Finding All Real Points of a Complex Algebraic Curve Charles Wampler General Motors R&D Center In collaboration with Ye Lu (MIT), Daniel Bates (IMA), & Andrew Sommese (University of Notre Dame)
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 2
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 y Isolated real points Example: [x(x-1)(x-2)] 2 +[y(y-1)y-2)] 2 =0 x 9 isolated points in one complex curve Example: an isolated and a 1-dim real pieces in the same complex curve y 2 -x 2 (x-1)=0 Real dimension = complex dimension 3
Conjugation Let κ : C C be the conjugation operation x is real iff κ(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* κ(x,y)=(conj(y),conj(x)) real part of V(xy-1) consists of pairs of complex conjugate points on the unit circle in C 1 Algorithms require only minor adjustments for different conjugation operators i.e., sweep with a real hyperplane y x y C 1 x 4
Conjugate components Let f(x) be polynomial system with real coefficients If f(z)=0, then f(κ(z))=0 Suppose f -1 (0) has several 1-dimensional irreducible components The components must either: Be self-conjugate, κ(z)=z, or Appear in conjugate pairs, κ(z)=z? Z 5
Real points of conjugate pairs If κ(z)=z? Z, the real points of Z and Z must be their intersection Z Z 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 6
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 z reg(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 + R N (sing(z)) 7
Sweeping out the curve Pick a general real projection π V(π(x)-t) as t varies along R is a sweep hyperplane For general t, V(π(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 8
Morse-like representation Let B* be the real points where the sweep will fail Note: sing(z) B* Let B = π Ζ 1 (π(b*)) R N Morse-like representation of Z consists of The generic real linear projection π The boundary points B={B 1,,B n }, The edges E={e 1,,e k }, where each edge is {x,l,r}: x is a point on an arc of Z R N (l,r) are pointers to B: B l is left endpoint of the arc or l=-8 B r is right endpoint of the arc or r=+8 9
Algorithm for self-conjugate curves 1. Define π 2. Find B* 3. Slice between B* 4. Track to find endpoints and extend B* to B Key: = B* = edge point 10
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 conjugate 3. 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 11
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 W Z, W Z 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 κ(z)=z iff κ(w Z )=W Z With probability 1 This identifies self-conjugacy too 12
Determining B* Let s consider a reduced component, Z Nonreduced introduces technicalities Easy and not so interesting Let a be the sweep direction: π(x)=a T x Find B, a basis for the sweep tangents [a B] is full rank, B T a=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 π x and π y, natural projections B* = π x ((Z P N-2 ) V(h)) B* are isolated ((Z P N-2 ) V(h)) can be any dimension 0,,N-2 Find by a witness cascade 13
Extra points in B* Suppose f(x):c N C m 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 Z Z in B* Not a problem, just makes extra edges Deflation of nonreduced components can similarly add extra points to B* 14
Examples Three easy One substantial A foldable Griffis-Duffy platform 15
Example 1 Suppose f(x)=x 2 +y 2 V(f) is two lines (x,y) = (u,ui) and (x,y)=(u,-ui), u C These are a conjugate pair The lines intersect in the real point (0,0) 16
Example 2 f ( x, y) x( y 1) ( y 1) = 2 (0,1) is an embedded point in V(y-1) 17
Example 3 f ( x, y) 2 2 = y x ( x 1)( x 2) A single irreducible self-conjugate complex curve Double point in B* 18
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? 19
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 Find real points 4 quartic curves Remark: 2x3+3x2+4x4=28 20
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 21
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.) 22
A projection of the real curves Curves live in P 7. Project onto a C 7 patch, then project down to C 2. Key: + = B* = B\B* = edge point 23
Foldable Griffis-Duffy views 24
More views 25
And finally 26
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 27