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5. 2: Let U ⊂ Cn be open and f : U → H be holomorphic. Suppose that 0 ∈ U. Show that you can write f (z) = ∑α cα zα , where cα ∈ H . 32) where C is the matrix of coefficients and Z is an infinite vector of all the monomials. Now Z does not map all of U into 2 and C may not even be a bounded operator. We have to be a little careful. We can always rescale the variables to ensure that the series for r(z, z¯) has a domain of convergence that includes the point z = (1, 1, . . , 1) in its interior. Then r converges on the polydisc ∆ = ∆1 (0) and complexifies into ∆ × ∆.

The equivalence of the induced maps is by a diagonal matrix. 2. If F : Bn → BN is a proper monomial map of degree 2 or less taking. Then F is spherically equivalent to one map of the form Dz ⊕ ( I − D2 z) ⊗ z , 0 t1 .. 22) with 0 ≤ t1 ≤ t2 ≤ . . ≤ tn ≤ 1. If F(0) = 0, then equivalence is by permuting . tn the variables and components of the map and post-composing with unitary matrix. Proof. Take F and construct p(x) as above. We obtain a polynomial p(x) of degree 2, such that p − 1 is divisible by (x1 + · · · + xn − 1), p(0) = 0, and all coefficients of p are nonnegative.

Tn ). We note that for the special case n = 2 a classification theorem was proved by Ji and Zhang. The fact that the maps are all equivalent to monomial maps is new however. Furthermore, the moduli space is much easier to see in our formulation. We also note one minor point from the proof. As long as the source is a sphere, we can let the target be any hyperquadric and we still obtain that such maps are equivalent to monomial maps except for perhaps two variables. 7 Monomial degree estimates Let us end the course with a quick tour of the degree estimates proof monomial maps.

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