Embedding of $\mathfrak{sl}(2,\mathbb{C})$ in matrix algebra

Question

Consider the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$ defined in terms of generators and relations by $$\langle x,y,h: [h,x]=2x, [h,y]=-2y, [x,y]=h\rangle.$$ If $X,Y,H$ denote the standard $2\times 2$ matrices giving a representation of $\mathfrak{sl}(2,\mathbb{C})$ then this fact can be interpreted as follows:

The $2\times 2$ matrices $X,Y,H$ generate an associative (sub)algebra inside $M_2(\mathbb{C})$, whose associated Lie algebra is $\mathfrak{sl}(2,\mathbb{C})$.

However this subalgebra of matrices satisfies some extra conditions - namely $X^2=0$, $Y^2=0$. Thus, this subalgebra is not universal enveloping algebra.

Q. Can we embed $\mathfrak{sl}(2,\mathbb{C})$ in some matrix algebra $M_n(\mathbb{C})$ so that the images $X$ and $Y$ of $x$ and $y$ satisfy $X^{n_1}=0$ and $X^{n_2}=0$ for $n_1,n_2$ smallest such integers that are bigger than $2$?

In other words, can we embed $\mathfrak{sl}(2,\mathbb{C})$ in bigger matrix algebra so that the nilpotncy of images of $x,y$ is bigger than $2$? If yes, can this nilpotency be arbitrarily large?

Answer 1

What about $\mathfrak{sl}(2,\mathbb{C})\hookrightarrow M_3(\mathbb{C})$ given by the adjoint representation of $\mathfrak{sl}(2,\mathbb{C})$? In the basis $\{x,y,h\}$, $\operatorname{ad}(x)$, $\operatorname{ad}(y)$, and $\operatorname{ad}(h)$ are given by the following matrices $$X=\begin{pmatrix}0&0&-2\\0&0&0\\0&1&0\end{pmatrix},Y=\begin{pmatrix}0&0&0\\0&0&2\\-1&0&0\end{pmatrix},H=\begin{pmatrix}2&0&0\\0&-2&0\\0&0&0\end{pmatrix}.$$ We have $X^2=\operatorname{ad}(x)^2\neq 0$ and $Y^2=\operatorname{ad}(y)^2\neq 0$, but $X^3=\operatorname{ad}(x)^3=0$ and $Y^3=\operatorname{ad}(y)^3=0$.

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In fact, you can embed $\mathfrak{sl}(2,\mathbb{C})$ into any $M_n(\mathbb{C})$ by equipping $\mathbb{C}^n$ with the simple $n$-dimensional module structure with respect to $\mathfrak{sl}(2,\mathbb{C})$. Then, the images $X$ and $Y$ of, respectively, $x$ and $y$ satisfy the condition that $X^k$ and $Y^k$ are nonzero for $k=1,2,\ldots,n-1$, but $X^n=Y^n=0$.

Answer 2

The Morozov-Jacobson theorem states, that for every nilpotent element $X$ of a semisimple lie algebra $\mathfrak g$ there exist $Y$ and $H$ that satisfy the relations $[H,X] = 2X$, $[H,Y] = -2Y$ and $[X,Y] = H$.

More concretely, you could take the matrices \begin{align*} X &= \begin{pmatrix}0 & 1\\ & 0 & 2\\ & & \ddots & \ddots\\ & & & \ddots & n-1\\ & & & & 0\end{pmatrix},\\ Y &= \begin{pmatrix}0\\ n-1 & 0\\ & n-2 & \ddots\\ & & \ddots & \ddots\\ & & & 1 & 0 \end{pmatrix},\\ H &= \begin{pmatrix}n-1\\ & n-3\\ & & n-5\\ & & &\ddots\\ & & & & -(n-3)\\ & & & & & -(n-1)\end{pmatrix} \end{align*} and check that they satisfy your relations.

Answer 3

In the finite-dimensionl setting, $x$ and $y$ are necessarily nilpotent with the same nilpotency index. In the infinite-dimensional setting, $x$ and $y$ may not be nilpotent at all. They can also behave differently in terms of nilpotent actions. While this answer does not directly answer the question, I think it is worthwhile to have it here. (I believe, however, that this holds: for any $\mathfrak{sl}_2(\mathbb{C})$-module $M$, $x$ acts nilpotently on $M$ if and only if $y$ also acts nilpotently on $M$. When this happens, $M$ is a direct sum of finite-dimensional simple $\mathfrak{sl}_2(\mathbb{C})$-modules. However, I would really love to see if there is a proof or a counterexample of this claim. I think this is equivalent to saying that the quotient algebra $\mathfrak{U}\big(\mathfrak{sl}_2(\mathbb{C})\big)/J_k$, where $J_k$ is the two-sided ideal generated by $x^k$ of the universal enveloping algebra $\mathfrak{U}\big(\mathfrak{sl}_2(\mathbb{C})\big)$, is a finite-dimensional associative $\mathbb{C}$-algebra.)

Consider the space $V:=\mathcal{C}^\infty(\mathbb{R}^2,\mathbb{C})$ of all complex-valued smooth functions on $\mathbb{R}^2$ and let $x,y,h$ act on $V$ as follows: $$(x\cdot f)(s,t):=s\,\frac{\partial}{\partial t}\,f(s,t)$$ $$(y\cdot f)(s,t):=t\,\frac{\partial}{\partial s}\,f(s,t)$$ and $$(h\cdot f)(s,t):=s\,\frac{\partial}{\partial s}\,f(s,t)-t\,\frac{\partial}{\partial t}\,f(s,t)\,,$$ for all $f\in V$ and $s,t\in\mathbb{R}$. Then, in the embedding $\mathfrak{sl}_2(\mathbb{C})$ into $\text{End}_\mathbb{C}(V)$ given by the representation above, neither $x$ nor $y$ is nilpotent.

However, if you consider the subspace $U$ of $V$ consisting of polynomial functions. Then, $x$ and $y$ are locally nilpotent, but not nilpotent. (In other words, for each $f\in U$, there exist integers $m_f>0$ and $n_f>0$, depending on $f$, such that $x^{m_f}\cdot f=0$ and $y^{n_f}\cdot f=0$.) However, as you can see, $$U\cong\bigoplus_{k\in\mathbb{Z}_{\geq 0}}\,\mathfrak{L}(k)\,,$$ where $\mathfrak{L}(k)$ is the simple $(k+1)$-dimensionsional representation of $\mathfrak{sl}_2(\mathbb{C})$. Ergo, in the embedding $\mathfrak{sl}_2(\mathbb{C})$ into $\text{End}_\mathbb{C}(U)$, both $x$ and $y$ are not nilpotent, but locally nilpotent.

Furthermore, there also exists an embedding $\mathfrak{sl}_2(\mathbb{C})$ into $\text{End}_\mathbb{C}(W)$ in which neither $x$ or $y$ is nilpotent, but only $x$ is locally nilpotent. Let $\mathfrak{L}(\lambda)$ be the simple highest-weight representation of $\mathfrak{sl}_2(\mathbb{C})$ with respect to the Borel subalgebra $\text{span}_{\mathbb{C}}\{x,h\}$ and the Cartan subalgebrta $\mathbb{C}h$, with $x$-highest $h$-weight $\lambda\in\mathbb{C}$. For $\lambda\notin\mathbb{Z}_{\geq 0}$, $\mathfrak{L}(\lambda)$ is not finite-dimensional. However, $x$ acts locally nilpotently on $\mathfrak{L}(\lambda)$, but $y$ is not a nilpotent nor locally nilpotent operator. Taking $W:=\mathfrak{L}(\lambda)$, you get an embedding of $\mathfrak{sl}_2(\mathbb{C})$ into an associative algebra in which $x$ is locally nilpotent but $y$ is not.