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  • $\begingroup$ Can you explain the notations $\mathfrak{sl}_2$ and $\mathfrak{gl}_{n+1}(\mathbb C)$, and also what "the adjoint action of $\mathrm{SL}_2$ on $\mathfrak{sl}_2$" is? $\endgroup$ Commented Apr 15, 2024 at 16:25
  • $\begingroup$ The Lie group $\mathrm{SL}_2$ has the tautological $2$-dimensional representation $V$, so $\mathrm{Sym}^nV$ is a $(n+1)$-dimensional representation. This is a group homomorphism $\mathrm{SL}_2\to\mathrm{GL}_{n+1}$, whose derivative is the Lie algebra homomorphism $\mathfrak{sl}_2\to \mathfrak{gl}_{n+1}$ in my answer. $\endgroup$ Commented Apr 15, 2024 at 21:01
  • $\begingroup$ Thank you for your response. However, I know next to nothing about Lie groups and algebras, and I don't know what $\text{Sym}^n V$ stands for. Can the eigenvectors here be simply described? $\endgroup$ Commented Apr 15, 2024 at 21:30
  • $\begingroup$ @IosifPinelis Think of your matrices as operators on the space of all homogeneous polynomials $h(x,y)$ of degree $n$. Then, in the monomial basis, $B_n$ is the operator $x\frac\partial{\partial y}+y\frac\partial{\partial x}$ (this is one possible realization of the $\mathfrak{sl}_2$-irrep). The $\lambda$-eigenfunctions for this operator are $\left(\frac{x+y}{x-y}\right)^{\frac\lambda2}F(x^2-y^2)$, with $F$ any function. So one just examines for which $\lambda$ can one obtain polynomials. $\endgroup$ Commented Apr 19, 2024 at 16:25
  • $\begingroup$ @მამუკაჯიბლაძე : Thank you for your comment. Can you expand its last sentence, maybe in a formal answer? I think it would be useful to know the eigenvectors more or less explicitly. $\endgroup$ Commented Apr 19, 2024 at 16:39