CMI Entrance Exam 2018 Question Paper Solution M.Sc Maths

CHENNAI MATHEMATICAL INSTITUTE Postgraduate Programme in Mathematics MSc/PhD Entrance Examination 15 May 2018 Part A (1) A, D. (2) B, C, D. (3) B, D. (4) A, B, D. (5) A, B, D. (6) A, C, D. (7) A, B, C. (8) C, D. (9) A, B, C. (10) 2. Part B (11) (A) Consider r : R −→ [0, 1]  0, r < 0;  r(x) = x, r ∈ [0, 1];  1, r > 1.  (B) No. Every retract Y of R must be connected because the map r is continuous. (C) Every retract Y of R is closed. To see this, consider φ : R −→ R × R, x 7→ (r(x), x). Then Y = φ−1 (diagonal). Since R is Hausdorff, the diagonal is closed, and so is Y . (12) Write ξn Z 1 gn (z) = dξ. 2πi Γ ξ − z Note that  1,  n = 0, z = 0 gn (z) = 0, n 6= 0, z = 0  n  z , z 6= 0 Hence n=N ( X a0 , z = 0; f (z) = an gn (z) = n=−N F (z), z 6= 0. R1 φ(t)tn dt (13) Write Fn = (−1)n 0 n! . Then ∞ Z 1 Z 1 ! X (−at) n 0= φ(t)e−at dt = φ(t) dt 0 0 n! n=0 N Z 1 ! X (−at)n = lim φ(t) dt N −→∞ 0 n! n=0 N Z 1 (−at)n X   = lim φ(t) dt N −→∞ 0 n! n=0 ∞ X = Fn an n=0 1

Since ∞ n P n=0 Fn a = 0 for every a ∈ R+ , we see that Fn = 0 for every n ≥ 0. (14) By way of contradiction assume that U ( R. Then, since U is open, U ( U . Pick x ∈ U r U and a sequence {xn } ⊆ U converging to x. Since h is uniformly continuous, {h(xn )} is a Cauchy sequence in R, it converges to y ∈ R. Hence {xn } converges go h−1 (y) ∈ U , a contradiction. (15) Since det A = −1, the characteristic polynomial of A is of the form X 2 + bX − 1 for some b ∈ R, so A has real eigenvalues, λ1 , λ2 . Let vi be an eigenvector for λi , i = 1, 2. For i = 1, 2, λ2i vit vi = λi vit λi vi = vit At Avi = vit vi , so λi is 1 or −1. Without loss of generality, λ1 = 1 and λ2 = −1. Then A gives a reflection about the line spanned by v1 sending v2 to −v2 . (16) (A) For any G, 0 is a characteristic subgroup. Let 0 6= H ⊆ Q be a characteristic subgroup. Let 0 6= x ∈ H and y ∈ Q. Then the map r 7→ ry/x is an automorphism of Q, and it takes x to y. Hence y ∈ H, so H = Q. Hence 0 and Q are the only characteristic subgroups of Q. (B) For any g ∈ G, the map G −→ G, g1 7→ gg1 g −1 is an isomprphism, so, for every characteristic subgroup H of G, gHg −1 = H. Hence H is normal. The converse is false: take H = Z inside G = Q. (17∗ ) (A) (At M A)t = At M t A = −At M A. (B) Check directly with v = e1 , e2 , e3 after noting tha the action of SOE (R) and Φ are k-linear. Also use the fact that the entries At = A−1 are the signed minors of A, coming from the adjoint of A. (C) Using (B) it is enough to show that there does not exist a nonzero v ∈ R3 whose span is stable under the action of SO3 . This is true: if v, w ∈ R3 are non-zero vectors of the same length, there exists A ∈ SO3 such that Av = w. (18∗ ) For positive integers n, write Un = {z ∈ C r {0} : |z| < n}. These form an open cover of C r {0}. Note that for every z1 ∼ z2 , |z1 | = |z2 |, so for every n, π −1 (π(Un )) = Un . Hence π(Un ), n ≥ 1 is an open cover of X. This does not have a finite sub-cover since the open cover Un , n ≥ 1 does not have a finite sub-cover. (19∗ ) First note that the minimal polynomial of g divides X |G| − 1. (A) If char k = 0 then for every g, g has distinct eigenvalues and hence is diagonalizable, so g = 1, contradicting the hypothesis that |G| > 1. (B) Let p = char k. Let g ∈ G and write its order as pe m e with m = 1 or m > 1 and p - m. The the minimal polynomial of g p is X m − 1 which e has distinct roots, so, again, by the above argument, g p = 1, so m = 1. (C) Hence G is a p-group. Use class equation. (20∗ ) The limit is f (1). This is true for xk , k ≥ 0, and hence also for polynomials. By Weierstrass’ theorem, it is true for all continuous functions. 2