FAMILIES OF IRREDUCIBLE REPRESENTATIONS OF S S 3 JAY TAYLOR We would like to consider the representation theory of the Weyl group of type B 3, which is isomorphic to the wreath product S S 3 = (S S S ) S 3 We will consider the elements of S to be {, x} throughout Also we let C = {} and C = {x} be the two conjugacy classes of S We will denote an element of the wreath product by (x, x, x 3 : π) where x i S and π S 3 Conjugacy Classes We recall from [JK09, Theorem 48 that the conjugacy classes in S S n are parameterised by their type We now define what we mean by type using an example First note that it is clear that for two entries in S S n to be conjugate we must have that the entries in the S n component are conjugate, ie of the same cycle type The notion of type for the wreath product is a generalisation of cycle type for S n Consider our case S S 3 and think of the 3 matrix [ 3 a a a 3 x a a a 3 Down the left hand side we give conjugacy class representatives for S and along the top row we give the lengths of possible cycles in a cycle decomposition in S 3 The entry a ik in the matrix corresponds to the number of cycle products, (see [JK09, 4), associated to a cycle of length k which are in the conjugacy class C i of S These matrices satisfy certain conditions [JK09, 43 and a complete list of such matrices for S S 3 is given by 3 0 0 0 0 0 0, 0 0 0 0 3 0 0 0 0 0 0 0, 0 0 0 0 0 0 0 0 0 From these matrices it is now possible to write down corresponding entries in the wreath product that are conjugacy class representatives
JAY TAYLOR (,, : ) (x,, : ) (x, x, : ) (x, x, x : ), (,, : ()) (,, x : ()) (x,, : ()) (x,, x : ()), (,, : (3)) (x,, : (3)) From left to right we label these conjugacy classes K,, K 0 By [JK09, Lemma 40 we have a simple way to calculate the order of the conjugacy class in the wreath product and hence the order of the centraliser We give this information below K K K 3 K 4 K 5 K 6 K 7 K 8 K 9 K 0 K i 3 3 6 6 6 6 8 8 C S S 3 (K i ) 48 6 6 48 8 8 8 8 6 6 From this it is clear that the conjugacy class K 4 must contain the longest element of the Weyl group as this is always in its own conjugacy class Obtaining Representations We let, ε denote the two irreducible representations of S where ε denotes the sign representation Recall from the theory of wreath products that N = S {} S S 3 is a normal subgroup isomorphic to S S S, (see [JK09, 45) A complete list of irreducible representations for this normal subgroup is given by ε ε ε, ε ε ε, ε ε ε ε ε ε, As N is a normal subgroup of S S 3 we can consider the action of this group on N by conjugation We then also have that S S 3 will act on the irreducible characters of N by composition with conjugation We consider the stabiliser of this action in S S 3 Clearly N is in the stabiliser and in fact we will have that the stabiliser is of the form S H for some subgroup H S 3 Let n be the number of copies of the trivial character in the representation of S 3 and n the number of copies of the sign character Then by [JK09, Theorem 437 the stabiliser is isomorphic to the group S (S n S n ) We call the stabiliser of this action the inertia group of the irreducible representation The component S n S n is called the inertia factor By [JK09, Theorem 443 we have that two irreducible representations χ, ψ of S 3 are in the same equivalence class under the conjugation action by S S 3 if and only if the representations contain the same number of copies of the trivial and sign representation Hence representatives of these equivalence classes are given by
FAMILIES OF IRREDUCIBLE REPRESENTATIONS OF S S 3 3 ε ε ε ε ε ε By [JK09, 438 we can extend these representations of N to its inertia group and we denote this extension χ Let [3, [,, [ 3 be the irreducible representations of S 3, [, [ be the irreducible representations of S and [ be the irreducible representation of S Then every irreducible representation of S S 3 is given by tensoring the extension of an irreducible representation of N with the inflation of an irreducible representation of its inertia factor and inducing to the whole group For example, letting [λ denote the inflation of [λ, we have the complete list of irreducible representations of this wreath product is given by (λ, µ) Representation χ() Inertia Factor Inertia Group (3, ) ( ) [3 S 3 S S 3 (, ) ( ) [, S 3 S S 3 ( 3, ) ( ) [ 3 S 3 S S 3 (, ) (ε ) ([ [ ) S S 3 3 S S S S (, ) (ε ) ([ [ ) S S 3 3 S S S S (, ) ( ε ε) ([ [ ) S S 3 3 S S S S (, ) ( ε ε) ([ [ ) S S 3 3 S S S S (, 3) (ε ε ε) [3 S 3 S S 3 (, ) (ε ε ε) [, S 3 S S 3 (, 3 ) (ε ε ε) [ 3 S 3 S S 3 Note that when the inertia factor is S 3 we don t need to induce to the whole group because we already have an irreducible representation of the wreath product Also we recall that the index of S S is given by [S S 3 : S S = S 3 S 3 S 3 S = S 3 S = 6 = 3 Hence this explains the dimensions of the representations We have listed 0 irreducible representations of the wreath product By a specialisation of [JK09, 436 we have that the number of irreducible representations of S S n is given by (n,n ) p(n )p(n ),
4 JAY TAYLOR such that n, n N 0 and n +n = n and p(n i ) is the number of partitions of n i Therefore applying this to our case we have the number of irreducible representations of S S 3 is p(3)p(0) + p()p() = 3 + = 6 + 4 = 0 Hence we are likely correct Indeed recalling the fact that S S 3 = S 3 S 3 = 3 3! = 48 we have from the dimensions of the irreducible representations that + + + 3 + 3 + 3 + 3 + + + = 48 and so we are almost certainly correct 3 The Character Table We now go about calculating the character table of S S 3 We note that in the table above we have indexed the characters in the standard way by bi-partitions A bi-partition is a pair (λ, µ) such that λ, µ n are partitions of n and λ + µ = n K K K 3 K 4 K 5 K 6 K 7 K 8 K 9 K 0 K i 3 3 6 6 6 6 8 8 C S S 3 (K i ) 48 6 6 48 8 8 8 8 6 6 χ (3, ) χ (, ) 0 χ ( 3, ) χ (,) 3 3 0 0 χ (,) 3 3 0 0 χ (,) 3 3 0 0 χ (, ) 3 3 0 0 χ (,3) χ (,) 0 χ (, 3 ) To calculate the first and last three characters in this table it is simply a case of multiplying the character values together as we already obtain an irreducible representation of the whole wreath product by tensoring In order to calculate the middle four characters we will need to induce the representations from the inertia group We note that the inertia group is S (3) = S S for all of these representations Therefore we will need to know
FAMILIES OF IRREDUCIBLE REPRESENTATIONS OF S S 3 5 how the conjugacy classes split upon restriction to this subgroup By sheer calculation we see that K S (3) = {(,, : )}, K S (3) = {(x,, : )} {(, x, : ), (,, x : )}, K 3 S (3) = {(x, x, : ), (x,, x : )} {(, x, x : )}, K 4 S (3) = {(x, x, x : )}, K 5 S (3) = {(,, : (3)), (, x, x : (3))}, K 6 S (3) = {(x,, : (3)), (x, x, x : (3))}, K 7 S (3) = {(, x, : (3)), (,, x : (3))}, K 8 S (3) = {(x, x, : (3)), (x,, x : (3))} This information gives us the structure of the conjugacy classes in S (3) We can now use this information to calculate the uninduced characters of this group K K K K 3 K 3 K 4 K 5 K 6 K 7 K 8 K i S (3) C S S 3 (K i ) 48 6 6 6 6 48 8 8 8 8 C S (3) (K i ) 6 8 6 8 6 6 8 8 8 8 (ε ) ([ [ ) (ε ) ([ [ ) ( ε ε) ([ [ ) ( ε ε) ([ [ ) This then makes it very easy to construct the induced characters in the table of S S 3 Alternatively one could use [GHL + 96 to produce the character table 4 Lusztig s Symbols We are interested in how the irreducible representations of S S 3 fall into families in the sense of [Lus84, 4 Lusztig gives a criterion [Lus84, 456 for two irreducible representations to be in the same family in a Weyl group of type B n This criterion is based upon certain combinatorial symbols that he associates to the representations Recall that to each representation we associated a bi-partition (λ, µ) of n The current way we have expressed the partitions λ, µ are as weakly decreasing partitions However it would make no different to consider them as weakly increasing This allows us
6 JAY TAYLOR to add as many 0 s as we like to the front of our partitions without affecting their meaning By doing this we can always arrange that λ has exactly one more entry than µ by adding a suitable number of zeroes to the front of the partitions We let this new partition be denoted (λ, µ ) To each bipartition (λ, µ) we can associate a combinatorial symbol [Λ using the bipartition (λ, µ ) Let λ i, µ i denote the entries of λ, µ then we define a new pair ( λ, µ) by λ = (λ, λ +, λ 3 +,, λ m+ + m) and µ = (µ, µ +, µ 3 +,, µ m + m ) Then Lusztig s combinatorial symbol, denoted [Λ, associated to (λ, µ) is given by [ λ λ λm λm+ [Λ = µ µ µ m In the following table we give the partitions (λ, µ ) and the Lusztig symbols [Λ for each of the partitions (λ, µ) (λ, µ) (λ, µ ) ( λ, µ) [ΛΛ χ (, 3 ) 3 0 3 (3, ) (3, ) (3, ) 3 3 0 (, ) (, 0) (3, 0) 0 3 3 0 ( 3, ) ( 3, 0 ) (3, 0) 0 3 0 3 0 3 (, ) (0, ) (03, ) 0 (, ) (, ) (, ) 0 (, ) (0, ) (0, ) 0 3 0 3 (, ) (0, ) (03, ) 0 3 (, 3) (0, 3) (0, 3) 3 0 0 3 (, ) (0 3, ) (0, 3) 3 0 [ 0 3 3 (, 3 ) (0 4, 3 ) (03, 3) 3 Now Lusztig s criterion, (see [Lus84, 456), states that two irreducible representations of the Weyl group of type B n are in the same family if and only if their symbols [Λ contain
FAMILIES OF IRREDUCIBLE REPRESENTATIONS OF S S 3 7 the same entries For example {χ (, ), χ (,), χ (,3) } is a family of irreducible characters in S S 3 because their associated symbols [Λ are 3 0 3 0 0 3 Therefore using this criterion we have the families of irreducible characters of S S 3 are {χ (3, ) } {χ (,)} {χ (,) } {χ (, 3 )} {χ (, ), χ (,), χ (,3) } {χ ( 3, ), χ (, ), χ (,) } Indeed using [GHL + 96 to calculate the truncated induction tables from parabolic subgroups one can verify that these are the families of irreducible characters in S S 3 Recall that χ (, 3 ) is the sign representation of S S 3 Now Lusztig prescribes a way to determine [Λ sgn in the case of a Weyl group of type B n Let [Λ be a symbol representing an irreducible representation of W (B n ) with associated partitions (λ, µ ) Then if t = max{λ i, µ j i m + and j m} we define the symbol for [Λ sgn in the following way The top row is given by (t j 0 j t and j µ,, µ m ) and the bottom row is given by (t i 0 i t and i λ,, λ m+ ) In the table we have calculated these symbols in our case 5 The Fourier Transform Matrix We wish to describe how, given a family F of irreducible characters of S S 3, we can obtain the associated Fourier transform matrix Let F be the family {χ ( 3, ), χ (, ), χ (,) } which have associated symbols [ 0 3 0 3 3 0 Now in each family there is a unique special symbol In the case of type B n the unique special symbol is the symbol satisfying the condition that λ µ λ µ µ m λ m+ Therefore the unique special symbol in the family F is 0 3 [Z = To the special symbol [Z we wish to define a set of symbols M F A symbol in M F will be of the form λ λ λ b, µ µ µ b such that b < b and b + b = m +, (in our case m + = 5), and 0 λ < λ < < λ b, 0 µ < µ < < µ b, where {λ,, λ b, µ,, µ n } is precisely the set of entries in [Z For example, in our case we would have the set of symbols M F is
8 JAY TAYLOR [ 0 3 0 3 3 0 0 3 It s clear to see that we can identify the symbols corresponding to F in M F as the set of all symbols such that b = b + Hence we have an embedding F M F We have any element of M F is of the form [ Z (Z M) if Z (Z M) > Z M, Z M [Λ M = [ Z M Z (Z M) if Z M > Z (Z M), where we define the sets Z, Z and M in the following way We let Z be the set of elements which appear in both rows of the special symbol [Z We let Z be the set of entries that appear only once in [Z and M is a subset of Z such that M d (mod ) where d is such that Z = d + For example, in our case we have Z = {}, Z = {0,, 3} and the respective sets M are {}, {3}, {0} and {0,, 3} Note that we can identify the special symbol [Z as [Λ M0 where M 0 is the set of elements of Z appearing in the second row of [Z Hence in our case M 0 = {} Now we associate to each set M a subset M # Z where M # = (M M 0 ) (M M 0 ) This gives us a bijection [Λ M M # between the set of symbols M F and the set V Z of all subsets of Z of even cardinality We consider the sets M # in our example [Λ M M M # 3 {0} {0, } 0 0 3 {} 0 {3} {, 3} 3 0 3 {0,, 3} {0, 3} It s easy to verify that V Z = {, {0, }, {, 3}, {0, 3}} is the set of all subsets of Z of even cardinality We can turn V Z into a vector space over the field F by defining addition to be the symmetric difference of sets Let A, B be two sets then we define their symmetric difference to be the set A B = (A B) (A B) For example {0, } {0, 3} = {, 3} This vector space is, in general, of dimension d and hence is of dimension in our case
FAMILIES OF IRREDUCIBLE REPRESENTATIONS OF S S 3 9 Let M #, M # be two elements of V Z then we define a nonsingular symplectic form, : V Z V Z F on V Z by M #, M # = M # M # (mod ) We wish to define a basis for V Z Let Z = {z,, z d+ } be such that z z z d+, then for each i d we define an element e i = {z i, z i+ } V Z For example in our case we will have two elements e = {0, } and e = {, 3} The elements e i will form a basis for the vector space V Z Let Γ be the span over F of the basis elements e, e 3, e 5, and Γ be the span over F of the basis elements e, e 4, e 6,, then we have V Z = Γ Γ For example in our case we have Γ = {, e } = {, {0, }} and Γ = {, e } = {, {, 3}} and it s clear that V Z = Γ Γ We now associate to the family F a finite group G(F) in the following way Let G(F) be the abelian group Γ In general this will be isomorphic to Z Z but in our case it is just isomorphic to Z Recall that M(G(F)) is the collection of all pairs (x, σ) such that x G(F) and σ is an irreducible character of C G(F) (x) Note that in the case of B n we have G(F) is always abelian and so C G(F) (x) = G(F), for all x G(F) Using the pairing, we have Γ = Hom(Γ, F ) via the map γ ϕ γ, where ϕ γ (δ) = ( ) δ,γ for each δ Γ and γ Γ Therefore we canonically have V Z = Γ Γ = M(G(F)) Now for each two pairs (x, σ), (y, τ) M(G(F)) we have their corresponding entry in the Fourier transform matrix will be / k σ(y)τ(x), where k is the number of copies of Z that appear in G(F) Therefore in our case we will have k = and hence the entry of the Fourier transform matrix will be /σ(y)τ(x) This means the Fourier transform matrix will be (, ϕ ) (, ϕ e ) (e, ϕ ) (e, ϕ e ) (, ϕ ) (, ϕ e ) (e, ϕ ) (e, ϕ e ) Finally we comment that we have an embedding F M(G(F)) given by the embedding F M F composed with the bijection M F V Z and the isomorphism V Z = M(G(F)) References [GHL + 96 M Geck, G Hiss, F Lübeck, G Malle, and G Pfeiffer CHEVIE A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras Appl Algebra Engrg Comm Comput, 7:75 0, 996 [JK09 Gordon James and Adalbert Kerber The Representation Theory of The Symmetric Group, volume 6 of Encylopedia of Mathematics and its Applications Cambridge University Press, paperback reissue edition, 009 [Lus84 George Lusztig Characters of Reductive Groups Over a Finite Field Number 07 in Annals of Mathematics Studies Princeton University Press, 984