Recall that the $q$-binomial coefficient $\big[\begin{smallmatrix}a\\b\end{smallmatrix}\big]$ is the Laurent polynomial in $q$ given by $$ \big[\begin{smallmatrix}a\\b\end{smallmatrix}\big]=\frac{[a]!}{[b]![a-b]!} $$ where $[n]!=[1][2][3]...[n]$, and $[i]=\tfrac{q^i-q^{-i}}{q-q^{-1}}=q^{i-1}+q^{i-3}+q^{i-5}+...+q^{-(i-1)}$.

Let $V_n$ be the $n$-dimensional irrep of $\mathfrak{sl}(2)$.

It is easy to check that there exists an isomorphism of $\mathfrak{sl}(2)$-representations $$ S^k V_{n+1} \cong \textstyle \bigwedge^k V_{n+k} $$ between the $k$-th symmetric power of $V_{n+1}$ and the $k$-th exterior power of $V_{n+k}$, as both have $\big[\begin{smallmatrix}n+k\\k\end{smallmatrix}\big]$ as their character

**Proof:** We exhibit a basis of weight vectors of $S^k V_{n+1}$, a basis of weight vectors of $\bigwedge^k V_{n+k}$, and a bijection between these bases which respects the weights.
Let $x_0,\ldots, x_n$ be the standard basis of $V_{n+1}$, and let
$y_1,\ldots y_{n+k}$ be the standard basis of $V_{n+k}$.
Then $x_{i_1}{\cdot}\, x_{i_2} \cdot... \cdot x_{i_k}\mapsto y_{i_1+1}\wedge y_{i_2+2}\wedge...\wedge y_{i_k+k}$ is the desired bijection. $\square$

The problem with the above proof is that it only proves the existence of an isomorphism, without actually constructing one.

Question:It there anaturalisomorphism of $\mathfrak{sl}(2)$-representations $S^k V_{n+1} \cong \textstyle \bigwedge^k V_{n+k}$ that one can write?