By Jiří Lebl

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**Extra info for Hermitian Forms Meet Several Complex Variables: Minicourse on CR Geometry Using Hermitian Forms**

**Example text**

32 CHAPTER 2. CR GEOMETRY In fact, if a linear operator on 2 is bounded then it is defined by a matrix, but we will not prove this fact although it is not hard with some basic functional analysis. See [AG]. Thus if C is as in the discussion before the proposition, the proposition says that C defines a bounded operator on H (note that C being absolutely summable implies it is square summable). 3. Let U ⊂ Cn be open. Suppose that F : U → 2 is holomorphic such that the series for F(z) 2 converges on a neighborhood of the closed unit polydisc ∆.

By setting p2 t = 0 we note that p2 is divisible by (x1 + · · · + xn ). Write a polynomial q = p1 + x1 +···+x . 3. MONOMIAL PROPER MAPS 43 polynomial q is of degree 1, q(0) = 0, and q − 1 is also divisible by (x1 + · · · + xn − 1). Hence, q = x1 + · · · + xn . So p was constructed from q by a partial tensoring operation, which in this language is just p2 multiplication of a part of q by (x1 + · · · + xn ). The p1 corresponds to Dz. Then obviously x1 +···+x n √ must correspond to I − D2 z, where I − D2 ≥ 0.

We can then simply f or g and use the proposition. Note that we will only consider V where the number of positive and negative eigenvalues are not equal. Otherwise we would also want to consider maps that take V to −V . The V we will consider defines the ball and always has only one negative eigenvalue. For the hyperquadric HQ(a, b) we let V be the matrix with a ones and b negative ones on the diagonal. Thus for a sphere, V will always have N ones and one −1. Similarly for J. Let U(N, 1) be the set of matrices M such that M ∗V M = V .