Let $V$ be a $\mathbb{C}$-vector space of dimension $N$, let $d$ be a positive integer, let $l \leq N$ be a positive integer, and let $U \subseteq S^d(V)$ be a linear subspace of codimension $k=\binom{l+d-1}{d}$. Does there exist a linear subspace $W \subseteq V$ of dimension $l$ for which $S^d(V)=U \oplus S^d(W)$?


I tried to prove yes by showing that the dimension of the subvariety of $Gr(k,S^d(V))$,
consisting of subspaces of the form $S^d(W)$ for an $l$-dimensional linear subspace $W \subseteq V$, is greater than the dimension of the subvariety of subspaces that intersect $U$ non-trivially, but this is emphatically not the case. The former has dimension $l(N-l)$, while the latter has dimension $k(\binom{N+d-1}{d}-k)$. (See [here](https://mathoverflow.net/questions/415732/what-is-the-dimension-of-this-subvariety-of-the-grassmannian))



Given a subspace $U \subseteq S^d(V)$, does there always exist a complement of the form $S^d(W)$?
Without any further hypothesis on $U$, the answer is no. Take $V = \mathbb{C}^3$, $d=2$, $l=2$ and $U = S^2(W\_1)$ where $W\_1$ is a $\mathbb{C}^2$ inside $V$. Then for any other $W\_2 \subset V$ of dimension $2$, we have:


$$S^2(W\_1 \cap W\_2) \subset U \cap S^2(W\_2)$$


and obiously $\dim S^2(W\_1 \cap W\_2) \geq 1$. On the other hand, I think the answer should be *yes* if you assume $U$ to be generic (and perhaps some adequate numerical conditions). This should follow from a dimension count bounding the dimension of the subvariety of $\mathrm{Gr}\left( \binom{l+d-1}{l},S^d V \right)$ representing linear spaces which intersect $S^d(W)$ non-transversally for all $W \in \mathrm{Gr}(l,V)$.


