2010首届丘成桐大学生数学竞赛个人赛解答

S.-T. Yau College Student Mathematics Contests 2010 Analysis and Diff erential Equations Individual Solution 1. a) Let xk, k = 1,.,n be real numbers from the interval (0,) and defi ne x = n X i=1 xi n . Show that n Y k=1 sinxk xk ?sinx x n . b) From Z 0 ex 2dx = 2 , calculate the integral R 0 sin(x2)dx. Proof. a) Apply inequality of convex functions to f(x) = log(sin(x)/x). 2. Let f : R R be any function. Prove that the set of points x in R where f is continuous is a countable intersection of open sets. Proof. Use the defi nition of a continuous function. 3. Consider the equation x = x+f(t,x), where |f(t,x)| (t)|x| for all (t,x) RR, R (t)dt 0, i = 1,2,.,r, thus UTAV = diag(1,2,.,p), where p = minm,n and 1 2 . p 0. (b) If k = 0, then kAk2= 1. Let k 0, A Ak= Pr i=k+1iuiv T i , and UT(A Ak)V = D k 0 00 ,Dk= 0 k+1 . r , thus kA Akk2= k+1. Since rank(Ak) = k, we have min rank(B)=kkA Bk2 kA Akk2= k+1. As rank(B) = k, it leads to dim(null(B) = n k, so we have null(B) spanv1,v2,.,vk+1 6= 0. Take x null(B) spanv1,v2,.,vk+1, such that x = k+1 X i=1 liviandkxk2 2 = k+1 X i=1 l2 i = 1, 5 where li R, i = 1,2,.,k + 1. We then have kA Bk2 2 k(A B)xk2 2 = kAxk2 2 = ? ? ? ? ? k+1 X i=1 liAvi ? ? ? ? ? 2 2 = ? ? ? ? ? k+1 X i=1 liiui ? ? ? ? ? 2 2 = k+1 X i=1 l2 i 2 i k+1 X i=1 l2 i 2 k+1 = 2 k+1. Therefore for any positive integer k r, we have min rank(B)=kkA Bk2 = k+1. Solutions to: S.-T. Yau College Student Mathematics Contest 2010 Geometry and Topology Individual 1. Let D= (x,y) R2| 0 x2+ y20 be a fi nite set of positive integers. For each integer n 0, defi ne an to be the number of all fi nite sequences (t1,.,tm) with m n, ti T for all i = 1,.,m and t1+ . + tm= n. Prove that the infi nite series 1 + X n1 anzn Cz is a rational function in z, and fi nd this rational function. Proof. The infi nite series f(z) := 1 + P n=1 anzn is by defi nition f(z) = X mN X tiT i=1,.,m zt1+.+tm= X mN bm(z). The m-th term bm(z) is equal to the evaluation of the polynomial (PtTUt)min the variables UttTat “Ut= ztt T”. So f(z) = X mN X tT zt !m = 1 X tT zt !1 . 6. Describe all the irreducible complex representations of the group S4 (the symmetric group on four letters). Proof. As with any symmetry group, there are two 1-dimensional rep- resentations: the trivial U and the alternating U0. For any g S4, g acting on any v U by g(v) = v, and on any v U0by g(v) = sgn(g)v, where sgn(g) is 1 or 1, depending on whether g is an even or odd per- mutation. Being 1-dimensional, U and U0are certainly irreducible. The standard representation V of S4is the 3-dimensional represen- tation V = z = (z1,.,z4) C4| z1+z2+z3+z4= 0, where the action of g S4on V is given by the standard permutation g(z) = (zg1(1),.,zg1(4). Take v = (1,1,0,0) V . Then g = (13) and g = (14) send v to v0= (0,1,1,0) and v00= (0,1,0,1), respectively. Clearly v,v0,v00 are linearly independent, so V is irreducible. Take the tensor product of V with the 1-dimensional representation U0, we get an irreducible representation of S4: V 0 = U0V . To see that V 0 is not isomorphic to V , let us consider the value of their character at the permutation g = (12): U0(g) = 1, V(g) = 1, so V0(g) = 1 thus V and V 0 cannot be isomorphic. 6 One more irreducible representation of S4is the 2-dimensional rep- resentation W that we will describe as the following.Consider the normal subgroup H S4of order 4: H = 1,(12)(34),(13)(24),(14)(23) It is easy to see that S4/H = S3. Let W be the standard representation of S3, namely, W = z C3| z1+ z2+ z3= 0,g(z) = (zg1(1),zg1(2),zg1(3) for any g S3. As in the case of V , it is easy to see that this 2- dimensional representation is irreducible. Then through the pull back S4 S4/H, W becomes an irreducible representation of S4