Finding a basis for the weight space of $(\mathbb C^{n+1})^{\otimes \ell}$ of weight $\omega_\ell$ Let $\mathfrak g = \mathfrak{sl}_{n+1}$ and let $V= \mathbb C^{n+1}$. This Lie algebra has a "natural" representation $\rho:\mathfrak g\rightarrow \mbox{End}(V)$ by setting $\rho(x)(v)= xv$, the right hand side of this equality understood as matrix multiplication and $v\in V$ as a column vector. Given $\ell \leq n$, this enables us to consider its tensor representation $\rho_T: \mathfrak g \rightarrow \mbox{End}(V^{\otimes \ell})$ acting on pure tensors as $\rho(x)(w_1\otimes\cdots\otimes w_\ell) = \sum_{j=1}^\ell w_1\otimes \cdots \otimes \rho(x)w_i\otimes \cdots \otimes w_\ell$.

Denote by $\omega_\ell$ the weight $\omega_\ell: \mathfrak h \rightarrow \mathbb C$ defined on the Cartan subalgebra $\mathfrak h$ by $\omega_\ell(h_j)=\delta_{j\ell}$. I want to find a basis or a generating set for the weight space $V_{\,\, \omega_\ell}^{\otimes \ell} = \{v\in V^{\otimes \ell}: \rho_T(h)(v) = \omega_\ell(h)v \; \forall \; h\in \mathfrak h\}$

Progress so far: denote by $\{v_1,\cdots, v_{n+1}\}$ the canonical basis of $V$. Write $v_0 = v_1\otimes \cdots \otimes v_\ell$ and denote by $S_\ell$ the symmetric group on $\ell$ elements. This group acts on $V^{\otimes \ell}$ by $\sigma(w_1\otimes \cdots \otimes w_\ell) = w_{\sigma(1)}\otimes \cdots \otimes w_{\sigma(\ell)}$. It seems that
$$\{\sigma\cdot v_0: \sigma \in S_\ell\} \mbox{ generates the weight space $V^{\otimes \ell}_{\,\, \omega_\ell}$.} $$
First of all,  the vector $v_0$ has weight $\omega_\ell$. Indeed, each basis vector $v_i$ has weight $\epsilon_i$ in $V$, where $\epsilon_i:\mathfrak h \rightarrow \mathbb C$ is the weight given by $\epsilon_i(\sum a_j e_{jj})=a_i$, where $e_{jj}$ denotes the diagonal matrix with 1 in the $(i,i)$ position. Thus, by the given action on the tensor product, $v_0$ has weight $\epsilon_1+\cdots+\epsilon_\ell = \omega_\ell.$ Now, since $\sigma$ just permutes the vectors $v_i$ appearing in $v_0$, then $\sigma \cdot v_0$ also has weight $\omega_\ell$. So the above set is a subset of $V^{\otimes \ell}_{\,\,\omega_\ell}$.
But I cant prove that it generates the whole weight space. Since $S_\ell$ acts on $V^{\otimes \ell}$, the above set is linearly independent. Thus, its dimension is at least $\ell!$. I was thinking about using Kostant formula to calculate the dimension of the weight space, and hence conclude that the set is in fact a basis. But this is getting way too complicated and I believe that there is an easier argument to prove it. Any insight will be helpful!
 A: I'll consider $\mathfrak{gl}_{n + 1}$ instead of $\mathfrak{sl}_{n + 1}$, since the difference does not matter in this case. Let $\mathfrak{d}_{n+1}$ denote the subalgebra of diagonal matrices: for $1 \leq \ell \leq n$, the weight $\omega_\ell$ is equal to the weight $\epsilon_1 + \cdots + \epsilon_\ell$, where $\epsilon_i \colon \mathfrak{d}_{n+1} \to \mathbb{C}$ takes a diagonal matrix to its $(i, i)$th entry.
Let $v_1, \ldots, v_{n+1}$ be the canonical basis of $\mathbb{C}^{n + 1}$, then we can write an arbitrary tensor $t \in (\mathbb{C}^{n + 1})^{\otimes \ell}$ uniquely in the form
$$ t = \sum_{s} t_s v_{s_1} \otimes \cdots \otimes v_{s_\ell},$$
where $s$ ranges over all words of length $\ell$ on the alphabet $\{1, \ldots, n + 1\}$, and $t_s \in \mathbb{C}$ records the coefficient of the basis vector $v_s := v_{s_1} \otimes \cdots \otimes v_{s_\ell}$.
Let $e_{ii} \in \mathfrak{gl}_{n + 1}$ be the matrix with a $1$ in the $(i, i)$th entry, and zeroes elsewhere. Consider the equation $$\rho_T(e_{ii})(t) = \omega_\ell(e_{ii})(t).$$
The action of $e_{ii}$ on the basis $(v_1, \ldots, v_{n + 1})$ is to keep $v_i$ and kill everything else. Accordingly, the action of $\rho_T(e_{ii})$ on the basis $\{v_s\}_{s}$ is to scale $v_s$ by the number of $i$'s appearing in the word $s$. Words that have no occurrence of $i$ get killed. We get
$$\rho_T(e_{ii})(t) = \sum_{s} t_s v_s [\text{# of times $i$ occurs in $s$}].$$
The basis $\{v_s\}_s$ is an eienbasis for the operator $\rho_T(e_{ii})$, where the eigenvalues are $\{0, 1, 2, \ldots, \ell\}$, and for example the eigenspace corresponding to the eigenvalue $3$ is spanned by all $v_s$ such that $i$ occurs $3$ times in the word $s$.
On the right hand side, we simply have
$$\omega_\ell(e_{ii})t = \begin{cases} t & \text{if } i \leq \ell, \\ 0 &\text{otherwise}, \end{cases}$$
and hence solutions to the equation $\rho_T(e_{ii})(t) = \omega_\ell(e_{ii})(t)$ are precisely either the $1$ or $0$-eigenvectors of $\rho_T(e_{ii})$, depending on whether $i \leq \ell$ or $i > \ell$.
We want the same $t$ to work for all $1 \leq i \leq n + 1$, and hence we need the intersection of the $1$-eigenspaces of $\rho_T(e_{ii})$ for $1 \leq i \leq \ell$, and the $0$-eigenspaces for the rest of the $i$. This simultaneous eigenspace is spanned by the $v_s$ such that:

*

*1 occurs once in $s$,

*...

*$\ell$ occurs once in $s$,

*$\ell + 1$ does not occur in $s$,

*...

*$n + 1$ does not occur in $s$.

Now it should be clear that this is exactly the $S_\ell$-orbit of $v_1 \otimes \cdots \otimes v_\ell$.
The key insight of this exercise is that in terms of the $\epsilon_i$ basis of the dual Cartan, the weight spaces are incredibly easy to classify, and in fact are all spanned by those straightforward $v_s$ tensors.
