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3.v. Proof in some cases

We have proofs of some cases of Conjecture 2. Most notably (and not given here, but rather in the paper [So00]) is that Conjecture 2 holds when p=2 and one Schubert condition is 1,m+1 (Theorem 3.8 of [So00]).

   We also have a number of specific cases proven, which we summarize in the following table. For this, we introduce the following notation: Given a Schubert condition a and a positive integer j, (a)j means that a is repeated j times. Also, set Ji to be the sequence
                    Ji  =   1, 2, ..., p-1, p+i .
The Schubert condition Ji is a Pieri-type condition, it is the condition that a general p-plane H meet a fixed m+1-i-plane nontrivially.

Cases Proven:

Schubert data  (J2)4   (J2)4, J1   (1,3,5)2, (J1)3   (1,3,5)4 
m,p    4,2       3,3          3,3      4,3
d      3         3            6        8

   These are all proven in the same way as the proof of Conjecture 1 when (m,p)=(2,3). That is, we consider an equivalent system of polynomials (as described in Section 3.iii) in the local coordinates Xa,b. From this, we compute a universal eliminant and then show that its discriminant is a sum of squares. The links in this table to the data m,p are to documented Maple V.5 scripts detailing these calculations, and the links to the number d are to outputs of these scripts.

   The case (J2)4, J1 is interesting in that the discriminant is not symmetric in the 3 parameters. The case (1,3,5)2, (J1)3 is the first we showed where the conditions were not just `Pieri-type'.

   The last case of (1,3,5)4 with (m,p) = (4,3) is special. Maple was completely unable to handle this on its own. The Maple script in the link generates a Singular input file, given here. This computes a Gröbner basis, but not an eliminant, as that was also too hard. This Gröbner basis is stored as a Maple V.5 file, and we have a Maple V.5 script that reads these equations, and (cleverly) computes the universal eliminant. This is the first case in which the universal eliminant factored---which suggests that the Galois group of this enumerative problem is not the full symmetric group on 8 elements.


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