On tempered representations

Lemma A.1.

Let λ:=(λ1,…,λn)∈C≤0n, let m:=(m1,…,mn)∈Z≥0, and let K be a compact subset of R≥0n. Assume that Re⁢(λ)=0 and λ∉(2⁢π⁢i)⁢Zn. We have

as r→+∞, where Q denote convex subsets.

Proof.

Let us re-order the variables, assuming that λ1∉2⁢π⁢i⁢ℤ. Let us write x=(x1,x′), where x′=(x2,…,xn) and analogously write m′, etc. Given a convex subset Q⊂K and x′∈ℝ≥0n-1, let us denote by Qx′⊂ℝ≥0 the subset consisting of x1 for which (x1,x′)∈Q (it is an interval). Let us enlarge K for convenience, writing it in the form K=K1×K′, where K1⊂ℝ≥0 is a closed interval and K′⊂ℝ≥0n-1 is the product of closed intervals.

We have

We have Qx′⊂K′ and it is elementary to see that

as r→+∞, where R denote intervals. Therefore we obtain, for some C>0 (not depending on Q) and all r≥1,

Since the expression in brackets is clearly bounded independently of r, we are done. ∎

Lemma A.2.

Let (λ,m)∈C≤0n×Z≥0n. Then the limit

exists, and it is equal to 0 if resJλ⁢(λ)∉2⁢π⁢i⋅ZJλ and otherwise equal to

(the sum converging absolutely).

Proof.

Let us abbreviate J:=Jλ. Let us denote λ′:=resJ⁢(λ) and λ′′:=resJc⁢(λ), and similarly for m. Given x′′∈ℤ≥0Jc, let us denote by P(<r)x′′⊂ℝ≥0J the subset consisting of y′ for which extJ⁢(r⁢y′)+extJc⁢(x′′)∈P<r.

We have

Let us assume first that λ′∉2⁢π⁢i⋅ℤJ. Then by Lemma A.1 there exists C>0 such that for all convex subsets Q⊂PJ and all r≥1 we have

Therefore

giving the desired.

Now we assume that λ′∈2⁢π⁢i⋅ℤJ. It is not hard to see that

Hence we have (by dominated convergence)

Claim A.3.

Let p≥1, let {(λ(ℓ),m(ℓ))}ℓ∈[p]⊂C≤0n×Z≥0n be a collection of pairwise different couples and let {c(ℓ)}ℓ∈[p]⊂C∖{0} be a collection of non-zero scalars. Denote

The limit

exists and is strictly positive.

Proof.

Let us break the integrand into a sum following

Using Lemma A.2, we see the that resulting limit breaks down as a sum, over (ℓ1,ℓ2)∈[p]2, of limits which exist, so the only thing to check is that the resulting limit is non-zero. It is easily seen that the limit at the (ℓ1,ℓ2) place is zero unless d⁢(λ(ℓ1),m(ℓ1))=d, d⁢(λ(ℓ2),m(ℓ2))=d, Jλ(ℓ1)=Jλ(ℓ2) and resJ(ℓ1)⁢(λ(ℓ2))-resJ(ℓ1)⁢(λ(ℓ1))∈2⁢π⁢i⋅ℤJ. We thus can reduce to the case when, for a given J⊂[n], we have Jλ(ℓ)=J for all ℓ∈[p], we have d⁢(λ(ℓ),m(ℓ))=d for all ℓ∈[p], and we have resJ⁢(λ(ℓ2))-resJ⁢(λ(ℓ1))∈2⁢π⁢i⋅ℤJ for all ℓ1,ℓ2∈[p]. We then obtain, using Lemma A.2, that our overall limit equals

It is therefore enough to check that

a function in (z,y)∈ℤ≥0Jc×PJ, is not identically zero. By the local linear independence of powers of y, we can further assume that resJ⁢(m(ℓ)) is independent of ℓ∈[p], and want to check that

a function in z∈ℤ≥0Jc, is not identically zero. Notice that, by our assumptions, the elements in the collection {(resJc⁢(λ(ℓ)),resJc⁢(m(ℓ)))}ℓ∈[p] are pairwise different. Thus the non-vanishing of our sum is clear (by linear algebra of generalized eigenvectors of shift operators on ℤJc). ∎

Lemma A.4.

Let λ:=(λ1,…,λn)∈C≤0n, let m:=(m1,…,mn)∈Z≥0, and let K be a compact subset of R≥0n. Assume that Re⁢(λ)=0 and λ≠0. We have

as r→+∞, where Q denote convex subsets.

Proof.

Let us re-order the variables, assuming that λ1≠0. Let us write x=(x1,x′), where x′=(x2,…,xn) and analogously write m′, etc. Given a convex subset Q⊂K and x′∈ℝ≥0n-1, let us denote by Qx′⊂ℝ≥0 the subset consisting of x1 for which (x1,x′)∈Q (it is an interval). Let us enlarge K for convenience, writing it in the form K=K1×K′, where K1⊂ℝ≥0 is a closed interval and K′⊂ℝ≥0n-1 is the product of closed intervals.

Using Fubini’s theorem,

We have Qx′⊂K′ and it is elementary to see that

as r→+∞, where R denote intervals. Therefore we obtain, for some C>0 and all r≥1,

as desired. ∎

Lemma A.5.

Let (λ,m)∈C≤0n×Z≥0n and let ϕ:B×R≥0n→C be a nice function. Then the limit

exists, and it is equal to 0 if resJλ⁢λ≠0 and otherwise equal to

(the double integral converging absolutely).

Proof.

Let us re-order the variables, assuming that J:=Jλ=[k]. Write x=(x′,x′′), where x′ consists of the first k components and x′′ consists of the last k components. Let us write analogously m′,λ′, etc.

First, let us notice that if k≠0, we can write

where ϕ0,ϕ1:B×ℝ≥0n→ℂ are nice functions and ϕ1 does not depend on x1. Dealing with e-x1⁢ϕ0⁢(b,x) instead of ϕ⁢(b,x) makes us consider λ with smaller set Jλ and thus (λ,m) with a smaller d⁢(λ,m) and from this, reasoning inductively, we see that we can assume that ϕ only depends on (b,x′′). Let us write ϕ′′:=resJc⁢ϕ.

Let us perform a change of variables y′:=1r⁢x′. Let P(<r)⊂ℝ≥0n denote the transform of P<r under this changes of variables (i.e. (x′,x′′)∈P<r if and only if (y′,x′′)∈P(<r)). We obtain

Given x′′∈ℝ≥0Jc, let us denote by P(<r)x′′⊂ℝ≥0J the set

Notice that P(<r1)x′′⊂P(<r2)x′′ for r1<r2 and ⋃rP(<r)x′′=PJ. Using Fubini’s theorem,

If λ′≠0, by Lemma A.4 there exists C>0 such that for all convex subsets Q⊂PJ and all r≥1 we have

We have therefore

and thus indeed the desired limit is equal to 0.

Now we assume λ′=0. Using Lebesgue’s dominated convergence theorem, we have

as desired. ∎

Claim A.6.

Let {(λ(ℓ),m(ℓ))}ℓ∈[p]⊂C≤0n×Z≥0n be a collection of pairwise different couples. Let {ϕ(ℓ)}ℓ∈[p] be a collection of nice functions B×R≥0n→C such that for every ℓ∈[p] the function b↦ϕ(ℓ)⁢(b,+∞) on B is not identically zero. Denote

The following limit exists and is strictly positive:

Proof.

Let us break the integrand into a sum following

Using Lemma A.5, we see the that resulting limit breaks down as a sum, over (ℓ1,ℓ2)∈[p]2, of limits which exist, so the only thing to check is that the resulting limit is non-zero. It is easily seen that the limit at the (ℓ1,ℓ2) place is zero unless d⁢(λ(ℓ1),m(ℓ1))=d, d⁢(λ(ℓ2),m(ℓ2))=d, Jλ(ℓ1)=Jλ(ℓ2) and resJ(ℓ1)⁢(λ(ℓ1))=resJ(ℓ1)⁢(λ(ℓ2)). We thus can reduce to the case when, for a given J⊂[n], we have Jλ(ℓ)=J for all ℓ∈[p], we have d⁢(λ(ℓ),m(ℓ))=d for all ℓ∈[p], and we have resJ⁢(λ(ℓ1))=resJ⁢(λ(ℓ2)) for all ℓ1,ℓ2∈[p]. We then obtain, using Lemma A.5, that our overall limit equals

It is therefore enough to check that

a function in (b,z,y)∈B×ℝ≥0Jc×PJ, is not identically zero. By the local linear independence of powers of y, we can further assume that resJ⁢(m(ℓ)) is independent of ℓ∈[p], and want to check that

a function in (b,z)∈B×ℝ≥0Jc, is not identically zero. Notice that, by our assumptions, the elements in the collection {(resJc⁢(λ(ℓ)),resJc⁢(m(ℓ)))}ℓ∈[p] are pairwise different and for every ℓ∈[p], the function b↦ϕ(ℓ)⁢(b,extJc⁢(+∞)) on B is not identically zero. Considering the partial order on ℂJc given by μ1≤μ2 if μ2-μ1∈ℤ≥0Jc, we can pick ℓ∈[p] for which resJc⁢(λ(ℓ)) is maximal among the {resJc⁢(λ(ℓ′))}ℓ′∈[p]. We can then pick b∈B such that ϕ(ℓ)⁢(b,extJc⁢(+∞))≠0. We then boil down to Lemma A.7 that follows. ∎

Lemma A.7.

Let {(λ(ℓ),m(ℓ))}ℓ∈[p]⊂Cn×Z≥0n be a collection of pairwise different couples. Let {ϕ(ℓ)}ℓ∈[p] be a collection of nice functions R≥0n→C (so here B={1}). Suppose that ϕ(ℓ)⁢(+∞)≠0 for some ℓ∈[p] for which λ(ℓ) is maximal among the {λ(ℓ′)}ℓ′∈[p] with respect to the partial order λ1≤λ2 if λ2-λ1∈Z≥0n. Then the function

on R≥0n is not identically zero.

Proof.

We omit the proof of the lemma – one develops the ϕ(ℓ) into power series in e-x1,…,e-xn and uses separation by generalized eigenvalues of the partial differentiation operators ∂x1,…,∂xn. ∎