On representations of $\mathfrak{sl}_{n+1}$: why $V(\ell\omega_1)\cong S^{\ell}(\mathbb C^{n+1})?$

Question

Given the lie algebra $\mathfrak g = \mathfrak{sl}_{n+1}$ and $\ell \in \mathbb Z$, denote by $V(\ell\omega_1)$ the irreducible $\mathfrak g$-module with highest weight $\ell\omega_1$, where $\omega_1\in \mathfrak h^*$ is given by $\omega_1(h_j) = \delta_{1j}$. I want to prove that $V(\ell\omega_1)\cong S^{\ell}(\mathbb C^{n+1})$ (the $\ell$-th symmetric power of $\mathbb C^{n+1}$)

What I already know is that $V(\omega_1)\cong \mathbb C^{n+1}$, with the highest-weight vector being $v_1$ from the standard basis. By the way $\mathfrak g$ acts on the tensor product, we have that $v = v_1\cdots v_1$ is a highest-weight vector for $S^{\ell}(\mathbb C^{n+1})$ of weight $\ell\omega_1$; but how to prove that the submodule generated by $v$ is the entire space? I could prove somehow that $S^{\ell}(\mathbb C^{n+1})$ is irreducible, but that doens't seem to be a good strategy. I also tried to work with dimensions, but it didn't help too. Any suggestions?

Answer 1

A computational way that seems to work would be the following.

We can view this space $V(\ell)$ as the subspace of homogeneous polynomials of degree $\ell$ in the polynomial algebra $\Bbb{C}[x_0,x_1,\ldots,x_n]$, where $x_0,x_1,\ldots,x_n$ are independent indeterminates.

• Then the Lie algebra $\mathfrak{sl}_{n+1}$ acts on $V(\ell)$ via partial differential operators in such a way that the basis matrix $e_{ij}$ acts by $x_i\dfrac\partial{\partial x_j}$. It is easy to check that these operators follow the same commutator rules as the matrices $e_{ij}$.
• The diagonal matrices $e_{ii}-e_{jj}$ operate as $$x_i\frac{\partial}{\partial x_i}-x_j\frac{\partial}{\partial x_j}.$$ This implies that all the monomials $P(a_0,a_1,\ldots,a_n):=\prod_{j=0}^nx_j^{a_j}$, $\sum_ja_j=\ell$, are shared eigenvectors of all the matrices in the Cartan subalgebra. In other words, they are weight vectors.
• Furthermore, $(e_{ii}-e_{jj})\cdot P(a_0,a_1,\ldots,a_n)=(a_i-a_j)P(a_0,a_1,\ldots,a_n)$, and then a moment of thought reveals that each monomial $P(a_0,a_1,\ldots,a_n)$ belongs to a weight space of it own. In other words, all the weight spaces of $V(\ell)$ are 1-dimensional.
• It is well known that any finite dimensional representation of $\mathfrak{sl}_{n+1}$ is a direct sum of its weight spaces. By the previous bullet any putative submodule $M$ of $V(\ell)$ is thus spanned by the monomials it contains.
• But, starting from any monomial $P(a_0,a_1,\ldots,a_n)$ you can reach (a non-zero scalar multiple of) the highest weight vector $x_0^\ell$ by applying a sequence of operators of the form $x_i\partial/\partial x_j$ to it. Similarly, you can reach a scalar multiple of any monomial from $x_0^\ell$. These observations prove that $V(\ell)$ is, indeed, irreducible.

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(Rust alert – $\color{#f00}{\mathrm{RED}}$)

Another way would come from the group side. The irreducible modules of $G=SL_{n+1}$ can be constructed as the spaces of sections of a sheaf on $G/B$ associated to a dominant weight $\lambda$ (viewed as a character of the Borel subgroup $B$). In the special case when $\lambda=\ell\omega_1$ is a multiple of the first fundamental weigt this process is reduced to looking at a sheaf on $G/P$ where $P$ is a certain parabolic subgroup that is also the stabilizer of a point when $G$ acts on the projective space $\Bbb{P}^n(\Bbb{C})$. In this case the sheaf is a shifted version of the structure sheaf of the projective space, and the global sections of such sheaves are exactly the spaces of homogeneous polynomials.