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3 Feb 2006: f(z) =a(z z0) bc(z z0) d. ,. for some a,b,c,d,z0 C. ... 1 < |z| < 2}, {0 < |z| < 1}, {0 < |z| < }. 8. (i) Let R and S be some Riemann surfaces, f : R S a continuous map, and p a pointin R.

10 Oct 2006: n}.How large can we make a family A of subsets of [n] subject to thecondition that for all A, A′ A, there is some B B with B A A′? ... n}.How large can we make a family A of subsets of [n] subject to thecondition that for all A, A′ A, there is some

8 Oct 2006: GEOMETRY AND GROUPS. These notes are to remind you of the results from earlier courses that we will need at some pointin this course. ... d(xn, ) 0 as n. A subsequence of the sequence (xn) is given by taking only some of the terms in the sequence.

13 Oct 2006: ON SOME -ADIC REPRESENTATIONS OF Gal(Q/Q)ATTACHED TO NONCONGRUENCE SUBGROUPS. A. J. ... Therefore for some a, b Z, azHT b is asemisimple element of g with exactly 2 eigenvalues.

10 May 2006: Earlierproofs of Theorem B for some geodesic and magnetic flows using Fourier analysis canbe found in [12, 18]. ... Systems. 4 (1984) 67–80.[12] V. Guillemin, D. Kazhdan, Some inverse spectral results for negatively curved 2-manifolds,.

13 May 2006: for some a b. If a > b then the first binary quartic has singularreduction above (x : z) = (0 : 1). ... ]for some a b c. If b > c then the first ternary cubic has singularreduction at (x : y : z) = (0 : 0 : 1).

21 Oct 2006: f(z) =a(z z0) bc(z z0) d. ,. for some a,b,c,d,z0 C. ... 1 < |z| < 2}, {0 < |z| < 1}, {0 < |z| < }. 8. (i) Let R and S be some Riemann surfaces, f : R S a continuous map, and p a pointin R.

8 Nov 2006: J : z 7 az bcz d. for some complex numbers a, b, c, d with ad bc = 1. ... For which choices of a, b, c, d is this map J aninvolution, that is J 2 = I?

16 May 2006: A V (G), |A| = m = |b(A)| f (m). for some function f. ... But |B| > |X(<k)|, so B contains some k-set, whencetr B k.

1 Jun 2006: that P1BT = cP2 for some c k. Then fixing A2 and varying B we. ... Then Φ1 = cBT Φ2B for. some B GLm(k) and c k.

8 Nov 2006: Let u =(. ac. ), v =. (bd. )for some integers a, b, c, d with ad bc = 1. ... Show that every vector. v Z Z can be written as mu nv for some integers m and n.

21 Mar 2006: neous cubic polynomial in the generalized Weierstrass normal form XZ24Y 3AX2Y BX3,for some complex constants A, B, A3 27B2 6= 0. ... You will need to recall appropriate results from several topics of the course (and some previousexample(s)).

1 Jun 2006: The numerical properties of anextremal Gorenstein ring were determined by Schenzel [S, Theorem B].We rewrite some of his expressions using the numbers β(r,n) definedin the introduction. ... r,n = {A Z/nZ : A B (1 B) for some |B| r}.

22 Nov 2006: Questions 7–10 are more challenging than others and some parts certainly go beyond limitsof the examination. ... neous cubic polynomial in the generalized Weierstrass normal form XZ24Y 3AX2Y BX3,for some complex constants A, B, A3 27B2 6= 0.

30 Nov 2006: of the first part. 1. The square matrices A and B over the field F are congruent if B = P tAP for some invertible matrixP over F. ... 7. Let S be a real symmetric matrix with Sk = I for some k 1.

4 May 2006: Prove in this case thata map f from the unit interval [0, 1] to B may be “lifted” to E: there is some g : [0, 1] Esuch that p g = f. ... c) From above, we know (x, P ) Cl(Γ) for some x X.

20 Feb 2006: Then γ(b) = γ(a), so exp h(b) = 1and we must have h(b) = 2N πi for some integer N. ... f (w) =. n=0. an(w z)n. that converges to f (w) on some disc B(z, R).

9 Oct 2006: F [A, B] = (det A)p(det B)qF. for some integers p, q. ... We must show that φ is properly equivalent to a Weierstrassmodel πn(0, 0, 0, A, B) for some unique A, B K.

2 Oct 2006: Let kL = kK (b) for some b with bq = a kK kpK , and let u oLbe any lift of b. ... Assume thereexists a surjection (K′/K) (oK′/pλ)d1 for some λ 0. Then.

13 May 2006: 2. Corollary 3.6. If ρR H then ρ = γ for some γ R. ... Ifρ1 = ρ2γ for some γ R then there is an isomorphism of K-algebras.

16 Feb 2006: Then γ(b) = γ(a), so exp h(b) = 1and we must have h(b) = 2N πi for some integer N. ... A domain D is a star domain if it is a star with some centre zo.

9 Mar 2006: Then γ(b) = γ(a), so exp h(b) = 1and we must have h(b) = 2N πi for some integer N. ... f (w) =. n=0. an(w z)n. that converges to f (w) on some disc B(z, R).

6 Apr 2006: Sk musthave the same colour; say cm = cn for some 0 m < n k. ... Now, suppose that we have already constructed Ui1 as required for some i,.

22 Oct 2006: The proof has three stages. First, we briefly review some facts about quasirandom. ... As promised, let us briefly review some of the standard theory of quasirandomness,.