Proper mappings and CR Geometry

Transcription

1 Proper mappings and CR Geometry Partially supported by NSF grant DMS John P. D Angelo University of Illinois at Urbana-Champaign August 5, / 71

2 Definition of proper map Assume X, Y are locally compact topological spaces. f : X Y is continuous. f is proper if K compact in Y implies f 1 (K) is compact in X. Alternate characterization connects with CR geometry. Consider the one-point compactifications. F : X Y with F ( ) =, otherwise F (x) = f (x) Then f is proper if and only if F is continuous. 2 / 71

3 Relationship with CR Geometry Assume X and Y are domains in complex manifolds, and their boundaries have some smoothness. If f : X Y is proper, holomorphic, and extends to bx, then its restriction to bx is a CR map. Thus studying proper holomorphic mappings is very close to studying CR maps. Note: For most hypersurfaces M and M, the only CR maps between them are constant. For spheres and hyperquadrics there are many maps. 3 / 71

4 I will focus on the unit sphere, where there are many maps, many results, and many directions for research. Hermitian analogues of Hilbert s 17-th problem Group-invariant examples and representation theory Algebraic combinatorics Variational problems for volumes Classification of rational proper maps 4 / 71

5 Proper maps between balls B n denotes unit ball. f : B n B N proper, holomorphic. Case 1. N < n. There are no examples. The inverse image of a point would be a positive dimensional compact complex analytic subvariety. 5 / 71

6 Proper maps between balls B n denotes unit ball. f : B n B N proper, holomorphic. Case 1. N < n. There are no examples. The inverse image of a point would be a positive dimensional compact complex analytic subvariety. Case 2. N = n. Completely understood. Blaschke products if n = 1, automorphisms if n 2. 6 / 71

7 Proper maps between balls B n denotes unit ball. f : B n B N proper, holomorphic. Case 1. N < n. There are no examples. The inverse image of a point would be a positive dimensional compact complex analytic subvariety. Case 2. N = n. Completely understood. Blaschke products if n = 1, automorphisms if n 2. Case 3. N > n. Non rational examples exist. If the map is smooth and n 2, then the map is rational (Forstnerič) and extends holomorphically past (Cima-Suffridge). My work goes toward classifying the rational maps and connections with the above topics. 7 / 71

8 Theorem A proper holomorphic self-map of the unit disk is a finite Blaschke product: f (z) = e iθ m j=1 z a j 1 a j z. (1) 8 / 71

9 Theorem A proper holomorphic self-map of the unit disk is a finite Blaschke product: f (z) = e iθ m j=1 z a j 1 a j z. (1) Replace each a j in (1) by (1 t)a j and replace θ by (1 t)θ. We get a one-parameter family of Blaschke products H t with H 0 = f and H 1 = z m. Each proper map is homotopic to the map z z m. Furthermore m = 1 f (z) 2πi f (z) dz. Note also: if f = p q, then (Schwarz reflection) p(z) = z m q(1/z). We will generalize to higher dimensions. 9 / 71

10 Theorem (almost known to Poincaré). Proved independently by H. Alexander and Pinchuk in the 1970 s) Assume n = N 2. Assume f : B n B n proper, holomorphic. Then f is an automorphism. 10 / 71

11 Theorem (almost known to Poincaré). Proved independently by H. Alexander and Pinchuk in the 1970 s) Assume n = N 2. Assume f : B n B n proper, holomorphic. Then f is an automorphism. What is Aut(B n )? It is transitive. Here U is unitary, a < 1, and L a (z) = U L a(z) a 1 z, a z, a a s sz. Here s = 1 a 2. Note that an automorphism is homotopic to the identity. 11 / 71

12 algebraic reformulation Write (p; q) for a polynomial map from C n to C N+1. Thus q is scalar-valued and not 0 on the closed unit ball. Without loss of generality p and q have no common factors. We want p 2 q 2 = 0 on sphere. Equivalently, there are polynomial maps f, g such that p 2 q 2 = ( f 2 g 2 )( z 2 1). (1) We call f 2 g 2 the quotient form. 12 / 71

13 algebraic reformulation Write (p; q) for a polynomial map from C n to C N+1. Thus q is scalar-valued and not 0 on the closed unit ball. Without loss of generality p and q have no common factors. We want p 2 q 2 = 0 on sphere. Equivalently, there are polynomial maps f, g such that p 2 q 2 = ( f 2 g 2 )( z 2 1). (1) We call f 2 g 2 the quotient form. Formula (1) is equivalent to saying that (f z) g 2 (g z) f 2 has signature pair (N, 1). In other words, as a Hermitian form, there are N positive and 1 negative eigenvalue. 13 / 71

14 We can also homogenize. Write F (z, t) for the homogenization of f (z), and so on. We get (F (z, t) z) (G(z, t)t) 2 (G(z, t) z) (F (z, t)t) 2 (2) = A(z, t) 2 B(z, t) 2 = P(z, t) 2 Q(z, t) 2. (3) Thus we want to consider all homogeneous polynomial maps F (z, t) and G(z, t) for which (2) has signature pair (N, 1). 14 / 71

15 There are many solutions. In fact, Corollary Let q be a (scalar-valued) polynomial with q 0 on closed ball. Then there is a (vector-valued) polynomial p such that p q is reduced to lowest terms and maps sphere to sphere. If q = 0 on sphere, conclusion is false, by Cima-Suffridge. This result is a kind of Schwarz reflection principle! 15 / 71

16 There are many solutions. In fact, Corollary Let q be a (scalar-valued) polynomial with q 0 on closed ball. Then there is a (vector-valued) polynomial p such that p q is reduced to lowest terms and maps sphere to sphere. If q = 0 on sphere, conclusion is false, by Cima-Suffridge. This result is a kind of Schwarz reflection principle! Corollary Let p q : Cn C K be an arbitrary rational map, reduced to lowest terms, with p q 2 < 1 on closed ball. Then there is a polynomial h : C n C M such that p h q maps sphere to sphere. Thus N = M + K. No bounds on N or deg(h) in terms of K, n and deg(p) possible! 16 / 71

17 Quotient forms p(z) = z has quotient form 1. The automorphism interchanging 0 and a has quotient form (1 a 2 ). p(z) = z m has quotient form 1 + z 2 + z z 2m / 71

18 Quotient forms p(z) = z has quotient form 1. The automorphism interchanging 0 and a has quotient form (1 a 2 ). p(z) = z m has quotient form 1 + z 2 + z z 2m 2. p(z, w) = (z 5, 5z 3 w, 5zw 2, w 5 ) has quotient form f 2 g 2 with signature pair (12, 3). 18 / 71

19 Quotient forms p(z) = z has quotient form 1. The automorphism interchanging 0 and a has quotient form (1 a 2 ). p(z) = z m has quotient form 1 + z 2 + z z 2m 2. p(z, w) = (z 5, 5z 3 w, 5zw 2, w 5 ) has quotient form f 2 g 2 with signature pair (12, 3). Behavior under composition is analogous to the chain rule. If p 2 1 = φ ( z 2 1) and f 2 1 = ψ ( ζ 2 1), then f p 2 1 = (ψ (φ p))( z 2 1). The quotient forms multiply. 19 / 71

20 Orthogonal homogenization Suppose f : C n C N is a rational proper map between balls. Let A be a subspace of C N. We define Ef as follows: Ef = (1 π A )f (π A f z). The map Ef is also proper, but to a higher dimensional target ball. A simple calculation shows that the quotient form for Ef is the quotient form for f plus π A (f ) 2. I have used this idea many times. 20 / 71

21 An isoperimetric inequality from ten years ago: Theorem Let p : B n B N be a proper holomorphic polynomial mapping between unit balls. Let V p denote the 2n-dimensional volume of the image p(b n ) with multiplicity counted. If p is of degree m, then 0 < V p πn m n. n! Equality holds if and only if p is homogeneous of degree m. 21 / 71

22 The idea of the proof. Assume p is not already homogeneous. Write p = p ν p m in terms of homogeneous parts. Condition that p maps sphere to sphere implies p ν is orthogonal to p m. (z e iθ z, Fourier series expansion) Let A be a subspace into which p ν maps, but p m A. Take a partial tensor product Ep = (1 π A )p (z π A p). Ep 2 = p 2 + π A p 2 ( z 2 1). ( ) Ep also maps sphere to sphere, also of degree m, but ν is increased: The step (*) necessarily increases the volume of the image. Keep on going until we get a homogeneous map. Compute the answer explicitly in the homogeneous case. 22 / 71

23 The penultimate step in the previous result relies on the following fact, proved independently by me and by Rudin: His proof used spherical harmonics, mine is elementary and shorter. Theorem Assume p proper between balls, p a homogeneous polynomial of degree m. Assume its components are linearly independent. Then there is a unitary U such that p(z) = Uz m. Equivalently, p(z) 2 = z 2m. ( ) Here the notation means symmetric tensor product. By (*), z m amounts to a list of all monomials of degree m with coefficients equal to square roots of multinomial coefficients. 23 / 71

24 Monge-Ampere variational inequality Let dv 2n denote the usual Euclidean volume form on C n. Let Ω be a bounded domain in C n, with boundary smooth enough to apply Stokes theorem. Consider C 2 functions F, G on the closure of Ω which agree on bω. We seek natural conditions on Ω, F, and G in order to have Φ(F ) = det(f zj z k )dv 2n det(g zj z k )dv 2n = Φ(G). Ω Two years ago, at Serra Negra: The geometry of an L 2 estimate. Hence only a few remarks. Ω 24 / 71

25 In the special case when F = f 2 is the squared norm of a holomorphic mapping, Φ(F ) equals the volume V f, with multiplicity counted, of the image of Ω under f. Note that det(f zj z k )dv 2n = c n Λ n ( F ). 25 / 71

26 In the special case when F = f 2 is the squared norm of a holomorphic mapping, Φ(F ) equals the volume V f, with multiplicity counted, of the image of Ω under f. Note that det(f zj z k )dv 2n = c n Λ n ( F ). Note also that Λ n ( F ) Λ n ( G) = (F G) η, (1) where η = dξ is exact. Finally (F G) = d (F G). The left-hand side of (1) is therefore d (F G) dξ. Hence it is exact and Φ(G) Φ(F ) equals a boundary integral. 26 / 71

27 Let Ω be a bounded domain with C 1 boundary. Let ρ be a defining function for bω, with dρ pointing outward. We say that a (1, 1) form ω is CR positive on bω if, for every non-negative function h, 0 h i ρ Λ n 1 ω. bω When f is holomorphic, the form i 2 ( f 2 ) is CR positive. 27 / 71

28 Theorem Let Ω be a bounded domain in C n, with boundary of class C 2. Assume that F and G are of class C 2 on the closed domain and that F = G on bω. Then Φ(F ) Φ(G) holds if all of the following are satisfied: G F on Ω. The (1, 1) form i F is CR positive. Ω has a defining function which is plurisubharmonic on bω. 28 / 71

29 Theorem Let Ω be a bounded domain in C n, with boundary of class C 2. Assume that F and G are of class C 2 on the closed domain and that F = G on bω. Then Φ(F ) Φ(G) holds if all of the following are satisfied: G F on Ω. The (1, 1) form i F is CR positive. Ω has a defining function which is plurisubharmonic on bω. Corollary Assume that Ω has a real-analytic plurisubharmonic defining function ρ. Suppose that f : Ω C N is holomorphic and extends smoothly to bω. If there is a non-negative function h for which f 2 + hρ is a squared norm g 2, then V g V f. If h > 0 somewhere on bω, then V g > V f. Proofs use Stokes s theorem. 29 / 71

30 New classification result Above we saw how to classify all polynomial proper mappings. For years I have wanted to do the same for all rational ones. I first illustrate when denominator is linear. Assume q = 1 + z, a. Again replace z by e iθ z. By analogy with polynomial case, apply tensor product operations to reduce to f = p q = p m+1 + p m 1 + z, a. ( ) 30 / 71

31 New classification result Above we saw how to classify all polynomial proper mappings. For years I have wanted to do the same for all rational ones. I first illustrate when denominator is linear. Assume q = 1 + z, a. Again replace z by e iθ z. By analogy with polynomial case, apply tensor product operations to reduce to Then (*) is proper if two things are met f = p q = p m+1 + p m 1 + z, a. ( ) p m+1, p m = z 2m z, a (1) p m z 2 p m 2 = z 2m+2 + z 2m z, a 2. (2) 31 / 71

32 We try: p m = ( z m v ) g 0 p m+1 = ( z, a z m v v 2 ) 0 h. Then (1) always holds. (Here v is a constant vector and g, h are vector-valued polynomials.) Two slides hence we find all possibilities. 32 / 71

33 We need to make (2) hold. After some computation we need h 2 + z 2 g 2 = z 2m (1 v 2 ) ( z 2 z, a 2 v 2 ). We need right-hand side to define positive semi-definite form. Using Cauchy-Schwarz, we can always make this work, even choosing v = a. (Need a v < 1.) Mention the subtlety here if time! 33 / 71

34 To find all solutions: there is an invertible linear L such that p m+1 = q 1 L(z m ) 0 h. p m = (L 1 ) (z m ) g 0 If L is unitary, the map is reducible and a polynomial. We will get a necessary and sufficient condition on the eigenvalues of L. 34 / 71

35 Theorem Let f be a proper rational mapping between balls with linear denominator 1 + q 1 and of degree m + 1. The following hold: There is a finite number of tensor operations such that E s... E 1 (f ) = p m+1 + p m 1 + q 1. ( ) There is an invertible linear L such that The Hermitian form defined by p m+1 = q 1 L(z m ) 0 h. p m = (L 1 ) (z m ) g 0 q 1 2( z 2m L(z m ) 2) + z 2( z 2m (L 1 ) (z m ) 2) is positive semi-definite. 35 / 71

36 The condition on the previous page amounts to saying: Put q 1 (z) = z, a. Let λ be an eigenvalue of L. Then we need 1 λ 2 1 a / 71

37 Similar ideas hold in general. Given q with q(z) 0 on the sphere, how do we construct p? What are all the possible p? Write q = 1 + q q k. 37 / 71

38 On the sphere we have k p m+j 2 = j=0 k q j 2 j=0 p m+j, p m+l = q j q l. j,l j,l Replace z by e iθ z and use homogeneity to get: p m+j, p m+l e iθ(j l) = q j q l e iθ(j l). j,l j,l Equate Fourier coefficients to get k + 1 equations, which hold on the sphere. Then homogenize, using z 2b = z b, z b. 38 / 71

39 For each a with 0 a k we get identities that hold everywhere: p m+k, p m = q k z 2m = q k z m, z m p m+k, p m+1 + p m+k 1, p m z 2 = q k q 1 z 2m + q k 1 z 2m+2 and finally (when a = 0) p m+l 2 z 2k 2l = q l 2 z 2m+2k 2l. l 39 / 71

40 Because pm+j 2 z 2k 2j = q j 2 z 2m+2k 2j, there is (a large dimensional!) unitary matrix U with p m+k... p m+j z (k j)... = U p m z k q k z m... q j z (m+k j).... q 0 z (m+k) Thus we could put p m+j = q j z m v j f m+j. For simplicity in the next slide we choose U in a simple way. The method does not construct all possible numerators. One must take additional subspaces into account. 40 / 71

41 Theorem Let f be a proper rational mapping between balls with denominator of degree k and of degree m + k. The following hold: There is a finite number of tensor operations such that E s... E 1 (f ) = k j=0 p m+j k j=0 q. ( ) j Given q we can construct p as follows. Choose v j with v i, v j = 1 for i j and p m+j = q j z m v j f m+j. The Hermitian form defined by z 2m( q j 2 (1 v j 2 ) z 2k 2j) is positive semi-definite. j 41 / 71

42 This work is connected with many classical things. Let us start with the Riesz-Fejer theorem. 42 / 71

43 Theorem (Riesz-Fejer) Let p(e iθ ) be a trig polynomial; assume p(e iθ ) 0 for all θ. Then there is a polynomial f (z) such that on the circle. p(z) = f (z) 2 43 / 71

44 Theorem (Riesz-Fejer) Let p(e iθ ) be a trig polynomial; assume p(e iθ ) 0 for all θ. Then there is a polynomial f (z) such that on the circle. p(z) = f (z) 2 If p is of degree d, then we may take f to be of degree d. p is allowed to have zeroes on the circle. Neither of these properties will hold in higher dimensions. 44 / 71

45 Basic question relating these topics Let X C n and let f be a non-negative function on X. Under what conditions on X and on f can we find a sequence of holomorphic functions {h j (z)} such that f (z, z) = h j (z) 2 = h(z) 2, z X? j=1 Thus f agrees with a Hermitian squared norm on X. This question is too general. 45 / 71

46 In analogy with Riesz-Fejer, assume X is an algebraic set and f (z, z) is a non-negative Hermitian symmetric polynomial. Can we find finitely many holomorphic polynomials h j such that f (z, z) = h(z) 2 = N h j (z) 2 j=1 on X? The answer is no, even for the sphere. The minimum N is the rank or the Hermitian length of h / 71

47 In analogy with Riesz-Fejer, assume X is an algebraic set and f (z, z) is a non-negative Hermitian symmetric polynomial. Can we find finitely many holomorphic polynomials h j such that f (z, z) = h(z) 2 = N h j (z) 2 j=1 on X? The answer is no, even for the sphere. The minimum N is the rank or the Hermitian length of h 2. Gaps in the possible Hermitian lengths. (Huang lemma, result of D-Lebl, recent work by Peter Ebenfelt.) If h 2 = 1 on sphere, then rank( h 2 ) can be 1, n, 2n 1, 3n 2,... but not values in between. 47 / 71

48 In analogy with Riesz-Fejer, assume X is an algebraic set and f (z, z) is a non-negative Hermitian symmetric polynomial. Can we find finitely many holomorphic polynomials h j such that f (z, z) = h(z) 2 = N h j (z) 2 j=1 on X? The answer is no, even for the sphere. The minimum N is the rank or the Hermitian length of h 2. Gaps in the possible Hermitian lengths. (Huang lemma, result of D-Lebl, recent work by Peter Ebenfelt.) If h 2 = 1 on sphere, then rank( h 2 ) can be 1, n, 2n 1, 3n 2,... but not values in between. Result with Lebl: No gaps beyond N n 2 2n / 71

49 Example: A non-negative polynomial need not agree with a Hermitian squared norm on the sphere: f (z, z) = ( z 1 2 z 2 2 ) 2. Then f vanishes on z 1 2 = z 2 2 = 1 2. This zero set would force f to vanish identically, if f agreed with a Hermitian squared norm. 49 / 71

50 Example: A non-negative polynomial need not agree with a Hermitian squared norm on the sphere: f (z, z) = ( z 1 2 z 2 2 ) 2. Then f vanishes on z 1 2 = z 2 2 = 1 2. This zero set would force f to vanish identically, if f agreed with a Hermitian squared norm. For ɛ > 0 consider f ɛ (z, z) = ɛ z 1 2 z ( z 1 2 z 2 2 ) 2. Then f ɛ is strictly positive on the sphere and f ɛ = h ɛ 2 there. Both the rank and the degree of h ɛ tend to as ɛ tends to / 71

51 Theorem (Catlin-D) A polynomial, positive on the unit sphere, agrees with a Hermitian squared norm there. Proof uses compact operator theory on A 2 of the unit ball. It can also be derived from a result of Quillen, which uses A 2 of C n with the Gaussian measure. 51 / 71

52 Theorem (Catlin-D) A polynomial, positive on the unit sphere, agrees with a Hermitian squared norm there. Proof uses compact operator theory on A 2 of the unit ball. It can also be derived from a result of Quillen, which uses A 2 of C n with the Gaussian measure. Theorem (Løw) A continuous function, positive on the boundary X (assumed of class C 2 ) of a bounded strongly pseudoconvex domain Ω, agrees with the squared norm h 2 of a holomorphic mapping there. Even when f is a polynomial, h need not be; it is holomorphic on the domain, but typically cannot be extended holomorphically past the boundary, even when X is algebraic. 52 / 71

53 I wish to discuss the distinction between non-negative and positive. 53 / 71

54 Corollary Assume f : C n C K is a polynomial and f 2 < 1 on closed ball. Then there is a polynomial map g : C n C M such that f g maps sphere to sphere. In other words, we can solve on the sphere. g(z) 2 = 1 f (z) 2 54 / 71

55 Corollary Let p q : Cn C K be an arbitrary rational map, reduced to lowest terms, with p q 2 < 1 on closed ball. Then there is a polynomial h : C n C M such that p h q maps sphere to sphere. Corollary Suppose q is a (scalar-valued) polynomial and q 0 on closed ball. Then there is a (vector-valued) polynomial p such that p q is reduced to lowest terms and maps sphere to sphere. If q = 0 on sphere, conclusion is false, by Cima-Suffridge. 55 / 71

56 In the above results we demand a non-strict inequality. What happens at the end-point? Situation is quite subtle. Example For any k the monomial p 0 (z) = z k n has p on the sphere, and equals 1 somewhere on the sphere. There is a p such that p 2 + p 0 2 = 1 on the sphere. Example Consider p 0 (z) = 4 z 1 z 2 2. Then p on the sphere, and equals 1 somewhere, but there is NO p with p 2 + p 0 2 = 1 on the sphere. 56 / 71

57 In the first example p 0 (z) 2 = 1 for z n 2 = 1. In the second example p 0 (z) 2 = 1 for z 1 2 = z 2 2 = 1 2. There is an obvious necessary condition. The intersection of the level-set p 0 2 = 1 with the unit sphere must agree with the intersection of a complex variety with the sphere. The example 4 z 1 z 2 2 does not satisfy this condition. It also illustrates the distinction between the Riesz-Fejer theorem and the theorem in higher dimensions. 57 / 71

58 Assume f, g : B n B N are proper holomorphic. They are spherically equivalent if there are automorphisms φ, χ such that f = χ g φ. They are homotopy equivalent if there is a one-parameter family of proper holomorphic maps H t such that H 0 = f and H 1 = g. We assume the map (z, t) H t (z) is continuous. 58 / 71

59 Assume f, g : B n B N are proper holomorphic. They are spherically equivalent if there are automorphisms φ, χ such that f = χ g φ. They are homotopy equivalent if there is a one-parameter family of proper holomorphic maps H t such that H 0 = f and H 1 = g. We assume the map (z, t) H t (z) is continuous. Consequence: If H t (z) = c α (t)z α, then each c α is continuous. Since H t (z) 2 = c α (t), c β (t) z α z β, the map to the space of Hermitian forms (with appropriate topology) is also continuous. 59 / 71

60 Spherical equivalence implies homotopy equivalence, converse is false. The target dimension matters. Assume f : B n B N1 and g : B n B N2 are proper holomorphic. They are homotopic in target dimension N if there is a homotopy (of proper maps) between f 0 and g 0. Lemma Any pair of proper holomorphic maps are homotopic in dimension N = N 1 + N 2. In fact, for N = max(n 1, N 2 ) + n. Difficult? Given f, g proper, find the minimum N in which they are homotopic in target dimension N. 60 / 71

61 Surprising fact. The degree is not a homotopy invariant. Example f, g : B 2 B 5. Each map has embedding dimension 5. These maps are of different degree but they are homotopic in target dimension 5. f (z, w) = (z, zw, zw 2, zw 3, w 4 ). g(z, w) = ( w 2, zw, zw 2, z 2 w, z 2 ). With c denoting cosine and s denoting sine, put H t (z, w) = (cz sw 2, zw, (cz sw 2 )(sz + cw 2 ), zw(sz + cw 2 ), (sz + cw 2 ) 2 ). When sin(θ) = t = 0 we obtain f and when t = 1 we obtain g. 61 / 71

62 Automorphisms are homotopic to the identity. But, as the target dimension rises, we get uncountably many spherical equivalence classes, whereas Theorem (D-Lebl) For n 2 and any N, the number of homotopy classes for rational proper maps from B n B N is finite. 62 / 71

63 Automorphisms are homotopic to the identity. But, as the target dimension rises, we get uncountably many spherical equivalence classes, whereas Theorem (D-Lebl) For n 2 and any N, the number of homotopy classes for rational proper maps from B n B N is finite. Sketch of Proof. Step 1 is a degree bound: If n 2, and f is a proper rational map from B n to B N, then N(N 1) d 2(2n 3). (This bound is not sharp. Long ago I conjectured certain bounds. Still open.) 63 / 71

64 Step 2. Let f = p q be proper. Put R = p 2 q 2. R is of degree at most d in z, of total degree at most 2d, and is divisible by z 2 1. R determines a Hermitian form on the finite-dimensional vector space of polynomials of degree at most d. This form has one negative eigenvalue. The (vector) coefficients of p and coefficients of q are bounded. (Assume q(0) = 1.) 64 / 71

65 Step 2. Let f = p q be proper. Put R = p 2 q 2. R is of degree at most d in z, of total degree at most 2d, and is divisible by z 2 1. R determines a Hermitian form on the finite-dimensional vector space of polynomials of degree at most d. This form has one negative eigenvalue. The (vector) coefficients of p and coefficients of q are bounded. (Assume q(0) = 1.) R must be divisible by z 2 1. Equate Fourier coefficients in n variables after replacing z by e iθ z. Obtain various linear conditions on the inner products of the vector coefficients. These generalize p ν p m from before. Putting it together gives finitely many polynomial inequalities on the coefficients. 65 / 71

66 Step 3. Conclusion: A proper rational map corresponds to the intersection of the unit sphere in a finite-dimensional real vector space with a set described by finitely many polynomial inequalities. Such a set is semi-algebraic and can have at most a finite number of components. Each component corresponds to a collection of homotopic rational proper mappings with target dimension at most N. 66 / 71

67 Theorem (D-Lebl) Let H t be a homotopy of rational proper maps between balls. Then either all the maps are spherically equivalent or there are uncountably many spherical equivalence classes. Proof. Brief sketch: First step is to show that the set of t in [0, 1] for which H t is spherically equivalent to H 0 is closed. Second step is to quote a result of Sierpinski: [0, 1] is not a countable union of (non-empty) disjoint closed sets. 67 / 71

68 Theorem (D-Lebl) Let H t be a homotopy of rational proper maps between balls. Then either all the maps are spherically equivalent or there are uncountably many spherical equivalence classes. Proof. Brief sketch: First step is to show that the set of t in [0, 1] for which H t is spherically equivalent to H 0 is closed. Second step is to quote a result of Sierpinski: [0, 1] is not a countable union of (non-empty) disjoint closed sets. Sierpinksi theorem: Let X be a compact, connected Hausdorff space. If {F n } is a closed cover of pairwise disjoint subsets, then one of these sets is X (and the others are empty). 68 / 71

69 The previous result generalizes the following. Theorem If f, g are spherically equivalent proper polynomial maps preserving 0, then they are unitarily equivalent. There exists a one-parameter family of spherically inequivalent quadratic polynomial mappings from B n B 2n. This result has been generalized by Aeryeong Seo to bounded symmetric domains. 69 / 71

70 The next two slides were not shown at the talk. Theorem (Pfister) In R n, each non-negative polynomial is a sum of 2 n squares of rational functions. Note that the number of terms is independent of the degree. Unlike in the real case, not every non-negative Hermitian polynomial r divides a squared norm. Suppose however that r does so; in other words, assume that there is a polynomial s such that rs = p 2. We naturally ask what bounds are possible on the rank of p 2. There is no bound independent of the degree of r. 70 / 71

71 Theorem (D-Lebl) Let M(n, d) denote the dimension of the vector space of polynomials of degree at most d in n variables. Set r(z, z) = (1 + z 2 ) d and let g be a nonzero multiple of r. Then rank(g) M(n, d) (and equality is possible). Corollary Put n = 1 and set r(z, z) = (1 + z 2 ) d. Assume that p 2 is a multiple of r and that p is not the zero polynomial. Then rank( p 2 ) d + 1, and equality is possible. Equality holds with r itself. Expanding by the binomial theorem writes r as a squared norm with rank d / 71