Given gq, degree g is m + 1. Assume q = 1 + hz, ai. Replace z by ei θz in the identity ||g (z)||2− |q(z)|2 = 0 on the sphere. Expand in terms of homogeneous parts. Equate Fourier coefficients. Apply various operations E . The final descendant is
p
q = pm+1+ pm
1 + hz, ai . (∗)
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The expression (*) is a rational sphere map if two things are met hpm+1, pmi = ||z||2mhz, ai (1)
||pm+1||2+ ||z||2||pm||2 = ||z||2m+2+ ||z||2m|hz, ai|2. (2) Polarize (1).
We try:
pm = M(z⊗m) ⊕ 0
pm+1= hz, aiL(z⊗m) ⊕ h.
To make (1) hold we need L = (M−1)∗. Here h is a vector-valued polynomial.
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We need to make (2) hold. After some computation we need
||h||2 = ||z||2 ||z||2m−||M(z⊗m)||2+|hz, ai|2 ||z||2m−||L(z⊗m||2.
In other words, the form on the right side is non-negative definite.
In terms of eigenvalues of M, assume Mv = λv and hence
||L(v )||2= ||v |||λ|22.
After some computation:
||h||2 = ||z||2m(1 − |λ|2) ||z||2−|hz, ai|2
|λ|2 . Using Cauchy-Schwarz, works if ||a|| ≤ |λ| ≤ 1.
We need to make (2) hold. After some computation we need
||h||2 = ||z||2 ||z||2m−||M(z⊗m)||2+|hz, ai|2 ||z||2m−||L(z⊗m||2.
In other words, the form on the right side is non-negative definite.
In terms of eigenvalues of M, assume Mv = λv and hence
||L(v )||2= ||v |||λ|22.
After some computation:
||h||2 = ||z||2m(1 − |λ|2) ||z||2−|hz, ai|2
|λ|2 .
Using Cauchy-Schwarz, works if ||a|| ≤ |λ| ≤ 1.
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To find all solutions: there is an invertible linear M such that pm+1= q1(M−1)∗(z⊗m) ⊕ h.
pm = M(z⊗m) ⊕ 0
If M is unitary, the map is reducible and a polynomial.
Necessary and sufficient condition on eigenvalues of M:
||a|| ≤ |λ| ≤ 1.
Thus: assume degree g is m + 1 and degree q is 1. Assume pq is a final descendant. Then the target space for pq contains an
isomorphic copy of V (n, m). The orthogonal complement contains excess; it is determined up to a unitary.
CN = V ⊕ W . V is isomorphic to V (n, m) and πV(pm) = pm. πV(pm+1) is divisible by q1.
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Theorem
Let f be a proper rational mapping between balls with linear denominator 1 + q1 and of degree m + 1. The following hold:
I There is a finite number of tensor operations such that Es ◦ ... ◦ E1(f ) = pm+1+ pm
1 + q1 . (∗)
I There is an invertible linear M such that pm+1= q1(M−1)∗(z⊗m) ⊕ h.
pm = M(z⊗m) ⊕ 0.
I The Hermitian form defined by
|q1|2 ||z||2m−||(M−1)∗(z⊗m)||2+||z||2 ||z||2m−||M(z⊗m)||2 is positive semi-definite.
Similar ideas hold in general. Given q with q(z) 6= 0 on the sphere, how do we construct p? What are all the possible p?
Write q = 1 + q1+ ... + qk.
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On the sphere we have
On the sphere we have
Equate Fourier coefficients to get k + 1 equations, which hold on the sphere. Then homogenize, using ||z||2b = hz⊗b, z⊗bi.
For each a with 0 ≤ a ≤ k we get identities that hold everywhere:
For a = k,
hpm+k, pmi = qk||z||2m = hqkz⊗m, z⊗mi
hpm+k, pm+1i + hpm+k−1, pmi||z||2= qkq1||z||2m+ qk−1||z||2m+2 and finally (when a = 0)
X
l
||pm+l||2 ||z||2k−2l =X
|ql|2 ||z||2m+2k−2l.
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These identities have a nice interpretation as upper trace identities.
Consider a matrix whose column vectors are A0, ...Ad. These identities say:
hA0, Adi = λd.
hA0, Ad −1i + hA1, Adi = λd −1 X
j
hAj, Aj +ki = λk
X
j
||Aj||2 = λ0.
In domain dimension 1, things simplify.
One can identify a homogeneous polynomial with a constant!
D-Huo-Xiao have used the upper trace identities to give a normal form for polynomial sphere maps from the unit disk, under unitary equivalence in the target.
I’ll just write it out for degree 2 and then for degree 4, the first case hard enough to illustrate the idea.
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Write f (z) = A0+ A1z + A2z2. The normal form becomes
|A0| −α|A2| 0
0 a11 0
0 α|A0| |A2|
where
(|A0|2+ |A1|2)(1 + |α|2) + |a11|2 = 1.
Here α ∈ C; using a source unitary we can make it real.
Now for degree four. Write
f (z) = A0+ A1z + A2z2+ A3z3+ A4z4.
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The condition hA0, A4i = 0 enables us to choose coordinates
Regard the ** as complex parameters. The rest of the identities become upper-trace identities on the three-by-three matrix in the center. Since everything depends only on the inner products, things are unitarily invariant. We can then use a unitary to put the middle matrix in upper triangular form.
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Return to Cn and general denominator.
Because
X||pm+j||2 ||z||2k−2j =X
|qj|2 ||z||2m+2k−2j, there is (a large dimensional!) unitary matrix U with
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.
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Theorem
Let f be a proper rational mapping between balls with denominator of degree k and of degree m + k. The following hold:
I There is a finite number of tensor operations such that
Es ◦ ... ◦ E1(f ) =
I Given q we can construct p as follows.
I Choose vj with hvi, vji = 1 for i 6= j and pm+j = qjz⊗m⊗ vj⊕ fm+j.
I The Hermitian form defined by
||z||2m X
j
|qj|2(1 − |vj|2)||z||2k−2j
is positive semi-definite.
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 Bn→ BN is finite.
Sketch of Proof.
Step 1 is a degree bound: If n ≥ 2, and f is a proper rational map from Bn to BN, then
d ≤ N(N − 1) 2(2n − 3).
(This bound is not sharp. Long ago I conjectured certain bounds. Still open.)
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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 Bn→ BN is finite.
Sketch of Proof.
Step 1 is a degree bound: If n ≥ 2, and f is a proper rational map from Bn to BN, then
d ≤ N(N − 1) 2(2n − 3).
(This bound is not sharp. Long ago I conjectured certain bounds.
Still open.)
Step 2. Let f =pq 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 ei θz. Obtain various linear conditions on the inner products of the vector coefficients. These generalize pν ⊥ pm from before.
Putting it together gives finitely many polynomial inequalities on the coefficients.
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Step 2. Let f =pq 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 ei θz.
Obtain various linear conditions on the inner products of the vector coefficients. These generalize pν ⊥ pm from before.
Putting it together gives finitely many polynomial inequalities on the coefficients.
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.
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Theorem
(D-Lebl) Let Ht 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.
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 Bn→ B2n.
This result has been generalized by Aeryeong Seo to bounded symmetric domains.