REPRESENTATION THEORYWEEK 51. Invariant formsRecall that a bilinear form on a vector space V is a mapB : V × V → ksatisfyingB (cv, dw) = cdB (v, w) , B (v1+ v2, w) = B (v1, w)+B (v2, w) , B (v, w1+ w2) = B ( v, w1)+B (v, w2) .One can also think ab out a bilinear form as a vector in V∗⊗ V∗or as a homo-morphism B : V → V∗given by the formula Bv(w) = B (v, w). A bilinear form issymme tr ic if B (v, w) = B (w, v) and skew-symmetric if B (v, w) = −B (w, v). Everybilinear form is a sum B = B++ B−of a symmetri c and a skew-symmetric form,B±(v, w) =B (v, w) ± B (w, v)2.Such decomposition corresponds to the decomposition(1.1) V∗⊗V∗= S2V∗⊕Λ2V∗.A form is non-degenerate if B : V → V∗is an isomorphism, in other words B (v, V ) =0 implies v = 0.Let ρ : G → GL (V ) be a re presentation. We say that a bilinear form B on V isG-invariant ifB (ρsv, ρsw) = B (v, w)for any v, w ∈ V , s ∈ G.The foll owing properties of an invariant form are easy to check(1) If W ⊂ V is an invariant subspace, then W⊥= {v ∈ V | B (v, W ) = 0} isinvariant. In particul ar, Ker B is invariant.(2) B : V → V∗is invariant iff B ∈ HomG(V, V∗).(3) If B is invariant, then B+and B−are invariant.Lemma 1.1. Let ρ be an irreducible representation of G, then any bilinear invariantform is non-degenerate. If¯k = k, then a bilinear form is unique up to multiplicationon a scalar.Proof. Follows from (2) and Schur’s lemma. Date: September 29, 2005.12 REPRESENTATION THEORY WEEK 5Corollary 1.2. A represe ntation ρ of G admits an invariant form iff χρ(s) = χρ(s−1)for any s ∈ G.Lemma 1.3. If¯k = k, then an invariant form on an irreducible representation ρ iseither symmetric or skew-symmetric. Letmρ=1|G|Xs∈Gχρs2.Then mρ= 1, 0 or −1. If mρ= 0, then ρ does not admi t an invariant form. Ifmρ= ±1, then mρadmits a symmetric (skew-symmetric) invariant form.Proof. Recall that ρ ⊗ρ = ρalt⊕ ρsym.(χsym, 1) =1|G|Xs∈Gχρ(s2) + χρ(s2)2,(χalt, 1) =1|G|Xs∈Gχρ(s2) − χρ(s2)2.Note that1|G|Xs∈Gχρs2= (χρ, χρ∗) .Therefore(χsym, 1) =(χρ, χρ∗) + mρ2, (χalt, 1) =(χρ, χρ∗) − mρ2If ρ does not have an invariant form, then (χsym, 1) = (χalt, 1) = 0, and χρ∗6= χρ,hence (χρ, χρ∗) = 0. Thus, mρ= 0.If ρ has a symmetric invariant form, t hen (χρ, χρ∗) = 1 and (χsym, 1) = 1. Thisimplies mρ= 1. Similarl y, if ρ admits a skew-symmetric invariant form, then mρ=−1. Let k = C. An irreducible representation is called real if mρ= 1, complex if mρ= 0and quaternionic if mρ= −1. Since χρ(s−1) = ¯χρ(s), then χρtakes only real valuesfor real and quaternionic representations. If ρ is complex then χρ(s) /∈ R at least forone s ∈ G.Example. Any irreducible representation of S4is real. A non-trivi al representa-tion of Z3is complex. The two-dimensional representation of quaternionic group isquaternionic.Exercise. Let |G| be odd. Then any non-trivial irreduc ible representation of Gover C is complex.REPRESENTATION THEORY WEEK 5 32. Some generalities about field extensionLemma 2.1. If char k = 0 and G is finite, then a representation ρ : G → GL (V ) isirreducible iff EndG(V ) is a divisi on ring.Proof. In one direction it is Schur’s Lemma. In the opposite dire ction if V is notirreducible, then V = V1⊕ V2, then the projectors p1and p2are intertwiners suchthat p1◦ p2= 0. For any extension F of k and a representation ρ : G → GL (V ) over k we defineby ρFthe re presentation G → GL (F ⊗kV ).For any re presentation ρ : G → GL (V ) we denote by VGthe subspace of G-invariants in V , i.e.VG= {v ∈ V | ρsv = v, ∀s ∈ G}.Lemma 2.2. (F ⊗kV )G= F ⊗kVG.Proof. The embedding F ⊗kVG⊂ (F ⊗kV )Gis trivial. On the other hand, VGisthe image of the operatorp =1|G|Xs∈Gτs,in particular dim VGequals the rank of p. Since rank p does not depend on a field,we havedim F ⊗kVG= dim (F ⊗kV )G. Corollary 2.3. Let ρ : G → GL (V ) and σ : G → GL (W ) be two representationsover k. ThenHomG(F ⊗kV, F ⊗kW ) = F ⊗ HomG(V, W ) .In particular,dimkHomG(V, W ) = di mFHomG(F ⊗kV, F ⊗kW ) .Proof.HomG(V, W ) = (V∗⊗ W )G. Corollary 2.4. Even i f a field is not algebraically closeddim HomG(V, W ) = (χρ, χσ) .A representation ρ : G → GL (V ) over k is called absolutely irreducible if it remainsirreducible after any extension of k. This is equi valent to (χρ, χρ) = 1. A field issplitting for a group G if any irreducible representation is absolutely irreducible. Itis not difficult to see that some finite extension of Q is a splitting field for a finitegroup G.4 REPRESENTATION THEORY WEEK 53. Representations over RA biline ar symmetric form B is positive definite if B (v, v) > 0 for any v 6= 0.Lemma 3.1. Every representation of a finite group over R admits positive-definiteinvariant symmetric form. Two invariant symmetric forms on an ir reducible r epre-sentation are proportional.Proof. Let B′be any p ositive definite form. DefineB (v, w) =1|G|Xs∈GB′(ρsv, ρsw) .Then B is positive definite and invariant.Let Q (v, w) b e another invariant symmetric form. Then from linear algebra weknow that they can be diagonalized i n the same basis. Then for some λ ∈ R,Ker (Q −λB) 6= 0. Si nce Ker (Q − λB) is invariant, Q = λB. Theorem 3.2. Let R ⊂ K be a division ring, finite-dimensi onal over R. Then R isisomorphic R, C or H (quaternions).Proof. If K is a field, then K∼=R or C, b ecause C =¯R and [C : R] = 2. Assumethat K is not commutative. For any x ∈ K\R, R [x] = C. There fore we have a chainR ⊂ C ⊂ K. Let f (x) = ixi−1. Obviously f is an automorphism of K and f2= id.Hence K = K+⊕K−, whereK±= {x ∈ K | f ( x) = ±x}.Moreover, K+K+⊂ K+, K−K−⊂ K+, K+K−⊂ K−, K−K+⊂ K−. If x ∈ K+,then C [x] is a finite extension of C. Therefore K+= C. For any nonzero y ∈ K−the left multiplic ation on y defines an isomorphism of K+and K−as vector spacesover R. In parti cular dimRK−= dimRK+= 2. For any y ∈ K−, x ∈ C, we havey¯x = xy, therefore y2∈ R. Moreover, y2< 0. (If y2> 0, then y2= b2for somereal b and (y − b) (y + b) = 0, which is impossible). Put j =y√−y2. Then we havek = ij = −ji, ki = (ij) i = j, K = R [i, j] is isomorphic to H. Lemma 3.3. Let ρ : G → GL (V ) be an ir r educible representation over R, then thereare three possibilities:(1) EndG(V ) = R and (χρ, χρ) = 1;(2) EndG(V )∼=C and (χρ, χρ) =