We continue to assume thatGis semisimple, split and simply connected and following Iwahori and Matsumoto we consider the structure of the Iwahori Hecke algebra.
It will be convenient to normalize the Haar measure so that J has volume 1.
ThenHJ is the ring of J-bi-invariant functions. The convolution is then normalized thus:
(φ1∗φ2)(g) = Z
G
φ1(gh−1)φ2(h)dg.
Since we have a set of J double cosets in the affine, let φw be the characteristic function ofBwB for w∈Waff.
We have an augmentation map ε:H −→C defined by ε(φ) =
Z
G
φ(g)dg.
Lemma 12 We have ε(φ1∗φ2) = ε(φ1)ε(φ2).
Proof Indeed ε(φ1 ∗φ2) = R
G
R
Gφ1(gh−1)φ2(h)dh dg. Interchanging the order of integration and substituting g −→gh, the integral factors as required.
Lemma 13 Let G be a group, H a subgroup, and x ∈ G. Then the cardinality of the coset space HxH/H is [H :H∩xHx−1].
Proof Now if K and H are arbitrary subgroups of G then the inclusion of K into KH induces a bijection K/(H ∩K) −→ KH/H. (The coset spaces here are not groups since we are not assuming that H is normal.) Indeed, the composition K −→ KH −→ KH/H is certainly surjective, and two cosets kH = k0H with k, k0 ∈K are equal if and only if k−1k0 ∈H; since k−1k0 ∈K, this is equivalent to k and k0 having the same image in K/(H∩K).
Left multiplication by x−1 commutes with right multiplication by elements ofH, hence induces a bijection HxH/H −→ x−1HxH/H = KH/K, where K = x−1Hx.
Therefore HxH/H is in bijection with x−1Hx/(H∩x−1Hx). Now conjugating by x, this is equivalently in bijection withH/(xHx−1∩H).
Proposition 53 Let w∈W. Then
ε(φw) = |J wJ/J|= [J∩wJ w−1].
Proof Since φw is the characteristic function of J wJ, it is clear that ε(φw) is the volume of this double coset. This equals the number of right cosets inJ wJ/J since
each of those cosets has volume 1.
We will say that an element of d ∈ Q∨ or its ambient vector space is dominant if hd, αi>0 for all α ∈Φ+, and antidominant if −d is dominant. More generally, if w∈Waff, we say thatwisdominant ifwFis contained in the positive Weyl chamber C, and antidominant if wF is contained in −C.
Lemma 14 Suppose that d∈Q∨ is dominant, and letw=$d. Thenε(φw) =ql(w). Note that $d is antidominant, since the embedding of Theorem 15 sends τ(d) to
$−d. Thus w actually corresponds to−d.
Proof We note that wU(o)w−1 ⊆ U(o) while wU−(p)w−1 ⊇ U−(p). Indeed, by (62) and our assumption thatd is dominant, we havewxα(o)w−1 ⊆xα(o) if α∈Φ+, while wxα(p)w−1 ⊇xα(p) ifα ∈Φ−.
Both groups J and wJ w−1 have Iwahori factorizations, J =U−(p)T(o)U(o) and wJ w−1 = (wU−(p)w−1)T(o)(wU(o)w−1).
It follows that
J∩wJ w−1 =U−(p)T(o)wU(o)w−1
Indeed, it is clear that the right-hand side is contained in the left-hand side. For the other inclusion, if we have an element g of J and write it as g = u−tu with u−∈U−(p),t∈T(o) andu∈U(o), and if it also equalsu0−t0u0 with u0−∈wU−(p)−1, t0 ∈ T(o) and u0 ∈ wU(o)w−1, then u−1− u0− = tu(t0u0)−1 ∈ U−(F) ∩B(F). The intersection of these two groups is trivial, so u− = u0−, and so g = u0−tu is in U−(p)T(o)wU(o)w−1.
We see that [J :J∩wJ w−1] = [U(o) :wU(o)w−1]. This index is, again by (62) Y
α∈Φ+
[xα(o) :wxα(o)w−1] = Y
α∈Φ+
[xα(o) :wxα(phd,αi)w−1] = Y
α∈Φ+
qhd,αi.
By Proposition 48, this equals ql(d).
Proposition 54 Let 06i6r. Then ε(φsi) =q.
Proof We leave it to the reader to check that if 1 6 i 6 r then siJ s−1i = U−(p)T(o)Ui(o) where
Ui(o) = Y
α∈Φ+
xα(p) if α=αi, xα(o) otherwise.
The index in J equals [U(o) : Ui(o)] = q. Similarly if i = 0 then siJ s−1i = U−0(p)T(o)U(o) where
U−0 (p) = Y
α∈Φ−
xα(p2) if α=α0, xα(p) otherwise,
and again the index is q.
Now let G be a group and H a subgroup. We will assume for all x ∈ G that i(x)<∞, wherei(x) = |HxH/H|= [H :H∩xHx−1].
Lemma 15 Let G be a group and H a subgroup. Define i(x) = H ∩ xHx−1 for x∈G. Suppose that x, y ∈G. Then i(xy)6i(x)i(y).
Proof We have i(y) = [H : H∩yHy−1] and conjugating by x, i(y) = [xHx−1 : xHx−1 ∩(xy)H(xy)−1]. Intersecting with H can only decrease the index, so
[H∩xHx−1 :H∩xHx−1∩(xy)H(xy)−1]6i(y).
Now
i(xy) = [H :H∩(xy)H(xy)−1] 6 [H :H∩xHx−1∩(xy)H(xy)−1] =
[H :H∩xHx−1][H∩xHx−1 :H∩xHx−1∩(xy)H(xy)−1] 6 i(x)i(y).
Theorem 20 Let w∈Waff. Then ε(φw) = ql(w).
Iwahori and Matsumoto deduce this quickly from Proposition 54. It seems to me that there is a gap in their proof, which I fill using Lemma 14.
Proof By Proposition 53 and Lemma 15 we have ε(φww0) 6 ε(φw)ε(φw0). Using this fact and Proposition 54 it follows that ε(φw) 6 ql(w), after factoring w into a product of simple reflections.
We claim that for every w ∈ Waff there exists a w0 ∈ Waff such that l(w0w) = l(w)+l(w0) andε(φw0w) = ql(w)+l(w0). By Lemma 14 it is sufficient to findw0 such that w0w∈Q∨, with w0w an antidominant element ofQ∨. It is easy to see geometrically that we may find a path fromwF to the negative Weyl chamber that does not cross any of the hyperplanes Hα,kbetween F and wF, and we may arrange that the path ends in an alcovew0wF that is aQ∨-translate of F. Since the path does not recross any hyperplane that it has already crossed, l(w0w) = l(w0) +l(w).
Now we have
ql(w)+l(w0) =ε(φw0w)6ε(φw0)ε(φw)6ql(w0)ε(φw),
soε(φw)>ql(w). We have proved both inequalities and the statement follows.
Corollary 1 If l(ww0) = l(w) +l(w0) then φww0 =φw∗φw0. Proof In the integral
(φw∗φw0)(g) = Z
G
φw(gh−1)φw0(h)dg,
the integrand vanishes unless gh−1 ∈ J wJ and h ∈ J w0J, which implies that g = gh−1·his inJ wJ0·J w0J. But by Proposition 51 we haveJ wJ·J w0J =J ww0J. Thus the convolution is supported on a single double coset and so φw ∗φw0 = cφww0 for some constant c. We apply ε and apply the Theorem to get ql(w)+l(w0)=cql(w)+l(w0),
soc= 1.
Now we need to know the quadratic relations. We will leave some of the work to the reader.
Exercise 16 Show that if 06i6r then J∪J siJ is a group. Hint: ifi6= 0 then this is U−i(p)iαi(SL(2,o))Ui(o)
where
U−i(p) = Y
α∈Φ− α6=−αi
xα(p), Ui(o) = Y
α∈Φ+ α6=αi
xα(o).
Show that this is closed under multiplication because iαi(SL(2,o)) normalizes the two unipotent groups.
Proposition 55 Let 06i6r. Then φ2s
i = (q−1)φsi+q.
Proof Since the support of φi is contained in J ∪J siJ, which is a group, φ2s
i is a linear combination ofφ1 (the identity element inHJ) and φsi. Thusφ2si =aφsi+bφ1 for some a and b. To compute b, evaluate at the identity. We have φ1(1) = 1 and φsi(1) = 0, while
(φsi∗φsi)(1) = Z
G
φsi(h)φsi(h−1)dh = Z
J siJ
1dh =|J siJ/J|=q
by Proposition 54. Thereforeb=q. We apply the homomorphismε:HJ −→Csuch thatε(φsi) = 1 and obtain the relation q2 =aq+b and sinceb =qwe get a=q−1.
Taking W to be the Coxeter group Waff, we have an Iwahori Hecke algebra Hq(Waff), with generators Ti in bijection with the si, subject to the quadratic rela-tions and the braid relarela-tions. Using Tits’ Theorem (see Theorem 14) we may define elements Tw for w ∈ Waff by Tw = Ti1· · ·Tik where si1· · ·sik = w is a reduced de-composition for any w ∈ Waff with l(w) = k. It is easy to see that the Tw span Hq(Waff).
Theorem 21 We have an isomorphism HJ −→ Hq(Waff) in whichφsi 7−→Ti. Proof We have checked that theφi satisfy the quadratic relations in Proposition 55.
The braid relations follow from Corollary 1. For example, suppose that the order n(i, j) of sisj is 3; then sisjsi = sjsisj. Let w denote sisjsi. Then l(w) = l(si) + l(sj) +l(si) and so φw = φsi ∗φsj ∗φsi. Similarly φw = φsj ∗φsi ∗φsj, and so the braid relation is satisfied. Hence there is a homomorphismHq(Waff)−→ HJ in which Ti −→φsi.
Since theTw spanHq(Waff), and their imagesφw are a basis ofHJ, it follows that
this homomorphism is a vector space isomorphism.