After the Bruhat decomposition was found in the 1950’s, it was extended in generality by Chevalley and Borel. Tits gave a fully axiomatic approach (1962). A completely unexpected instance of Bruhat’s axioms was found by Iwahori and Matsumoto, in a p-adic group. The (noncompact) Borel subgroup B is replaced by the (compact) Iwahori subgroup, and the (finite) Weyl group W is replaced by the (infinite) affine Weyl group Waff.
Let G be a split, simply-connected semisimple affine algebraic group over the nonarchimedean local field F. Notations will be as in the preceding section. The Iwahori subgroup J, we have noted, consists of k ∈ G(o) such that the image of k in G(Fq) lies in the Borel subgroup B(Fq). Let I0 = {s0, s1,· · · , sr} be the set of simple reflections, together with the “affine” reflections0.
Theorem 16 (Iwahori and Matsumoto)LetGbe a split, simply-connected semisim-ple affine algebraic group over the nonarchimedean local fieldF. The data(J, N(T(F), I0) are a Tits’ system. Therefore
G(F) = [
w∈Waff
J wJ (disjoint).
The proof will occupy the present section. The main thing to be verified is Axiom TS3 (Section 7).
Let us consider how this will change if G is not simply-connected. In this case, we have already explained, there is still a homomorphism from the coroot latticeQ∨ to T(F)/T(o), but this homomorphism is no longer surjective. It must be supple-mented by another subgroup Ω of N(T(F)), as at the end of the last section. IfG is semisimple, then Ω is isomorphic to the fundamental group π1(G), which can be viewed as a group of automorphisms of the extended Dynkin diagram.
The general case where Gis reductive was considered by Bruhat and Tits.
For the rest of the section we will assume that Gis split, semisimple and simply-connected.
LetU be the unipotent algebraic subgroup generated by thexα(u) withα∈Φ+. LetU− be the unipotent subgroup generated by the xα(u) with α ∈Φ−. If a is any fractional ideal let U(a) be the group generated by xα(u) with u ∈ Φ+ and α ∈ a, and similarly forU−(a).
The following fact is very important.
Proposition 50 (Iwahori Factorization) We have
J =U−(p)U(o)T(o).
That is, the multiplication map U−(p)×U(o)×T(o) −→ J is bijective. The three factors can be written in any order.
Proof Rather than prove this in general we prove it for SL3 in order to make the ideas clear. Let
g =
t1 u1 u2
v1 t2 u3 v2 v3 t3
∈J.
We will show that g ∈U−(p)B(o). The ui ∈ o, vi ∈ p and since g is invertible, the ti are units. Now we may multiply on the left by
x−α1−α2(−v2t−11 ) =
1
1
−vt2
1 1
which is inU−(p); this annihilatesv2. For the purpose of showing thatg ∈U−(p)B(o), we see that we may assumev2 = 0. Next we may multiply on the left byx−α1(−v1t−11 ) and arrange thatv1 = 0. Finally, we usex−α2(−v3t−12 ) to arrange that v3 = 0.
Now we know thatJ =U−(p)B(o), we may use the fact thatB(o) =T(o)U(o) = U(o)T(o) to see thatJ =U−(p)T(o)U(o) =U−(p)U(o)T(o). We also haveU−(p)T(o) = T(o)U−(p), a semidirect product withU−(p) normal. This allows us to put theU−(p) in the middle if we want. We see that the factors may be in any order.
This same proof works for general G, and we leave this to the reader. See
Lemma 11 for a similar situation.
Lemma 9 If u∈U−(p) and w∈W then wuw−1 ∈J. If t ∈T(o) then wtw−1 ∈J. There is an abuse of notation here, since wis actually a coset of N(T(F))/T(o).
The truth of wtw−1 ∈J is independent of the choice of w.
Proof We may assume that u=xα(v) with v ∈p. Thenwuw−1 =xw(α)(εv) where ε is a unit. If w(α) ∈ Φ+ then this is in U(o). If w(α) ∈ Φ− it is in U−(p). Either way it is in J. Also wT(o)w−1 =T(o) since we chose the representatives w of W to
be in the normalizer ofT(o) in G(o).
Lemma 10 Suppose that α ∈ Φ, α 6= α0. Suppose that u = xα(v) where either α∈Φ+ and v ∈o or α ∈Φ− and v ∈p. Then s−10 us0 ∈J.
Proof We write
s0 =$α∨0i−α0
1
−1
=$α∨0rα0.
Then
s−10 us0 =rα−10xα($−hα∨0,αiv)rα0 =xrα
0(α)($−hα∨0,αiv).
By abuse of notation we are using the notationrα0 to denote both the reflectionrα0 inW and its representativei−α0
1
−1
in T(F)/T(o).
First assume that α ∈Φ+ and v ∈o. Since −α0 is the highest root,hα∨0, αi6 0 by Proposition 41. If hα∨0, αi < 0 then this is in J by Lemma 9. On the other hand if hα∨0, αi = 0 then r−1α
0(α) = α since α0 and α are orthogonal roots, and so s−10 us0 =u∈J.
Next assume that α ∈ Φ− and that α 6= α0. In this case we are also assuming that v ∈p. Then hα∨0, αi>0 by Proposition 41. Indeed, by Proposition 41 and our assumption thatα 6=α0 we have hα∨0, αi<hα∨0, α0i= 2. Since hα∨0, αiis an integer, hα∨0, αi 6 1. if hα∨0, αi6 0 then $−hα∨0,αiv ∈ p and s0us0 =xrα
0(α)($−hα∨0,αiv)∈ J by Lemma 9. We are left with the case hα∨0, αi = 1. In this case $−hα∨0,αiv ∈ o.
Moreover rα0(α) = α− hα∨0, αiα0 = α−α0. This cannot be a negative root since
−α0 is the highest root, and soxα0(α)($−hα∨0,αiv)∈U(o)⊂J.
Iwahori and Matsumoto showed that the Iwahori subgroup satisfies Bruhat’s axioms, giving rise to a Bruhat decomposition based on the affine Weyl group. The next result is a first step towards this goal.
Proposition 51 If w∈Waff and l(wsi) = l(w) + 1 then wJ si ⊆J wsiJ.
Proof First assume that 16i6r.
Using the Iwahori factorization, we may write an element of wJ si as wu+u−tsi where u− ∈ U−(p), t ∈ T(o) and u+ ∈ U(o). By Lemma 9 u−tsi ∈ siJ, so we may assume that the element is of the form wu+si. Now we write u+ as a product of elements of the formxα(v) with v ∈o. If α 6=αi we have s−1i xα(v)si =xsi(α)(v) and si(α) ∈Φ+, so this is in J. Therefore we may handle all the xα(v) this way except only one root α=αi. It is thus sufficient to show that wxαi(v)w−1 ∈J w.
We may write w = $−dw0 where w0 ∈ W and d ∈ Q∨. By (62) we have wxαi(v)w−1 = $−dxw0(αi)(v)$d = xw0(αi)($h−d,w0(αi)iv). By Proposition 47 we have either hw0(αi), di < 0 or hw0(αi), di = 0 and w0(αi) ∈ Φ+. Assume first that hw(αi), di < 0. Then $h−d,w0(αi)iv ∈ p and xw0(αi)($h−d,w0(αi)iv) ∈ J regardless of whetherw0(αi) is a positive or negative root. On the other hand ifhw0(αi), di= 0, we are guaranteed thatw0(α)∈Φ+. In the second case we also havexw0(αi)($h−d,w0(αi)iv) = xw0(αi)(v)∈J.
It remains for us to treat the case i = 0. We may use the Iwahori factorization again to write an element of wJ s0 as wu−u+ts0 where u− ∈ U−(p), t ∈ T(o) and
u+∈U+(o). We have, as beforets0 ∈s0J, and we may writeu− andu+ as products
Proof First assume that 1 6 i6 r. We may write an arbitrary element of J as a product of factors of the form t ∈ T and xα(v) where either v ∈ o and α ∈ Φ+ or v ∈ p and α ∈ Φ−. Except in the case α =αi we have s−1i xα(v)si = xsi(α)(v) ∈ J, because if α ∈ Φ+ and α 6= αi then si(α) ∈ Φ+, while if α ∈ Φ− then v ∈ p and so s−1i xα(v)si ∈ J by Lemma 9. Also s−1i tsi ∈ T(o) ∈ J. In conclusion, every one of the factors except one may be moved across si. We are left with showing that sixαi(v)si ∈J∪J siJ. We havev ∈o. Ifv ∈pthen we may use Lemma 9. Therefore
This concludes the proof when 16i6r.
Next assume that i = 0. We leave it to the reader to check that if α 6= α0 and
Applying i−α0 shows thats−10 xα0($ε)s0 ∈C(s0).
Exercise 14 Verify the claim in the above proof that ifi= 0 andα6=α0 and eitherv∈o and α∈Φ+ orv∈pand α∈Φ−, thens−10 xα(v)s0 ∈J.
We will use the notationJ wJ =C(w) as in Section 7. Then the content of Propo-sition 51 may be writtenC(w)C(si) = C(wsi) ifl(wsi) = l(w)+1, and Proposition 52 may be writtenC(si)C(si)⊂C(1)∪C(si).
Theorem 17 If w∈Waff and 06i6r then
wJ si ⊆J wsiJ∪J wJ.
Proof This may be written C(w)C(si)⊆C(wsi)∪C(w). If l(wsi) =l(w) + 1, this follows from Proposition 51.
Therefore we assume that l(wsi) = l(w)−1. Let w0 = wsi. Then l(w0si) = l(w0) + 1 and so by Proposition 51 we haveC(w0)C(si) =C(w0si) = C(w). Therefore
C(w)C(si) = C(w0)C(si)C(si).
By Proposition 52 this is contained in
C(w0)C(si)∪C(w0)C(1) =C(w)∪C(w0) = C(w)∪C(wsi),
as required.
This gives us Axiom TS3. The other axioms we leave to the reader.
Exercise 15 Verify the remaining axioms of a Tits System. Hint: For Axiom TS5, you must show that G is generated by J and N(T(F)). Show that conjugates of U(o) by elements of T(F) contain all of U(F), and so the group generated by J and N(T(F)) containsB(F).
This concludes the proof of Theorem 16.
Lemma 11 If u∈U−(F) then u∈G(o)B(F).
Proof We write
u= Y
α∈Φ−
xα(uα), uα ∈F.
We order the roots so that if β comes after α in the product, then either β−α is not a root or β −α ∈ Φ+. We may accomplish this by taking the negative roots α such that the inner productsh−ρ, αi are in nonincreasing order.
Now we modify u by left multiplications by elements of G(o) and right multipli-cation by elements of B(F) so that until all uα are 0. Let α be the first root such that xα(uα) 6= 0. If uα ∈ o, we left multiply by xα(−uα). If uα ∈ o then we left multiply byi−α
1
−1 uα
. We have i−α
1
−1 uα
xα(u) = i−α
u 1 u−1
∈B(F).
Conjugating the remaining xβ(uβ) by this element of B(F) produces commutators that are inB(F) by the way we have ordered the roots. In either case, we are able to replace uα by 0. Continuing, eventually all uα are zero.
Theorem 18 (Iwasawa Decomposition)We haveG(F) =G(o)B(F) =B(F)G(o).
Proof Using the Iwahori-Bruhat decomposition, it is sufficient to show thatJ wJ ⊆ G(o)B(F), where w ∈ Waff may be written w =w0t with w0 ∈ W and t ∈ T. Now J w0 ∈ G(o) so what we must show is that tJ ⊆ G(o)B(F). Using the Iwahori factorization, we write a typical element of J as u−b with u− ∈U−(p) andb ∈B(o).
Now tu−t−1 ∈G(o)B(F) by the Lemma, sotu−b = (tu−t−1)tb∈G(o)B(F).
The following decomposition is called the Cartan decomposition by analogy with the corresponding decomposition in Lie groups. However in this p-adic context the result is actually due to Bruhat.
We will say that an element of d∈Q∨ or its ambient vector space is dominant if hd, αi>0 for allα∈Φ+. Then (because we are assumingGto be simply-connected) the $d with d ∈ Q∨ dominant form a fundamental domain for the action of W on T(F)/T(o).
Theorem 19 (Cartan Decomposition) We have
G(F) = [
d∈Q∨ ddominant
G(o)$dG(o) (disjoint).
Proof We have
G(F) = [
w∈Waff
J wJ = [
d∈Q∨ w∈W
J w$dJ,
where $−d ∈ T(F) and w ∈ G(o). This shows that G(F) = S
d∈Q∨G(o)$dG(o).
Since G(o) contains representatives for W, we may conjugate $d by W and assume that d is dominant.
We will omit the proof that these double cosets are disjoint. For GLn, we proved it in Proposition 35. For the general case, see Bruhat, Sur une classe du sous-groupes compacts maximaux des groupes de Chevalley sur un corpsp–adique. (French) Inst.
Hautes ˜Atudes Sci. Publ. Math. No. 23 1964 45–74, Th ˜Aor ˜Am 12.2.