Fano Schemes for Generic Sums of Products of Linear Forms
We study the Fano scheme of -planes contained in the hypersurface cut out by a generic sum of products of linear forms. In particular, we show that under certain hypotheses, linear subspaces of sufficiently high dimension must be contained in a coordinate hyperplane. We use our results on these Fano schemes to obtain a lower bound for the product rank of a linear form. This provides a new lower bound for the product ranks of the Pfaffian and permanent, as well as giving a new proof that the product and tensor ranks of the determinant equal five. Based on our results, we formulate several conjectures.
Given an embedded projective variety , its Fano scheme is the fine moduli space parametrizing projective -planes contained in . Such Fano schemes have been considered extensively for the case of sufficiently general hypersurfaces [AK77, BVdV79, Lan97] but less so for particular hypersurfaces [HMP98, Beh06, CI15]. In this article, we study the Fano schemes for the special family of irreducible hypersurfaces
for any . We exclude the case since this is a smooth quadric hypersurface with significantly different behaviour.
In [IT16, §3], Z. Teitler and the first author considered the Fano scheme . With the help of a computer-assisted calculation, they observed the curious fact that every -plane of is either contained in a coordinate hyperplane, or there exist such that is contained in . This motivates the following definition:
Definition 1.1 (-splitting).
Consider . A -plane contained in admits a -splitting if there exist such that is contained in
We say that is -split if every -plane of admits a -splitting for some .
The above-mentioned observation from [IT16] can now be rephrased as the statement that is two-split.
We make two conjectures regarding the splitting behaviour of these Fano schemes:
Conjecture 1.2 (One-Splitting).
Assume and . The Fano scheme is one-split if and only if
Conjecture 1.3 (Two-Splitting).
Assume is even and . The Fano scheme is two-split if
We show in Example 3.1 that the bound on of Conjecture 1.2 is indeed necessary for one-splitting. However, the sufficiency of the conditions of Conjectures 1.2 and 1.3 for one- and two-splitting is less obvious.
Consider a degree homogeneous equation of the form
where the are pairwise coprime squarefree monomials, the are degree products of linear forms, and the are degree products of linear forms.
Definition 1.4 (Property ).
We say that is true if, for any equation of the form (1) satisfying for all , it follows that there is some for which .
Our first main result relates the above definition to our two conjectures:
Secondly, we use this to prove our conjectures in some special cases.
Our analysis of Equation (1) makes use of relatively elementary methods. However, a more sophisticated approach should also be possible. Equation (1) posits that is a point in the th secant variety of a Chow variety parametrizing degree products of linear forms. Equations for the Chow variety are classical, going back to Brill and Gordan [GKZ08]. More recently, Y. Guan has provided some equations for secant varieties of Chow varieties [Gua15, Gua16] . It would be interesting to see if these equations shed light on the vanishing of the from Equation (1).
Our motivation for studying is twofold. Firstly, we wish to add to the body of examples of varieties for which one understands the geometry of . If the Fano scheme is -split for some , -dimensional linear subspaces of can be understood in terms of linear subspaces of for certain . We illustrate this by describing the irreducible components of for whenever or , see Examples 3.5 and 3.7. We also characterize when is connected, see Theorem 3.4.
Secondly, we may use our results to obtain lower bounds on the product rank of certain linear forms. Recall that the product rank (also known as Chow rank) of a degree form is the smallest number such that we can write
where the are products of linear forms. We denote the product rank of by . Note that product rank may be used to give a lower bound on tensor rank, see [IT16, §1.3] for details. Generalizing [IT16, Theorem 3.1], we prove the following:
Let be an irreducible degree form in variables such that is covered by -planes, and let with .
If is one-split, then .
If is even, , and
are two-split, then .
Applying this to the determinant of a generic matrix, we recover that its product and tensor ranks are five [IT16]. Note that we have replaced the computer-aided computation of with a conceptual proof. We may also apply Theorem 1.7 to the determinant of a generic matrix to obtain ; this is equal to the lower bound one obtains from Derksen and Teitler’s lower bound on the Waring rank [DT15]. In Example 4.4 we apply the theorem to the Pfaffian of a generic skew-symmetric matrix to obtain , beating the previous lower bound of . Finally, in Example 4.5 we use a slightly different argument to obtain that the product rank (and tensor rank) of the permanent is at least , beating the previous lower bound of .
The rest of the paper is organized as follows. In §2, we study equations of the form (1). We use this in §3 to show our splitting results for the Fano schemes , as well as studying several cases in more detail. Finally, we prove Theorem 1.7 in §4 and apply our results to a number of examples including the Pfaffian and permanent.
For simplicity, we will be working over an arbitrary algebraically closed field . Note however that all our main results clearly hold for arbitrary fields simply by restricting from to any subfield.
2. Special Sums of Products of Linear Forms
In this section we will prove that property holds for or . We will obtain this result by using induction arguments. These arguments involve a refined version of the property. Consider an equation of the form
with . As before, the are pairwise coprime squarefree monomials and the are degree products of linear forms. We now assume simply that are degree forms, no longer requiring that they be products of linear forms. It will be convenient to order the summands on the right hand side so that
We will maintain this ordering convention throughout all of §2. Similar to the property we make the following definition.
Definition 2.1 (Property ).
We say that is true, if for any equation of the form (2) satisfying
it follows that there are for which .
Note that by definition implies .
Fix and and assume that holds for every . Then holds for every .
We may argue by induction on . Obviously, the hypotheses for imply those for . Hence, by the induction hypothesis we have the vanishing of of the , with . Using this we end up with the hypotheses for being fulfilled, and we see that another of the with has to vanish. ∎
We will first analyze the case . Therefore, we consider an equation of the form
with forms of degree . As before, we order indices such that .
Remark 2.3 (Cancellation).
Assume we are given a variable which divides , one monomial , and all for . Setting
where we have reduced from forms of degree to degree . We call this the cancellation of (5) by .
Let be a linear form dividing , where the are forms of degree .
If , then for all , divides or .
If and is a monomial, then for all , divides or .
We first prove the second statement. We have
for some variable and form . Expanding the left hand side as a sum of monomials, we see that the degree condition ensures that no terms from cancel with for . But every monomial on the right hand side is divisible by , hence also on the left hand side. The claim follows.
For the first statement, we reduce to the second by performing a change of coordinates taking to a monomial. This can be achieved while preserving all variables in the with at most one exception, say in . After factoring out this one linear form from , the pairwise of sum degrees is still at least and we may apply the second claim. ∎
Suppose for , forms of degree . Let be the number of distinct factors of . If
then there is a variable dividing both and one of the . This is true if , in particular if .
For each , choose some variable dividing it. Setting will result in the equality , hence one factor of can only depend on . There are possible ways to choose the , and factors of , so there must be one factor of which depends only on the for
different choices. On the other hand, the intersection of more than
choices of the contains at most one variable. Hence, if the above inequality is satisfied, the claim follows. ∎
with forms of degree .
If , then either or vanishes.
If is not squarefree and , then either or vanishes.
In particular, property holds for every .
For the first case, note that the hypothesis implies in particular that and . Hence, Lemma 2.5 implies the existence of a variable dividing both and one of the . But then is divisible by for , hence divides . Cancelling by , we may proceed by induction on the degree .
For the second case, we proceed with a similar argument. The inequality
implies that , which is larger than or equal to the number of distinct factors of . Thus, we again find a variable dividing both and one of the . If in fact divides , then after factoring out one power of from and , Lemma 2.4 guarantees that divides as well. Dividing , , and by , we reduce to the first case.
If does not divide , we may cancel by as in the first case, maintaining that is still not squarefree. To finish, we again proceed by induction on the degree . ∎
It is clear that the degree bounds in Proposition 2.6 cannot be improved upon. If and are variables dividing respectively, then setting gives
Likewise, if there are non-trivial degree syzygies between and , so we cannot expect the second claim to hold.
We next prove a stronger version of .
Suppose for ,
with forms of degree . Then if , some with must vanish.
In particular, holds for every , .
Set and . Note that it suffices to prove the proposition in the case that and . Indeed, we may absorb variables from into the corresponding to reduce to this case. Henceforth we will assume we are in such a situation.
We begin by proving the claim when and , that is, is a single variable . Using Proposition 2.6(1), we see that modulo either or vanish. But since , and are both just constants, hence one must vanish outright.
Next, we consider the case when and . First, we show that some must vanish, with no restriction on its degree. We apply Lemma 2.5 to find a variable dividing some monomial and . Applying Lemma 2.4, we may conclude that divides for . Cancelling by we may reduce the degree by one and conclude by induction on degree that for some . Now we show that we can impose the desired degree restriction on . Indeed, if , or and , this is automatic. If instead and , then we have satisfying the hypotheses of Proposition 2.6(1), from which the claim follows.
It remains to consider the cases when . We will now induct on . First assume that . By setting any variable in equal to zero for , we reduce to an equation of the form (6) with one fewer summand on the right hand side, yet , , and the same. Hence, by induction, divides or . Now, there are variables appearing in the for , yet
Thus, if then either or must vanish.
If instead and , we may again apply Lemma 2.5 followed by Lemma 2.4 to find a variable dividing some and for . Note that we may reorder the monomials such that , since all have the same degree . Cancelling by , we again find ourselves in the situation of (6), but now with , so by the above, or vanishes, thus or does as well. ∎
We now move to the case of :
with forms of degree . If , then either , , or vanishes.
In particular, holds for every .
We will prove the statement by induction on the degree . Note that if we can show that have a common factor , then we are done. Indeed, by Lemma 2.4, must divide either each or . Pulling out of each where we can, and out of in at most one position, allows us to “cancel by ” in a fashion similar to Remark 2.3. We thus reduce the degree and the claim follows by induction.
In the following, we will assume that no common factor of and exists, and that all are non-zero. For simplicity, we may assume that , since this case implies the more general one. We denote by , and by . Our hypothesis on degrees is now simply .
Consider any factor of or . Setting , we reduce to the case of Proposition 2.6(1) (if divides some ) or Proposition 2.8 (by absorbing into some a variable of made linear dependent modulo ). In either case, we see that modulo , some must vanish, that is, is a factor of . We may proceed to do this for all distinct divisors of and . But since
we conclude that together and have at most
distinct factors. It follows that either both and contain a square, or else that the non-squarefree product has at most distinct factors.
Assume first that is squarefree, and fix some factor . We now argue in a similar fashion to the proof of Lemma 2.5. For each , fix a variable so that and all remaining variables are linearly independent. For each , choose some variable dividing . Setting will result in the equality , hence one factor of is zero modulo . There are possible ways to choose the , and at most factors of , so there must be one fixed factor of which is zero mod for
different choices. On the other hand, the intersection of more than choices of the contains no variable. Hence, since it follows that there is a factor of which is zero modulo , that is, agrees with it.
We now instead assume that both and contain factors with multiplicity at least two. Consider any factor of or . As long as is not a variable in or , we may set and conclude that divides or . Indeed, if divides this follows from Proposition 2.6. Otherwise we may absorb into some a variable of made linear dependent modulo , and then apply Proposition 2.8 followed by Proposition 2.6(2) to conclude that two of , , and vanish modulo .
If at most one factor of , divides or but not or , we thus obtain that have at most distinct factors. But then either or has at most distinct factors, so an argument similar to the case squarefree above shows that and would have to possess a common factor.
So we now finally consider the case that at least two distinct factors of are variables found in and , neither dividing or . It follows by Proposition 2.6 that each such factor must divide . Without loss of generality, we assume that divides and . We obtain
so has factors which only depend on and a single variable of .
If divides and , then setting , we obtain , a contradiction. If instead divides and we obtain
and has factors which only depend on and a single variable of . Since has factors, one must also just be a variable of . So in this case, we conclude that a variable of divides . If instead divides , we see by setting that must divide , so we can take to produce as above.
We thus may assume that we are in the situation of variables with dividing and , and dividing . By Proposition 2.6, divides for or . Now let be such that . We thus obtain
hence has factors which depend on and a single variable of .
The right hand side of Equation (7) clearly contains monomials divisible by . But the left hand side cannot: while each monomial of has degree at least in the variables of and , and each monomial of has degree at least in the variables of and , the part of relatively prime to has degree at least . The inequality
then shows that this impossible. We conclude that in fact some must equal zero. ∎
2.4. Property for
We now prove a lemma that will help us with the degree four case:
Fix , , and and let . Assume that holds whenever , or whenever and . In Equation (2), consider any linear form dividing of the summands on the left hand side. Then must also divide of the with .
Assume divides some factor of . Setting , we now have that the hypothesis for is fulfilled. Since we have assumed that is true, of the with must vanish modulo .
If does not divide any , we still may set , modifying the right hand side of the equation to replace one factor of some by a linear form which is no longer a monomial. Now we have to distinguish two cases. Let us assume first that . Then, since the degree of drops by one, we are in the situation of . As before by our assumption, of the with must vanish modulo .
If then the fact that the degree of drops may violated condition (4). However, we may bring the summand to the left hand side of the equation. This leaves us in the situation of and our assumption again provides the vanishing of of the , with . ∎
We now use this lemma to show for arbitrary and :
If , property holds for arbitrary and . In particular, holds for .
We will use induction on and . Note, that for and arbitrary, follows from Theorem 2.8. Lemma 2.2 then provides for arbitrary . Assume we have proven property is true whenever , or whenever and . We must prove that property is true as well, which then again implies for by Lemma 2.2.
For we have . Either one of the has to vanish, which would prove our claim, or by Lemma 2.12, all the factors of the occur as one of the (at most) linear factors of the for . Let us partition the multiset of these linear factors in such a way that two of them are coprime exactly if they belong to different subsets. We say one of the subsets covers if is divisible by the elements of . In this way every, all the have to be covered by one of the subsets of the partition. On the other hand, we have seen by Lemma 2.12 that every may cover at most of the . Hence, there can be at most one which is covered by more than one subset of the partition. This implies, we have for all but one of the summands on the left hand side.
For this implies that we can write the left hand side as the sum of only products, since we can write as a product of linear forms. Then using the induction hypothesis for concludes the proof for the case .
To conclude, we consider the case . If none of the (with ) vanishes, we have seen that at most three linear factors , and can occur on the left hand side. We choose and set . Now, the left hand side depends only on two linear forms. In this situation the left hand side is actually a product of linear forms (since is algebraically closed). In the same way as in the proof of Lemma 2.12 (when setting one of the to zero) we see by the induction hypothesis that one of the has to be divisible by . We have only finitely many choices for the linear factors of the , but we have infinitely many choices for . Hence, there are with dividing , which implies either or . In any case, one of the summands or has to vanish, and we are in the case of . ∎
The following lemma derives a consequence of the property which will be used later.
Consider an equation of the form
where the are degree products of linear forms, and the are pairwise relatively prime squarefree monomials of degree . If property is true, then there is a permutation such that for all .
Consider any factor of some . If does not divide any , we may set , modifying the right hand side of Equation (8) to replace one factor of some by a linear form which is no longer a monomial. But this equation still satisfies the hypotheses necessary for , as long as , so in fact, must have divided one of the all along.
We thus see that every factor of each is just a variable, up to scaling. By comparing the monomials on both sides of (8), we find the desired permutation. ∎
We may interpret the above lemma geometrically as saying that, if is true, then the subgroup of taking to itself is generated by the semidirect product of the torus
with the copy of the symmetric group permuting the indices of , and the copies of permuting the indices of for some fixed .
3. Fano Schemes and Splitting
3.1. Main results
In this section, we will prove Theorems 1.5 and 1.6. For , consider projective space with coordinates , and . Let be a -dimensional linear subspace of . We may represent as the rowspan of a full rank matrix , with rows indexed by and column corresponding to the homogeneous coordinates on . We define linear forms in by
along with degree forms
The condition that is contained in is equivalent to the condition
The condition that is one-split is equivalent to the condition that some vanishes.
Example 3.1 (-planes which are not one-split).
For and , let be any -plane with all linearly independent for , and . Then clearly is contained in , but is not one-split (although it is two-split).
For , consider forms satisfying and all linearly independent, for , for , and . Let be the corresponding -plane. Clearly is contained in , but is not one-split.
We thus see that the bound on in Conjecture 1.2 is sharp.
We henceforth assume that , that is, that , and that none of the vanish, that is, is not one-split. Without loss of generality, we may inductively reorder the forms as follows: given , we take to be any form such that the dimension of the vector space spanned by the is maximal. We may then choose a new basis
for the degree one piece of with the property that each is a factor of , and each factor of is in the span of
By the way we have ordered the forms , this implies that .
We now will assume that
For , suppose that . If is even, is even and , or is odd and , then .
We have that
Using our assumption on we thus have
Suppose that . If is even, then , and thus for all . But then
contradicting (10), since .
Assume instead that is odd. Then for all , so
But this contradicts (10) if is even and , or if is odd and . ∎
Proof of Theorem 1.5.
First note that . Indeed, if not, then contains a factor which is not in the span of the factors of the for , so it is impossible to satisfy Equation (9). Suppose that we have inductively shown that for some . Then by Lemma 3.2, we have that . If , we set for and use property applied to
to conclude that some for vanishes modulo . But by our construction of the , this is impossible, and we conclude that .
We proceed in this fashion until we obtain for
If is odd, we conclude again by Lemma 3.2 that