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. . . . . . . . . . . . . . 107 INTRODUCTION Very roughly speaking, representation theory studies symmetry in linear spaces. It is a beautiful mathematical subject which has many applications, ranging from number theory and combinatorics to geometry, probability theory, quantum mechanics and quantum field theory. Re...
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representation theory of groups, Lie algebras, and quivers. We mostly follow [FH], with the exception of the sections discussing quivers, which follow [BGP]. We also recommend the comprehensive textbook [CR]. The notes should be accessible to students with a strong background in linear algebra and a basic knowledge ...
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include algebras defined by generators and relations, such as group algebras and universal enveloping algebras of Lie algebras. ab, a, b � � · · A representation of an associative algebra A (also called a left A-module) is a vector space EndV , i.e., a linear map preserving the multiplication V equipped with a ho...
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− − This means that the problem of finding, say, N -dimensional representations of A reduces to solving a bunch of nonlinear algebraic equations with respect to a bunch of unknown N by N matrices, for example system (1) with respect to unknown matrices h, e, f . It is really striking that such, at first glance hopeles...
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the same thing as a representation of a certain algebra PQ called the path algebra of Q. Thus one may ask: what are the indecomposable finite dimensional representations of Q? ⊃ More specifically, let us say that Q is of finite type if it has finitely many indecomposable representations. We will prove the following str...
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fascinating chapters of representation theory. In this theory, one considers representations of the G and multiplication group algebra A = C[G] of a finite group G – the algebra with basis ag, g law agah = agh. We will show that any finite dimensional representation of A is a direct sum of irreducible representation...
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elements. Definition 1.3. An associative algebra over k is a vector space A over k together with a bilinear map A ab, such that (ab)c = a(bc). A, (a, b) A × ⊃ �⊃ Definition 1.4. A unit in an associative algebra A is an element 1 A such that 1a = a1 = a. � Proposition 1.5. If a unit exists, it is unique. Proo...
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A is commutative if ab = ba for all a, b A. � For instance, in the above examples, A is commutative in cases 1 and 2, but not commutative in cases 3 (if dim V > 1), and 4 (if n > 1). In case 5, A is commutative if and only if G is commutative. Definition 1.8. A homomorphism of algebras f : A f (x)f (y) for all x, ...
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that δ(a)b = ab (the usual product). This representation is called the regular representation of A. Similarly, one can equip A with a structure of a right A-module by setting δ(a)b := ba. ⊃ 3. A = k. Then a representation of A is simply a vector space over k. 4. A = k x1, ..., xn� ◦ . Then a representation of A is j...
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1 : V2 V1 (check it!). linear operator θ− ⊃ ⊃ V2 is an isomorphism of representations then so is the Two representations between which there exists an isomorphism are said to be isomorphic. For practical purposes, two isomorphic representations may be regarded as “the same”, although there could be subtleties re...
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1, V2 be representations of an algebra A over any field V2 be a nonzero homomorphism of F (which need not be algebraically closed). Let θ : V1 representations. Then: ⊃ (i) If V1 is irreducible, θ is injective; 8 ⇒ (ii) If V2 is irreducible, θ is surjective. Thus, if both V1 and V2 are irreducible, θ is an isomo...
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as an R-algebra), and V = A. Proof. Let ∂ be an eigenvalue of θ (a root of the characteristic polynomial of θ). It exists since k is V , which an algebraically closed field. Then the operator θ is not an isomorphism (since its determinant is zero). Thus by Proposition 1.16 this operator is zero, hence the result. ...
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representations. As we discussed above, they are defined by a single operator δ(x). In the 1-dimensional case, this is just a number from k. So all the irreducible representations of A are V� = k, ∂ k, in which the action of A defined by δ(x) = ∂. Clearly, these representations are pairwise non-isomorphic. � The cl...
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representation of A is the same thing as Aut(V ), a representation of G, i.e., a vector space V together with a group homomorphism δ : G whre Aut(V ) = GL(V ) denotes the group of invertible linear maps from the space V to itself. ⊃ Problem 1.20. Let V be a nonzero finite dimensional representation of an algebra A....
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is again called the central character of V . ⊃ � (c) Does δ(z) in (b) have to be a scalar operator? Problem 1.22. Let A be an associative algebra, and V a representation of A. By EndA(V ) one V . Show that EndA(A) = Aop, denotes the algebra of all homomorphisms of representations V the algebra A with opposite mul...
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for all a ∧ � A such that aI I for all a A. Similarly, a right A. A two-sided ideal is a � Left ideals are the same as subrepresentations of the regular representation A. Right ideals are the same as subrepresentations of the regular representation of the opposite algebra A op. Below are some examples of ide...
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) = β(ab) − − because (a� a)b Ib � ∧ − I = ker β, as I is a right ideal; similarly, if β(b) = β(b�) then β(ab�) = β(ab + a(b� b)) = β(ab) + β(a(b� b)) = β(ab) − − because a(b� b) � − aI ∧ I = ker β, as I is also a left ideal. Thus, A/I is an algebra. Similarly, if V is a representation of A, and W V is ...
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if and only if it is isomorphic to A/I, where I is a left ideal in A. (c) Give an example of an indecomposable representation which is not cyclic. Hint. Let A = C[x, y]/I2, where I2 is the ideal spanned by homogeneous polynomials of degree be the space of linear functionals on A, with the action 2 (so A has a bas...
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A is (ii) A basis for the q-Weyl algebra Aq is xiyj , i, j { � 11 0 . } ⊂ xiyj , i, j { Z . } ⇒ ⇒ Proof. (i) First let us show that the elements xiyj are a spanning set for A. To do this, note that any word in x, y can be ordered to have all the x on the left of the y, at the cost of intercha...
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Qj (t)a(a − 1)...(a − j + 1)ta − j . This must be zero, so we have leading term in a, we get Qr(t) = 0, a contradiction. r Qj (t)a(a j=0 − 1)...(a − (ii) Any word in x, y, x− 1, y− ⎨ 1 can be ordered at the cost of multiplying it by a power of q. This j + 1)t− j = 0 in k[a][t, t− 1]. Taking the easily imp...
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als in A? Hint. For the first question, use the fact that for two square matrices B, C, Tr(BC) = Tr(CB). For the second question, show that any nonzero two-sided ideal in A contains a nonzero polynomial in x, and use this to characterize this ideal. Suppose for the rest of the problem that chark = p. (b) What is th...
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c) of the previous problem. 1.8 Quivers Definition 1.28. A quiver Q is a directed graph, possibly with self-loops and/or multiple edges between two vertices. Example 1.29. • • � � � � � � • • We denote the set of vertices of the quiver Q as I, and the set of edges as E. For an edge h let h�, h�� denote the source ...
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concatenation of paths: ab is the path obtained by first tracing b and then a. If two paths cannot be concatenated, the product is defined to be zero. � Remark 1.32. It is easy to see that for a finite quiver pi = 1, so PQ is an algebra with unit. I i ⎨ � Problem 1.33. Show that the algebra PQ is generated by pi for ...
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: Vh⊗ Vh⊗⊗ be the composition of the operators corresponding to the edges occurring in p (and the action of this operator on the other Vi is zero). i Vi, let pi : V Vi � ⊃ ⊃ m 1 It is clear that the above assignments V Vi are inverses of each other. Thus, we have a bijection between isomorphism classes of r...
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is not isomorphic to a direct sum of two nonzero representations. ∞ xh for all h −⊃ �h⊗ = �h⊗⊗ Definition 1.37. Let (Vi, xh) and (Wi, yh) be representations of the quiver Q. A homomorphism � : (Vi) Wi such that yh (Wi) of quiver representations is a collection of maps �i : Vi � 0A[n], and A[n] A[m] A[n + m]...
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the defining relations xixj + xj xi = 0 and xi degree). √k[x1, ..., xm], generated over some field k by 2 = 0 for all i, j (the grading is by (d) A is the path algebra PQ of a quiver Q (the grading is defined by deg(pi) = 0, deg(ah) = 1). Hint. The closed answer is written in terms of the adjacency matrix MQ of Q. 1...
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A which satisfy Remark 1.41. Derivations are important because they are the “infinitesimal version” of automor­ phisms (i.e., isomorphisms onto itself). For example, assume that g(t) is a differentiable family of φ, φ) such that automorphisms of a finite dimensional algebra A over R or C parametrized by t g(0) = Id. T...
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15 3. The Heisenberg Lie algebra It has the basis of matrices H 0 ⊕ ⊕ 0 0 ⊕ 0 0 0 � � 0 0 0 0 0 1 0 0 0 � ⎝ x = ⎧ � y, x 0 0 1 0 0 0 0 0 0 � ⎝ y = ⎧ � ] = 0. with relations [ ] = and [ c y, c ] = [ x, c 0 0 1 0 0 0 0 0 0 � ⎝ c = ⎧ � 4. The algebra aff(1) of matrices ( ⊕ ⊕0 0 ) Its basis consists...
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V with δ = 0 (the trivial representation). 3. The adjoint representation V = g with δ(a)(b) := [a, b]. That this is a representation follows from Equation (2). Thus, the meaning of the Jacobi identity is that it is equivalent to the existence of the adjoint representation. It turns out that a representation of a Li...
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c yx − yc − cy = 0 − ) by the relation c = 1. xc cx = 0. Note that the Weyl algebra is the quotient of ( H U 16 • (v1 + v2) w � − W, a v1 � k. where v V, w � � � w The elements v there are elements of V � � � 1.10 Tensor products In this subsection we recall the notion of tensor produ...
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� W which are not pure tensors. V, w � � W are called pure tensors. Note that in general, This allows one to define the tensor product of any number of vector spaces, V1 that this tensor product is associative, in the sense that (V1 with V1 V3). (V2 � � V2) � � Vn. Note � V3 can be naturally identified ...
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B in V , which change according to a certain rule when the basis B is changed. Here it is important to distinguish upper and lower indices, since lower indices of T correspond to V and upper ones to V ⊕. The physicists don’t write the sum sign, but remember that one should sum over indices that repeat twice - once a...
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{ V ⊃ is a basis of V and � � (c) Construct a natural isomorphism V ⊕ dimensional (“natural” means that the isomorphism is defined without choosing bases). W Hom(V, W ) in the case when V is finite � ⊃ (d) Let V be a vector space over a field k. Let S nV be the quotient of V � n (n-fold tensor product n, and s is...
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SnV nW which are defined in ⊃ an obvious way. Suppose V = W and has dimension N , and assume that the eigenvalues of A are ∂1, ..., ∂N . Find T r(SnA), T r( SnW , n : V � nA). nA : W � ⊃ √ nV ⊃ √ √ n √ (g) Show that N A = det(A)Id, and use this equality to give a one-line proof of the fact that det(AB) = det...
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both a left A-module structure and a right B-module structure which satisfy (av) b = a (vb) for B. Note that both the notions of ”left A-module” and ”right A- any v module” are particular cases of the notion of bimodules; namely, a left A-module is the same as an (A, k)-bimodule, and a right A-module is the same as...
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right C-module by (v �B w) c = v bimodule structure, then V c V and w C, v W . � � � If V is an (A, B)-bimodule and W is a (B, C)-bimodule, then these two structures on V can be combined into one (A, C)-bimodule structure on V �B W . �B W (a) Let A, B, C, D be four algebras. Let V be an (A, B)-bimodule, W...
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c HomA (V, W ) and v C, f B, f � � � � � Let A, B, C, D be four algebras. Let V be a (B, A)-bimodule, W be a (C, B)-bimodule, and X a �B V, X) as (A, D)-bimodules. W f (v) w) for all v (C, D)-bimodule. Prove that HomB (V, HomC (W, X)) ∪= HomC (W The isomorphism (from left to right) is given by f �⊃ and f...
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v v, v, w V . w w − � � (ii) The exterior algebra � V of V is the quotient of T V by the ideal generated by v � v, v V . � √ (iii) If V is a Lie algebra, the universal enveloping algebra U(V ) of V is the quotient of T V by the ideal generated by v w w � v − − � [v, w], v, w V . � It is easy ...
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the tensor product any polyhedron A let us attach its “Dehn invariant” D(A) in V = R of Q-vector spaces). Namely, � D(A) = l(a) � α(a) , β � a where a runs over edges of A, and l(a), α(a) are the length of a and the angle at a. (a) Show that if you cut A into B and C by a straight cut, then D(A) = D(B) + D(C)....
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(x)⊕. − to a representation V of a Lie algebra g is the dual It is easy to check that these are indeed representations. Problem 1.54. Let V, W, U be finite dimensional representations of a Lie algebra g. Show that the space Homg(V W, U ) is isomorphic to Homg(V, U W ⊕). (Here Homg := Hom (g)). � U � 1.14 Repre...
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generalized eigenspace corresponding to ∂. Show that E |V¯ (�) = 0. W be a nonzero vector such that Ew = 0. For any k > 0 find a polynomial Pk(x) of degree k such that E kF kw = Pk(H)w. (First compute EF kw, then use induction in k). � (c) Let v V (∂) be a generalized eigenvector of H with eigenvalue ∂. Show that...
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operators E, F, H in this basis.) � Denote the ∂ + 1-dimensional irreducible representation from (f) by V�. Below you will show that any finite dimensional representation is a direct sum of V�. (g) Show that the operator C = EF + F E + H 2/2 (the so-called Casimir operator) commutes +2) Id on V�. with E, F, H and ...
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F x = 0 and Hx = µx then Cx = µ(µ − 2 n + 1, 0 j i ∗ ∗ ∗ ∂). − (k) Define Wi = span(vi, F vi, ..., F �vi). Show that Vi are subrepresentations of V and derive a contradiction with the fact that V cannot be decomposed. (l) (Jacobson-Morozov Lemma) Let V be a finite dimensional complex vector space and A : V ⊃ V ...
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i.e., Jn(0)ei = ei normal form of A using (l),(m). − � � IdN + IdM JN (0), where Jn(0) is the Jordan block 1, i = 2, ..., n, and Jn(0)e1 = 0). Find the Jordan � 1.15 Problems on Lie algebras Problem 1.56. (Lie’s Theorem) The commutant K(g) of a Lie algebra g is the linear span g. This is an ideal in g (i.e., i...
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([x, a]) = 0 for x U containing v and invariant under x. This subspace is invariant under K(g) and any a � acts with trace dim(U )ν(a) in this subspace. In particular 0 = Tr([x, a]) = dim(U )ν([x, a]).). g and a ⊃ � � � Problem 1.57. Classify irreducible finite dimensional representations of the two dimensional Lie...
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irreducible representations. Example. Let V be an irreducible representation of A of dimension n. Then Y = End(V ), with action of A by left multiplication, is a semisimple representation of A, isomorphic to nV (the direct sum of n copies of V ). Indeed, any basis v1, ..., vn of V gives rise to an isomorphism of re...
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1, ..., vri ) = (v1, ..., vri )Xi. V is a direct sum of inclusions θi : riVi ni, and the inclusion θ : W ⊃ ⊃ ∗ � ∗ m Proof. The proof is by induction in n := i=1 ni. The base of induction (n = 1) is clear. To perform the induction step, let us assume that W is nonzero, and fix an irreducible subrepresentation P ...
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of niVi (namely, it is P gi), nmVm is the kernel of the projection hence W gi = Vi − of W gi to the first summand Vi along the other summands. Thus the required statement follows from the induction assumption. W �, where W � n1V1 → 1)Vi (ni ... ... � � � � � � Remark 2.3. In Proposition 2.2, it is not i...
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A, and v1, ..., vn V there exists an element a be any linearly independent vectors. Then for any w1, ..., wn such that avi = wi. � V A � � Proof. Assume the contrary. Then the image of the map A (av1, ..., avn) is a proper subrepresentation, so by Proposition 2.2 it corresponds to an r-by-n matrix X, r < n....
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be a basis of V , and wi = cvi. By Corollary 2.4, there exists a B, and we are done. Then a maps to c, so c End(V ), A such that avi = wi. � � � (ii) Let Bi be the image of A in End(Vi), and B be the image of A in r End(Vi) is semisimple: it is isomorphic to i=1 r End(Vi). Recall that as i=1 r diVi, where ...
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action (f a)(v) := f (av). · Proof of Theorem 2.6. First, the given representations are clearly irreducible, as for any v = 0, w Vi, there exists a A. Then, X ⊕ isomorphism �(X) = X T , as (BC)T = C T BT . n-dimensional representation of A. Define � A such that av = w. Next, let X be an n-dimensional representation...
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The goal of this exercise is to give an alternative proof of Theorem 2.6, not using any of the previous results of Chapter 2. Let A1, A2, ..., An be n algebras with units 11, 12, ..., 1n, respectively. Let A = A1 Clearly, 1i1j = ζij1i, and the unit of A is 1 = 11 + 12 + ... + 1n. A2 � ... � An. � For every repres...
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sum of copies of kd . � 2, let Eij � { 1, 2, ..., d } Hint: For every (i, j) Matd(k) be the matrix with 1 in the ith row of the jth column and 0’s everywhere else. Let V be a finite dimensional representation of Mat d(k). Show Ei1v is an isomorphism for that V = E11V . Prove that S (v) every i is a subrepres...
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→ → − Proof. The proof is by induction in dim(V ). The base is clear, and only the induction step needs V , and consider the representation to be justified. Pick an irreducible subrepresentation V1 U = V /V1. Then by the induction assumption U has a filtration 0 = U0 U1 1 = U − such that Ui/Ui 1 under the ... Vn = V...
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i) Let V be an irreducible representation of A. Let v tation. If Iv = 0 then Iv = V so there is x Thus Iv = 0, so I acts by 0 in V and hence I Rad(A). � V . Then Iv V is a subrepresen­ I such that xv = v. Then x = 0, a contradiction. n→ � → A1 (ii) Let 0 = A0 ... An = A be a filtration of the regular representati...
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representations V1, V2, . . . , Vr. By Theorem 2.5, the homomorphism δi : A � i −⊃ � i End Vi is surjective. So r irreducible representations (at most dim A). i dim End Vi ∗ ∗ ⎨ dim A. Thus, A has only finitely many non-isomorphic Now, let V1, V2, . . . , Vr be all non-isomorphic irreducible finite dimensional ...
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the irreducible representations of A are Vi, i = 1, ..., n, which are 1-dimensional, and any matrix x acts by xii. So the radical Rad(A) is the ideal of strictly upper triangular matrices (as it is a nilpotent ideal and contains the radical). A similar result holds for block-triangular matrices. Definition 2.15. A fi...
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� ≥ Next, (3) (4) by Theorem 2.6. Clearly (4) i niVi. Consider EndA(A) (endomorphisms of A as a representation of A). As the Vi’s are pairwise non- isomorphic, by Schur’s lemma, no copy of Vi in A can be mapped to a distinct Vj . Also, again by = Aop by Problem Schur’s lemma, EndA (Vi) = k. Thus, EndA(A) ∪ = 1.2...
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17. (i) Characters of (distinct) irreducible finite-dimensional representations of A are linearly independent. (ii) If A is a finite-dimensional semisimple algebra, then these characters form a basis of (A/[A, A])⊕. 27 Proof. (i) If V1, . . . , Vr are nonisomorphic irreducible finite-dimensional representations of A...
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can write A = Matd1 (k) Matdr (k). Then [A, A] = sld1 (k) � · · · � sldr (k), and A/[A, A] ∪ kr .= By Theorem 2.6, there are exactly r irreducible representations of A (isomorphic to kd1 , . . . , kdr , respectively), and therefore r linearly independent characters on the r-dimensional vector space A/[A, A]. Th...
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2.17, the charac­ ters of irreducible representations are linearly independent, so the multiplicity of every irreducible representation W of A among Wi and among Wi� are the same. This implies the theorem. 3 Second proof (general). The proof is by induction on dim V . The base of induction is clear, so let us prove ...
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..., W � . We are done. m The Jordan-H¨older theorem shows that the number n of terms in a filtration of V with irre­ ducible successive quotients does not depend on the choice of a filtration, and depends only on 3This proof does not work in characteristic p because it only implies that the multiplicities of Wi and Wi...
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Vm = V1� � V1 s n χs = 1. Now maps associated to these decompositions. Let χs = p1i p i1 : V1 � � s s s=1 we need the following lemma. Vs, p� : V s ⊃ ⊃ V1. We have ⊃ ... ... ⊃ � � � s n s ⎨ Lemma 2.20. Let W be a finite dimensional indecomposable representation of A. Then (i) Any homomorphism χ : W ⊃ W...
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χ1 + ... + χ− 1 is an − 1χn − By the lemma, we find that for some s, χs must be an isomorphism; we may assume that is indecomposable, we get that s = 1. In this case, V1� = Im(p1� i1) Ker(p1i1� ), so since V1� � V1 are isomorphisms. V1� and g := p1i1� : V1� ⊃ f := p�1i1 : V1 = �j>1Vj�; then we have V = V1 ⊃ ...
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algebra of real-valued continuous functions on R which are periodic with period 1. Let M be the A-module of continuous functions f on R which are antiperiodic with period 1, i.e., f (x + 1) = f (x). − (i) Show that A and M are indecomposable A-modules. (ii) Show that A is not isomorphic to M but A � A is isomorp...
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What is the necessary and sufficient condition on f (a) under which δU (a) is a repre­ sentation? Maps f satisfying this condition are called (1-)cocycles (of A with coefficients in Homk(W, V )). They form a vector space denoted Z 1(W, V ). (b) Let X : W V be a linear map. The coboundary of X, dX, is defined to be the f...
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are finite dimensional irreducible representations of A. For any f � Ext1(W, V ), let Uf be the corresponding extension. Show that Uf is isomorphic to Uf ⊗ as repre­ are proportional. Thus isomorphism classes (as representations) sentations if and only if f and f � V ) are parametrized by the of nontrivial extensio...
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. Let A be an algebra, and V a representation of A. Let δ : A deformation of V is a formal series ⊃ EndV . A formal δ˜ = δ0 + tδ1 + ... + tnδn + ..., End(V ) are linear maps, δ0 = δ, and δ˜(ab) = δ˜(a)˜δ(b). where δi : A ⊃ If b(t) = 1 + b1t + b2t2 + ..., where bi End(V ), and ˜ δ is a formal deformation of δ,...
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� xixj + xjxi = 2aij , x i 2 = aii. Thus, if (, ) = 0, Cl(V ) = V . √ (i) Show that if (, ) is nondegenerate then Cl(V ) is semisimple, and has one irreducible repre­ sentation of dimension 2n if dim V = 2n (so in this case Cl(V ) is a matrix algebra), and two such representations if dim(V ) = 2n + 1 (i.e., in thi...
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elements. 1)degree or by ( − − √ − (ii) Show that Cl(V ) is semisimple if and only if (, ) is nondegenerate. If (, ) is degenerate, what is Cl(V )/Rad(Cl(V ))? 2.10 Representations of tensor products Let A, B be algebras. Then A a1a2 b1b2. � B is also an algebra, with multiplication (a1 b1)(a2 b2) = � � �...
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End(V the map A End V B � ⊃ � ⊃ W ) is surjective. Thus, V � ⊃ � End V and B End W are surjective. Therefore, W is irreducible. (ii) First we show the existence of V and W . Let A�, B� are finite dimensional algebras, and M is a representation of A� A�, B� without loss of generality that A and B are finite...
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ducible M = V representation of B/Rad(B), and V, W are uniquely determined by M (as all of the algebras involved are direct sums of matrix algebras). � � � 32 3 Representations of finite groups: basic results Recall that a representation of a group G over a field k is a k-vector space V together with a GL(V ). ...
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ii), and to prove (i), it is sufficient to show that if V is a finite-dimensional representation of G and W V is any subrepresentation, then there exists a → W � subrepresentation W � as representations. V such that V = W → � Choose any complement Wˆ of W in V . (Thus V = W Wˆ as vector spaces, but not necessarily...
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Theorem 3.1(i) also holds. Proposition 3.2. If k[G] is semisimple, then the characteristic of k does not divide G . | | 33 Proof. Write k[G] = trivial one-dimensional representation. Then � r i=1 End Vi, where the Vi are irreducible representations and V1 = k is the k[G] = k r � � i=2 End V...
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= (x − − Problem 3.4. Let G be a group of order pn . Show that every irreducible representation of G over a field k of characteristic p is trivial. 3.2 Characters If V is a finite-dimensional representation of a finite group G, then its character νV : G k is defined by the formula νV (g) = tr V (δ(g)). Obviously, νV...
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Fun(G, k) f (gh) = f (hg) gh hg − ker � g, h | | ⊕ G } � g, h ⊕ , G } � Corollary 3.6. The number of isomorphism classes of irreducible representations of G equals the number of conjugacy classes of G (if G = 0 in k). | | ⇒ 34 Exercise. Show that if G = 0 in k then the number of isomor...
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1(g)δ2(g) and δ1(g)− of G. | is an abelian group: if δ1, δ2 : G Znk . Let G∗ 1 . G∗ ⊃ | | | For given n ⊂ so Z∗ = Zn. In general, 1, define δ : Zn n ∪ C× by δ(m) = e2νim/n. Then Z∗n = δk : k = 0, . . . , n { − 1 } , ⊃ (G1 G2 × × · · · × Gn)∗ = G1 ∗ G∗2 × · · · × × G∗n , = G for any finite abelian g...
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= 6, so S3 must have representations over C. If their dimensions are d1, d2, d3, then d1 two 1-dimensional and one 2-dimensional representations. The 1-dimensional representations are the trivial representation C+ given by δ(ε) = 1 and the sign representation C given by δ(ε) = ( 2 +d3 2 +d2 1)ε . − − The 2-dimensio...
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− 35 The 5 conjugacy classes are , so there are 5 different irreducible , {± representations, the sum of the squares of whose dimensions is 8, so their dimensions must be 1, 1, 1, 1, and 2. , {± , {− , {± 1 } k { } } } } 1 j i 1 The center Z(Q8) is representations of Z2 quotient map, and δ ...
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xg)). ⊃ − ∞ − × × Z2 is = S4/Z2 Z2, where Z2 e, (12)(34), (13)(24), (14)(23) { 4. The symmetric group S4. The order of S4 is 24, and there are 5 conjugacy classes: e, (12), (123), (1234), (12)(34). Thus the sum of the squares of the dimensions of 5 irreducible representations is 24. As with S3, there are two of...
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of 4 elements (on which S4 acts by permutations) with zero sum of values. = 1 while det g 1)3 = ( = − − 3 + 3 − + } + − − C C | | 3.4 Duals and tensor products of representations If V is a representation of a group G, then V ⊕ is also a representation, via δV � (g) = (δV (g)⊕)− 1 = (δV (g)− 1)⊕ = δV (g− 1)⊕. ...
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ality of characters We define a positive definite Hermitian inner product on Fc(G, C) (the space of central functions) by (f1, f2) = f1(g)f2(g). 1 G � G g � The following theorem says that characters of irreducible representations of G form an orthonormal basis of Fc(G, C) under this inner product. | | Theorem 3...
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� Theorem 3.8 gives a powerful method of checking if a given complex representation V of a finite group G is irreducible. Indeed, it implies that V is irreducible if and only if (νV , νV ) = 1. Exercise. Let G be a finite group. Let Vi be the irreducible complex representations of G. For every i, let ξi = dim Vi ...
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νV (h) = νV � (h), so the left hand side equals (using Maschke’s theorem): νV (g)νV � (h) = Tr |∈ � V V � (g V V � (h⊕)− 1) = � Tr |∈ V EndV (x gxh− 1) = Tr | C[G](x �⊃ �⊃ gxh− 1). If g and h are not conjugate, this trace is clearly zero, since the matrix of the operator x in the basis of group elements has...
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complex finite dimensional vector space V over C equipped with a G-invariant positive definite Hermitian form4 (, ), i.e., such that δV (g) are unitary operators: (δV (g)v, δV (g)w) = (v, w). 4We agree that Hermitian forms are linear in the first argument and antilinear in the second one. 38 ⇒ ⇒ Theor...
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addition and the same action of G, but complex conjugate action of scalars) is isomorphic to the dual representation V ⊕. is obviously the same thing as an invariant Indeed, a homomorphism of representations V ⊃ sesquilinear form on V (i.e. a form additive on both arguments which is linear on the first one and ant...
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a finite group G, and v1, v2, . . . , vn be an orthonormal basis of V under the invariant Hermitian form. The matrix elements of V are tV (x) = (δV (x)vi, vj ). ij Proposition 3.14. (i) Matrix elements of nonisomorphic irreducible representations are orthog­ onal in Fun(G, C) under the form (f, g) = (ii) (tV ij , tV...
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j )(xwi⊕⊗ , wj⊕⊗ ) = (P (vi � G x � � wi⊕⊗ ), vj wj⊕⊗ ) � If V = W, this is zero, since P projects to the trivial representation, which does not occur in vj⊕⊗ ). We have a G-invariant decomposition vi⊕⊗ ), vj V W ⊕. If V = W, we need to consider (P (vi � � V ⊕ V � � L C = span( C � = L = spana: vk ...
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character table lists representatives of conjugacy classes, the second one the numbers of elements in the conjugacy classes, and the other rows list the values of the characters on the conjugacy classes. Due to Theorems 3.8 and 3.9 the rows and columns of a character table are orthonormal with respect to the appropr...
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i ). where φ = exp( 2 3 The last row can be computed using the orthogonality of rows. Another way to compute the last row is to note that C3 is the representation of A4 by rotations of the regular tetrahedron: in this case (123), (132) are the rotations by 1200 and 2400 around a perpendicular to a face of the tetr...
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(12)(34) 15 1 -1 -1 0 1 (12345) 12 1 1+⊗5 2 ⊗5 − 2 -1 0 1 1 (13245) 12 1 ⊗5 − 2 ⊗5 1+ 2 -1 0 Indeed, the computation of the characters of the 1-dimensional representations is straightfor­ ward. The character of the 2-dimensional representation of Q8 is obtained from the explicit formula (3) for ...
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. is found by 3C multiplying the character of + Now the character of C3 − by the character of the sign representation. Finally, we explain how to obtain the character table of A5 (even permutations of 5 items). The group A5 is the group of rotations of the regular icosahedron. Thus it has a 3-dimensional “rotation...
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the 1-dimensional trivial representation (constant functions). The former at an element g equals to the number of items among 1,2,3,4,5 which are fixed by g. 1, 2, 3, 4, 5 } { The representation C5 is realized on the space of functions on pairs of opposite vertices of the icosahedron which has zero sum of values (che...
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C2 3C − C3 − 3C − C − C+ � � 42 C3 −3C − C5 � C5 � C4 C � C3 + 3 C+ C+ 3 C5 � C � 3 C+ A5 C C C C3 + C3 −4C C5 3.10 Problems C4 C4 C4 C4 C C3 � −3C + � C3 − � C5 C5 C4 � � � C3 + � C5 C5 C3 + � C3 + � C3 + � C3 C3 C4 � − C3 C4 � − 2C5 C3 � − 3C − �...
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order is p3). It is called the Heisenberg group. For any complex number z such that zp = 1 we define a representation of G on the space V of complex functions on Fp, by 1 1 0 0 1 0 � 0 0 1 ⎝ (δ ⎧ � f )(x) = f (x − 1), (note that zx makes sense since zp = 1). 1 0 0 0 1 1 � 0 0 1 ⎝ (δ ⎧ � f )(x) = z xf (x). (a) S...
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in V . Find a diagonal element H of GL(V ) such that δ(H) has distinct eigenvalues. (where δ is one of the above representations). This shows that if W is a subrepresentation, then it is spanned by a subset S of a basis of eigenvectors of δ(H). Use the invariance of W under the operators δ(1 + Eij ) (where Eij is de...
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representations (i.e., find the multiplicities of occurrence of all irreducible representations). (b) Do the same for the representation of G on the space of functions on the set of faces and the set of edges of the icosahedron. Problem 3.22. Let Fq be a finite field with q elements, and G be the group of nonconstant ...
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an algebra with division. (c) Find a basis 1, i, j, k of H such that 1 is the unit and i2 = j2 = k2 = ji = k, jk = 1, ij = − of invertible elements ik = j. Thus we have that Q8 is a subgroup of the group H× kj = i, ki = − − of H under multiplication. − The algebra H is called the quaternion algebra. (d) For q ...
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1 , q { − } Problem 3.24. It is known that the classification of finite subgroups of SO(3) is as follows: 1) the cyclic group Z/nZ, n ⊂ 1, generated by a rotation by 2β/n around an axis; 2) the dihedral group Dn of order 2n, n a plane containing a regular n-gon5; ⊂ 2 (the group of rotational symmetries in 3-space...
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= G . Now let P1, ..., Pk be the poles representing all the orbits of G on the set of poles, and m1, ..., mk be their orders. By counting nontrivial elements of G, show that | | 2(1 − 1 n ) = (1 − 1 mi ). � i Then find all possible mi and n that can satisfy this equation and classify the corresponding groups. (b...
https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/24d8b3fa2ce48e48ee6c2d8d5e3562f6_MIT18_712F10_replect.pdf