Geometry of LMIs and determinantal representations of algebraic - - PowerPoint PPT Presentation

Geometry of LMIs and determinantal representations of algebraic plane curves Didier HENRION LAAS-CNRS Toulouse, FR Czech Tech Univ Prague, CZ April 2007

Outline

• 1. LMIs and SDP
• 2. Geometry of LMI sets
• 3. Cubic curves
• 4. Contact curves
• 5. Resultants

LMI Linear matrix inequality F(x) = F0 +

n

• i=1

xiFi 0 where Fi are given symmetric real matrices and constraint 0 means positive semidefinite (all eigenvalues real nonnegative) Arise in control theory (Lyapunov 1890, Willems 1971, Boyd et al. 1994), combinatorial optimization, finance, structural me- chanics, and many other areas Key property = convex in x

SDP Decision problem minx

• i cixi

s.t. F0 +

i xiFi 0

Optimization over LMIs = semidefinite programming, versatile generalization of linear (and convex quadratic) programming to the convex cone of positive semidefinite matrices At given accuracy can be solved in polynomial time using interior- point methods (Nesterov, Nemirovski 1994) Many public-domain solvers available

Outline

• 1. LMIs and SDP
• 2. Geometry of LMI sets
• 3. Cubics
• 4. Contact curves
• 5. Resultants

Geometry of LMI sets How does an LMI set F = {x ∈ Rn : F(x) = F0 +

n

• i=1

xiFi 0} look like in Euclidean space ? Matrix F(x) is PSD iff its diagonal minors fi(x) are nonnegative So the LMI set can be described as F = {x ∈ Rn : fi(x) ≥ 0, i = 1, 2, . . .} a convex closed basic semialgebraic set

Semialgebraic formulation For an d-by-d matrix F(x) we have 2d − 1 diagonal minors A simpler criterion follows from the fact that a poly t → f(t) =

• k fd−ktk =

k(t − tk) which has only real roots satisfies tk ≤ 0

iff fk ≥ 0 Apply to characteristic poly f(t, x) = det(tId+F(x)) = d

k=0 fd−k(x)tk

which is monic, i.e. f0(x) = 1 Only d poly ineqs fk(x) ≥ 0 to be checked Polys fk(x) are sums of principal minors of F(x) of order k

• r equivalently sums of k-term-products of eigenvalues of F(x)

Example of 2D LMI feasible set F(x) =

  

1 − x1 x1 + x2 x1 x1 + x2 2 − x2 x1 1 + x2

   0

System of 3 polynomial inequalities fi(x) ≥ 0 The first one is f1(x) = trace F(x) = 4 − x1 ≥ 0

f2(x) = 5 − 3x1 + x2 − 2x2

1 − 2x1x2 − 2x2 2 ≥ 0

x1 x2 −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4

f3(x) = detF(x) = 2−2x1+x2−3x2

1−3x1x2−2x2 2−x1x2 2−x3 2 ≥ 0

x1 x2 −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4

LMI set = intersection of level-sets fk(x) ≥ 0, k = 1, 2, 3

−4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 x1 x2

Boundary of LMI region shaped by determinant Other polys only isolate convex connected component

LMI set or not ?

0.5 1 1.5 2 1 2 3 4 5 6 7 8 9 10 x1 x2

x1x2 ≥ 1 and x1 ≥ 0

LMI x1x2 ≥ 1 and x1 ≥ 0 ⇐ ⇒

• x1

1 1 x2

LMI set or not ?

−1 −0.5 0.5 1 −0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x1 x2

x2 ≥ x2

1

LMI x2 ≥ x2

1

⇐ ⇒

• 1

x1 x1 x2

LMI set or not ?

x1 x2 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1

x2

1 + x2 2 ≤ 1

LMI x2

1 + x2 2 ≤ 1

⇐ ⇒

• 1 + x1

x2 x2 1 − x1

LMI set or not ?

−1 −0.5 0.5 1 −0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x1 x2

{x ∈ R2 : t2 + 2x1t + x2 ≥ 0, ∀t ∈ R}

NOT LMI: not basic semialgebraic

−1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 −0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x1 x2

x2 ≥ x2

1 or x1, x2 ≥ 0

LMI set or not ?

−1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

1 − 2x1 − x2

1 − x2 2 + 2x3 1 ≥ 0

NOT LMI: not connected

−1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

1 − 2x1 − x2

1 − x2 2 + 2x3 1 ≥ 0

LMI set or not ?

−1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

1 − 2x1 − x2

1 − x2 2 + 2x3 1 ≥ 0 and x1 ≤ 1 2

LMI 1 − 2x1 − x2

1 − x2 2 + 2x3 1 ≥ 0 and x1 ≤ 1 2

⇐ ⇒

  

1 x1 x1 1 x2 x2 1 − 2x1

   0

LMI set or not ?

−1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 x1 x2

x4

1 + x4 2 ≤ 1

NOT LMI but projection of an LMI

         

1 + u1 u2 u2 1 − u1 1 x1 x1 u1 1 x2 x2 u2

         

with two liftings u1 and u2

Key questions Which convex closed basic semialgebraic sets are LMI sets ? Which convex closed semialgebraic sets are projections of LMI sets ? In this talk we only address the first question: projections are not allowed

Determinantal representation Consider the non-empty semialgebraic set F = {x ∈ Rn : f(x) ≥ 0} where f(x) is a given polynomial of degree d Without loss of generality, assume that we are given a point e (typically the origin) satisfying f(e) = 1 Since the boundary of an LMI set is shaped by a determinant, can we find symmetric real matrices Fi such that F(x) = F0 +

n

• i=1

xiFi, det F(x) = f(x) So we would like to find a linear symmetric determinantal representation (a symmetric pencil) for polynomial f(x)

Definite determinantal representation = LMI Once we have det F(x) = f(x), we would like to know whether F = closure {x ∈ Rn : det F(x) > 0} ∋ e = {x ∈ Rn : F(x) ≥ 0} Since f(e) = 1, it holds e ∈ int F and F(e) ≻ 0 so the represen- tation must be definite for F to be expressed as an LMI Under which conditions on f can we find such a definite representation ? Define the algebraic curve C = {x ∈ Rn : f(x) = 0} containing the boundary of F

Rigid convexity Necessary condition for F to have a definite symmetric linear determinantal, or LMI representation: Any line passing through an interior point of F must intersect C exactly d times (counting multiplicities and points at infinity) Rigid convexity implies convexity Strong result by Helton and Vinnikov (2002): the condition is also sufficient in the plane, i.e. for n = 2 Also sufficient for n > 2 ?

Cartesian ovals

x1 x2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0.5 1 1.5 2

(3((x1+1)2+x2

2+1)−10(x1+1))2−10((x1+1)2+x2 2+1)+12(x1+1)+1 ≥ 0

Cartesian ovals

x1 x2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0.5 1 1.5 2

Inner oval is rigidly convex hence LMI representable

Constructive methods Checking rigid convexity amounts to checking positive semidefi- niteness of the Hermite matrix of polynomial p(x) for all x Given f(x) and e, once we know that the set F = {x ∈ Rn : f(x) ≥ 0} ∋ e is rigidly convex, how can we systematically build symmetric ma- trices Fi such that F = {x ∈ Rn : F(x) = F0 +

n

• i=1

xiFi 0} and so f(x) = det F(x) ? When/how can we enforce F0 = I ? In the sequel we focus exclusively on the plane case (n = 2)

Vinnikov’s construction (1993) In principle, matrices Fi can be built from polynomial f(x) using complex Riemann surface theory Pencil F(x) can be obtained via the construction of a basis for a complete linear system of the algebraic curve f(x) = 0 Determinantal representations are characterized by line bundles parametrized in the Jacobian variety of the curve Coefficients of Fi can be obtained via the period matrix of the curve and transcendental theta functions evaluated by numerical integration No numerical implementation attempted so far..

Outline

• 1. LMIs and SDP
• 2. Geometry of LMI sets
• 3. Cubics
• 4. Contact curves
• 5. Resultants

Cubics When deg f(x) = 3 the genus of f(x) can be 0 (rational, or singular cubic) or 1 (elliptic, or smooth cubic) Homogeneize f(x0, x1, x2) = x3

0f(x1 x0, x2 x0), define Hessian matrix

H(f(x)) =

• ∂2f(x)

∂xi∂xj

• i,j=0,1,2

which is real symmetric linear, and Hessian h(x) = det H(f(x)) which is cubic An elliptic curve has 9 flexes x∗ (3 of which are real) satisfying f(x∗) = h(x∗) = 0

Parametrized Hessian f(x) and h(x) share the same flexes and we know a symmetric linear determinantal representation for h(x), so use linear homo- topy to find one for f(x) (thanks to Fr´ ed´ eric Han) For real t define parametrized Hessian g(x, t) = det H(h(x) + tf(x)) and find t∗ satisfying g(x∗, t∗) = f(x∗) at a real flex x∗ by solving a cubic equation Three distinct pencils not equivalent by congruence transforma- tion, one of which is definite hence LMI

Elliptic curve Let us find symmetric pencils for f(x) = x3

1 − x2 2 − x1

First build Hessian h(x) = det H(f(x)) = 8(x3

0 + 3x0x2 1 − 3x1x2 2)

Parametrized Hessian g(x, t) = detH(h(x)+tf(x)) = 24t3x0x2

1−576t2x2 0x1+· · ·+110592x3 1

matches f(x) at flex x∗

0 = 0 for

t∗ ∈ {0, 24, −24} yielding the following three representations...

Elliptic curve F 1(x) =

  

1 −x2 x1 −x2 −x1 x1 1

  

F 2(x) = 4−1

3

  

1 + 3x1 −x2 −1 + x1 −x2 −1 − x1 −x2 −1 + x1 −x2 1 − x1

  

F 3(x) = 4−1

3

  

1 − 3x1 −x2 1 + x1 −x2 1 − x1 x2 1 + x1 x2 1 + x1

  

Outline

• 1. LMIs and SDP
• 2. Geometry of LMI sets
• 3. Cubics
• 4. Contact curves
• 5. Resultants

Dixon’s construction

• A. C. Dixon described in 1902 an explicit construction of a sym-

metric pencil of a plane algebraic curve based on the knowledge

• f a contact curve

Given a curve f(x) = 0 of degree d, a contact curve g(x) = 0 is a curve of degree d − 1 touching f(x) = 0 at 1

2d(d − 1) points

(intersection multiplicity of two or more) Once g(x) is given, a determinantal representation for f(x) follows by simple linear algebra Better illustrated with an example..

Adjoint matrix Consider the elliptic curve f(x) = 1 − 2x1 − x2

1 − x2 2 + 2x3 1 = 0

with (known) pencil F(x) =

  

1 x1 x1 1 x2 x2 1 − 2x1

  

Consider its adjoint (or adjugate) matrix V (x) = F −1(x)f(x) =

  

1 − 2x1 − x2

2

x1(−1 + 2x1) x1x2 x1(−1 + 2x1) 1 − 2x1 −x2 x1x2 −x2 1 − x2

1

  

which contains quadratic cofactors

Quotient ring Since F(x)V (x) = f(x)I, the cofactors in a column of V (x) generate a basis for the quotient ring R[x]/f(x)

  

1 x1 x1 1 x2 x2 1 − 2x1

     

1 − 2x1 − x2

2

x1(−1 + 2x1) x1x2 x1(−1 + 2x1) 1 − 2x1 −x2 x1x2 −x2 1 − x2

1

   =   

f(x) f(x) f(x)

  

Net of contact curves For all v ∈ R3 the conic g(x, v) = vTV (x)v = 0 is a contact curve touching f(x) = 0 at 3 points Here are some random contact conics for our cubic..

Contact curves

−2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

Contact curves

−2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

Contact curves

−2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

Contact curves

−2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

Contact parabola

−2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0.5 1 1.5 2 x1 x2

Filling in the adjoint V (x) =

  

1 − 2x1 − x2

2

? ? ? ? ? ? ? ?

  

Let (1,1) be the contact parabola

Filling in the adjoint V (x) =

  

1 − 2x1 − x2

2

? ? x1(−1 + 2x1) ? ? x1x2 ? ?

  

Build basis for quotient ring R[x]/f(x)

Filling in the adjoint V (x) =

  

1 − 2x1 − x2

2

x1(−1 + 2x1) x1x2 x1(−1 + 2x1) ? ? x1x2 ? ?

  

All these contact conics share common touching points

Filling in the adjoint V (x) =

  

1 − 2x1 − x2

2

x1(−1 + 2x1) x1x2 x1(−1 + 2x1) 1 − 2x1 ? x1x2 ? ?

  

(2,2) shares contact points with (1,1) and (2,1)

Filling in the adjoint V (x) =

  

1 − 2x1 − x2

2

x1(−1 + 2x1) x1x2 x1(−1 + 2x1) 1 − 2x1 ? x1x2 ? 1 − x2

1

  

(3,3) shares contact points with (1,1) and (3,1)

Filling in the adjoint V (x) =

  

1 − 2x1 − x2

2

x1(−1 + 2x1) x1x2 x1(−1 + 2x1) 1 − 2x1 −x2 x1x2 −x2 1 − x2

1

  

(3,2) shares contact points with (2,2) and (3,3)

Inverting the adjoint f(x)V −1(x) =

  

1 x1 x1 1 x2 x2 1 − 2x1

  

Invert adjoint matrix to recover linear representation

Dixon’s construction Given f(x) of degree d, and a contact curve g(x) of degree d−1:

• let g(x) be a diagonal entry in V (x)
• derive the whole row and column in V (x) by generating a basis
• f the quotient ring R[x]/f(x)
• derive the remaining entries in V (x) by solving linear equations

in the quotient ring

• compute F(x) = fd−2(x)V (x)−1

A key issue remains: how can we algebraically and systematically find a contact curve g(x) ? Impact of the choice of the contact points ? Definiteness of the representation ?

Outline

• 1. LMIs and SDP
• 2. Geometry of LMI sets
• 3. Cubic curves
• 4. Contact curves
• 5. Resultants

Rational curves An algebraic plane curve {x ∈ R2 : f(x) = 0}

• f genus zero, that is, with a maximal number of singularities,

admits a rational parametrization x1(u) = f1(u) f0(u), x2(u) = f2(u) f0(u) with fi(u) real polys of real indeterminate u Degrees of fi do not exceed degree of f Coeffs of fi chosen in (typically small) algebraic extension

• f the coeff field of f

B´ ezoutian The resultant of the two polys g1(x1, u) = f0(u) − f1(u)x1 g2(x2, u) = f0(u) − f2(u)x2 vanishes whenever they have a common root as polynomials in u (variable to be eliminated) It is equal to the determinant of the B´ ezoutian matrix arising in the quadratic form g1(u)g2(v) − g1(v)g2(u) u − v =

• a,b

[B]a,buavb By multilinearity, F(x) = B(x) is symmetric and linear in x and det F(x) = f(x)

Capricorn curve

x1 x2 −1 −0.5 0.5 1 0.2 0.4 0.6 0.8 1 1.2

f(x) = x2

1(x2 1 + x2 2) − 2(x2 1 + x2 2 − x2)2

Capricorn LMI F(x) =

    

1960 − 868x − 1924y −952 − 940x + 740y −952 − 940x + 740y 776 + 540x + 476y −168 + 180x + 180y −8 − 36x − 84y 56 − 4x − 52y −72 + 20x + 52y −168 + 180x + 180y 56 − 4x − 52y −8 − 36x − 84y −72 + 20x + 52y 40 + 60x + 92y 8 + 20x − 28y 8 + 20x − 28y 8 − 4x − 4y

    

Definite around e = (0, 1/2) hence LMI representable

Bean curve

x1 x2 0.2 0.4 0.6 0.8 1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8

f(x) = x4

1 + x2 1x2 2 + x4 2 − x1(x2 1 + x2 2)

Bean determinant F(x) =

    

x1 x2 x1 x2 x2 1 x2 1 − x1 x1 x2 x2 1 − x1 1 − x1

    

Indefinite around e = (1/2, 0) Not rigidly convex hence not LMI representable

Multivariate B´ ezoutian Given three bivariate polynomial equations g1(x1, x2) = g2(x1, x2) = g3(x1, x2) =

• A. L. Dixon (1908) described a determinantal resultant vanishing

in the ideal I =< g1, g2, g3 > A symmetric linear determinantal representation of a bivariate polynomial f(x1, x2) (sometimes multiplied by an extraneous fac- tor) follows via the parametrized B´ ezoutian of g1(x1, x2, u1, u2) = x1 − u1 g2(x1, x2, u1, u2) = x2 − u2 g3(x1, x2, u1, u2) = f(u1, u2)

• btained by eliminating variables u1 and u2

Multivariate B´ ezoutian Define the discrete differentials ∂1g(u, v) = g(u1, u2) − g(v1, u2) u1 − v1 ∂2g(u, v) = g(v1, u2) − g(v1, v2) u2 − v2 and the quadratic form det

  

g1 ∂1g1 ∂2g1 g2 ∂1g2 ∂2g2 g3 ∂1g3 ∂2g3

   = det   

x1 − u1 −1 x2 − u2 −1 p(u) ∂1p(u, v) ∂2p(u, v)

   =

• a,b[B]a,buavb

Then B(x) is a symmetric pencil (by multilinearity) satisfying det B(x) = f(x)e(x) with (hopefully) e(x) constant

Multivariate B´ ezoutian for quartic For f(x) = 1 − x4

1 − x4 2, using Bernard Mourrain’s multires

package for Maple, we obtained B(x) =

           

−1 x1 x2 x1 −1 x2 −1 x1 −1 x2 −1 x1 −1 x2 −1

           

and det B(x) = −f(x) i.e. e(x) = −1 This 7 × 7 pencil can be reduced to a 4 × 4 pencil via linear algebra in quotient ring R[u, v]/ < g1, g2, g3 > ..but we do not know how to preserve linearity in x

Multivariate B´ ezoutian for cubic For p(x) = 1 − 2x1 − x2

1 − x2 2 + 2x3 1, we obtain

B(x) =

    

1 − 2x1 −x1 −x2 2x1 −x1 1 + 2x1 −2 −x2 1 2x1 −2

    

but we know that for F(x) =

  

1 x1 x1 1 x2 x2 1 − 2x1

  

we have det B(x) = det F(x) = p(x) How can B(x) be reduced to F(x) ?

Open problems For genus zero curves:

• how can we detect/enforce definiteness ?

For positive genus curves:

• how can we find a contact curve ?
• how can we detect/enforce definiteness ?
• for quartics, should we use the bitangents (Edge 1935) ?
• how can we reduce B´

ezoutian pencils ? This is about LMI curves.. ..what about LMI surfaces ?

LMI surfaces Cayley cubic

1 u0 + 1 u1 + 1 u2 + 1 u3 = 0

u0u1u2 + u0u1u3 + u0u2u3 + u1u2u3 = 0 Under involutary linear mapping

    

x0 x1 x2 x3

     = 1

2

    

1 1 1 1 1 1 −1 −1 1 −1 1 −1 1 −1 −1 1

         

u0 u1 u2 u3

    

the dehomogenized (x0 = 1) algebraic equation becomes 1 − x2

1 − x2 2 − x2 3 − 2x1x2x3 = det

  

1 x1 x2 x1 1 x3 x2 x3 1

  

the 3x3 moment matrix of the MAXCUT LMI relaxation

Cayley cubic with 4 nodes

Smooth cubic surfaces

• A. Buckley and T. Koˇ

sir (2006) use tritangent planes to parametrize constructively self-adjoint determinantal representations They prove that the number of non-equivalent representations depends on the Segre type (number and class of real lines among the 27 complex lines of the surface) Segre type pencils definite F1 F2 2 F3 4 F4 6 F5 24 16

Only F5 surfaces have definite pencils (convex component)