We work over the field of complex number, using the Grothendieck projectivization, that is we have $$ \mathbb P(\cdot):=\operatorname{Proj} \bigoplus_{m \ge 0} \operatorname{Sym}^m(\cdot). $$
Let $X$ be the Grassmannian of $3$-planes in $\mathbb C^6$, that is $X=G(3,6)\subset \mathbb P(V)= \mathbb P^{19}$. We can reformulate the embedding in the projective space as a map $$ \operatorname{Proj} \bigoplus_{m \ge 0} H^0(X, \mathcal O(m)) \hookrightarrow \operatorname{Proj} \bigoplus_{m \ge 0} \operatorname{Sym}^m V. $$ This means that for all $m \ge 0$ there exists a surjective map $$ f_m: \operatorname{Sym}^m V=H^0(\mathbb P(V), \mathcal O(m)) \to H^0(X, \mathcal O(m)), $$ which I suppose is just the restriction of sections on $\mathbb P(V)$ to sections on $X$. The kernel of $f_m$ is, then, just the sections of $\mathbb P(V)$ that vanish on $X$. In particular, since $X$ is homogeneous we have $H^0(\mathbb P(V), \mathcal O(1))=V=H^0(X, \mathcal O(1))$, then $\ker f_m$ is not trivial for $m \ge 2$. The ideal $I$ given by $$ 0 \to I:=\bigoplus_{m \ge 0} \ker f_m \hookrightarrow \bigoplus_{m \ge 0} H^0(\mathbb P(V), \mathcal O(m)) \to \bigoplus_{m \ge 0} H^0(X, \mathcal O(m)) \to 0 $$ is just the ideal of $X$, that is $X=Z(I)$.
On the other hand, we have the semisimple group $G:=\operatorname{SL}_6(\mathbb C)$ acting on $X \subset \mathbb P(V)$. In fact, $H^0(X, \mathcal O(m))$ is an irreducible $G$-representation, while $\operatorname{Sym}^m V$ is not an irreducible one when $m\ge 2$.
Question. As a $G$-representation we have $\operatorname{Sym}^m V= H^0(X,\mathcal O(m)) \oplus \ker f_m$. Is there someone who have studied the decomposition of $\ker f_m$ in irreducible $G$-representations? I have tried to compute $\ker f_m$ for small $m$ using SAGE, but I cannot find a nice way to write the decomposition (the computations show that it is something involving the adjoint representation of $G$).