Homework set 5 - Solutions
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1 Homework set 5 - Solutions Math 469 Renato Feres 1. Hall s textbook, Exercise 4.9.2, page 105. Show the following: (a) The adjoint representation and the standard representation are isomorphic representations of the Lie algebra so(3). (b) The adjoint and standard representations of the group SO(3) are isomorphic. Solution. (a) Let V 1 = R 3 and V 2 = so(3). We want to find a linear isomorphism ϕ : V 1 V 2 which is an intertwining map for the two representations; that is, such that for all X so(x ) and u R 3 ϕ(x u) = [X,ϕ(u)]. Recall the map ω : R 3 so(3) from homework assignments 3 and 4: ω : a = a 1 a 2 a 3 ω(a) = 0 a 3 a 2 a 3 0 a 1 a 2 a 1 0 If we write X = ω(x) and let ϕ = ω, then X u = x u and ω(x u) = ω(x u) = [ω(x),ω(u)] = [X,ω(u)]. (b) Note: if ϕ : V 1 V 2 is an intertwining map for two representations Π : G GL(V 1 ) and Σ : G GL(V 2 ) of a matrix Lie group G, then for all X g ϕ e tπ(x ) = ϕ Π(e t X ) = Σ(e t X ) ϕ = e tσ(x ) ϕ, from which it follows by differentiating in t at t = 0 that ϕ π(x ) = σ(x ) ϕ. Here we used the fact that ϕ is a linear map, so that it is equal to its (Jacobian) derivative map. This suggests trying ϕ = ω as the intertwining map for the standard and adjoint representations of SO(3). In other words, we would like to show that ω(au) = Aω(u)A 1 for all A SO(3) and u R 3. In other words, we wish to show that for all u, v R 3 and A SO(3) (Au) v = A(u A 1 v).
2 Equivalently, we wish to show that A(u v) = (Au) (Av). But this property is immediate from the geometric definition of the cross-product: if u and v are not collinear, then w = u v is the unique vector perpendicular to the plane spanned by u and v, of length equal to the area of the parallelogram generated by u and v, and such that u, v, w form a positive basis of R 3. These defining properties of the cross product are all invariant under an orientation preserving isometry of Euclidean space. 2. Hall s textbook, Exercise 4.9.5, page 105. Consider the standard representation of the Heisenberg group, acting on C 3. Determine all subspaces of C 3 which are invariant under the action of the Heisenberg group. Is this representation completely reducible? Solution. The Heisenberg group, in dimension 3, is the group of real 3 3 upper triangular matrices of the form 1 a c 0 1 b under matrix multiplication. In higher dimensions, the Heisenberg group H n is the group of (n + 2) (n + 2) upper triangular matrices of the form 1 a c H(a,b,c) = 0 I n b where a is a row vector in dimension n, is a column vector in dimension n and c R. The standard (complex) representation is the representation of the Heisenberg group by matrix multiplication on column vectors in C n+2. I will describe the invariant subspaces for the general n case. Note that the special case in dimension 3 corresponds to n = 1. Let e 0,e 1,...,e n+1 be the standard basis elements of C n+2 regarded as column vectors. We may write a general vector in C n as w = w 0 e 0 + w + w n+1 e n+1, where w is in the linear span of e 1,...,e n. Let V C n+2 be a nonzero invariant subspace and let w V be a nonzero vector. Let us consider the following cases (I regard a and b as in the span of e 1,...,e n ): (a) Suppose w n+1 0. Then it is clear that by taking (complex) linear combinations of H(a,b,c)w = (w 0 + a w + cw n+1 )e 0 + w + w n+1 b + w n+1 e n+1 for different choices of a,b,c, we can generated all of C n+2. Therefore, a proper invariant subspace must have w n+1 = 0. (b) Now suppose that w n+1 = 0. Then H(a,b,c)w = (w 0 + a w)e 0 + w. By taking linear combinations of this vector for different choices of we generate the invariant vector space Ce 0 Cw. 2
3 (c) Finally, suppose that w n+1 = 0 and w = 0. Then it is clear that H(a,b,c)w = w 0 e 0 spans Ce 0. We conclude that if V C n+2 is an invariant subspace for the standard representation of the Heisenberg group, then V is one of {0}, Ce 0, Ce 0 Cw for some w C n, C n+2. Here I identify C n with the n-dimensional subspace of C n+2 with the first and last coordinates equal to 0. For the 3-dimensional Heisenberg group, the invariant subspaces are {0}, span{e 0 }, span{e 0,e 1 }, C 3. Note in particular that the invariant subspace Ce 0 cannot have an invariant complement since any nonzero invariant subspace must contain this one. Therefore, the standard representation of the Heisenberg group is not completely reducible. 3. Hall s textbook, Exercise 4.9.9, page 106. Suppose V is a finite-dimensional representation of a group or Lie algebra and that W is a nonzero invariant subspace of V. Show that there exists a nonzero irreducible invariant subspace for V that is contained in W. Solution. If there exists no proper invariant subspace in W, then W itself is irreducible. Suppose this is not the case. Let W 1 be a proper (non-zero) invariant subspace of W. In particular, dimw 1 < dimw. If W 1 is irreducible, we are done. Otherwise, we continue in this way until we arrive at a proper, irreducible subspace of W which is invariant for the representation. This induction must terminate after finitely many steps because the representation space V is finite dimensional. 4. Hall s textbook, Exercise , page 106. Suppose that V 1 and V 2 are nonisomorphic irreducible representations of a group or Lie algebra, and consider the associated representation V 1 V 2. Regard V 1 and V 2 as subspaces of V 1 V 2 in the obvious way. Following the outline below, show that V 1 and V 2 are the only nontrivial invariant subspaces of V 1 V 2. (a) First assume that U is a nontrivial irreducible invariant subspace. Let P 1 : V 1 V 2 V 1 be the projection onto the first factor and let P 2 be the projection onto the second factor. Show that P 1 and P 2 are intertwining maps. Show that U = V 1 or U = V 2. (b) Using Exercise 3 above, show that V 1 and V 2 are the only nontrivial invariant subspaces of V 1 V 2. Solution. I will consider the problem for representations of matrix Lie groups. The proof is similar for Lie algebras. Let Π i : G GL(V i ), for i = 1,2, be irreducible representations of the Lie group G. The direct sum representation is given by Π(g )(v 1, v 2 ) = (Π 1 (g )v 1,Π 2 (g )v 2 ). The projection P i : V 1 V 2 V i is clearly an intertwining map: P i Π(g ) = Π i (g ) P i. 3
4 Let U be a irreducible invariant subspace of V 1 V 2 for the direct sum representation and consider the restriction R i = P i U. Then, as U is invariant, it is immediate that R i Π(g ) = Π i (g ) R i. By Shur s lemma, R i is either an isomorphism or 0. Because R 1 + R 2 is the identity on U, they cannot be both 0. Without loss of generality, say that R 2 = 0. It follows that U is a nontrivial invariant subspace of V 1. But V 1 is irreducible, so U = V 1. We conclude that V 1,V 2 are the only nontrivial invariant subspaces of V 1 V Hall s textbook, Exercise , page 106. Suppose that V is an irreducible finite-dimensional representation of a group or Lie algebra over C, and consider the associated representation V V. Show that every nontrivial invariant subspace U of V V is isomorphic to V and is of the form U = {(λ 1 v,λ 2 v) : v V } for some constants λ 1,λ 2, not both zero. Solution. Let Π : G GL(V ) be the irreducible representation on V and Π = (Π,Π) the direct sum representation on V V. Consider the projection P i : V V V, i = 1,2, onto the first and second V factor. Let U be a nontrivial invariant subspace of the direct sum. Just as in the previous exercise, the restriction of P i to U is intertwining for the representation of Π on U and of Π on V. It is not difficult to conclude that P i U is an invariant subspace of V. Because V is irreducible, P i U is either {0} or V. It cannot be {0} for both i = 1,2. If this space is {0} for i = 1 and V for i = 2, then U = {(0, v) : v V }. If P 2 U = {0}, then similarly U = {(v,0) : v V }. These cases correspond to U being the space of vectors of the form (λ 1 v,λ 2 v) where either λ 1 = 0 and λ 2 = 1, or vice-versa. Now suppose the remaining case: P 1 U = P 2 U = V. Consider the invariant subspaces K i = ker(p i U ) of U. The image of K 1 under P 2, by the argument already used, is either {0} or V. If P 2 K 1 = {0}, then as P 1 K 1 = {0} by definition, we conclude that K 1 = {0} and so P 1 : U V is an isomorphism. If P 2 K 1 = V, then K 1 = {(0, v) : v V }. This implies that U = V V. In fact, for any u V there is (u, v) U because P 1 U is surjective. Thus (u, v) (0, v) = (u,0) also belongs to U for all v and so (u, v) belongs to U for all u, v V. This is not possible under the assumption that U is nontrivial. So we are left with the following: P 1 : U V is an isomorphism. Similarly, P 2 : U V is also an isomorphism. This gives an isomorphism L := (P 2 U ) 1 P 1 U : V V. By Schur s lemma, L = λi for some nonzero λ C. We conclude that U = {(u,λu) : u V }. 6. Hall s textbook, Exercise , page 106. Recall the spaces V m introduced in Section 4.2 of Hall s book, viewed as representations of the Lie algebra sl(2,c). In particular, consider the space V 1 (which has dimension 2). (a) Regard V 1 V 1 as a representation of sl(2,c), as in Definition Show that this representation is not irreducible. (b) Now, view V 1 V 1 as a representation of sl(2,c) sl(2,c), as in definition Show that this representation is irreducible. 4
5 Solution. Recall That V 1 is the space of homogeneous complex polynomials of degree 1 in two complex variables: V 1 = {a 0 z 1 + a 1 z 2 : a 0, a 1 C}. In other words, V 1 is the space of complex linear maps from C 2 to C and the representation on V 1 is simply the dual of the standard representation of sl(2,c). Thus, for concreteness, I will identify V 1 with C 2 (column vectors) and write π(x ) = X t. (a) The tensor product C 2 C 2 is not irreducible. In fact, consider the subspaces S and A of C 2 C 2 defined as follows: S consists of all the complex linear combinations of tensor elements of the form u v + v u and A consists of all the linear combinations of elements of the form u v v u. It is clear that these two subspaces are invariant because (π π)(x )(u v ± v u) = π(x )u v + u π(x )v ± (π(x )v u + v π(x )u) = π(x )u v ± π(x )v u + u π(x )v ± v π(x )u = (π(x )u v ± v π(x )u) + (u π(x )v ± π(x )v u). Also C 2 C 2 is the direct sum of S and A. (b) We now view C 2 C 2 as the representation space for sl(2,c) sl(2,c). In this case, by definition, (π π)(x,y )u v = π(x )u v + u π(y )v. Let W be an invariant subspace of C 2 C 2 and u v a nonzero element of W. Note that (π π)(x,0)u v = π(x )u v. It is not difficult to see that the standard representation on C 2 and its dual representation are both irreducible. It follows that W must contain u v for all u C 2. In particular, e 1 v and e 2 v lie in W, where e i are the standard basis elements of C 2. Similarly, (π π)(0,y )e i v = e i π(y )v for all Y sl(2,c). Again by irreducibility of π we conclude that e i v W for all v C 2. Thus e i e j W for all i, j. Since {e i e j : i, j {1,2}} is a basis for C 2 C 2, we conclude that the tensor product representation is irreducible. 5
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