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Approximating coarse Ricci curvature on submanifolds of Euclidean space

  • Antonio G. Ache und Micah W. Warren EMAIL logo
Veröffentlicht/Copyright: 15. April 2022
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Abstract

For an embedded submanifold Σ ⊂ ℝN, Belkin and Niyogi showed that one can approximate the Laplacian operator using heat kernels. Using a definition of coarse Ricci curvature derived by iterating Laplacians, we approximate the coarse Ricci curvature of submanifolds Σ in the same way. For this purpose, we derive asymptotics for the approximation of the Ricci curvature proposed in [2]. Specifically, we prove Proposition 3.2 in [2].

MSC 2010: 53C23

A Integrals of monomials

We have encountered some integrals of products of monomials and Gaussian functions on balls centered at the origin of ℝd. It is important to point out that these integrals will be computed with respect to the Lebesgue measure of the Euclidean metric. We now explain how to compute these integrals in a systematic way.

Lemma A.1

Let z = (z1, …, zd) ∈ ℝdand consider a subset {i1, …, ip} of {1, …, d}. Given positive integers αi1, …, αip we have

1(2π)d/2Rdezd2/2zi1αi1zipαipdz=ωαi1,,αip(A.1)

where ωαi1,…,αipis defined by ωαi1αip = 0 if some αij is odd, andωαi1αip=j=1p(αij1)!!if all αij are even; here (αij − 1)!! is the odd factorial, that is the product of all odd numbers 1, 3, …, αij − 1.

A similar computation that we will need is

Rdzdkezd2/2dz=volSd12k+d22Γk+d2

where Γk+d12 denotes the Gamma function and vol(Sd−1) denotes the volume of the sphere Sd−1 with respect to the round metric. Observe that

vol(Sd1)=(2π)d/22(d2)/2Γ(d/2)

which implies

Rdzdkezd22dz=(2π)d/22k/2Γk+d2Γ(d2)

and by scaling we have the identity

1(2πt)d/2Rdzdkezd22tdz=tk/22k/2Γ((k+d)/2)Γ(d/2).

Note also that using z=2w we can compute

Rdzdkezd24tdz=Rd2wdkewd22t12ddw=(2πt)d/2tk/222kd/2Γ((k+d)/2)Γ(d/2).(A.2)

From the above we observe that if f(z)=O(zdk and f(z) has polynomial growth as ∥zd → ∞, then

|Rdf(z)ezd22tdz|=|Bε(0)f(z)ezd22tdz+RdBε(0)f(z)ezd22tdz|C0(2πt)d/2tk/22k/2Γ((k+d)/2)Γ(d/2)+O(eε2t)(A.3)

for some constant C0 > 0, and similarly

|Rdf(z)ezd24tdz|(2πt)d/2tk/22(2kd)/2Γ((k+d)/2)Γ(d/2)+O(eε4t).(A.4)

We now state the main lemma in this section.

Lemma A.2

(Odd-cancellation). Let {i1, …, ip} be a subset of {1, …, d} and let αi1, …, αip be positive integers and let α = αi1 + … + αip. Then for any ε > 0 we have

  1. If all integers αij, …, αip are even then

    1(2πt)d/2Bεd(0)ezd22tzi1αi1zipαipdz=tα/2ωαi1,,αip+O(tα/2eε24t).
  2. If at least one integer αii, …, αip is odd then1(2πt)d/2Bεd(0)ezd22tzi1αi1zipαipdz=0.

A.1 Proofs of the two lemmas

Proof of Lemma A.1

The computation of the integral on the left-hand side of (A.1) is in fact equivalent to computing a moment of order α = αi1 + … + αip of a multivariate normal random vector. More specifically, let ξ1, …, ξd be independent and identically distributed (i.i.d.) random variables and assume that the common distribution of these random variables is 𝓝(0, 1), i.e., a normal random variable with mean 0 and variance 1. In this case, the random vector ξ = (ξ1, …, ξd) has a multivariate normal distribution with mean vector μ = (0, …, 0) and co-variance matrix Id. Using that the components of the random vector ξ are independent we have

1(2π)d/2Rdezd2/2zi1αi1zipαipdz=E(ξi1αi1ξipαip)=E(ξi1αi1)E(ξipαip)

where 𝔼 is the expectation with respect to the multivariate normal distribution 𝓝(μ, Id). In particular we may view each factor E(ξjαij) as an expectation with respect to the one-dimensional normal random variable 𝓝(0, 1). Now, for j = 1, …, d, we have

E(ξi1αi1)=12πex2/2xαijdx.

Observe that by symmetry, if αij is odd we have E(ξi1αi1)=0. If αij is even we write αij = 2k so that the computation of E(ξi1αi1) reduces to the integral of x2kex2/2, or in other words, the 2kth moment of an univariate normal random variable with mean zero and variance 1. Even though this computation is well-known, we illustrate a simple way to compute this moment for the reader's convenience. For the standard normal random variable we have (2π)1/2ex2/2dx=1, and by scaling, if we consider a parameter s > 0 we have

esx2/2dx=2πs1/2

and differentiating both sides k times and evaluating at s = 1 we obtain

(1)k2kx2kex2/2dx=(1)k2π12322k12=(1)k2k2π(2k1)!!.

We obtain E(ξijαij)=(αij1)!! which clearly implies the rest of the lemma. □

Proof of Lemma A.2

We start by writing

1(2πt)d/2Bεd(0)ezd22tzi1αi1zipαipdz=1(2πt)d/2Rdezd22tzi1αi1zipαipdz+1(2πt)d/2RdBεd(0)ezd22tzi1αi1zipαipdz=I+II.

By scaling, i.e. using z=tw, we obtain

I=1(2πt)d/2Rdezd22tzi1αi1zipαipdz=1(2πt)d/2Rdewd22(tw)i1αi1(tw)ipαiptd/2dw=tα2(2π)d/2Rdewd22wi1αi1wipαipdw=tα/2ωαi1,,αip(A.5)

noting that (A.5) follows from Lemma A.1. We now estimate II. Observe that from equation (A.2) we have

|II|=1(2πt)d/2|RdBεd(0)ezd22tzi1αi1zipαipdz|=1(2πt)d/2|RdBεd(0)ezd24tzd24tzi1αi1zipαipdz|eε4t(2πt)d/2Rdezd24tzαdz=eε4t2(2α+d)/2tα/2(Γ((k+α)/2)Γ(α/2))+O(eε4t).

This proves the lemma. □

B Proof of Expansion Lemma 3.12

Proof of Lemma 3.12

Recall the following expansions, cf. equation (3.18):

eϕ(z)2t=1+p4(ϕ,x)[z]+p5(ϕ,x)[z]2t+ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z]

and

dμ=(1+j=2mpj(dμ,x)[z]+ϱm+1(dμ,x)[z])dz.

Then

1(2πt)d/2Bεd(0)exexpxzN22tw~(x,z)dμ(z)=1(2πt)d/2Bεd(0)ezd22teϕ(z)2tw~(x,z)dμ(z)(B.1a)

by definition of ϕ(z). We now look at the expansion of the integrand, Taylor expanding (x, z) in terms of z at z = 0:

eϕ(z)2tw~(x,z)dμ=(1+p4(ϕ,x)[z]+p5(ϕ,x)[z]2t+ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z])×(j=04Dzjw~(x,0)+ϱ5(w~,x)[z])×(1+j=24pj(dμ,x)[z]+ϱ5(dμ,x)[z])dz.

We group these by order of polynomial in z, and order of remainder. We distinguish between terms that have the factor 1/(2t) and those that do not. First,

P0(x,z)=w~(x,0)(B.2)
P1(x,z)=Dz1w~(x,0)[z](B.3)
P2(x,z)=Dz2w~(x,0)[z]+w~(x,0)p2(dμ,x)[z](B.4)
P3(x,z)=Dz3w~(x,0)[z]+Dz1w~(x,0)p2(dμ,x)[z]+w~(x,0)p3(dμ,x)[z](B.5)
P4(x,z)=Dz4w~(x,0)[z]+Dz2w~(x,0)p2(dμ,x)[z]+Dz1w~(x,0)p3(dμ,x)[z]+w~(x,0)p4(dμ,x)[z](B.6)
P5(x,z)=Dz3w~(x,0)p2(dμ,x)[z]+Dz2w~(x,0)p3(dμ,x)[z]+Dz1w~(x,0)p4(dμ,x)[z](B.7)
P6(x,z)=Dz4w~(x,0)p2(dμ,x)[z]+Dz3w~(x,0)p3(dμ,x)[z]+Dz2w~(x,0)p4(dμ,x)[z](B.8)
P7(x,z)=Dz4w~(x,0)p3(dμ,x)[z]+Dz3w~(x,0)p4(dμ,x)[z](B.9)
P8(x,z)=Dz4w~(x,0)p4(dμ,x)[z].(B.10)

Next, we list all homogeneous terms containing the factor 1/(2t)

P~4(x,z)=w~(x,0)p4(ϕ,x)[z]2t(B.11)
P~5(x,z)=w~(x,0)p5(ϕ,x)[z]2t+Dz1w~(x,0)p4(ϕ,x)[z](B.12)
P~6=12tw~(x,0)p4(ϕ,x)p2(dμ,x)[z]+p4(ϕ,x)Dz2w~(x,0)[z]+12tp5(ϕ,x)Dz1w~(x,0)[z](B.13)
P~7=12tw~(x,0)p4(ϕ,x)p3(dμ,x)[z]+p4(ϕ,x)Dz3w~(x,0)[z]+p4(ϕ,x)p2(dμ,x)Dzw~(x,0)[z]+p5(ϕ,x)Dz2w~(x,0)+p5(ϕ,x)p2(dμ,x)[z](B.14)
P~8=12t(p4(ϕ,x)+p5(ϕ,x))(j=04Dzjw~(x,0))(1+j=24pj(dμ,x))[z]i=07P~i(z)(B.15)

and the remainder terms

Rϕ(x,z,t)=[ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z]](j=04Dzjw~(x,0))(1+j=24pj(dμ,x))[z](B.16)
Rw(x,z,t)=ϱ5(w~,x)[z](1+p4(ϕ,x)+p5(ϕ,x)2t)(1+j=24pj(dμ,x))[z](B.17)
Rμ(x,z,t)=ϱ5(dμ,x)[z](1+p4(ϕ,x)+p5(ϕ,x)2t)(j=04Dzjw~(x,0))[z](B.18)
Rϕw(x,z,t)=[ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z]]ϱ5(w~,x)[z](1+j=24pj(dμ,x))[z](B.19)
Rϕμ(x,z,t)=[ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z]]ϱ5(dμ,x)[z](j=04Dzjw~(x,0))[z](B.20)
Rwμ(x,z,t)=ϱ5(w~,x)[z]ϱ5(dμ,x)[z](1+p4(ϕ,x)[z]+p5(ϕ,x)[z]2t)(B.21)
Rϕwμ(x,z,t)=[ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z]]ϱ5(w~,x)[z]ϱ5(dμ,x)[z].(B.22)

With this setup (B.1a) becomes

1(2πt)d/2Bεd(0)ezd22ti=08Pi(x,z)+i=47P~i(x,z)+P~8(x,z)+{ϕ,w.μ,ϕw,wμ,ϕμw}R(x,z,t)dz.(B.23)

We deal with these in turn. First, applying Lemma A.2 we have

1(2πt)d/2Bεd(0)ezd22t(i=08Pi(x,z))dz=1(2π)d/2Rdeζd22w~(x,0)+Dz2w~(x,0)[ζ]+w~(x,0)p2(dμ,x)[ζ]t+Dz4w~(x,0)[ζ]+Dz2w~(x,0)p2(dμ,x)[ζ]+Dz1w~(x,0)p3(dμ,x)[ζ]+w~(x,0)p4(dμ,x)[ζ]t2+Dz4w~(x,0)p2(dμ,x)[ζ]++Dz3w~(x,0)p3(dμ,x)[ζ]+Dz4w~(x,0)p2(dμ,x)[ζ]t3+Dz4w~(x,0)[ζ]p4(dμ,x)[ζ]t4dζ(B.24)
+α=08Otα/2eε24t;Σ,J4w,J4dμ.(B.25)

We repeat this for

1(2πt)d/2Bεd(0)ezd22ti=47P~i(x,z)dz=1(2π)d/2Rdeζd22w~(x,0)p4(ϕ,x)[ζ]2tt2+p4(ϕ,x)2tw~(x,0)p2(dμ,x)[ζ]+Dz2w~(x,0)[ζ]+p5(ϕ,x)2tDzw~(x,0)[ζ]t3dζ(B.26)
+12tα=46Otα/2eε24t;Σ,J2w,J2dμ,J5ϕ(B.27)
=1(2π)d/2Rd(0)12eζd22p4(ϕ,x)w~(x,0)[ζ]t+p4(ϕ,x)w~(x,0)p2(dμ,x)[ζ]+Dz2w~(x,0)[ζ]+p5(ϕ,x)Dzw~(x,0)[ζ]t2dζ(B.28)
+α=24O(tα/2eε24t;Σ,J2w,J2dμ,J5ϕ).(B.29)

At this point we pause and, collecting terms, observe the fact that

1(2πt)d/2Bεd(0)ezd22t{i=08Pi(x,z)+i=47P~i(x,z)}dz==1(2π)d/2Rdeζd2/2{w~(x,0)+W1(x,ζ)t+W2(x,ζ)t2+W3(x,ζ)t3+W4(x,ζ)t4}dζ=+O(teε24t;Σ,J2w,J2dμ,J5ϕ)+O(eε24t;Σ,J4w,J4dμ)=1(2π)d/2Rdeζd2/2{w~(x,0)+W1(x,ζ)t+W2(x,ζ)t2+W3(x,ζ)t3+W4(x,ζ)t4}dζ=+O(eε24t;Σ,J4w,J4dμ,J5ϕ).

In other words, going back to (B.23), what is needed to complete the proof is to show that

1(2πt)d/2Bεd(0)ezd22t{P~8(x,z)+{ϕ,w.μ,ϕw,wμ,ϕμw}R(x,z,t)}dz(B.30)
=w~(x,0)Φ(t2;ϕ)+O(t5/2;Σ,J5w,J4dμ,J5ϕ).(B.31)

First we argue that

1(2πt)d/2Bεd(0)ezd22tP~8(x,z)dz=O(t3;Σ,J4w,J4dμ,J5ϕ).(B.32)

Note that the expansion of 8 involves 40 terms, before subtracting the lower order terms, so we elect to not express these explicitly. However, for fixed x, 8 is a polynomial in z with a factor of 1/t, which can be made explicit in terms of J4w, J5ϕ and J4. Each term has order at least 8 in z. Using the change of variable z=tζ as in the proof of Lemma A.2 gives that each term has order at least 4 in t. Since there was an initial factor of 1/t we conclude each term is bounded by a term of order O(t3; J4w, J4, J5ϕ). From Lemma A.2 the remainder not yet accounted for is of order O(t3eε2/4t; J4w, J4, J5ϕ). We combine to conclude (B.32).

Now we turn to the remainder terms:

1(2πt)d/2Bεd(0)ezd22t{{ϕ,w.μ,ϕw,wμ,ϕμw}R(x,z,t)}dz(B.33)
=w~(x,0)Φ(t2;ϕ)(x)+O(t5/2;J5w,J5dμ,J5ϕ).(B.34)

We will explicitly deal with Rϕ(x, z, t), the remaining 7 terms can be dealt with in a similar fashion:

1(2πt)d/2Bεd(0)ezd22tRϕ(x,z,t)dz=1(2πt)d/2Bεd(0)ezd22t[ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z]]×(j=04Dzjw~(x,0))(1+j=24pj(dμ,x))[z]dz.

Recall

ϱ6(ϕ,x)[z]=O(z6;Σ,J6ϕ)(B.35)

and (3.17)

ezd22t|ϱ(ϕ,x,t)[z]|CΣ2zd88t2ezd24t.(B.36)

We then get

1(2πt)d/2Bεd(0)ezd22t[ϱ6(ϕ,x)[z]2t+ϱ(ϕ,x,t)[z]]×(j=04Dzjw~(x,0))(1+j=24pj(dμ,x))[z]dz1(2πt)d/2Bεd(0)ezd22t[Cϕz6](j=04Dzjw~(x,0))(1+j=24pj(dμ,x))[z]dz+1(2πt)d/2Bεd(0)CΣ2zd88t2ezd24t(j=04Dzjw~(x,0))(1+j=24pj(dμ,x))[z]dz

where Cϕ is a constant that depends on bounds on J6ϕ. Scaling, we obtain the following upper bound:

1(2π)d/2Bε/td(0)eζd2/2[Cϕζ6t3]×(j=04Dzjw~(x,0)tj/2)(1+j=24pj(dμ,x)tj/2)[ζ]dζ(B.37)
+1(2π)d/2Bε/td(0)CΣ2ζd8t28eζd2/4×(j=04Dzjw~(x,0)tj/2)(1+j=24pj(dμ,x)tj/2)[ζ]dζ.(B.38)

Now clearly, the first term (in the sense by Lemma A.2) is a sum of terms with order at least 3 in t and dependence on J4w and J4. The second term has a single term of order 2 in t, namely

1(2π)d/2Bε/td(0)CΣ2ζd8t28eζd2/4w~(x,0)dζ=w~(x,0)Φ(t2;ϕ).

The remainder of the terms (again using the estimates in Lemma A.2) are terms of higher order. We have shown (B.33). The analysis of each of the remainder terms of the form

1(2πt)d/2Bεd(0)ezd22tR(x,z,t)dz(B.39)

is similar: In each case we can use bounds such as (B.35) or (B.36) and in others we can use bounds such as

ϱ5(w~,x)[z]=O(zd5;Σ,J5w~(x,z)).

We note that after integration, the terms (B.16), (B.17) and (B.18) determine the least decaying terms. In fact, each of these terms decays like O(t5/2), but (B.16) contains a term of order 6 in z with a factor of 1/(2t) which of course comes down to a term of order O(t2) after rescaling. This is why this term needed special consideration and why we see the term (x, 0)Φ(t2; ϕ) in the expansion. In fact, the term (B.16) is equivalent to (x, 0)Φ(t2; ϕ) in the sense of inequality (3.21). □

Acknowledgements

The authors are grateful to the anonymous referees who provided input and comments that greatly helped to improve the presentation of this article.

  1. Communicated by: F. Duzaar

  2. Funding: The first author was partially supported by a postdoctoral fellowship of the National Science Foundation, award No. DMS-1204742. The second author is partially supported by NSF Grant DMS-1438359.

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Received: 2018-03-22
Revised: 2020-12-21
Published Online: 2022-04-15
Published in Print: 2022-04-26

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