TITLE: Which representations of $\mathfrak{sl}(2)$ are homomorphic images of the tensor product of finitely many copies of $\mathbb{C}^2$?
QUESTION [9 upvotes]: My questions may turn out to be related to Schur functors.
If $\mathfrak{g}$ is a complex semisimple Lie algebra and $\lambda$ is the highest weight of an irreducible representation $V$ of $\mathfrak{g}$, I am interested in restricting the representation to an $\mathfrak{sl}(2)$ subalgebra of $\mathfrak{g}$. For example, if $\mathfrak{g} = \mathfrak{sl}(n)$, then I am interested in $V = \mathbb{C}^n$ being the standard representation. If $\mathfrak{g} = \mathfrak{so(n)}$, I am interested in $V = \mathbb{C}^n$, i.e. in the vector representation. Similarly if $\mathfrak{g} = Sp(m, \mathbb{C})$, I am interested in $V = \mathbb{C}^{2m}$ being its standard representation. I am also interested in the $5$ exceptional cases and their fundamental representations.
As for the $\mathfrak{sl}(2)$ subalgebra of $\mathfrak{g}$, I am mostly interested in "the" so-called "regular" $\mathfrak{sl}(2)$ subalgebra (it is unique up to conjugation in $\mathfrak{g}$), which was studied for instance by Kostant and others.
I can now formulate my questions. From now on, consider the representation $V$, but view it as a representation of a fixed $\mathfrak{sl}(2)$ subalgebra of $\mathfrak{g}$ (for the sake of this post, one may for instance assume it to be a regular $\mathfrak{sl}(2)$ subalgebra).

Question 1: Does there always exist a positive integer $m$, and a $\mathfrak{sl}(2)$ intertwining (meaning the corresponding notion at
the Lie algebra level) linear map
$$ \mathbb{C}^2 \otimes \cdots \otimes \mathbb{C}^2 \to V, $$
which is onto, where the domain is the tensor product of $m$ copies of
$\mathbb{C}^2$?


Question 2: Assuming the answer to question 1 is yes, can one write down such a linear map using weights and roots? For example,
what would $m$ be in terms of Lie-theoretic data, etc?

Note: I did edit my post quite a bit, so some of the comments may not make sense now to a new reader, because of my edits. The new reader should keep this in mind. However, these are now the precise questions that I would like to see answered. I apologize for the little mess that I have made, with my latest edits. I think though that my questions are more focused now, in this version.

REPLY [8 votes]: $\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\sl{\mathfrak{sl}}$In the comments to hm2020's answer, the OP explains that they want a surjection from $(\mathbb{C}^2)^{\otimes m}$, not from a direct sum $\bigoplus_i (\mathbb{C}^2)^{\otimes m_i}$. In that case, there is one obstruction: The weights of the representation must be either all even or all odd, and then the answer is yes.
Notation: Let $h = \left[ \begin{smallmatrix} 1&0 \\0&-1 \end{smallmatrix} \right]$. Recall that the eigenvalues of $h$ on any (finite dimensional) $\sl(2)$ representation are integers. We write $V_i$ for the $i$-eigenspace of $h$ on a representation $V$, and write $\chi_V = \sum_i \dim V_i q^i$ for a formal variable $q$. We call $\chi_V$ the character of $V$.
We put $R = \mathbb{C}^2$ with the obvious $\sl(2)$ action, and put $S_k:=\operatorname{Sym}^k R$ (so $R = S_1$).
It is well known that the finite dimensional representation theory of $\sl(2)$ is semisimple, and that the simple representations are the $S_k$ with character $q^k + q^{k-2} + \dotsb + q^{-k}$. Meanwhile, the character of $R^{\otimes m}$ is $(q+q^{-1})^m$. We thus see that $\chi_{R^{\otimes m}}$ is either an even or odd Laurent polynomial in $q$, according to the parity of $m$, and that $\chi_{S_k}$ is either even or odd according to the parity of $k$. So $S_k$ can only be a summand of $R^{\otimes m}$ if $m \equiv k \bmod 2$.
Thus, a representation $\bigoplus S_{k_i}$ is only a summand of $R^{\otimes m}$ if all the $k_i$ are congruent modulo $2$.
This is also a sufficient criterion. The multiplicity of $S_k$ in $R^{\otimes (k+2 \ell)}$ is $\binom{k+2\ell}{\ell} - \binom{k+2\ell}{\ell-1}$, which goes to infinity as $\ell$ goes to infinity. So, for any fixed list of $k_i$ which are all congruent to a fixed value $k$ modulo $2$, the multiplicities of $S_{k_i}$ in $R^{\otimes k+2 \ell}$ will get arbitrarily large. Thus, any finite dimensional representation obeying this partiy condition will eventually be a summand of $R^{\otimes m}$.
I find this more intuitive with the Lie group $SL(2)$ instead of the Lie algebra: $\chi_V(q)$ is the trace of the Lie group element $\left[ \begin{smallmatrix} q & 0 \\ 0 & q^{-1} \\ \end{smallmatrix} \right]$, and the parity condition is that $\left[ \begin{smallmatrix} -1 & 0 \\ 0 & -1 \\ \end{smallmatrix} \right]$ must act either $1$ or by $-1$.