The Volume of the Fundamental Domain for Some Arithmetical Subgroups of Chevalley Groups

by

R.P. Langlands

Let 𝔤_{Q} be a split semisimple Lie algebra of linear transformations of the finite dimensional vector space V_{Q} over Q. Let 𝔥_{Q} be a split Cartan subalgebra of 𝔤_{Q} and choose for each root α of 𝔥_{Q} a root vector X_{α} so that if [X_{α}, X_{−α}] = H_{α} then α(H_{α}) = 2 and so that there is an automorphism θ of 𝔤_{Q} with θ(X_{α}) = −X_{−α}. Let L be the set of weights of 𝔥_{Q} and if λ ∈ L let

V_{Q}(λ) = {v ∈ V_{Q} | Hv = λ(H)v for all H ∈ 𝔥_{Q}};

let H_{1}, …, H_{p} be a basis over Z of

{H | λ(H) ∈ Z if V_{Q}(λ) ≠ 0}.

As usual, there is associated to 𝔤_{Q} a connected algebraic group G of linear transformations of V_{C} = V_{Q} ⊗_{Q}C. If H is some lattice in V_{Q} satisfying

(i) M = ∑_{λ∈L} M ∩ V (λ), (ii) (X_{α}^{n}/n!)M ⊆ M for all α, then we let G_{Z} = {g ∈ G | gM = M}. Let ω be a left invariant form on G_{R} of highest degree which takes the value ±1 on (∧_{i=1}^{p}H_{1}) ∧ (∧_{α>0}X_{α}) and let [dg] be the Haar measure associated to ω. Our purpose now is to show the following.

If ξ(·) is the Riemann zeta function, Π_{i=1}^{p}(t^{2a_{i}−1} + 1) is the Poincaré polynomial of G_{C}, and c is the order of the fundamental group of G_{C} then

∫_{GZ/GR}[dg] = cΠ_{i=1}^{p}ξ(a_{i}).

The method to be used to find the volume of G_{Z}\G_{R} is not directly applicable to [dg]. So it is necessary to introduce another Haar measure on the group G_{R}. Let U be the connected subgroup of G whose Lie algebra is spanned over R by {X_{α} − X_{−α}, i(X_{α} + X_{−α}), iH_{α} | α a root} and let K = G_{R} ∩ U. Choose an order on the roots and let N = N_{R} be the set of real points on the connected algebraic subgroup of G_{C} with the Lie algebra ∑_{α>0} CX_{α}. Let A_{R} be the normalizer of 𝔥_{C} in G_{R}. Let dn be the Haar measure on N defined by a form which takes the value ±1 on ∧_{i=1}^{p}H_{i}. Let dk be the Haar measure on K such that the total volume of K is one. Let ρ = \frac{1}{2} ∑_{α>0} α and let ξ_{2ρ}(a) be the character of AC associated to 2ρ. Finally let dg be such that

∫_{GR} ϕ(g)dg = ∫_{N×AR×K} |ξ_{2ρ}(a)|^{−1}ϕ(nak)dn da dk.

If N^{−} is the set of real points on the group associated to ∑_{α<0} CX_{α} define dn^{−} in the same way as we defined dn. It is easy to see that

∫_{G} ϕ(g)[dg] = ∫_{N} dn ∫_{AR} da ∫_{N^{−}} dn^{−}|ξ_{2ρ}(a)|^{−1}ϕ(nan^{−}).

Suppose ϕ(gk) = ϕ(g) for all g ∈ G_{R} and all k ∈ K. Then

∫_{G} ϕ(g)dg = ∫_{N×AR} dn da|ξ_{2ρ}(a)|^{−1}ϕ(na).

On the other hand, if n^{−} = n(n^{−})a(n^{−})k(n^{−}),

\begin{aligned}∫ ϕ(g)[dg] & = ∫_{N^{−}} dn^{−} \{∫_{A} da ∫_{N} dn|ξ_{2ρ}(a)|^{−1}ϕ(nan(n^{−})a(n^{−})k(n^{−}))\} \\ & = \{∫_{A} da ∫_{N} dn|ξ_{2ρ}(a)|^{−1}ϕ(na)\} \{∫_{N^{−}} |ξ_{2ρ}a(n^{−})|dn^{−}\} . \end{aligned}

It follows from a formula of Gindikin and Karpelevich that the second factor equals

\begin{aligned}∏_{α>0} \frac{π^{+ \frac{1^{b}}{2}} Γ(ρ(H_{α})/2)}{Γ((ρ(H_{α}) + 1)/2)} & = ∏_{α>0} \frac{π^{−ρ(H_{α})/2}Γ(ρ(H_{α})/2)}{π^{−(ρ(H_{α})+1)/2}Γ((ρ(H_{α}) + 1)/2)} \\ & = \frac{∏′_{α>0} π^{−ρ(H_{α})/2}Γ(ρ(H_{α})/2)}{∏_{α>0} π^{−(ρ(H_{α})+1)/2}Γ((ρ(H_{α}) + 1)/2)}, \end{aligned}

since when α is simple ρ(H_{α}) = 1 and

π^{− \frac{1}{2}} Γ(\frac{1}{2}) = 1.

The product in the numerator is taken over the positive roots which are not simple. By a wellknown result, the numbers, with multiplicities, in the set

{ρ(H_{α}) + 1 | α > 0}

are just the numbers ρ(H_{α}) with α positive and not simple, together with the numbers a_{1}, …, a_{p}. So if

ξ(s) = π^{−s/2}Γ (\frac{s}{2}) ξ(s),

we have to show that

∫_{GZ/GR} dg = \frac{cΠ_{α>0}ξ(ρ(H_{α}) + 1)}{Π′_{α>0}ξ(ρ(H_{α}))}.

By the way, it is well to keep in mind that ρ(H_{α}) > 1 if α is not simple.

Let A be the connected component of A_{R} and let M be the points of finite order in A_{R}. Certainly A_{R} = AM. Moreover, by Iwasawa, G = NAK. If g = nak and a = exp H, we set H = H(g).

If ϕ is an infinitely differentiable function with compact support on N\G such that ϕ(gk) = ϕ(g) for all g in G and all k in K we can write ϕ as a Fourier integral.

ϕ(g) = \frac{1}{(2π)^{p}} ∫_{Reλ=λ_{0}} exp(λ(H(g)) + ρ(G(g))Φ(λ)|dλ|;

λ is the symbol for an element of the dual of 𝔥_{C}; Φ(λ) is an entire complex-valued function of λ; and dλ = dz_{1} ∧ … ∧ dz_{p} with z_{i} = λ(H_{i}). As in the lectures on Eisenstein series we can introduce

ϕ̂(g) = ∑_{γ∈GZ∩NM\GZ} ϕ(γg).

Our evaluation of the volume of G_{Z}\G_{R} will be based on the simple relation

(ϕ̂, 1)(1, ψ̂) = (1, 1)(Πϕ̂, Πψ̂).

The inner products are taken in L^{2}(G_{Z}\G_{R}) with respect to dg and Π is the orthogonal projection on the space of constant functions. Since

(1, 1) = ∫_{GZ\GR} dg,

it is enough to find an explicit formula for the other three terms. Now

(ϕ̂, 1) = ∫_{GZ∩NM\GR} ϕ(g)dg
= μ(G_{Z} ∩ NM\NM) ∫_{A} |ξ_{2ρ}(a)|^{−1}ϕ(a)da
= Φ(ρ)

since μ(G_{Z} ∩ NM\NM) = 1. To see the latter we have to observe that M ⊆ G_{Z} and that, as follows from results stated in Cartier’s talk, μ(G_{Z} ∩ N\N) = 1. It is also clear that (1, ψ̂) = Ψ̄(ρ). The nontrivial step is to evaluate

(Πϕ̂, Πψ̂).

From the theory of Eisenstein series we know that

(ϕ̂, ψ̂ = \frac{1}{(2π)^{p}} ∫_{Reλ=λ_{0}} ∑_{s∈Ω} M(s, λ)Φ(λ)Ψ̄(−sλ̄)|dλ|.

Ω is the Weyl group, λ_{0} is any point such that λ_{0}(H_{α}) > 1 for every simple root, and

M(s, λ) = Π_{α>0} \frac{ξ(1 + sλ(H_{α}))}{ξ(1 + λ(H_{α}))} = Π_{α>0;sα<0} \frac{ξ(λ(H_{α}))}{ξ(1 + λ(H_{α}))}.

In the lectures on Eisenstein series I introduced an unbounded self-adjoint operator A on the closed subspace of L^{2}(G_{Z}\G_{R}) generated by the functions ϕ̂ with ϕ of the form indicated above. Comparing the definition of A with the formula for (ϕ̂, 1) we see that

(Aϕ̂, 1) = (ρ, ρ)(ϕ̂, 1).

Since the constant functions are in this space A1 = (ρ, ρ) · 1. As a consequence, if E(x), −∞ < x < ∞, is the spectral resolution of A the constant functions are in the range of E((ρ, ρ)) − E((ρ, ρ) − 0) = E. We show that this range consists precisely of the constant functions and compute (Eϕ̂, ψ̂) = (Πϕ̂, Πψ̂).

Suppose a > (ρ, ρ) > b and a − b is small. According to a well-known formula,

\frac{1}{2}\{(E(a)ϕ̂, ψ̂) + (E(a − 0)ϕ̂, ψ̂)\} − \frac{1}{2}\{(E(b)ϕ̂, ψ̂) + (E(b − 0)ϕ̂, ψ̂)\}

is equal to

(a)  lim_{δ↓0} \frac{1}{2πi} ∫_{C(a,b,c,δ)}(R(μ, A)ϕ̂, ψ̂)dμ

if C(a, b, c, δ) is the following contour.

Recall that, if Reμ > (λ_{0}, λ_{0}),

(R(μ, A)ϕ̂, ψ̂) = ∑_{s∈Ω} \frac{1}{(2πi)^{p}} ∫_{Reλ=λ_{0}} \frac{1}{μ − (λ, λ)}M(s, λ)Φ(λ)Ψ̄(−sλ̄)dλ.

If w = (w_{1}, …, w_{p}) belongs to C^{p} let λ(w) be such that λ(H_{α_{i}}) = w_{i}, where α_{1}, …, α_{p} are the simple roots. Set

\begin{aligned}ϕ_{p}(w, s) & = M(s, λ(w))Φ(λ(w))Ψ̄(−sλw̄)), \\ Q_{p}(w) & = (λ(w), λ(w)), \end{aligned}

then (a) is equal to

\frac{1}{c} ∑_{s∈Ω} lim_{δ↓0} \frac{1}{2πi} ∫_{C(a,b,c,δ)} dμ \{\frac{1}{(2πi)^{p}} ∫_{Rew=w_{0}} \frac{1}{μ − Q_{p}(w)}ϕ_{p}(w, s)dw_{1} … dw_{p}\}

provided each of these limits exist.^{2} The coordinates of w_{0} must all be greater than one. We shall consider the limits individually.

Let w^{q} = (w_{1}, …, w_{q}) and define ϕ_{q}(w^{q}; s) inductively for 0 ≦ q ≦ p by

ϕ_{q}(q_{1}, …, w_{q}; s) = Residue_{w_{q+1}=1} ϕ_{q+1}(w_{1}, …, w_{q+1}; s).

2 The inner integral is defined for Reμ > Q_{p}(w_{0}). However, as can be seen from the discussion to follow, the function of μ it defines can be analytically continued to a region containing C(a, b, c, δ).

It is easily seen that ϕ_{q}(w^{q}; s) has no singularities in the region defined by the inequalities Re w_{i} > 1, 1 ≦ i ≦ q; that ϕ_{q}(w^{q}; s) goes to zero very fast when the imaginary part of w^{q} goes to infinity and its real part remains in a compact subset of this region; and that there is a positive number ϵ such that the only singularities of ϕ_{q}(w^{q}; s) in

{(w_{1}, …, w_{q}) | |Re w_{i} − 1| < ϵ, 1 ≦ i ≦ q}

lie on the hyperplanes w_{i} = 1 and are at most simple poles. ϕ_{0}(s) is of course a constant. Set Q_{q}(w^{q}) = Q_{p}(w_{1}, …, w_{q}, 1, …, 1).

Let us show by induction that the given limit equals

(b)  lim_{δ↓0} \frac{1}{2πi} ∫_{C(a,b,c,δ)} dμ \{\frac{1}{2πi)^{q}} ∫_{Re w^{q}=w_{0}^{q}} \frac{1}{μ − Q_{q}(w^{q})}ϕ_{q}(w^{q}; s)dw_{1} … dw_{q}\}

if w_{0}^{q} = (w_{0,1}, …, w_{0,q}) with w_{0,i} > 1, 1 ≦ i ≦ q. Of course, the above expression is independent of the choice of such a point w_{0}^{q}. Take w_{0}^{q} = (1 + u, …, 1 + u, 1 + v), with u and v positive but small, and w_{0}^{q−1} = (1 + u, …, 1 + u). If Λ_{1}, …, Λ_{q} are such that Λ_{i}(H_{αj}) = δ_{ij}, then (Λ_{i}, Λ_{j}) ≧ 0. As a consequence, if u is much smaller than v, then

Q_{q}(1 + u, …, 1 + u, 1 − v) < (ρ, ρ).

Choose (b) to be larger than the number on the left. Also

Re Q_{q}(w^{q}) = Q_{q}(Re w^{q}) − Q_{p}(Imw_{1}, …, Imw_{q}, 0, …, 0).

Thus there is a constant N such that if either Re w_{i} = 1 + u, 1 ≦ i ≦ q − 1 and Re w_{q} = 1 − v or Re w_{i} = 1 + u, 1 ≦ i ≦ p and |Re w_{q} − 1| ≦ v and |Imw_{q}| > N, then

Re Q_{q}(w^{q}) < b − 1/N.

In (b) we may perform the integrations in any order. Integrate first with respect to w_{q}. If C is the indicated contour, the result is the sum of (b) with q replaced by q − 1 and

lim_{δ↓0} \frac{1}{(wπi)^{q}} ∫_{Re w^{q−1}=w_{0}^{q−1}} dw_{1} . . . . . . dw_{q−1} ∫_{C} dw_{q}ϕ_{q}(w^{q}, s) \{\frac{1}{2πi} ∫_{C(a,b,c,,δ)} \frac{1}{μ − Q_{q}(w^{q})}dμ\}

which is obviously zero.

Taking q = 0 in (b) we get

lim_{δ↓0} \frac{ϕ_{0}(s)}{2πi} ∫_{C(a,b,c,δ)} \frac{1}{μ − (ρ, ρ)}dμ = ϕ_{0}(s).

It is clear that ϕ_{0}(s) is zero unless s sends every positive root to a negative root but that for the unique element of the Weyl group which does this

ϕ_{0}(s) = \frac{Π′_{a>0}ξ(ρ(H_{α}))Φ(ρ)\overline{Ψ(ρ)}}{Π_{α>0}ξ(ρ(H_{α}) + 1)}

since sρ = −ρ. This is the result required.

Finally, I remark that although the method just described for computing the volume of Γ\G has obvious limitations, it can be applied to other groups. In particular it works for Chevalley groups over a numberfield.