By E. Askwith

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**Example text**

9 which says that the transcendence degree of the function field of a compact complex manifold is at most the dimension of the manifold. ��: · · · Let E := (5, . . , 5) and let f be a holomorphic function on an open neighbourhood of the closure of the polydisc B"(O) . e. in the power series expansion non-trivial monomials of degree < k do not occur. If lf(z) l for z E B" (O) can be bounded from above by C, then Proposition 1 . 36 (Schwarz lemma) � (1;1) k lf(z) l C for all z E B"(O) . � Proof.

L�t j = k = 0 and a = 1, then we obtain *1 = � n . Ln 1 = -"'-,n . Thus, vol = -"'-,n . 22). For k = 0, a = 1, and j = 1, the proposition yields *W = (n2 1 ) 1w n-1 . e. 1) ! wn - 2 /\ a. 33 Since L, A, and H are of pure type (1, 1), ( -1, - 1) and (0, 0), respectively, the Lefschetz decomposition is compatible with the bidegree de composition. Thus, P� = EB + = k pp,q, where pp,q = P� n 1\p,q V * . Since A p q and L are real, one also has pp,q = pq,p. 34 In particular, 1\ 0 VC. = P0 • 0 = Pg =

E. w E A 1 ' 1 V,C . 0 Note that two of the three structures { ( , ) , I, w} determine the remaining one. Following a standard procedure, the scalar product and the fundamental form are encoded by a natural hermitian form. 30 1 Local Theory Let (V, ( , ) ) be an euclidian vector space endowed with a compatible complex structure. The form ( , ) := ( , ) - i · w is a positive hermitian form on (V, I) . Lemma 1 . 2 . 1 5 Proof. - linear and (v, v) = (v, v) Moreover, (v, w) (w, v) and = (I(v), w) = = = > 0 for 0 =f.