The following question makes sense in a more general setting but for sake of simplicity let me stick to a particular case.

Consider the degree three Veronese embedding $V\subset\mathbb{P}^9$ of $\mathbb{P}^2$ via the complete linear system of plane cubics.

Does there exist a point $p\in\mathbb{P}^9$ not lying on any secant line to $V$ (properly secant, if $p$ is on a tangent line it is ok), but lying on infinitely many planes that intersects $V$ in three points in linear general position?

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    $\begingroup$ In every characteristic other than characteristic $3$ the non-existence of such $p$ follows from the fact that an elliptic curve has only finitely many $3$-torsion points. $\endgroup$ – Jason Starr Jan 3 at 0:28
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    $\begingroup$ I am clarifying that in my comment, I am assuming that the "three points in linear general position" on $V$ are non-collinear points in the projective plane $V=\mathbb{P}^2$ (this is my own understanding of "three points in linear general position"). If the three points are allowed to be collinear, then of course there are solutions, as Zach Teitler correctly points out. $\endgroup$ – Jason Starr Jan 3 at 13:00

Here is an answer in terms of power sum decompositions of polynomials. A point $p \in \mathbb{P}^9$ corresponds to a homogeneous polynomial $P$ of degree $3$ in $3$ variable, defining a plane cubic. Points of the Veronese $q \in V$ correspond to pure powers of linear forms, $q = \ell^3$. A point $p$ lies in the span of points $q_1,\dotsc,q_r \in V$ if and only if the corresponding polynomial $P$ is a linear combination of powers, $P = c_1 \ell_1^3 + \dotsc + c_r \ell_r^3$, where the $c_i$ are scalars. Such an expression is often called a power sum decomposition. The least number of terms in any power sum decomposition of a polynomial is called the polynomial's Waring rank. In these terms you are asking whether there exists a plane cubic of rank $3$, with infinitely many power sum decompositions in three terms.

The polynomial $x^2 y$ has this property (assuming the characteristic is not $2$ or $3$). A power sum decomposition with $3$ terms is given by $$ x^2 y = \frac{1}{6}(y+x)^3 + \frac{1}{6}(y-x)^3 - \frac{2}{6} y^3 . $$ More generally, $$ x^2 y = \frac{1}{6a^2}(y+ax)^3 + \frac{1}{6a^2}(y-ax)^3 - \frac{2}{6a^2} y^3 , $$ giving infinitely many power decompositions. In your terms, the planes $\operatorname{span}\{(y+ax)^3, (y-ax)^3, y^3\}$ are distinct, and meet $V$ at those three points, which are linearly independent. Those aren't all of the decompositions. I can try to write the rest if you need. But there are infinitely many, anyway.

To see that the planes in this family are pairwise distinct, note that $(y+bx)^3$ can't lie in the span of $(y+ax)^3,(y-ax)^3,y^3$ unless $b \in \{a,-a,0\}$, because the matrix whose rows are the coefficients of those cubic forms is a Vandermonde matrix, so they are linearly independent unless $b$ has one of those values.

The polynomial $x^2 y$ does lie on a tangent line to the Veronese: $$ x^2 y = \lim_{t \to 0} \frac{ (x+ty)^3 - x^3 }{ 3t } , $$ a limit of secant lines through $(x+ty)^3$ and $x^3$ (a tangent line through $x^3$).

But $x^2 y$ doesn't lie on any proper secant line to $V$, meaning we can't write $x^2 y$ as $\ell_1^3 + \ell_2^3$. The first one factors with a repeated factor, while the second one factors as $$ (\ell_1 + \ell_2)(\ell_1 + \omega \ell_2)(\ell_1 + \omega^2 \ell_2), $$ $\omega$ a $3$rd root of unity, which has distinct factors.

Edit: The other decompositions of $x^2 y$ actually aren't that hard to describe. Let $B$ (for "binary") be the span of $x^3,x^2 y, x y^2, y^3$ in $\mathbb{P}^9$, so $B$ is a $\mathbb{P}^3$, parametrizing the binary cubics in the variables $x,y$. The intersection $B \cap V$ is a twisted cubic, the pure powers $(ax+by)^3$. Now the point is that any $2$-plane passing through $x^2 y$ and contained in $B$ will cut that twisted cubic in three points, giving a three-term power sum decomposition.

And the converse holds: any three-term power sum decomposition of $x^2 y$ doesn't involve $z$, that is, the three terms are all linear forms in just $x$ and $y$. It's not too hard to prove this "hands-on" for this case. More generally, Buczyński and Landsberg write about this in terms of "rank preserving pairs", https://arxiv.org/abs/0909.4262.

In a comment it is observed that $x^2 y$ depends essentially on only $2$ variables in the sense that its second derivatives span a $2$ dimensional subspace of the linear forms, and there is the question whether there is a form $P$ with the same rank properties (rank $3$ with infinitely many rank decompositions) but depending essentially on all variables ("concise"). The answer is no. Ranks of plane cubics have been worked out by many authors (it's been a bit of a running example, used to demonstrate new approaches) including B. Segre, Reznick, Comon and Mourrain, Kleppe, Landsberg and myself, plenty of others. The only plane cubics of rank $3$ are (up to linear change of coordinates) $x^2 y$ and $x^3 + y^3 + z^3$. The derivatives of $P = x^3 + y^3 + z^3$ do span all the linear forms, and it has rank $3$. But it doesn't have infinitely many rank decompositions. In fact the rank decomposition is unique.

Briefly, suppose $x^3 + y^3 + z^3 = \ell_1^3 + \ell_2^3 + \ell_3^3$. Taking first derivatives, each of $x^2$, $y^2$, $z^2$ lies in the span of $\{\ell_1^2,\ell_2^2,\ell_3^2\}$. In fact $\{x^2,y^2,z^2\}$ must have the same span. So each $\ell_i^2 = a x^2 + b y^2 + c z^2$. The only such squares are just scalar multiples of the $x^2,y^2,z^2$ (i.e., two of the coefficients $a,b,c$ vanish). So the $\ell_i$ are the same as $x,y,z$, up to order and scalar multiple.

For more on homogeneous polynomials with infinitely many rank decompositions, look for keywords like "identifiable polynomials" (or non-identifiable, really). Good luck!

  • $\begingroup$ The OP specifies that the 3 points on $V=\mathbb{P}^2$ should be in general position. In what sense are $3$ collinear points on $\mathbb{P}^2$ in linear general position? $\endgroup$ – Jason Starr Jan 3 at 12:57
  • $\begingroup$ Yeah. I wrote a really long, detailed proof that the cubics of rank $3$ are $x^2y$ and Fermat, and the classification of their decompositions (collinear and unique resp.), in complete agreement with you. But on reflection I think OP meant the simpler, linearly general in $\mathbb{P}^9$. I think this “intersecting $V$ in three linearly general points” is just saying “a plane spanned by points of $V$”. For that matter “plane” seems to mean $2$-plane. But to be sure, @Jack can you please say if I’ve misunderstood the question? $\endgroup$ – Zach Teitler Jan 3 at 14:13
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    $\begingroup$ I agree that $3$ distinct points in the projective plane map to non-collinear points in $\mathbb{P}^9$. The OP should clarify what is meant by "linearly general position", $\endgroup$ – Jason Starr Jan 3 at 14:23
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    $\begingroup$ Thank you very much for your answer. Yes, I just meant that the $2$-plane intersects $V$ in three points spanning the $2$-plane. If I got it correctly your are looking at the image of $z=0$ which is a twisted cubic $C$ in $V$. The polynomial $x^2y$ lies on a tangent line of $C$ but not on a secant line. The polynomial $x^2y$ does not depend on $z$. Do you know if there exist a polynomial with the same properties of the one you proposed but whose three partial derivatives are linearly independent (in the space of plane conics)? $\endgroup$ – Jack Jan 3 at 19:18
  • $\begingroup$ No, that doesn’t exist. I can try to write it tonight. $\endgroup$ – Zach Teitler Jan 3 at 21:49

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