Associative $\mathbb{C}$-algebra with 3 generators, $2 \times 2$ matrices of action of generators in basis $x, y$.

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Let $\text{U}$ be an associative $\mathbb{C}$-algebra with three generators $E$, $H$, $F$, and three defining relations$$HE - EH = 2E,\text{ }HF - FH = -2F, \text{ }EF - FE = H.$$The formulas$$E(f) := x{{\partial f}\over{\partial y}},\text{ }H(f) := x{{\partial f}\over{\partial x}} - y{{\partial f}\over{\partial y}},\text{ }F(f) := y{{\partial f}\over{\partial x}}$$give the vector space $\mathbb{C}[x, y]$ the structure of an $\text{U}$-module. For each $m \ge 0$, the space $\mathbb{C}^m[x, y]$, of homogeneous polynomials of degree $m$, is a simple $\text{U}$-submodule of $\mathbb{C}[x, y]$.

My question is, for $m=1$, what are the $2 \times 2$ matrices of the action of the elements $E$, $H$, $F$ in the basis $x$, $y$? Much thanks in advance.

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Note that $U$ is the universal enveloping algebra of the Lie algebra $\mathfrak{sl}_2$. The given action is a way to define (unique) $m+1$-dimensional simple modules over this algebra.

The computation, even given matrices in the general form, is for example spelled out in these notes (see Chapter 7, and 7.3 for the matrices):

http://www-groups.mcs.st-and.ac.uk/~neunhoef/Teaching/liealg/liealgchap3.pdf

(similar computations should also appear in most introductory texts on Lie algebras, but maybe not in that much detail).

Actually, you can easily compute this yourselves (especially in the case $m-1$), and a very big hint is already given in the comment by David Hill, you need to understand:

  1. what is the matrix displaying a linear transformation with respect to a certain basis (linear algebra)

  2. How to differentiate polynomials (basic analysis)

  3. What is an action?

For 3. note that an action is a morphism of algebras $$U\rightarrow \operatorname{End}(V),$$ where $V=\mathbb{C}^m[x,y]$ in this example.

It is already indicated in the comment by David Hill what the action of $E$ in the case $m=1$ is. From this one should be able to derive a matrix easily.