A bijection between reflections and roots

In this part, we try to link the reflections in W(sl_(C)) with the roots ofsl_n(C). More precisely, Theorem3.1.13 gives the required bijection.

Construction 3.1.9. Given an expressionsi1si2. . . sin ∈S

∗_{, we continue to introduce} a couple of definitions: for 1≤j≤n,

tij := (si1si2. . . sij−1)sij(si1si2. . . sij−1)

−1_∈T,

the ordered n-tuple

b

T(si1si2. . . sin) := (ti1, ti2, . . . , tin),

and

n(si1si2. . . sin :t) := the number of times t appears in Tb(si1si2. . . sin).

There are few interesting properties pertaining to ti such as

tijsi1si2. . . sin =si1. . .scij. . . sin

with sij omitted and

si1si2. . . sij =tijtij−1· · ·ti1.

Moreover, we want to study the group S(Ω) of all permutations of the set Ω =

Definition 3.1.10. Define a bijective functionπs fromR to itself for eachs∈S such that πs(t, ) := (sts, η(s:t)) where η(s:t) := ( −1, if s=t, +1, if s6=t.

Proposition 3.1.11. If si1si2. . . sin, withn minimal, then for all 1≤j < k≤n,

tij 6=tik ∈Tb(si1si2. . . sin).

Proof. Suppose, for contrary,tij =tik for somej < k. But,

si1si2. . . sin =tijtikw=tijtiksi1si2. . . sin =si1. . .scij. . .scik. . . sin

contradicting the minimality ofsi1si2. . . sin.

QED Lemma 3.1.12. (i) The mappings7→πsextends uniquely to an injective homomor-

phism w7→πw from W to S(Ω).

(ii) For allt∈T, πt(t, ) = (t,−).

Proof. (i) First, we want to utilise the universal property. Suppose we have (s, s) with

m(s, s0)6=∞. Then,

π_s2(t, ) =πs(sts, η(s:t)) = (sstss, η(s:t)η(s:sts)) = (t, )

which implies that π_s2 =idR and πs∈S(Ω).

Next, suppose (s, s0) withm(s, s0) =p6=∞. For the sake of notation simplicity, we denote

si = (

s0, ifi is odd, s, ifii is even.

We want to show that

(πsπs0)p =id_R.

Now, we claim that

(πsπs0)p(t, ) = (s_2p. . . s₁ts₁. . . s_2p,

2p Y

i=1

The first component is clear since s2p. . . s1 = (ss0)p =eby assumption. Note that for

the second component,

2p Y

i=1

η(si:si−1. . . s1ts1. . . si−1) =(−1)n(s1s2...s2p:t).

To make sure that it is equal to, we need to shown(s1s2. . . s2p :t) is even. Consider b

T(s0ss0s . . . s0s

| {z }

2p

) and we found that

tp+i=ti

for 1≤i≤pas (s, s0)p =e. Thus, iftappears inTb(s0ss0s . . . s0s

| {z }

2p

), it must appear twice, as desired.

By the universal property, we can extend uniquely the mapping s 7→ πs to a ho- momorphism w 7→ πw from W to S(Ω). Let w = sinsin−1. . . si1 be any arbitrary expression. We see that

πw(t, ) =πsinπsin−1. . . πsi1(t, ) =  sin. . . si1tsi1. . . sin, n Y j=1 η(sij :sij−1. . . si1tsi1. . . sij−1)   = (wtw−1, (−1)n(si₁si₂...sin:t)_).

This is suggesting us to define

η(w−1 :t) := (−1)n(si1si2...sin:t)

which is well-defined since we just showed that n(s1s2. . . sn : t) depends only on

w and t. Hence, we can write our new homomorphism as

πw(t, ) = (wtw−1, η(w−1:t)).

This homomorphism is injective. Suppose w 6= e. Write w = sinsin−1. . . si1 with

nminimal. Then, by Proposition 3.1.11, we know that

n(si1si2. . . sin :ti) = 1 sinceTb(s_i1s_i2. . . s_in) = (t₁, t₂, . . . , t_n) andt_k 6=t_l fork6=l. It follows that the homomorphism is injective since

(ii) We will prove by induction. Suppose t=si1. Obviously,

πt(t, ) = (ttt−1, η(t:t)) = (t,−).

Now, lett=si1si2. . . sin. . . si2si1. Then, by inductive hypothesis together with part (i),

πt(t, ) =πsi₁si₂...sin...si₂si₁(si1si2. . . sin. . . si2si1, ) =πsi₁πsi₂...sin...si₂πsi₁(si1si2. . . sin. . . si2si1, ) =πsi₁πsi₂...sin...si₂(si1si1si2. . . sin. . . si2si1si1, η(si1 :si1si2. . . sin. . . si2si1)) =πsi₁πsi₂...sin...si₂(si2. . . sin. . . si2, η(si1 :si1si2. . . sin. . . si2si1)) =πsi1πsi2...sin...si2(si2. . . sin. . . si2, ) =πsi₁(si2. . . sin. . . si2,−) = (si1si2. . . sin. . . si2si1,−η(si1 :si2. . . sin. . . si2)) = (t,−), as desired. QED Theorem 3.1.13. Suppose Ω =T × {+1,−1} and πw : Ω→Ω and let ∆ be the root system of sln(C).The map φ: Ω→∆ defined by

(t,+1)→ γ ∈∆+ such that tγ=t, (t,−1)→ γ ∈∆− such that tγ=t. is bijective satisfying for all (t, )∈Ω andw∈W,

φ(πw(t, )) =w(φ(t, )).

Proof. The mapφis surjective because every root is associated with a reflection. Fur- thermore, the mapφ is injective because the adjoint representation is faithful.

To check that φ(πw(t, )) = w(φ(t, )) for all (t, ) ∈ Ω and w ∈ W, it suffices to check it on the generatorsi, that is, for allsi,

φ(πsi(t, )) =si(φ(t, )). Note that φ(πsi(t, )) =φ(sitsi, η(si :t)) =si(γ)∈∆ η(si:t) and si(φ(t, )) =si(γ)

forγ ∈∆.By definition, it is equivalent to see that η(si:t) = ( −1, ift=si, 1, ift6=si, and si(γ)∈ ( ∆−, ifγ =αi, ∆, ifγ 6=αi. But,γ =αi if and only if t=si.Thus, we obtain

φ(πsi(t, )) =si(φ(t, )).

for all si since the left hand side and the right hand side has the same image. QED Remark 3.1.14. This theorem gives the precise bijection between signed reflections and the set of roots, which intertiwes the action of W on both sets.