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Let $Sec_r(V)$ be the $r$-secant variety of a Veronse variety $V\subset\mathbb{P}^N$, that is $$Sec_r(V) = \bigcup_{p_1,...,p_r\in V}\left\langle p_1,...,p_r\right\rangle\subset\mathbb{P}^N$$ where $V$ is the image of $\mathbb{P}^n$ via the embedding induced by $\mathcal{O}_{\mathbb{P}^n}(d)$.

Is it true that if $Sec_r(V)\neq\mathbb{P}^N$ then any polynomial on $\mathbb{P}^N$ vanishing on $Sec_r(V)$ has degree greater or equal than $r+1$ ?

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    $\begingroup$ Firstly, you need to add some non-degeneracy hypothesis. Linear spaces violate your question. I thought the answer to your question was 'yes' and was proved in a paper by JM Landsberg, possibly with co-authors.; However, as this link to a previous MO question makes clear, the answer is a little more subtle- mathoverflow.net/questions/114854/… $\endgroup$
    – meh
    Dec 15, 2016 at 22:53
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    $\begingroup$ As the above link, as well as the link to right to "Equations of the secant variety" make clear, your statement is known in some cases. In some sense one needs the embedding of X is 'sufficiently ample'. Even then, off the top of my head, I am not sure. What I can say is that in that case, the equations are discussed in this arxiv.org/abs/1012.3563 paper. $\endgroup$
    – meh
    Dec 15, 2016 at 22:58
  • $\begingroup$ Of course I meant $d\geq 2$. $\endgroup$
    – user101890
    Dec 15, 2016 at 23:09
  • $\begingroup$ The results you are citing are very precise. Indeed in these papers there are interesting in finding equations cutting out secant varieties. I am looking for something much more rough: the minimal degree of a polynomial in the ideal of a secant variety. $\endgroup$
    – user101890
    Dec 15, 2016 at 23:13
  • $\begingroup$ @James: in some cases it is easy to find set-theoretic equations while what you call "something much more rough", i.e., finding the degree of minimal generators of the ideal is extremely hard. For example, take the Chow variety of forms $F=L_1\cdots L_d$ instead of the Veronese. $\endgroup$
    – Abdelmalek Abdesselam
    Dec 17, 2016 at 18:20

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The answer by JM Landsberg in the link from aginensky's comment precisely says that the $r+1$ lower bound on the degree of generators is true. See top of page 2 in the article "Prolongations and computational algebra" by Sidman and Sullivant where they point to the article "On the ideal of an embedded join" by Ulrich and Simis.

Edit: Having thought about the question a bit more, I realized this is trivial (for the Veronese, but not for secants of general varieties as in the above articles) if one knows the symbolic method from 19th century invariant theory.

Consider the secant $\sigma_r(v_d(\mathbb{P}^n))$ for the degree $d$ Veronese embedding. Let $C(F)$ be a homogeneous polynomial of degree $r$ in the coefficients of a generic form $F$ of degree $d$ in $n+1$ variables. If $C(F)=0$ for all $F$'s which can be written $F=L_1^d+\cdots +L_r^d$ for some linear forms $L_i$, then the polynomial $C$ vanishes identically.

Indeed, $$ C(F)=\left.M(F_1,\ldots,F_r)\right|_{F_1=\cdots=F_r=F} $$ where $M$ is the associated symmetric multilinear form given by $$ M(F_1,\ldots,F_r)=\frac{1}{r!}\frac{\partial^r}{\partial t_1\cdots \partial t_r} \ C(t_1 F_1+\cdots+t_r F_r)\ . $$ From the hypothesis (and working over a field like $\mathbb{C}$ which contains $d$-th roots of unity) one immediately gets $$ M(L_1^d,\ldots,L_r^d)=0 $$ for linear all forms $L_1,\ldots,L_r$.

Finally, one has the identity $$ C(F)=\frac{1}{d!^r}\ F(\frac{\partial}{\partial L_1})\cdots F(\frac{\partial}{\partial L_r})\ M(L_1^d,\ldots,L_r^d) $$ where the point coordinates $x_1,\ldots,x_{n+1}$ are replaced by differential operators in the coefficients of the linear forms $L_1,\ldots,L_r$. As a result, $C(F)$ is identically zero. The last identity is what makes the classical symbolic method work. For more on this, please see this article (or here for the preprint version).

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  • $\begingroup$ I thought the bound was strict, but the answer makes it appear as if there could be an inequality i.e $> r+1$ $\endgroup$
    – meh
    Dec 16, 2016 at 7:03
  • $\begingroup$ @aginensky: I am not sure I understood your comment correctly, but there are plenty of examples with equations of degree exactly $r+1$. These come from catalecticants. $\endgroup$
    – Abdelmalek Abdesselam
    Dec 16, 2016 at 20:00
  • $\begingroup$ Agreed about your example. If not the 'generic case' the "most understood" case is that what the OP calls $Sec_r$ has equations in degree $r+1$. What I am saying is that I don't know for a fact that it is 4 $\endgroup$
    – meh
    Dec 16, 2016 at 20:49
  • $\begingroup$ @aginensky: in your last sentence "What I am saying is that I don't know for a fact that it is 4" I still don't understand "it" and "4". Are you asking for examples where the are no equations in degree $r+1$? If so, please look at my article with Ikenmeyer and Royle that I referred to which is about a case where $r=7$ and there are no equations before degree 15. $\endgroup$
    – Abdelmalek Abdesselam
    Dec 16, 2016 at 21:20
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    $\begingroup$ I don't know what happened. My sentence should end "for a fact that there are equations in degree r+1 " as per your guess. I will certainly have a look at your article !. $\endgroup$
    – meh
    Dec 18, 2016 at 14:13

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