Let $V \cong \mathbb{C}^2$ be the standard representation of $\mathfrak{sl}_2 \mathbb{C} = \text{Mat}_2(\mathbb{C})$. We follow the convention from Fulton & Harris' Representation Theory book and choose as generators for $\mathfrak{sl}_2 \mathbb{C}$ the matrices $H, X,Y$ given on page 147 and so we obtain the eigenvalue/ weight decomposition $V= V_{-1} \bigoplus V_1= \mathbb{C} \cdot y \oplus \mathbb{C} \cdot x$ such that $H(x)=x$ and $H(y)=-y$ and $X,Y$ are the ladder operators.
Below the Exercise 11.34 (Hermite Reciprocity) on page 160, which asks the reader to use the eigenvalues of $H$ to show that
$$ \text{Sym}^k(\text{Sym}^n V)) \cong \text{Sym}^n(\text{Sym}^k V)) $$
there is a text below which provides a rather interesting geometric picture I would like understand better:
Note that in the examples of the exercise we have seen, it seems completely coincidental: for example, the fact that the representations
$\text{Sym}^3(\text{Sym}^4 V)) $ and $ \text{Sym}^4(\text{Sym}^3 V))$ both contain a trivial representation corresponds
to the facts that the tangential developable of the twisted cubic in
$\mathbb{P}^3$ has degree $4$, while the chordal variety of the rational normal quartic in $\mathbb{P}^4$ has degree $3$.
a remark: if we think about $V$ as isomorphic dual space $V^*$ and the generators as monomials $x,y$, then $\text{Sym}^n V$ can be regarded as homogeneous polynomials in $x$ and $y$ of degree $n$.
Moreover, the rational normal curve of degree $n$ can be realized as image under the map $V \to \text{Sym}^n V, v \mapsto v^n$, ie $n$-th power on $v \in V$.
Question:
I not understand why the fact that $\text{Sym}^3(\text{Sym}^4 V)) $ and $ \text{Sym}^4(\text{Sym}^3 V))$ both contain a trivial representation $V_0 \cong \mathbb{C}$ corresponds
to the facts that the tangential developable of the twisted cubic in
$\mathbb{P}^3$ has degree $4$, while the chordal variety of the rational normal quartic in $\mathbb{P}^4$ has degree $3$. How to see this geometric interesting connection?
I'm not looking for a formal proof, just a sketchy idea providing this geometric picture would make me happy.
Best Answer
In general, a copy of the trivial representation in $\operatorname{Sym}^k(V)$ corresponds to an invariant, degree $k$, (symmetric) polynomial function on $V^*$ (up to scale). In the case of $\mathfrak{sl}_2$ all representations are self-dual so we can translate this to polynomial functions on $V$.
Then we can see that the rational normal curve must lie in the variety of this polynomial. It certainly lies in an orbit of the $\operatorname{SL}_2$ action since $\operatorname{SL}_2$ acts transitively on $\mathbb{P}(V)$ and the polynomial is invariant. Thus we only need to test 1 element.
Upgrading this to the tangent developable/chordal variety is probably a case by case thing but I don't know this last bit off the top of my head.