The Generic Hilbert-Burch matrix AMS meeting at Georgia Southern University March, 2011 Andy Kustin University of South Carolina I have put a copy of this talk on my website 1
The talk consists of: The set-up The questions The people involved The motivation The answer to Question 1 is yes: precise statement and proof The answer to Question 2 is yes: precise statement and proof Extensions Other interpretations 2
The set up: Let k be a field, B=k[x,y] be a polynomial ring in two variables over k, c and d be positive integers with d = 2c, B c be the vector space of homogeneous forms of degree c in B, H d be the affine space of 3 2 matrices with entries from B c, (SoH d is an affine space of dimension 6c+6) and BalH d ={M H d ht(i 2 (M))=2} 3
The set up, page 2: Let A d be the affine space B d B d B d, (Each element g ofa d is a 3-tuple g=(g 1,g 2,g 3 ), with g i B d So,A d is an affine space of dimension 3d+ 3=6c+3), and Φ :H d A d be the morphism P 1 P 2 Φ P 3 P 4 = P 3 P 4 P 5 P 6, P 1 P 2 P 5 P 6, P 1 P 2 P 3 P 4 P 5 P 6 4
The set up, page 3: Notice that: if M is in BalH d, then M is the Hilbert-Burch matrix for Φ(M), 0 B( 3c) B( 3c) M B( 2c) 3 Φ(M) B B/I 2 (M) 0 is a free resolution of B/I 2 (M) and M is a Balanced Hilbert Burch Matrix in the sense that the degrees of the columns of M are as close as possible namely, the column degrees are equal Summary: If g is in Φ(BalH d ), then the ideal generated by g has height two and the Hilbert-Burch matrix for the row vector g is Balanced 5
The Questions: Question 1 Can one separate Φ(BalH d ) from its complement A d \ Φ(BalH d ), in a polynomial manner That is, do there exist polynomials {F i } in 6c+3 variables such that if g is ina d, then g A d \ Φ(BalH d ) F i (the coefficients of g)=0 for all i? Question 2 Does the morphism Φ : BalH d Φ(BalH d ) admit a local section? That is, does there exist an open cover {U j } of Φ(BalH d ) such that, for each index j there exists a morphism σ j : U j A d with the composition U j σ j BalH d Φ A d equal to the identity map on U j for all j 6
The people involved: The original work on the Generic Hilbert-Burch matrix is part of the project with David Cox, Claudia Polini, and Bernd Ulrich Today s talk is part of section 5 of A study of singularities on rational curves via syzygies, which we recently posted on the arxiv Very recently, I asked Brett Barwick to explore various questions about Generic Hilbert-Burch matrices The extensions part of the talk is Brett s work 7
The Motivation: David, Claudia, Bernd, and I are in the business of studying singularities on rational plane curves We fix a parameterization g for the curve and we use information obtained from the Hilbert-Burch matrix for g to describe the singularities of the curve We have results that say When the coefficients of the Hilbert-Burch matrix satisfy all of these polynomials; but not all of those polynomials, then the singularities xxx The coefficients of the parameterization are more natural as data than the coefficients of the Hilbert-Burch matrix The Generic Hilbert-Burch matrix allows us to express our results in terms of the more natural data the coefficients of the parameterization 8
The Answer to both questions is YES The precise answer to Question 1 If g=(g 1,g 2,g 3 ) is ina d with g j = z 0, j x 0 y d + z 1, j x 1 y d 1 + +z d, j x d y 0, then g A d \ Φ(BalH d ) detw = 0, where W is 9
W is the 3c 3c matrix: z 0,1 0 0 z 0,2 0 0 z 0,3 0 0 z 1,1 z 0,1 0 z 1,2 z 0,2 0 z 1,3 z 0,3 0 z 2,1 z 1,1 0 z 2,2 z 1,2 0 z 2,3 z 1,3 0 z d 1,1 z d 2,1 z d 1,2 z d 2,2 z d 1,3 z d 2,3 z d,1 z d 1,1 z d,2 z d 1,2 z d,3 z d 1,3 0 z d,1 0 z d,2 0 z d,3 0 0 z d,1 0 0 z d,2 0 0 z d,3 Each block of columns has c columns for a total of 3c columns There are (d+ 1)+(c 1)=d+ c=3c rows 10
Proof of the answer to Question 1 Here is the significance of W If q is a 3 1 matrix with entries from B c 1 and b is the column vector of coefficients of q, then [g 1 g 2 g 3 ] q=0 W b=0 because = = So, [g 1 g 2 g 3 ] q ] yc 1 xc 1 [g 1 g 2 g 3 y [y d+c 1 x d+c 1 ] W b c 1 xc 1 b yc 1 xc 1 11
g A d \ Φ(BalH d ) there exists a non-zero 3 1 matrix q of forms of degree c 1 ] from B with [g 1 g 2 g 3 q=0 there exists a non-zero 3c 1 matrix b of constants with W b=0 detw = 0 THIS is the critical step 12
The precise answer to Question 2 If g is in Φ(BalH d ), then detw 0 and z 0, j 0 for some j {1,2,3} (Otherwise, the ideal (g 1,g 2,g 3 ) is contained in the ideal (x) and hence has the wrong height) We exhibit σ 1, a section of Φ on the open subset U 1 ={g Φ(BalH d ) z 0,1 0} (The other two members of the open cover of Φ(BalH d ) and the other two σ j are defined similarly) 13
Consider the(3c+1) (3c+3) matrix A: z 0,1 0 0 z 0,2 0 0 z 0,3 0 0 z 1,1 z 0,1 0 z 1,2 z 0,2 0 z 1,3 z 0,3 0 z 2,1 z 1,1 0 z 2,2 z 1,2 0 z 2,3 z 1,3 0 z d 1,1 z d 2,1 z d 1,2 z d 2,2 z d 1,3 z d 2,3 z d,1 z d 1,1 z d,2 z d 1,2 z d,3 z d 1,3 0 z d,1 0 z d,2 0 z d,3 0 0 z d,1 0 0 z d,2 0 0 z d,3 Each block of columns has c+1 columns for a total of 3c+3 columns There are (d+ 1)+(c)=3c+1 rows 14
Here is the significance of A If q Mat 3 (B c ) and b is the column vector of coefficients of q, then [g 1 g 2 g 3 ] q=0 Ab=0 To describe a Hilbert-Burch matrix for g, we need only to produce two linearly independent relations q Mat 3 (B c ) We carefully choose two Eagon-Northcott relations on A THIS is the critical observation 15
Cross out column c+2 of A (This is the FIRST column of the SECOND block of columns The maximal minors of the resulting (3c + 1) (3c + 2) matrix become the relation q 2 = y c + xy c 1 + + x c 0y c + xy c 1 + + x c y c + xy c 1 + + x c on[g 1,g 2,g 3 ], where is the determinant of A with columns c+2 and 2c+3 removed Cross out column 2c+3 of A This is the FIRST column of the THIRD block of columns) The resulting relation is y c + xy c 1 + + x c q 3 = ( 1) c+1 y c + xy c 1 + + x c 0y c + xy c 1 + + x c We have =z 0,1 detw 0 We see that q 2 and q 3 are 0 and lin indept 16
The precise answer to Question 2 Recall the open subset U 1 ={g Φ(BalH d ) z 0,1 0} of Φ(BalH d ) Theorem If σ 1 : U 1 H d is defined by [ ] 1 σ 1 (g)= q z 0,1 (detw) 2 2 q 3, then Φ σ 1 is the identity map on U 1 17
Extensions: Question What happens when one considers Φ : Mat n+1,n (B c ) Mat 1,n+1 (B nc ) instead of Φ : Mat 3,2 (B c ) Mat 1,3 (B 2c )? Answer (Brett Barwick) One gets the same answer Question What happens when d is odd? Say d = 2c + 1 Can one find a local section of the map m i,1 B c, Φ : M =(m i, j ) Mat 3,2 m i,2 B c+1, Im Mat 1,3 (B 2c+1 )? ht(i 2 (M))=2 18
Question 2, repeated Suppose d = 2c+1 Can one find a local section of the map m i,1 B c, Φ : M =(m i, j ) Mat 3,2 m i,2 B c+1, Im Mat 1,3 (B 2c+1 )? ht(i 2 (M))=2 Work in progress (Brett Barwick) One can again describe explicitly the equations which define the complement of Im There is promising evidence that one can again define a local section One must use a much larger open cover of Im 19
Other Interpretations: One can build G j = d z i, j x i y d i, for 1 j 3, i=0 in S=Z[{z i, j }][x,y] One can also build W, q 1, q 2, and q 3 Let w=detw Theorem The following three statements hold (1) (G 1,G 2,G 3 )S w is a perfect height two ideal of S w (2) If F is the complex 0 S( 3c, 3c 3) d 3 S( 3c, 3c 2) 2 d 2 S( 2c, 1) 3 d 1 S, with d 3 = z 0,1 z 0,2, d 2 = [q 1 q 2 q 3 ], d 1 = [G 1,G 2,G 3 ], z 0,3 then F w is a free resolution of S w /(G 1,G 2,G 3 )S w 20
d 3 (3) We split S w S w from F w to produce a Universal Projective Resolution for the graded Betti numbers: (The base ring for this UPR is S w ) Some consequences of 0 B( 3c) 2 B( 2c) 3 B (1) (G 1,G 2,G 3 )S w is a perfect height two ideal of S w are (a) S w /(G 1,G 2,G 3 )S w is a Cohen-Macaulay ring, (b) grade Sw (G 1,G 2,G 3 )=pd Sw S w /(G 1,G 2,G 3 )S w = 2, (c) (The Persistence of Perfection Principal) if N is a noetherian S w -algebra and(g 1,G 2,G 3 )N is a proper ideal of N of grade at least 2, then (G 1,G 2,G 3 )N is a perfect ideal of grade equal to 2 and F w Sw N is a resolution of N/(G 1,G 2,G 3 )N 21