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23 Jan 2019: As described above, by a theorem of Borel[Bor74] the maps (3.1) are isomorphisms, so ch4kd(ξH) is equal to some poly-nomial p(σ1,σ2,. , ... Hi(B; Q) 0.Now s(τc)|E = 0 so (τc)|E = u (E)x for some x H|c|d(B; Q).

14 Oct 2019: 2. Show that the following sets of subsets of R all generate the same σ-algebra:(a) {(a,b) : a < b}, (b) {(a,b] : a < b}, (c) {(,b] : b R}. ... 10. Let (E,E,µ) be a measure space. Call a subset N E null if N B for some B E withµ(B) = 0.

13 Nov 2019: What happensif Θ = [a,b] for some 0 < a < b <? 9. Let X be multivariate normal N(θ,I) where θ Θ = Rp,p 3, and where I is the ppidentity ... a) Show that the James-Stein estimator θ̃(1) dominates all estimators θ̃(c),c 6= 1.b) Let θ̂ be the

5 Feb 2019: Prove that there exists some c [0, 1]with f(c) = c. ... Show that thereis some c (a,b) with f ′(c) = k.

11 Nov 2019: Some more terminology: B is called the base and E the total space of this vector bundle. ... The ψβα’s can be taken from some vector bundle E over B, then P will be‘constructed from E’.

31 Oct 2019: Suppose for some θ0 in the interior of Θ and every ε > 0 small enough we haveS(θ0 ε) < 0 < S(θ0 ε), and that Sn has exactly one zero θ̂n ... f(,φ) : φ = Φ(θ) for some θ Θ} equals Φ(θ̂MLE).b) Now consider a mapping Φ that is not

2 Nov 2019: Theorem 1.27. Every (smooth) vector field on S2m vanishes at some point. ... A contradiction. Some page references. to Warner, and Guillemin–Pollack, and Gallot–Hulin–Lafontaine.N.B.

26 Feb 2019: 71). Find an element of OK with norm 2a 3b 5 for some. ... a,b 0. Hence prove that the class group of K is cyclic and find its order.

4 Dec 2019: The square matrices A and B over the field F are congruent if B = PTAP for some invertible matrixP over F. ... amk50@cam.ac.uk - 1 - November 2019. for some λ,µ,ν R (which should turn out in this example to be integers).Does there exist a linear

1 May 2019: By replacing L by πmL for some positive integer m, we mayassume that Z is πnL-accessible for all n > 0 (see [4, Proposition 7.6]).Let B be an affine ... By definition, there exists some Zariskiopen affine subscheme Ux = Spec B(x) X containing x such