## Varieties cut by quadrics

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Is there a characterization of the class of varieties which can be described as an intersection of quadrics, apart from the taulogical one?

Lots of varieties arise in this way (my favorite examples are the Grassmanianns and Schubert varieties and some toric varieties) and I wonder how far can one go.

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A Google search turned up Remark 1 on page 131 here: books.google.com/… – Qiaochu Yuan Jan 12 2010 at 1:45

## 5 Answers

In fact the answer is in some sense tautological: every projective variety can be realized as a scheme-theoretic intersection of quadrics! See e.g.

D. Mumford, "Varieties defined by quadratic equations", Questions on algebraic varieties, C.I.M.E. Varenna, 1969 , Cremonese (1970) pp. 29–100,

for quantitative refinements of this question.

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Ah! Quite interesting. I knew hypersurfaces always arise this way, but this is considerably more than what I would have wanted :P – Mariano Suárez-Alvarez Jan 12 2010 at 1:56

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As Pete already indicated, Mumford's theorem says that for any projective variety $X\subset \mathbb P^n$, its Veronese emberdding $v_d(X)\subset \mathbb P^N$ is cut out by quadrics, for $d\gg0$. So a reasonable question is for the variety with a fixed projective embedding (such as Grassmannians in the Plucker embedding and not in some other random embedding).

For this latter more meaningful question, many "combinatorial" rational varieties, such as Grassmannians, Schubert varieties (as you pointed out), flag varieties, determinantal varieties, etc., are cut out by quadrics.

For the "non-combinatorial", non-rational varieties, the most classical result is Petri's theorem: a smooth non-hyperelliptic curve of genus $g\ge 4$ in its canonical embedding is cut out by quadrics, with the exceptions of trigonal curves and plane quintics.

There is a vast generatization of this property: $X\subset \mathbb P^n$ satisfies property $N_p$ if the first syzygy of its homogeneous ideal $I_X$ is a direct sum of $\mathcal O(2)$, the second syzygy is a direct sum of $\mathcal O(3)$, etc., the $p$-th syzygy is a direct sum of $\mathcal O(p+1)$. In this language, $X$ is cut out by quadrics is equivalent to the property $N_1$.

A 1984 Green's conjecture is that a smooth nonhyperelliptic curve satisfies $N_p$ for $p=Cliff(X)-1$, where $Cliff(X)$ is the Clifford of $X$. This has been proved for generic curves of any genus by Voisin (in characteristic 0; it is false in positive characteristic).

Another notable case: the ideal of $2\times 2$ minors of a $p\times q$ matrix has property $N_{p+q-3}$.

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I meant canonical curves; I made the correction, thanks. – VA Jan 12 2010 at 2:31
By the way, the canonical image of hyperelliptic curves, being rational normal curves, are also cut out by quadrics. – jvp Jan 12 2010 at 2:33

As Pete L. Clark says, if you can Veronese your line bundle then the answer is "all of them".

So a more interesting question may be: for which varieties M does every ample line bundle give an embedding defined by quadrics?

The best sufficient condition I know is that MxM have a Frobenius splitting w.r.t. which the diagonal is compatibly split. See Brion and Kumar's book about Frobenius splitting, and Sam Payne's article, which discusses the toric case. The first shows that it's true for flag manifolds, and the second that it's not true for all Schubert varieties.

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That is an embedding by a COMPLETE linear system, right? And isn't a singular toric variety compatibly split (but the ideal may not be generated by quadrics?) – VA Jan 12 2010 at 19:59
I had never seen Veronese used as a verb. Cute :) – Mariano Suárez-Alvarez Jan 12 2010 at 20:05
And isn't a singular toric variety compatibly split  No. A singular TV M is indeed split, but the issue is not splitting M itself, but splitting MxM with the diagonal M being split. If I understood right Sam shows you can't do this for M = F_1, which occurs as a Schubert variety. – Allen Knutson Jan 12 2010 at 20:23
BTW there's another variant of this question: let M > N be a pair of varieties, and ask that M be cut out be quadratics, and N further cut from M by linear conditions. Again, this occurs for any (M,N) once one Veroneses enough, and occurs for (M = flag manifold, N = Schubert variety) for any line bundle by Frobenius-splitting results of Ramanathan. – Allen Knutson Jan 12 2010 at 23:52

It seems impossible to give a better answer than Pete L. Clark's but those passing by might appreciate a sketchy proof.

If $X \subset \mathbb P^n$ is a projective variety then it is the intersection of finitely many hypersurfaces of degree at most $k$. If we consider the natural morphism from $\mathbb P^n$ to $\mathbb P H^0(\mathbb P^n,\mathcal O_{\mathbb P^n}(k))^{\ast}$ then the image of $X$ will be the intersection of the image of $\mathbb P^n$ with a finite number of hyperplanes.

Being the image of $\mathbb P^n$ itself a intersection of quadrics, it follows that any projective variety can be expressed as the intersection of quadrics and hyperplanes.

A more condensed version of the argument above can also be found here.

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Here is an easy hands-on way to define any variety as being cut out by quadrics. Write your variety as the zero locus of a bunch of polynomials. Mark each monomial that occurs in any one of these polynomials and which has degree higher than two. We are going to add new variables and new quadratic equations to turn that monomial into a degree two monomial. For example if the monomial $x y^2$ occurs in one of your polynomials , add a new variable $z$ and add the equation $z = xy$. Then rewrite all occurrences of $xy^2$ in any of your polynomials as $zy$. To take another example, if the monomial $x^5$ occurs then add two new variables, say $u$ and $v$, together with the two new quadric equations $u = x^2$ $v = u^2$ and the replace all occurrences of $x^5$ by $xv$. Just keep on tacking new variables, and new equations until you've turned all of your original monomials into either quadratic or linear monomials. Now you've got your variety sitting in a much bigger space due to all the new variables, but its defining polynomials are now all quadratic plus linear.
(If you don't like the linear terms arising , play the homogenization game.)

I learned this trick in some two page paper by a Russian showing every compact manifold is
diffeomorphic to one defined by quadratic equations. That paper in turn is related to a theorem attributed to Milnor (or Thurston) asserting that every manifold can be realized by planar linkages. Sorry I don't have the references.

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Skolem used this trick to show that to decide the solvability of arbitrary polynomial equations in integers, it would suffice to do it for those of total degree 4: see page 3 of math.mit.edu/~poonen/papers/h10_notices.pdf . (Skolem's work was in the 1920s, which presumably predates the two-page paper you are referring to. And it wouldn't surprise me if the trick were known even before Skolem!) – Bjorn Poonen Feb 8 2010 at 7:29
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