v_{n}^{*}$ is a basis of weight vectors of $V^{*}$ with weights $-\lambda_{1},\ldots,-\lambda_{n}$ . Proof: we have

$Hv_{i}^{*}(v_{j})=-v_{i}^{*}(Hv_{j})=-\lambda_{j}\delta_{ij}$

so that

$Hv_{i}^{*}=-\lambda_{i}v_{i}^{*}$

as required.
2.

Since the weights of every irreducible representation of $\mathfrak{sl}_{2,\mathbb{C}}$ are symmetrical about the origin, and every representation is a direct sum of irreducible representations, the weights of $V$ are symmetrical about the origin. Thus the weights of $V^{*}$ are the same as the weights of $V$ . Since the weights determine the representation up to isomorphism, we have $V^{*}\cong V$ .

Solution to (24).

We will show that $\mathcal{C}$ commutes with each of $\pi(X)$ , $\pi(Y)$ and $\pi(H)$ . Note that, since $\pi$ is a Lie algebra representation, we have

	$\displaystyle\pi(X)\pi(Y)$	$\displaystyle=\pi(Y)\pi(X)+\pi(H)$
	$\displaystyle\pi(H)\pi(X)$	$\displaystyle=\pi(X)\pi(H)+2\pi(X)$
	$\displaystyle\pi(H)\pi(Y)$	$\displaystyle=\pi(Y)\pi(H)-2\pi(Y).$

For instance, the first equation follows from $[\pi(X),\pi(Y)]=\pi([X,Y])=\pi(H)$ . So we get

	$\displaystyle\mathcal{C}\pi(X)$	$\displaystyle=\pi(X)\pi(Y)\pi(X)+\pi(Y)\pi(X)^{2}+\frac{1}{2}\pi(H)^{2}\pi(X)$
		$\displaystyle=2\pi(X)\pi(Y)\pi(X)-\pi(H)\pi(X)+\frac{1}{2}\pi(H)\pi(X)\pi(H)+% \pi(H)\pi(X)$
		$\displaystyle=2\pi(X)\pi(Y)\pi(X)+\frac{1}{2}\pi(H)\pi(X)\pi(H)$

and therefore

	$\displaystyle\pi(X)\mathcal{C}$	$\displaystyle=\pi(X)^{2}\pi(Y)+\pi(X)\pi(Y)\pi(X)+\frac{1}{2}\pi(X)\pi(H)^{2}$
		$\displaystyle=2\pi(X)\pi(Y)\pi(X)+\pi(X)\pi(H)+\frac{1}{2}\pi(H)\pi(X)\pi(H)-% \pi(X)\pi(H)$
		$\displaystyle=2\pi(X)\pi(Y)\pi(X)+\frac{1}{2}\pi(H)\pi(X)\pi(H)$
		$\displaystyle=\mathcal{C}\pi(X).$

Similarly, $\mathcal{C}$ commutes with $\pi(Y)$ . Finally,

\mathcal{C}\pi(H)=\pi(H)\mathcal{C}=\pi(X)\pi(H)\pi(Y)+\pi(H)\pi(Y)\pi(H)+2\pi% (X)\pi(Y)+2\pi(Y)\pi(X)+\frac{1}{2}\pi(H)^{3}.

If $V$ is irreducible, then since $\mathcal{C}$ commutes with all elements of $\mathfrak{sl}_{2,\mathbb{C}}$ it is a $\mathfrak{sl}_{2,\mathbb{C}}$ homomorphism $V\rightarrow V$ and so is scalar by Schur’s lemma.

2.

The representation $V=V^{(n)}$ is irreducible, and so $\mathcal{C}$ is a scalar. To find the scalar, we just need to evaluate $\mathcal{C}$ on a single element of $V$ ; I will use the highest weight vector $e_{1}^{n}$ . We have

$\displaystyle\mathcal{C}e_{1}^{n}$ $\displaystyle=\pi(X)ne_{1}^{n-1}e_{-1}+\pi(Y)0+\frac{1}{2}n^{2}e_{1}^{n}$

$\displaystyle=(n+\frac{1}{2}n^{2})e_{1}^{n}$

so $\mathcal{C}$ acts as the scalar $\frac{n^{2}+2n}{2}$ on $V^{(n)}$ . Here we see why we might want to use $1+2\mathcal{C}$ instead: then it acts as $(n+1)^{2}$ .

Recall that $X$ acts as $-y\frac{\partial}{\partial x}$ , $Y$ acts as $-x\frac{\partial}{\partial y}$ , $H$ acts as $y\frac{\partial}{\partial y}-x\frac{\partial}{\partial x}$ . We see that

	$\displaystyle\mathcal{C}$	$\displaystyle=y\frac{\partial}{\partial x}(x\frac{\partial}{\partial y})+x% \frac{\partial}{\partial y}(y\frac{\partial}{\partial x})+\frac{1}{2}(y\frac{% \partial}{\partial y}-x\frac{\partial}{\partial x})(y\frac{\partial}{\partial y% }-x\frac{\partial}{\partial x})$
		$\displaystyle=y\frac{\partial}{\partial y}+yx\frac{\partial^{2}}{\partial x% \partial y}+x\frac{\partial}{\partial x}+xy\frac{\partial^{2}}{\partial x% \partial y}+\frac{1}{2}\left(y\frac{\partial}{\partial y}+y^{2}\frac{\partial^% {2}}{\partial y^{2}}-2xy\frac{\partial^{2}}{\partial x\partial y}+x\frac{% \partial}{\partial x}+x^{2}\frac{\partial^{2}}{\partial x^{2}}\right)$
		$\displaystyle=\frac{1}{2}\left(3x\frac{\partial}{\partial x}+3y\frac{\partial}% {\partial y}+x^{2}\frac{\partial^{2}}{\partial x^{2}}+2xy\frac{\partial^{2}}{% \partial x\partial y}+y^{2}\frac{\partial^{2}}{\partial y^{2}}\right).$

If we apply this to a monomial $x^{a}y^{b}$ of degree $n=a+b$ , we find that

	$\displaystyle\mathcal{C}x^{a}y^{b}$	$\displaystyle=\frac{1}{2}\left(3(a+b)+a(a-1)+2ab+b(b-1)\right)x^{a}y^{b}$
		$\displaystyle=\frac{1}{2}\left(2n+n^{2}\right)x^{a}y^{b}.$

We can explain this as follows: the space of homogeneous polynomial functions of degree $n$ is isomorphic to $V^{(n)}$ and so the calculation from the previous part applies!

Solution to (25).

The idea is to use the formulae we found for the actions of $\rho(Y)$ and $\rho(X)$ on a weight basis for a representation of $\mathfrak{sl}_{2,\mathbb{C}}$ in the course of proving their classification to just define a representation with highest weight $\lambda$ .

Let $V$ be an infinite-dimensional vector space with basis $v_{0},v_{1},v_{2},\ldots$ . Define a representation $\rho=\rho_{\lambda}$ of $\mathfrak{sl}_{2,\mathbb{C}}$ on $V$ by

	$\displaystyle\rho(H)v_{i}$	$\displaystyle=(\lambda-2i)v_{i}$
	$\displaystyle\rho(Y)v_{i}$	$\displaystyle=v_{i+1}$
	$\displaystyle\rho(X)v_{i}$	$\displaystyle=i(\lambda-i+1)v_{i-1}$

for all $i\geq 0$ (so $\rho(X)v_{0}=0$ ). To check that this does define a representation, we must check that $[\rho(A),\rho(B)]=\rho([A,B])$ for all $A,B\in\{X,Y,H\}$ (it suffices to consider each pair $(A,B)$ of distinct elements in one particular order). And indeed we have:

	$\displaystyle[\rho(X),\rho(Y)]v_{i}$	$\displaystyle=\left((i+1)(\lambda-i)-i(\lambda-i+1)\right)v_{i}$
		$\displaystyle=(\lambda-2i)v_{i}$
		$\displaystyle=\rho(H)v_{i}$
	$\displaystyle[\rho(H),\rho(X)]v_{i}$	$\displaystyle=i(\lambda-i+1)v_{i-1}\times(\lambda-2i+2-(\lambda-2i))$
		$\displaystyle=2i(\lambda-i+1)v_{i-1}$
		$\displaystyle=2\rho(X)v_{i}$
	$\displaystyle[\rho(H),\rho(Y)]v_{i}$	$\displaystyle=(\lambda-2i-2-(\lambda-2i))v_{i+1}$
		$\displaystyle=-2\rho(Y)v_{i}$

as required.

Finally, $v_{0}$ is killed by $\rho(X)$ and is a weight vector of weight $\lambda$ , so $V$ has highest weight $\lambda$ .

Advanced remark: $(V,\rho_{\lambda})$ will be irreducible unless $\lambda$ is a nonnegative integer, in which case the vector $v_{\lambda+1}$ generates a subrepresentation $W$ isomorphic to $(V,\rho_{-\lambda-2})$ and the quotient representation $V/W$ is exactly the finite-dimensional irreducible representation with highest weight $\lambda$ .

Solution to (26).

We use the notation $v_{2k-n}=e_{1}^{k}e_{-1}^{n-k}$ for the usual weight basis of $\operatorname{Sym}^{n}(\mathbb{C}^{2})$ , so $v_{2k-n}$ has weight $2k-n$ , for $0\leq k\leq n$ . We have the formulas $Xv_{2k-n}=(n-k)v_{2k+2-n}$ and $Yv_{2k-n}=kv_{2k-2-n}$ . We also abbreviate $\operatorname{Sym}^{n}=\operatorname{Sym}^{n}(\mathbb{C}^{2})$ .

1.

The weights of $\operatorname{Sym}^{2}(\mathbb{C}^{2})$ are $\{2,0,-2\}$ . To obtain the weights of $\operatorname{Sym}^{2}(\operatorname{Sym}^{2}(\mathbb{C}^{2}))$ we add together all possible (unordered, possibly equal) pairs of these and get:

$\{4,2,0,0,-2,-4\}.$

Thus

$\operatorname{Sym}^{2}(\operatorname{Sym}^{2}(\mathbb{C}^{2}))\cong% \operatorname{Sym}^{4}(\mathbb{C}^{2})\oplus\mathbb{C}.$

A highest weight vector of weight 4 is $v_{2}^{2}$ (clear as it is a symmetric product of highest weight vectors). To get a highest weight vector of weight 2 we must take a linear combination of $v_{0}^{2}$ and $v_{2}v_{-2}$ which is killed by $X$ . Since $Xv_{0}^{2}=2v_{0}v_{2}$ and $Xv_{2}v_{-2}=2v_{2}v_{0}$ we see that

$v_{0}^{2}-v_{2}v_{-2}$

is a weight vector of weight 0 killed by $X$ , so a highest weight vector of weight 0.
2.

This time we must take all sums of unordered pairs of /distinct/ elements of $\{2,0,-2\}$ . This gives

$\{2,0,-2\}$

so that $\Lambda^{2}(\operatorname{Sym}^{2}(\mathbb{C}^{2}))\cong\operatorname{Sym}^{2}% (\mathbb{C}^{2})$ . A highest weight vector of weight $2$ is $v_{2}\wedge v_{0}$ .
3.

We have to add together all pairs of weights from $\{3,1,-1,-3\}$ and $\{2,0,-2\}$ , giving

$\{5,3,3,1,1,1,-1,-1,-1,-3,-3,-5\}$

as the weights of $\operatorname{Sym}^{3}\otimes\operatorname{Sym}^{2}$ . Thus the decomposition is

$\operatorname{Sym}^{5}\oplus\operatorname{Sym}^{3}\oplus\operatorname{Sym}^{1}.$

A highest weight vector of weight 5 is $v_{3}\otimes v_{2}$ . We can now apply $Y$ repeatedly (and divide out by constant factors where possible to keep the numbers small) to obtain a weight basis of the copy of $\operatorname{Sym}^{5}$ in the representation, as shown in the table.

$\begin{array}[]{r|r}\text{weight}&\text{vector}\\ \hline\cr 5&v_{3}\otimes v_{2}\\ 3&3v_{1}\otimes v_{2}+2v_{3}\otimes v_{0}\\ 1&3v_{-1}\otimes v_{2}+6v_{1}\otimes v_{0}+v_{3}\otimes v_{-2}\\ -1&3v_{1}\otimes v_{-2}+6v_{-1}\otimes v_{0}+v_{-3}\otimes v_{2}\\ -3&3v_{-1}\otimes v_{-2}+2v_{-3}\otimes v_{0}\\ -5&v_{-3}\otimes v_{-2}\end{array}$

We have $X(v_{3}\otimes v_{0})=v_{3}\otimes v_{2}$ and $X(v_{1}\otimes v_{2})=v_{3}\otimes v_{2}$ so that

$v_{3}\otimes v_{0}-v_{1}\otimes v_{2}$

is a highest weight vector of weight 3. We apply $Y$ repeatedly (and divide out scalars where possible) to obtain a weight basis of the copy of $\operatorname{Sym}^{3}$ in the representation:

$\begin{array}[]{r|r}\text{weight}&\text{vector}\\ \hline\cr 3&v_{3}\otimes v_{0}-v_{1}\otimes v_{2}\\ 1&v_{3}\otimes v_{-2}+v_{1}\otimes v_{0}-2v_{-1}\otimes v_{2}\\ -1&v_{-3}\otimes v_{2}+v_{-1}\otimes v_{0}-2v_{1}\otimes v_{-2}\\ -3&v_{-3}\otimes v_{0}-v_{-1}\otimes v_{-2}\end{array}$

Notice that we can ‘cheat’ and obtain just the weight vectors with nonpositive weight, and then apply the symmetry sending $v_{i}$ to $v_{-i}$ to obtain those of nonnegative weight.

Finally, we have $X(v_{3}\otimes v_{-2})=2v_{3}\otimes v_{0}$ , $X(v_{1}\otimes v_{0})=v_{3}\otimes v_{0}+v_{1}\otimes v_{2}$ , and $X(v_{-1}\otimes v_{2})=2v_{1}\otimes v_{2}$ so that

$v_{3}\otimes v_{-2}-2v_{1}\otimes v_{0}+v_{-1}\otimes v_{2}$

is a highest weight vector of weight 1. Applying $Y$ (or the symmetry discussed above) we see that this vector together with

$v_{-3}\otimes v_{2}-2v_{-1}\otimes v_{0}+v_{1}\otimes v_{-2}$

is a weight basis for the copy of $\mathbb{C}^{2}$ .
4.

We must add together all unordered triples of (not necessarily distinct) elements of $\{2,0,-2\}$ . We get that the weights of $\operatorname{Sym}^{3}(\operatorname{Sym}^{2}(\mathbb{C}^{2}))$ are:

$\{6,4,2,2,0,0,-2,-2,-4,-6\}$

so that

$\operatorname{Sym}^{3}(\operatorname{Sym}^{2}(\mathbb{C}^{2}))\cong% \operatorname{Sym}^{6}(\mathbb{C}^{2})\oplus\operatorname{Sym}^{2}(\mathbb{C}^% {2}).$

A highest weight vector of weight $6$ is $v_{2}^{3}$ . We have $X(v_{2}^{2}v_{-2})=2v_{2}^{2}v_{0}$ and $X(v_{2}v_{0}^{2})=2v_{2}^{2}v_{0}$ so that

$v_{2}^{2}v_{-2}-v_{2}v_{0}^{2}$

is a highest weight vector of weight 2.

Solution to (27).

1.
The weights of $\operatorname{Sym}^{a}\mathbb{C}^{2}$ are $a,a-2,\ldots,2-a,-a$ and the weights of $\operatorname{Sym}^{b}\mathbb{C}^{2}$ are $b,b-2,\ldots,2-b,-b$ . Without loss of generality, $a\geq b$ . Adding these lists together, remembering multiplicity, we see that in the tensor product:
- •
  
  For weights $a+b,a+b-2,\ldots,a-b=a+b-2b$ , each $a+b-2k$ occurs $k+1$ times as
  
  $a+(b-2k),(a-2)+(b-2k+2),\ldots,(a-2k)+b$
  
  ; the same holds for their negatives.
- •
  
  Each weight $a-b,a-b-2,\ldots,b-a+2,b-a$ occurs $b+1$ times; specifically, $a-b-2k$ occurs as
  
  $(a-2k)+(-b),(a-2k-2)+(2-b),\ldots,(a-2k-2b)+b.$
This agrees with the weights of

$\operatorname{Sym}^{a+b}\mathbb{C}^{2}\oplus\operatorname{Sym}^{a+b-2}\mathbb{% C}^{2}\oplus\ldots\oplus\operatorname{Sym}^{a-b}\mathbb{C}^{2}$

and so this is the decomposition of $\operatorname{Sym}^{a}\mathbb{C}^{2}\otimes\operatorname{Sym}^{b}\mathbb{C}^{2}$ into irreducibles.
2.

Omitted (for now).

Solution to (28).

Let $\mathcal{I}=\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}$ , $\mathcal{J}=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}$ , $\mathcal{K}=\begin{pmatrix}0&i\\ i&0\end{pmatrix}$ be a basis for $\mathfrak{su}_{2}$ , so that $[\mathcal{I},\mathcal{J}]=2\mathcal{K}$ etc. Let $A=x\mathcal{I}+y\mathcal{J}+z\mathcal{K}$ be a general element of $\mathfrak{su}_{2}$ . Then we compute that the matrix for the adjoint action of $A$ with respect to the basis $\mathcal{I},\mathcal{J},\mathcal{K}$ is:

2\begin{pmatrix}0&-z&y\\ z&0&-x\\ -y&x&0\end{pmatrix}.

This has characteristic polynomial $T(T^{2}+4(x^{2}+y^{2}+z^{2}))$ which has distinct roots (unless $x=y=z=0$ in which case $A=0$ ) so the adjoint action of $A$ is diagonalizable. So, for every $A\in\mathfrak{su}_{2}$ , the map $B\mapsto[A,B]$ is diagonalizable. This is not true for $\mathfrak{sl}_{2,\mathbb{R}}$ , since the action of $X=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$ is, wrt the standard basis $X, H, Y$ ,

\begin{pmatrix}0&-2&0\\ 0&0&1\\ 0&0&0\end{pmatrix}

which is not diagonalizable.

Here is another, related solution. For every finite-dimensional (nonzero) representation $(\rho,V)$ of $\mathfrak{sl}_{2,\mathbb{R}}$ , $\rho(X)$ is not diagonalizable. But in the standard representation of $\mathfrak{su}_{2}$ , every element is diagonalizable. Indeed, if $A\in\mathfrak{su}_{2}$ then $i A$ is Hermitian, and hence diagonalizable, so $A$ is diagonalizable. So $\mathfrak{su}_{2}$ cannot be isomorphic to $\mathfrak{sl}_{2,\mathbb{R}}$ .

Solution to (29).

The eigenvalues of $\begin{pmatrix}e^{it}&0\\ 0&e^{-it}\end{pmatrix}$ on the weight basis $\{e_{1}^{k}e_{-1}^{n-k}:0\leq k\leq n\}$ for $\operatorname{Sym}^{n}\mathbb{C}^{2}$ are $e^{kit}e^{-(n-k)it}=e^{(2k-n)it}$ for $0\leq k\leq n$ . The trace is then (using the formula for the sum of a GP)

	$\displaystyle\chi_{n}(t)$	$\displaystyle=\sum_{k=0}^{n}e^{(2k-n)it}$
		$\displaystyle=e^{-nit}\frac{e^{(2n+2)it}-1}{e^{2it}-1}$
		$\displaystyle=\frac{e^{(n+1)it}-e^{-(n+1)it}}{e^{it}-e^{-it}}$
		$\displaystyle=\frac{\sin((n+1)t)}{\sin(t)}.$

Solution to (30).

1.

This follows from Schur’s lemma, since $Z$ is in the centre of $\mathfrak{gl}_{2,\mathbb{C}}$ .
2.

Suppose otherwise. Then there is a proper, nonzero subspace $W\subset V$ preserved by $\rho(g)$ for all $g\in\mathfrak{sl}_{2,\mathbb{C}}$ . But $W$ is also preserved by $\rho(Z)$ as $\rho(Z)$ is scalar. So $W$ is preserved by

$\mathfrak{sl}_{2,\mathbb{C}}\oplus\mathbb{C}\cdot Z=\mathfrak{gl}_{2,\mathbb{C}}$

so it is a $\mathfrak{gl}_{2,\mathbb{C}}$ -subrepresentation, contradicting that $V$ is irreducible.
3.

Let $V^{(n)}$ be the irreducible representation of $\mathfrak{sl}_{2,\mathbb{C}}$ of dimension $n+1$ . To extend the action of $\mathfrak{sl}_{2,\mathbb{C}}$ to an action of $\mathfrak{gl}_{2,\mathbb{C}}$ we have to send $Z$ to a map $V^{(n)}\rightarrow V^{(n)}$ commuting with the action of $\mathfrak{sl}_{2,\mathbb{C}}$ . Since $V^{(n)}$ is irreducible, we must send $Z$ to a scalar map, and any scalar $\lambda$ is possible.
4.

If the representation $\rho$ from the third part exponentiates to a representation of $\operatorname{GL}_{2}(\mathbb{C})$ , then we must have that

$\exp(\pi i\lambda I)=\exp(\rho(\pi iZ))=\rho(\exp(\pi iZ))=\rho(\begin{pmatrix% }-1&0\\ 0&-1\end{pmatrix})=(-1)^{n}I.$

This implies that $\lambda$ is an integer of the same parity as $n$ . Moreover, if $\lambda$ is such an integer then we get a representation of $\operatorname{GL}_{2}(\mathbb{C})$ with the required derivative by taking

$\operatorname{Sym}^{n}(\mathbb{C}^{2})\otimes\det^{\frac{\lambda-n}{2}}$

where $\mathbb{C}^{2}$ is the standard representation of $\operatorname{GL}_{2}(\mathbb{C})$ and $\det^{k}$ is the one dimensional representation on which $g$ acts as $\det(g)^{k}$ .

Thus the irreducible finite-dimensional representations of $\operatorname{GL}_{2}(\mathbb{C})$ are parametrised by pairs $(n,\lambda)$ where $n\geq 0$ is an integer and $\lambda$ is an integer of the same parity as $n$ .

Solution to (31).

We have $J_{x}=z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}$ and so

	$\displaystyle J_{x}^{2}$	$\displaystyle=z\frac{\partial}{\partial y}z\frac{\partial}{\partial y}-z\frac{% \partial}{\partial y}y\frac{\partial}{\partial z}-y\frac{\partial}{\partial z}% z\frac{\partial}{\partial y}+y\frac{\partial}{\partial z}y\frac{\partial}{% \partial z}$	$\displaystyle=z^{2}\frac{\partial^{2}}{\partial y^{2}}-yz\frac{\partial^{2}}{% \partial y\partial z}-z\frac{\partial}{\partial z}-y\frac{\partial}{\partial y% }-yz\frac{\partial^{2}}{\partial z\partial y}+y^{2}\frac{\partial^{2}}{% \partial z^{2}}$
		$\displaystyle=z^{2}\frac{\partial^{2}}{\partial y^{2}}+y^{2}\frac{\partial^{2}% }{\partial z^{2}}-2yz\frac{\partial^{2}}{\partial y\partial z}-z\frac{\partial% }{\partial z}-y\frac{\partial}{\partial y}.$

Summing over $x$ , $y$ and $z$ we see that

\sum J_{x}^{2}=(\sum x^{2})(\sum\frac{\partial^{2}}{\partial x^{2}})-\sum x^{2% }\frac{\partial^{2}}{\partial x^{2}}-2\sum x\frac{\partial}{\partial x}-2\sum yz% \frac{\partial^{2}}{\partial y\frac{\partial}{\partial z}}.

Note that

(\sum x\frac{\partial}{\partial x})^{2}=\sum x^{2}\frac{\partial^{2}}{\partial x% ^{2}}+\sum x\frac{\partial}{\partial x}+2\sum yz\frac{\partial^{2}}{\partial y% \frac{\partial}{\partial z}}

since

x\frac{\partial}{\partial x}x\frac{\partial}{\partial x}=x^{2}\frac{\partial^{% 2}}{\partial x^{2}}+x\frac{\partial}{\partial x}.

Thus

\sum J_{x}^{2}=r^{2}\Delta-(\sum x\frac{\partial}{\partial x})^{2}-(\sum x% \frac{\partial}{\partial x}).

On $\mathcal{P}^{\ell}$ , $\sum x\frac{\partial}{\partial x}$ acts as multiplication by $\ell$ (Euler’s identity), and so

\sum J_{x}^{2}=r^{2}\Delta-\ell^{2}-\ell

as required.

2.

The isomorphism $\mathfrak{sl}_{2,\mathbb{C}}\to\mathfrak{so}_{3}$ sends

$\displaystyle X=\mathcal{I}-i\mathcal{J}$ $\displaystyle\mapsto J_{x}-iJ_{y}$

$\displaystyle Y=-(\mathcal{I}+i\mathcal{J})$ $\displaystyle\mapsto-(J_{x}+iJ_{y})$

$\displaystyle H=-2i\mathcal{K}$ $\displaystyle\mapsto-2iJ_{z}.$

The Casimir for $\mathfrak{sl}_{2,\mathbb{C}}$ is

$\mathcal{C}=XY+YX+\tfrac{1}{2}H^{2}$

(where this multiplication is not as $2\times 2$ matrices but as operators on a fixed representation!), and under the isomorphism above this goes to

$-(J_{x}-iJ_{y})(J_{x}+iJ_{y})-(J_{x}+iJ_{y})(J_{x}-iJ_{y})-2J_{z}^{2}$

which multiplies out to

$-2(J_{x}^{2}+J_{y}^{2}+J_{z}^{2}).$

Thus, as operators on $\mathcal{P}^{\ell}$ ,

$\mathcal{C}=2(\ell^{2}+\ell-r^{2}\Delta).$

Solution to (32).

1.

We first check that $(x-iy)^{\ell}$ is a weight vector for $J_{z}=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$ :

$\displaystyle J_{z}(x-iy)^{\ell}$ $\displaystyle=y\ell(x-iy)^{\ell-1}+i\ell x(x-iy)^{\ell-1}$

$\displaystyle=\ell(y+ix)(x-iy)^{\ell-1}$

$\displaystyle=i\ell(x-iy)^{\ell}$

so that it is a weight vector of weight $i\ell$ .

Recall that the raising operator is

$J_{x}-iJ_{y}=z(\frac{\partial}{\partial y}+i\frac{\partial}{\partial x})-(y+ix% )\frac{\partial}{\partial z}.$

Thus

$(J_{x}-iJ_{y})(x-iy)^{\ell}=\ell z(-i+i)(x-iy)^{\ell-1}=0$

so that $(x-iy)^{\ell}$ is a highest weight vector as required.
2.

The lowering operator is (up to sign)

$J_{x}+iJ_{y}=z(\frac{\partial}{\partial y}-i\frac{\partial}{\partial x})-(y-ix% )\frac{\partial}{\partial z}.$

Applying this once to $(x-iy)^{\ell}$ we find that

$\ell z(-i-i)(x-iy)^{\ell-1}$

is a weight vector for $J_{z}$ of weight $i(\ell-1)$ ; we may throw away the constants to get just

$z(x-iy)^{\ell-1}$

(for $\ell\geq 1$ ; for $\ell=0$ we just get $0$ !). Applying the lowering operator to this, we get

$-2i(\ell-1)z^{2}(x-iy)^{\ell-2}-(y-ix)(x-iy)^{\ell-1}$

which simplifies to

$i(x^{2}+y^{2}-2(\ell-1)z^{2})(x-iy)^{\ell-2},$

a weight vector of weight $i(\ell-2)$ (for $\ell\geq 2$ ; for $\ell=1$ we just get $-(y-ix)=i(x+iy)$ here, while for $\ell=0$ we get $0$ ).
3.

When $\ell=1$ we apply the procedure from the previous part to get a weight basis

$(x-iy),z,(x+iy).$

When $\ell=2$ the procedure from the previous part gives weight vectors

$(x-iy)^{2},z(x-iy),(x^{2}+y^{2}-2z^{2}).$

We can now either continue applying the lowering operator, or observe that the substitution $y\mapsto-y$ preserves harmonic polynomials and sends weight vectors for $J_{z}$ of weight $i k$ to weight vectors for $J_{z}$ of weight $-ik$ so that we get a weight basis

$(x-iy)^{2},z(x-iy),(x^{2}+y^{2}-2z^{2}),z(x+iy),(x+iy)^{2}.$

(Proof: let $f(x,y,z)$ be a harmonic polynomial and weight vector for $J_{z}$ , and let $g(x,y,z)=f(x,-y,z)$ . Then $\Delta g=\Delta f=0$ , so $g$ is harmonic, while

$J_{z}g=(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y})g=-(-y)(% \frac{\partial}{\partial x}f)(x,-y,z)+x(\frac{\partial}{\partial y}f)(x,-y,z)=% -(J_{z}f)(x,-y,z)$

and so $g$ is a weight vector with weight minus that of $f$ .)

Solution to (33).

1.

Note that

$\frac{\partial}{\partial x}(r^{2}f)=r^{2}\frac{\partial}{\partial x}f+2xf$

and so

$\frac{\partial^{2}}{\partial x^{2}}(r^{2}f)=r^{2}\frac{\partial^{2}}{\partial x% ^{2}}f+4x\frac{\partial}{\partial x}f+2f.$

Summing over $x, y, z$ and using Euler’s identity

$\sum x\frac{\partial}{\partial x}f=\ell f$

for $f\in\mathcal{P}^{(\ell)}$ gives

$\Delta r^{2}=r^{2}\Delta+4\ell+6$

on $\mathcal{P}^{(\ell)}$ , as required.
2.

Notice that, if $f\in\mathcal{P}^{(\ell)}$ , then $r^{2k-2}f\in\mathcal{P}^{(\ell+2k-2)}$ and so

$\Delta(r^{2k}f)=r^{2}\Delta(r^{2k-2}f)+2(2\ell+4k-1)r^{2k-2}f.$

Iterating, we see that

$\displaystyle\Delta(r^{2k}f)$ $\displaystyle=r^{2k}\Delta(f)+\sum_{i=1}^{k}2(2\ell+4i-1)r^{2k-2}f$

$\displaystyle=r^{2k}\Delta(f)+2k(2\ell+2k+1)r^{2k-2}f.$
3.

Suppose $f\in\mathcal{H}^{\ell}\cap r^{2}\mathcal{P}^{\ell-2}$ is nonzero. Then we may write $f=r^{2k}g$ where $\ell/2\geq k\geq 1$ and $g$ is not divisible by $r^{2}$ . Then

$0=\Delta f=\Delta(r^{2k}g)=r^{2k}\Delta(g)+2k(2(\ell-2k)+2k+1)r^{2k-2}g$

by the previous part and the fact that $f$ is harmonic. Since $2k(2(\ell-2k)+2k+1)\neq 0$ , we see that

$r^{2k-2}g=cr^{2k}\Delta g$

and so $g=cr^{2}\Delta g$ for some constant $c$ . But this contradicts our assumption that $g$ is not divisible by $r^{2}$ !

Solution to (34).

1.

For the first part, we first find a basis of weight vectors for $V$ (with respect to $J_{z}=\begin{pmatrix}0&-1&0\\ 1&0&0\\ 0&0&0\end{pmatrix}$ ). This amounts to finding the eigenvectors for $J_{z}$ ; we see that $e_{3}$ is an eigenvector (weight vector) of eigenvalue (weight) 0 and that $e_{1}\mp ie_{2}$ are eigenvectors of eigenvalue $\pm i$ (note the signs!!).

It follows that a weight basis of $\operatorname{Sym}^{2}(V)$ is given by taking the product of any two of $\{e_{1}\mp ie_{2},e_{3}\}$ with the weights being the corresponding sums of two of $\{\pm i,0\}$ . We find that the weights are

$\{2i,i,0,0,-i,-2i\}.$

A weight vector of weight $\pm 2i$ is given by $(e_{1}\mp ie_{2})^{2}=e_{1}^{2}\mp 2ie_{1}e_{2}-e_{2}^{2}$ . A weight vector of weight $\pm i$ is given by $(e_{1}\mp ie_{2})e_{3}=e_{1}e_{3}\mp ie_{2}e_{3}$ . Finally, a basis of weight vectors of weight 0 is given by $(e_{1}+ie_{2})(e_{1}-ie_{2})=e_{1}^{2}+e_{2}^{2}$ and $e_{3}^{2}$ .

Since the weights of $\operatorname{Sym}^{2}(V)$ agree with the weights of $V^{(2)}\oplus V^{(0)}$ — notation as in Theorem 3.29 — this is the decomposition into irreducible representations. We now have to find $V^{(0)}$ and $V^{(2)}$ as subrepresentations of $\operatorname{Sym}^{2}(V)$ .

To find the copy of $V^{(0)}$ , we need to look for a weight vector of weight 0 preserved by $J_{x}$ , $J_{y}$ and $J_{z}$ . Note that

$J_{x}e_{3}^{2}=2(J_{x}e_{3})e_{3}=-2e_{2}e_{3},$

$J_{y}e_{3}^{2}=2e_{1}e_{3}$

and $J_{z}e_{3}^{2}=0$ . Similar calculations show $J_{x}(e_{1}^{2}+e_{2}^{2})=0+2e_{2}e_{3}$ and $J_{y}(e_{1}^{2}+e_{2}^{2})=-2e_{1}e_{3}$ . It follows that $e_{1}^{2}+e_{2}^{2}+e_{3}^{2}$ is killed by each of $J_{x}$ , $J_{y}$ and $J_{z}$ and therefore spans a copy of the trivial representation.

The copy of $V^{(2)}$ inside $\operatorname{Sym}^{2}(V)$ is spanned by the weight vectors of nonzero weights together with a weight vector of weight 0 which may be obtained by applying the lowering operator to the weight vector of weight $i$ . This gives

$(J_{x}+iJ_{y})(e_{1}e_{3}-ie_{2}e_{3})=-e_{1}e_{2}-ie_{3}^{2}+ie_{2}^{2}+i(-e_% {3}^{2}+e_{1}^{2}-ie_{1}e_{2})$

simplifying to $i(e_{1}^{2}+e_{2}^{2}-2e_{3}^{2})$ . Thus a basis of weight vectors for the copy of $V^{(2)}$ in $\operatorname{Sym}^{2}(V)$ is

$(e_{1}\mp ie_{2})^{2},(e_{1}\mp ie_{2})e_{3},(e_{1}^{2}+e_{2}^{2}-2e_{3}^{2}).$

An alternative approach is to notice that if we write $x=e_{1}$ , $y=e_{2}$ and $z=e_{3}$ , then $V$ is simply isomorphic to the representation of $\mathfrak{so}_{3}$ on homogeneous linear polynomials. Then the symmetric square of $V$ is exactly the space $\mathcal{P}^{(2)}$ of homogeneous quadratic polynomials, which we know decomposes

$\mathcal{H}^{(2)}\oplus\left\langle r^{2}\right\rangle\mathcal{H}^{(0)}.$

The basis for $\mathcal{H}^{(2)}$ we wrote down above then corresponds exactly to that in example 3.40, while the copy of the trivial representation is spanned by

$r^{2}=x^{2}+y^{2}+z^{2}=e_{1}^{2}+e_{2}^{2}+e_{3}^{2}.$
2.

The $J_{z}$ weights of $\mathcal{H}^{(2)}$ are $2i,i,0,-i,-2i$ . Thus the $J_{z}$ weights of $\mathcal{H}^{(2)}\otimes\mathcal{H}^{(2)}$ are obtained by adding all possible ordered pairs of these, giving

$\{\pm 4i,\pm 3i,\pm 3i,\pm 2i,\pm 2i,\pm 2i,\pm i,\pm i,\pm i,\pm i0,0,0,0,0\}$

This agrees with the multiset of weights of

$\mathcal{H}^{(4)}\oplus\mathcal{H}^{(3)}\oplus\mathcal{H}^{(2)}\oplus\mathcal{% H}^{(1)}\oplus\mathcal{H}^{(0)}$

so that this is the required decomposition (because representations of $\mathfrak{so}_{3}$ are determined by their $J_{z}$ weights). Note that, thankfully, we aren’t asked to find these as subrepresentations!

Solution to (35).

This is just matrix multiplication:

\begin{pmatrix}a_{1}&0&0\\ 0&a_{2}&0\\ 0&0&a_{3}\end{pmatrix}E_{ij}=a_{i}E_{ij}

and

E_{ij}\begin{pmatrix}a_{1}&0&0\\ 0&a_{2}&0\\ 0&0&a_{3}\end{pmatrix}=a_{j}E_{ij}

whence

[\begin{pmatrix}a_{1}&0&0\\ 0&a_{2}&0\\ 0&0&a_{3}\end{pmatrix},E_{ij}]=(a_{i}-a_{j})E_{ij}.

Similarly $E_{12}E_{23}=E_{13}$ and $E_{23}E_{12}=0$ so that

[E_{12},E_{23}]=E_{13}.

Solution to (36).

For $\mathfrak{sl}_{2}$ we have $\mathfrak{h}=\left\langle H\right\rangle$ and a weight, a linear map $\lambda:\mathfrak{h}\to\mathbb{C}$ , is uniquely determined by $\lambda(H)$ . Thus we identify weights with complex numbers (actually integers, for finite-dimensional representations).

With this identification, the roots are $\{\pm 2\}$ and the root spaces are $\left\langle X\right\rangle$ and $\left\langle Y\right\rangle$ .

Solution to (37).

Let $\rho^{*}$ be the dual representation. If $A\in\mathfrak{sl}_{3,\mathbb{C}}$ , then the matrix of $\rho^{*}(A)$ with respect to the dual basis is $-A^{T}$ . From this, we see that $\rho^{*}(E_{ij})e_{3}^{*}=0$ if $i\neq 3$ , while $\rho^{*}(E_{31})e_{3}^{*}=-e_{1}^{*}$ and $\rho^{*}(E_{32})e_{3}^{*}=-e_{2}^{*}$ .

Moreover, if $H$ is diagonal with entries $a_{1},a_{2},a_{3}$ , then $\rho^{*}(H)e_{3}^{*}=-a_{3}e_{3}^{*}$ . Thus $e_{3}^{*}$ is a weight vector with weight $-L_{3}$ . Since $e_{3}^{*}$ is killed by $E_{12}$ and $E_{23}$ , it is a highest weight vector.

It is worth thinking about how you derive the formula for the matrix of $\rho^{*}(A)$ with respect to the dual basis. It is defined so that, for $v^{*}\in V^{*}$ and $w\in V$ , and $A=(a_{ij})\in\mathfrak{g}$ ,

(\rho^{*}(A)v^{*})(w)=-v^{*}(\rho(A)w).

We apply this with $v^{*}=e_{i}^{*}$ , recalling that $e_{i}^{*}e_{k}=\delta_{ik}$ . Then

(\rho^{*}(A)e_{i}^{*})(e_{j})=-e_{i}^{*}(\rho(A)e_{j})=-e_{i}^{*}(\sum_{k}a_{% kj}e_{k})=-a_{ij}.

This implies that

\rho^{*}(A)e_{i}^{*}=-\sum_{j}a_{ij}e^{*}_{j}

which exactly says that the matrix of $\rho^{*}(A)$ with respect to the dual basis is minus the transpose of the matrix of $\rho(A)$ with respect to the original basis.

Solution to (38).

By definition $\Lambda_{W}$ consists of elements of $\mathfrak{h}^{*}$ that may be written as an integer linear combination of the $L_{i}$ . Any such linear combination clearly satisfies the given condition. Conversely, if $a_{1}L_{1}+a_{2}L_{2}+a_{3}L_{3}$ satisfies $a_{1}-a_{2}\in\mathbb{Z}$ and $a_{2}-a_{3}\in\mathbb{Z}$ , then

a_{1}L_{1}+a_{2}L_{2}+a_{3}L_{3}=(a_{1}-a_{2})L_{1}-(a_{2}-a_{3})L_{3}\in% \Lambda_{W}

since $L_{1}+L_{2}+L_{3}=0$ .

The $a_{i}$ need not be integers; indeed, $zL_{1}+zL_{2}+zL_{3}=0\in\Lambda_{W}$ for any $z\in\mathbb{C}$ .

Solution to (39).

1.

See Figure 12.

Figure 12: Root lattice (blue) inside weight lattice (black).
2.

A complete set of coset representatives for $\Lambda_{W}/\Lambda_{R}$ is $0$ , $L_{1}$ , $2L_{1}$ (please check this!), so the index is three.

Alternatively, there is a homomorphism $\Lambda_{W}\rightarrow\mathbb{Z}/3\mathbb{Z}$ sending $aL_{1}+bL_{2}+cL_{3}$ to $a+b+c\mod 3$ , and one can show that the kernel of this homomorphism is $\Lambda_{R}$ .
3.

Here $\mathfrak{h}=\mathbb{C}\left\langle H\right\rangle$ . We have an isomorphism $\mathfrak{h}^{*}\cong\mathbb{C}$ taking $\alpha$ to $\alpha(H)$ . The weight lattice is then $\mathbb{Z}\subset\mathbb{C}$ . The root lattice is $2\mathbb{Z}\subset\mathbb{Z}$ since the eigenvalues of $H$ on the adjoint representation are $0,\pm 2$ . The index is therefore two.
4.

If $v\in V$ is a weight vector of weight $\alpha$ then the subrepresentation generated by $v$ must be all of $V$ . Thus $V$ is spanned by

$\{X_{1}\ldots X_{n}v:X_{i}\in\{E_{ij}\}\cup\mathfrak{h}\}$

and we see from the fundamental weight calculation that $V$ is spanned by weight vectors all of whose weights differ from $\alpha$ by an integer linear combination of roots, as required.

Solution to (40).

The weights of $\operatorname{Sym}^{3}(\mathbb{C}^{3})$ are $a_{1}L_{1}+a_{2}L_{2}+a_{3}L_{3}$ with $a_{1},a_{2},a_{3}$ non-negative integers summing to 3. By Weyl symmetry it is enough to find the dominant weights. Since

a_{1}L_{1}+a_{2}L_{2}+a_{3}L_{3}=(a_{1}-a_{2})L_{1}-(a_{2}-a_{3})L_{3}

these are those with $a_{1}\geq a_{2}\geq a_{3}$ . The only possibilities for $(a_{1},a_{2},a_{3})$ are then $(3,0,0),(2,1,0)$ and $(1,1,1)$ corresponding to

3L_{1},2L_{1}+L_{2}(=L_{1}-L_{3}),L_{1}+L_{2}+L_{3}(=0).

Applying Weyl symmetry we see that the weights are

\{3L_{1},3L_{2},3L_{3},L_{i}-J_{j}\text{ for $i\neq j$},0\}.

It is left to you to draw these; for a similar picture of $\operatorname{Sym}^{5}(\mathbb{C}^{3})$ see 11.

Solution to (41).

The representation $\mathbb{C}^{3}\otimes(\mathbb{C}^{3})^{*}$ has a highest weight vector $e_{1}\otimes e_{3}^{*}$ of weight $L_{1}-L_{3}$ . It therefore contains a subrepresentation isomorphic to $V^{(1,1)}=\mathfrak{sl}_{2,\mathbb{C}}$ (which will be the subrepresentation generated by $e_{1}\otimes e_{3}^{*}$ ).

The weights of $\mathbb{C}^{3}\otimes(\mathbb{C}^{3})^{*}$ are $\{L_{i}-L_{j}:i\neq j\}\cup\{0,0,0\}$ (as we can write $0=L_{i}-L_{i}$ in three ways). Therefore, looking at weights, we have

\mathbb{C}^{3}\otimes(\mathbb{C}^{3})^{*}\cong\mathfrak{sl}_{2,\mathbb{C}}\oplus W

where $W$ is a one-dimensional representation with a single weight, 0. Therefore $W$ is trivial and

\mathbb{C}^{3}\otimes(\mathbb{C}^{3})^{*}\cong\mathfrak{sl}_{2,\mathbb{C}}% \oplus\mathbb{C}.

To find the trivial representation $\mathbb{C}$ inside $\mathbb{C}^{3}\otimes(\mathbb{C}^{3})^{*}$ , we look for a HWV of weight 0. The following works:

e_{1}\otimes e_{1}^{*}+e_{2}\otimes e_{2}^{*}+e_{3}\otimes e_{3}^{*}.

For a conceptual proof, if $V$ and $W$ are any representations of a Lie algebra $\mathfrak{g}$ then we can define a representation on $\operatorname{Hom}(V,W)$ by

(XT)(v)=X(Tv)-T(Xv)

for $X\in\mathfrak{g},v\in V,T\in\operatorname{Hom}(V,W)$ . Then we have a map

V^{*}\otimes W\to\operatorname{Hom}(V,W)

sending

\lambda\otimes w\mapsto T_{\lambda,w}

where $T_{\lambda,w}\in\operatorname{Hom}(V,W)$ is defined by

T_{\lambda,w}(v)=\lambda(v)w.

One can check that this is a $\mathfrak{g}$ -isomorphism (if $V$ and $W$ are finite-dimensional). In the case at hand we get

(\mathbb{C}^{3})^{*}\otimes\mathbb{C}^{3}\cong\operatorname{Hom}(\mathbb{C}^{3% },\mathbb{C}^{3})=\mathfrak{gl}_{3,C}

where the right hand side is a representation of $\mathfrak{sl}_{3,\mathbb{C}}$ by the same formula defining the adjoint representation. Then

\mathfrak{gl}_{3,\mathbb{C}}=\mathbb{C}\oplus\mathfrak{sl}_{3,\mathbb{C}}

as representations of $\mathfrak{sl}_{3,\mathbb{C}}$ , with $\mathbb{C}$ corresponding to the subspace of scalar matrices and $\mathfrak{sl}_{3,\mathbb{C}}\subset\mathfrak{gl}_{3,\mathbb{C}}$ in the obvious way.

An intuitive way to see the isomorphism $W\otimes V^{*}\mapsto\operatorname{Hom}(V,W)$ is that $V^{*}$ is ‘row vectors of length $m$ ’, $W$ is ‘column vectors of length $n$ ’, and $\operatorname{Hom}(V,W)$ is ’ $m\times n$ matrices’, and the isomorphism takes a column vector $(a_{1},\ldots,a_{m})^{T}$ tensored with a row vector $(b_{1},\ldots,b_{n})$ to the matrix $(a_{i}b_{j})_{i,j}$ .

Solution to (42).

First I claim that if $\tilde{\rho}$ is a representation of a Lie group $G$ on $V$ with derivative $\rho$ , then

\tilde{\rho}(g)\rho(H)v=\rho(\operatorname{Ad}_{g}(H))\tilde{\rho}(g)v.

Indeed, we may assume that $G\in GL(V)$ when this just becomes $gH=(gHg^{-1})g$ . Alternatively, the left hand side is the derivative at $t=0$ of

\tilde{\rho}(g)\tilde{\rho}(\exp(tH))=\tilde{\rho}(g\exp(tH))=\tilde{\rho}(% \exp(t\operatorname{Ad}_{g}(H))g)=\tilde{\rho}(\exp(t\operatorname{Ad}_{g}(H))% )\tilde{\rho}(g)

which gives the result.

Let $v\in V_{\alpha}$ . By the above, we have that, for $H\in\mathfrak{h}$ ,

\rho(H)(\tilde{\rho}(\sigma_{3})v)=\tilde{\rho}(\sigma_{3})\rho(\operatorname{% Ad}_{\sigma_{3}}(H))v=\alpha(\operatorname{Ad}_{\sigma_{3}}H)\tilde{\rho}(% \sigma_{3})v.

If $\alpha=a_{1}L_{1}+a_{2}L_{2}+a_{3}L_{3}$ and $H$ is diagonal with entries $x, y, z$ , then $\alpha(H)=a_{1}x+a_{2}y+a_{3}z$ . Since $\operatorname{Ad}_{\sigma_{3}}(H)$ is diagonal with entries $y, x, z$ ,

\alpha(\operatorname{Ad}_{\sigma_{3}}H)=a_{1}y+a_{2}x+a_{3}z=(s_{3}\alpha)(H)

where $s_{3}\alpha=a_{2}L_{1}+a_{1}L_{2}+a_{3}L_{3}$ .

This shows that $\tilde{\rho}(\sigma_{3})$ maps $V_{\alpha}$ to $V_{s_{3}\alpha}$ ; since the inverse provides a map in the other direction, this is an isomorphism.

This shows that the weights of a representation, with their multiplicities, are preserved under swapping $L_{1}$ and $L_{2}$ . Similarly for any transposition, and hence for any permutation of the $L_{i}$ .

Solution to (43).

It is easy to see that its weight is $aL_{1}-bL_{3}$ . Moreover, as $e_{1}$ and $e_{3}^{*}$ are highest weight vectors, so are $e_{1}^{a}$ and $(e_{3}^{*})^{b}$ , and so is their tensor product.

(General lemmas: if $v\in V$ is a highest weight vector, then $Ev^{n}=n(Ev)v^{n-1}=0$ for each positive root vector $E$ , so $v^{n}$ is a highest weight vector in $\operatorname{Sym}^{n}V$ (you should also check it is a weight vector!). If $v,w\in V,W$ are highest weight vectors, then

E(v\otimes w)=(Ev)\otimes w+v\otimes(Ew)=0

for each positive root vector $E$ so $v\otimes w$ is a highest weight vector (you should also check it is a weight vector!).)

Solution to (44).

Suppose that $V$ is reducible, so that there is a nonzero proper subrepresentation $W\subsetneq V$ . Then by complete reducibility, there is another subrepresentation $W^{\prime}$ with $V=W\oplus W^{\prime}$ . Then $W$ and $W^{\prime}$ both have nonzero highest weight vectors, which must be linearly independent from each other, contradicting the assumption on $V$ .

For the standard representation, the weight spaces are spanned by $e_{1}$ , $e_{2}$ , and $e_{3}$ respectively and only $e_{1}$ (or a scalar multiple of it) is a highest weight vector. So it is irreducible. Similarly for the dual representation.

For the adjoint representation, out of the nonzero weights only $\alpha_{1}-\alpha_{3}$ is a highest weight, with unique highest weight vector. We have to check there are no highest weight vectors of weight zero. Such a vector would be a nonzero element $H\in\mathfrak{h}$ such that $[E_{ij},H]=0$ for all $i<j$ . This would imply that $H$ is scalar, but since $H$ has trace zero this is impossible.

Remark: this is not the simplest way to see that the standard representation is irreducible; indeed, the action of $\mathfrak{sl}_{3,\mathbb{C}}$ on $\mathbb{C}^{3}$ is transitive, which implies irreducibility. Similarly for the dual. Can you prove that the adjoint representation is irreducible without using weights?

Solution to (45).

1.

The weights of $\operatorname{Sym}^{2}(\mathbb{C}^{3})$ are $2L_{1},2L_{2},2L_{3},L_{1}+L_{2},L_{1}+L_{3},L_{2}+L_{3}$ while the weights of $(\mathbb{C}^{3})^{*}$ are $-L_{1}$ , $-L_{2}$ , $-L_{3}$ . Adding everything from the first list to everything from the second, we see that $\operatorname{Sym}^{2}(\mathbb{C}^{3})\otimes(\mathbb{C}^{3})^{*}$ has weights

$L_{1},L_{1},L_{1},L_{2},L_{2},L_{2},L_{3},L_{3},L_{3},\\ 2L_{1}-L_{2},2L_{1}-L_{3},2L_{2}-L_{1},2L_{2}-L_{3},2L_{3}-L_{1},2L_{3}-L_{2},% \\ L_{1}+L_{2}-L_{3},L_{2}+L_{3}-L_{1},L_{3}+L_{1}-L_{2}.$

The weight diagram is shown in figure 13.

Figure 13: Weights for $\operatorname{Sym}^{2}\mathbb{C}^{3}\otimes(\mathbb{C}^{3})^{*}$ .
2.

Since $e_{1}^{2}$ is a weight vector of weight $2L_{1}$ and $e_{1}^{*}$ is a weight vector of weight $-L_{1}$ , their tensor product is a weight vector of weight $2L_{1}-L_{1}=L_{1}$ . Similarly, the other terms are also weight vectors of weight $L_{1}$ . Therefore their sum is also a weight vector of weight $L_{1}$ .

We can hit it with $E_{12}$ and $E_{23}$ , using $E_{ij}e_{k}^{*}=-\delta_{ik}e_{j}^{*}$ :

$\displaystyle E_{12}(e_{1}^{2}\otimes e_{1}^{*}+e_{1}e_{2}\otimes e_{2}^{*}+e_% {1}e_{3}\otimes e_{3}^{*})=e_{1}^{2}\otimes(-e_{2}^{*})+e_{1}^{2}\otimes e_{2}% ^{*}$

$\displaystyle=0,$

and

$\displaystyle E_{23}(e_{1}^{2}\otimes e_{1}^{*}+e_{1}e_{2}\otimes e_{2}^{*}+e_% {1}e_{3}\otimes e_{3}^{*})=e_{1}e_{2}\otimes(-e_{3}^{*})+e_{1}e_{2}\otimes e_{% 3}^{*}$

$\displaystyle=0$

so it is a highest weight vector.
3.

We find $E_{21}v=2e_{1}e_{2}\otimes e_{3}^{*}$ whence

$E_{32}E_{21}v=2e_{1}e_{3}\otimes e_{3}^{*}-2e_{1}e_{2}\otimes e_{2}^{*},$

while $E_{32}v=-e_{1}^{2}\otimes e_{2}^{*}$ so

$E_{21}E_{32}v=e_{1}^{2}\otimes e_{1}^{*}-2e_{1}e_{2}\otimes e_{2}^{*}.$

These are linearly independent, as $e_{1}^{2}\otimes e_{1}^{*}$ only appears in the second while $e_{1}e_{3}\otimes e_{3}^{*}$ only appears in the first.
4.

Let $V=\operatorname{Sym}^{2}\mathbb{C}^{3}\otimes(\mathbb{C}^{3})^{*}$ . Note that $V$ is completely reducible, and the possible irreducible constituents are $V^{(2,1)}$ , $V^{(0,2)}$ , and $V^{(1,0)}$ , since these are the only weights of $V$ that are dominant. Since $V$ has a highest weight vector of weight $2L_{1}-L_{3}$ , and this is not a weight of $V^{(0,2)}$ or $V^{(1,0)}$ , $V$ must have a subrepresentation $V_{1}\cong V^{(2,1)}$ . As $e_{1}^{2}\otimes e_{3}^{*}$ is the unique highest weight vector of weight $2L_{1}-L_{3}$ , it must occur in $V$ . Similarly, since (by part 2) $V$ has a highest weight vector of weight $L_{1}$ , $V$ must have a subrepresentation $V_{2}\cong V^{(1,0)}\cong\mathbb{C}^{3}$ . In fact, one can check that $V_{2}$ has basis

$e_{1}^{2}\otimes e_{1}^{*}+e_{1}e_{2}\otimes e_{2}^{*}+e_{1}e_{3}\otimes e_{3}% ^{*}$

together with the two similar vectors obtained by permuting the roles of $e_{1},e_{2},e_{3}$ .

So we have $V_{1}\oplus V_{2}\subset V$ and we want to show equality. Note that $v=e_{1}^{2}\otimes e_{3}^{*}\in V_{1}$ is a highest weight vector in $V_{1}$ of weight $2L_{1}-L_{3}$ , and $E_{21}v$ is a nonzero weight vector in $V_{1}$ of weight $-2L_{3}$ . Moreover, part (3) implies that $L_{1}$ occurs with multiplicity at least two in $V_{1}$ . We also have that $L_{1}$ occurs in $V_{2}$ with multiplicity one. All the dominant weights of $V$ are now accounted for by $V_{1}\oplus V_{2}$ with the correct multiplicities. Thus, by Weyl symmetry, the weights of $V$ agree with the weights of $V_{1}\oplus V_{2}$ and we have equality. We see that the multiplicities of $L_{1}$ , $L_{2}$ , $L_{3}$ are exactly two in $V_{1}$ and all other weights in $V_{1}$ occur with multiplicity one, and we have

$V=V_{1}\oplus V_{2}\cong V^{(2,1)}\oplus\mathbb{C}^{3}.$

The weight diagram of $V_{1}\cong V^{(2,1)}$ is obtained from that of $V$ (see part (1)) by removing one circle around each of $L_{1}$ , $L_{2}$ , and $L_{3}$ .

Solution to (46).

1.

We have that

$\displaystyle H(e_{1}^{a}e_{2}^{b}e_{3}^{c})$ $\displaystyle=aL_{1}(H)e_{1}e_{1}^{a-1}e_{2}^{b}e_{3}^{c}+e_{1}^{a}bL_{2}(H)e_% {2}e_{2}^{b-1}e_{3}^{c}+e_{1}^{a}e_{2}^{b}cL_{3}(H)e_{3}e_{3}^{c-1}$

$\displaystyle=(aL_{1}+bL_{2}+cL_{3})(H)e_{1}^{a}e_{2}^{b}e_{3}^{c}.$

So $e_{1}^{a}e_{2}^{b}e_{3}^{c}$ is a weight vector, and we know that these are a basis for $\operatorname{Sym}^{n}(\mathbb{C}^{3})$ .

To see that they are distinct, suppose that $a+b+c=n$ and $a^{\prime}+b^{\prime}+c^{\prime}=n$ for nonnegative integers $a,a^{\prime}$ , etc., and that

$aL_{1}+bL_{2}+cL_{3}=a^{\prime}L_{1}+b^{\prime}L_{2}+c^{\prime}L_{3}.$

Then

$(a-a^{\prime})L_{1}+(b-b^{\prime})L_{2}+(c-c^{\prime})L_{3}=0$

which implies that $a-a^{\prime}=b-b^{\prime}=c-c^{\prime}$ . But as $a+b+c=a^{\prime}+b^{\prime}+c^{\prime}$ , this common difference must be zero, so $a=a^{\prime}$ , $b=b^{\prime}$ and $c=c^{\prime}$ as required.
2.

Suppose that $v$ is a highest weight vector. Since all the weights from the first part are distinct, the weight spaces are one-dimensional, so (after scaling) $v=e_{1}^{a}e_{2}^{b}e_{3}^{c}$ for some $a, b, c$ . We have

$E_{12}v=be_{1}^{a+1}e_{2}^{b-1}e_{3}^{c}=0$

as $v$ is a highest weight vector, so $b=0$ . Similarly, $c=0$ . Thus $v=e_{1}^{n}$ (up to scalar) is the unique highest weight vector.
3.

By part 2 and the previous question, $\operatorname{Sym}^{n}(\mathbb{C}^{3})$ has a unique highest weight vector (up to scalar), and so is irreducible. Its highest weight is the weight of $e_{1}^{n}$ , which is $nL_{1}$ .

	$\displaystyle(H\phi)\left((x,y)^{T}\right)$	$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\left.\phi\left(\exp(-tH)(x,y)^{T}% \right)\right\|_{t=0}$
		$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\left.\phi\left((e^{-t}x,e^{t}y)^{% T}\right)\right\|_{t=0}$
		$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}(e^{t}y)\|_{t=0}\frac{\partial}{% \partial y}(\phi)-\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}x)\|_{t=0}\frac{\partial% }{\partial x}(\phi)$
		$\displaystyle=y\frac{\partial}{\partial y}(\phi)-x\frac{\partial}{\partial x}(\phi)$

	$\displaystyle\exp\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&t\\ \hidden@noalign{}\hfil\textstyle-t&0\end{pmatrix}$	$\displaystyle=(1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\ldots)I+(t-\frac{t^{3}}{3!% }+\frac{t^{5}}{5!}-\ldots)\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&1% \\ \hidden@noalign{}\hfil\textstyle-1&0\end{pmatrix}$
		$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle\cos(t)&\sin(t)\\ \hidden@noalign{}\hfil\textstyle-\sin(t)&\cos(t)\end{pmatrix}.$

	$\displaystyle\mathcal{C}e_{1}^{n}$	$\displaystyle=\pi(X)ne_{1}^{n-1}e_{-1}+\pi(Y)0+\frac{1}{2}n^{2}e_{1}^{n}$
		$\displaystyle=(n+\frac{1}{2}n^{2})e_{1}^{n}$

	$\displaystyle X=\mathcal{I}-i\mathcal{J}$	$\displaystyle\mapsto J_{x}-iJ_{y}$
	$\displaystyle Y=-(\mathcal{I}+i\mathcal{J})$	$\displaystyle\mapsto-(J_{x}+iJ_{y})$
	$\displaystyle H=-2i\mathcal{K}$	$\displaystyle\mapsto-2iJ_{z}.$

	$\displaystyle J_{z}(x-iy)^{\ell}$	$\displaystyle=y\ell(x-iy)^{\ell-1}+i\ell x(x-iy)^{\ell-1}$
		$\displaystyle=\ell(y+ix)(x-iy)^{\ell-1}$
		$\displaystyle=i\ell(x-iy)^{\ell}$

	$\displaystyle\Delta(r^{2k}f)$	$\displaystyle=r^{2k}\Delta(f)+\sum_{i=1}^{k}2(2\ell+4i-1)r^{2k-2}f$
		$\displaystyle=r^{2k}\Delta(f)+2k(2\ell+2k+1)r^{2k-2}f.$

	$\displaystyle E_{12}(e_{1}^{2}\otimes e_{1}^{}+e_{1}e_{2}\otimes e_{2}^{}+e_% {1}e_{3}\otimes e_{3}^{})=e_{1}^{2}\otimes(-e_{2}^{})+e_{1}^{2}\otimes e_{2}% ^{*}$
		$\displaystyle=0,$

	$\displaystyle E_{23}(e_{1}^{2}\otimes e_{1}^{}+e_{1}e_{2}\otimes e_{2}^{}+e_% {1}e_{3}\otimes e_{3}^{})=e_{1}e_{2}\otimes(-e_{3}^{})+e_{1}e_{2}\otimes e_{% 3}^{*}$
		$\displaystyle=0$

	$\displaystyle H(e_{1}^{a}e_{2}^{b}e_{3}^{c})$	$\displaystyle=aL_{1}(H)e_{1}e_{1}^{a-1}e_{2}^{b}e_{3}^{c}+e_{1}^{a}bL_{2}(H)e_% {2}e_{2}^{b-1}e_{3}^{c}+e_{1}^{a}e_{2}^{b}cL_{3}(H)e_{3}e_{3}^{c-1}$
		$\displaystyle=(aL_{1}+bL_{2}+cL_{3})(H)e_{1}^{a}e_{2}^{b}e_{3}^{c}.$