The heat kernel on the torus

Jordan Bell
October 7, 2014

1 Heat kernel on 𝕋

For t>0, define kt:ℝ→(0,∞) by11 1 Most of this note is my working through of notes by Patrick Maheux. http://www.univ-orleans.fr/mapmo/membres/maheux/InfiniteTorusV2.pdf

kt⁒(x)=(4⁒π⁒t)-1/2⁒exp⁑(-x24⁒t),xβˆˆβ„.

For t>0, define gt:ℝ→(0,∞) by

gt⁒(x)=2β’Ο€β’βˆ‘kβˆˆβ„€kt⁒(x+2⁒π⁒k),xβˆˆβ„,

which one checks indeed converges for all xβˆˆβ„. Of course, gt⁒(x+2⁒π⁒k)=gt⁒(x) for any kβˆˆβ„€, so we can interpret gt as a function on 𝕋, where 𝕋=ℝ/2⁒π⁒℀.

Let m be Haar measure on 𝕋: d⁒m⁒(x)=(2⁒π)-1⁒d⁒x, and so m⁒(𝕋)=1. With βˆ₯fβˆ₯1=βˆ«π•‹|f|⁒𝑑m for f:𝕋→ℂ, we have, because gt>0,

βˆ₯gtβˆ₯1=βˆ‘kβˆˆβ„€βˆ«π•‹kt⁒(x+2⁒π⁒k)⁒𝑑x=βˆ«β„kt⁒(x)⁒𝑑x=1.

Hence gt∈L1⁒(𝕋). For ΞΎβˆˆβ„€, we compute

g^t⁒(ΞΎ) = βˆ«π•‹gt⁒(x)⁒e-i⁒ξ⁒x⁒𝑑m⁒(x)
= βˆ‘kβˆˆβ„€βˆ«π•‹kt⁒(x+2⁒π⁒k)⁒e-i⁒ξ⁒x⁒𝑑x
= βˆ‘kβˆˆβ„€βˆ«π•‹kt⁒(x+2⁒π⁒k)⁒e-i⁒ξ⁒(x+2⁒π⁒k)⁒𝑑x
= βˆ«β„kt⁒(x)⁒e-i⁒ξ⁒x⁒𝑑x
= k^t⁒(ΞΎ2⁒π)
= e-ξ2⁒t.
Lemma 1.

For t>0 and x∈R,

gt⁒(x)=Ο€t⁒exp⁑(-x24⁒t)⁒(1+2β’βˆ‘kβ‰₯1exp⁑(-Ο€2⁒k2t)⁒cosh⁑(π⁒k⁒xt)).
Proof.

Using the definition of gt,

gt⁒(x) = 2β’Ο€β’βˆ‘kβˆˆβ„€kt⁒(x+2⁒π⁒k)
= 2β’Ο€β’βˆ‘kβˆˆβ„€(4⁒π⁒t)-1/2⁒exp⁑(-(x+2⁒π⁒k)24⁒t)
= Ο€t⁒exp⁑(-x24⁒t)β’βˆ‘kβˆˆβ„€exp⁑(-π⁒k⁒xt)⁒exp⁑(-Ο€2⁒k2t)
= Ο€t⁒exp⁑(-x24⁒t)
(1+βˆ‘kβ‰₯1(exp⁑(π⁒k⁒xt)+exp⁑(-π⁒k⁒xt))⁒exp⁑(-Ο€2⁒k2t)),

which gives the claim, using cosh⁑y=ey+e-y2. ∎

Definition 2.

For x∈R, let βˆ₯xβˆ₯=inf{|x-2Ο€k|:k∈Z}.

For kβˆˆβ„€, βˆ₯x+2⁒π⁒kβˆ₯=βˆ₯xβˆ₯, so it makes sense to talk about βˆ₯xβˆ₯ for xβˆˆπ•‹.

Theorem 3.

For t>0 and x∈R,

exp⁑(-βˆ₯xβˆ₯24⁒t)⁒gt⁒(0)≀gt⁒(x)≀exp⁑(-βˆ₯xβˆ₯24⁒t)⁒(Ο€t+gt⁒(0)).
Proof.

Let x=2⁒π⁒m+ΞΈ with |ΞΈ|≀π, so that βˆ₯xβˆ₯=βˆ₯ΞΈβˆ₯=|ΞΈ|, and gt⁒(x)=gt⁒(ΞΈ). Using Lemma 1 and the fact that cosh⁑yβ‰₯1, we get

gt⁒(ΞΈ)β‰₯exp⁑(-ΞΈ24⁒t)⁒πt⁒(1+2β’βˆ‘kβ‰₯1exp⁑(-Ο€2⁒k2t))=exp⁑(-ΞΈ24⁒t)⁒gt⁒(0),

hence

gt⁒(x)β‰₯exp⁑(-βˆ₯xβˆ₯24⁒t)⁒gt⁒(0),

the lower bound we wanted to prove.

Write

S=1+2β’βˆ‘kβ‰₯1exp⁑(-Ο€2⁒k2t)⁒cosh⁑(π⁒k⁒θt).

For any kβ‰₯1, using |ΞΈ|≀π,

2⁒cosh⁑(π⁒k⁒θt)≀2⁒cosh⁑(Ο€2⁒kt)=exp⁑(Ο€2⁒kt)+exp⁑(-Ο€2⁒kt)≀1+exp⁑(Ο€2⁒kt).

Hence

S ≀ 1+βˆ‘kβ‰₯1exp⁑(-Ο€2⁒k2t)⁒(1+exp⁑(Ο€2⁒kt))
= 1+βˆ‘kβ‰₯1exp⁑(-Ο€2⁒k2t)+exp⁑(-Ο€2⁒k⁒(k-1)t)
≀ 1+βˆ‘kβ‰₯1exp⁑(-Ο€2⁒k2t)+exp⁑(-Ο€2⁒(k-1)2t)
= 2+2β’βˆ‘kβ‰₯1exp⁑(-Ο€2⁒k2t)
= 1+tπ⁒gt⁒(0).

But gt⁒(ΞΈ)=Ο€t⁒exp⁑(-ΞΈ24⁒t)⁒S, so

gt⁒(ΞΈ)≀exp⁑(-ΞΈ24⁒t)⁒(Ο€t+gt⁒(0))=exp⁑(-βˆ₯xβˆ₯24⁒t)⁒(Ο€t+gt⁒(0)),

the upper bound we wanted to prove. ∎

Applying Lemma 1 with x=0 gives gt⁒(0)β‰₯Ο€t, and using this with the above theorem we obtain

gt⁒(x)≀2⁒exp⁑(-βˆ₯xβˆ₯24⁒t)⁒gt⁒(0). (1)
Theorem 4.

For t>0,

Ο€t≀gt⁒(0)≀1+Ο€t

and

2⁒e-t≀gt⁒(0)-1≀2⁒e-t1-e-t.
Proof.

Using Lemma 1 we have

gt⁒(0)β‰₯Ο€t.

For each xβˆˆβ„ we have

gt⁒(x)=βˆ‘kβˆˆβ„€g^t⁒(k)⁒ei⁒k⁒x=βˆ‘kβˆˆβ„€e-k2⁒t⁒ei⁒k⁒x=1+2β’βˆ‘kβ‰₯1e-k2⁒t⁒cos⁑(k⁒x).

Writing ϕ⁒(t)=βˆ‘kβ‰₯1e-k2⁒t, we then have

gt⁒(0)=1+2⁒ϕ⁒(t).

But as e-x2⁒t is positive and decreasing, bounding a sum by an integral we get

ϕ⁒(t)β‰€βˆ«0∞e-x2⁒t⁒𝑑x=1t⁒∫0∞e-x2⁒𝑑x=12⁒πt,

hence

gt⁒(0)=1+2⁒ϕ⁒(t)≀1+Ο€t.

Moreover, because ϕ⁒(t)β‰₯e-t (lower bounding the sum by the first term), we have

gt⁒(0)=1+2⁒ϕ⁒(t)β‰₯1+2⁒e-t.

Finally, because e-t⁒k2≀e-t⁒k for kβ‰₯1,

ϕ⁒(t)β‰€βˆ‘kβ‰₯1e-t⁒k=e-t⁒11-e-t,

thus

gt⁒(0)≀1+2⁒e-t1-e-t.

∎

Taking tβ†’0 and tβ†’βˆž in the above theorem gives the following asymptotics.

Corollary 5.
gt⁒(0)βˆΌΟ€t,tβ†’0

and

gt⁒(0)-1∼2⁒e-t,tβ†’βˆž.

2 Heat kernel on 𝕋n

Fix nβ‰₯1, and let π’œ=(a1,…,an), ai positive real numbers. We define gtπ’œ:ℝnβ†’(0,∞) by

gtπ’œβ’(x)=∏k=1ngak⁒t⁒(xk),x=(x1,…,xn)βˆˆβ„n.

For xβˆˆβ„n and ΞΎβˆˆβ„€n we have

gtπ’œβ’(x+2⁒π⁒ξ)=∏k=1ngak⁒t⁒(xk+2⁒π⁒ξk)=∏k=1ngak⁒t⁒(xk)=gtπ’œβ’(x),

so gtπ’œ can be interpreted as a function on 𝕋n.

Let mn be Haar measure on 𝕋n:

d⁒mn⁒(x)=∏k=1nd⁒m⁒(xk)=∏k=1n(2⁒π)-1⁒d⁒xk=(2⁒π)-n⁒d⁒x,

which satisfies mn⁒(𝕋n)=1. Define ΞΌtπ’œ to be the measure on 𝕋n whose density with respect to mn is gtπ’œ:

d⁒μtπ’œ=gtπ’œβ’d⁒mn.

We now calculate the Fourier coefficients of gtπ’œ. For ΞΎβˆˆβ„€n,

ℱ⁒(gtπ’œ)⁒(ΞΎ) = βˆ«π•‹ngtπ’œβ’(x)⁒e-i⁒ξ⋅x⁒𝑑mn⁒(x)
= βˆ«π•‹n∏k=1ngak⁒t⁒(xk)⁒e-i⁒ξ1⁒x1-β‹―-i⁒ξn⁒xn⁒d⁒mn⁒(x)
= ∏k=1nβˆ«π•‹gak⁒t⁒(xk)⁒e-i⁒ξk⁒xk⁒𝑑m⁒(xk)
= ∏k=1ng^ak⁒t⁒(ξk)
= ∏k=1ne-ξk2⁒ak⁒t
= e-t⁒q⁒(ξ),

where

q⁒(ΞΎ)=βˆ‘k=1nak⁒ξk2,ΞΎβˆˆβ„€n.
Definition 6.

For x=(x1,…,xn)∈Rn we define

βˆ₯xβˆ₯π’œ2=1a1⁒βˆ₯x1βˆ₯2+β‹―+1an⁒βˆ₯xnβˆ₯2,

with A=(a1,…,an).

For ΞΎ=(ΞΎ1,…,ΞΎn)βˆˆβ„€n, because βˆ₯xk+2⁒π⁒ξkβˆ₯=βˆ₯xkβˆ₯, we have βˆ₯x+2⁒π⁒ξβˆ₯π’œ=βˆ₯xβˆ₯π’œ, so it makes sense to talk about βˆ₯β‹…βˆ₯π’œ on 𝕋n.

Using Theorem 3 and (1) we get the following.

Theorem 7.

For t>0 and x∈Rn,

exp⁑(-βˆ₯xβˆ₯π’œ24⁒t)⁒gtπ’œβ’(0)≀gtπ’œβ’(x)≀2n⁒exp⁑(-βˆ₯xβˆ₯π’œ24⁒t)⁒gtπ’œβ’(0).

Combining this with Theorem 4 we obtain the following. The first inequality is appropriate for tβ†’0+ and the second inequality for tβ†’βˆž.

Theorem 8.

For t>0 and x∈Rn,

exp⁑(-βˆ₯xβˆ₯π’œ24⁒t)⁒∏k=1nΟ€ak⁒t≀gtπ’œβ’(x)≀2n⁒exp⁑(-βˆ₯xβˆ₯π’œ24⁒t)⁒∏k=1n(1+Ο€ak⁒t)

and

exp⁑(-βˆ₯xβˆ₯π’œ24⁒t)⁒∏k=1n(1+2⁒e-ak⁒t)≀gtπ’œβ’(x)≀2n⁒exp⁑(-βˆ₯xβˆ₯π’œ24⁒t)⁒∏k=1n(1+2⁒e-ak⁒t1-e-ak⁒t).

3 The infinite-dimensional torus

π•‹βˆž with the product topology is a compact abelian group. Let m∞ be Haar measure on π•‹βˆž:

d⁒m∞⁒(x)=∏k=1∞d⁒m⁒(xk),x=(x1,x2,…)βˆˆπ•‹βˆž,

where m is Haar measure on 𝕋.

For t>0, let ΞΌt be the measure on 𝕋 whose density with respect to Haar measure m is gt:

d⁒μt=gt⁒d⁒m.

This is a probability measure on 𝕋.

Let π’œ=(a1,a2,…)βˆˆβ„•βˆž. For t>0 we define

ΞΌtπ’œ=∏k=1∞μak⁒t.

This is a probability measure on π•‹βˆž.22 2 Christian Berg determines conditions on π’œ and t so that ΞΌtπ’œ is absolutely continuous with respect to Haar measure m∞ on π•‹βˆž: Potential theory on the infinite dimensional torus, Invent. Math. 32 (1976), no. 1, 49–100.

The Pontryagin dual of π•‹βˆž is the direct sum βŠ•k=1βˆžβ„€, which we denote by β„€(∞), which is a discrete abelian group. For ΞΎβˆˆβ„€(∞) and xβˆˆπ•‹βˆž, we write

eξ⁒(x)=exp⁑(iβ’βˆ‘k=1∞ξk⁒xk).

The Fourier transform of ΞΌtπ’œ is ℱ⁒(ΞΌtπ’œ):β„€(∞)β†’β„‚ defined by

ℱ⁒(ΞΌtπ’œ)⁒(ΞΎ)=βˆ«π•‹βˆže-ξ⁒(x)⁒𝑑m∞⁒(x),ΞΎβˆˆβ„€(∞),

which is

βˆ«π•‹βˆže-ξ⁒(x)⁒𝑑m∞⁒(x) =βˆ«π•‹βˆžexp⁑(-iβ’βˆ‘k=1∞ξk⁒xk)⁒𝑑μtπ’œβ’(x)
=βˆ«π•‹βˆžβˆk=1∞exp⁑(-i⁒ξk⁒xk)⁒d⁒μtπ’œβ’(x)
=∏k=1βˆžβˆ«π•‹exp⁑(-i⁒ξk⁒xk)⁒gak⁒t⁒(xk)⁒𝑑m⁒(xk)
=∏k=1∞g^ak⁒t⁒(ξk)
=∏k=1∞exp⁑(-ξk2⁒ak⁒t)
=exp⁑(-tβ’βˆ‘k=1∞ak⁒ξk2).

4 Convergence of infinite products

If ckβ‰₯0, then for any n,

1+βˆ‘k=1nckβ‰€βˆk=1n(1+ck)≀exp⁑(βˆ‘k=1nck).

Thus, the limit of ∏k=1n(1+ck) as nβ†’βˆž exists if and only if

βˆ‘k=1∞ck<∞.

For the second inequality in Theorem 8, the limit of ∏k=1n(1+2⁒e-ak⁒t) as nβ†’βˆž exists if and only if

βˆ‘k=1∞2⁒e-ak⁒t<∞.