Does passing through a point in general position cut down the dimension by one?

Question

Let $\mathcal{D} \approx P^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three variables (up to scaling), where $\delta_d = \frac{d(d+3)}{2}$. A point $p\in \mathbb{P}^2$ gives us a hyperplane $H_p \subset \mathcal{D}$, i.e it is the space of degree $d$ polynomials vanishing at $p$.

Define $H^*_p \subset H_p$ to be the space of degree d polynomials $f$ that vanish at $p$, but whose derivative at $p$ does not, i.e. $$ H^*_p := \{f \in \mathcal{D}: f(p)=0, \nabla f|_p \neq 0 \} .$$

Let $\mathcal{A} \subset \mathcal{D}$ be a non-singular algebraic variety (not necessarily closed) of dimension $k$. Define $$ \partial H_p := H_p - H_p^*.$$ I have four questions:

1) Is it true that $H_p^*$ intersects $\mathcal{A}$ transversally for a generic choice of $p$?

2) Is $\partial{H_p} \cap \mathcal{A}$ an algebraic variety for generic choices of $p$?

3) Is there any reasonable condition on $\mathcal{A}$ so that the ''dimension'' of $\partial H_p \cap \mathcal{A}$ is less than or equal to $k-3$?

4) Is the dimension of $\partial H_p \cap \mathcal{A}$ at most $k-2$? I am assuming it can not be more than $k-1$?

When I say a statement is true for a generic choice of $p$ I mean that the set of $p$ for which it is true forms an open dense subset of $\mathbb{P}^2$.

Answer 1

Q1: In general no: take $X$ to be the subvariety of $P^{\delta_d\vee}$, the dual of $P^{\delta_d}$, formed by all $H_p$'s. Note that $X$ is just the image of the Veronese map $\mathbb{P}(V)\to\mathbb{P}(Sym^d(V^\vee))$ for $V=\mathbb{C}^3$. So $X$ lies on a quadric $Q$ given by $x_i x_j=x_kx_l$ with $i,j,k,l$ pairwise disjoint. Here $x_i,x_j,x_k,x_l$ are coordinate functions on $(Sym^d \mathbb{C}^3)^\vee$. Set $\mathcal{A}=Q^\vee$, the dual of $Q$; this is a 2-dimensional quadratic surface in $\mathcal{D}$. By the biduality theorem, $\mathcal{A}^\vee=Q$. So $X\subset Q$ is contained in the dual of $\mathcal{A}$, which means that for every $H_p\in X$ the intersection $H_p\cap \mathcal{A}$ (and hence also $H^*_p\cap\mathcal{A}$) is not transversal.

Q2: Yes, for any $p$, since both $\mathcal{A}$ and $\partial H_p$ are (quasiprojective) algebraic varieties.

Q3: In general no: take $\mathcal{A}$ to be the smooth part of the discriminant hypersurface $\Delta\subset P^{\delta_d}$ formed by projectivizing the set of all homogeneous polynomials $f$ such that the gradient of $f$ vanishes at some point of $\mathbb{P}^2$. Then for any $p$ we have $\dim \partial H_p\cap \mathcal{A}=\dim\partial H_p=\delta_d-3\neq k-3=\delta_d-4$.

Re questions 3 and 4: let me describe informally what one should expect. One of the two things can happen. Either $\mathcal{A}\subset\Delta$, in which case $\dim \mathcal{A}\cap\Delta=\dim\mathcal{A}=k$ and for generic $p$ one would expect $\dim \partial H_p\cap \mathcal {A}$ to be $k-2$ since $\partial H_p$'s form a 2-parametric family of projective subspaces. Or $\mathcal{A}\not\subset\Delta$, in which case $\dim \mathcal{A}\cap\Delta=\dim\mathcal{A}=k-1$ and for generic $p$ one would expect $\dim \partial H_p\cap \mathcal {A}$ to be $k-3$