Weeks 6 and 7. November 24, 2013

Transcription

1 Weeks 6 and 7 November 4, 03 We start by calculating the irreducible representation of S 4 over C. S 4 has 5 congruency classes: {, (, ), (,, 3), (, )(3, 4), (,, 3, 4)}, so we have 5 irreducible representations. We start with the trivial representation, g v = v for every v. Other representation of dimension that we already know is the sign representation g v = sign g v. Now Let use the following observation: Lemma. S n has a n dimensional faithful representation. Proof. We start with n dimensional representation: g (x 0,..., x n ) = (x g(0),..., x g(n ) ). This representation is reducible, since g (,..., ) = (,... ) for every g S n. Let restrict the action of S n to the subspace Sp{e 0 e, e 0 e,..., e 0 e n }, and claim that this representation is irreducible. Let x k n such that there is i, j, x i x j. Then, for g = (i, j), g x x = (x i x j ) (e j e i ). It will be convenience to use the reducible representation as above in order to calculate the characters (by using the linearity of the characters). Denition. The tensor product of to G-modules, U k V is dened to be G-module by g(u v) := gu gv When V is one-dimensional, it is called the twisted product. Remark. This is not so trivial that this denition is well dened, and we actually using the fact that G is a group in order to identify (k[g]) op with k[g op ] with in turn is isomorphic to k[g] using the map g g. Remark. If U is irreducible representation and V is one dimensional representation, then the tensor product U V is is irreducible. This is true since we can always tensor with the dual to V and use the distributivity of the tensor product with the direct sum to get decomposition of U. So we have so far 4 irreducible representation: U 0 - the trivial representation, U - the sign representation, U - the faithful representation of dimension 3 and U 3 = U U - the twisted product. U 3 is indeed dierent representation since it has dierent character than U.

2 We have one representation left, and its dimension must be, since = 4 =. We claim that the one representation left is induced from the faithful representation of S 3. This will follow from the following observation: Claim. K = {, (, )(3, 4), (, 3)(, 4), (, 4)(, 3)} is normal subgroup of S 4 and S 3 = S 4 /K. Proof. The rst claim is direct computation. The other follows from the fact that there is only one non-abelian group of order 6. We can show that S4 /K is not abelian by observing that if it was we would have too many irreducible representations, since it would give us 6 one dimensional non-isomorphic representations. Schur indicator We will present a method to classify some properties of representations over the complex eld using there characters. We will follow [ in this part. Denition. Let V be a representation of G over C. V is called real representation if there is a representation of G over R, V 0, such that V 0 R C = V. V is called quaternion representation if there is a conjugate linear mapping ψ : V V (i.e. ψ(λv) = λψ(v)) such that ψ = Id V. A representation is complex is it's neither real or quaternion. Remark 3. If V is quaternion representation, we can actually think of V as H[G]-module, by jv = ψ(v), kv = iψ(v). The following two lemmas show how to classify whether a representation is real, complex or quaternion: Lemma 3. Let V be irreducible representation. If V is real then V posses a G-invariant symmetric non-degenerated bi-linear form. Proof. First note that there is always G-invariant bi-linear form - just start with arbitrary form H 0 on V and average it: H(u, v) = H0 (gu, gv) The question is whether this form can be symmetric and non-degenerated. Since V is real, we can choose a R[G]-module V 0 such that V = V 0 C, and choose some symmetric, non-degenerated, bi-linear, G-invariant form H on V 0. Since we are in the real eld, we may assume that this form is positivedenite. Now, we extent H to V 0 C by H(v λ, u µ) = λµh(v, u). Using the usual universal property argument (or direct calculation) we show that H is non-degenerated.

3 Lemma 4. Let V be irreducible representation. If V is quaternion representation then V posses a G-invariant skew-symmetric non-degenerated bi-linear form. Proof. Let start with a regular inner product on V and average it in order to get G-invariant non-degenerated form H 0. Note that H 0 is not bi-linear since H 0 (λv, µu) = λ µh 0 (v, u). Let H(v, u) = H 0 (v, ψ(u)). H is bi-linear, G-invariant, and non degenerated. We need to show that H is skew-symmetric. Let H (u, v) = H(v, u). H and H induce a homomorphism between V and V (we send u V to the functional x H(x, u)), so by Schur's lemma there are the same up to multiplication by some λ C. It's obvious that λ = ±, since H is not degenerated. H 0 (ψ(u), ψ(v)) = H(ψ(u), v) = λh (ψ(u), v) = λh(v, ψ(u)) = λh 0 (v, ψ (u)) = λh 0 (v, u). Now choosing v = u 0 we get that λ =, since H 0 (x, x) 0 for every x Note that this proof actually shows more: since the bi-linear form is unique (is exists) up to multiplication by scalar, V cannot be real and quaternion in the same time, since otherwise we will have two non-degenerated forms, symmetric and skew-symmetric and those cannot be multiplication by scalar of each other. We state it in the following theorem: Theorem. Let V be irreducible representation of nite group G over C. Then V is real i it has a non-degenerated symmetric bi-linear form, quaternion i it has non-degenerated skew-symmetric bi-linear form and complex i it has no non-degenerated bi-linear forms. Proof. We showed already some of those implications. We need to show that if V has a non-zero bi-linear G-invariant form, then V is real or quaternion. Let H be non-zero bi-linear G-invariant form. By Schur's lemma, H is non-degenerated (and actually H must be symmetric or skew-symmetric). We want to dene a G-invariant conjugate linear isomorphism ψ. Let x a G-invariant inner product of V and use it to dene a R[G]-modules isomorphism between V and V. Using rst the isomorphism produced by H and then the inverse of the isomorphism that was produced by the inner product, we'll get ψ - a conjugate G-linear map. Since ψ is a isomorphism of V as C[G]-module, it must be λ Id. By the similar arguments to the ones in the previous proof, λ must be real, and it's positive when H is symmetric and negative when H is skew-symmetric. If ψ = Id, then thinking of V as a linear space over R, ψ is diagonalizable and V = V + V where ψ V + = Id V+, ψ V = Id V. Since ψ is conjugate linear, i V + = V. V + is clearly R[G]-module so we get V = V + C, i..e. V is real. If ψ = Id V is quaternion by denition. 3

4 We want to use this fact in order to prove a useful formula that enable us to calculate whether a given representation is real, complex or quaternion. First let state a general fact: Denition 3. Let V be module. V S V = Sym (V ) is symmetric product, which can be seen as the sub-module of V V generated by {u v + v u}. V A V = Alt (V ) is the alternating product, which may be seen as the submodule of V V by the module generated by {u v v u}. Remark 4. Sym (V ) has the universal property that every symmetric bi-linear form factors uniquely through a linear map on it. The same hold for Alt (V ) with skew-linear forms instead. Remark 5. The module of all bi-linear forms on V can be naturally identied with V V. Under this isomorphism, the module of symmetric bi-linear forms is Sym (V ) and the module of skew-symmetric forms is Alt (V ). The fact that (assuming char k ) every bi-linear form is the sum of symmetric form and skew-symmetric form in a unique way translate to: V V = Sym (V ) Alt (V ). If V is G-module, and χ is its character, it is easy to show that character of the representation Sym (V ) is χ sym (g) = ( χ(g) + χ(g )) and of Alt (V ) is χ alt (g) = ( χ(g) χ(g )) (it can be done, for example, using the density of the diagonalizable matrices or by noting that ρ(g) is always diagonalizable). Now, by Schur's lemma, if V is irreducible representation, then if there is some G-invariant bi-linear form (i.e. an element of (V V ) G ), it must be 0 or non-degenerated, and if it's non-degenerated, there can be at most one (up to multiplication by scalar), and therefore it must be in Sym (V ) G or in Alt (V ) G. Now let χ be the character of V (in the end of this subsection we will see that the same formulas hold with the character of V ). We can extract the dimension of Sym (V ) G by the inner product of χ sym with χ trivial, since (by orthogonality of the characters) it is exactly how many copies of the trivial representation Sym (V ) has. So we get: dim Sym (V ) G = [χ(g) + χ(g )] = χ(g) + χ(g) dim Alt(V ) G = [χ(g) χ(g )] = χ(g) χ(g) dim(v V ) G = χ(g) Now, if the representation is not complex, dim(v V ) G calculation will give us: =, so easy χ(g ) = where V is real χ(g ) = where V is quaternion When the representation is complex then all term are zero, so we easily get: 4

5 χ(g ) = 0 where V is complex Now we're going to explain more closely the terms real, complex and quaternion representation. Let V be simple R[G]-module. V R C is a C[G]-module but it might be reducible. In the next lemmas we'll analyze the possibilities for reductivity of V R C and connect it to the previous denitions. Lemma 5. Let V be irreducible representation of G over R. If V R C is reducible, then it is the sum of at most two simple C[G]-modules. Proof. Assume W V C be irreducible G-invariant subspace. It's easy to see that W = { w w W } is G-invariant (where v λ := v λ). There are two possibilities (since W is irreducible): W W = {0} and therefore U = (W W ) (V ) is non-zero R[G]- module, since for every w W w + w V, and if w + w = 0 it's easy to see that iw W W. By the irreducibility of V, U = V. W W = W and therefore U = W (V ) is R[G]-module. The same reasoning will give that U {0}, and by the irreducibility of V, U = V. This proof shows us that if V C if reducible, it decomposes into two conjugate spaces. We will show next that those spaces are isomorphic i V C is quaternion, and this will provide another explanation to the dierence between real, complex and quaternion representations. Lemma 6. Assume V is simple R[G]-module, and that V C = W W. Then W is quaternion if and only if W = W. Proof. Note that we have a G-invariant conjugate map v λ v λ from W to W, but it's of course not C-linear. Assume that there is φ : W W C[G]-modules isomorphism. Since W is irreducible, by Schur's lemma there is only one such φ up to scalar from C. Now, let dene ψ : W W by ψ(x) = φ(x). This is obviously a conjugate linear G-invariant map, and ψ is an isomorphism of W, so ψ = µ Id. Multiply φ by µ to get ψ = Id. On the other hand, if ψ : W W is conjugate linear G-invariant isomorphism, then φ(x) = ψ(x) is isomorphism between W and W (since those spaces have equal dimension). Let conclude with the following observation: Theorem. Let V be simple R[G]-module. There are three possibilities: If V C is real, then End R[G] V = R. If V C is the direct sum of two none isomorphic simple representations, then they are both complex and End R[G] V = C. 5

6 If V C is the direct sum of two isomorphic simple representations, then they are both quaternion and End R[G] V = H. Proof. The theorem will follow from the following observation: End C[G] (V R C) = End R[G] (V ) R C for every simple R[G]-module V. There is a C-algebras homomorphism: η : End R[G] (V ) R C End C[G] (V R C), dened by η(f c)(v λ) = f(v) (cλ) on the generators. It's routine to verify that this is well dened C-linear map, that respects multiplication and unit. I claim that η is one to one. Otherwise, there would be non zero f, g End R[G] (V ) such that η(f + g i) = 0. By applying this function on some non zero v and remembering that f, g are one to one, we get a contradiction. Let dene a map in the opposite direction: ξ : End C[G] (V C) End R[G] (V ) C. For g End C[G] (V C), let dene ḡ(v λ) = g(v λ). The restriction of (g + ḡ) to the copy of V in V C is an element of End R[G](V ), since for every w = v λ V C, w + w = v (λ + λ) V. So (g + ḡ) is a candidate for the real part of g. In the same spirit, (g ḡ) is a candidate from the imaginary part, meaning that i (g ḡ) V End R[G](V ). We denote the restriction map, together with the identication of the restricted element with element in End R[G] (V ) by r. Let dene: ξ(g) = r( (g + ḡ)) r( i (g ḡ)) i End R[G](V ) C We have to show that ξη = Id, ηξ = Id. For every g End C[G] (V ), ηξ(g) = η(r( (g + ḡ)) r( i (g ḡ)) i) = (g + ḡ) id C + ( i (g ḡ)) (i id C) = g For every f λ End R[G] (V ) C, ξη(f λ) = ξ(f (λ id C )). Note that f (λ id C )(v µ) = f(v) λµ, so it's clear that ξη(f λ) = f λ. 6