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5 Nov 2006: Let f : X Y be a smooth map that extendssmoothly to all W. ... a (global) trivializationover all of B.[Hint: covariant-constant sections.]. 13. Verify that if M is a submanifold of RN then the Euclidean inner product restricts to define aRiemannian

3 Feb 2006: Show, directly from the definition of holomorphic maps, that if f is holomorphic onR {p} then f is in fact holomorphic on all of R.(ii) Suppose that each of A = { ... algebraically related’: there is a polynomial Q in two variables, so that Q(f(z),g(z))

8 Oct 2006: b) Symmetry: d(x, y) = d(y, x) for all x, y X;. ... z for all z C {}. This occurs if and only if a = d and b = c = 0, so(.

21 Oct 2006: Show, directly from the definition of holomorphic maps, that if f is holomorphic onR {p} then f is in fact holomorphic on all of R.(ii) Suppose that each of A = { ... algebraically related’: there is a polynomial Q in two variables, so that Q(f(z),g(z))

27 Oct 2006: 0|f|. is a normed space. b) Is this norm equivalent to the uniform norm ‖ ‖? c) Is the space of all (Riemann) integrable functions with norm ‖f‖ = 1. ... iii) {x : there exists N such that xn = 0 for all n > N} in l,.

10 Nov 2006: iii) X = Q, d(x,x) = 0 for all x, otherwise d(x,y) = 3n if x y = 3na/b where a,b are prime to 3(and n may be ... 0). Show that all functions f : (R,d′) (R,d) are continuous. What are thecontinuous functions from (R,d) (R,d′).

28 Nov 2006: a) has all directional derivatives at zero but is not continuous at zero. ... b) has all directional derivatives at zero, is continuous everywhere, but is not differentiable at.

13 Feb 2006: Recall that A is orthogonal if AT A = I).b) Determine all finite groups which have a faithful representation on a two dimensional real vector space. ... 1.12 Question. Do all the exercises set in lectures! 1.13 Question.

7 Feb 2006: reparametrizations of the form f : [0,1] [0,1],with f′(t) > 0 for all t [0,1]. ... Prove that the Gaus-sian curvatures at the points (a,0,0),(0,b,0),(0,0,c) are all equal if and only if S is asphere.

8 Nov 2006: 5. Find all of the Möbius transformations that commute with Mk for a fixed k. ... Are these all inversions? 9. How many square roots of a Möbius transformation are there?

30 Nov 2006: 3. Show that the function ψ(A,B) = tr(ABt) is a symmetric positive-definite bilinear form on the spaceMn(R) of all n n real matrices.Show that the map ... 15. (i) Show that On(R), the group of all real orthogonal n n matrices, has a normal subgroup SOn(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

4 May 2006: c) For all A X, f (cl(A)) cl(f (A)). 2. 12. ... 19. Prove that in a Hausdorff space X, the intersection of all open sets containing a pointx X is just the singleton set {x}.

13 Feb 2006: a) Show that ker ρ = {g G | χ(g) = d}.(b) Show that |χ(g)| d for all g G, and that if |χ(g)| = d then ρ(g) = λI, where ... Prove that the number of irreducible characters which take only real values is equal to the number ofself-inverse conjugacy

9 Oct 2006: You should attempt these questions, but may not want to write out all parts in much detail.](a) The set C of continuous functions.(b) The set {f C : |f(t)| ... ne1. 5. Let P denote the space of all polynomial functions R R.

16 May 2006: 3. • |N (B)| |N (A)|; and. • B is i-compressed for all i. ... Thus |N (B)| |N (A)|. //Among all B [k]n with |B| = |A| and |N (B)| |N (A)|, choose one with.

18 Sep 2006: coloured,. 2. there are positive integers a, b, a λb S all the same colour, but such that Scontains no solution to the equation x y = z. ... ad) [n]d and b = (b1, b2,. , bd) [n]d, we write a bto mean ai bi for all i = 1, 2,. ,

13 May 2006: We must choose A = (aij). Let b OK be anS-unit which is p-adically small for all p S. ... This is possible by thefiniteness of the class group.) We set ai i1 = b for all 1 i n1.

16 Feb 2006: Thenn(γ; wo) = 0 for all points wo / B(zo, R). ... One component containsC B(0, R), so it is the unique unbounded component that contains all points of sufficiently largemodulus.

22 Oct 2006: Then there exist a A, b B and c C with ab = c.In particular, this is true if all of A, B and C have size greater than 2n8/9. ... All that remains to prove the theorem is to show that (i) implies (iii).