Examples of numerics in commutative algebra and algebraic ...conf/mcaag/talk_slides/... · and...

NumericsResidual Intersection ExampleSections of a Sheaf Example

”Syzygy” Example

Examples of numerics in commutative algebraand algebraic geometry

MCAAG - JuanFest

Chris Peterson

Colorado State University

May 16, 2016

Chris Peterson Examples of numerics in commutative algebra and algebraic geometry



Portions of this talk include joint work with:

Sandra Di RoccoDavid EklundMichael KirbyDan BatesAndrew SommeseHirotachi AboThomas KahleBrent Davis

It has also, no doubt, benefited from discussions and inspirationsfrom many others




Overview

I Homotopy continuation

I Examples: Residual Intersections, Sections of bundles,“Syzygies”




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.




Homotopy Continuation:

In homotopy continuation, a polynomial ideal, I , is cast as amember of a parameterized family of polynomial ideals one ofwhich has known isolated solutions.

Each of the known isolated solutions can be tracked through thefamily to a point which lies numerically close to the algebraic setV (I ) determined by I .

These points can then be later refined to lie within a prescribedtolerance of V (I ).




Example:

t ∗ c ∗ (x5 − 1) + (1− t) ∗ (x5 + 5x + 7)

Start with roots of x5 − 1 and deform to roots of (x5 + 5x + 7) byletting t go from 1 to 0.

c is a random complex number.

One real root x = −1.090942496




Key Aspects:

You can numerically sample generic points from each irreduciblecomponent of an algebraic set over C.

You can determine the degree of each non-embedded irreduciblecomponent of a scheme.




Example involving residual intersections:

Intersection theory provides a connection between the Chernnumbers of an algebraic variety and the equivalence of thealgebraic variety in a complete intersection.

Let X1, . . . ,Xr be hypersurfaces in Pr and let Z be a smoothconnected component of X1 ∩ . . . ∩ Xr .

Fulton defines (X1 · . . . ·Xr )Z , the equivalence of Z for X1 · . . . ·Xr ,as the part of the intersection product supported on Z .




One expression of this equivalence is:

Proposition

(Fulton) Let X1, . . . ,Xr be hypersurfaces in Pr and let Z be asmooth connected component of X1 ∩ . . . ∩ Xr . Let Ni be therestriction of NXi

Pr to Z . Then

(X1 · . . . · Xr )Z = {(Πri=1c(Ni ))c(TPr |Z )−1c(TZ ) ∩ [Z ]}0.




Let ni = degXi , dim(Z ) = n, c0, . . . , cn the Chern classes of Z .

σk denotes the kth elementary symmetric function in n1, . . . , nr .

If we let

ak =n−k∑i=0

(−1)i(r + i

i

)σn−k−i

then we can write: deg (X1 · . . . · Xr )Z =∑n

i=0 aideg ci .




Key Point:

The equivalence of a component, Z , in an intersection ofhypersurfaces can be expressed in terms of the Chern numbers ofZ and the degrees of the hypersurfaces.




Suppose X1 ∩ . . . ∩ Xr is supported on a finite set S . Let mp

denote the intersection multiplicity of X1 · . . . · Xr at p ∈ S . Theclassical version of Bezout’s theorem states that

deg (X1 · . . . · Xr ) =∑p∈S

mp =r∏

i=1

ni .

A refined version of Bezout’s theorem is the following:

Suppose X1 ∩ . . . ∩ Xr consists of a connected component Z and afinite set S then

deg (X1 · . . . · Xr )Z +∑p∈S

mp =r∏

i=1

ni .




Let F1, . . . ,Fr be homogenous polynomials in r + 1 variables.

Suppose (F1, . . . ,Fr ) = I ∩ J with:

I defining a smooth n-dimensional scheme Z

J defining a zero-scheme S (disjoint from Z )

thenn∑

i=0

aideg ci =r∏

i=1

ni − deg S




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




With Homotopy continuation:

We can determine numerically the dimension and degree of eachalgebraic variety appearing in the decomposition of an algebraic set.

We can determine numerically the multiplicity of a non-embeddedcomponent of a non-reduced scheme.

In other words, we can determine the zero dimensional componentsof an ideal and can numerically compute their degrees as schemes.So we can compute

∑p∈S mp.




Case of curves:

If Z is a smooth curve of genus g in Pr then the first Chern classof Z is −KZ and deg (−KZ ) = 2− 2g .

If I = (F1,F2, . . . ,Fr ) ⊂ C[z0, z1, . . . , zr ] is a homogeneous idealwith I = IZ ∩ IS where Z is a curve and S is a zero-scheme then

deg (X1 · . . . · Xr )Z = (n1 + . . .+ nr − (r + 1))degZ + 2− 2g

Which leads to the expression∏i

ni − deg S = (n1 + . . .+ nr − (r + 1))degZ + 2− 2g .




Example 1: Twisted cubic

Let I = (x2 − wy , y2 − xz ,wz − xy) ⊂ C[w , x , y , z ].

If we choose F1,F2,F3 ∈ I of degrees (2, 2, 2) then thecorresponding scheme consists of a degree 3 curve and noadditional points (i.e. S is empty).

The numerical conditions give the relation

2 · 2 · 2− 0 = (2 + 2 + 2− (3 + 1)) · 3 + 2− 2g

which we can solve to get g = 0.




If we choose F1,F2,F3 ∈ I of degrees (2, 3, 4) then we obtain adegree 3 curve and 7 additional points.

The numerical conditions give the relation

2 · 3 · 4− 7 = (2 + 3 + 4− (3 + 1)) · 3 + 2− 2g

again leading to g = 0.

If we did not know the degree of Z , the two computations in thisexample would yield

2 · 2 · 2− 0 = (2 + 2 + 2− (3 + 1))degZ + 2− 2g

2 · 3 · 4− 7 = (2 + 3 + 4− (3 + 1))degZ + 2− 2g

The unique solution to these equations is degZ = 3 and g = 0.




Example 2: Determinantal threefold

Let I be the ideal defined by the 4× 4 minors of a 4× 5 matrix ofgeneral linear forms in C[x0, x1, . . . , x5] and let Z be thecorresponding threefold. The following table shows the degree of Sand the equivalence of Z for various choices.

(n1, n2, n3, n4, n5) Degree of S Equivalence of Z

(4, 4, 4, 4, 4) 0 1024(4, 4, 4, 4, 5) 1 1279(4, 4, 4, 5, 5) 6 1594(4, 4, 5, 5, 5) 21 1979




The formulas from before lead to:

deg c3 + 14 deg c2 + 61 deg c1 + 44 degZ = 1024deg c3 + 15 deg c2 + 71 deg c1 + 65 degZ = 1279deg c3 + 16 deg c2 + 82 deg c1 + 92 degZ = 1594deg c3 + 17 deg c2 + 94 deg c1 + 126 degZ = 1979.

Solution: degZ = 10, deg c1 = 0, deg c2 = 45, deg c3 = −46.




Similar ideas can be used to compute things like c31 , c1c2




Summary:

Intersection theory provides a framework for using residualintersections 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 variouszero-schemes. This can be computed symbolically or numerically.




Example involving sections of a sheaf

The Euler sequence for P2 is the short exact sequence

0→ OP2(−1)

xyz

−−−→ 3OP2 → TP2(−1)→ 0




This arises as a piece of the Koszul complex

0→ OP2(−1)

xyz

−−−→ 3OP2

y −x 0z 0 −x0 z −y

−−−−−−−−−−−→ 3OP2(1)

The columns of the second matrix are a basis for the vector spaceH0(P2, TP2(−1)).




A spanning set for H0(P2, TP2) can be determined by multiplying

each of the columns of

y −x 0z 0 −x0 z −y

by the monomials x , y , z .

This gives the following spanning set for H0(P2, TP2):xy y2 yz −x2 −xy −xz 0 0 0xz yz z2 0 0 0 −x2 −xy −xz0 0 0 xz yz z2 −xy −y2 −yz

Not a basis (but almost): Column 1 + Column 5 + Column 9 = 0.




Matrices and sections of the Tangent Bundle:

Let M denote the vector space of (n + 1)× (n + 1) matrices andlet v = [x0 x1 . . . xn]T

There is a map M → H0(Pn, TPn) given by A→ ∧2[Av |v ].

This map is linear and surjective.

Let SA denote the section corresponding to a matrix A.

It is not hard to check that the kernel of this map is spanned bythe identity matrix. I.e. SA = SB if and only if A = B + λI forsome λ ∈ C.




Example: Consider the matrix A =

1 0 00 2 00 0 3

. Let v =

xyz

then

Av ∧ v =

1 0 00 2 00 0 3

xyz

∧xyz

=

x2y3z

∧xyz

=

−xy−2xz−yz

So SA =

−xy−2xz−yz

= C1 + 2C5 + 3C9.




In general, if A =

a b cd e fg h i

then

SA =

xy y2 yz −x2 −xy −xz 0 0 0xz yz z2 0 0 0 −x2 −xy −xz0 0 0 xz yz z2 −xy −y2 −yz

abcdefghi

.




Summary:

Every square (n + 1)× (n + 1) matrix, A, determines a section, SA,of H0(Pn, TPn).

Two matrices determine the same section if and only if theirdifference is a multiple of the identity matrix.




Eigenschemes

The zero locus of SA determines an ideal IA and a scheme VA. Werefer to VA as the eigenscheme determined by A.

Each connected component of VA 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 structurecorresponds to the generalized eigenvectors.




Remarks:

A is diagonalizable iff IA is a radical ideal.

VA is a zero scheme iff each eigenvalue of A has exactly oneJordan block.

VA is a reduced zeroscheme iff A has n distinct eigenvalues.




Example:

Consider the matrices A =

1 0 00 2 00 0 3

,B =

1 0 00 1 00 0 2

,

C =

1 1 00 1 00 0 1

,D =

1 1 00 1 00 0 2

,E =

1 1 00 1 10 0 1

.

(SA)0 = (ab, ac , bc) = (a, b) ∩ (a, c) ∩ (b, c) =⇒ three points.

(SB)0 = (ac, bc) = (c) ∩ (a, b) =⇒ a line and a point

(SC )0 = (b2, bc) = (b) ∩ (b2, c) =⇒ line with an embedded pt

(SD)0 = (b2, ac, bc) = (a, b) ∩ (b2, c) =⇒ a pt and a double pt

(SE )0 = (b2 − ac, bc, c2) =⇒ a triple point




Example:

Consider the matrices A =

1 0 00 2 00 0 3

,B =

1 0 00 1 00 0 2

,

C =

1 1 00 1 00 0 1

,D =

1 1 00 1 00 0 2

,E =

1 1 00 1 10 0 1

.

(SA)0 = (ab, ac , bc) = (a, b) ∩ (a, c) ∩ (b, c) =⇒ three points.

(SB)0 = (ac, bc) = (c) ∩ (a, b) =⇒ a line and a point

(SC )0 = (b2, bc) = (b) ∩ (b2, c) =⇒ line with an embedded pt

(SD)0 = (b2, ac, bc) = (a, b) ∩ (b2, c) =⇒ a pt and a double pt

(SE )0 = (b2 − ac, bc, c2) =⇒ a triple pointChris Peterson Examples of numerics in commutative algebra and algebraic geometry



Eigenschemes for tensors

Let V be an n + 1-dimensional vector space.

The previous slides describe how to identify elements of V ⊗ Vwith elements of H0(Pn, TPn).

In a similar manner, one can identify elements of Sk+1V ⊗ V withelements of H0(Pn, TPn(k)) and eigenschemes can again be defined.

Fast determination of the zero loci can be achieved by homotopycontinuation through the space of sections of H0(Pn, TPn(k)).




Example: Finding almost syzygies in sets of points:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

Figure: Bunch of points in the plane




Let (xi , yi ) 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

|Axi + Byi + C |√A2 + B2

So if A2 + B2 = 1 then the distance from the point to the line is

|Axi + Byi + C |




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 leastsquares line of best fit.

When p = 1 we get the “median line” of best fit.

What about when p < 1?




We would like to turn the problem of minimizing ‖D‖p with p < 1into a problem involving polynomials.

One way to do this is to find a polynomial P(x) that behaves likexp on the interval [0, a].

If this is accomplished then we can try to find values for A,B,Cthat minimize

n∑i=1

P(|Axi + Byi + C |)

subject to the constraint that A2 + B2 = 1.




That absolute value sign is troublesome so we instead minimize

n∑i=1

P((Axi + Byi + C )2)


However, now we must find a polynomial P(x) that behaves likexp/2 on the interval [0, a].

So how can we find such a polynomial?




Recall that given an inner product space, V , and a set of vectorsv1, . . . , vk in V we can apply Gram Schmidt orthogonalization.

This leads to orthonormal vectors e1, . . . , ek and we can projectelements in V to the space spanned by these vectors by

v → < v , e1 > e1 + · · ·+ < v , ek > ek

We will use the inner product

< f , g > =

∫ 1

0f g dx

and project xp to the span of

1, x , . . . , x11




Here is what the orthogonal basis looks like:

Figure: Orthogonal Basis functions




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

Figure: Degree 11 approximation to x .1 on [0,1]




Lagrange multipliers can be used to generate a polynomial systemwhose solution is the set of critical points of the function

n∑i=1

P((Axi + Byi + C )2)


Numerical methods determines a few hundred critical points.

If we sift through these points, we can find the critical point whichminimizes the function

This can be interpreted as a line of best fit




Recall this picture:

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2




Here is a Line of best fit

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

SolutionData on y = x + 0.1Noise

Figure: Line of best fit




Thank you Juan for being aninspiration, an amazing advisor, and

a great friend


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