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 (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 (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 (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 (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 (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 (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 (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 (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 12.

Figure 12: 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 $nL_{1}$ .

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

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

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

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

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

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