Math 593: Problem Set Feng Zhu, edited by Prof Smith Hermitian inner-product spaces (a By conjugate-symmetry and linearity in the first argument, f(v, λw = f(λw, v = λf(w, v = λf(w, v = λf(v, w. (b We may verify that, for any x, y C n and λ C, λx, y = i λx i y i = λ i x i y i = λ x, y and x, y = i λx i y i = i y i x i = y, x and x, x = i x i x i = i x i 2 with equality iff x =, so that, as defined is indeed a Hermitian inner product on C n. (c Write A := [f] B. Then A ii = f(v i, v i = f(v i, v i by conjugate-symmetry, so that A ii = f(v i, v i R. If i j, then A ji = f(v j, v i = f(v i, v j = A ij. Hence A = A, i.e. A = [f] B is Hermitian. (d Write v = n i= a iv i and w = n j= b jv j (so that [v] B = (a,..., a n and [w] B = (b,..., b n. Then by sesquilinearity, f(v, w = i,j n a if(v i, v j b j = [v] T B [f] B[w] B, and we may re-arrange this as f(v, w = n n a i f(v i, v j b j = a i f(v i, v j b j = [v] B, A[w] B or as f(v, w = i,j n i,j n a i f(v i, v j b j = i= j= ( n n a i f(v i, v j b j = A T [v] B, [w] B j= Collaborated, directly or indirectly, with Umang Varma, Daniel Irvine, Joe Kraisler, Alex Vargo, Punya Satpathy. Was also helped by discussions with Sachi Hashimoto. i=
(e Note we have [v] B = P [v] B and [w] B = P [w] B. Then and so [f] B = P T [f] B P. f(v, w = [v] T B[f] B [w] B = [v] T B P T [f] B P [w] B = [v] B [f] B [w] B (f If f(v, v = for all v, then f is an alternating symmetric form, i.e. it is the zero form; then the matrix of f w.r.t. any basis B is diagonal (is zero. So now suppose we can find v s.t. f(v, v. We now induce on the dimension n of our vector space V. For n = the statement is clear. More generally, define the linear map L v : C n C by L v (w = f(v, w. Note L v (v = im L v = C, and so W := ker L v has (C-dimension n. Now f W W is a Hermitian form on W, so inductively there is some basis B for W s.t. f W W has a diagonal matrix w.r.t. B ; then if we take B = {v} B, then [f] B is a diagonal matrix. If f is positive-definite, then all of the entries ( f(v i, v i along the diagonal of [f] B are positive, and so, if we normalize the basis B by B := v f(v,v,..., v n, then [f] B is the n n f(vn,v n identity matrix. (g If A is a n n Hermitian matrix, it has n eigenvalues counted with multiplicty (since its characteristic polynomial splits completely; in particular, it has an eigenvalue λ with associated eigenvector v. Define the Hermitian form f : C n C n by f(x, y = x, Āy. Since f(v, v = v, Āv = v, λv = λ v, v R and v, v R, we have λ R. Hence we may conclude that A has a real eigenvalue. Now let v = v, the eigenvector above, and let W = (C v, where the indicates the vectors in C n orthogonal to v under the standard Hermitian inner product on C n. For any w W, Aw, v = w, A v = w, Av = λ w, v =, and so W is A-invariant. Hence we may induce on the dimension n (the base case is supplied by n =, for which the desired result is trivially true. 2 Inner-product spaces and Gram-Schmidt (a Fix v, w V, and consider f(t = v + tw, v + tw = v, v + ( v, w + w, v t + w, w t 2 = v, v + 2R( v, w t + w, w t 2 (here assuming t R. Note that all of these coefficients are real, and that v + tw, v + tw with equality iff v = tw by the properties of the inner product. Considering the discriminant of this quadratic equation, we obtain 4(R( v, w 2 4 v, v w, w, i.e. (R( v, w 2 v 2 w 2 ; taking square roots, we obtain R( v, w v w u which suffices to show the Cauchy-Schwartz inequality in a real inner-product space. In a Hermitian inner-product space, let w = e iθ w, where θ ( π, π] is chosen s.t. v, w R (specifically, if v, w = λ, we take e iθ = λ/ λ. Then we have, from sesquilinearity and from the above, v, w = v, w v w = v w 2
(b Noting that v 2 = v, v, we have v + w 2 = v + w, v + w = v, v + w, w + v, w + w, v by bilinearity / sesquilinearity = v, v + w, w + 2R( v, w v, v + w, w + 2 v, w v 2 + w 2 + 2 v w = ( v + w 2 from Cauchy-Schwartz and, taking square roots on both sides, we obtain the triangle inequality. (c V is a 3-dimensional R-vector subspace of R 4 ; the dot product induces an inner product on R 4 (i.e. a positive-definite symmetric bilinear form R 4 R 4 R and this descends to a positive-definite symmetric bilinear form V V R, i.e. an inner product on the vector subspace V R 4. Letting e, e 2, e 3, e 4 denote the unit vectors in the positive x, y, z, w directions resp. We note that (, 2, 3, 4 is a normal vector to V in R 4, so that we may take (e.g. {(2,,,, (3,,,, (4,,, } to be a basis for V. Now we apply the Gram-Schmidt orthogonalisation process to obtain v = (2,,, = (2,,, (2,,, 5 u 2 = (3,,, v, (3,,, v = (3,,, 6 (2,,, = (.6,.2,, 5 v 2 = u 2 u 2 = (.6,.2,, 2.8 u 3 = (4,,, v, (4,,, v v 2, (4,,, v 2 = (4,,, 8 2.4 (2,,, (.6,.2,, 5 2.8 = (.8 3.6/7,.6 7.2/7, 6/7, = (2, 4, 6, 7 7 v 3 = 7 (2, 4, 6, 7 5 ( and then (v, v 2, v 3 = 5 (2,,,, 2.8 (.6,.2,,, 7 (2, 4, 6, 7 is an orthonormal 5 basis for V. (d It is clear that V is a R-vector subspace of R[x] (since it is closed under the vector addition and scalar multiplication on R[x]. We may verify that f + f 2, g = (f + f 2 g dx = f g + f 2 g dx = f g dx + f 2g dx = f, g + f 2, g and similarly (by linearity of the integral f, g + g 2 = f, g + f, g 2, λf, g = f, λg = λ f, g for all λ R and f, g, f, f 2, g, g 2 V. Moreover g, f := gf dx = fg dx =: f, g ; hence f, g as defined endows V with an inner product structure. To find an orthonormal basis, we start with the basis, x, x 2 and apply the Gram-Schmidt 3
procedure: v = u 2 = x x, = x = x 2 v 2 = ( u 2 u 2 = = ( 2 x 2 x dx u 3 = x 2 x 2, x 2, v 2 v 2 ( ( 2 = x 2 x 2 dx = x 2 3 ( 2 4 6 = x 2 3 x + 2 2 2 = x 2 x + ( 2 2 2 3 x 2 x + /2 ( 4 dx x 2 ( x 2 and so we have an orthonormal basis for V given by (v, v 2, v 3 = x 3 ( 2 x2 dx x 2 (, 2 ( x 2, x 2 2 x + ( 2 2 3. 3 Hessian (a We note that for a smooth function f we have 2 f x y = 2 f y x, and so the Hessian H(a, b at each (a, b is a symmetric matrix, and hence defines a symmetric bilinear form R 2 R 2 R by (x, y x T H(a, by. (b The second derivative test states that if (a, b is a critical point of f, then f has a local maximum at (a, b if the Hessian H(a, b is negative-definite, i.e. has signature (, 2; f has a local minimum at (a, b if the Hessian H(a, b is positive-definite, i.e. has signature (2, ; f has a saddle point at (a, b if the Hessian H(a, b is indefinite, i.e. has signature (,. The test is inconclusive if the Hessian H(a, b fails to have full rank. 4 Duals and double duals (a Suppose we have f M with λf = for some nonzero λ R. Then im f ker(x λx. Now if R is a domain and λ, then ker(x λx =, and so we may conclude f =. Hence there are no (nonzero torsion elements in M. 4
(b We may define a map M M by x e x, where e x (f = f(x (any element of x is sent to the evaluation at x homomorphism on M. This is an R-linear map: to check that e rx+y = re x + e y, we need only check that these maps take the same values on all f M. But e rx+y (f = f(rx + y = rf(x + f(y = re x (f + e y (f, by R-linearity of f. (c Given a finitely-generated free module M = R m, let e : M M be the map above. We already know this is an R-module map, so it suffices to show that it is bijective. For this, it suffices to show that the basis B := (e,..., e m is taken to a basis of M. Denote by e i the image of e i under e, and the set of all the e i by B. Before starting, we observe that because M is free, the choice of the basis B establishes an isomorphism of M with R m. So elements of M are uniquely determined by where they send the basis elements e i, which in turn, can be sent to any element of R (by the universal property of free modules. That is, M can be thought of as the module of m matrices, acting on column vectors in R m by multiplication. The elements e i in M corresponding to the unit rows (consisting of all zeros except for a in the i-th spot are a basis for M (since the unit rows are a basis for the R-module of row vectors. Note that e i (e j = δ ij (Kronecker delta. Note also, by definition of e : M M, that e(e i M is the map sending e j to e j (e i = δ ij. In particular e i (e j = δ ij. We now prove that the set B is a free generating set for M. First we show that B is linearly-independent: if m i= r ie i is the zero map M R, then in particular, it sends each of the elements e j M to zero. But then = m i= r ie i (e j = m i= r ie j (e i = m i= r iδ ij = r j, establishing linear independence. We next show that B spans M. Let Ψ be any element in M. Let r i = Ψ(e i. We claim that Ψ = m i= r ie i. To check this, it suffices to check that these maps take the same value on every element f M, and by R-linearity in fact, it is enough to check this on the generators e j. But m i= r ie i (e j = m i= r ie j (e i = m i= r iδ ij = r j, which is equal to Ψ(e j by definition. So Ψ = m i= r ie i. QED. (d If R = k is a field, then all R-modules are free; hence we have already shown that M finitelygenerated = M reflexive in the previous part. It remains to show that M reflexive = M finitely-generated in this case. Suppose M is not finitely-generated as a R-module. We claim that the map M M is injective, but not surjective in this case. Let (v α α J M be a basis for M, and let vα M be evaluation at v α. Now any v M for which v is not an evaluation map (at some element of M is not in the span of the vα ; we can construct such a v explicitly by letting v (vα = ( α, which is well defined by the universal property of freeness, where vα are the dual elements defined by vα(v β = δ αβ, (which are easily seen to be independent as above even though there are infinitely many. Now v thus defined cannot be an evaluation map at any element of M, for any element of M is a (finite R-linear combination of the v α, and hence any evaluation map must take v α to zero for all but finitely many α. (e Consider the Z-module Q. Since Q = Hom Z (Q, Z =, Q =, but this is clearly not isomorphic to Z. Hence Q is not reflexive (as a Z-module. Alternatively, consider any torsion Z-module (i.e. any finite abelian group Γ. From (a Γ and Γ are torsion-free, and so certainly not isomorphic to Γ (in particular, any R-linear map Γ Z must have ϕ( =, since there are no nonzero torsion elements in Z, so Γ = Hom Z (Γ, Z =. More generally, consider the R-module R/I for I some proper ideal of R. We must have (R/I = Hom R (R/I, R =, since any map R-linear map ϕ : R/I R sends to some element annihilated 5
by I, but since R is a domain the only such element is. So (R/I = (R/I =, and this is not isomorphic to the nonzero module R/I. 5 Adjoint functors (a We wish to show that for any R-modules C, D, there is a natural bijection Ψ : Hom R (C, Hom R (M, D Hom R (C R M, D. This is given by (c (m d (c m d. We verified in the previous homework that this was a well-defined bijection. (b Let F (S denote the free R-module on the set S. Now, for any set S, Hom(S, S = Hom(F (S, F (S via the bijection given by Hom(S, S ϕ ϕ Hom(F (S, F (S, where vp is the morphism of the free R-modules obtained by extending ϕ : S S (which is just any function of sets linearly. Since morphisms of free R-modules are determined entirely by where the generators are sent, it is clear that this map is indeed injective and surjective, and hence a bijection, as claimed. Hence the forgetful functor F (S S and the functor S F (S are adjoint. (c We should have, for all C C and D and for all γ Hom(C, G(D, C C γ G(D ϕ γ = and similarly, for all γ Hom(F (C, D, F (C F (C Ψ(γ F (ϕ Ψ(γ D D ϕ D γ F (C γ = G(D G(D Ψ (γ C G(ϕ Ψ (γ (where the vertical arrows in the diagrams on the right change direction if F or G are contravariant rather than covariant. 6