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19 Oct 2010: Hint: argue by contradiction.). (6) Let f : B B be a continuous map from the open disc B = {(x, y) R2 : x2 y2 < 1} into it-self. ... b) Assume n = 1. Note that in this case, we may, for any point x = {x1, x2,. ,

29 Jan 2010: 1/2 it is the subspace spanned by the functionφ: bk 7 δ(b) (for b B, k K), and the linear form in (1.2) can be nor-malised so ... hence b = c since (,) is symmetric. Likewise b = a = c hence a = b =c = 0.

12 Oct 2010: Let B = B(u, v) be the bilinear form on K3 uniquely determined by. ... M =. α2 α1α3 α2. . Using that B has rank 1 we compute.

1 Nov 2010: B}. Show thatif A and B are both closed and one of them is bounded then A B is closed. ... b) Is the space R[0, 1] of integrable functions on [0, 1], equipped with the uniform norm, complete?

29 May 2010: γ = γa,b,n : nCa,b nCa,b1. the alternating sum of the Gysin maps. ... Assume:. (a) RH(d 1) holds;(b) YK and ZK satisfy the weight-monodromy conjecture.

29 Jul 2010: H1(R,A) ={a A : a σ(a) = 0}{b σ(b) : b A}. ... 3ab3 b4. f6(a,b) = a6 2. 3a5b 5a4b2 5a2b4 2. 3ab5 b6.

29 Jan 2010: denotes the set of rational primes. Hypothesis (b) includes a weak form of. ... Moreover we have. ords=0 L(M′, s) = 0. Now by hypothesis (b).

17 Feb 2010: φ̂(bng) = |b|fφ̂(g). for all b BQ, n NẐ, g Gf. ... Now if Im(z) > 0 then. η̄ ω =(b>0. Wη. ((b 00 1.

29 Jan 2010: ghi Ý É j k l } m ô n o p q r } ó s t i u < v i w x y z } { | } È » ß k $ ù B B ... d Ç L ü Q ' % ¡ ¢ G % £ ¤ ¥ P ( ª « 6 Õ b b d Q ÿ Æ µ d ù ä ý d º » b ¿ À Á Â?

29 Jan 2010: x a u w t b w , - U k T V U 4 Q ( 8 g _ x ] ] w b _ b c e g h O - 0 4 6 1 k V ... c g h i j k@ A B @ B C B B B B D B E F G D H @ B C I J C @ BB B B D K