Given a subspace $U \subseteq S^d(V)$ of a particular form, does there always exist a complement of the form $S^d(W)$?

Question

Let $V$ be a $\mathbb{C}$-vector space of dimension $N \geq 2$, let $d$ be a positive integer, let $l < N$ be a positive integer, and let $U \subseteq S^d(V)$ be a linear subspace of codimension $k=\binom{l+d-1}{d}$. Suppose that there exists a basis of $U$ of the form $\{u_1^{\vee d},\dotsc, u_r^{\vee d}\}$ such that $$ \operatorname{span}\{u_1^{\vee d-1},\dotsc, u_r^{\vee d-1}\}=S^{d-1}(V), $$ where $r=\binom{N+d-1}{d}-k$. Does there exist a linear subspace $W \subseteq V$ of dimension $l$ for which $S^d(V)=U \oplus S^d(W)$?

Here, $\vee$ denotes the symmetric product.

This is a variant of my previous question Given a subspace $U \subseteq S^d(V)$, does there always exist a complement of the form $S^d(W)$?, which did not assume any particular form for $U$.

Answer 1

The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}\cdot t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \operatorname{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim S^2V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2{\left(\mathbb C^3 \cap W\right)} \subset \left(U \cap S^2W \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.