Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 20 August 2013

Lecture 10

Proposition 2.6 If 𝒮m is defined in terms of diagrams, with si i i+1 then {s1,,sm-1} generate 𝒮m.


We will examine the permutation σ=(14)(365)𝒮6.

σ = = s3 s4 s5 s1 s2 s3 s1

Thus σ=s3s4s5s1s2s3s1.

Definition 2.7 A reduced word for σ𝒮m is an expression σ=si1sip of minimal length. The number p is the length of σ, often denoted (σ).

Recall (or note) that the length of a permutation σ is equal to the number of crossings in the diagram for σ (with suitable conventions on the diagram – no superfluous crossings, e.g.). The length is also equal to the number of inversions of the permutation. We will also need in what follows the fact that every presentation of a permutation of a product of si has its number of generators in the presentation congruent to (σ) modulo two.

Definitions 2.8 The set of inversions of σ𝒮m is the set inv(σ)= { (i,j)|1i <jm,σ(i)> σ(j) } The sign is ε(σ)=(-1)(σ).

Note that in 𝒮m each permutation has the same cycle type as its inverse, and therefore for any character χ one has χ(σ-1)=χ(σ)=χ(γτ(σ)). By an abuse of notation, we will sometimes denote χ(γμ) by χ(μ).

For the trivial representation of 𝒮m, we have V(m):𝒮mM1():σ(1). The associated character is χ(m):𝒮m:σ1. One has that χ(m)(γμ)=1.

Now, let us consider the alternating representation of 𝒮m. Denoted by V(1m), it is generated by sending each si to the matrix (-1). One has V(1m): 𝒮m M1() si (-1) σ=si1 si(σ) (-1)(σ) (Again, note that V(1m)(σ) is independent of the choice of presentation of σ.) This representation gives rise to the character χ(1m): 𝒮m si -1 σ (-1)(σ) Since characters are class functions, we only need consider their behavior on cycle structures; we have that χ(1m)(γμ)= (-1)(γμ)= (-1)m-(μ), since (γμ) = ( γμ1× γμ2×× γμk ) = (μ1-1)+ (μ2-1) (μk-1) = m-(μ)

Recall that the minimal central idempotent for the trivial representation was z(m)=1m! σ𝒮mσ For the alternating representation, we have z(1m) = 1m!σ𝒮m d(1m)χ(1m) (σ-1)σ = 1m!σ𝒮m (-1)(σ)σ = 1m!σ𝒮m ε(σ)σ

Homework Problem 2.9 Find analogs of the above representations, characters, and idempotents for the Hecke algebras.

We have seen the two one-dimensional representations of 𝒮m. Now we will consider the permutation representation, wherein V:𝒮mMm() sends each permutation to the appropriate permutation matrix. This representation (unlike the trivial and alternating representations) is not irreducible; in fact VV(m)V(m-1,1). Note that the determinant of the permutation matrix σ is (-1)(σ)=ε(σ), so by taking the composition of the permutation representation and the determinant map one (essentially) obtains the alternating representation.

Now we consider the representation of 𝒮m on words of length m. Let v1,,vn be n letters (i.e. indeterminate symbols), and let Vm=-span of { vi1vim |1i1, ,imn } 𝒮m acts on Vm from the right by vi1vim· σ=viσ(1) viσ(m)

Definition 2.10 Let x1,,xn be commuting variables. Define the weight of vi1vim to be wt(vi1vim)=xi1xim.

Note, then, that upon re-ordering, wt(vi1vim)=x1λ1xnλn, where λj is the number of occurrences of the letter vj in the word vi1vim. Note also that wt(vi1vimσ)=wt(vi1vim) for σ𝒮m.

Definition 2.11 A sequence λ=(λ1,,λn) such that λj0 and λj=m is called a composition of m, denoted λm.

Definition 2.12 Fix a composition λm. The following are all submodules of Vm. 1s𝒮m= -span of { vi1 vim| wt(vi1vim) =xλ }

The fact that the above are submodules follows from the stability of the weight operator under the action of 𝒮m.

Definition 2.13 The weighted trace of a permutation σ𝒮m, denoted wtr(σ), is wtr(σ)= vi1vim (vi1vimσ|vi1vim) xi1xim

We would like to compute the characters of the representations 1s𝒮m. Note that trλ(σ) = vi1vimwt(vi1vim)=xλ vi1vimσ |vi1vim = wtr(σ)|xλ Now, since the trace tr of a permutation is completely determined by the trace of the associated cycle structure, the same is true of the weighted trace wtr. Thus to compute the traces trλ, it suffices to compute the weighted traces wtr(γμ).

Lemma 2.14 If μ=(μ1,,μk)m, so γμ=γμ1××γμk, then wtr(γμ)=wtr(γμ1)wtr(γμ2)wtr(γμk).

wtr(γμ) = 1i1,imm (vi1vim) (γμ1×γμ2××γμk) |vi1vim xi1xim = ( i1,,iμ1 (vi1viμ1) γμ1 |vi1viμ1 xi1xiμ1 ) × × ( iμ1+1 ,,iμ2 (viμ1+1viμ2) γμ2 |viμ1+1viμ2 xiμ1+1 xiμ2 ) = wtr(γμ1) wtr(γμ2)

The above works since γμ1 only acts on the first μ1 many characters, γμ2 only acts on the μ1+1 through μ2 characters, etc..

Proposition 2.15 If r>0, and r, then wtr(γr)=x1r++xnr.


Recall that vi1vi2virγr=virvi1vir-1. Thus wtr(γr) = 1i1,,irn vi1vir·γr |vi1vir xi1xir = 1i1,,irn virvi1vir-1 |vi1vir xi1xir = ir=i1=i2==ir-1 xi1xir = 1i1n xi1r = x1r+x2r++ xnr.

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

page history