Green--Wasserstein Inequality on Compact Surfaces

This paper resolves a question posed by Steinerberger by proving that the $\sqrt{\log n}$ factor in the two-dimensional Green–Wasserstein inequality is necessary and cannot be removed while maintaining a uniform $O(n^{-1/2})$ remainder for point sets on compact connected surfaces.

The Gist

Here is an explanation of the paper using simple language and creative analogies.

The Big Picture: Arranging a Party on a Curved Surface

Imagine you have a curved surface, like a smooth, round ball or a donut (mathematicians call this a "compact surface"). You want to scatter $n$ guests (points) evenly across this surface so that no spot is too crowded and no spot is too empty.

Mathematicians have a special tool called the Wasserstein distance to measure how "messy" or "uneven" your arrangement is. The smaller the number, the better the arrangement.

There is also a mysterious "energy score" called the Green Energy. Think of this as a social tension meter. If two guests stand too close to each other, the tension (energy) spikes. If they are far apart, the tension is low. The Green function ($G$) is the formula that calculates this tension based on distance.

The Question: Can We Simplify the Formula?

For years, mathematicians knew a rule (an inequality) that connected the "messiness" of the arrangement to the "tension" between guests.

In 3D space (like a room), the formula was clean and simple. But in 2D space (like the surface of a balloon), there was a messy extra term: $\sqrt{\log n}$.

• The Formula: Messiness $\approx$ (Tension) + $\sqrt{\log n}$.
• The Problem: As the number of guests ($n$) gets huge, that $\sqrt{\log n}$ term grows. It acts like a "tax" you have to pay just for being in 2D.

A mathematician named Steinerberger asked a bold question: "Can we get rid of this tax? Can we make the formula just 'Messiness $\approx$ (Tension)' without the extra $\sqrt{\log n}$ term?"

He hoped that if we just looked at the tension between guests, it would perfectly predict how well they are arranged, even in 2D.

The Answer: No, You Can't Remove the Tax

This paper, written by Maja Gwóźdz, says a firm "No."

The author proves that on any curved 2D surface, you cannot remove that $\sqrt{\log n}$ factor. If you try to write a rule that ignores it, the rule will eventually break when you have enough guests.

How Did They Prove It? (The Detective Story)

The author used a clever "proof by contradiction." Here is the step-by-step logic, translated into a story:

1. The Hypothetical Scenario
Imagine, for a moment, that Steinerberger was right. Imagine there is a perfect rule where Messiness is just a tiny bit of "random noise" ($1/\sqrt{n}$) plus the Tension.

2. The Random Experiment
The author decides to test this rule by throwing the guests onto the surface completely at random (like throwing darts blindfolded).

• In a random scattering, the "Tension" (Green Energy) usually averages out to zero because the highs and lows cancel each other.
• However, because of the nature of random numbers, the Tension doesn't stay perfectly at zero; it wiggles around.

3. The Two Clues
The author looked at two different mathematical clues about what happens when you throw points randomly:

• Clue A (The Author's Calculation): If the "Perfect Rule" (without the tax) were true, then the average messiness of a random party should shrink very fast as you add more guests. Specifically, it should shrink at a rate of $1/n$.
• Clue B (The Known Fact): Mathematicians already knew (from a different study by Ambrosio and Glaudo) that when you scatter points randomly on a 2D surface, the messiness actually shrinks much slower. It shrinks at a rate of $(\log n) / n$.

4. The Showdown
The author compared the two clues:

• The "Perfect Rule" predicts: Messiness $\approx 1/n$.
• Reality says: Messiness $\approx (\log n) / n$.

As $n$ gets huge, $(\log n) / n$ is much bigger than $1/n$. The "Perfect Rule" claims the party gets perfectly tidy much faster than reality allows.

5. The Conclusion
Because the "Perfect Rule" predicts a result that contradicts known reality, the rule must be false. Therefore, the $\sqrt{\log n}$ tax is unavoidable. You cannot simplify the formula by removing it.

The Takeaway

Think of the $\sqrt{\log n}$ factor as a friction coefficient specific to 2D surfaces.

• In 3D, you can slide things around easily.
• In 2D, there is a specific kind of "static electricity" or "friction" that makes it slightly harder to arrange points perfectly.

This paper proves that this friction is a fundamental law of geometry. You can't wish it away or hide it in the math; it is always there, ensuring that the "Green Energy" alone isn't enough to tell the whole story of how well points are arranged.

Technical Summary

Here is a detailed technical summary of the paper "Green–Wasserstein Inequality on Compact Surfaces" by Maja Gwóźdź.

1. Problem Statement and Context

The paper addresses a specific question regarding the Green–Wasserstein inequality on compact, connected, two-dimensional Riemannian manifolds $(M, g)$ without boundary.

• Background: Steinerberger previously established that for $d \geq 3$, the quadratic Wasserstein distance $W_2$ between an empirical measure $\mu_n$ (based on $n$ points) and the normalized volume measure $dx$ is bounded by the unrenormalized off-diagonal Green energy term plus a remainder of order $n^{-1/d}$.
• The 2D Case: In dimension $d=2$, Steinerberger's estimate includes an additional factor of $\sqrt{\log n}$ in the remainder term:
  $$W_2(\mu_n, dx) \lesssim_M \frac{1}{n} \left| \sum_{i \neq j} G(x_i, x_j) \right|^{1/2} + \sqrt{\frac{\log n}{n}}$$
  where $G$ is the mean-zero Green function of the Laplacian.
• The Core Question (Problem 1): Can the $\sqrt{\log n}$ factor be removed in dimension 2 while retaining the unrenormalized off-diagonal Green energy term? Specifically, does there exist a constant $C_M$ such that for all $n$ and all point sets $\{x_1, \dots, x_n\}$:
  $$W_2(\mu_n, dx) \leq C_M \left( \frac{1}{\sqrt{n}} + \frac{1}{n} \left| \sum_{i \neq j} G(x_i, x_j) \right|^{1/2} \right)?$$

2. Methodology

The author answers this question in the negative using a proof by contradiction that combines probabilistic estimates with asymptotic matching theory. The methodology proceeds in three main stages:

A. Setup and Definitions

• Green Function: The paper utilizes the symmetric, mean-zero Green function $G(x, y)$ for the Laplace–Beltrami operator $-\Delta$. It is noted that $G$ has a logarithmic singularity along the diagonal ($x=y$) but is smooth elsewhere.
• Integrability: A crucial preliminary step (Lemma 1) proves that despite the logarithmic singularity, $G \in L^2(M \times M, dx \otimes dx)$. This ensures that the second moment of the Green energy is finite.

B. Probabilistic Analysis (Second-Moment Estimate)

The author considers a random configuration where points $X_1, \dots, X_n$ are independent and identically distributed (i.i.d.) according to the volume measure $dx$.

• Green Energy Variable: Define the random variable $S_n = \sum_{i \neq j} G(X_i, X_j)$.
• Expectation: Using the mean-zero property of $G$ and Fubini's theorem, it is shown that $\mathbb{E}[S_n] = 0$.
• Variance: The paper calculates the second moment $\mathbb{E}[S_n^2]$. By analyzing the independence of terms and the symmetry of $G$, it is proven that:
  $$\mathbb{E}[S_n^2] = 2n(n-1)\sigma^2$$
  where $\sigma^2 = \iint G(x,y)^2 dx dy$. Consequently, $\mathbb{E}[|S_n|] \leq \sqrt{2}\sigma n$.

C. Asymptotic Matching Theory

The proof relies on a known result by Ambrosio and Glaudo regarding the semi-discrete random matching problem on compact surfaces.

• Lower Bound: Ambrosio–Glaudo established that for random points, the expected squared Wasserstein distance behaves asymptotically as:
  $$\mathbb{E}[W_2(\mu_n, dx)^2] \sim \frac{\text{vol}(M)}{4\pi} \frac{\log n}{n}$$
  Specifically, the term $\frac{\log n}{n}$ is the dominant lower bound, with error terms of order $O(\frac{\sqrt{\log n \log \log n}}{n})$.

3. Key Contributions and Results

The main result is Theorem 1, which provides a negative answer to Steinerberger's problem.

• Theorem 1: There exists no constant $C_M$ such that the inequality in Problem 1 holds uniformly for all $n$ and all point sets on a compact surface.
• Proof Logic:
  1. Assume the inequality holds with a $1/\sqrt{n}$ remainder.
  2. Apply this inequality to the random configuration $X_1, \dots, X_n$.
  3. Square the inequality and take expectations. Using the bound on $\mathbb{E}[|S_n|]$ derived in Section 2, the right-hand side scales as $O(1/n)$.
  4. However, the left-hand side, $\mathbb{E}[W_2(\mu_n, dx)^2]$, is known to scale as $\Omega(\frac{\log n}{n})$ due to the Ambrosio–Glaudo asymptotics.
  5. Contradiction: For large $n$, $\frac{\log n}{n}$ grows strictly faster than $\frac{1}{n}$. Therefore, the assumed inequality cannot hold.

4. Significance and Implications

• Optimality of the $\sqrt{\log n}$ Factor: The paper rigorously proves that the $\sqrt{\log n}$ factor in the 2D Green–Wasserstein inequality is necessary if one wishes to keep the Green energy term in its unrenormalized form.
• Distinction from Renormalization: The author clarifies in Remark 4 that this result does not rule out $n^{-1/2}$ rates for specific deterministic point sets, nor does it rule out bounds involving a renormalized Green energy (where the singularity is subtracted). The impossibility arises specifically from the interaction between the unrenormalized sum and the random fluctuations of the Wasserstein distance.
• Methodological Insight: The work demonstrates the power of combining deterministic geometric inequalities with probabilistic second-moment estimates and random matching asymptotics to establish sharp bounds in geometric analysis.

In summary, the paper resolves an open problem by showing that the logarithmic correction in the 2D Green–Wasserstein inequality is intrinsic to the unrenormalized formulation and cannot be eliminated.