Examples of numerics in commutative algebra and algebraic geo

Transcription

1 Examples of numerics in commutative algebra and algebraic geometry MCAAG - JuanFest Colorado State University May 16, 2016

2 Portions of this talk include joint work with: Sandra Di Rocco David Eklund Michael Kirby Dan Bates Andrew Sommese Hirotachi Abo Thomas Kahle Brent Davis It has also, no doubt, benefited from discussions and inspirations from many others

3 Overview Homotopy continuation Examples: Residual Intersections, Sections of bundles, Syzygies

4 Why Numerics? In some settings, numerics is faster than exact computation. Exactness can sometimes be recovered from approximate answers. For some problems, numerical answers are preferred. Numerics extends the use of AG/CA.

5 Homotopy Continuation: In homotopy continuation, a polynomial ideal, I, is cast as a member of a parameterized family of polynomial ideals one of which has known isolated solutions. Each of the known isolated solutions can be tracked through the family to a point which lies numerically close to the algebraic set V (I ) determined by I. These points can then be later refined to lie within a prescribed tolerance of V (I ).

6 Example: t c (x 5 1) + (1 t) (x 5 + 5x + 7) Start with roots of x 5 1 and deform to roots of (x 5 + 5x + 7) by letting t go from 1 to 0. c is a random complex number. One real root x =

7 Key Aspects: You can numerically sample generic points from each irreducible component of an algebraic set over C. You can determine the degree of each non-embedded irreducible component of a scheme.

8 Example involving residual intersections: Intersection theory provides a connection between the Chern numbers of an algebraic variety and the equivalence of the algebraic variety in a complete intersection. Let X 1,..., X r be hypersurfaces in P r and let Z be a smooth connected component of X 1... X r. Fulton defines (X 1... X r ) Z, the equivalence of Z for X 1... X r, as the part of the intersection product supported on Z.

9 One expression of this equivalence is: Proposition (Fulton) Let X 1,..., X r be hypersurfaces in P r and let Z be a smooth connected component of X 1... X r. Let N i be the restriction of N Xi P r to Z. Then (X 1... X r ) Z = {(Π r i=1c(n i ))c(t P r Z ) 1 c(t Z ) [Z]} 0.

10 Let n i = deg X i, dim(z) = n, c 0,..., c n the Chern classes of Z. σ k denotes the k th elementary symmetric function in n 1,..., n r. If we let then we can write: n k a k = i=0 ( 1) i ( r + i i ) σ n k i deg (X 1... X r ) Z = n i=0 a ideg c i.

11 Key Point: The equivalence of a component, Z, in an intersection of hypersurfaces can be expressed in terms of the Chern numbers of Z and the degrees of the hypersurfaces.

12 Suppose X 1... X r is supported on a finite set S. Let m p denote the intersection multiplicity of X 1... X r at p S. The classical version of Bézout s theorem states that deg (X 1... X r ) = p S m p = r n i. i=1 A refined version of Bézout s theorem is the following: Suppose X 1... X r consists of a connected component Z and a finite set S then deg (X 1... X r ) Z + p S m p = r n i. i=1

13 Let F 1,..., F r be homogenous polynomials in r + 1 variables. Suppose (F 1,..., F r ) = I J with: I defining a smooth n-dimensional scheme Z J defining a zero-scheme S (disjoint from Z) then n r a i deg c i = n i deg S i=0 i=1

14 In order to compute the Chern numbers of a smooth variety, determine enough linearly independent relations that they satisfy.

15 With Homotopy continuation: We can determine numerically the dimension and degree of each algebraic variety appearing in the decomposition of an algebraic set. We can determine numerically the multiplicity of a non-embedded component of a non-reduced scheme. In other words, we can determine the zero dimensional components of an ideal and can numerically compute their degrees as schemes. So we can compute p S m p.

16 Case of curves: If Z is a smooth curve of genus g in P r then the first Chern class of Z is K Z and deg ( K Z ) = 2 2g. If I = (F 1, F 2,..., F r ) C[z 0, z 1,..., z r ] is a homogeneous ideal with I = I Z I S where Z is a curve and S is a zero-scheme then deg (X 1... X r ) Z = (n n r (r + 1))deg Z + 2 2g Which leads to the expression n i deg S = (n n r (r + 1))deg Z + 2 2g. i

17 Example 1: Twisted cubic Let I = (x 2 wy, y 2 xz, wz xy) C[w, x, y, z]. If we choose F 1, F 2, F 3 I of degrees (2, 2, 2) then the corresponding scheme consists of a degree 3 curve and no additional points (i.e. S is empty). The numerical conditions give the relation = ( (3 + 1)) g which we can solve to get g = 0.

18 If we choose F 1, F 2, F 3 I of degrees (2, 3, 4) then we obtain a degree 3 curve and 7 additional points. The numerical conditions give the relation = ( (3 + 1)) g again leading to g = 0. If we did not know the degree of Z, the two computations in this example would yield = ( (3 + 1))deg Z + 2 2g = ( (3 + 1))deg Z + 2 2g The unique solution to these equations is deg Z = 3 and g = 0.

19 Example 2: Determinantal threefold Let I be the ideal defined by the 4 4 minors of a 4 5 matrix of general linear forms in C[x 0, x 1,..., x 5 ] and let Z be the corresponding threefold. The following table shows the degree of S and the equivalence of Z for various choices. (n 1, n 2, n 3, n 4, n 5 ) Degree of S Equivalence of Z (4, 4, 4, 4, 4) (4, 4, 4, 4, 5) (4, 4, 4, 5, 5) (4, 4, 5, 5, 5)

20 The formulas from before lead to: deg c deg c deg c deg Z = 1024 deg c deg c deg c deg Z = 1279 deg c deg c deg c deg Z = 1594 deg c deg c deg c deg Z = Solution: deg Z = 10, deg c 1 = 0, deg c 2 = 45, deg c 3 = 46.

21 Similar ideas can be used to compute things like c 3 1, c 1c 2

22 Summary: Intersection theory provides a framework for using residual intersections to compute Chern numbers of a smooth variety. This is done by producing linear constraints on the Chern numbers.. With enough constraints, you can determine the Chern numbers. The constraints involve the determination of the degree of various zero-schemes. This can be computed symbolically or numerically.

23 Example involving sections of a sheaf The Euler sequence for P 2 is the short exact sequence x y z 0 O P 2( 1) 3O P 2 T P 2( 1) 0

24 This arises as a piece of the Koszul complex x y z 0 O P 2( 1) 3O P 2 y x 0 z 0 x 0 z y 3O P 2(1) The columns of the second matrix are a basis for the vector space H 0 (P 2, T P 2( 1)).

25 A spanning set for H 0 (P 2, T P 2) can be determined by multiplying y x 0 each of the columns of z 0 x by the monomials x, y, z. 0 z y This gives the following spanning set for H 0 (P 2, T P 2): xy y 2 yz x 2 xy xz xz yz z x 2 xy xz xz yz z 2 xy y 2 yz

26 A spanning set for H 0 (P 2, T P 2) can be determined by multiplying y x 0 each of the columns of z 0 x by the monomials x, y, z. 0 z y This gives the following spanning set for H 0 (P 2, T P 2): xy y 2 yz x 2 xy xz xz yz z x 2 xy xz xz yz z 2 xy y 2 yz Not a basis (but almost): Column 1 + Column 5 + Column 9 = 0.

27 Matrices and sections of the Tangent Bundle: Let M denote the vector space of (n + 1) (n + 1) matrices and let v = [x 0 x 1... x n ] T There is a map M H 0 (P n, T P n) given by A 2 [Av v].

28 Matrices and sections of the Tangent Bundle: Let M denote the vector space of (n + 1) (n + 1) matrices and let v = [x 0 x 1... x n ] T There is a map M H 0 (P n, T P n) given by A 2 [Av v]. This map is linear and surjective.

29 Matrices and sections of the Tangent Bundle: Let M denote the vector space of (n + 1) (n + 1) matrices and let v = [x 0 x 1... x n ] T There is a map M H 0 (P n, T P n) given by A 2 [Av v]. This map is linear and surjective. Let S A denote the section corresponding to a matrix A.

30 Matrices and sections of the Tangent Bundle: Let M denote the vector space of (n + 1) (n + 1) matrices and let v = [x 0 x 1... x n ] T There is a map M H 0 (P n, T P n) given by A 2 [Av v]. This map is linear and surjective. Let S A denote the section corresponding to a matrix A. It is not hard to check that the kernel of this map is spanned by the identity matrix. I.e. S A = S B if and only if A = B + λi for some λ C.

31 1 0 0 x Example: Consider the matrix A = Let v = y z then x x x x xy Av v = y y = 2y y = 2xz z z 3z z yz xy So S A = 2xz = C 1 + 2C 5 + 3C 9. yz

32 a b c In general, if A = d e f then g h i a b xy y 2 yz x 2 c xy xz d S A = xz yz z x 2 xy xz e xz yz z 2 xy y 2 yz f g h i

33 Summary: Every square (n + 1) (n + 1) matrix, A, determines a section, S A, of H 0 (P n, T P n). Two matrices determine the same section if and only if their difference is a multiple of the identity matrix.

34 Eigenschemes The zero locus of S A determines an ideal I A and a scheme V A. We refer to V A as the eigenscheme determined by A. Each connected component of V A is supported on a linear space. These linear spaces correspond to the eigenspaces of A. If the scheme is non-reduced then the additional scheme structure corresponds to the generalized eigenvectors.

35 Remarks: A is diagonalizable iff I A is a radical ideal. V A is a zero scheme iff each eigenvalue of A has exactly one Jordan block. V A is a reduced zeroscheme iff A has n distinct eigenvalues.

36 Example: Consider the matrices A = 0 2 0, B = 0 1 0, C = 0 1 0, D = 0 1 0, E =

37 Example: Consider the matrices A = 0 2 0, B = 0 1 0, C = 0 1 0, D = 0 1 0, E = (S A ) 0 = (ab, ac, bc) = (a, b) (a, c) (b, c) = three points.

38 Example: Consider the matrices A = 0 2 0, B = 0 1 0, C = 0 1 0, D = 0 1 0, E = (S A ) 0 = (ab, ac, bc) = (a, b) (a, c) (b, c) = three points. (S B ) 0 = (ac, bc) = (c) (a, b) = a line and a point

39 Example: Consider the matrices A = 0 2 0, B = 0 1 0, C = 0 1 0, D = 0 1 0, E = (S A ) 0 = (ab, ac, bc) = (a, b) (a, c) (b, c) = three points. (S B ) 0 = (ac, bc) = (c) (a, b) = a line and a point (S C ) 0 = (b 2, bc) = (b) (b 2, c) = line with an embedded pt (S D ) 0 = (b 2, ac, bc) = (a, b) (b 2, c) = a pt and a double pt (S E ) 0 = (b 2 ac, bc, c 2 ) = a triple point

40 Eigenschemes for tensors Let V be an n + 1-dimensional vector space. The previous slides describe how to identify elements of V V with elements of H 0 (P n, T P n). In a similar manner, one can identify elements of S k+1 V V with elements of H 0 (P n, T P n(k)) and eigenschemes can again be defined. Fast determination of the zero loci can be achieved by homotopy continuation through the space of sections of H 0 (P n, T P n(k)).

41 Example: Finding almost syzygies in sets of points: Figure: Bunch of points in the plane

42 Let (x i, y i ) be a point in the plane. Consider the line Ax + By + C = 0. The distance from the point to the line can be shown to be Ax i + By i + C A 2 + B 2 So if A 2 + B 2 = 1 then the distance from the point to the line is Ax i + By i + C

43 Let (x 1, y 1 ),..., (x n, y n ) be a bunch of points in the plane. We can consider the vector of distances as a function of A, B, C: D = [ Ax 1 + By 1 + C, Ax 2 + By 2 + C,..., Ax n + By n + C ] We would like to consider various functions of the entries of D. For instance we can define n D p = ( i=1 Ax i + By i + C p ) 1/p

44 When p = 2 this is the 2-norm or length of the vector. The values of A, B, C which minimize D 2 leads to a least squares line of best fit. When p = 1 we get the median line of best fit. What about when p < 1?

45 We would like to turn the problem of minimizing D p with p < 1 into a problem involving polynomials. One way to do this is to find a polynomial P(x) that behaves like x p on the interval [0, a]. If this is accomplished then we can try to find values for A, B, C that minimize n P( Ax i + By i + C ) i=1 subject to the constraint that A 2 + B 2 = 1.

46 That absolute value sign is troublesome so we instead minimize n i=1 P((Ax i + By i + C) 2 ) subject to the constraint that A 2 + B 2 = 1. However, now we must find a polynomial P(x) that behaves like x p/2 on the interval [0, a]. So how can we find such a polynomial?

47 Recall that given an inner product space, V, and a set of vectors v 1,..., v k in V we can apply Gram Schmidt orthogonalization. This leads to orthonormal vectors e 1,..., e k and we can project elements in V to the space spanned by these vectors by v < v, e 1 > e < v, e k > e k We will use the inner product < f, g > = and project x p to the span of 1 0 1, x,..., x 11 f g dx

48 Here is what the orthogonal basis looks like:

49 Here is a degree 11 approximation (in red) to x.1 (in blue) on [0,1]:

50 Lagrange multipliers can be used to generate a polynomial system whose solution is the set of critical points of the function n i=1 P((Ax i + By i + C) 2 ) subject to the constraint that A 2 + B 2 = 1. Numerical methods determines a few hundred critical points. If we sift through these points, we can find the critical point which minimizes the function This can be interpreted as a line of best fit

51 Recall this picture: Numerics

52 Here is a Line of best fit Solution Data on y = x Noise

53 Thank you Juan for being an inspiration, an amazing advisor, and a great friend