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 $\mathbb{P}^2$.

  • $\begingroup$ A small remark: since $\mathcal{D}$ is the projectivized space of polynomials (and not the dual thereof), $H_p$ is a subset of $\mathcal{D}$, not an element. $\endgroup$
    – algori
    Oct 25, 2011 at 0:23
  • $\begingroup$ .. and also, I presume in question 4 you mean the dimension of $\mathcal{A}\cap \partial H_p$ to be $k-2$, not the dimension of $\mathcal{A}$. $\endgroup$
    – algori
    Oct 25, 2011 at 2:56
  • $\begingroup$ I am sorry for these mistakes I keep making! $\endgroup$
    – Ritwik
    Oct 25, 2011 at 3:38

1 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 $k-3$

  • $\begingroup$ Thank you for your answer. Is it true that the intersection of algebraic varieties is again a variety? Secondly, I am a bit confused with the second answer. I am assuming A is the set of polynomials that have a specific point (say p0) as a singular point. I assume that the polynomial also vanishes at the point p0 (in addition to the gradient vanishing). I agree that for p = p0 your claim is correct. But why is it true for every p? Having an extra singular point should impose an extra condition. $\endgroup$
    – Ritwik
    Oct 24, 2011 at 23:54
  • $\begingroup$ By Euler's relation, if the gradient of a homogeneous poly vanishes then the poly vanishes as well. The locus $\Delta$ is the locus of all singular curves, so it contains $\partial H_p$ for every $p$. $\endgroup$ Oct 25, 2011 at 0:05
  • $\begingroup$ Ritwik -- re "Is it true ..." -- yes, provided both are quasi-projective in the same projective space. Re "I am a bit confused ..." -- in your posting $\mathcal{A}$ is an arbitrary smooth quasi-projective subvariety of $\mathcal{D}$. In your comment you say that "I am assuming A is the set of polynomials that have a specific point (say p0) as a singular point". So what is $\mathcal{A}$ exactly? $\endgroup$
    – algori
    Oct 25, 2011 at 0:19
  • $\begingroup$ Ritwik -- I've read Jack Huizenga's comment and now I see what probably caused the confusion. In my example $\mathcal{A}$ is obtained by first taking the union of all $\partial H_p$'s and then taking the smooth part of the resulting variety. In other words, $\mathcal{A}$ is formed by all curves that have one simple node and no other singularities. $\endgroup$
    – algori
    Oct 25, 2011 at 0:48
  • $\begingroup$ Thank you, I think I understand the example. Its certainly a counter example. In that I case I will modify my question and ask 1) Is my claim true under any reasonable assumption on $\mathcal{A}$, such as being the zero locus of $k$ polynomials (and certain non vanishing conditions.......my set $\mathcal{A}$ need not be closed). 2) Does the dimension at least drop by 2? I assume it has to drop at least by 1? $\endgroup$
    – Ritwik
    Oct 25, 2011 at 2:22

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