REPRESENTATION THEORY.WEEK 3VERA SERGANOVA1. Examples.Example 1. Let G = S3. There are three conjugacy classes in G, which wedenote by some element in a class:1,(12),(123). Therefore there are three irreduciblerepresentations, denote their characters by χ1, χ2and χ3. It is not difficult to seethat we have the following table of chara cters1 (12) (123 )χ11 1 1χ21 −1 1χ32 0 −1The characters of one-dimensional representations are given in the first and the secondrow, the last character χ3can be obtained by using the identity(1.1) χperm= χ1+ χ3,where χpermstands fo r the character of the permutation representat io n.Example 2. Let G = S4. In this case we have the following character table (inthe first row we write the number of elements in each conjugacy class).1 6 8 3 61 (12) (123) (12) (34) (1234)χ11 1 1 1 1χ21 −1 1 1 −1χ33 1 0 −1 −1χ43 −1 0 −1 1χ52 0 −1 2 0First two rows are characters of one-dimensional representations. The third can beobtained again from (1.1), χ4= χ2χ3, the correspo nding representation is obtainedas the tensor product ρ4= ρ2⊗ ρ3. The last character can be found fro m theorthogonality relation. Alternative way to describe ρ5is to consider S4/V4, whereV4= {1, (12) (34) , (13) (24) , (14) (23)}Date: February 1, 2011.12 VERA SERGANOVAis the Klein subgroup. Observe that S4/V4∼=S3, and therefore the two-dimensionalrepresentation σ of S3can be extended to the representation of S4by the formulaρ5= σ ◦ p,where p : S4→ S3is the natural projection.Solution of the marcian problem. Recall that S4is isomorphic to the groupof rotations of a cube. Hence it acts on the set of faces F , and therefore we have arepresentationρ: S4→ GL (C (F )) .It is not difficult to calculate χρusing the formulaχρ(s) = |{x ∈ F | s (x) = x}|.We obtainχρ(1) = 6, χρ((12)) = χρ((123)) = 0 , χρ((12) (34)) = χρ((1234)) = 2.Furthermore,(χρ, χ1) = (χρ, χ4) = (χρ, χ5) = 1, (χρ, χ2) = (χρ, χ3) = 0.Hence χρ= χ1+ χ4+ χ5, and C (F ) = W1⊕ W2⊕ W3the sum of three invariantsubspaces. The intertwining operator T : C (F ) → C (F ) of food redistribution mustbe a scalar operator on each Wiby Schur’s Lemma. Note thatW1=(Xx∈Fax | a ∈ C),W2=(Xx∈Faxx | ax= −axop),W3=(Xx∈Faxx |Xax= 0, ax= axop),where xopdenotes the face opposite to the f ace x. A simple calculation shows thatT|W1= Id, T|W2= 0, T|W3= −12Id. Therefore Tn(v) approaches a vector in W1asn → ∞, and eventually everybody will have the same amount of food.Example 3. Now let G = A5. There are 5 irreducible representation of G overC. Here is the character table1 20 15 12 121 (123) (12 ) (34) (12345) (12354)χ11 1 1 1 1χ24 1 0 −1 −1χ35 −1 1 0 0χ43 0 −11+√521−√52χ53 0 −11−√521+√52REPRESENTATION THEORY. WEEK 3 3To obtain χ2use the permutation representation and (1.1) again. Let χsymandχaltbe the characters of the second symmetric and the second exterior powers of ρ2respectively. Then we obtain1 (123) (12 ) (34) (12345) (12354)χsym10 1 2 0 0χalt6 0 −2 1 1It is easy to check t hat(χsym, χsym) = 3, (χsym, χ1) = (χsym, χ2) = 1 .Thereforeχ3= χsym− χ1− χ2is the character of an irreducible representation.Furthermore,(χalt, χalt) = 2, (χalt, χ1) = (χalt, χ2) = (χalt, χ3) = 0.Therefore χalt= χ4+ χ5is the sum of two irreducible characters. First we find thedimensions of ρ4and ρ5using12+ 42+ 52+ n24+ n25= 60.We obtain n4= n5= 3. The equations(χ4, χ1+ χ2) = 0, (χ4, χ3) = 0implyχ4((123)) = 0, χ4((12) (34)) = −1.The same argument showsχ5((123)) = 0, χ5((12) (34)) = −1.Finally denotex = χ4((12345)) , y = χ4((12354))and write down the equation (χ4, χ4) = 1. It is1609 + 15 + 12x2+ 12y2= 1,or(1.2) x2+ y2= 3.On the other hand, (χ4, χ1) = 0, that gives3 − 15 + 12 (x + y) = 0,or(1.3) x + y = 1.4 VERA SERGANOVAOne can solve (1.2) and (1.3)x =1 +√52, y =1 −√52.The second solutionx =1 −√52, y =1 +√52will give the last character χ5.2. ModulesLet R be a ring, usually we assume that 1 ∈ R. An abelian group M is calleda (left) R-module if there is a map α : R × M → M, (we write α (a, m) = am )satisfying(1) (ab) m = a (bm);(2) 1m = m;(3) (a + b) m = am + bm;(4) a (m + n) = am + an.Example 1. If R is a field then any R-module M is a vector space over R.Example 2. If R = k (G) is a group algebra, then every R-module defines therepresentation ρ: → GL (V ) by the formulaρsv = svfor any s ∈ G ⊂ k (G), v ∈ V . Conversely, every representat io n ρ: G → GL (V ) in avector space V over k defines on V a k (G)-module structure by Xs∈Gass!v =Xs∈Gasρsv.Thus, representations of G over k are k (G)-modules.A submodule is a subgroup invariant under R-a ction. If N ⊂ M is a submodulethen the quotient M/N has the natural R-module structure. A module M is simpleor (irreducible) if any submodule is either zero or M. A sum and an intersection ofsubmodules is a submo dule.Example 3. If R is an arbitrary ring, then R is a left module over itself, wherethe action is given by the left multiplication. Submodules are left ideals.For any two R-modules M and N one can define an abelian group HomR(M, N)and a ring EndR(M) in the manner similar to the group case. Schur’s Lemma holdsfor simple modules in the following form.Lemma 2.1. Let M and N be simple modules and ϕ ∈ HomR(M, N), then eitherϕ is an isomorphism or ϕ = 0. For a simple module M, EndR(M) is a division ring.REPRESENTATION THEORY. WEEK 3 5Recall that for every ring R one defines Ropas the same abelian group with newmultiplication * given bya ∗ b = ba.Lemma 2.2. The ring EndR(R) is isomorphic to Rop.Proof. For each a ∈ R define ϕa∈ End (R) by the formulaϕa(x) = xa.It is easy to check that ϕa∈ EndR(R) and ϕba= ϕa◦ ϕb. Hence we constructed ahomomorphism ϕ : Rop→ EndR(R). To prove that ϕ is injective let ϕa= ϕb. Thenϕa(1) = ϕb(1), i.e. a = b. To prove surjectivity of ϕ, note that for any γ ∈ EndR(R)one hasγ (x) = γ (x1) = xγ (1) .Therefore γ = ϕγ(1). Lemma 2.3. Let ρi: G → GL (Vi), i = 1, . . . , l be pairwise non-isomorphic irre-ducible representations of a gro up G over a lgebraically closed field k, andV = V⊕m11⊕ ··· ⊕V⊕mll.ThenEndG(V )∼=Endk(km1) × ··· × Endk(kml) .Proof. First, note that the Schur’s Lemma implies that ϕV⊕mii⊂ V⊕miifor anyϕ ∈ EndG(V ), i = 1, . . . , l. HenceEndG(V )∼=EndGV⊕m11× ··· ×