Finding All Real Points of a Complex Algebraic Curve

Charles WamplerGeneral Motors R&D Center

In collaboration withYe Lu (MIT), Daniel Bates (IMA), & Andrew Sommese (University of Notre Dame)

2

Outline

Real points in a curve Relationship between a complex curve and its

real points Morse-like representation for real curves

Numerical algorithm Find isolated real points Find boundary points, B, of real curve arcs Find arcs and how they connect to B Find how components meet each other

Examples A Griffis-Duffy platform with a curious real

motion

3

Real Solutions & Complex Dimension

At complex dimension 0, a point is either real or not

Just check the imaginary part A complex curve may contain

Real arcs Isolated real points Example:

[x(x-1)(x-2)]2+[y(y-1)y-2)]2=0 9 isolated points in one complex curve

Example: an isolated and a 1-dim real pieces in the same complex curve

y2-x2(x-1)=0

Real dimension ≤ complex dimension

y

x

4

Conjugation Let : C C be the conjugation operation

x is real iff (x)=x Complex conjugation conj(a+bi)=a-bi

Real part of V(xy-1) is a unit hyperbola Other conjugation operations

Hermitian transpose of a square matrix, A* (x,y)=(conj(y),conj(x))

“real” part of V(xy-1) consists of pairs of complex conjugate points on the unit circle in C1

Algorithms require only minor adjustments for different conjugation operators i.e., sweep with a “real” hyperplane

y

x

C1

y

x

5

Conjugate components

Let f(x) be polynomial system with real coefficients If f(z)=0, then f((z))=0

Suppose f-1(0) has several 1-dimensional irreducible components

The components must either: Be self-conjugate, (Z)=Z, or Appear in conjugate pairs, (Z)=Z’≠Z

6

Real points of conjugate pairs

If (Z)=Z’≠Z, the real points of Z and Z’ must be their intersection ZZ’ These points must be isolated

We can numerically intersect any two algebraic sets Use a diagonal homotopy

This is all we need to handle conjugate pairs.

Self-conjugate components are another matter

7

Local smoothness

Let Z be Quasiprojective complex algebraic set Reduced Dimension 1

Let sing(Z) be its singular points Let reg(Z)=Z\sing(Z) Suppose zreg(Z) is real,

Then there is an open complex neighborhood of z such that the real points of Z in the neighborhood form a smooth connected curve.

Upshot: real points = real smooth arcs + RN(sing(Z))

8

Sweeping out the curve

Pick a general real projection V((x)-t) as t varies along R is a “sweep hyperplane” For general t, V((x)-t) Z consists of isolated

regular points Witness set of size d=deg Z On an open real interval of t, we can numerically sweep out

arcs Sweep fails where:

Tangent to curve lies inside the sweep plane Jacobian matrix at point on curve is corank ≥ 2

9

Morse-like representation

Let B* be the real points where the sweep will fail Note: sing(Z) B*

Let B = B*)) RN

Morse-like representation of Z consists of The generic real linear projection The boundary points B={B1,…,Bn}, The edges E={e1,…,ek}, where each edge is

{x,l,r}: x is a point on an arc of Z RN

(l,r) are pointers to B: Bl is left endpoint of the arc or l=-∞ Br is right endpoint of the arc or r=+∞

10

Algorithm for self-conjugate curves

1. Define 2. Find B*3. Slice between

B*4. Track to find

endpoints and extend B* to B

Key: = B* = edge point

11

Full algorithm in outline

For polynomial system f(x)1. Find the 1-dimensional components

Numerical irreducible decomposition Deflate any nonreduced components

2. Test to find each component’s conjugate3. Intersect conjugate pairs

Use diagonal intersection Result is isolated real points

4. Compute the Morse-like representation for the real part of each self-conjugate component

Result is boundary points B and edges E Some boundary points B* may be isolated

5. (Optional) Find where components not related by conjugation meet

Let’s look at some steps in more detail

12

Determining conjugate pairs

Numerical irreducible decomposition gives us witness sets for the components Let Z,Z’ be components

Move the slicing plane (for all components) to the same generic real plane Let WZ, WZ’ be the witness sets for Z,Z’

Considering all 1-dim components of f-1(0), witness points must appear in conjugate pairs

We have (Z)=Z’ iff (WZ)=WZ’ With probability 1 This identifies self-conjugacy too

13

Determining B*

Let’s consider a reduced component, Z Nonreduced introduces technicalities

Easy and not so interesting Let a be the sweep direction: (x)=aTx Find B , a basis for the sweep tangents

[a B] is full rank, BTa=0 Let J(x) be the Jacobian matrix of f(x) Define h(x,y)=J(x)B y

y parameterizes the tangent space of the sweep Define x and y, natural projections

B* = x((ZPN-2)V(h)) B* are isolated ((ZPN-2)V(h)) can be any dimension 0,…,N-2 Find by a witness cascade

14

Extra points in B*

Suppose f(x):CNCm has m≥N-1 equations

For a 1-dim’l component Z “square up” f(x) as g(x)=Qf(x),

Q is (N-1)m This can introduce an extraneous

component, Z’ Z,Z’ V(g)

We may get extra points ZZ’ in B* Not a problem, just makes extra edges Deflation of nonreduced components can

similarly add extra points to B*

15

Examples

Three easy One substantial

A foldable Griffis-Duffy platform

16

Example 1

Suppose f(x)=x2+y2

V(f) is two lines (x,y) = (u,ui) and (x,y)=(u,-ui), uC

These are a conjugate pair

The lines intersect in the real point (0,0)

17

Example 2

(0,1) is an embedded point in V(y-1)

2)1(

)1(),(

y

yxyxf

18

Example 3

A single irreducible self-conjugate complex curve

)2)(1(),( 22 xxxyyxf

Double point in B*

19

Example 4: Foldable Griffis-Duffy

Griffis-Duffy platform with Equal triangles Joints at

midpoints of sides

Leg lengths all equal to altitude of the triangles

Deg 28 irreducible

motion in Study Coords

(legs not equal)

What about this one?

20

Foldable Platform: Step 1

Numerical Irreducible Decomposition gives 1 double quadric surface

Nonphysical, throw away 12 nonphysical lines (throw these away) 3 double lines (deflate these) 3 quadric curves 4 quartic curves Remark: 2x3+3x2+4x4=28

Find real points

21

Foldable platform (cont)

All 10 components are self-conjugate Each of the 3 lines are completely

determined by their witness point and the tangent at it. Do nothing more to these

Run the algorithm on the other 7 pieces 3 real points show up in the B* set for several

components Each quadric passes through 2 of these

Three quartics have: no real B* no real points on a random real slice hence no real points.

The other quartic has a double point in its B* set

22

The special B* points

An example of how a numerical result can lead one to see an exact result

(The numerics were done to 12 digits. Only 4 are shown.)

23

A projection of the real curves

Curves live in P7.

Project onto a C7 patch,

then project down to C2.

Key: = B* = B\B* = edge point

24

Foldable Griffis-Duffy views

25

More views

26

And finally

27

Summary We have described a method to find the

real points in a complex curve Uses the following operations of numerical

algebraic geometry Irreducible decomposition Deflation Intersection of components Tracking on a component

The foldable Griffis-Duffy is fully analyzed 1 quartic, 3 quadrics, 3 double lines All connect via 4 singularities