Representation of $sl(2,R)$.

Question

I am interested in the unique (up to isomorphism) $5$-dimensional representation of the Lie algebra $sl(2,R)$.

I understand that one can choose the module $V_4 = \text{span}\{x^4,x^3y,x^2y^2,xy^3,y^4\}$, but I'd like to work with a more natural module, like a collection of matrices. I suppose this comes down to finding a nice isomorphism of $V_4$ into a nice space.

I believe for each $n \geq 1$, $Sym(n)$, the space of $n \times n$ symmetric matricies is an irreducible $SL(2,\mathbb{R})$ via the action $g.v := gvg^t$. I'd like something similar for the group $SL(2,\mathbb{R})$ or equivalently the Lie algebra $sl(2,\mathbb{R})$.

Answer 1

Let $V_n=<v_0,v_1,\ldots,v_n>$ be a vector space and and define the $3$ operators $d, \hat d $ and $e$ as follows $$ d(v_i) = v_{i-1}, \hat d(v_i)=(i+1)(n-i)v_{i+1}, e(v_i)=(n-2i) v_i. $$ Then $V_n$ turns into an irreducible $sl_2$-module. I hope you may find now the matrix representation for $V_4$.

Answer 2

The $5$-dimensional irreducible representation $\rho:\mathfrak{sl}_2(K)\rightarrow \mathfrak{gl}_5(K)$, with respect to the standard basis $(e,f,h)$ of $\mathfrak{sl}_2(K)$, can be given as follows: $$ \rho(e)=\begin{pmatrix} 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0\end{pmatrix}, \quad \rho(f)=\begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0\end{pmatrix}, \quad \rho(h)=\begin{pmatrix} 4 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 & -4\end{pmatrix}. $$ In general, the $n$-dimensional irreducible representation $V=\langle v_0,\ldots ,v_{n-1}\rangle$ can be given by the formulas $\rho(h)v_i=(n-1-2i)v_i$, $\rho(e)v_i=(n-i)v_{i-1}$ and $\rho(f)v_i=(i+1)v_{i+1}$.