# X and Y example

## Review

So far we have seen that, given a complex representation $R\colon SU(2)\to GL(V)$ , we get:

• a (real linear) representation $R_{*}\colon\mathfrak{su}(2)\to\mathfrak{gl}(V)$ of the real Lie algebra $\mathfrak{su}(2)$

• a (complex linear) representation $R_{*}^{\mathbf{C}}\colon\mathfrak{sl}(2,\mathbf{C})\to\mathfrak{gl}(V)$ of its complexification $\mathfrak{sl}(2,\mathbf{C})=\mathfrak{su}(2)\otimes\mathbf{C}$ .

Define the basis $H=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}$ , $X=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$ and $Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ of $\mathfrak{su}(2)$ . We saw that $V=\bigoplus W_{m}$ where $W_{m}=\{v\in V\ :\ R_{*}^{\mathbf{C}}(H)v=mv\}$ and saw that $X$ and $Y$ act on weight spaces as follows:

• $X$ takes vectors in $W_{m}$ to vectors in $W_{m+2}$ ,

• $Y$ takes vectors in $W_{m}$ to vectors in $W_{m-2}$ .

• We will usually omit the superscript $\mathbf{C}$ from $R_{*}^{\mathbf{C}}$ and actually, we'll usually omit the $R_{*}$ altogether.

In this video, I want to see how this actually works in practice for the example $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ . But first, we will need to understand how Lie algebras act on symmetric powers and, more generally, tensor products.

## Leibniz rule

Recall that $R^{\otimes 2}(g)(v_{1}\otimes v_{2})=(R(g)v_{1})\otimes(R(g)v_{2})$ . What is the Lie algebra representation corresponding to this? In other words, what is $(R^{\otimes 2})_{*}$ ?

Lemma:

If $R\colon G\to GL(V)$ is a representation and $X\in\mathfrak{g}$ (this is not necessarily the $X$ in $\mathfrak{sl}(2,\mathbf{C})$ ) we have $(R^{\otimes n})_{*}(X)(v_{1}\otimes\cdots\otimes v_{n})=(R_{*}(X)v_{1})\otimes v% _{2}\otimes\cdots\otimes v_{n}+v_{1}\otimes(R_{*}(X)v_{2})\otimes\cdots\otimes v% _{n}+\cdots+v_{1}\otimes\cdots\otimes R_{*}(X)v_{n}.$ In other words, we use the "Leibniz/product rule".

For now, we will assume this lemma, work out the example and then justify the lemma.

## Example: Sym2(C2)

Example:

Let $V=\mathrm{Sym}^{2}(\mathbf{C}^{2})$ . If $e_{1}$ and $e_{2}$ form a basis for $\mathbf{C}^{2}$ then $e_{1}^{2}=e_{1}\otimes e_{1}$ , $e_{1}e_{2}=e_{1}\otimes e_{2}+e_{2}\otimes e_{1}$ and $e_{2}^{2}=e_{2}\otimes e_{2}$ form a basis for $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ . The Lie algebra elements $H$ , $X$ and $Y$ act on $e_{1}$ and $e_{2}$ as follows:

• Since $H=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}$ , we have $He_{1}=e_{1}$ , $He_{2}=-e_{2}$ ,

• Since $X=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$ , we have $Xe_{1}=0$ , $Xe_{2}=e_{1}$ ,

• Since $Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ , we have $Ye_{1}=e_{2}$ , $Ye_{2}=0$ .

$\mathrm{Sym}^{2}(H)$ therefore sends $e_{1}^{2}=e_{1}\otimes e_{1}$ to $(He_{1})\otimes e_{1}+e_{1}\otimes(He_{1})=2e_{1}^{2}$ by the product rule. Similarly, $\mathrm{Sym}^{2}(H)(e_{1}e_{2})=(He_{1})e_{2}+e_{1}(He_{2})=e_{1}e_{2}-e_{1}e_% {2}=0$ and $\mathrm{Sym}^{2}(H)(e_{2}^{2})=-2e_{2}^{2}.$ The weight spaces, that is the eigenspaces of $H$ , are spanned by $e_{1}^{2}$ (weight 2), $e_{1}e_{2}$ (weight 0) and $e_{2}^{2}$ (weight $-2$ ).

Let's compute $\mathrm{Sym}^{2}(X)$ :

• $\mathrm{Sym}^{2}(X)(e_{1}^{2})=(Xe_{1})e_{1}+e_{1}(Xe_{1})=0.$

• $\mathrm{Sym}^{2}(X)(e_{1}e_{2})=(Xe_{1})e_{2}+e_{1}(Xe_{2})=e_{1}^{2}.$

• $\mathrm{Sym}^{2}(X)(e_{2}^{2})=(Xe_{2})e_{2}+e_{2}(Xe_{2})=2e_{1}e_{2}.$

• Notice that in this last example we used the fact that $e_{1}e_{2}=e_{2}e_{1}$ (which holds because it's a symmetric tensor).

In terms of our weight diagram, we can see that $\mathrm{Sym}^{2}(X)$ is taking $e_{2}^{2}\in W_{-2}$ to $2e_{1}e_{2}\in W_{0}$ (increasing the weight by 2) and $e_{1}e_{2}\in W_{0}$ to $e_{1}^{2}\in W_{2}$ (again, increasing the weight by 2). Finally, it takes $e_{1}^{2}\in W_{2}$ to $0\in W_{4}$ (necessarily, since $W_{4}=0$ ).

It's an exercise to calculate $\mathrm{Sym}^{2}(Y)$ .

## Proof of lemma

We still need to prove that $(R^{\otimes n})_{*}$ can be evaluated using the Leibniz rule, as stated in the lemma.

We have $(R^{\otimes n})(\exp tX)=\exp(t(R^{\otimes n})_{*}X)$ by our usual formula for Lie algebra representations.

Let's apply both sides to $v_{1}\otimes\cdots\otimes v_{n}$ and expand both sides in powers of $t$ . The right-hand side is $(I+t(R^{\otimes n})_{*}X+\mathcal{O}(t^{2}))(v_{1}\otimes\cdots\otimes v_{n})=% v_{1}\otimes\cdots\otimes v_{n}+t(R^{\otimes n})_{*}X(v_{1}\otimes\cdots% \otimes v_{n})+\mathcal{O}(t^{2}).$ The left-hand side is $R(\exp(tX))v_{1}\otimes\cdots\otimes R(\exp(tX))v_{n}$

We have $R(\exp(tX))v_{i}=(I+tR_{*}X+\mathcal{O}(t^{2}))v_{i}=v_{i}+tR_{*}Xv_{i}+% \mathcal{O}(t^{2})$ , so multiplying out the brackets, the left-hand side becomes: $v_{1}\otimes\cdots\otimes v_{n}+t(R_{*}Xv_{1}\otimes v_{2}\otimes\cdots\otimes v% _{n}+v_{1}\otimes R_{*}Xv_{2}\otimes\cdots\otimes v_{n}+v_{1}\otimes\cdots% \otimes R_{*}Xv_{n})+\mathcal{O}(t^{2}).$

By comparing the terms of order $t$ , we get the desired formula for $(R^{\otimes n})_{*}(X)$ .

## Pre-class exercise

Exercise:

Compute the action of $\mathrm{Sym}^{2}(Y)$ on the basis $e_{1}^{2}$ , $e_{1}e_{2}$ , $e_{2}^{2}$ .