Stable Complete Intersections - math.kth.se fileM. Torrente University of Genova (joint work with L....

Stable Complete Intersections

M. TorrenteUniversity of Genova

(joint work with L. Robbiano)

MEGA 2011, Stockholm

M. Torrente (University of Genova) Stable Complete Intersections MEGA 2011 1 / 22

Introduction

Fact: A complete intersection of n polynomials in n indeterminates has onlya finite number of zeros.

Main question: how do the zeros change when the coefficients of thepolynomials are perturbed?

Motivation: problem of (”almost”) interpolating a set of points whosecoordinates are perturbed by errors.

Our main question can be rephrased as follows: is it possible to...

... devise a strategy to make the situation reasonably stable?

... change the generating polynomials s.t. the stability of the zeros increases?

Aim of the talk is to address these problems using a mixture of methodsfrom classical algebraic geometry and numerical algebra.

Introduction

Motivating example - The Linear Case

I = (1.001 x + y − 1, x + y + 1) I = (x − 2000, x + y + 1)

Solution = (2000,−2001)

The first representation is ill-conditioned (almost parallel lines).

The second representation is much better (though not optimal).

Perturbations:I1 = (1.002 x + y − 1, x + y + 1) I2 = (x − 2000, x + y + 1.001)Solution = (1000,−1001) Solution = (2000,−2001.001).

Note that:

Only one solution has to be stabilized.

It is well-known that for a linear system with n equations and n unknowns,the most stable situation occurs when the coefficient matrix is orthonormal(the 2-norm condition number of the coefficient matrix is 1).

So another question arises: is there an analogue to orthonormality whenwe deal with polynomial systems?

Introduction

I = (1.001 x + y − 1, x + y + 1) I = (x − 2000, x + y + 1)

Solution = (2000,−2001)

Note that:

Introduction

I = (1.001 x + y − 1, x + y + 1) I = (x − 2000, x + y + 1)

Solution = (2000,−2001)

Note that:

Introduction

I = (1.001 x + y − 1, x + y + 1) I = (x − 2000, x + y + 1)

Solution = (2000,−2001)

Note that:

Introduction

I = (1.001 x + y − 1, x + y + 1) I = (x − 2000, x + y + 1)

Solution = (2000,−2001)

Note that:

Introduction

I = (1.001 x + y − 1, x + y + 1) I = (x − 2000, x + y + 1)

Solution = (2000,−2001)

Note that:

Introduction

Plan of the talk

The talk can be divided into two parts: an algebraic part and a numerical part.

The algebraic part:

Problem: given a zero-dimensional smooth complete intersection determinehow far the polynomials coefficients can be perturbed so that the zerosremain smooth and their number does not change.

� discriminant (Gelfand et al.), resultants, tropical geometry, ....

Philosophy: embed the polynomial system f(x) in a family of polynomialsystems F (a, x).

The numerical part:

Problem: introduce a measure of stability of the zeros, and, based on it,determine how to manipulate the equations so that the the stability of thezeros increases.

� condition number Shub-Smale....

Philosopy: introduce a measure of stability which is a generalization of theclassical notion of condition number for linear systems.

Introduction

Plan of the talk

The algebraic part:

The numerical part:

Introduction

Plan of the talk

The algebraic part:

The numerical part:

Introduction

Plan of the talk

The algebraic part:

The numerical part:

Introduction

Plan of the talk

The algebraic part:

The numerical part:

The algebraic part

Notations

K field (K = C or K = R)

x = (x1, . . . , xn) unknowns and a = (a1, . . . , am) parameters

f(x) = {f1(x), . . . , fn(x)} ⊂ K [x] zero-dimensional complete intersection

I ⊂ K [x] ideal generated by f(x)

F (a, x) = {F1(a, x), . . . ,Fn(a, x)} ⊂ K [a, x]

I (a, x) ⊂ K [a, x] ideal generated by F (a, x)

S = AmK scheme of the a-parameters

Φ : Spec(K [a, x]/I (a, x)) −→ S morphism of schemes

The algebraic part

Freeness

Let F (a, x) be a generically zero-dimensional family which contains f(x).

DefinitionA dense Zariski-open subscheme U of S such that Φ−1(U) −→ U is free,is said to be an I -free subscheme of S.

PropositionLet σ be a term ordering on Tn

let G (a, x) be the reduced σ-Grobner basis of I (a, x)K (a)[x]let d(a) be the l.c.m. of all denominators of the polys coefficients of G (a, x)let T = Tn \ LTσ(I (a, x)K (a)[x]). Then:

(a) The open subscheme U of S defined by d(a) 6= 0 is I -free.

(b) The multiplicity of each fiber over U coincides with the cardinality of T.

The algebraic part

Freeness

The algebraic part

Freeness

The algebraic part

Smoothness

DefinitionA dense Zariski-open subscheme U of S such that Φ−1(U) −→ U is smooth,i.e. all the fibers of Φ−1(U) −→ U are zero-dimensional smooth completeintersections, is said to be an I -smooth subscheme of S.

TheoremLet D(a, x) = det(JacF (a, x)) be the det of the Jacobian of F (a, x) ∈ K [a][x]let J(a, x) = I (a, x) + (D(a, x)) in K [a, x]let H = J(a, x) ∩ K [a]. Then:

(a) There exists an I -smooth subscheme of S if and only if H 6= (0).

(b) Let 0 6= h(a) ∈ H; the open subscheme U of S defined by h(a) 6= 0 isI -smooth.

The algebraic part

Smoothness

DefinitionA dense Zariski-open subscheme U of S such that Φ−1(U) −→ U is smooth,i.e. all the fibers of Φ−1(U) −→ U are zero-dimensional smooth completeintersections, is said to be an I -smooth subscheme of S.

TheoremLet D(a, x) = det(JacF (a, x)) be the det of the Jacobian of F (a, x) ∈ K [a][x]let J(a, x) = I (a, x) + (D(a, x)) in K [a, x]let H = J(a, x) ∩ K [a]. Then:

(a) There exists an I -smooth subscheme of S if and only if H 6= (0).

(b) Let 0 6= h(a) ∈ H; the open subscheme U of S defined by h(a) 6= 0 isI -smooth.

The algebraic part

Freeness + Smoothness = Optimality

DefinitionA dense Zariski-open subscheme U of S such Φ−1(U) −→ Uis free and smooth is said to be I -optimal.

TheoremThere is an algorithm which computes an I -optimal subscheme of S.

The algebraic part

An example

We consider the ideal I = (xy + 1, x2 + y2 − 5) ∈ R[x , y ] embedded into thefamily I (a, x , y) = (xy + a1x + 1, x2 + y2 + a2).

Freeness

We compute the reduced Lex-Grobner basis of I (a, x)K (a)[x , y ]

g1 = x − y3 − a1y2 − a2y − a1a2,

g2 = y4 + 2a1y3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1)

There is no condition for the free locus.

Smoothness

We compute D(a, x , y) = det(JacF (a, x , y)) = −2x2 + 2y2 + 2a1y ,J(a, x , y) = I (a, x , y) + (D(a, x , y)) and H = J(a, x , y) ∩ K [a] = (h(a))

h(a) = a61a2 + 3a41a22 + a41 + 3a21a

32 + 20a21a2 + a42 − 8a22 + 16

Optimality

An I -optimal subscheme is U = {α ∈ A2R : h(α1, α2) 6= 0}.

The algebraic part

An example

Freeness

g1 = x − y3 − a1y2 − a2y − a1a2,

g2 = y4 + 2a1y3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1)

Smoothness

h(a) = a61a2 + 3a41a22 + a41 + 3a21a

32 + 20a21a2 + a42 − 8a22 + 16

Optimality

The algebraic part

An example

Freeness

g1 = x − y3 − a1y2 − a2y − a1a2,

g2 = y4 + 2a1y3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1)

Smoothness

h(a) = a61a2 + 3a41a22 + a41 + 3a21a

32 + 20a21a2 + a42 − 8a22 + 16

Optimality

The algebraic part

An example

Freeness

g1 = x − y3 − a1y2 − a2y − a1a2,

g2 = y4 + 2a1y3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1)

Smoothness

h(a) = a61a2 + 3a41a22 + a41 + 3a21a

32 + 20a21a2 + a42 − 8a22 + 16

Optimality

The algebraic part

What about real zeros?

Let f(x) be a zero-dimensional complete intersection in R[x]let F (a, x) ∈ R[a, x] be generically zero-dimensional family containing f(x).

TheoremAssume that there exists an I -optimal subscheme U of Am

R ;let αI ∈ U be the point in the parameter space corresponding to I = (f(x))let µR,I be the number of real points in the fiber over αI (= zeroes of I )

Then:there exists an open semi-algebraic subscheme V of U s.t.∀ α ∈ V the number of real points in the fiber over α is µR,I .

� proof Shape Lemma + theorem Basu-Pollack-Roy

The algebraic part

An example: continuation 1

Going back to the example ...

An I -optimal subscheme isU = {α ∈ A2

R : h(α) 6= 0}Therefore, for h(a) 6= 0 each fiberis smooth and has multiplicity 4hence it consists of 4 distinct complexpoints.

What about real points?

αI = (0,−5) ∈ U and µR,I = 4

The real curve h(a) = 0 is the unionof two branches and point (0, 2).

Region R1 (with the exception of (0, 2)) : four complex non-real points;

Regions R2 and R3: two complex non-real points and two real points;

Region R4: four real points.

The algebraic part

αI = (0,−5) ∈ U and µR,I = 4

The algebraic part

αI = (0,−5) ∈ U and µR,I = 4

The algebraic part

αI = (0,−5) ∈ U and µR,I = 4

The algebraic part

αI = (0,−5) ∈ U and µR,I = 4

The algebraic part

To describe the four regions algebraically, we usethe Sturm-Habicht sequence ofg2 = y4 + 2a1y

3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1) ∈ R(a)[y ].

The leading monomials of the sequence are

y4, 4y3, 4r(a)y2, −8`(a)y , 16h(a)

where r(a) = a21 − 2a2 and `(a) = a41a2 + 2a21a22 + 2a21 + a32 − 4a2.

To get the total number of real roots we countthe sign changes in the sequence at −∞ and +∞;in particular, we observe that in the parameter spacethe ideal I corresponds to the point αI = (0,−5) ∈ R4.

We getR4 = {α ∈ R2 | r(α) > 0, `(α) < 0, h(α) > 0}

which is semi-algebraic open, not Zariski-open.

The algebraic part

3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1) ∈ R(a)[y ].

y4, 4y3, 4r(a)y2, −8`(a)y , 16h(a)

We getR4 = {α ∈ R2 | r(α) > 0, `(α) < 0, h(α) > 0}

The algebraic part

3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1) ∈ R(a)[y ].

y4, 4y3, 4r(a)y2, −8`(a)y , 16h(a)

We getR4 = {α ∈ R2 | r(α) > 0, `(α) < 0, h(α) > 0}

The algebraic part

3 + (a21 + a2)y2 + 2a1a2y + (a21a2 + 1) ∈ R(a)[y ].

y4, 4y3, 4r(a)y2, −8`(a)y , 16h(a)

We getR4 = {α ∈ R2 | r(α) > 0, `(α) < 0, h(α) > 0}

The numerical part

Admissible Perturbations

Let f(x) be a zero-dimensional smooth complete intersection in R[x]let µR,I be the number of real solutions of f(x) = 0

DefinitionLet ε(x) = {ε1(x), . . . , εn(x)} be a set of polynomials in R[x].Let V ⊂ Am

R be an open semi-algebraic subset of U such that

αI ∈ V∀ α ∈ V the number of real roots of F (α, x) = 0 is µR,I .

If there exists α ∈ V such that (f + ε)(x) = F (α, x),then ε(x) is called an admissible perturbation of f(x) .

The numerical part

Admissible Perturbations

Let f(x) be a zero-dimensional smooth complete intersection in R[x]let µR,I be the number of real solutions of f(x) = 0

DefinitionLet ε(x) = {ε1(x), . . . , εn(x)} be a set of polynomials in R[x].Let V ⊂ Am

R be an open semi-algebraic subset of U such that

αI ∈ V∀ α ∈ V the number of real roots of F (α, x) = 0 is µR,I .

If there exists α ∈ V such that (f + ε)(x) = F (α, x),then ε(x) is called an admissible perturbation of f(x) .

The numerical part

Example of admissible perturbation

Let f = {f1, f2} ∈ R[x , y ] and I = (f1, f2) where

f1 = xy − 6f2 = x2 + y2 − 13

and ZR(f) = {(−3,−2), (3, 2), (−2,−3), (2, 3)}

We embed f(x) in F (a, x) = {xy + a1, x2 + a2y

2 + a3}.The semi-algebraic open set

V = {α ∈ R3 | α23 − 4α2

1α2 > 0, α2 > 0, α3 < 0}

is a subset of the I -optimal scheme U = {α ∈ A3R | α2(α2

3 − 4α21α2) 6= 0}

αI = (−6, 1,−13) ∈ Vthe fiber over each α ∈ V consists of 4 real points

The polynomial set ε(x , y) = {2, 54y

2} is an admissible perturbation of f(x , y).The real roots of (f + ε)(x) = 0 are

ZR(f + ε) ={(−3,− 4

),(3, 43

), (−2,−2), (2, 2)

The numerical part

f1 = xy − 6f2 = x2 + y2 − 13

and ZR(f) = {(−3,−2), (3, 2), (−2,−3), (2, 3)}

2 + a3}.

The semi-algebraic open set

V = {α ∈ R3 | α23 − 4α2

1α2 > 0, α2 > 0, α3 < 0}

3 − 4α21α2) 6= 0}

ZR(f + ε) ={(−3,− 4

),(3, 43

), (−2,−2), (2, 2)

The numerical part

f1 = xy − 6f2 = x2 + y2 − 13

and ZR(f) = {(−3,−2), (3, 2), (−2,−3), (2, 3)}

V = {α ∈ R3 | α23 − 4α2

1α2 > 0, α2 > 0, α3 < 0}

3 − 4α21α2) 6= 0}

ZR(f + ε) ={(−3,− 4

),(3, 43

), (−2,−2), (2, 2)

The numerical part

f1 = xy − 6f2 = x2 + y2 − 13

and ZR(f) = {(−3,−2), (3, 2), (−2,−3), (2, 3)}

V = {α ∈ R3 | α23 − 4α2

1α2 > 0, α2 > 0, α3 < 0}

3 − 4α21α2) 6= 0}

αI = (−6, 1,−13) ∈ V

the fiber over each α ∈ V consists of 4 real points

ZR(f + ε) ={(−3,− 4

),(3, 43

), (−2,−2), (2, 2)

The numerical part

f1 = xy − 6f2 = x2 + y2 − 13

and ZR(f) = {(−3,−2), (3, 2), (−2,−3), (2, 3)}

V = {α ∈ R3 | α23 − 4α2

1α2 > 0, α2 > 0, α3 < 0}

3 − 4α21α2) 6= 0}

ZR(f + ε) ={(−3,− 4

),(3, 43

), (−2,−2), (2, 2)

The numerical part

f1 = xy − 6f2 = x2 + y2 − 13

and ZR(f) = {(−3,−2), (3, 2), (−2,−3), (2, 3)}

V = {α ∈ R3 | α23 − 4α2

1α2 > 0, α2 > 0, α3 < 0}

3 − 4α21α2) 6= 0}

ZR(f + ε) ={(−3,− 4

),(3, 43

), (−2,−2), (2, 2)

}M. Torrente (University of Genova) Stable Complete Intersections MEGA 2011 14 / 22

The numerical part

Local Condition Number

Let p be a nonsingular real solution of f(x) = 0let ε(x) be an admissible perturbation of f(x)

DefinitionThe number κ(f, p) = ‖ Jacf(p)−1‖‖ Jacf(p)‖is called the local condition number of f(x) at p.

CorollaryLet p + ∆p be the real solution of (f + ε)(x) = 0 corresponding to p. Then

0 = (f + ε)(p + ∆p) ≈ ε(p) + Jacf+ε(p) ∆p (1)

and the approximate solution of (1) is denoted by ∆p1 = − Jacf+ε(p)−1ε(p).

Theorem (Local Condition Number)Under the condition ‖ Jacf(p)−1 Jacε(p)‖ < 1, we have

‖∆p1‖‖p‖

≤ Λ(f, ε, p) κ(f, p)

(‖ Jacε(p)‖‖ Jacf(p)‖

+‖ε(0)− ε≥2(0, p)‖‖f(0)− f≥2(0, p)‖

where Λ(f, ε, p) = 1/(1− ‖ Jacf(p)−1 Jacε(p)‖).

The numerical part

0 = (f + ε)(p + ∆p) ≈ ε(p) + Jacf+ε(p) ∆p (1)

‖∆p1‖‖p‖

≤ Λ(f, ε, p) κ(f, p)

+‖ε(0)− ε≥2(0, p)‖‖f(0)− f≥2(0, p)‖

The numerical part

0 = (f + ε)(p + ∆p) ≈ ε(p) + Jacf+ε(p) ∆p (1)

‖∆p1‖‖p‖

≤ Λ(f, ε, p) κ(f, p)

+‖ε(0)− ε≥2(0, p)‖‖f(0)− f≥2(0, p)‖

The numerical part

0 = (f + ε)(p + ∆p) ≈ ε(p) + Jacf+ε(p) ∆p (1)

‖∆p1‖‖p‖

≤ Λ(f, ε, p) κ(f, p)

+‖ε(0)− ε≥2(0, p)‖‖f(0)− f≥2(0, p)‖

where Λ(f, ε, p) = 1/(1− ‖ Jacf(p)−1 Jacε(p)‖).M. Torrente (University of Genova) Stable Complete Intersections MEGA 2011 15 / 22

The numerical part

Generalization of the Linear Case

The notion of local condition number is a generalization of the classical notion ofcondition number of linear systems.

f(x) linear implies f(x) = Ax− b with A ∈ Matn(R) invertibleJacf(x) = A

and so κ(f, p) = ‖ Jacf(p)−1‖‖ Jacf(p)‖ = ‖A−1‖‖A‖

Further, using the perturbation ε(x) = ∆Ax−∆b, relation (2) becomes

‖∆p‖‖p‖

1− ‖A−1‖ ‖∆A‖‖A−1‖ ‖A‖

(‖∆A‖‖A‖

+‖∆b‖‖b‖

)which is the relation that quantifies the sensitivity of the Ax = b problem.

The numerical part

‖∆p‖‖p‖

1− ‖A−1‖ ‖∆A‖‖A−1‖ ‖A‖

(‖∆A‖‖A‖

+‖∆b‖‖b‖

The numerical part

‖∆p‖‖p‖

1− ‖A−1‖ ‖∆A‖‖A−1‖ ‖A‖

(‖∆A‖‖A‖

+‖∆b‖‖b‖

The numerical part

‖∆p‖‖p‖

1− ‖A−1‖ ‖∆A‖‖A−1‖ ‖A‖

(‖∆A‖‖A‖

+‖∆b‖‖b‖

The numerical part

Further properties and Optimization

Further properties

‖ · ‖ is and induced matrix norm ⇒ κ(f, p) ≥ 1

κ2(f, p) = σmax(Jacf (p))σmin(Jacf (p))

κ2(f, p) = 1 ⇐⇒ Jacf(p) orthonormal

κ(f, p) invariant under multiplication by unique nonzero scalar γ ∈ R

PropositionSuppose that deg(f1) = . . . = deg(fn)let C = (cij) ∈ Matn(R) invertible, and g be defined by gtr = C · ftr .Then the following conditions are equivalent:

(a) κ2(g, p) = 1, the minimum possible

(b) C tC = (Jacf(p) Jacf(p)t)−1

The numerical part

Further properties and Optimization

Further properties

‖ · ‖ is and induced matrix norm ⇒ κ(f, p) ≥ 1

κ2(f, p) = σmax(Jacf (p))σmin(Jacf (p))

κ2(f, p) = 1 ⇐⇒ Jacf(p) orthonormal

κ(f, p) invariant under multiplication by unique nonzero scalar γ ∈ R

PropositionSuppose that deg(f1) = . . . = deg(fn)let C = (cij) ∈ Matn(R) invertible, and g be defined by gtr = C · ftr .Then the following conditions are equivalent:

(a) κ2(g, p) = 1, the minimum possible

(b) C tC = (Jacf(p) Jacf(p)t)−1

Experiments

Experiments I

We consider f = {f1, f2} ∈ R[x , y ] and I = (f1, f2) where

f1 = 14x

2y + xy2 + 14y

3 + 15x

2 − 58xy + 13

40y2 + 9

40x −35y + 1

f2 = x3 + 1413xy

2 + 5752x

2 − 2552xy + 8

13y2 − 11

52x −413y −

and the point p = (0, 1) ∈ ZR(f).

the system g = {f1, 6316 f1 −6516 f2} is an alternative representation of I

Experiments

Experiments I

f1 = 14x

2y + xy2 + 14y

3 + 15x

2 − 58xy + 13

40y2 + 9

40x −35y + 1

f2 = x3 + 1413xy

2 + 5752x

2 − 2552xy + 8

13y2 − 11

52x −413y −

Experiments

Experiments I

-0,5 -0,25 0 0,25 0,5

Experiments

Experiments I

f1 = 14x

2y + xy2 + 14y

3 + 15x

2 − 58xy + 13

40y2 + 9

40x −35y + 1

f2 = x3 + 1413xy

2 + 5752x

2 − 2552xy + 8

13y2 − 11

52x −413y −

q ∈ ZR(f + ε) and r ∈ ZR(g + ε) are perturbations of pfor different admissible perturbations ε

κ2(f, p) ‖q−p‖2‖p‖2

8 0.000097

κ2(g, p) ‖r−p‖2‖p‖2

1 0.000023

Experiments

Experiments II

We consider f = {f1, f2, f3} ∈ R[x , y , z ] and I = (f1, f2, f3) where

f1 = 617x

2 + xy − 2485x −

885y −

f2 = 3989x

2 + 7089xy + yz − 39

89x + 1089y

f3 = y2 + 2xz + z2 − z

and the point p = (1, 0, 0) ∈ ZR(f).

the system g = {f1, 7564123 f1 −7565123 f2, f3} is an alternative representation of I

κ2(f, p) ‖q−p‖2‖p‖2

123 0.0436

κ2(g, p) ‖r−p‖2‖p‖2

1 0.0221

Experiments

Experiments II

f1 = 617x

2 + xy − 2485x −

885y −

f2 = 3989x

2 + 7089xy + yz − 39

89x + 1089y

f3 = y2 + 2xz + z2 − z

κ2(f, p) ‖q−p‖2‖p‖2

123 0.0436

κ2(g, p) ‖r−p‖2‖p‖2

1 0.0221

Experiments

Experiments II

f1 = 617x

2 + xy − 2485x −

885y −

f2 = 3989x

2 + 7089xy + yz − 39

89x + 1089y

f3 = y2 + 2xz + z2 − z

κ2(f, p) ‖q−p‖2‖p‖2

123 0.0436

κ2(g, p) ‖r−p‖2‖p‖2

1 0.0221

Future work

Optimization in the case of arbitrary degrees

Definition of global condition number

Thank you!

Future work

Optimization in the case of arbitrary degrees

Definition of global condition number

Thank you!

Stable Complete Intersections - math.kth.se fileM. Torrente University of Genova (joint work with L....

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