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A then for any Z(A), the operator δ(z) by which z acts in V has only one eigenvalue νV (z), equal to the k is a z scalar by which z acts on some irreducible subrepresentation of V . Thus νV : Z(A) homomorphism, which is again called the central character of V . ⊃ � (c) Does δ(z) in (b) have to be a scalar opera...
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from this that C(θ) is uncountably dimensional and obtain a contradiction. V is a scalar operator. → 1.4 Ideals A left ideal of an algebra A is a subspace I ideal of an algebra A is a subspace I subspace that is both a left and a right ideal. ∧ ∧ A such that Ia ∧ I for all a ∧ � A such that aI I for all ...
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algebra and I a two-sided ideal in A. Then A/I is the set of (additive) cosets of I. A/I be the quotient map. We can define multiplication in A/I by β(a) β(b) := β(ab). Let β : A This is well defined because if β(a) = β(a�) then ⊃ · β(a�b) = β(ab + (a� a)b) = β(ab) + β((a� a)b) = β(ab) − − because (a� a)b Ib �...
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posable representation of A. ⊂ V is cyclic if it Problem 1.25. Let V = 0 be a representation of A. We say that a vector v generates V , i.e., Av = V . A representation admitting a cyclic vector is said to be cyclic. Show that � (a) V is irreducible if and only if all nonzero vectors of V are cyclic. (b) V is cyc...
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= 0, . . . , fm = , we say that the algebra 1.7 Examples of algebras 1. The Weyl algebra, k x, y ◦ / yx ◦ − � xy − 2. The q-Weyl algebra, generated by x, x− x− 1x = yy− 1 = y− 1y = 1. 1 . � 1, y, y− 1 with defining relations yx = qxy and xx− 1 = Proposition. (i) A basis for the Weyl algebra A is (ii) A basis fo...
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f = dt (where Suppose now that we have a nontrivial linear relation 1] (here ta is just a formal symbol, so really E = k[a][t, t− cij x yj = 0. Then the operator := (a + n)ta+n d(ta+n) dt df − i acts by zero in E. Let us write L as L = i cij t � j ⎨ d dt � � L = r � j=0 Qj(t) � j , d dt � where Qr = 0...
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of the Weyl algebra, if k has characteristic zero (check it!), but not in characteristic p, where (d/dt)pQ = 0 for any polynomial Q. However, the representation E = tak[a][t, t− 1], as we’ve seen, is faithful in any characteristic. Problem 1.26. Let A be the Weyl algebra, generated by two elements x, y with the relat...
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, and A be the q-Weyl algebra over C generated by x± 1 with defining relations xx− 1y = 1, and xy = qyx. 1x = 1, yy− 1 and y± 1 = x− 1 = y− (a) What is the center of A for different q? If q is not a root of unity, what are the two-sided ideals in A? (b) For which q does this algebra have finite dimensional representat...
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sense that for each quiver Q, there exists a certain algebra PQ, called the path algebra of Q, such that a representation of the quiver Q is “the same” as a representation of the algebra PQ. We shall first define the path algebra of a quiver and then justify our claim that representations of these two objects are “the...
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piah = 0 for i = h�� We now justify our statement that a representation of a quiver is the same thing as a represen­ tation of the path algebra of a quiver. Let V be a representation of the path algebra PQ. From this representation, we can construct a ph⊗⊗ V representation of Q as follows: let Vi = piV, and for any...
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representation (Vi, xh) of a quiver Q is a representation Wh⊗⊗ for (Wi, x�h) where Wi E. all h Wh⊗⊗ and x�h = xh|Wh⊗ : Wh⊗ I and where xh(Wh⊗ ) Vi for all i −⊃ ∧ ∧ � � Definition 1.36. The direct sum of two representations (Vi, xh) and (Wi, yh) is the representation (Vi Wi, xh yh). � � As with represe...
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of A[n]). Often this series converges to a rational function, and the answer is written in the form of such function. For example, if A = k[x] and deg(xn) = n then �n E. −⊃ ⎨ ∞ ∧ hA(t) = 1 + t + t2 + ... + tn + ... = 1 t 1 − Find the Hilbert series of: (a) A = k[x1, ..., xm] (where the grading is by degree of ...
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1.39. (g, [ , ]) is a Lie algebra if [ , ] satisfies the Jacobi identity Example 1.40. Some examples of Lie algebras are: [a, b] , c [b, c] , a + � � + � [c, a] , b = 0. � (2) � � 1. Any space g with [ , ] = 0 (abelian Lie algebra). 2. Any associative algebra A with [a, b] = ab − ba . 3. Any subspace U of an a...
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we see that Lie algebras arise as spaces of infinitesimal automorphisms (=derivations) of associative algebras. In fact, they similarly arise as spaces of derivations of any kind of linear algebraic structures, such as Lie algebras, Hopf algebras, etc., and for this reason play a very important role in algebra. Here...
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and Y = ( 0 1 ), with [X, Y ] = Y . 0 0 0 0 5. so(n), the space of skew-symmetric n × n matrices, with [a, b] = ab − ba. Exercise. Show that Example 1 is a special case of Example 5 (for n = 3). Definition 1.42. Let g1, g2 be Lie algebras. A homomorphism � : g1 linear map such that �([a, b]) = [�(a), �(b)]. −⊃...
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defined by [xi, xj ] = universal enveloping algebra defining relations xixj k cij xk. The (g) is the associative algebra generated by the xi’s with the U k k cij xk. xjxi = ⎨ k − ⎨ Remark. This is not a very good definition since it depends on the choice of a basis. Later we will give an equivalent definition whic...
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formal symbols v � W of vector spaces V and W over a field k is the quotient W , by the subspace V , w w, v � � � (v1 + v2) w v1 � − V, w w v2 − � W, a � k. � � where v � w, v � (w1 + w2) − v � w1 v � − w2, av � w − a(v � w), v � aw − a(v � w), Exercise. Show that V group...
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, E := V � tensors of type (m, n) on V . For instance, tensors of type (0, 1) are vectors, of type (1, 0) - linear functionals (covectors), of type (1, 1) - linear operators, of type (2, 0) - bilinear forms, of type (2, 1) - algebra structures, etc. (V ⊕)� � � n n = V If V is finite dimensional with basis ei, i = ...
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the Einstein summation, and it also stipulates that if an index appears once, then there is no summation over it, while no index is supposed to appear more than once as an upper index or more than once as a lower index. One can also define the tensor product of linear maps. Namely, if A : V are linear maps, then one...
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, and s is some transposition. V � of V ) by the subspace spanned by the tensors T n by the subspace spanned by the tensors T such that s(T ) = T nV be the quotient of V � Also let √ for some transposition s. These spaces are called the n-th symmetric, respectively exterior, power nV ? If dimV = m, what are their...
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= det(A)Id, and use this equality to give a one-line proof of the fact that det(AB) = det(A) det(B). √ Remark. Note that a similar definition to the above can be used to define the tensor product �A W �A W , where A is any ring, V is a right A-module, and W is a left A-module. Namely, V V is the abelian group which...
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a (k, A)-bimodule. A and b V , a � � � Let B be a k-algebra, W a left B-module and V a right B-module. We denote by V v k-vector space (V a pure tensor v � tensor product V vb �k W ) / ◦ � w (with v �B W is the one defined in the Remark after Problem1.49.) bw W, b V, w W ) onto the space V B � �...
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)-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 be a (B, C)-bimodule, �C X) as (A, D)-bimodules. �B (W V , �C x) for all v �B (w v �⊃ and X a (C, D)-bimodule. Prove that (V The isomorphism (from left to right) is given by (v w �C X ∪ V = �B w) ...
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�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 HomB (V, HomC (W, X)). �B v V , w (w �⊃ � � � 1.11 The tensor algebra The notion of tensor product allows us to give more conceptual...
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ii) 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 to see that a choice of a basis x1, ..., xN in V identifies SV with the polynomial algebra √k(x1, ..., xN ), and the universal enveloping alge...
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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). (b) Show that ϕ = arccos(1/3)/β is not a rational number. Hint. Assume that ϕ = 2m/n, f...
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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 Representations of sl(2) This subsection is devoted to ...
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and w 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 ∂. Sh...
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..., F �v is a basis of V , and compute the matrices of the 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 ...
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and therefore form a basis basis, show that F j vi, 1 ∗ 2) x and hence µ = of V (establish that if 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 decom...
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+ νW (x) and νV C. Show that νV � � (n) Let V = CM CN , and A = JM (0) of size n with eigenvalue zero (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....
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common eigenspaces of K(g) (for this you will need to K(g). To prove this, consider the smallest vector subspace K(g) common eigenvector v in V , that is there is a linear function ν : K(g) for any a show that ν([x, a]) = 0 for x U containing v and invariant under x. This subspace is invariant under K(g) and an...
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dimension. Construct explicitly a basis of this algebra. 22 2 General results of representation theory 2.1 Subrepresentations in semisimple representations Let A be an algebra. Definition 2.1. A semisimple (or completely reducible) representation of A is a direct sum of irreducible representations. Example. Le...
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representations of A, and W be a subrepresentation of V = i=1niVi. Then W is isomorphic to niVi given �i=1riVi, ri by multiplication of a row vector of elements of Vi (of length ri) by a certain ri-by-ni matrix Xi with linearly independent rows: θ(v1, ..., vri ) = (v1, ..., vri )Xi. V is a direct sum of inclusi...
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of subrepresentations of V , preserving the property we need to establish: namely, under the action of gi, the matrix Xi goes to Xigi, while Xj , j = i don’t change. Take gi Gi such that (q1, ..., qni )gi = (1, 0, ..., 0). Then W gi contains the first summand Vi of niVi (namely, it is P gi), nmVm is the kernel of ...
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summand V1. Then K is a subrepresentation of (n1 − 1)V1 � ... � nmVm, which is nonzero since W is not irreducible, so K contains an irreducible subrepresentation by the induction assumption. 23 ⇒ ⇒ 2.2 The density theorem Let A be an algebra over an algebraically closed field k. Corollary 2.4. Let V be an irreduc...
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. Then the map δ : A EndV is surjective. ⊃ Vr, where Vi are irreducible pairwise nonisomorphic finite dimensional (ii) Let V = V1 ... � � representations of A. Then the map r δi : A i=1 � ⊃ � r End(Vi) is surjective. i=1 Proof. (i) Let B be the image of A in End(V ). We want to show that B = End(V ). Let c ...
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r = kdr , and any finite dimensional representation of A is a direct sum of copies of V1, . . . , Vr. i Matdi (k) for any field k. � � In order to prove Theorem 2.6, we shall need the notion of a dual representation. Definition 2.7. (Dual representation) Let V be a representation of any algebra A. Then the is the re...
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is clearly surjective, as k A. Thus, the dual map θ⊕ ∪= An as representations of A (check it!). Hence, Im θ⊕ where yi} is injective. But An tation of An . Next, Matdi (k) = diVi, so A = Hence by Proposition 2.2, X = : X An ∪= X is a subrepresen­ r i=1ndiVi, as a representation of A. i=1diVi, An = r −⊃ � � → ⊕ ⊕...
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1v1, a2v2, ..., anvn)). �⊃ � i � { then V1 on V1 1, 2, ..., n } V2 � ... � V2 � � � (a) Show that a representation V of A is irreducible if and only if 1iV is an irreducible repre­ , while 1iV = 0 for all the other i. Thus, classify the sentation of Ai for exactly one i � { irreducible representations of A ...
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), where S (v2) E22V � { ... � � � � � � � � � � v1, v2, ..., vk} { (c) Conclude Theorem 2.6. 2.4 Filtrations Let A be an algebra. Let V be a representation of A. A (finite) filtration of V is a sequence of subrepresentations 0 = V0 ... Vn = V . V1 → → → Lemma 2.8. Any finite dimensional representation V of an...
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2 ⊃ → → → − 25 2.5 Finite dimensional algebras Definition 2.9. The radical of a finite dimensional algebra A is the set of all elements of A which act by 0 in all irreducible representations of A. It is denoted Rad(A). Proposition 2.10. Rad(A) is a two-sided ideal. Proof. Easy. Proposition 2.11. ...
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Ai+1/Ai by zero, so x maps Ai+1 to Ai. This implies that Rad(A)n = 0, as desired. → → → � Theorem 2.12. A finite dimensional algebra A has only finitely many irreducible representations Vi up to isomorphism, these representations are finite dimensional, and A/Rad(A) ∪ = End Vi. � i Proof. First, for any irreducible ...
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by definition, is exactly Rad(A). Corollary 2.13. i (dim Vi)2 ∗ ⎨ dim A, where the Vi’s are the irreducible representations of A. Proof. As dim End Vi = (dim Vi)2, Theorem 2.12 implies that dim A 0, (dim Vi)2 . As dim Rad(A) (dim Vi)2 dim A. − dim Rad(A) = dim End Vi = i ⎨ i ⎨ ⊂ i ⎨ ∗ 26 ⇒ ⇒ ⇒ ...
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⎨ 3. A ∪= 4. Any finite dimensional representation of A is completely reducible (that is, isomorphic to a Matdi (k) for some di. � i direct sum of irreducible representations). 5. A is a completely reducible representation of A. Proof. As dim A 0. Thus, (1) − (2). ⊆ dim Rad(A) = (dim Vi)2, clearly dim A = i ⎨...
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(k))op = � (5). To see that (5) (k), as desired. (3), let A = Matni Matni � ≥ ≥ i (k). Thus, A ∪ = ( i i � � = ∪ i � 2.6 Characters of representations Let A be an algebra and V a finite-dimensional representation of A with action δ. Then the character of V is the linear function νV : A k given by ⊃ If [A...
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) = 0 for all a EndkVi. But each tr(Mi) can range independently over k, so it must be that ∂1 = ∂iTr(Mi) = 0 for all Mi � = ∂r = 0.) A, then End V1 δVr : A � · · · � ⊃ � ⎨ ⎨ · · · (ii) First we prove that [Matd(k), Matd(k)] = sld(k), the set of all matrices with trace 0. It is sld(k). If we denote by Eij the...
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Jordan-H¨older theorem We will now state and prove two important theorems about representations of finite dimensional algebras - the Jordan-H¨older theorem and the Krull-Schmidt theorem. Theorem 2.18. (Jordan-H¨older theorem). Let V be a finite dimensional representation of A, ... Vm� = V be filtrations of V , such t...
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assumption the theorem holds for V /W1. So assume W1 = W1�. In this case W1 W1, W 1� are irreducible), so we have an embedding f : W1 W1�), and 1 (it exists by 0 = U0 U1 → → Lemma 2.8). Then we see that: ... Up = U be a filtration of U with simple quotients Zi = Ui/Ui ∈ V . Let U = V /(W1 W1� ⊃ → � � − 1) V /W...
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Wi ⊗ are the same modulo p, which is not sufficient. In fact, the character of the representation pV , where V is any representation, is zero. 28 ⇒ ⇒ V . This number is called the length of V . It is easy to see that n is also the maximal length of a filtration of V in which all the inclusions are str...
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⊃ � � � s n s ⎨ Lemma 2.20. Let W be a finite dimensional indecomposable representation of A. Then (i) Any homomorphism χ : W ⊃ W is either an isomorphism or nilpotent; (ii) If χs : W ⊃ W , s = 1, ..., n are nilpotent homomorphisms, then so is χ := χ1 + ... + χn. Proof. (i) Generalized eigenspaces of χ are subre...
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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 ⊃ �j>1Vj , B� defined as a composition of the natural maps B B�. Consider the map attached to these h : B B � decompositions. We claim that h is an is...
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A and M are indecomposable A-modules. (ii) Show that A is not isomorphic to M but A � A is isomorphic to M M . � 29 Remark. Thus, we see that in general, the Krull-Schmidt theorem fails for infinite dimensional modules. However, it still holds for modules of finite length, i.e., modules M such that any filtration...
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(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 function A ⊃ Homk(W, V ) given by dX(a) = δV (a)X only if X is a homomorphism of representations. Thus coboundaries form a subspace B 1(W, ...
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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 extensions of W by V (i.e., those not isomorphic to W projective space PExt1(W, V ). In particular, every extension is trivial if and only...
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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, then b˜ δb− 1 is also a deformation of δ, which is ...
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= 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 this case Cl(V ) is a direct sum ...
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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) = � � � Exercise. ...
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Proof. (i) By the density theorem, the maps A End W = 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� ...
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(A W , where V is an irreducible representation of A/Rad(A) and W is an irreducible 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 representati...
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and one has dim(Vi)2 . G = | | � i (the “sum of squares formula”). Proof. By Proposition 2.16, (i) implies (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 ...
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= W ker P = W �. Thus, W � W � � is invariant under the action of G and is therefore a subrepre­ is the desired decomposition into subrepresentations. � The converse to Theorem 3.1(i) also holds. Proposition 3.2. If k[G] is semisimple, then the characteristic of k does not divide G . | | 33 ...
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representation of G over k is trivial (so k[Z/pZ] indeed is not semisimple). Indeed, an irreducible representation of this group is a 1-dimensional space, on which the generator acts by a p-th root of unity, and every p-th 1)p over k. root of unity in k equals 1, as xp 1 = (x − − Problem 3.4. Let G be a group o...
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a basis of (A/[A, A])⊕, where A = k[G]. It suffices to note that, as vector spaces over k, (A/[A, A])⊕ = ∪ = ∪ which is precisely Fc(G, k). � { f { � � Homk(k[G], k) � 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 irreducibl...
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of irreducible representations of G. Every element of G forms a conjugacy class, so G∗ = G . Recall that all irreducible representations over C (and algebraically closed fields in general) of commutative algebras and are irreducible C× groups are one-dimensional. Thus, G∗ is called the dual or character group re...
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symmetric group S3. In Sn, conjugacy classes are determined by cycle decomposition sizes: two permutations are conjugate if and only if they have the same number of cycles of each length. For S3, there are 3 conjugacy classes, so there are 3 different irreducible 2 = 6, so S3 must have representations over C. If the...
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the span of a subset of the eigenvectors of δ((123)), which are different vectors. Thus, V must be either C2 or 0. 3. The quaternion group Q8 = 1, i, i = jk = kj, − {± ± j = ki = ± j, ± ik, , with defining relations k } k = ij = ji, − − 1 = i2 = j2 = k2 . − 35 The 5 conjugacy classes are...
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∀ − − , 1 � δ(k) = 0 ∀ − � − 1 − 1 ∀ − 0 � . (3) These are the Pauli matrices, which arise in quantum mechanics. Exercise. Show that the 2-dimensional irreducible representation of Q8 can be realized in 1f (g) (the action of G is by right the space of functions f : Q8 multiplication, g C such that f (gi) = ∀ ...
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of rotations of a cube acting by permuting the interior diagonals (or, equivalently, on a regular octahedron permuting pairs of opposite faces); this gives the 3-dimensional representation C3 +. The last 3-dimensional representation is C3 , the product of C3 with the sign representation. C3 and C3 are different, fo...
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1 = ∂−i ∂i = ∂i = νV (g). � as representations (not just as vector spaces) if and only if νV (g) � � In particular, V ∪= V ⊕ g G. � R for all � 36 If V, W are representations of G, then V W is also a representation, via � W (g) = δV (g) δV � δW (g). � Therefore, νV W (g) = νV (g)νW (...
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� � G g � � G g � W � (P ), νV (g)νW � (g) 1 where P = G � | | representation of G then G g � g ⎨ Z(C[G]). (Here Z(C[G]) denotes the center of C[G]). If X is an irreducible P X = | � Id, if X = C, 0, X = C. Therefore, for any representation X the operator P X is the G-invariant projector onto the subspace | X...
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element of k [G], and thus acts on Vj as an intertwining operator. Corollary 1.17 thus yields that ξi acts on Vj as a scalar. Compute this scalar by taking its trace in Vj . Here is another “orthogonality formula” for characters, in which summation is taken over irre­ ducible representations rather than group elemen...
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�⊃ 1 Remark. Another proof of this result is as follows. Consider the matrix U whose rows are labeled by irreducible representations of G and columns by conjugacy classes, with entries U V,g = is the number of elements νV (g)/ conjugate to G. Thus, by Theorem 3.8, the rows of the matrix U are orthonormal. This mea...
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ne another form B as follows: B(v, w) = B(δV (g)v, δV (g)w) � G g � Then B is a positive definite Hermitian form on V, and δV (g) are unitary operators. If V is an irreducible representation and B1, B2 are two positive definite Hermitian forms on V, then B1(v, w) = B2(Av, w) for some homomorphism A : V V (since a...
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­ ducible. Proof. Let W be a subrepresentation of V . Let W � under the Hermitian inner product. Then W � This implies that V is completely reducible. be the orthogonal complement of W in V W �. is a subrepresentation of W , and V = W � Theorems 3.11 and 3.12 imply Maschke’s theorem for complex representations ...
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Thus, matrix elements of irreducible representations of G form an orthogonal basis of Fun(G, C). Proof. Let V and W be two irreducible representations of G. Take of V and { forms. Let wi⊕ � to be an orthonormal basis to be an orthonormal basis of W under their positive definite invariant Hermitian be the linear ...
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), � � k,l and P projects to the first summand along the second one. The projection of vi is thus � vi⊕⊗ to C C L → � This shows that ζii⊗ dim V vk vk⊕ � � (P (vi vi⊕⊗ ), vj vj⊕⊗ ) = ζii⊗ ζjj⊗ dim V � which finishes the proof of (i) and (ii). The last statement follows immediately from the sum of squar...
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1 2 C − C2 (12) 3 1 -1 0 (123) 2 1 1 -1 It is obtained by explicitly computing traces in the irreducible representations. For another example consider A4, the group of even permutations of 4 items. There are three one-dimensional representations (as A4 has a normal subgroup Z2 Z2 = Z3). Since there are f...
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to two opposite edges. Example 3.15. The following three character tables are of Q8, S4, and A5 respectively. Q8 # C++ C+ C C −− C2 − + − 1 1 1 1 1 1 2 -1 1 1 1 1 1 -2 i 2 1 1 -1 -1 0 j 2 1 -1 1 -1 0 k 2 1 -1 -1 1 0 S4 # C+ C − C2 C3 + C3 − A5 # C C3 + C3 − C4 C5 Id ...
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Q8 is obtained from the explicit formula (3) for this representation, or by using the orthogonality. For S4, the 2-dimensional irreducible representation is obtained from the 2-dimensional irre­ S3, which allows to obtain its ducible representation of S3 via the surjective homomorphism S4 character from the charac...
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in which (12)(34) is the rotation by 1800 around an axis perpendicular to two opposite edges, (123) is the rotation by 1200 around an axis perpendicular to two opposite faces, and (12345), (13254) are the rotations by 720, respectively 1440, around axes going through two opposite vertices. The character of this repr...
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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 (check that it is irreducible!). The character of this representation is computed similarly to the character of C4, or from the orthogonality formula. 3.9 Co...
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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 − � � � 2C4 � �+ � C5 C5 C4 ...
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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) Show that ...
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ei} Hint: Choose a basis 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...
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G on Fun(I). | | (a) Decompose this representation in a direct sum of irreducible 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. Probl...
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EndR(V ) consisting of endomorphisms of V as a real representation. Show that H is 4-dimensional and closed under multiplication. Show that every nonzero element in H is invertible, i.e., H is 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 =...
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rotations of the three-dimensional Euclidean space. Show that this homomorphism is surjective and that its kernel is H spanned by i, j, k, by x 1, ⊃ ⊃ → � . 1 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...
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G fixes a unique pair of opposite poles, and that the subgroup of G fixing a particular pole P is cyclic, of some order m (called the order of P). Thus the orbit of P has n/m elements, where n = 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. ...
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on V ⊕ SV Show that this map is surjective and use this to deduce the desired result. whose stabilizer in G is 1. Now define the map to the function fu on G given by fu(g) = f (gu). ⊃ � Problem 3.27. This problem is about an application of representation theory to physics (elasticity theory). We first describe the...
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, i.e., an element of S 2V . This matrix is called the deformation tensor at P . ⊃ Now we define the stress tensor, which characterizes stress. Let v be a small nonzero vector in . Let Fv be the force with which V , and ε a small disk perpendicular to v centered at P of area the part of the material on the v-side o...
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(V ) admits a decomposition R W , where R is the trivial representation, V is the standard 3-dimensional representation, and W is a 5-dimensional representation of SO(3). Show that S2V = R W V � � � (b) Show that V and W are irreducible, even after complexification. Deduce using Schur’s W one has f (x + y) = Kx ...
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V is C for V of complex type, Mat2(R) for V of real type, and H for V of quaternionic type, which motivates the names above. Hint. Show that the complexification VC of V decomposes as V V ⊕. Use this to compute the EndR[G] V , prove the result dimension of EndR[G] V in all three cases. Using the fact that C in th...
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(b) Show that V is of real type if and only if V is the complexification of a representation V R over the field of real numbers. Example 4.2. For Z/nZ all irreducible representations are of complex type, except the trivial one 1)m, which are of real type. For S3 all three ( and, if n is even, the “sign” representati...
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∗ ⎨ Proof. Let A : V ⊃ V have eigenvalues ∂1, ∂2, . . . , ∂n. We have Tr S2V (A Tr �2V (A | | A) = ∂i∂j A) = � i j → � i<j ∂i∂j � � Thus, Tr |S2V (A � A) − Tr |�2V (A � A) = ∂2 = Tr(A2). i � 1 i n → → 47 Thus for g � G we have Therefore, νV (g 2) = νS2V (g) − ν�2V (g) 1νV ( − G | | g...
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in G. Exercise. Show that any nontrivial finite group of odd order has an irreducible representation which is not defined over R (i.e., not realizable by real matrices). 4.2 Frobenius determinant Enumerate the elements of a finite group G as follows: g1, g2, . . . , gn. Introduce n variables indexed with the elements...
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is ⎨ − − ( · Now we are ready to proceed to the proof of Theorem 4.7. 48 Proof. Let V = C[G] be the regular representation of G. Consider the operator-valued polynomial L(x) = xgδ(g), � G g � where δ(g) � EndV is induced by g. The action of L(x) on an element h G is � L(x)h = xgδ(g)h = xggh...
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and algebraic integers We are now passing to deeper results in representation theory of finite groups. These results require the theory of algebraic numbers, which we will now briefly review. Definition 4.9. z root of a monic polynomial with rational (respectively, integer) coefficients. C is an algebraic number (resp...
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. � ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ Since z is a root of the characteristic polynomial of this matrix, it is an eigenvalue of this matrix. The set of algebraic numbers is denoted by Q, and the set of algebraic integers by A. Proposition 4.12. (i) A is a ring. (ii) Q is a field. Namely, it is an algebraic closure of the field of rational ...
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. + an 1x + an, − z = p q � Q, gcd(p, q) = 1. Notice that the leading term of p(x) will have qn in the denominator, whereas all the other terms / Z, a contradiction. Thus, will have a lower power of q there. Thus, if q = � Z is a root of x z Z. The reverse inclusion follows because n 1, then p(z) n. Q ...
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