# Ricci Curvature on Polyhedral Surfaces via Optimal Transportation

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## Abstract

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**MSC**05C10; 68U05; 90B06

## 1. The Coarse Ricci Curvature of Ollivier

- (1)
- Let ${\mu}_{x}$ be the Dirac measure, ${\delta}_{x}$, i.e., ${\delta}_{x}(y)$ equals one when $y=x$ and zero elsewhere. Then, there is only one coupling, ξ, between ${\delta}_{x}$ and ${\delta}_{{x}^{\prime}}$, and it satisfies $\xi (y,z)=1$ if $y=x$ and $z={x}^{\prime}$ and vanishes elsewhere. Obviously, ${W}_{1}(x,{x}^{\prime})=d(x,{x}^{\prime})$.
- (2)
- Consider now ${\mu}_{x}^{1},{\mu}_{{x}^{\prime}}^{1}$, the uniform measures on unit spheres around x and ${x}^{\prime}$, respectively; then, a coupling, ξ, vanishes on $(y,{y}^{\prime})$, whenever y lies outside the sphere, ${S}_{x}$, or ${y}^{\prime}$ outside ${S}_{{x}^{\prime}}$. So the (a priori infinite) matrix, ξ, has at most ${d}_{x}{d}_{{x}^{\prime}}$ nonzero terms, and we can focus on the ${d}_{x}\times {d}_{{x}^{\prime}}$ submatrix, $\xi {(y,{y}^{\prime})}_{y\in {S}_{x},{y}^{\prime}\in {S}_{{x}^{\prime}}}$, whose lines sum to $\frac{1}{{d}_{x}}$ and columns to $\frac{1}{{d}_{{x}^{\prime}}}$. For instance, ξ could be the uniform coupling:$$\xi =\frac{1}{{d}_{x}{d}_{{x}^{\prime}}}\left(\begin{array}{cccc}1& \cdots & \cdots & 1\\ \vdots & & & \vdots \\ 1& \cdots & \cdots & 1\end{array}\right)$$
- (3)
- A variant from the above measure is the measure uniform on the ball ${B}_{x}=\left\{x\right\}\cup {S}_{x}$.

## 2. A Linear Programming Problem

**Lemma 1.**

**Proof.**

**Remark 1.**

**Proposition 2.**

**Proof.**

**Remark 2.**

**Theorem 3.**

**Proof.**

- the entries of A lie in $\{-1,0,1\}$;
- A has no more than two nonzero entries on each column;
- its rows can be partitioned into two sets ${I}_{1}=\{1,\cdots ,d+2\}$ and ${I}_{2}=\{d+3,\cdots ,d+{d}^{\prime}+2\}$, such that if a column has two entries of the same sign, their rows are in different sets.

## 3. Curvature of Discrete Surfaces

**Theorem 4**

**.**If $ric$ is bounded below on all edges by a positive constant, ρ, then S is finite, and its diameter is bounded above by $2/\rho $.

**Proof.**

**Table 1.**Ollivier–Ricci (asymptotic and discrete at time 1 for the Platonic solids, along Forman’s version of Ricci curvature (divided by three, to be comparable). Values in bold are sharp with respect to Myers’ theorem.

Tetrahedron | Cube | Octahedron | Dodecahedron | Icosahedron | |
---|---|---|---|---|---|

$ric$ | $4/3$ | $\mathbf{2}/\mathbf{3}$ | $\mathbf{1}$ | 0 | $2/5$ |

${\kappa}^{1}$ | $2/3$ | 0 | $1/2$ | $-1/3$ | $1/5$ |

$\frac{1}{3}$ Forman | $4/3$ | $\mathbf{2}/\mathbf{3}$ | $2/3$ | 0 | 0 |

**Table 2.**Asymptotic Ollivier–Ricci curvature, $ric(x,{x}^{\prime})$, according to respective degrees of x and ${x}^{\prime}$, compared to the Time 1 Ollivier–Ricci ${\kappa}^{1}$, as well as Forman’s Ricci curvature (divided by three for comparison purposes); ${\kappa}^{1}\ge $ refers to the estimates of Jost and Liu.

$\mathbf{(}\mathbf{d}\mathbf{,}{\mathbf{d}}^{\mathbf{\prime}}\mathbf{)}$ | $\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}$ | $\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{4}\mathbf{)}$ | $\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{5}\mathbf{)}$ | $\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{6}\mathbf{)}$ | $\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{)}$ | $\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{6}\mathbf{)}$ | Others |
---|---|---|---|---|---|---|---|

$ric$ | $\frac{4}{3}$ | $\frac{3}{4}$ | $\frac{11}{20}$ | $\frac{1}{3}$ | $\frac{2}{5}$ | $\frac{2}{15}$ | $\frac{4}{d}+\frac{8}{{d}^{\prime}}-2$ |

${\kappa}^{1}$ | $\frac{2}{3}$ | $\frac{1}{2}$ | $\frac{7}{20}$ | $\frac{1}{4}$ | $\frac{1}{5}$ | $\frac{1}{10}$ | $\frac{4}{d}+\frac{8}{{d}^{\prime}}-2$ |

${\kappa}^{1}\ge $ | $\frac{2}{3}$ | $\frac{1}{2}$ | $\frac{1}{5}$ | $0$ | $0$ | $-\frac{1}{5}$ | $\left\{\begin{array}{cc}{\displaystyle \frac{5}{{d}^{\prime}}-\frac{2}{3}}\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}d=3\hfill \\ {\displaystyle \frac{4}{d}+\frac{6}{{d}^{\prime}}-2}\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}d\ge 4\hfill \end{array}\right.$ |

$\frac{1}{3}\phantom{\rule{0.166667em}{0ex}}\text{Forman}$ | $\frac{4}{3}$ | $\frac{2}{3}$ | $\frac{1}{3}$ | $0$ | $0$ | $-\frac{1}{3}$ | $\frac{10-d-{d}^{\prime}}{3}$ |

`Sage`[20] is attached to the article). The next results however are of a more general nature, with variable degrees, and cannot be obtained by simple computations. We consider adjacent vertices $x,{x}^{\prime}$ on a triangulated surface with the following genericity hypotheses:

- (
**B**) - $x,{x}^{\prime}$ are not on the boundary;
- (
**G**) - for any $y\in \text{star}(x)$ and ${y}^{\prime}\in \text{star}({x}^{\prime})$, there is a geodesic of length $d(y,{y}^{\prime})$ in $\text{star}(x)\cup \text{star}({x}^{\prime})$.

**Theorem 5.**

**Proof.**

**Remark 3.**

- 2.
- Zero Ollivier–Ricci curvature is attained only with degrees $(6,6)$ (regular triangular tiling), $(4,8)$ and $(3,12)$.

## 4. Varying Edge Lengths

- (a)
- ${\Delta}_{xy}>0$ whenever $x\sim y$;
- (b)
- ${\Delta}_{xy}=0$ whenever $x\ne y$ and $x\nsim y$ (locality property);
- (c)
- ${\sum}_{y}{\Delta}_{xy}=0$, which implies that ${\Delta}_{xx}<0$ (note that the sum is finite, due to the previous assumption and the local finiteness of S).

**Proposition 6.**

**Proof.**

**Remark 4.**

**Remark 5.**

## 5. Appendix: Solutions for the Linear Programming Problem on Generic Triangulated Surfaces

`Sage`program attached.

#### 5.1. The Regular Tetrahedron

**Remark 6.**

#### 5.2. Generic Triangulated Surfaces

- ${d}^{\prime}\ge 6$ and $d\ge 5$:$$\Delta =\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 1& 1\\ 2& 0& 2& 2\\ 1& 0& 0& 2\\ 1& 0& 2& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 1\\ 1& \vdots & & \vdots & 1\\ 1& \vdots & & \vdots & 0\\ 0& 0& \cdots & 0& 1\end{array}\\ \begin{array}{cccc}0& 0& 0& 1\\ \vdots & \vdots & 1& \vdots \\ \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & 1\\ 0& 0& 1& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 0\\ \vdots & \vdots & & \vdots & 1\\ \vdots & \vdots & & \vdots & \vdots \\ 1& \vdots & & \vdots & \vdots \\ 0& 0& \cdots & 0& 1\end{array}\end{array}\right)\phantom{\rule{-0.166667em}{0ex}},{\tilde{\xi}}^{t}=\left(\begin{array}{cc}\begin{array}{cccc}\frac{t}{{d}^{\prime}}& {x}_{t}& 0& 0\\ 0& \frac{t}{{d}^{\prime}}& 0& 0\\ 0& 0& \frac{t}{{d}^{\prime}}& 0\\ 0& 0& 0& \frac{t}{{d}^{\prime}}\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\\ 0& {y}_{t}& \cdots & {y}_{t}& 0\\ 0& {y}_{t}& \cdots & {y}_{t}& 0\\ 0& {y}_{t}& \cdots & {y}_{t}& 0\end{array}\\ \begin{array}{cccc}0& \cdots & \cdots & 0\\ \vdots & & & \vdots \\ \vdots & & & \vdots \\ \vdots & & & \vdots \\ 0& \cdots & \cdots & 0\end{array}& \begin{array}{ccccc}0& {y}_{t}& \cdots & {y}_{t}& \frac{t}{{d}^{\prime}}\\ \vdots & {z}_{t}& \cdots & {z}_{t}& 0\\ \vdots & \vdots & & \vdots & \vdots \\ 0& {z}_{t}& \cdots & {z}_{t}& \vdots \\ \frac{t}{{d}^{\prime}}& {y}_{t}& \cdots & {y}_{t}& 0\end{array}\end{array}\right)$$$${x}_{t}=1-t-\frac{t}{{d}^{\prime}},\phantom{\rule{1.em}{0ex}}{y}_{t}=\frac{t}{{d}^{\prime}-5}\left(\frac{1}{d}-\frac{1}{{d}^{\prime}}\right),\phantom{\rule{1.em}{0ex}}{z}_{t}=\frac{t}{d({d}^{\prime}-5)}$$
- ${d}^{\prime}\ge 6$ and $d=4$:The distance matrix, D, has only five rows, and its submatrix is slightly different:$$\tilde{D}=\left(\begin{array}{cc}\begin{array}{cccc}0& 1& 1& 1\\ 1& 0& 1& 1\\ 1& 1& 0& 2\\ 1& 1& 2& 0\end{array}& \begin{array}{ccccc}2& \cdots & \cdots & \cdots & 2\\ 1& \cdots & \cdots & \cdots & 1\\ 2& \cdots & \cdots & 2& 1\\ 1& 2& \cdots & \cdots & 2\end{array}\\ \begin{array}{cccc}1& 2& 1& 1\end{array}& \begin{array}{ccccc}2& 3& \cdots & 3& 2\end{array}\end{array}\right),\phantom{\rule{1.em}{0ex}}\Delta =\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 1& 1\\ 2& 0& 2& 2\\ 1& 0& 0& 2\\ 1& 0& 2& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 1\\ 1& 0& \cdots & 0& 1\\ 1& 0& \cdots & 0& 0\\ 0& 0& \cdots & 0& 1\end{array}\\ \begin{array}{cccc}0& 0& 0& 0\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\end{array}\end{array}\right)$$$${\tilde{\xi}}^{t}=\left(\begin{array}{cc}\begin{array}{cccc}\frac{t}{{d}^{\prime}}& {x}_{t}& 0& 0\\ 0& \frac{t}{{d}^{\prime}}& 0& 0\\ 0& 0& \frac{2t}{3{d}^{\prime}}& 0\\ 0& 0& 0& \frac{2t}{3{d}^{\prime}}\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\\ 0& {y}_{t}& \cdots & {y}_{t}& 0\\ 0& {z}_{t}& \cdots & {z}_{t}& \frac{2t}{3{d}^{\prime}}\\ \frac{2t}{3{d}^{\prime}}& {z}_{t}& \cdots & {z}_{t}& 0\end{array}\\ \begin{array}{cccc}0& 0& \frac{t}{3{d}^{\prime}}& \frac{t}{3{d}^{\prime}}\end{array}& \begin{array}{ccccc}\frac{t}{3{d}^{\prime}}& {z}_{t}& \cdots & {z}_{t}& \frac{t}{3{d}^{\prime}}\end{array}\end{array}\right)$$$${x}_{t}=1-t-\frac{t}{{d}^{\prime}},\phantom{\rule{1.em}{0ex}}{y}_{t}=\frac{t}{{d}^{\prime}-5}\left(\frac{1}{4}-\frac{1}{{d}^{\prime}}\right),\phantom{\rule{1.em}{0ex}}{z}_{t}=\frac{t}{{d}^{\prime}-5}\left(\frac{1}{4}-\frac{4}{3{d}^{\prime}}\right),\phantom{\rule{1.em}{0ex}}t\le \frac{{d}^{\prime}}{{d}^{\prime}+1}$$
- ${d}^{\prime}\ge 6$ and $d=3$:Similarly, we have:$$\tilde{D}=\left(\begin{array}{cc}\begin{array}{cccc}0& 1& 1& 1\\ 1& 0& 1& 1\\ 1& 1& 0& 2\\ 1& 1& 2& 0\end{array}& \begin{array}{ccccc}2& \cdots & \cdots & \cdots & 2\\ 1& \cdots & \cdots & \cdots & 1\\ 2& \cdots & \cdots & 2& 1\\ 1& 2& \cdots & \cdots & 2\end{array}\end{array}\right),\phantom{\rule{1.em}{0ex}}\Delta =\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 1& 1\\ 2& 0& 2& 2\\ 1& 0& 0& 2\\ 1& 0& 2& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 1\\ 1& 0& \cdots & 0& 1\\ 1& 0& \cdots & 0& 0\\ 0& 0& \cdots & 0& 1\end{array}\end{array}\right)$$$${\tilde{\xi}}^{t}=\left(\begin{array}{cc}\begin{array}{cccc}\frac{t}{{d}^{\prime}}& {x}_{t}& 0& 0\\ 0& \frac{t}{{d}^{\prime}}& 0& 0\\ 0& 0& \frac{t}{{d}^{\prime}}& 0\\ 0& 0& 0& \frac{t}{{d}^{\prime}}\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\\ 0& {y}_{t}& \cdots & {y}_{t}& 0\\ 0& {z}_{t}& \cdots & {z}_{t}& \frac{t}{{d}^{\prime}}\\ \frac{t}{{d}^{\prime}}& {z}_{t}& \cdots & {z}_{t}& 0\end{array}\end{array}\right)$$$${x}_{t}=1-t-\frac{t}{{d}^{\prime}},\phantom{\rule{1.em}{0ex}}{y}_{t}=\frac{t}{{d}^{\prime}-5}\left(\frac{1}{3}-\frac{1}{{d}^{\prime}}\right),\phantom{\rule{1.em}{0ex}}{z}_{t}=\frac{t}{{d}^{\prime}-5}\left(\frac{1}{3}-\frac{2}{{d}^{\prime}}\right),\phantom{\rule{1.em}{0ex}}t\le \frac{{d}^{\prime}}{{d}^{\prime}+1}$$

#### 5.3. ${\kappa}^{1}$ Computation

- ${d}^{\prime}\ge 6$ and $d\ge 6$The matrix, Δ, is the same as above, but because of the additional constraint, an optimal coupling is now given by:$${\tilde{\xi}}^{1}=\frac{1}{d{d}^{\prime}}\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& d& 0\\ 0& 0& 0& d\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\\ 0& \frac{{d}^{\prime}}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}}{{d}^{\prime}-5}& 0\\ 0& \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& 0\\ 0& \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& 0\end{array}\\ \begin{array}{cccc}0& 0& \cdots & 0\\ d& \vdots & & \vdots \\ 0& \vdots & & \vdots \\ \vdots & \vdots & & \vdots \\ \vdots & \vdots & & \vdots \\ 0& 0& \cdots & 0\end{array}& \begin{array}{ccccc}0& \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& d\\ 0& \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& 0\\ \vdots & \frac{{d}^{\prime}}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}}{{d}^{\prime}-5}& 0\\ \vdots & \vdots & & \vdots & \vdots \\ 0& \frac{{d}^{\prime}}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}}{{d}^{\prime}-5}& \vdots \\ d& \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-d}{{d}^{\prime}-5}& 0\end{array}\end{array}\right)$$
- $d=5$ and ${d}^{\prime}\ge 7$$$\tilde{D}=\left(\begin{array}{cc}\begin{array}{cccc}0& 1& 1& 1\\ 1& 0& 1& 1\\ 1& 1& 0& 2\\ 1& 1& 2& 0\end{array}& \begin{array}{ccccc}2& \cdots & \cdots & \cdots & 2\\ 1& \cdots & \cdots & \cdots & 1\\ 2& \cdots & \cdots & 2& 1\\ 1& 2& \cdots & \cdots & 2\end{array}\\ \begin{array}{cccc}1& 2& 1& 2\\ 1& 2& 2& 1\end{array}& \begin{array}{ccccc}3& \cdots & \cdots & 3& 2\\ 2& 3& \cdots & \cdots & 3\end{array}\end{array}\right),\phantom{\rule{1.em}{0ex}}\Delta =\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 1& 1\\ 2& 0& 2& 2\\ 1& 0& 0& 2\\ 1& 0& 2& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 1\\ 1& \vdots & & \vdots & 1\\ 1& \vdots & & \vdots & 0\\ 0& 0& \cdots & 0& 1\end{array}\\ \begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 0& 0& \cdots & 0& 1\end{array}\end{array}\right)$$$${\tilde{\xi}}^{1}=\frac{1}{5{d}^{\prime}}\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 5\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\\ 0& \frac{{d}^{\prime}}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}}{{d}^{\prime}-5}& 0\\ 0& \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& 5\\ 1& \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& 0\end{array}\\ \begin{array}{cccc}3& 0& 3& 0\\ 2& 0& 0& 0\end{array}& \begin{array}{ccccc}0& \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& 0\\ 4& \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& 0\end{array}\end{array}\right)\phantom{\rule{1.em}{0ex}}{W}_{1}=\frac{11{d}^{\prime}-40}{5{d}^{\prime}}=3-\frac{4}{5}-\frac{8}{{d}^{\prime}}$$
- $d=4$ and ${d}^{\prime}\ge 7$$$\tilde{D}=\left(\begin{array}{cc}\begin{array}{cccc}0& 1& 1& 1\\ 1& 0& 1& 1\\ 1& 1& 0& 2\\ 1& 1& 2& 0\end{array}& \begin{array}{ccccc}2& \cdots & \cdots & \cdots & 2\\ 1& \cdots & \cdots & \cdots & 1\\ 2& \cdots & \cdots & 2& 1\\ 1& 2& \cdots & \cdots & 2\end{array}\\ \begin{array}{cccc}1& 2& 1& 1\end{array}& \begin{array}{ccccc}2& 3& \cdots & 3& 2\end{array}\end{array}\right),\phantom{\rule{1.em}{0ex}}\Delta =\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 1& 1\\ 2& 0& 2& 2\\ 1& 0& 0& 2\\ 1& 0& 2& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 1\\ 1& \vdots & & \vdots & 1\\ 1& \vdots & & \vdots & 0\\ 0& 0& \cdots & 0& 1\end{array}\\ \begin{array}{cccc}0& 0& 0& 0\end{array}& \begin{array}{ccccc}1& 0& \cdots & 0& 0\end{array}\end{array}\right)$$$${\tilde{\xi}}^{1}=\frac{1}{4{d}^{\prime}}\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 3& 0\\ 0& 0& 0& 3\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\\ 0& \frac{{d}^{\prime}}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}}{{d}^{\prime}-5}& 0\\ 0& \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& 4\\ 4& \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& 0\end{array}\\ \begin{array}{cccc}4& 0& 1& 1\end{array}& \begin{array}{ccccc}0& \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-6}{{d}^{\prime}-5}& 0\end{array}\end{array}\right)\phantom{\rule{1.em}{0ex}}{W}_{1}=\frac{1}{4{d}^{\prime}}(8{d}^{\prime}-32)=3-\frac{4}{4}-\frac{8}{{d}^{\prime}}$$
- $d=3$ and ${d}^{\prime}\ge 8$$$\tilde{D}=\left(\begin{array}{cc}\begin{array}{cccc}0& 1& 1& 1\\ 1& 0& 1& 1\\ 1& 1& 0& 2\\ 1& 1& 2& 0\end{array}& \begin{array}{ccccc}2& \cdots & \cdots & \cdots & 2\\ 1& \cdots & \cdots & \cdots & 1\\ 2& \cdots & \cdots & 2& 1\\ 1& 2& \cdots & \cdots & 2\end{array}\end{array}\right)$$$$\begin{array}{ccc}\hfill \Delta & =& \tilde{D}+2{L}_{1}+2{L}_{2}+{L}_{3}+{L}_{4}\hfill \\ & & -2{C}_{1}-2{C}_{2}-{C}_{3}-{C}_{4}-2{C}_{5}-3({C}_{6}+\cdots +{C}_{{d}^{\prime}})-2{C}_{{d}^{\prime}+1}\hfill \\ & =& \left(\begin{array}{cc}\begin{array}{cccc}0& 1& 2& 2\\ 1& 0& 2& 2\\ 0& 0& 0& 2\\ 0& 0& 2& 0\end{array}& \begin{array}{ccccc}2& 1& \cdots & 1& 2\\ 1& 0& \cdots & 0& 1\\ 1& 0& \cdots & 0& 0\\ 0& 0& \cdots & 0& 1\end{array}\end{array}\right)\hfill \end{array}$$$${\tilde{\xi}}^{1}=\frac{1}{3{d}^{\prime}}\left(\begin{array}{cc}\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 2& 0& 3& 0\\ 1& 0& 0& 3\end{array}& \begin{array}{ccccc}0& \cdots & \cdots & \cdots & 0\\ 0& \frac{{d}^{\prime}}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}}{{d}^{\prime}-5}& 0\\ 0& \frac{{d}^{\prime}-8}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-8}{{d}^{\prime}-5}& 3\\ 3& \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& \cdots & \frac{{d}^{\prime}-7}{{d}^{\prime}-5}& 0\end{array}\end{array}\right)\phantom{\rule{2.em}{0ex}}{W}_{1}=\frac{1}{3{d}^{\prime}}(5{d}^{\prime}-21)=3-\frac{4}{3}-\frac{7}{{d}^{\prime}}$$

#### 5.4. The Rectangular Parallelepiped

## Acknowledgments

## Author Contributions

## Conflicts of Interest

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Loisel, B.; Romon, P.
Ricci Curvature on Polyhedral Surfaces via Optimal Transportation. *Axioms* **2014**, *3*, 119-139.
https://doi.org/10.3390/axioms3010119

**AMA Style**

Loisel B, Romon P.
Ricci Curvature on Polyhedral Surfaces via Optimal Transportation. *Axioms*. 2014; 3(1):119-139.
https://doi.org/10.3390/axioms3010119

**Chicago/Turabian Style**

Loisel, Benoît, and Pascal Romon.
2014. "Ricci Curvature on Polyhedral Surfaces via Optimal Transportation" *Axioms* 3, no. 1: 119-139.
https://doi.org/10.3390/axioms3010119