Ternary forms with lots of zeros

Transcription

1 Ternary forms with lots of zeros Bruce Reznick University of Illinois at Urbana-Champaign Conference/Workshop on Inverse Moment Problems, IMS National University of Singapore, Dec. 19, 2013

2 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman)

3 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman) Relation to cones

4 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman) Relation to cones Relation to moments

5 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman) Relation to cones Relation to moments Hilbert 1893

6 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman) Relation to cones Relation to moments Hilbert 1893 Robinson 1969

7 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman) Relation to cones Relation to moments Hilbert 1893 Robinson 1969 w/ Choi-Lam 1980

8 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman) Relation to cones Relation to moments Hilbert 1893 Robinson 1969 w/ Choi-Lam 1980 Arxiv paper 2007

9 17 Outline of the talk Statement of the problem and the conjectures (joint with Greg Blekherman) Relation to cones Relation to moments Hilbert 1893 Robinson 1969 w/ Choi-Lam 1980 Arxiv paper 2007 New material (joint with Greg Blekherman and??)

10 In this table, we see, in order: 2r, r 2, (2r 1)(2r 2) 2, their maximum, and then 3 2r(r 1) + 1. The smaller of the last two is the upper bound on Z(p) for p P 3,2r.

11 Let I(n, d) = {(i 1,..., i n ) : 0 i k Z, i k = d} and for i I(n, d) write the multinomial coefficient c(i) = d! i 1!...i. Write n! x i = x i xn in and for a real form p of degree d in n variables, scale the coefficients by: p = c(i)a(p; i)x i i I(n,d) Then the Fischer inner product is defined by [p, q] = c(i)a(p; i)a(q; i). i I(n,d) For α R n define (α ) d = (α x) d, then it is easy to see that [p, (α ) d ] = p(α).

12 It is not so hard to see from this that P n,2r and Q n,2r are dual cones. The dual cone to Σ n,2r is easily understood. If p Σ n,2r, then [p, h 2 ] 0 for all h of degree r. Write a general form of degree r with the coefficients as variables h(x) = a(p; l+l )t(l)t(l ) := H p (t) l I (n,r) t(l)x l = [p, h 2 ] = l This construction was done by Sylvester for ternary quartics; he called it the catalecticant. It s important to note that ( H (α ) 2r (t) = α l t(l) and so the rank of H p is a lower bound on the width of p. When P n,2r = Σ n,2r, the duals are equal and the two are the same. In general, this doesn t happen for the other cases. l l ) 2

13 Suppose p(x, y, z) = i I (3,d) d! i!j!k! a(p; (i, j, k))x i y j z k. Then N N p(x, y, z) = (a t x +b t y +c t z) d a(p; (i, j, k)) = atb i tc j t k t=1 t=1

14 Suppose p(x, y, z) = i I (3,d) d! i!j!k! a(p; (i, j, k))x i y j z k. Then N N p(x, y, z) = (a t x +b t y +c t z) d a(p; (i, j, k)) = atb i tc j t k t=1 t=1 If c t 0, this can be rewritten as a(p; (i, j, k)) = N ct d t=1 ( at ) i ( bt c t c t ) j

15 Suppose p(x, y, z) = i I (3,d) d! i!j!k! a(p; (i, j, k))x i y j z k. Then N N p(x, y, z) = (a t x +b t y +c t z) d a(p; (i, j, k)) = atb i tc j t k t=1 t=1 If c t 0, this can be rewritten as a(p; (i, j, k)) = N t=1 c d t or a(p; (i, j, k)) = ( at ) i ( bt c t c t X i Y j dµ, ) j where µ has N masses of weight c d t at ( at c t, bt c t ).

16 ContourPlot x^2 1 ^2 x^2 9 ^2 y^2 1 ^2 y^2 9 ^2 10, x, 4, 4, y, 4,

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19 Out[751]= 6075 x x 4 y x 2 y 6 In[752]:= Out[752]= 23 23

20 In[756]:= ContourPlot 751. z , x, 4, 4, 4 2 Out[756]=

21 Out[748]= 25 x 8 72 x 6 y x 5 y x 4 y x 3 y y z 9720 x z 1440 x y z 9720 y z 8100 z

22 In[750]:= ContourPlot 748. z 1 100, x, 4, 4, 4 2 Out[750]=

23 Thank you for your patience.