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Ricci Curvature Formula: Applications to Bonnet-Myers Sharp Irregular Graphs

Yupei Li University of South Carolina, Columbia, SC 29208, ([email protected])    Linyuan Lu University of South Carolina, Columbia, SC 29208, ([email protected]). This author was supported in part by NSF grant DMS 2038080.
Abstract

In this paper, we establish a simple formula for computing the Lin-Lu-Yau Ricci curvature on graphs. For any edge xyxy in a simple locally finite graph GG, the curvature κ(x,y)\kappa(x,y) can be expressed as a cost function of an optimal bijection between two blow-up sets of the neighbors of xx and yy. Utilizing this approach, we derive several results including a structural theorem for the Bonnet-Myers sharp irregular graphs of diameter 33 and a theorem on C3C_{3}-free Bonnet-Myers sharp graphs.

1 Introduction

In Riemannian geometry, the renowned Bonnet–Myers theorem [Mye41] asserts that if the Ricci curvature of a complete Riemannian manifold MM is bounded below by (n1)κ>0(n-1)\kappa>0, then its diameter is at most πκ\frac{\pi}{\sqrt{\kappa}}. Over the past three decades, there have been numerous efforts to extend the concept of Ricci curvature from Riemannian geometry to a discrete setting. There are different definitions of Ricci curvature defined on graphs, see references [Jos17, CY96, LY10, Oll07]. For instance, Ollivier [Oll09] introduced a notion of Ricci curvature in metric spaces equipped with a measure of a random walk. Lin-Lu-Yau [LLY11] defined a variation of Ollivier-type Ricci curvature on graphs. These notions allow for the generalization of several classical theorems associated with positive Ricci curvature, such as the Lichnerowicz theorem, the Bonnet–Myers theorem, and many others. Numerous properties and implications of the Ollivier Ricci curvature and Lin-Lu-Yau curvature have been explored, see [BCL+17, BL12, BM13, Liu14, Smi14], etc. These curvatures has been applied in various research areas such as network analysis[NLLG19, SJB19], quantum computation, dynamic Networks[GHJ+16], etc.

In this paper, we consider locally finite graphs, where each vertex has a finite degree, but the total number of vertices may be countably infinite. Let G=(V,E)G=(V,E) be a simple locally finite graph with the vertex set VV and edge set EE. (Here “simple” refers to the absence of self-loops and multi-edges.) For any vertex vV(G)v\in V(G), let NG(v)N_{G}(v) denote the neighborhood of vv in GG, i.e., NG(v)={u:vuE(G)}N_{G}(v)=\{u:vu\in E(G)\}, and let NG[v]:=NG(v){v}N_{G}[v]:=N_{G}(v)\cup\{v\} denote the closed neighborhood of vv. The degree of a vertex vv, denoted by dvd_{v}, is the number of the neighbors of vv, i.e., dv=|NG(v)|d_{v}=|N_{G}(v)|. A graph GG is called dd-regular if dv=dd_{v}=d for any vertex vv. We call GG irregular if it is not regular. For SV(G)S\subseteq V(G), NG(S)={uV(G)S:usE(G) for some sS}N_{G}(S)=\{u\in V(G)\setminus S:us\in E(G)\textrm{ for some }s\in S\}. When GG is clear under context, we will omit the subscript GG in the notation.

A uvuv-walk PP is a sequence of vertices u=u0,u1,,uk=vu=u_{0},u_{1},\ldots,u_{k}=v such that uiu_{i} and ui+1u_{i+1} are adjacent for i=0,,k1i=0,\ldots,k-1. Sometime we write uPvuPv to indicate the walk is from uu to vv. Given two walks uP1vuP_{1}v and vP2wvP_{2}w, the concatenation of two walk is a new walk, written by uP1vP2wuP_{1}vP_{2}w, that walks from uu to vv via P1P_{1} and then continues from vv to ww via P2P_{2}. A walk PP is called uvuv-path if all internal vertices are distinct. A walk is called closed if u=vu=v. A cycle is a closed walk so that all internal vertices are distinct. We say GG is connected if for any pair of vertices uu and vv, there is a path (or walk) uPvuPv.

For any two vertices u,vV(G)u,v\in V(G), the distance from uu to vv in GG, denoted by dG(u,v)d_{G}(u,v), is the number of edges of a shortest path from uu to vv in GG. The diameter of a graph GG is defined to be diam(G)=sup{dG(u,v):u,vV(G)}diam(G)=\sup\{d_{G}(u,v):u,v\in V(G)\}.

A mass distribution mm (over the vertex set V=V(G)V=V(G)) is a mapping m:V[0,)m:V\to[0,\infty). The total mass of mm is the L1L_{1} norm, m1=vVm(v)\|m\|_{1}=\sum_{v\in V}m(v). A probability distribution is a mass distribution with total mass equal to 1.

Let m1m_{1} and m2m_{2} be two mass distributions on VV with equal total masses. A coupling between m1m_{1} and m2m_{2} is a mapping A:V×V[0,1]A:V\times V\to[0,1] such that

yVA(x,y)=m1(x) and xVA(x,y)=m2(y).\displaystyle\sum_{y\in V}A(x,y)=m_{1}(x)\textrm{ and }\displaystyle\sum_{x\in V}A(x,y)=m_{2}(y). (1)

The cost of AA, denoted by C(A)C(A), is given by

C(A)=u,vA(u,v)d(u,v).C(A)=\sum_{u,v}A(u,v)d(u,v). (2)

The transportation distance between the two probability distributions m1m_{1} and m2m_{2} is defined as follows:

W(m1,m2)=infAC(A),W(m_{1},m_{2})=\inf_{A}C(A), (3)

where the infimum is taken over all coupling AA between m1m_{1} and m2m_{2}. By the duality theorem of a linear optimization problem, the transportation distance can also be expressed as follows:

W(m1,m2)=supfxVf(x)(m1(x)m2(x)),W(m_{1},m_{2})=\sup_{f}\displaystyle\sum_{x\in V}f(x)\left(m_{1}(x)-m_{2}(x)\right), (4)

where the supremum is taken over all 11-Lipschitz functions ff.

A random walk mm on G=(V,E)G=(V,E) is defined as a family of probability measures {mv()}vV\{m_{v}(\cdot)\}_{v\in V} such that mv(u)=1dvm_{v}(u)=\frac{1}{d_{v}} for any uN(v)u\in N(v) and 0 otherwise. The Ricci curvature κ:(V(G)2)\kappa:\binom{V(G)}{2}\to\mathbb{R} of GG can then be defined as follows:

Definition 1.

Given a connected locally finite graph G=(V,E)G=(V,E), a random walk m={mv()}vVm=\{m_{v}(\cdot)\}_{v\in V} on GG and two vertices x,yVx,y\in V, define the Ollivier Ricci Curvature

κ0(x,y)=1W(mx,my)d(x,y).\kappa_{0}(x,y)=1-\frac{W(m_{x},m_{y})}{d(x,y)}.

For 0α<10\leq\alpha<1, the α\alpha-lazy random walk mxαm_{x}^{\alpha} (for any vertex xx), is defined as

mxα(v)={α if v=x,(1α)/dx if vN(x),0 otherwise.m_{x}^{\alpha}(v)=\begin{cases}\alpha&\textrm{ if $v=x$,}\\ (1-\alpha)/d_{x}&\textrm{ if $v\in N(x)$,}\\ 0&\textrm{ otherwise.}\end{cases}

In [LLY11], Lin, Lu, and Yau defined the Ricci curvature of graphs based on the α\alpha-lazy random walk as α\alpha goes to 11. More precisely, for any x,yVx,y\in V, they defined the α\alpha-Ricci-curvature κα(x,y)\kappa_{\alpha}(x,y) to be

κα(x,y)=1W(mxα,myα)d(x,y)\kappa_{\alpha}(x,y)=1-\frac{W(m_{x}^{\alpha},m_{y}^{\alpha})}{d(x,y)}

and the Lin-Lu-Yau Ricci curvature κLLY\kappa_{\textrm{LLY}} of GG to be

κLLY(x,y)=limα1κα(x,y)(1α).\kappa_{\textrm{LLY}}(x,y)=\displaystyle\lim_{\alpha\to 1}\frac{\kappa_{\alpha}(x,y)}{(1-\alpha)}.

This limit always exists since κα\kappa_{\alpha} is concave in α[0,1]\alpha\in[0,1] for any two vertices x,yx,y (see [LLY11]).

The parameter α\alpha is called the idleness of the random walk. When α=0\alpha=0, mxαm_{x}^{\alpha} becomes classical random walk. The curvature κ0\kappa_{0} is Ollivier’s original definition of Ricci curvature on graphs (see [Oll09]). It has been shown in [LLY11] that

κLLY(x,y)κ0(x,y)\kappa_{LLY}(x,y)\geq\kappa_{0}(x,y) (5)

for any pair of vertices x,yx,y.

In this paper, we only consider the Lin-Lu-Yau curvature and simply write κLLY(x,y)\kappa_{LLY}(x,y) as κ(x,y)\kappa(x,y) for the rest of the paper. Although the Ricci curvature κ(x,y)\kappa(x,y) is defined for all pairs x,yV(G)x,y\in V(G), it suffices to consider only κ(x,y)\kappa(x,y) for xyE(G)xy\in E(G) due to the following lemma.

Lemma 1.

[LLY11, Oll09] Let GG be a connected graph. If κ(x,y)k\kappa(x,y)\geq k for any edge xyE(G)xy\in E(G), then κ(x,y)k\kappa(x,y)\geq k for any pair of vertices (x,y)(x,y).

Bourne-Cushing-Liu-Münch-Pyerimhoff proved [BCL+18] the following result:

Theorem 1.

(see [BCL+18], Theorem 1.1) Let G=(V,E)G=(V,E) be a locally finite graph. For any edge xyxy, the function ακα(x,y)\alpha\to\kappa_{\alpha}(x,y) is concave and piece-wise linear over [0,1][0,1] with at most 3 linear parts. Furthermore κα(x,y)\kappa_{\alpha}(x,y) is linear on the intervals [0,1lcm(dx,dy)+1][0,\frac{1}{{\rm lcm}(d_{x},d_{y})+1}] and [1max(dx,dy)+1,1)[\frac{1}{\max(d_{x},d_{y})+1},1) Thus, if we have the further condition dx=dyd_{x}=d_{y}, then κα(x,y)\kappa_{\alpha}(x,y) has at most two linear parts.

Their result implies

κ(x,y)=1(1α)κα, for α[1max{dx,dy}+1,1).\kappa(x,y)=\frac{1}{(1-\alpha)}\kappa_{\alpha},\hskip 28.45274pt\mbox{ for }\alpha\in\left[\frac{1}{\max\{d_{x},d_{y}\}+1},1\right). (6)

Münch and Wojciechowski [MW19] gave another limit-free formulation of the Lin-Lu-Yau Ricci curvature using graph Laplacian. For a graph G=(V,E)G=(V,E), the (negative) combinatorial graph Laplacian Δ\Delta is defined as:

Δf(x)=1dxyN(x)(f(y)f(x)).\Delta f(x)=\frac{1}{d_{x}}\sum\limits_{y\in N(x)}(f(y)-f(x)).
Theorem 2.

[MW19] (Curvature via the Laplacian) Let GG be a simple graph and let xyV(G)x\neq y\in V(G). Then

κ(x,y)=inffLip(1)yxf=1xyΔf,\kappa(x,y)=\inf_{\begin{subarray}{c}f\in Lip(1)\\ \nabla_{yx}f=1\end{subarray}}\nabla_{xy}\Delta f, (7)

where xyf=f(x)f(y)d(x,y)\nabla_{xy}f=\frac{f(x)-f(y)}{d(x,y)}.

It was also proved (see [MW19, BCL+17, CK19]) that the optimal solution ff in (7) can be chosen to be an integer-valued function.

Bai, Huang, Lu, and Yau prove a dual theorem for a limit-free definition for the Lin-Lu-Yau Ricci curvature. For any two vertices xx and yy, a \ast-coupling between mx0m^{0}_{x} and my0m^{0}_{y} is a mapping B:V×VB:V\times V\to\mathbb{R} with finite support such that

  1. 1.

    0<B(x,y)0<B(x,y), but all other values B(u,v)0B(u,v)\leq 0.

  2. 2.

    u,vVB(u,v)=0\sum\limits_{u,v\in V}B(u,v)=0.

  3. 3.

    vVB(u,v)=mx0(u)\sum\limits_{v\in V}B(u,v)=-m^{0}_{x}(u) for all uu except xx.

  4. 4.

    uVB(u,v)=my0(v)\sum\limits_{u\in V}B(u,v)=-m^{0}_{y}(v) for all vv except yy.

Theorem 3.

[BHLY21] (Curvature via the \ast-Coupling function) For any two vertex x,yV(G)x,y\in V(G), we have

κ(x,y)=1d(x,y)supBu,vVB(u,v)d(u,v),\kappa(x,y)=\frac{1}{d(x,y)}\sup\limits_{B}\sum\limits_{u,v\in V}B(u,v)d(u,v), (8)

where the superemum is taken over all \ast-coupling BB between mx0m^{0}_{x} and my0m^{0}_{y}.

In this paper, we will derive a straightforward formula to calculate κ(x,y)\kappa(x,y) for any edge xyE(G)xy\in E(G). We use the following notation. Let lcm(dx,dy){\rm lcm}(d_{x},d_{y}) denote the least common multiplier of dxd_{x} and dyd_{y}. Let cxc_{x} and cyc_{y} are a pair of relative prime integers such that

lcm(dx,dy)=cxdx=cydy.{\rm lcm}(d_{x},d_{y})=c_{x}d_{x}=c_{y}d_{y}.

Without loss of generality, we assume dxdyd_{x}\leq d_{y}. Equivalently, we have cxcyc_{x}\geq c_{y}. Let μx\mu_{x} be a mass distribution defined as

μx(u)={cxcy if uN(x)N[y],cx if uN(x)\N[y],0otherwise.\mu_{x}(u)=\begin{cases}c_{x}-c_{y}&\mbox{ if }u\in N(x)\cap N[y],\\ c_{x}&\mbox{ if }u\in N(x)\backslash N[y],\\ 0&\mbox{otherwise.}\end{cases}

Let μy\mu_{y} be a mass distribution defined as

μy(u)={cy if uN(y)\N[x],0otherwise.\mu_{y}(u)=\begin{cases}c_{y}&\mbox{ if }u\in N(y)\backslash N[x],\\ 0&\mbox{otherwise.}\end{cases}

For any coupling σ\sigma between μx\mu_{x} and μy\mu_{y}, recall the cost function C(σ)C(\sigma) is

C(σ)=u,vσ(u,v)d(u,v).C(\sigma)=\sum_{u,v}\sigma(u,v)d(u,v). (9)

In the expression of the cost function C(σ)C(\sigma) above, it suffices to sum up only for uN(x)u\in N(x) and vN(y)N[x]v\in N(y)\setminus N[x] because σ(u,v)=0\sigma(u,v)=0 if (u,v)N(x)×(N(y)N[x])(u,v)\not\in N(x)\times(N(y)\setminus N[x]).

We have the following theorem.

Theorem 4.

For any edge xyE(G)xy\in E(G), assuming dxdyd_{x}\leq d_{y}, we have

κ(x,y)=1+1dyminσC(σ)lcm(dx,dy).\kappa(x,y)=1+\frac{1}{d_{y}}-\frac{\min_{\sigma}C(\sigma)}{{\rm lcm}(d_{x},d_{y})}. (10)

Here the minimum is taken over all integer-valued couplings σ\sigma between μx\mu_{x} and μy\mu_{y}.

For example, consider the following graph in Figure 1. We have dx=3d_{x}=3 and dy=4d_{y}=4. Thus, lcm(dx,dy)=12{\rm lcm}(d_{x},d_{y})=12, cx=4c_{x}=4, and cy=3c_{y}=3. The mass distribution μx\mu_{x} is labelled in red color while μy\mu_{y} is labelled in black color. Any coupling between μx\mu_{x} and μy\mu_{y} has a cost of at least 1414 since transferring 44 units from x1x_{1} to y1y_{1} or y2y_{2} costs at least 4×3=124\times 3=12 and moving other two units has the cost of at least 22. Clearly, an optimal coupling with cost 1414 exists. For example, one can transfer one unit from zz to y1y_{1}, one unit from yy to y2y_{2}, two units from x1x_{1} to y1y_{1}, and two units from x1x_{1} to y2y_{2}. So we have

minσC(σ)=14.\min_{\sigma}C(\sigma)=14.

By Theorem 4, we get

κ(x,y)=1+141412=112.\kappa(x,y)=1+\frac{1}{4}-\frac{14}{12}=\frac{1}{12}.
xxyy11zz11y1y_{1}33y2y_{2}33x1x_{1}44
Figure 1: A example: Mass distributions μx\mu_{x} (in red color) and μy\mu_{y} (in black color).

The integer-valued coupling can be viewed as a bijection between the blowup of N(x)N(x) and N(y)N[x]N(y)\setminus N[x]. More discussions will be given in Section 3. It is useful for understanding the structural meaning of Ricci curvature κ(x,y)\kappa(x,y). The special case of dx=dyd_{x}=d_{y} was considered by Bai and Lei [LB22]. They derived a formula using the number of (almost)-disjoint C3C_{3}, C4C_{4}, and C5C_{5} containing both xx and yy. The formula can be shown to equivalent to Equation (10).

Lin, Lu, and Yau [LLY11] and Ollivier [Oll09] proved the Bonnet-Myers type Theorem on Graphs:

Lemma 2.

If for every edge xyE(G)xy\in E(G), κ(x,y)κ1>0\kappa(x,y)\geq\kappa_{1}>0, then the diameter of the graph GG

diam(G)2κ1.\textrm{diam}(G)\leq\frac{2}{\kappa_{1}}.

Let κmin(G)=minxyE(G)κ(x,y)\kappa_{min}(G)=\min_{xy\in E(G)}\kappa(x,y). We say a connected graph GG is Bonnet-Myers sharp if diam(G)=2κmindiam(G)=\frac{2}{\kappa_{min}}. A pair of vertices (x,y)(x,y) in an Bonnet-Myers sharp graph is called poles if d(x,y)=diam(G)d(x,y)=diam(G). In this case, we call xx (and yy) is a pole.

Cushing-Kamtue-Koolen-Liu-Münch-Peyerimhoff[CKK+20] studied Bonnet-Myers sharp regular graphs. They called a graph GG is (D,L)-Bonnet-Myers sharp if GG is DD-regular, diam(G)=Ldiam(G)=L, and is Bonnet-Myers sharp. A graph G=(V,E)G=(V,E) is called self-centered if, for every vertex xVx\in V, there exists a vertex x¯V\overline{x}\in V such that d(x,x¯)=diam(G)d(x,\overline{x})={\rm diam}(G). Here is their main result.

Theorem 5.

[CKK+20] Self-centered (D,L)(D,L)-Bonnet-Myers sharp graphs are precisely the following graphs:

  1. 1.

    hypercubes QnQ_{n}, n1n\geq 1;

  2. 2.

    cocktail party graphs CP(n)CP(n), n3n\geq 3;

  3. 3.

    the Johnson graphs J(2n,n)J(2n,n), n3n\geq 3;

  4. 4.

    even-dimensional demi-cubes Q2n(2)Q_{2n}^{(2)}, n3n\geq 3;

  5. 5.

    the Gosset graph;

and Cartesian products of above graphs satisfying

D1L1=D2L2==DkLk.\frac{D_{1}}{L_{1}}=\frac{D_{2}}{L_{2}}=\cdots=\frac{D_{k}}{L_{k}}. (11)

Kamtue proved that a (D,3)(D,3)-Bonnet-Myers sharp graph is automatically self-centered (see [Kam20]). Thus, the condition of being self-centered in Theorem 5 can be removed for graphs with diameter 33.

Most known results assume that GG is DD-regular. The smallest irregular and not-self-centered Bonnet-Myers sharp graph is P3P_{3}. We have κmin(P3)=1\kappa_{min}(P_{3})=1 and diameter L=2L=2.

In this paper, we will omit the conditions of being regular and self-centered. To the best of our knowledge, there is only one other paper that has studied Bonnet-Myers sharp irregular graphs, providing a classification for those with a diameter of 22 and some necessary conditions for anti-trees being Bonnet-Myers sharp (see [CS24]). They showed that a graph GG is a diameter 22 Bonnet-Myers sharp graph if and only if GG is a complete graph with a matching removed.

The case for diameter 3 (of irregular Bonnet-Myers sharp graphs) remains entirely unexplored. In this paper, we establish the following structural theorem.

Theorem 6.

Suppose GG is a Bonnet-Myers sharp irregular graph on nn vertices with diameter 3, then we have:

  1. 1.

    GG has a unique pair of poles, say x,yx,y. We have dx=dy=n22d_{x}=d_{y}=\frac{n-2}{2}.

  2. 2.

    For any vertex uN(y)u\in N(y) (or vN(x)v\in N(x)), the induced graph G[N(u)N(x)]G[N(u)\cap N(x)] (or G[N(v)N(y)]G[N(v)\cap N(y)]) is a complete graph with a matching removed, respectively.

  3. 3.

    There exists two integers rr and tt satisfying 1tr2+21\leq t\leq\frac{r}{2}+2 such that dx=dy=2r+td_{x}=d_{y}=2r+t and du=3(r+1)d_{u}=3(r+1) for all other vertices uxu\not=x or yy.

Corollary 1.

If GG is an irregular Bonnet-Myers sharp graph of diameter 33, then GG has a unique pair of poles.

We have constructed an infinite number of Bonnet-Myers sharp irregular graphs with a diameter of 3. Specifically, there exists a Bonnet-Myers sharp graph of diameter 33 for t=1,2t=1,2 and all r1r\geq 1. (We suspect that tt is limited to a few specific values. Determining all possible pairs of (r,t)(r,t) appears to be a challenging problem). On the contrast, there are only four Bonnet-Myers sharp regular graphs of diameter 33: the hypercube Q3Q_{3}, the Johnson graph J(6,3)J(6,3), the demi-cubes Q6(2)Q_{6}^{(2)}, and the Gosset graph.

We additionally prove the following result concerning C3C_{3}-free Bonnet-Myers sharp graphs.

Theorem 7.

Suppose GG is a C3C_{3}-free Bonnet-Myers sharp graph with diameter LL. If GG contains a pole xx with dx>23Ld_{x}>\frac{2}{3}L, then GG must be the hypercube QLQ_{L}.

When dx23Ld_{x}\leq\frac{2}{3}L, Bonnet-Myers sharp irregular graphs may exist. See examples in Table 6.

The remainder of the paper is organized as follows: In Section 2, we establish a theorem on the existence of integer-valued optimal couplings and a lemma on complementary slackness. In Section 3, we present the proof of Theorem 4. Section 4 is dedicated to proving several lemmas for Bonnet-Myers sharp graphs. In Section 5, we prove Theorem 6 and also construct many diameter 3 Bonnet-Myers-sharp irregular graphs. Finally, in the last section, we examine the C3C_{3}-free Bonnet-Myers-sharp graphs.

2 Notation and Lemmas on Ricci Curvature

A mass distribution (over the vertex set V=V(G)V=V(G)) is a mapping m:V[0,)m:V\to[0,\infty). The total mass of mm is m1=vVm(v)\|m\|_{1}=\sum_{v\in V}m(v). A probability distribution can be viewed as a mass distribution with a total mass of 1. The concepts of coupling (Equation (1)) and transportation distance (Equations (3) and (4)) can be easily extended to two mass distributions with equal total mass.

For any positive constant cc, let cmcm (or cAcA) be the distribution (or coupling) scaled by a factor of cc. It is straightforward to verify

W(cm1,cm2)=cW(m1,m2)W(cm_{1},cm_{2})=cW(m_{1},m_{2}) (12)

We first prove a lemma on the existence of integer-valued optimal coupling.

Lemma 3.

If m1m_{1} and m2m_{2} are both integer-valued mass distributions (with finite supports), then there exists an integer-valued coupling (with finite supports) between m1m_{1} and m2m_{2}, A:V×VA:V\times V\to\mathbb{Z} such that AA is an optimal solution of Equation (3).

Proof.

Let A0A_{0} be an optimal solution of Equation 3. Suppose A0A_{0} is not integer-valued. Let L={uV:m1(u)>0}L=\{u\in V:m_{1}(u)>0\} and R={vV:m2(v)>0}R=\{v\in V:m_{2}(v)>0\}. We consider a bipartite graph H0:=H(A0)H_{0}:=H(A_{0}) with vertex bi-partition LRL\sqcup R (the disjoint union of LL and RR, i.e., common vertices are duplicated), and the edge set

E(H0)={(u,v):andA0(u,v)}.E(H_{0})=\{(u,v):\mbox{and}\ A_{0}(u,v)\notin\mathbb{Z}\}.

Since m1m_{1} and m2m_{2} are both integer-valued, the minimum degree of H0H_{0} is at least 22 so H0H_{0} contains a cycle CC. Since H0H_{0} is bipartite, this cycle CC must have even length, say 2k2k. Let u1,v1,,uk,vku_{1},v_{1},\ldots,u_{k},v_{k} be the vertices of CC under a cyclic orientation.

Here u1,u2,,ukLu_{1},u_{2},\ldots,u_{k}\in L while v1,v2,,vkRv_{1},v_{2},\ldots,v_{k}\in R. (For convenience, let uk+1=u1u_{k+1}=u_{1} and vk+1=v1v_{k+1}=v_{1}.) Without loss of generality, we can assume

i=1kd(ui,vi)i=1kd(ui+1,vi).\sum_{i=1}^{k}d(u_{i},v_{i})\leq\sum_{i=1}^{k}d(u_{i+1},v_{i}). (13)

(Otherwise, simply reverse the orientation of CC and relabel its vertices.)

For 1ik1\leq i\leq k, define

ϵi\displaystyle\epsilon_{i} =A0(ui,vi)A0(ui,vi),\displaystyle=\lceil A_{0}(u_{i},v_{i})\rceil-A_{0}(u_{i},v_{i}),
ϵi\displaystyle\epsilon^{\prime}_{i} =A0(ui+1,vi)A0(ui+1,vi).\displaystyle=A_{0}(u_{i+1},v_{i})-\lfloor A_{0}(u_{i+1},v_{i})\rfloor.

Here x\lceil x\rceil represents the ceil of xx, which is the smallest integer that is greater than or equal to xx. Similarly, x\lfloor x\rfloor represents the floor of xx, which is the greatest integer that is less than or equal to xx.

Let ϵ0\epsilon_{0} be the minimum value among ϵ1,,ϵk,ϵ1,,ϵk\epsilon_{1},\ldots,\epsilon_{k},\epsilon^{\prime}_{1},\ldots,\epsilon^{\prime}_{k}. By the definition of H0H_{0}, we have ϵ0>0\epsilon_{0}>0.

Now we define a new coupling A1A_{1}:

A1(u,v)={A0(u,v)+ϵ0if(u,v)=(ui,vi)E(C);A0(u,v)ϵ0if(u,v)=(ui+1,vi)E(C);A0(u,v)if(u,v)E(C).A_{1}(u,v)=\begin{cases}A_{0}(u,v)+\epsilon_{0}&\mbox{if}\ (u,v)=(u_{i},v_{i})\in E(C);\\ A_{0}(u,v)-\epsilon_{0}&\mbox{if}\ (u,v)=(u_{i+1},v_{i})\in E(C);\\ A_{0}(u,v)&\mbox{if}\ (u,v)\notin E(C).\\ \end{cases}

Observe that all entries of A1A_{1} are still non-negative. In addition, we have

vA1(u,v)=vA0(u,v)=m1(u),\displaystyle\sum_{v}A_{1}(u,v)=\sum_{v}A_{0}(u,v)=m_{1}(u),
uA1(u,v)=uA0(u,v)=m2(v).\displaystyle\sum_{u}A_{1}(u,v)=\sum_{u}A_{0}(u,v)=m_{2}(v).

Therefore A1A_{1} is still a coupling between m1m_{1} and m2m_{2}. Now we have

C(A1)\displaystyle C(A_{1}) =x,yVA1(x,y)d(x,y)\displaystyle=\sum_{x,y\in V}A_{1}(x,y)d(x,y)
=x,yVA0(x,y)d(x,y)+ϵ0i=1kd(ui,vi)ϵ0i=1kd(ui+1,vi)\displaystyle=\sum_{x,y\in V}A_{0}(x,y)d(x,y)+\epsilon_{0}\sum_{i=1}^{k}d(u_{i},v_{i})-\epsilon_{0}\sum_{i=1}^{k}d(u_{i+1},v_{i})
=C(A0)+ϵ0(i=1kd(ui,vi)i=1kd(ui+1,vi))\displaystyle=C(A_{0})+\epsilon_{0}\left(\sum_{i=1}^{k}d(u_{i},v_{i})-\sum_{i=1}^{k}d(u_{i+1},v_{i})\right)
C(A0).\displaystyle\leq C(A_{0}).

Since A0A_{0} is optimal, we must have C(A1)=C(A0)C(A_{1})=C(A_{0}). Thus, A1A_{1} is also optimal. If some values of A1(u,v)A_{1}(u,v) are not integers, we can repeat this process to define a new bipartite graph H1=H(A1)H_{1}=H(A_{1}). Note that H1H_{1} will have at least one fewer edge than H0H_{0}. This process will terminate after a finite number of iterations, ultimately yielding an integer-valued optimal coupling AA. ∎

Note that Equations (7) and (8) are dual to each other. We have the following lemma, which is the special case of the complementary slackness theorem in the linear programming.

Lemma 4.

Suppose ff is an optimal solution of Equation (7) and BB is an optimal solution of Equation (8). Then we have

  1. 1.

    If f(v)f(u)d(u,v)f(v)-f(u)\not=d(u,v), then B(u,v)=0B(u,v)=0.

  2. 2.

    If B(u,v)0B(u,v)\not=0, then f(v)f(u)=d(u,v)f(v)-f(u)=d(u,v).

Proof.

We have

κ(x,y)\displaystyle\kappa(x,y) =xyΔf\displaystyle=\nabla_{xy}\Delta f
=Δf(x)Δf(y)\displaystyle=\Delta f(x)-\Delta f(y)
=uμx(u)(f(u)f(x))vμy(v)(f(v)f(y))\displaystyle=\sum_{u}\mu_{x}(u)(f(u)-f(x))-\sum_{v}\mu_{y}(v)(f(v)-f(y))
=u,vB(u,v)(f(x)f(u))u,vB(u,v)(f(y)f(v))\displaystyle=\sum_{u,v}B(u,v)(f(x)-f(u))-\sum_{u,v}B(u,v)(f(y)-f(v))
=u,vB(u,v)(f(x)f(y))+u,vB(u,v)(f(v)f(u))\displaystyle=\sum_{u,v}B(u,v)(f(x)-f(y))+\sum_{u,v}B(u,v)(f(v)-f(u))
=u,vB(u,v)(f(v)f(u))\displaystyle=\sum_{u,v}B(u,v)(f(v)-f(u))
=B(x,y)+(u,v)(x,y)B(u,v)(f(v)f(u))\displaystyle=B(x,y)+\sum_{(u,v)\not=(x,y)}B(u,v)(f(v)-f(u))
B(x,y)+(u,v)(x,y)B(u,v)d(u,v)\displaystyle\geq B(x,y)+\sum_{(u,v)\not=(x,y)}B(u,v)d(u,v)
=u,vB(u,v)d(u,v)\displaystyle=\sum_{u,v}B(u,v)d(u,v)
=κ(x,y).\displaystyle=\kappa(x,y).

The equality holds if and only if for any pair of vertices (u,v)(x,y)(u,v)\not=(x,y),

B(u,v)(f(v)f(u)d(u,v))=0.B(u,v)(f(v)-f(u)-d(u,v))=0.

Thus, the assertion holds for all (u,v)(x,y)(u,v)\not=(x,y). On the other hand, the assertion is trivial for (u,v)=(x,y)(u,v)=(x,y). ∎

Lemma 5.

Suppose ff is an optimal solution of Equation (7) and σ\sigma is an optimal solution of Equation (10). Then for any pair of distinct vertices (u,v)(x,y)(u,v)\not=(x,y), we have

  1. 1.

    If f(v)f(u)d(u,v)f(v)-f(u)\not=d(u,v), then σ(u,v)=0\sigma(u,v)=0.

  2. 2.

    If σ(u,v)0\sigma(u,v)\not=0, then f(v)f(u)=d(u,v)f(v)-f(u)=d(u,v).

Proof.

We define a \ast-coupling BσB_{\sigma} between mx0m_{x}^{0} and my0m_{y}^{0} as follows. For any u,vV(G)u,v\in V(G), we have

Bσ(u,v)={1+1dy if (u,v)=(x,y);1dy if u=vN[x]N[y];σ(u,v)lcm(dx,dy) if (u,v)N[x]×(N(y)N[x]);0 otherwise.B_{\sigma}(u,v)=\begin{cases}1+\frac{1}{d_{y}}&\mbox{ if }(u,v)=(x,y);\\ -\frac{1}{d_{y}}&\mbox{ if }u=v\in N[x]\cap N[y];\\ -\frac{\sigma(u,v)}{{\rm lcm}(d_{x},d_{y})}&\mbox{ if }(u,v)\in N[x]\times(N(y)\setminus N[x]);\\ 0&\mbox{ otherwise.}\end{cases} (14)

Now we verify that BσB_{\sigma} is a \ast-coupling between mx0m^{0}_{x} and my0m^{0}_{y} by checking the four conditions in its definition. Condition (1) is satisfied by the definition. For Condition (2), we have

u,vV(G)B(u,v)\displaystyle\sum\limits_{u,v\in V(G)}B(u,v) =1+1dy|N[x]N[y]|dyu,vσ(u,v)lcm(dx,dy)\displaystyle=1+\frac{1}{d_{y}}-\frac{|N[x]\cap N[y]|}{d_{y}}-\frac{\sum_{u,v}\sigma(u,v)}{{\rm lcm}(d_{x},d_{y})}
=1+1dy|N(x)N(y)|+2dyμy1lcm(dx,dy)\displaystyle=1+\frac{1}{d_{y}}-\frac{|N(x)\cap N(y)|+2}{d_{y}}-\frac{\|\mu_{y}\|_{1}}{{\rm lcm}(d_{x},d_{y})}
=1+1dy|N(x)N(y)|+2dy(dy1|N(x)N(y)|)cylcm(dx,dy)\displaystyle=1+\frac{1}{d_{y}}-\frac{|N(x)\cap N(y)|+2}{d_{y}}-\frac{(d_{y}-1-|N(x)\cap N(y)|)c_{y}}{{\rm lcm}(d_{x},d_{y})}
=0.\displaystyle=0.

Both sides of the equation in Condition (3) are zeros unless uN[x]u\in N[x]. Since uxu\not=x, we may assume uN(x)u\in N(x). We have

vV(G)B(u,v)\displaystyle\sum\limits_{v\in V(G)}B(u,v) =𝟏uN(x)N[y]dyvN(y)N[x]σ(u,v)lcm(dx,dy)\displaystyle=-\frac{\mathbf{1}_{u\in N(x)\cap N[y]}}{d_{y}}-\frac{\sum\limits_{v\in N(y)\setminus N[x]}\sigma(u,v)}{{\rm lcm}(d_{x},d_{y})}
=cy𝟏uN(x)N[y]lcm(dx,dy)μx(u)lcm(dx,dy)\displaystyle=-\frac{c_{y}\mathbf{1}_{u\in N(x)\cap N[y]}}{{\rm lcm}(d_{x},d_{y})}-\frac{\mu_{x}(u)}{{\rm lcm}(d_{x},d_{y})}
=cxlcm(dx,dy)\displaystyle=-\frac{c_{x}}{{\rm lcm}(d_{x},d_{y})}
=1dx\displaystyle=-\frac{1}{d_{x}}
=mx0(u).\displaystyle=-m_{x}^{0}(u).

Now we verify Condition (4). Similarly, we may assume vN(y)v\in N(y). We divide it into two cases. Case 1: vN(y)N[x]v\in N(y)\cap N[x]. We have

uV(G)B(u,v)\displaystyle\sum\limits_{u\in V(G)}B(u,v) =1dy\displaystyle=-\frac{1}{d_{y}}
=my0(v).\displaystyle=-m^{0}_{y}(v).

Case 2: vN(y)N[x]v\in N(y)\setminus N[x]. We have

uV(G)B(u,v)\displaystyle\sum\limits_{u\in V(G)}B(u,v) =uN[x]σ(u,v)lcm(dx,dy)\displaystyle=-\frac{\sum\limits_{u\in N[x]}\sigma(u,v)}{{\rm lcm}(d_{x},d_{y})}
=μy(v)lcm(dx,dy)\displaystyle=-\frac{\mu_{y}(v)}{{\rm lcm}(d_{x},d_{y})}
=cylcm(dx,dy)\displaystyle=-\frac{c_{y}}{{\rm lcm}(d_{x},d_{y})}
=1dy\displaystyle=-\frac{1}{d_{y}}
=my0(u).\displaystyle=-m_{y}^{0}(u).

All four conditions in the definition of \ast-coupling are verified.

From our construction, we observe that for uvu\not=v and (u,v)(x,y)(u,v)\not=(x,y) we have σ(u,v)>0\sigma(u,v)>0 if and only if Bσ(u,v)0B_{\sigma}(u,v)\not=0. The conclusion follows from Lemma 4. ∎

3 Proof of Theorem 4

Before we prove Theorem 4, we would like to give an structural interpretation.

Assume dxdyd_{x}\leq d_{y}. For any uN(x)N(y)u\in N(x)\cup N(y), we define a blowup set SuS_{u} (depending on the vertex uu) as follows:

Su={ if u=x;cxcy copies of u if uN(x)N[y]cx copies of u if uN(x)N[y];cy copies of u if uN(y)N[x].S_{u}=\begin{cases}\emptyset&\mbox{ if }u=x;\\ c_{x}-c_{y}\mbox{ copies of }u&\mbox{ if }u\in N(x)\cap N[y]\\ c_{x}\mbox{ copies of }u&\mbox{ if }u\in N(x)\setminus N[y];\\ c_{y}\mbox{ copies of }u&\mbox{ if }u\in N(y)\setminus N[x].\end{cases}

Define two multi-sets X:=X(x,y)=uN(x)SuX:=X(x,y)=\cup_{u\in N(x)}S_{u} and Y:=Y(x,y)=uN(y)N[x]SuY:=Y(x,y)=\cup_{u\in N(y)\setminus N[x]}S_{u}. Then we construct an auxiliary complete bipartite edge-weighed graph H:=H(x,y)H:=H(x,y) with vertex-partition XYX\cup Y. The weight of the edge (ui,vj)X×Y(u_{i},v_{j})\in X\times Y is set to d(ui,vj)=d(u,v)d(u_{i},v_{j})=d(u,v) if uiSuu_{i}\in S_{u} and vjSvv_{j}\in S_{v}.

We have

|X|\displaystyle|X| =(cxcy)(|N(x)N(y)|+1)+cx|N(x)N[y]|\displaystyle=(c_{x}-c_{y})(|N(x)\cap N(y)|+1)+c_{x}|N(x)\setminus N[y]|
=cx(|N(x)N(y)|+1+|N(x)N[y]|)cy(|N(x)N(y)|+1)\displaystyle=c_{x}(|N(x)\cap N(y)|+1+|N(x)\setminus N[y]|)-c_{y}(|N(x)\cap N(y)|+1)
=cxdxcy(|N(x)N(y)|+1)\displaystyle=c_{x}d_{x}-c_{y}(|N(x)\cap N(y)|+1)
=cydycy(|N(x)N(y)|+1)\displaystyle=c_{y}d_{y}-c_{y}(|N(x)\cap N(y)|+1)
=cy(dy|N(x)N(y)|1)\displaystyle=c_{y}(d_{y}-|N(x)\cap N(y)|-1)
=cy|N(y)N[x]|\displaystyle=c_{y}|N(y)\setminus N[x]|
=|Y|.\displaystyle=|Y|.

Any integer-valued coupling σ\sigma between μx\mu_{x} and μy\mu_{y} can be viewed as a perfect matching in HH. The cost C(σ)C(\sigma) is the sum of the edge-weights of the perfect matching.

Let H1:=H1(x,y)H_{1}:=H_{1}(x,y) be a bipartite graph on (X,Y)(X,Y) whose edge set consisting of all edges with weight 11 in HH. The graph H1H_{1} can be viewed as a blow-up graph of the bipartite induced graph G[N(x),N(y)N[x]]G[N(x),N(y)\setminus N[x]]. We have the following corollary.

Corollary 2.

For any edge xyE(G)xy\in E(G), assume dxdyd_{x}\leq d_{y}. We have

κ(x,y)|N(x)N(y)|+2dy.\kappa(x,y)\leq\frac{|N(x)\cap N(y)|+2}{d_{y}}.

The equality holds if and only if there is a perfect matching in H1(x,y)H_{1}(x,y).

Proof.
κ(x,y)\displaystyle\kappa(x,y) =1+1dyminσC(σ)lcm(dx,dy)\displaystyle=1+\frac{1}{d_{y}}-\frac{\min_{\sigma}C(\sigma)}{{\rm lcm}(d_{x},d_{y})}
1+1dyμy1lcm(dx,dy)\displaystyle\leq 1+\frac{1}{d_{y}}-\frac{\|\mu_{y}\|_{1}}{{\rm lcm}(d_{x},d_{y})}
=1+1dycy(dy1|N(x)N(y)|)lcm(dx,dy)\displaystyle=1+\frac{1}{d_{y}}-\frac{c_{y}(d_{y}-1-|N(x)\cap N(y)|)}{{\rm lcm}(d_{x},d_{y})}
=1+1dydy1|N(x)N(y)|dy\displaystyle=1+\frac{1}{d_{y}}-\frac{d_{y}-1-|N(x)\cap N(y)|}{d_{y}}
=2+|N(x)N(y)|dy.\displaystyle=\frac{2+|N(x)\cap N(y)|}{d_{y}}.

If the equality holds, then minσC(σ)=μy1\min_{\sigma}C(\sigma)=\|\mu_{y}\|_{1}. That is, d(u,v)=1d(u,v)=1 whenever σ(u,v)>0\sigma(u,v)>0. Equivalently, there is a perfect matching in H1(x,y)H_{1}(x,y). ∎

We have the following corollary of the Hall theorem.

Corollary 3.

The graph H1(x,y)H_{1}(x,y) has a perfect matching if and only if

cy|N(T1T2)(N(y)N[x])|cx|T1|+(cxcy)|T2|,c_{y}|N(T_{1}\cup T_{2})\cap(N(y)\setminus N[x])|\geq c_{x}|T_{1}|+(c_{x}-c_{y})|T_{2}|, (15)

holds (in GG) for any T1N(x)N[y]T_{1}\subseteq N(x)\setminus N[y] and T2N(x)N[y]T_{2}\subseteq N(x)\cap N[y].

Proof.

On one direction, if H1H_{1} has a perfect matching, then |NH1(K)||K||N_{H_{1}}(K)|\geq|K|, for any KXK\subseteq X. In particular, it holds for K=uT1T2SuK=\cup_{u\in T_{1}\cup T_{2}}S_{u}. We have

cx|T1|+(cxcy)|T2|=|K||NH1(K)|=cy|N(T1T2)(N(y)N[x])|.c_{x}|T_{1}|+(c_{x}-c_{y})|T_{2}|=|K|\leq|N_{H_{1}}(K)|=c_{y}|N(T_{1}\cup T_{2})\cap(N(y)\setminus N[x])|.

On the other direction, suppose Inequality (15) holds for any T1T_{1} and T2T_{2}. We will verify the Hall’s condition for H1H_{1}. For any KXK\subseteq X, define T={uN(x):KSu}T=\{u\in N(x)\colon K\cap S_{u}\not=\emptyset\}. Let T1=T(N(x)N[y])T_{1}=T\cap(N(x)\setminus N[y]) and T2=T(N(x)N[y])T_{2}=T\cap(N(x)\cap N[y]). Let T~=uTSu\tilde{T}=\cup_{u\in T}S_{u} and N(T)~=uN(T)(N(y)N[x])Su\widetilde{N(T)}=\cup_{u\in N(T)\cap(N(y)\setminus N[x])}S_{u}.

Observe NH1(T~)=NH1(K)=N(T)~N_{H_{1}}(\tilde{T})=N_{H_{1}}(K)=\widetilde{N(T)} and KT~K\subseteq\tilde{T}. We have

|NH1(K)|\displaystyle|N_{H_{1}}(K)| =|NH1(T~)|\displaystyle=|N_{H_{1}}(\tilde{T})|
=|N(T)~|\displaystyle=|\widetilde{N(T)}|
=cy|N(T1T2)(N(y)N[x])|\displaystyle=c_{y}|N(T_{1}\cup T_{2})\cap(N(y)\setminus N[x])|
cx|T1|+(cxcy)|T2|\displaystyle\geq c_{x}|T_{1}|+(c_{x}-c_{y})|T_{2}|
=|T~|\displaystyle=|\tilde{T}|
|K|.\displaystyle\geq|K|.

By the Hall theorem, H1H_{1} has a perfect matching. ∎

Now we are ready to prove Theorem 4.

Proof of Theorem 4.

Applying Equation (6) with α=1dy+1\alpha=\frac{1}{d_{y}+1}, we have

κ(x,y)\displaystyle\kappa(x,y) =κα1α\displaystyle=\frac{\kappa_{\alpha}}{1-\alpha}
=11α11αW(mxα,myα)\displaystyle=\frac{1}{1-\alpha}-\frac{1}{1-\alpha}W(m_{x}^{\alpha},m_{y}^{\alpha})
=1+1dy1lcm(dx,dy)W(lcm(dx,dy)(1α)mxα,lcm(dx,dy)(1α)myα)\displaystyle=1+\frac{1}{d_{y}}-\frac{1}{{\rm lcm}(d_{x},d_{y})}W\left(\frac{{\rm lcm}(d_{x},d_{y})}{(1-\alpha)}m_{x}^{\alpha},\frac{{\rm lcm}(d_{x},d_{y})}{(1-\alpha)}m_{y}^{\alpha}\right)
=1+1dy1lcm(dx,dy)W(μ~x,μ~y).\displaystyle=1+\frac{1}{d_{y}}-\frac{1}{{\rm lcm}(d_{x},d_{y})}W(\tilde{\mu}_{x},\tilde{\mu}_{y}). (16)

Here μ~x\tilde{\mu}_{x} and μ~y\tilde{\mu}_{y} are short notations for lcm(dx,dy)(1α)mxα\frac{{\rm lcm}(d_{x},d_{y})}{(1-\alpha)}m_{x}^{\alpha} and lcm(dx,dy)(1α)myα\frac{{\rm lcm}(d_{x},d_{y})}{(1-\alpha)}m_{y}^{\alpha}, respectively. It is easy to calculate

μ~x(u)={cy if u=x,cx if uN(x),0 otherwise;\displaystyle\tilde{\mu}_{x}(u)=\begin{cases}c_{y}&\mbox{ if }u=x,\\ c_{x}&\mbox{ if }u\in N(x),\\ 0&\mbox{ otherwise;}\end{cases} (17)
μ~y(u)={cy if u=y,cy if uN(y),0 otherwise.\displaystyle\tilde{\mu}_{y}(u)=\begin{cases}c_{y}&\mbox{ if }u=y,\\ c_{y}&\mbox{ if }u\in N(y),\\ 0&\mbox{ otherwise.}\end{cases} (18)

Now we prove W(μ~x,μ~y)=W(μx,μy)W(\tilde{\mu}_{x},\tilde{\mu}_{y})=W(\mu_{x},\mu_{y}).

First we show W(μ~x,μ~y)W(μx,μy)W(\tilde{\mu}_{x},\tilde{\mu}_{y})\leq W(\mu_{x},\mu_{y}). Suppose that σ\sigma is an optimal coupling between μx\mu_{x} and μy\mu_{y}, i.e., W(μx,μy)=C(σ)W(\mu_{x},\mu_{y})=C(\sigma).

Now define a map σ~:V×V[0,)\tilde{\sigma}\colon V\times V\to[0,\infty) as follows.

σ~(u,v)={σ(u,v)+cy if u=vN[x]N[y];σ(u,v) otherwise.\tilde{\sigma}(u,v)=\begin{cases}\sigma(u,v)+c_{y}&\mbox{ if }u=v\in N[x]\cap N[y];\\ \sigma(u,v)&\mbox{ otherwise}.\end{cases}

It is straightforward to verify σ~\tilde{\sigma} is indeed a coupling between μ~x\tilde{\mu}_{x} and μ~y\tilde{\mu}_{y}. We have

C(σ~)=C(σ)+uN[x]N[y]cyd(u,u)=C(σ).C(\tilde{\sigma})=C(\sigma)+\sum_{u\in N[x]\cap N[y]}c_{y}d(u,u)=C(\sigma).

Thus,

W(μ~x,μ~y)C(σ~)=C(σ)=W(μx,μy).W(\tilde{\mu}_{x},\tilde{\mu}_{y})\leq C(\tilde{\sigma})=C(\sigma)=W(\mu_{x},\mu_{y}).

Next, we will show W(μ~x,μ~y)W(μx,μy)W(\tilde{\mu}_{x},\tilde{\mu}_{y})\geq W(\mu_{x},\mu_{y}). Since both μ~x\tilde{\mu}_{x} and μ~y\tilde{\mu}_{y} are integer-valued, by Lemma 3, there exists an integer-valued optimal coupling σ~0\tilde{\sigma}_{0} between μ~x\tilde{\mu}_{x} and μ~y\tilde{\mu}_{y}. We would like to construct a new integer-valued optimal coupling σ~\tilde{\sigma} satisfying the following property.

Property A: For any zN[x]N[y]z\in N[x]\cap N[y], σ~(z,z)=cy\tilde{\sigma}(z,z)=c_{y}.

Suppose σ~0\tilde{\sigma}_{0} does not have Property A. There is a z0N[x]N[y]z_{0}\in N[x]\cap N[y] such that σ~0(z0,z0)cy\tilde{\sigma}_{0}(z_{0},z_{0})\not=c_{y}. Since σ~0(z,z)μ~y(z)=cy\tilde{\sigma}_{0}(z,z)\leq\tilde{\mu}_{y}(z)=c_{y}, we must have

σ~0(z0,z0)<cy.\tilde{\sigma}_{0}(z_{0},z_{0})<c_{y}.

There exists a pair of vertices u0u_{0} and v0v_{0} with σ~0(u0,z0)>0\tilde{\sigma}_{0}(u_{0},z_{0})>0 and σ~0(z0,v0)>0\tilde{\sigma}_{0}(z_{0},v_{0})>0.

Let c=min{σ~0(u0,z0),σ~0(z0,v0)}c=\min\{\tilde{\sigma}_{0}(u_{0},z_{0}),\tilde{\sigma}_{0}(z_{0},v_{0})\}. Now define a mapping σ~1:V×V[0,)\tilde{\sigma}_{1}:V\times V\to[0,\infty) as follows.

σ~1(u,v)={σ~0(u0,z0)c if (u,v)=(u0,z0);σ~0(z0,v0)c if (u,v)=(z0,v0);σ~0(z0,z0)+c if u=v=z0;σ~0(u0,v0)+c if (u,v)=(u0,v0);σ~0(u,v) otherwise.\tilde{\sigma}_{1}(u,v)=\begin{cases}\tilde{\sigma}_{0}(u_{0},z_{0})-c&\mbox{ if }(u,v)=(u_{0},z_{0});\\ \tilde{\sigma}_{0}(z_{0},v_{0})-c&\mbox{ if }(u,v)=(z_{0},v_{0});\\ \tilde{\sigma}_{0}(z_{0},z_{0})+c&\mbox{ if }u=v=z_{0};\\ \tilde{\sigma}_{0}(u_{0},v_{0})+c&\mbox{ if }(u,v)=(u_{0},v_{0});\\ \tilde{\sigma}_{0}(u,v)&\mbox{ otherwise}.\end{cases}

It is straightforward to verify σ~1\tilde{\sigma}_{1} is still a coupling between μ~x\tilde{\mu}_{x} and μ~y\tilde{\mu}_{y}. Since σ~0\tilde{\sigma}_{0} is integer-valued, σ~0(u0,z0)\tilde{\sigma}_{0}(u_{0},z_{0}), σ~0(z0,v0)\tilde{\sigma}_{0}(z_{0},v_{0}), and cc are integers. Thus, σ~1\tilde{\sigma}_{1} is integer-valued. Furthermore, we have

C(σ~1)=C(σ~0)c(d(u0,z0)+d(z0,v0)d(u0,v0))C(σ~0).C(\tilde{\sigma}_{1})=C(\tilde{\sigma}_{0})-c\left(d(u_{0},z_{0})+d(z_{0},v_{0})-d(u_{0},v_{0})\right)\leq C(\tilde{\sigma}_{0}).

Since σ~0\tilde{\sigma}_{0} is optimal, we must have C(σ~1)=C(σ~0)C(\tilde{\sigma}_{1})=C(\tilde{\sigma}_{0}). Thus, σ~1\tilde{\sigma}_{1} is also optimal.

We can iterate this process to construct a series of integer-valued optimal coupling σ~1\tilde{\sigma}_{1}, σ~2\tilde{\sigma}_{2}, \ldots, until Property A is satisfied. Let f(σ~i)=zN[x]N[y]σ~i(z,z)f(\tilde{\sigma}_{i})=\sum_{z\in N[x]\cap N[y]}\tilde{\sigma}_{i}(z,z). Observe that f(σ~i)>f(σ~i1)f(\tilde{\sigma}_{i})>f(\tilde{\sigma}_{i-1}) during the process. The process must terminate in a finite number of steps because ff is an integer-valued increasing function and is bounded by cy|N[x]N[y]|c_{y}|N[x]\cap N[y]|.

Upon termination of the process, we obtain an integer-valued optimal coupling σ~\tilde{\sigma} satisfying Property A. Now we define a map σ:V×V[0,)\sigma\colon V\times V\to[0,\infty) as follows.

σ(u,v)={0 if u=vN[x]N[y];σ~(u,v) otherwise.\sigma(u,v)=\begin{cases}0&\mbox{ if }u=v\in N[x]\cap N[y];\\ \tilde{\sigma}(u,v)&\mbox{ otherwise.}\end{cases}

It is easy to confirm that σ(u,v)\sigma(u,v) is an integer-valued coupling between μx\mu_{x} and μy\mu_{y}. Note that altering the values on the diagonal does not affect the cost. We have

C(σ~0)==C(σ~)=C(σ).C(\tilde{\sigma}_{0})=\cdots=C(\tilde{\sigma})=C(\sigma).

Therefore,

W(μx,μy)C(σ)=C(σ~0)=W(μ~x,μ~y).W(\mu_{x},\mu_{y})\leq C(\sigma)=C(\tilde{\sigma}_{0})=W(\tilde{\mu}_{x},\tilde{\mu}_{y}).

Thus, we proved

W(μ~x,μ~y)=W(μx,μy).W(\tilde{\mu}_{x},\tilde{\mu}_{y})=W(\mu_{x},\mu_{y}).

Plugging to Equation (16), we get

κ(x,y)=1+1dyW(μx,uy)lcm(dx,dy)=1+1dyminσC(σ)lcm(dx,dy).\kappa(x,y)=1+\frac{1}{d_{y}}-\frac{W(\mu_{x},\,u_{y})}{{{\rm lcm}(d_{x},d_{y})}}=1+\frac{1}{d_{y}}-\frac{\min_{\sigma}C(\sigma)}{{\rm lcm}(d_{x},d_{y})}.

Here the minimum is achieved by some integer-valued coupling σ\sigma between μx\mu_{x} and μy\mu_{y}. ∎

4 Notation and Lemmas on Bonnet-Myers sharp graphs

From now on, we assume G=(V,E)G=(V,E) is a simple connected finite graph so it has a finite diameter. We always assume diam(G)=Ldiam(G)=L unless LL being specified. For any two vertices u,vV(G)u,v\in V(G), a geodesic from uu to vv is a shortest path from uu to vv. We use an interval [u,v][u,v] to denote the set of all vertices lying on geodesics from uu to vv:

[u,v]={wV(G):d(u,w)+d(w,v)=d(u,v)}.[u,v]=\{w\in V(G):d(u,w)+d(w,v)=d(u,v)\}.

Fix a pair of poles (x,y)(x,y) of GG. We use Ni(x)={uV(G):d(x,u)=i}N_{i}(x)=\{u\in V(G):d(x,u)=i\} to denote the ii-th neighbor of xx. For any 11-Lipschitz function f:Vf\colon V\to\mathbb{Z} with f(x)=0f(x)=0 and f(y)=Lf(y)=L, we define

Nf+(u)\displaystyle N_{f}^{+}(u) ={vN(u):f(v)=f(u)+1};\displaystyle=\{v\in N(u)\colon f(v)=f(u)+1\};
Nf0(u)\displaystyle N_{f}^{0}(u) ={vN(u):f(v)=f(u)};\displaystyle=\{v\in N(u)\colon f(v)=f(u)\};
Nf(u)\displaystyle N_{f}^{-}(u) ={vN(u):f(v)=f(u)1},\displaystyle=\{v\in N(u)\colon f(v)=f(u)-1\},

If ff is clear under context, we will drop the subscript ff.

For this section, we assume GG is a Bonnet-Myers sharp graph with diameter LL, and x,yx,y is a pair of poles.

Lemma 6.

Let f:Vf\colon V\to\mathbb{Z} be a 11-Lipschitz function with f(x)=0f(x)=0 and f(y)=Lf(y)=L. For any shortest xyxy-path x=x0,x1,,xL1,xL=yx=x_{0},x_{1},\ldots,x_{L-1},x_{L}=y, we have

  1. 1.

    For i=1,2,,Li=1,2,\ldots,L, we have κ(xi1,xi)=2L\kappa(x_{i-1},x_{i})=\frac{2}{L}. Moreover, ff is an optimal solution of Equation (7) for computing κ(xi1,xi).\kappa(x_{i-1},x_{i}).

  2. 2.

    We have

    |Nf+(xi)||Nf(xi)|=(12iL)dxi,|N_{f}^{+}(x_{i})|-|N^{-}_{f}(x_{i})|=(1-\frac{2i}{L})d_{x_{i}}, (19)

    for 0iL0\leq i\leq L. In particular, 2idxi2id_{x_{i}} is divisible by LL.

Proof.

We have

2=Lκmin\displaystyle 2=L\kappa_{min} i=1Lκ(xi1,xi)\displaystyle\leq\sum_{i=1}^{L}\kappa(x_{i-1},x_{i})
i=1L(Δf(xi1)Δf(xi))\displaystyle\leq\sum_{i=1}^{L}(\Delta f(x_{i-1})-\Delta f(x_{i}))
=Δf(x0)Δf(xL)\displaystyle=\Delta f(x_{0})-\Delta f(x_{L})
1(1)\displaystyle\leq 1-(-1)
=2.\displaystyle=2.

Therefore, all of the inequalities occurred above are indeed equalities. Specifically, for 1iL1\leq i\leq L, we have

2L=κmin=κ(xi1,xi)=Δf(xi1)Δf(xi).\frac{2}{L}=\kappa_{min}=\kappa(x_{i-1},x_{i})=\Delta f(x_{i-1})-\Delta f(x_{i}).

Thus, ff is an optimal solution of Equation (7) for computing κ(xi1,xi)\kappa(x_{i-1},x_{i}).

Since Δf(x0)=1\Delta f(x_{0})=1, we have

Δf(xi)=12iL, for 0iL.\Delta f(x_{i})=1-\frac{2i}{L},\mbox{ for }0\leq i\leq L.

Equation (19) follows from the fact that

Δf(xi)=|Nf+(xi)||Nf(xi)|dxi.\Delta f(x_{i})=\frac{|N_{f}^{+}(x_{i})|-|N_{f}^{-}(x_{i})|}{d_{x_{i}}}.

Lemma 6 leads directly to the next lemma, which provides a formula for the ratio of |Nf+(u)||N_{f}^{+}(u)| and |Nf(u)||N_{f}^{-}(u)| for a vertex uu when |Nf0(u)|=0|N_{f}^{0}(u)|=0. This lemma states that the ratio depends solely on the distance between uu and xx.

Lemma 7.

Let f:Vf\colon V\to\mathbb{Z} be a 11-Lipschitz function with f(x)=0f(x)=0 and f(y)=Lf(y)=L. Then for uNi(x)u\in N_{i}(x) with |Nf0(u)|=0|N_{f}^{0}(u)|=0, we have

|Nf+(u)||Nf(u)|=Li1,\frac{|N_{f}^{+}(u)|}{|N_{f}^{-}(u)|}=\frac{L}{i}-1,

for i=1,,Li=1,\ldots,L.

Proof.

By Lemma 6, we have

|Nf+(u)||Nf(u)|=(12Li)du.|N_{f}^{+}(u)|-|N_{f}^{-}(u)|=\left(1-\frac{2}{L}i\right)d_{u}.

Since du=|Nf+(u)|+|Nf(u)|+|Nf0(u)|=|Nf+(u)|+|Nf(u)|d_{u}=|N_{f}^{+}(u)|+|N_{f}^{-}(u)|+|N_{f}^{0}(u)|=|N_{f}^{+}(u)|+|N_{f}^{-}(u)|, we have

|Nf+(u)||Nf(u)|=(12Li)(|Nf+(u)|+|Nf(u)|).|N_{f}^{+}(u)|-|N_{f}^{-}(u)|=\left(1-\frac{2}{L}i\right)\left(|N_{f}^{+}(u)|+|N_{f}^{-}(u)|\right).

It implies

|Nf+(u)||Nf(u)|=Li1.\frac{|N_{f}^{+}(u)|}{|N_{f}^{-}(u)|}=\frac{L}{i}-1.

Lemma 8.

We have [x,y]=V(G)[x,y]=V(G).

Proof.

We will prove by contradiction. Suppose there are some vertex in V[x,y]V\setminus[x,y]. Let CC be a connected component of the induced subgraph of GG on V[x,y]V\setminus[x,y]. Let f(v)=d(x,v)f(v)=d(x,v) be the distance function from xx to vv. Clearly ff is a 11-Lipschitz function with f(x)=0f(x)=0 and f(y)=Lf(y)=L. Now we define a new function f~\tilde{f}:

f~(v)={f(v)1 if vV(C),f(v)otherwise.\tilde{f}(v)=\begin{cases}f(v)-1&\mbox{ if }v\in V(C),\\ f(v)&\mbox{otherwise}.\end{cases}

We claim that f~\tilde{f} is also a 11-Lipschitz function with f(x)=0f(x)=0 and f(y)=Lf(y)=L. If not, there exists a pair of vertex (u,v)(u,v) such that

|f~(u)f~(v)|>d(u,v).|\tilde{f}(u)-\tilde{f}(v)|>d(u,v).

This can only occur when one vertex is in CC and the other vertex is in [x,y][x,y], say vV(C)v\in V(C) and u[x,y]u\in[x,y]. It is also necessary to have f(u)>f(v)f(u)>f(v) and

f~(u)f~(v)d(u,v)+1.\tilde{f}(u)-\tilde{f}(v)\geq d(u,v)+1.

This implies f(u)f(v)=d(u,v)f(u)-f(v)=d(u,v). Since u[x,y]u\in[x,y], we have d(u,y)=f(y)f(u)d(u,y)=f(y)-f(u). Hence,

d(x,v)+d(v,u)+d(u,y)=f(v)f(x)+f(u)f(v)+f(y)f(u)=f(y)f(x)=d(x,y).d(x,v)+d(v,u)+d(u,y)=f(v)-f(x)+f(u)-f(v)+f(y)-f(u)=f(y)-f(x)=d(x,y).

Thus xvuyx-v-u-y is a shortest path from xx to yy. Contradiction!

Consider v0v_{0} as a vertex in CC that attains the minimum value of ff, say f(v0)=i+1f(v_{0})=i+1 for some ii. Along a shortest path from xx to v0v_{0}, there exists a vertex, denoted as xix_{i}, which is located within the interval [x,y][x,y] and satisfies the condition f(xi)=if(x_{i})=i. Let x=x0,x1,,xi,,xL=yx=x_{0},x_{1},\ldots,x_{i},\ldots,x_{L}=y be a shortest xyxy-path passing through xix_{i}. Applying Lemma 6 to xix_{i} with respect to ff, we have

|Nf+(xi)||Nf(xi)|=(12iL)dxi.|N_{f}^{+}(x_{i})|-|N^{-}_{f}(x_{i})|=(1-\frac{2i}{L})d_{x_{i}}. (20)

Now Applying Lemma 6 to xix_{i} with respect to f~\tilde{f}, we have

|Nf~+(xi)||Nf~(xi)|=(12iL)dxi.|N_{\tilde{f}}^{+}(x_{i})|-|N^{-}_{\tilde{f}}(x_{i})|=(1-\frac{2i}{L})d_{x_{i}}. (21)

By the choice of xix_{i}, we have

|Nf~(xi)|=|Nf(xi)| but |Nf~+(xi)|<|Nf+(xi)|.|N^{-}_{\tilde{f}}(x_{i})|=|N^{-}_{f}(x_{i})|\mbox{ but }|N^{+}_{\tilde{f}}(x_{i})|<|N^{+}_{f}(x_{i})|. (22)

Equations (20), (21), and (22) conflict to one another. Contradiction!

Therefore [x,y]=V(G)[x,y]=V(G). ∎

This implies that for any pole xx, there is a unique vertex yy, such that d(x,y)=Ld(x,y)=L. (Otherwise, there is another vertex, say yy^{\prime}, satisfying d(x,y)=Ld(x,y^{\prime})=L. Then yy^{\prime} is not in the interval [x,y][x,y]. Contradiction!) We call yy is the anti-pole of xx.

Corollary 4.

The distance function from xx, i.e., f(z):=d(x,z)f(z):=d(x,z) (for zV(G)z\in V(G)), is the unique 11-Lipschitz function satisfying f(x)=0f(x)=0 and f(y)=Lf(y)=L.

From now on, unless specified otherwise, ff is always referred to the distance function d(x,)d(x,\bullet).

Corollary 5.

Consider an edge uvuv with d(x,u)<d(x,v)d(x,u)<d(x,v). Let σ\sigma be any optimal coupling of Equation (10) for computing κ(u,v)\kappa(u,v). For any wX(u,v)w\in X(u,v) and zY(u,v)z\in Y(u,v) with σ(w,z)>0\sigma(w,z)>0, we have

d(x,w)<d(x,z).d(x,w)<d(x,z).
Proof.

By Lemma 6, d(x,)d(x,\bullet) is an optimal function of Equation (7). Then applying Lemma 5, we have

d(x,z)d(x,w)=d(z,w)>0.d(x,z)-d(x,w)=d(z,w)>0.

We typically draw the graph GG such that the function d(x,)d(x,\bullet) increases from left to right. For simplicity, we say that σ\sigma transfers masses from left to right.

We define du+=|Nf+(u)|d^{+}_{u}=|N_{f}^{+}(u)|, du0=|Nf0(u)|d_{u}^{0}=|N_{f}^{0}(u)|, and du=|Nf(u)|d_{u}^{-}=|N_{f}^{-}(u)|.

Lemma 9.

Let u,vu,v be two vertices on a shortest xyxy-path, i.e. d(x,y)=d(x,u)+d(u,v)+d(v,y)d(x,y)=d(x,u)+d(u,v)+d(v,y). Then, κ(u,v)=2L\kappa(u,v)=\frac{2}{L}.

Proof.

By definition, we have

κ(u,v)\displaystyle\kappa(u,v) 1d(u,v)(Δf(u)Δf(v))\displaystyle\leq\frac{1}{d(u,v)}\left(\Delta f(u)-\Delta f(v)\right)
=1d(u,v)((12Ld(x,u))(12Ld(x,v)))\displaystyle=\frac{1}{d(u,v)}\left(\left(1-\frac{2}{L}d(x,u)\right)-\left(1-\frac{2}{L}d(x,v)\right)\right)
=1d(u,v)(2L(d(x,v)d(x,u)))\displaystyle=\frac{1}{d(u,v)}\left(\frac{2}{L}\left(d(x,v)-d(x,u)\right)\right)
=1d(u,v)(2Ld(u,v))\displaystyle=\frac{1}{d(u,v)}\left(\frac{2}{L}d(u,v)\right)
=2L.\displaystyle=\frac{2}{L}.

The second line comes from Lemma 6 since any vertex in GG lies on a geodesic from xx to yy. Since κ(u,v)κmin=2L\kappa(u,v)\geq\kappa_{min}=\frac{2}{L} by Lemma 1, we have κ(u,v)=2L\kappa(u,v)=\frac{2}{L}. ∎

Remark: Cushing-Kamtue-Koolen-Liu-Münch-Peyerimhoff [CKK+20] proved similar results to our Lemma 6, 8, and 9 on D-regular Bonnet-Myers sharp graphs. See Theorem 5.4, Theorem 5.5, and Lemma 5.3 in [CKK+20] respectively.

The next lemma gives a degree upper bound for the poles in a Bonnet-Myers sharp graph.

Lemma 10.

For any vertex uN(x)u\in N(x), we have dxdud_{x}\leq d_{u}.

Proof.

Take any uN(x)u\in N(x). Now we define a new function f~\tilde{f}:

f~(z)={f(z) if z=xoru,f(z)1otherwise.\tilde{f}(z)=\begin{cases}f(z)&\mbox{ if }z=x\ \mbox{or}\ u,\\ f(z)-1&\mbox{otherwise}.\end{cases}

Clearly, f~\tilde{f} is a 11-Lipschitz function. We have

Δf~(x)Δf(x)\displaystyle\Delta\tilde{f}(x)-\Delta f(x) =dx1dx,\displaystyle=-\frac{d_{x}-1}{d_{x}},
Δf~(u)Δf(u)\displaystyle\Delta\tilde{f}(u)-\Delta f(u) =du1du.\displaystyle=-\frac{d_{u}-1}{d_{u}}.

From the proof of Lemma 6, we know that ff is an optimal solution of equation (7). Thus,

κ(x,u)\displaystyle\kappa(x,u) =xuΔf\displaystyle=\nabla_{xu}\Delta f
=Δf(x)Δf(u)\displaystyle=\Delta f(x)-\Delta f(u)
=Δf~(x)Δf~(u)+1du1dx.\displaystyle=\Delta\tilde{f}(x)-\Delta\tilde{f}(u)+\frac{1}{d_{u}}-\frac{1}{d_{x}}.

By the definition of Lin-Lu-Yau curvature, we have κ(x,u)Δf~(x)Δf~(u)\kappa(x,u)\leq\Delta\tilde{f}(x)-\Delta\tilde{f}(u). Hence,

1du1dx0,\displaystyle\frac{1}{d_{u}}-\frac{1}{d_{x}}\leq 0,
dxdu.\displaystyle d_{x}\leq d_{u}.

Lemma 11.

For any edge uvuv, we have

du(|N(u)N(v)|+22)L.d_{u}\leq\left(\frac{|N(u)\cap N(v)|+2}{2}\right)L.

Moreover, if du=(|N(u)N(v)|+22)Ld_{u}=\left(\frac{|N(u)\cap N(v)|+2}{2}\right)L for some vN(u)v\in N(u) with dvdud_{v}\leq d_{u}, then there is a perfect matching in H1(v,u)H_{1}(v,u).

Proof.

Let vN(u)v\in N(u). We can assume dvdud_{v}\leq d_{u} since otherwise we can let uN(v)u\in N(v) and use the vertex vv instead. By Corollary 2, we have

κ(v,u)|N(u)N(v)|+2du.\kappa(v,u)\leq\frac{|N(u)\cap N(v)|+2}{d_{u}}.

Since κ(v,u)κmin=2L\kappa(v,u)\geq\kappa_{min}=\frac{2}{L}, we have

du(|N(u)N(v)|+22)L.d_{u}\leq\left(\frac{|N(u)\cap N(v)|+2}{2}\right)L.

It is easy to see that when du=(|N(u)N(v)|+22)Ld_{u}=\left(\frac{|N(u)\cap N(v)|+2}{2}\right)L, the upper bound of κ(v,u)\kappa(v,u) is achieved and equals to 2L\frac{2}{L}. The conclusion follows from Corollary 2. ∎

Lemma 12.

For any uN(x)u\in N(x), there is a perfect matching in H1(x,u)H_{1}(x,u).

Proof.

We have dxdud_{x}\leq d_{u} by Lemma 10. From Lemma 6, we have du+du=(12L)dud^{+}_{u}-d^{-}_{u}=\left(1-\frac{2}{L}\right)d_{u}. Since du=1d^{-}_{u}=1, we have du+=(12L)du+1d^{+}_{u}=\left(1-\frac{2}{L}\right)d_{u}+1. Then, we have

|N(x)N(u)|=du0=dududu+=2Ldu2.|N(x)\cap N(u)|=d^{0}_{u}=d_{u}-d^{-}_{u}-d^{+}_{u}=\frac{2}{L}d_{u}-2.

By Corollary 2, we have

κ(x,u)\displaystyle\kappa(x,u) |N(x)N(u)|+2du\displaystyle\leq\frac{|N(x)\cap N(u)|+2}{d_{u}}
=2Ldu2+2du\displaystyle=\frac{\frac{2}{L}d_{u}-2+2}{d_{u}}
=2L.\displaystyle=\frac{2}{L}.

On the other hand, since uu lies on a geodesic from xx to yy, we have κ(x,u)=2L\kappa(x,u)=\frac{2}{L} by Lemma 9. Hence, the conclusion follows from Corollary 2 since the upper bound is achieved. ∎

Lemma 13.

For any vertex vN2(x)v\in N_{2}(x), we have |(N(x)N(v))N[u]|dvdu|(N(x)\cap N(v))\setminus N[u]|\leq\frac{d_{v}}{d_{u}} for any uN(x)N(v)u\in N(x)\cap N(v).

Proof.

We define a new function f~\tilde{f}:

f~(z)={f(z)+1 if z=xorz(N(x)N(v))N[u],f(z)otherwise.\tilde{f}(z)=\begin{cases}f(z)+1&\mbox{ if }z=x\ \mbox{or}\ z\in(N(x)\cap N(v))\setminus N[u],\\ f(z)&\mbox{otherwise}.\end{cases}

Clearly f~\tilde{f} is a 11-Lipschitz function. We have

Δf~(u)Δf(u)\displaystyle\Delta\tilde{f}(u)-\Delta f(u) =1du,\displaystyle=\frac{1}{d_{u}},
Δf~(v)Δf(v)\displaystyle\Delta\tilde{f}(v)-\Delta f(v) =|(N(x)N(v))N[u]|dv.\displaystyle=\frac{|(N(x)\cap N(v))\setminus N[u]|}{d_{v}}.

Taking a difference of the two equations, we have

(Δf~(u)Δf~(v))(Δf(u)Δf(v))=1du|(N(x)N(v))N[u]|dv.(\Delta\tilde{f}(u)-\Delta\tilde{f}(v))-(\Delta f(u)-\Delta f(v))=\frac{1}{d_{u}}-\frac{|(N(x)\cap N(v))\setminus N[u]|}{d_{v}}.

By definition, κ(u,v)Δf~(u)Δf~(v)\kappa(u,v)\leq\Delta\tilde{f}(u)-\Delta\tilde{f}(v). Since ff is the optimal solution of equation (7), κ(u,v)=Δf(u)Δf(v)\kappa(u,v)=\Delta f(u)-\Delta f(v). We have

1du|(N(x)N(v))N[u]|dv0.\frac{1}{d_{u}}-\frac{|(N(x)\cap N(v))\setminus N[u]|}{d_{v}}\geq 0.

It implies

|(N(x)N(v))N[u]|dvdu.|(N(x)\cap N(v))\setminus N[u]|\leq\frac{d_{v}}{d_{u}}.

4.1 Lichnerowicz sharp graphs

The Ricci curvature is known to be related to the first non-zero eigenvalue, λ1\lambda_{1}, of the normalized combinatorial Laplacian L=ID12AD12L=I-D^{-\frac{1}{2}}AD^{-\frac{1}{2}}. Here DD represents the diagonal matrix of degrees, while AA denotes the adjacency matrix. It is important to note that LL and Δ-\Delta are similar matrices, sharing the same set of eigenvalues.

Lemma 14 (Lichnerowicz type Theorem on graphs).

[LLY11, Oll09] If for every edge xyE(G)xy\in E(G), κ(x,y)κ1>0\kappa(x,y)\geq\kappa_{1}>0, then we have

λ1(G)κ1.\lambda_{1}(G)\geq\kappa_{1}.

We say a connected graph GG is Lichnerowicz sharp if λ1(G)=κmin\lambda_{1}(G)=\kappa_{min}. Cushing-Kamtue-Koolen-Liu-Münch-Peyerimhoff [CKK+20] proved the following theorem.

Theorem 8.

[CKK+20] Every (D,L)(D,L)-Bonnet-Myers sharp graph G=(V,E)G=(V,E) is Lichnerowicz sharp with Laplace eigen-function g()=d(x,)L2g(\bullet)=d(x,\bullet)-\frac{L}{2}, where xx is a pole of GG.

We extend their result to all Bonnet-Myers sharp graphs.

Theorem 9.

Every Bonnet-Myers sharp graph G=(V,E)G=(V,E) is Lichnerowicz sharp with Laplace eigen-function g()=d(x,)L2g(\bullet)=d(x,\bullet)-\frac{L}{2}, where xx is a pole of GG.

Proof.

Since λ1κmin=2L\lambda_{1}\geq\kappa_{min}=\frac{2}{L}, it suffices to show 2L\frac{2}{L} is an eigenvalue for Δ-\Delta.

Let f()=d(x,)f(\bullet)=d(x,\bullet) denote the distance function to xx. Applying Lemma 6 and Corollary 4, we have

Δg(u)\displaystyle\Delta g(u) =1duvN(u)(g(v)g(u))\displaystyle=\frac{1}{d_{u}}\sum_{v\in N(u)}(g(v)-g(u))
=1duvN(u)(f(v)f(u))\displaystyle=\frac{1}{d_{u}}\sum_{v\in N(u)}(f(v)-f(u))
=1du(du+du)\displaystyle=\frac{1}{d_{u}}(d^{+}_{u}-d^{-}_{u})
=12d(x,u)L\displaystyle=1-\frac{2d(x,u)}{L}
=12(g(u)+L2)L\displaystyle=1-\frac{2(g(u)+\frac{L}{2})}{L}
=2Lg(u).\displaystyle=-\frac{2}{L}g(u).

Therefore gg is the eigenvector of the Laplace Δ\Delta with eigenvalue κmin-\kappa_{min}.

5 Bonnet-Myers sharp graphs of diameter 3

Cushing-Kamtue-Koolen-Liu-Münch-Peyerimhoff completely classified self-centered Bonnet-Myers sharp graphs of diameter 33 in [CKK+20]. There are only four graphs: the hypercube Q3Q_{3}, the Johnson graph J(6,3)J(6,3), the demi-cubes Q(2)6Q_{(2)}^{6}, and the Gosset graph. In this section, we discard the assumptions of self-centeredness and regularity, and outline the necessary conditions for a general diameter 33 graph to be Bonnet-Myers sharp. We will show that an irregular Bonnet-Myers sharp graph of diameter 3 has a single unique pair of poles; otherwise, it is regular. Kamtue demonstrated in [Kam20] that every regular Bonnet-Myers sharp graph of diameter 33 is self-centered. Thus, GG must be one of the four graphs if it is regular. We will conclude this section by presenting infinitely many examples of irregular Bonnet-Myers sharp graphs of diameter 33.

In this section, we assume GG is a Bonnet-Myers sharp graph of diameter 33 and (x,y)(x,y) is a pair of poles.

Lemma 15.

For any uN(x)u\in N(x) and vN(y)v\in N(y), we have

du0\displaystyle d^{0}_{u} =2du+4,\displaystyle=2d^{+}_{u}-4, (23)
dv0\displaystyle d^{0}_{v} =2dv4.\displaystyle=2d^{-}_{v}-4. (24)

In particular, du0d^{0}_{u} and dv0d^{0}_{v} are divisible by 22.

Proof.

By Lemma 6, we have

du+du=13du.d^{+}_{u}-d^{-}_{u}=\frac{1}{3}d_{u}. (25)

Observe that

du=du++du+du0.d_{u}=d^{+}_{u}+d^{-}_{u}+d^{0}_{u}. (26)

Since du=1d^{-}_{u}=1, combining the Equations 25 and 26, we have

du0=2du+4d^{0}_{u}=2d^{+}_{u}-4

The proof for the second item follows a symmetric argument. ∎

Lemma 16.

For any uN(x)u\in N(x) and vN(y)v\in N(y) such that uvE(G)uv\in E(G), we have du=dvd_{u}=d_{v}.

Proof.

Suppose dudvd_{u}\neq d_{v}. Without loss of generality, assume du<dvd_{u}<d_{v}. Let lcm(du,dv)=cudu=cvdv{\rm lcm}(d_{u},d_{v})=c_{u}d_{u}=c_{v}d_{v}. We have cu>cvc_{u}>c_{v} since du<dvd_{u}<d_{v}. Let σ\sigma be an optimal coupling between μu\mu_{u} and μv\mu_{v}.

xxcuc_{u}uuvvyycvc_{v}cucvc_{u}-c_{v}cuc_{u}cvc_{v}cucvc_{u}-c_{v}cvc_{v}cuc_{u}𝐍(𝐱)\bf N(x)𝐍(𝐲)\bf N(y)
Figure 2: Mass distributions μu\mu_{u} (in red color) and μv\mu_{v} (in black color).

For zN+(u)z\in N^{+}(u), if σ(z,w)0\sigma(z,w)\neq 0, then by Lemma 5 we have f(w)f(z)=d(z,w)f(w)-f(z)=d(z,w). Since f(z)=2f(z)=2 and d(z,w)1d(z,w)\geq 1, we have f(w)=3f(w)=3 so we must have w=yw=y. This implies that the masses from N+(u)N^{+}(u) are all transported to yy. We have

cu|(N(u)N[v])N(y)|+(cucv)(du+|(N(u)N[v])N(y)|)cv.c_{u}|(N(u)\setminus N[v])\cap N(y)|+(c_{u}-c_{v})(d^{+}_{u}-|(N(u)\setminus N[v])\cap N(y)|)\leq c_{v}.

Since cu>cvc_{u}>c_{v}, we must have |(N(u)N[v])N(y)|=0|(N(u)\setminus N[v])\cap N(y)|=0. Thus, we have

(cucv)du+cv.(c_{u}-c_{v})d^{+}_{u}\leq c_{v}.

It implies

(dvdu1)du+1.\left(\frac{d_{v}}{d_{u}}-1\right)d^{+}_{u}\leq 1. (27)

By Lemma 6, we have

13du\displaystyle\frac{1}{3}d_{u} =du+du=du+1,\displaystyle=d^{+}_{u}-d^{-}_{u}=d^{+}_{u}-1, (28)
13dv\displaystyle\frac{1}{3}d_{v} =dvdv+=dv1.\displaystyle=d^{-}_{v}-d^{+}_{v}=d^{-}_{v}-1. (29)

It implies

dvdu=dv1du+1.\frac{d_{v}}{d_{u}}=\frac{d^{-}_{v}-1}{d^{+}_{u}-1}. (30)

Plugging Equation (30) into Equation (27), we get

(dv1du+11)du+1.\left(\frac{d^{-}_{v}-1}{d^{+}_{u}-1}-1\right)d^{+}_{u}\leq 1. (31)

Solving dvd^{-}_{v}, we get

dvdu++du+1du+<du++1.d^{-}_{v}\leq d^{+}_{u}+\frac{d^{+}_{u}-1}{d^{+}_{u}}<d^{+}_{u}+1.

On the other hand, since dv>dud_{v}>d_{u}, we have

dv1du+1>1.\frac{d^{-}_{v}-1}{d^{+}_{u}-1}>1.

This gives

dv>du+.d^{-}_{v}>d^{+}_{u}.

Since both du+d^{+}_{u} and dvd^{-}_{v} are integers, it is impossible to have du+<dv<du++1d^{+}_{u}<d^{-}_{v}<d^{+}_{u}+1. Contradiction! Hence, we have du=dvd_{u}=d_{v}.

Lemma 17.

All vertices in V(G){x,y}V(G)\setminus\{x,y\} have the same degree.

Proof.

By Lemma 16, the degree of vertices are the same for any connected component of G{x,y}G\setminus\{x,y\}. It suffices to show G{x,y}G\setminus\{x,y\} is connected. For any vertex u1N(x)u_{1}\in N(x), by Lemma 12, there is a perfect matching in H1(x,u1)H_{1}(x,u_{1}). This implies that for any u2N(x)u_{2}\in N(x) there exists a vertex vN(u1)N(y)v\in N(u_{1})\cap N(y) such that u2vu_{2}v is an edge. Hence, G{x,y}G\setminus\{x,y\} is connected. ∎

Proof of theorem 6.

Proof of Item 1: For uN(x)u\in N(x), we have du+du=13dud^{+}_{u}-d^{-}_{u}=\frac{1}{3}d_{u} by Lemma 6. We observe that du=1d^{-}_{u}=1 so du=3du+3d_{u}=3d^{+}_{u}-3. Similarly, we have dv=3dv3d_{v}=3d^{-}_{v}-3 for vN(y)v\in N(y). By Lemma 17, we have du=dvd_{u}=d_{v} so du+=dvd^{+}_{u}=d^{-}_{v}. Consider the edges between N(x)N(x) and N(y)N(y). On one hand, each vertex in N(x)N(x) has du+d^{+}_{u} neighbors in N(y)N(y). On the other hand, each vertex in N(y)N(y) has dvd^{-}_{v} neighbors in N(x)N(x). Thus, we have

dxdu+=dydv.d_{x}d^{+}_{u}=d_{y}d^{-}_{v}.

Hence, we must have dx=dy=n22d_{x}=d_{y}=\frac{n-2}{2}.

Now we will show that GG only has a unique pair of poles x,yx,y. If GG has another pair of poles, say x,yV(G){x,y}x^{\prime},y^{\prime}\in V(G)\setminus\{x,y\}, then we have that dx=dy=n22d_{x^{\prime}}=d_{y^{\prime}}=\frac{n-2}{2} using the same argument above. Since all vertices in V(G){x,y}V(G)\setminus\{x,y\} have the same degree, we have GG is n22\frac{n-2}{2}-regular. Contradiction to GG being irregular.

Proof of Item 2: Let vN(y)v\in N(y). We observe that it is sufficient to show that |(N(x)N(v))N[u]|1|(N(x)\cap N(v))\setminus N[u]|\leq 1 for any uN(x)N(v)u\in N(x)\cap N(v). Let uN(x)N(v)u\in N(x)\cap N(v). Then from Lemma 17, we have du=dvd_{u}=d_{v}. By Lemma 13, we have

|(N(x)N(v))N[u]|dvdu=1.|(N(x)\cap N(v))\setminus N[u]|\leq\frac{d_{v}}{d_{u}}=1.

Proof of Item 3: We can only consider a vertex uN(x)u\in N(x) because all vertices in V(G){x,y}V(G)\setminus\{x,y\} have the same degree. From Lemma 15, we have du0=2du+4d^{0}_{u}=2d^{+}_{u}-4 and du0d^{0}_{u} is divisible by 22. Thus, we have du0=2rd^{0}_{u}=2r and du+=r+2d^{+}_{u}=r+2 for some rr\in\mathbb{N}. Then we have du=du++du0+du=r+2+2r+1=3(r+1)d_{u}=d^{+}_{u}+d^{0}_{u}+d^{-}_{u}=r+2+2r+1=3(r+1). Let t=|N(x)N(u)|t=|N(x)\setminus N(u)|. Then dx=2r+td_{x}=2r+t.

Now we will show that rr and tt must satisfy 1tr2+21\leq t\leq\frac{r}{2}+2. Let uN(x)u\in N(x). Since GG is irregular, by Lemma 10 we must have dx<dud_{x}<d_{u}. Let lcm(dx,du)=cxdx=cudu{\rm lcm}(d_{x},d_{u})=c_{x}d_{x}=c_{u}d_{u}. We have cx>cuc_{x}>c_{u}. We consider the curvature of the edge xuxu. By Lemma 12, there is a perfect matching in H1(x,u)H_{1}(x,u). Suppose N(x)N[u]N(x)\setminus N[u]\neq\emptyset. Each vertex in N(x)N[u]N(x)\setminus N[u] transfers its masses to at least two vertices in N(u)N[x]=N+(u)N(u)\setminus N[x]=N^{+}(u) because of cx>cuc_{x}>c_{u}. This implies each vertex in N(x)N[u]N(x)\setminus N[u] is adjacent to at least two vertices in N+(u)N^{+}(u). We observe that for any two different vertices u1,u2N(x)N[u]u^{\prime}_{1},u^{\prime}_{2}\in N(x)\setminus N[u], the two sets N(u1)N+(u)N(u^{\prime}_{1})\cap N^{+}(u) and N(u2)N+(u)N(u^{\prime}_{2})\cap N^{+}(u) must be disjoint. If not, then there exists a vN+(u)v\in N^{+}(u) such that |(N(x)N(v))N[u]|2|(N(x)\cap N(v))\setminus N[u]|\geq 2. This is impossible by item 2. Therefore, we have

2(t1)=2|N(x)N[u]|uN(x)N[u]|N(u)N+(u)||N+(u)|=r+2.2(t-1)=2|N(x)\setminus N[u]|\leq\sum_{u^{\prime}\in N(x)\setminus N[u]}|N(u^{\prime})\cap N^{+}(u)|\leq|N^{+}(u)|=r+2.

This gives us the upper bound of tt. The case of N(x)N[u]=N(x)\setminus N[u]=\emptyset gives us the lower bound. ∎

5.1 Construction of irregular diameter 3 Bonnet-Myers sharp graphs

Theorem 6 states that all irregular diameter 3 Bonnet-Myers sharp graphs have certain structures. In this section, we will show such graphs GG exist for t=1,2t=1,2 and any r1r\geq 1. Let xx and yy be a pair of poles of GG. We label the neighbors of xx by {u0,u1,..,u2r+t1}\{u_{0},u_{1},.....,u_{2r+t-1}\}, and the neighbors of yy by {v0,v1,..,v2r+t1}\{v_{0},v_{1},.....,v_{2r+t-1}\}. Here we assume that indexes are in 2r+t\mathbb{Z}_{2r+t}, i.e., u2r+t+i=uiu_{2r+t+i}=u_{i} and v2r+t+j=vjv_{2r+t+j}=v_{j}. We further assume uivjE(G)u_{i}v_{j}\in{\rm E}(G) if iji+r+1i\leq j\leq i+r+1. Let G[N(x)]G[N(x)] (and G[N(y)]G[N(y)]) be a 2r2r-regular graph as follows:

t=1t=1

: We set G[N(x)]G[N(x)] (and G[N(y)]G[N(y)]) being the complete graph K2r+1K_{2r+1}.

t=2t=2

: We set G[N(x)]G[N(x)] (and G[N(y)]G[N(y)]) being the complete graph K2r+2K_{2r+2} with a perfect matching M1M_{1} (and M2M_{2}) removed, respectively. Here the matching M1M_{1} in N(x)N(x) is given by uiui+r+1u_{i}u_{i+r+1} for i=0,1,2,,ri=0,1,2,\ldots,r while the matching M2M_{2} in N(y)N(y) is given by vivi+r+1v_{i}v_{i+r+1} for i=0,1,2,,ri=0,1,2,\ldots,r.

Proposition 1.

The graphs GG constructed above for t=1,2t=1,2, and any r1r\geq 1 are diameter 3 Bonnet-Myers sharp graphs.

Proof.

The graph GG clearly has diameter 33. It suffices to show κ(x,y)23\kappa(x,y)\geq\frac{2}{3} for any edge xyxy in GG. (This implies κmin23\kappa_{min}\geq\frac{2}{3} and diam(G)2κmin3diam(G)\leq\frac{2}{\kappa_{min}}\leq 3. Since diam(G)=3diam(G)=3, we must have κmin=23.\kappa_{min}=\frac{2}{3}.)

Graph GG has three type of edges.

  1. 1.

    xuixu_{i} and yvjyv_{j}, for i,j2r+ti,j\in\mathbb{Z}_{2r+t}.

  2. 2.

    uivju_{i}v_{j}, for iji+r+1i\leq j\leq i+r+1.

  3. 3.

    ui1ui2u_{i_{1}}u_{i_{2}} and vj1vj2v_{j_{1}}v_{j_{2}}.

For the first type of edges, by the symmetry of GG, it suffices to show κ(x,u0)23\kappa(x,u_{0})\geq\frac{2}{3}. We have dx=2r+td_{x}=2r+t and du0=3r+3d_{u_{0}}=3r+3. Since du0<2dxd_{u_{0}}<2d_{x}, we have cx<2cu0c_{x}<2c_{u_{0}} and

μx(u)\displaystyle\mu_{x}(u) ={cx if t=2 and u=ur+1;cxcu0uN(x) but not the first case;0 otherwise.\displaystyle=\begin{cases}c_{x}&\mbox{ if }t=2\mbox{ and }u=u_{r+1};\\ c_{x}-c_{u_{0}}&u\in N(x)\mbox{ but not the first case};\\ 0&\mbox{ otherwise.}\end{cases}
μu0(v)\displaystyle\mu_{u_{0}}(v) ={cu0 if v=vj for 0jr+1;0 otherwise.\displaystyle=\begin{cases}c_{u_{0}}&\mbox{ if $v=v_{j}$ for }0\leq j\leq r+1;\\ 0&\mbox{ otherwise.}\end{cases}

We use the following greedy algorithm to assign the coupling σ\sigma. Assume uiu_{i} has μx(ui)\mu_{x}(u_{i}) units of masses at the beginning of the algorithm while for each 0jr+10\leq j\leq r+1, vjv_{j} has an empty bin of capacity cu0c_{u_{0}}.

Algorithm:
iri\leftarrow-r
j0j\leftarrow 0
while (not complete){
Transfer as many masses from uiu_{i} to vjv_{j} as possible
if uiu_{i} is empty, ii+1i\leftarrow i+1
if vjv_{j} is full, jj+1j\leftarrow j+1
}
Table 1: A greedy algorithm to transfer masses from μx\mu_{x} to μu0\mu_{u_{0}}.

We claim that during the process, we always have 0jir+10\leq j-i\leq r+1.

The claim is true at the beginning. Since cxcu0cu0c_{x}-c_{u_{0}}\leq c_{u_{0}}, ii increases faster that jj. Thus jij-i decreases during the process. When the algorithm stops, we have i=r+t1i=r+t-1 and j=r+1j=r+1. Thus ji=2t0j-i=2-t\geq 0. The proof of the claim is finished. Therefore, uivju_{i}v_{j} is always an edge of GG.

Therefore, we show there is a coupling σ\sigma with cost (r+2)cu0(r+2)c_{u_{0}}. By Theorem 4, we have

κ(x,ui)\displaystyle\kappa(x,u_{i}) 1+1du0C(σ)lcm(dx,du0).\displaystyle\geq 1+\frac{1}{d_{u_{0}}}-\frac{C(\sigma)}{{\rm lcm}(d_{x},d_{u_{0}})}.
=1+1du0cu0(r+2)lcm(dx,du0)\displaystyle=1+\frac{1}{d_{u_{0}}}-\frac{c_{u_{0}}(r+2)}{{\rm lcm}(d_{x},d_{u_{0}})}
=1+13(r+1)r+23(r+1)\displaystyle=1+\frac{1}{3(r+1)}-\frac{r+2}{3(r+1)}
=23.\displaystyle=\frac{2}{3}.

Now consider second kind of edges. By symmetry, it suffices to show κ(u0,vj)23\kappa(u_{0},v_{j})\geq\frac{2}{3} for 0jr+10\leq j\leq r+1. Since du0=dvj=3r+3d_{u_{0}}=d_{v_{j}}=3r+3, we have cu0=cvj=1c_{u_{0}}=c_{v_{j}}=1. Note that for any uN(u0)N[vj]u\in N(u_{0})\setminus N[v_{j}], have μu0(u)=1\mu_{u_{0}}(u)=1. For any vN(vj)N[u0]v\in N(v_{j})\setminus N[u_{0}], we have μvj(v)=1\mu_{v_{j}}(v)=1.

For the case of t=1t=1, we have

X(u0,vj)\displaystyle X(u_{0},v_{j}) =N(u0)N[vj]={x}{uj+1,uj+2,,uj+r1},\displaystyle=N(u_{0})\setminus N[v_{j}]=\{x\}\cup\{u_{j+1},u_{j+2},\ldots,u_{j+r-1}\},
Y(u0,vj)\displaystyle Y(u_{0},v_{j}) =N(vj)N[u0]={y}{vr+2,vr+3,,v2r}.\displaystyle=N(v_{j})\setminus N[u_{0}]=\{y\}\cup\{v_{r+2},v_{r+3},\ldots,v_{2r}\}.

Transfer 11 unit of mass from uj+ku_{j+k} to vr+1+kv_{r+1+k} for k=1,2,,r1k=1,2,\ldots,r-1. Since

0(r+1+k)(j+k)r+1,0\leq(r+1+k)-(j+k)\leq r+1,

each move has cost 11. Finally, transfer 11 unit of mass from xx to yy with cost 3.

The total cost is

C(σ)=(r1)+3=r+2.C(\sigma)=(r-1)+3=r+2.

For the case of t=2t=2, we have the following three cases:

  1. 1.

    For 1jr1\leq j\leq r, we have

    X(u0,vj)\displaystyle X(u_{0},v_{j}) =N(u0)N[vj]={x}{uj+1,uj+2,,uj+r}{ur+1},\displaystyle=N(u_{0})\setminus N[v_{j}]=\{x\}\cup\{u_{j+1},u_{j+2},\ldots,u_{j+r}\}\setminus\{u_{r+1}\},
    Y(u0,vj)\displaystyle Y(u_{0},v_{j}) =N(vj)N[u0]={y}{vr+2,vr+3,,v2r+1}{vj+r+1}.\displaystyle=N(v_{j})\setminus N[u_{0}]=\{y\}\cup\{v_{r+2},v_{r+3},\ldots,v_{2r+1}\}\setminus\{v_{j+r+1}\}.

    Transfer 11 unit of mass from the vertex in set X(u0,vj){x}X(u_{0},v_{j})\setminus\{x\} with the lowest non-empty index to the vertex in set Y(u0,vj){y}Y(u_{0},v_{j})\setminus\{y\} with the lowest non-full index. Continue this process until all vertices in X(u0,vj){x}X(u_{0},v_{j})\setminus\{x\} are empty and all vertices in Y(u0,vj){y}Y(u_{0},v_{j})\setminus\{y\} are full. Observe that the difference of indexes during this process is r+1j+εr+1-j+\varepsilon where ε{1,0,1}\varepsilon\in\{-1,0,1\} depending on the relative positions of the exclusive vertices ur+1u_{r+1} and vj+r+1v_{j+r+1} in the sequence. Since

    0r+1j+εr+1,0\leq r+1-j+\varepsilon\leq r+1,

    each move has cost 1. By the end, a total of r1r-1 units of masses will have been moved from X(u0,vj){x}X(u_{0},v_{j})\setminus\{x\} to Y(u0,vj){y}Y(u_{0},v_{j})\setminus\{y\}. Finally, transfer 11 unit of mass from xx to yy with cost 33. The total cost is

    C(σ)=r1+3=r+2.C(\sigma)=r-1+3=r+2.
  2. 2.

    For j=0j=0, we have

    X(u0,v0)\displaystyle X(u_{0},v_{0}) =N(u0)N[v0]={x}{u1,u2,,ur}{vr+1},\displaystyle=N(u_{0})\setminus N[v_{0}]=\{x\}\cup\{u_{1},u_{2},\ldots,u_{r}\}\cup\{v_{r+1}\},
    Y(u0,v0)\displaystyle Y(u_{0},v_{0}) =N(v0)N[u0]={y}{vr+2,vr+3,,v2r+1}{ur+1}.\displaystyle=N(v_{0})\setminus N[u_{0}]=\{y\}\cup\{v_{r+2},v_{r+3},\ldots,v_{2r+1}\}\cup\{u_{r+1}\}.

    Transfer 11 unit of mass from uku_{k} to vr+1+kv_{r+1+k} for k=1,2,,rk=1,2,\ldots,r. Transfer 11 unit of mass from xx to ur+1u_{r+1}. Transfer 11 unit of mass from vr+1v_{r+1} to yy. The total cost is

    C(σ)=r+1+1=r+2.C(\sigma)=r+1+1=r+2.
  3. 3.

    For j=r+1j=r+1, we have

    X(u0,vr+1)\displaystyle X(u_{0},v_{r+1}) =N(u0)N[vr+1]={x}{ur+2,ur+3,,u2r+1}{v0},\displaystyle=N(u_{0})\setminus N[v_{r+1}]=\{x\}\cup\{u_{r+2},u_{r+3},\ldots,u_{2r+1}\}\cup\{v_{0}\},
    Y(u0,vr+1)\displaystyle Y(u_{0},v_{r+1}) =N(vr+1)N[u0]={y}{vr+2,vr+3,,v2r+1}{ur+1}.\displaystyle=N(v_{r+1})\setminus N[u_{0}]=\{y\}\cup\{v_{r+2},v_{r+3},\ldots,v_{2r+1}\}\cup\{u_{r+1}\}.

    Transfer 11 unit of mass from ur+1+ku_{r+1+k} to vr+1+kv_{r+1+k} for k=1,2,,rk=1,2,\ldots,r. Transfer 11 unit of mass from xx to ur+1u_{r+1}. Transfer 11 unit of mass from v0v_{0} to yy. The total cost is

    C(σ)=r+1+1=r+2.C(\sigma)=r+1+1=r+2.

By Theorem 4, we have

κ(u0,vj)\displaystyle\kappa(u_{0},v_{j}) 1+1dvjC(σ)lcm(du0,dvj)\displaystyle\geq 1+\frac{1}{d_{v_{j}}}-\frac{C(\sigma)}{{\rm lcm}(d_{u_{0}},d_{v_{j}})}
=1+13(r+1)r+23(r+1)\displaystyle=1+\frac{1}{3(r+1)}-\frac{r+2}{3(r+1)}
=23.\displaystyle=\frac{2}{3}.

Now consider third kind of edges. By symmetry, it suffices to show κ(u0,ui)23\kappa(u_{0},u_{i})\geq\frac{2}{3}. We can further assume i2r+t21i\leq\frac{2r+t}{2}-1. (If i>2r+t21i>\frac{2r+t}{2}-1, we can rotate the pair of vertices by 2r+ti2r+t-i so that u0uiu_{0}u_{i} is a shorter arc on the (2r+t)(2r+t)-cycle.)

Since du0=dui=3r+3d_{u_{0}}=d_{u_{i}}=3r+3, we have cu0=cui=1c_{u_{0}}=c_{u_{i}}=1. Note that for any uN(u0)N[ui]u\in N(u_{0})\setminus N[u_{i}], have μu0(u)=1\mu_{u_{0}}(u)=1. For any vN(ui)N[u0]v\in N(u_{i})\setminus N[u_{0}], we have μui(v)=1\mu_{u_{i}}(v)=1.

For the case of t=1t=1, we have

X(u0,ui)\displaystyle X(u_{0},u_{i}) =N(u0)N[ui]={v0,v1,,vi1},\displaystyle=N(u_{0})\setminus N[u_{i}]=\{v_{0},v_{1},\ldots,v_{i-1}\},
Y(u0,ui)\displaystyle Y(u_{0},u_{i}) =N(ui)N[u0]={vr+2,vr+3,,vi+r+1}.\displaystyle=N(u_{i})\setminus N[u_{0}]=\{v_{r+2},v_{r+3},\ldots,v_{i+r+1}\}.

For j=0,1,,i1j=0,1,\ldots,i-1, we transfer 11 unit of mass for vjv_{j} to vr+2+jv_{r+2+j}. Since G[N(y)]G[N(y)] is a complete graph, each move has cost 11. Since i2r+121=r12i\leq\frac{2r+1}{2}-1=r-\frac{1}{2} and ii is an integer, we have ir1i\leq r-1. We have

C(σ)ir1<r+2.C(\sigma)\leq i\leq r-1<r+2.

For the case of t=2t=2, we have

X(u0,ui)\displaystyle X(u_{0},u_{i}) =N(u0)N[ui]={ui+r+1,v0,v1,,vi1},\displaystyle=N(u_{0})\setminus N[u_{i}]=\{u_{i+r+1},v_{0},v_{1},\ldots,v_{i-1}\},
Y(u0,ui)\displaystyle Y(u_{0},u_{i}) =N(ui)N[u0]={ur+1,vr+2,vr+3,,vi+r+1}.\displaystyle=N(u_{i})\setminus N[u_{0}]=\{u_{r+1},v_{r+2},v_{r+3},\ldots,v_{i+r+1}\}.

For j=0,1,,i1j=0,1,\ldots,i-1, we transfer 11 unit of mass from vjv_{j} to vr+2+jv_{r+2+j}. Since r+2+jj=r+2r+2+j-j=r+2, vjvr+2+jv_{j}v_{r+2+j} is an edge. Each move has cost 11.

Finally, we transfer 11 unit of mass from ui+r+1u_{i+r+1} to ur+1u_{r+1}. Since i+r+1(r+1)=ii+r+1-(r+1)=i and i2r+221=ri\leq\frac{2r+2}{2}-1=r, ui+r+1ur+1u_{i+r+1}u_{r+1} is an edge. The cost of this move is 11. We have

C(σ)i+1r+1<r+2.C(\sigma)\leq i+1\leq r+1<r+2.

In either case, we have

κ(u0,ui)\displaystyle\kappa(u_{0},u_{i}) 1+1du0C(σ)lcm(du0,dui)\displaystyle\geq 1+\frac{1}{d_{u_{0}}}-\frac{C(\sigma)}{{\rm lcm}(d_{u_{0}},d_{u_{i}})}
>1+13(r+1)r+23(r+1)\displaystyle>1+\frac{1}{3(r+1)}-\frac{r+2}{3(r+1)}
=23.\displaystyle=\frac{2}{3}.

The proof of this proposition is finished. ∎

6 C3C_{3}-free Bonnet-Myers sharp graphs

In this section, we assume GG is a C3C_{3}-free Bonnet-Myers sharp graph with diam(G)=Ldiam(G)=L, and xx is a pole of GG.

Lemma 18.

We have

  1. 1.

    For any vertex uV(G)u\in V(G), we have duLd_{u}\leq L. If du=Ld_{u}=L, then for any vN+(u)N(u)v\in N^{+}(u)\cup N^{-}(u), there is a perfect matching in H1(v,u)H_{1}(v,u).

  2. 2.

    For any uN(x)u\in N(x), du=Ld_{u}=L.

  3. 3.

    For any uNi(x)u\in N_{i}(x), we have duid^{-}_{u}\leq i. Moreover, if du=id^{-}_{u}=i, then du0=0d^{0}_{u}=0 and du+=Lid^{+}_{u}=L-i.

Proof.

We first prove item 1. Since GG is triangle-free, we have |N(u)N(v)|=0|N(u)\cap N(v)|=0 for any edge uvuv. Applying Lemma 11, we have

du(|N(u)N(v)|+22)L=L.d_{u}\leq\left(\frac{|N(u)\cap N(v)|+2}{2}\right)L=L.

The equality holds if there is a perfect matching in H1(v,u)H_{1}(v,u).

Now we prove item 2. For uN(x)u\in N(x), we have du=1d^{-}_{u}=1. Since GG is triangle-free, we must have du0=0d^{0}_{u}=0. Applying Lemma 7, we get du+=L1d^{+}_{u}=L-1. Thus du=Ld_{u}=L.

Finally, we prove item 3. By Lemma 6, we have

du+du=(12iL)du.d^{+}_{u}-d^{-}_{u}=\left(1-\frac{2i}{L}\right)d_{u}.

Since du=du++du0+dud_{u}=d^{+}_{u}+d^{0}_{u}+d^{-}_{u}, we have

2du+du0=du(du+du)=2iLdu2i.2d^{-}_{u}+d^{0}_{u}=d_{u}-(d^{+}_{u}-d^{-}_{u})=\frac{2i}{L}d_{u}\leq 2i.

Therefore, we have duid^{-}_{u}\leq i. If du=id^{-}_{u}=i, then du0=0d^{0}_{u}=0. By Lemma 7, we have du+=Lid^{+}_{u}=L-i and du=Ld_{u}=L. ∎

Lemma 19.

We have

  1. 1.

    If dx>12Ld_{x}>\frac{1}{2}L, then for any uN2(x)u\in N_{2}(x), we have du=2d^{-}_{u}=2.

  2. 2.

    If dx>23Ld_{x}>\frac{2}{3}L, then for any two vertices u1,u2N2(x)u_{1},u_{2}\in N_{2}(x), we have N(u1)N(x)N(u2)N(x)N(u_{1})\cap N(x)\not=N(u_{2})\cap N(x).

Proof.

For item 1, let uN2(x)u\in N_{2}(x), we have du2d^{-}_{u}\leq 2 from Lemma 18. It is sufficient to show du2d^{-}_{u}\geq 2. We will prove it by contradiction. Otherwise we have du=1d^{-}_{u}=1. Let vv be the only vertex in N(u)N(x)N(u)\cap N(x). Consider the curvature of the edge xvxv. Let lcm(dx,dv)=cxdx=cvdv{\rm lcm}(d_{x},d_{v})=c_{x}d_{x}=c_{v}d_{v}. Since dx(12L,L]d_{x}\in\left(\frac{1}{2}L,L\right] and dv=Ld_{v}=L, we have cvcx<2cvc_{v}\leq c_{x}<2c_{v}. By Lemma 12, there exists a perfect matching in H1(x,v)H_{1}(x,v), i.e., all non-zero entries of the optimal coupling σ\sigma are equal to 1. However, the masses from vv are not sufficient to fill up the vertex uu because of cxcv<cvc_{x}-c_{v}<c_{v} (see Figure 3 (i)). Contradiction!

xxvvcxcvc_{x}-c_{v}uucvc_{v}......𝐍(𝐱)\bf N(x)𝐍𝟐(𝐱)\bf N_{2}(x)
xxv1v_{1}cxcv1c_{x}-c_{v_{1}}v2v_{2}cxc_{x}u1u_{1}cv1c_{v_{1}}u2u_{2}cv1c_{v_{1}}......𝐍(𝐱)\bf N(x)𝐍𝟐(𝐱)\bf N_{2}(x)
Figure 3: Two impossible cases: (i) N(u)N(x)={v}N(u)\cap N(x)=\{v\}. (ii) N(u1)N(x)=N(u2)N(x)={v1,v2}N(u_{1})\cap N(x)=N(u_{2})\cap N(x)=\{v_{1},v_{2}\}.

Now we prove item 2 by contradiction. Suppose that there exist two vertices u1,u2N2(x)u_{1},u_{2}\in N_{2}(x), such that N(u1)N(x)=N(u2)N(x)={v1,v2}N(u_{1})\cap N(x)=N(u_{2})\cap N(x)=\{v_{1},v_{2}\} (see Figure 3 (ii)).

Consider the curvature of the edge xv1xv_{1}. Let lcm(dx,dv1)=cxdx=cv1dv1{\rm lcm}(d_{x},d_{v_{1}})=c_{x}d_{x}=c_{v_{1}}d_{v_{1}}. Since dx(23L,L]d_{x}\in\left(\frac{2}{3}L,L\right] and dv1=Ld_{v_{1}}=L, we have cv1cx<32cv1c_{v_{1}}\leq c_{x}<\frac{3}{2}c_{v_{1}}. By Lemma 12, there is a perfect matching in H1(x,v1)H_{1}(x,v_{1}). The matching edges are from left to right (see Corollary 5). Since the masses from v1,v2v_{1},v_{2} need to be sufficient to fill up u1,u2u_{1},u_{2}, we have

cxcv1+cx=|Sv1Sv2||Su1Su2|=2cv1.c_{x}-c_{v_{1}}+c_{x}=|S_{v_{1}}\cup S_{v_{2}}|\geq|S_{u_{1}}\cup S_{u_{2}}|=2c_{v_{1}}.

It implies cx32cv1c_{x}\geq\frac{3}{2}c_{v_{1}}. Contradiction! ∎

Proof of Theorem 7.

We first show dx=Ld_{x}=L. By Lemma 19, all vertices in N2(x)N_{2}(x) have exactly two neighbors in N(x)N(x), and these pairs of neighbors are distinct. Thus, we have

|N2(x)|(dx2).|N_{2}(x)|\leq\binom{d_{x}}{2}. (32)

Now consider the number of edges between N(x)N(x) and N2(x)N_{2}(x). On one hand, each vertex in N(x)N(x) has L1L-1 neighbors in N2(x)N_{2}(x). On the other hand, each vertex in N2(x)N_{2}(x) has 22 neighbors in N(x)N(x). We have

dx(L1)=2|N2(x)|.d_{x}(L-1)=2|N_{2}(x)|. (33)

Combining Equations (32) and (33), we get

dx(L1)2(dx2).d_{x}(L-1)\leq 2\binom{d_{x}}{2}.

It implies LdxL\leq d_{x}. By Lemma 18, dxLd_{x}\leq L. Therefore, we must have dx=Ld_{x}=L.

We will establish an isomorphism gg between GG and QLQ_{L}. Label the vertices of QLQ_{L} by the subsets of [L][L]. First, we define g(x)=g(x)=\emptyset. Denote the set of neighbors of xx by u1,u2,,uLu_{1},u_{2},\dots,u_{L} and define g(uk)={k}g(u_{k})=\{k\} for 1kL1\leq k\leq L.

By Lemma 19, every vertex in N2(x)N_{2}(x) has a distinct pair of neighbors in N(x)N(x). For vN2(x)v\in N_{2}(x), we define g(v)=g(u1)g(u2)g(v)=g(u_{1})\cup g(u_{2}) whenever N(v)N(x)={u1,u2}N(v)\cap N(x)=\{u_{1},u_{2}\}. By Equation 33, we have |N2(x)|=(L2)|N_{2}(x)|=\binom{L}{2} so every distinct pair of vertices in N(x)N(x) have a common neighbor in N2(x)N_{2}(x). Thus, gg is an isomorphism between G[k=02Nk(x)]G\left[\cup_{k=0}^{2}N_{k}(x)\right] and QL[k=02([L]k)]Q_{L}\left[\cup_{k=0}^{2}\binom{[L]}{k}\right].

Now we will continue to define the map gg from Nk(x)N_{k}(x) to ([L]k)\binom{[L]}{k} for 3kL3\leq k\leq L by induction on kk. Suppose that the isomorphism gg from G[k=0i1Nk(x)]G\left[\cup_{k=0}^{i-1}N_{k}(x)\right] to QL[k=0i1([L]k)]Q_{L}\left[\cup_{k=0}^{i-1}\binom{[L]}{k}\right] has already defined. In particular, the induced bipartite graph on (Ni2(x),Ni1(x))(N_{i-2}(x),N_{i-1}(x)) is C4C_{4}-free. (Such a C4C_{4} is called a butterfly graph. There is no butterfly graph between two consecutive levels in the hypercube.) Now we need to define the map gg from Ni(x)N_{i}(x) to ([L]i)\binom{[L]}{i}.

First, we will show that dv=id^{-}_{v}=i for any vNi(x)v\in N_{i}(x). By Lemma 18, it suffices to show dvid^{-}_{v}\geq i. Let wN(v)Ni1(x)w\in N^{-}(v)\subseteq N_{i-1}(x). Take any zN(w)Ni2(x)z\in N^{-}(w)\subseteq N_{i-2}(x). Since the isomorphism gg from G[k=0i1Nk(x)]G[\cup_{k=0}^{i-1}N_{k}(x)] to QL[k=0i1([L]k)]Q_{L}[\cup_{k=0}^{i-1}\binom{[L]}{k}] has already defined, we must have dw=i1d^{-}_{w}=i-1 and dz=i2d^{-}_{z}=i-2. Then by Lemma 18, we have dw=dz=Ld_{w}=d_{z}=L. Consider the curvature of the edge zwzw. Let lcm(dz,dw)=czdz=cwdw{\rm lcm}(d_{z},d_{w})=c_{z}d_{z}=c_{w}d_{w}. We have cz=cw=1c_{z}=c_{w}=1. Since dw=dz=Ld_{w}=d_{z}=L, we know there is a perfect matching in H1(z,w)H_{1}(z,w) by Lemma 18. The matching edges are from left to right (see Corollary 5). In particular, a vertex w:=w(z,w,v)N+(z)Ni1(x)w^{\prime}:=w^{\prime}(z,w,v)\in N^{+}(z)\subseteq N_{i-1}(x) is matched to vv where www^{\prime}\neq w, for each zN(w)z\in N^{-}(w).

Now we will show that different zz give different choices of ww^{\prime}. Suppose z1,z2z_{1},z_{2} where z1z2z_{1}\neq z_{2} give the same choice of ww^{\prime}. Then we get a butterfly graph z1,z2,w,wz_{1},z_{2},w^{\prime},w (see Figure 4 (i)). However, QLQ_{L} does not contain the butterfly subgraph. Contradiction! Since there are i1i-1 choices of zz, we get i1i-1 choices of ww^{\prime}. These i1i-1 vertices together with ww are the neighbors of vv in Ni1(x)N_{i-1}(x). Therefore, dvid^{-}_{v}\geq i. Thus, dv=id^{-}_{v}=i. This implies dv=Ld_{v}=L by Lemma 18.

z1z_{1}z2z_{2}wwww^{\prime}vv............𝐍𝐢𝟐(𝐱)\bf N_{i-2}(x)𝐍𝐢𝟏(𝐱)\bf N_{i-1}(x)𝐍𝐢(𝐱)\bf N_{i}(x)
zzzz^{\prime}w1w_{1}w2w_{2}w3w_{3}vv............𝐍𝐢𝟐(𝐱)\bf N_{i-2}(x)𝐍𝐢𝟏(𝐱)\bf N_{i-1}(x)𝐍𝐢(𝐱)\bf N_{i}(x)
Figure 4: Two impossible cases: (i) z1,z2,w,wz_{1},z_{2},w^{\prime},w forms a butterfly subgraph. (ii) z,z,w1,w3z,z^{\prime},w_{1},w_{3} forms a butterfly graph.

For vNi(x)v\in N_{i}(x), we define g(v)=k=1ig(wk)g(v)=\cup_{k=1}^{i}g(w_{k}) whenever N(v)={wk:1ki}N^{-}(v)=\{w_{k}:1\leq k\leq i\}. We claim that |g(v)|=i|g(v)|=i. We need to show that any two distinct vertices in N(v)Ni1(x)N^{-}(v)\subseteq N_{i-1}(x) have a common neighbor in Ni2(x)N_{i-2}(x) and no three distinct vertices in N(v)Ni1(x)N^{-}(v)\subseteq N_{i-1}(x) have a common neighbor in Ni2(x)N_{i-2}(x). Let w1,w2N(v)Ni1(x)w_{1},w_{2}\in N^{-}(v)\subseteq N_{i-1}(x) where w1w2w_{1}\neq w_{2}. Consider the curvature of the edge w1vw_{1}v. Since dw1=dv=Ld_{w_{1}}=d_{v}=L, there is a perfect matching in H1(w1,v)H_{1}(w_{1},v) by Lemma 18. The matching edges are from left to right (see Corollary 5). In particular, a vertex denoted by z:=z(w1,v,w2)z:=z(w_{1},v,w_{2}), in N(w1)Ni2(x)N^{-}(w_{1})\subseteq N_{i-2}(x), is matched to w2w_{2}. Thus, zz is a common neighbor of w1w_{1} and w2w_{2} in Ni2(x)N_{i-2}(x). Suppose zz is a common neighbor of w1,w2,w3N(v)Ni1(x)w_{1},w_{2},w_{3}\in N^{-}(v)\subseteq N_{i-1}(x) where w1,w2,w3w_{1},w_{2},w_{3} are distinct (see Figure 4 (ii)). We observe that zz can only be matched to one of w2w_{2} and w3w_{3} in the perfect matching in H1(w1,v)H_{1}(w_{1},v) because of cw1=cv=1c_{w_{1}}=c_{v}=1. If zz is matched to w2w_{2}, then another vertex, denoted by z=:z(w1,v,w3)z^{\prime}=:z^{\prime}(w_{1},v,w_{3}), in N(w1)Ni2(x)N^{-}(w_{1})\subseteq N_{i-2}(x), must be matched to w3w_{3}. Then we get a butterfly graph z,z,w1,w3z,z^{\prime},w_{1},w_{3} (see Figure 4 (ii)). However, QLQ_{L} does not contain the butterfly subgraph. Contradiction! Hence, any two distinct vertices in N(v)Ni1(x)N^{-}(v)\subseteq N_{i-1}(x) have a common neighbor in Ni2(x)N_{i-2}(x) and no three distinct vertices in N(v)Ni1(x)N^{-}(v)\subseteq N_{i-1}(x) have a common neighbor in Ni2(x)N_{i-2}(x). Then this property implies |g(v)|=i|g(v)|=i. Hence, gg is an isomorphism from G[k=0iNk(x)]G\left[\cup_{k=0}^{i}N_{k}(x)\right] to QL[k=0i([L]k)]Q_{L}\left[\cup_{k=0}^{i}\binom{[L]}{k}\right]. The inductive proof is finished.

Use the same technique, we have completely classified C3C_{3}-free Bonnet-Myers sharp graphs with diameter L=2,3,4,5L=2,3,4,5 (see table 6). We omit the proof here. The second graph in the row where L=4L=4 and dx=2d_{x}=2 is the first incidence featuring exactly two pairs of poles. All of these graphs are bipartite. Additionally, we note that the maximum degree between any two adjacent vertices always matches the diameter and we conjecture this pattern holds for all C3C_{3}-free Bonnet-Myers sharp graphs. Based on these observations, we propose the following conjecture to conclude this paper.

Conjecture 1.

Suppose GG is an C3C_{3}-free and Bonnet-Myers sharp graph with diameter LL. Then,

  1. 1.

    GG is a bipartite graph.

  2. 2.

    For any edge uvuv, we have max{du,dv}=L\max\{d_{u},d_{v}\}=L.

dx=1d_{x}=1 P3P_{3}
L=2L=2 dx=2d_{x}=2 Q2Q_{2}
dx=1d_{x}=1 none
L=3L=3 dx=2d_{x}=2

dx=3d_{x}=3 Q3Q_{3} dx=1d_{x}=1 L=4L=4 dx=2d_{x}=2 dx=3d_{x}=3 none dx=4d_{x}=4 Q4Q_{4} dx=1d_{x}=1 none L=5L=5 dx=2d_{x}=2 dx=3d_{x}=3 none dx=4d_{x}=4 none dx=5d_{x}=5 Q5Q_{5}

Table 2: The list of C3C_{3}-free Bonnet-Myers Sharp Graphs with L=2,3,4,5.L=2,3,4,5.

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