The Filling Area Conjecture

The Filling Area Conjecture

Definitions and mathematical statement

Let $M$ be a compact orientable Riemannian surface whose boundary is isometric to the circle $C$ of circumference $2\pi$. Write $d_M$ for intrinsic distance allowing paths through $M$, and $d_C$ for the shorter-arc distance along the boundary circle. Suppose no interior path shortens any boundary distance: \[ d_M(x,y)=d_C(x,y)\qquad(x,y\in C). \] Gromov's filling area assertion is $\operatorname{Area}(M)\ge2\pi$. The unit round hemisphere has this boundary distance and area, fixing the constant. The surface is not required to be a disk: arbitrary orientable topology is allowed. The equality of boundary distances is stronger than prescribing boundary length alone, which would permit very small-area fillings.

Formulation and concept references: [S1].

Short English statement

Must every orientable surface filling a circle without shortening distances between boundary points have at least the area of a round hemisphere?

Research attempt: a two-distance calibration and its quantitative limit

Status. The sharp area bound $2\pi$ is UNRESOLVED by this attempt. We prove an elementary lower bound $\pi^2/2$ for orientable fillings, calculate the exact limitation of using two boundary-distance coordinates, and verify the hemisphere and several failed shortcuts. The bound is a partial calibration calculation; no novelty claim is made.

Source audit. The inspected primary manuscript [S1], arXiv:2602.17859v1, states the orientable filling conjecture and proves in Theorem 1.2 a lower bound $\sqrt3\pi^2/4$ for arbitrary Riemannian fillings, explicitly without requiring orientability. That broader scope matters: the signed Stokes argument below does use orientability. The manuscript also distinguishes the disk and genus-one cases from arbitrary topology. Neither its discrete lower bound nor its continuous theorem reaches the selected constant $2\pi$ in unrestricted orientable genus. We do not equate graph distances in an arbitrary triangulation with Riemannian intrinsic distances without quantitative control.

We may work with the connected component containing the boundary circle. Any other closed component only adds nonnegative area. Choose an orientation so that the boundary parameter $s\in\mathbb R/(2\pi\mathbb Z)$, traversed in the positive direction, is positively oriented.

1. Scaling and the model hemisphere

For a boundary circle of circumference $L$, rescaling the metric by the square of $2\pi/L$ changes all lengths by $2\pi/L$ and all areas by its square. Thus the conjectural inequality is \[ \operatorname{Area}(M)\ge \frac{L^2}{2\pi}. \tag{1} \] Every lower bound should respect this scaling; an additive dimensional constant would be inconsistent.

The round unit hemisphere has area $2\pi$ and equator length $2\pi$. For equatorial points $x,y$, their distance in the full sphere is the angle between their unit position vectors, namely the shorter equatorial arc length in $[0,\pi]$. A path confined to the hemisphere is also a sphere path, so it cannot be shorter. Conversely the shorter equatorial arc is allowed in the hemisphere and has that length. Hence the two distances agree exactly, including antipodal points. This verifies both the admissibility and the conjectural constant of the model.

The flat unit disk has the same boundary length but fails the distance condition: antipodal boundary points are joined by a diameter of length two, less than $\pi$. Boundary length alone therefore does not control the problem.

2. A general area estimate from two Lipschitz functions

Let $u,v:M\to\mathbb R$ be Lipschitz functions on a compact oriented Riemannian surface with boundary. Suppose their almost-everywhere gradient norms are at most one. At differentiability points, \[ |du\wedge dv|=|\det(\nabla u,\nabla v)|\,dA \le |\nabla u|\,|\nabla v|\,dA\le dA. \tag{2} \] Stokes' formula for Lipschitz functions then gives \[ \operatorname{Area}(M)\ge \left|\int_Mdu\wedge dv\right| =\left|\int_{\partial M}u\,dv\right|. \tag{3} \] The right side is the signed area enclosed by the planar boundary curve $(u,v)$, counted with multiplicity. Injectivity of $(u,v)$ in the interior is unnecessary.

Here is a way to justify the low regularity without assuming that distance functions are smooth at their cut loci. Approximate $u,v$ uniformly by smooth functions $u_\epsilon,v_\epsilon$ whose Lipschitz constants are at most $1+\epsilon$. Such approximation follows by smoothing in finitely many coordinate charts, using charts in which the metric varies by an arbitrarily small amount, and a partition of unity; the extra gradient terms from the partition are controlled by choosing uniform approximation errors smaller than the prescribed gradient error. Boundary charts are handled by extending across a collar before smoothing.

For the smooth approximants, Stokes and $2$ give $|\int_{\partial M}u_\epsilon\,dv_\epsilon|\le(1+\epsilon)^2\operatorname{Area}(M)$. Their boundary integrals converge to $\int u\,dv$. Indeed $\int(u_\epsilon-u)\,dv_\epsilon\to0$ by uniform convergence and the uniformly bounded total variation of $v_\epsilon$ on the finite-length boundary. For the other difference, integration by parts on the closed boundary gives $\int u\,d(v_\epsilon-v)=-\int(v_\epsilon-v)\,du\to0$. Passing to the limit proves $3$. This uses only the usual Lipschitz approximation and one-dimensional integration-by-parts facts; it never differentiates a distance function classically at a cut point.

The metric hypothesis enters only through the unit gradient bounds. Orientability enters through the signed integral and Stokes' theorem. Handles do not create boundary terms because the surface itself has only the prescribed boundary circle.

3. Boundary distance functions trace an explicit parallelogram

Choose two boundary points $a,b$ at shorter-arc separation $\delta$, with $0<\delta<\pi$, taking $a$ at $s=0$ and $b$ at $s=\delta$. Define \[ u(x)=d_M(a,x),\qquad v(x)=d_M(b,x). \] The triangle inequality makes each function 1-Lipschitz. The no-shortcuts condition determines their boundary values exactly: \[ u(s)=\min(s,2\pi-s),\qquad v(s)=\min(|s-\delta|,2\pi-|s-\delta|),\quad0\le s\le2\pi. \tag{4} \] Consequently $(u,v)$ traces successively the four vertices \[ (0,\delta),\quad(\delta,0),\quad(\pi,\pi-\delta), \quad(\pi-\delta,\pi),\quad(0,\delta). \tag{5} \] Between the successive parameter values $0,\delta,\pi,\pi+\delta,2\pi$, both functions are affine. The resulting curve is a simple parallelogram, not a multiply traversed or self-cancelling loop.

Its adjacent edge vectors can be taken as $(\delta,-\delta)$ and $(\pi-\delta,\pi-\delta)$. The signed area is their determinant, \[ \int_{\partial M}u\,dv=2\delta(\pi-\delta). \tag{6} \] One can also check $6$ by integrating the four linear formulas in $4$, avoiding any inference from a drawing. Combining $3$ and $6$ gives the family of inequalities \[ \operatorname{Area}(M)\ge2\delta(\pi-\delta). \] The quadratic is maximized at $\delta=\pi/2$, with value \[ \boxed{\operatorname{Area}(M)\ge\pi^2/2.} \tag{7} \] For a boundary of length $L$, the corresponding lower bound is $L^2/8$. Since $\pi<4$, $7$ is strictly weaker than the target $2\pi$. It leaves a quantitative gap $2\pi-\pi^2/2>0$ even though it applies to every orientable genus.

The antipodal choice $\delta=\pi$ is useless for this method: then $u+v=\pi$ along the boundary, so the planar curve collapses to a line segment and the signed integral vanishes. The two points must not be selected merely for their maximal mutual distance.

4. What the calculation says on the hemisphere

On the unit hemisphere, choose $a=(1,0,0)$ and $b=(0,1,0)$. For an interior point $(x,y,z)$, $z\ge0$, \[ u=\arccos x,\qquad v=\arccos y. \] The boundary image is the diamond in $5$ for $\delta=\pi/2$. Away from the exceptional poles, the gradients of both distance functions have norm one, but they are not everywhere perpendicular. Their inner product is \[ \langle\nabla u,\nabla v\rangle =-\frac{xy}{\sqrt{1-x^2}\sqrt{1-y^2}}. \tag{8} \] To derive this, project the ambient coordinate gradients onto the tangent plane of the sphere: $\nabla x=e_1-x(x,y,z)$ and $\nabla y=e_2-y(x,y,z)$, whose inner product is $-xy$, then apply the derivatives of arccos.

The determinant bound in $2$ is strict on an open set where $xy\ne0$ and the gradients are defined. Therefore the integral of $du\wedge dv$ can be $\pi^2/2$ while the hemisphere's area is $2\pi$. The loss is not caused by singularities or by genus; it already occurs in the proposed equality model. This prevents the two-function estimate from being sharpened to the target merely by claiming equality in the Jacobian bound.

A different pair of boundary points does not repair this loss: $6$ shows that every such pair gives at most $\pi^2/2$. Averaging these separate inequalities also cannot improve their maximum, since each spends the entire area bound in $3$. A successful multi-function argument would need a new pointwise inequality for a combined differential expression, not a sum of inequalities that each uses the same area independently.

5. Why orientability and all boundary distances matter

For a nonorientable filling there is no global signed two-form $du\wedge dv$ to integrate with a chosen orientation on $M$. Passing to the oriented double cover does not immediately restore $7$ for the original surface. Its boundary lifts to two components, and the induced orientations make their contributions cancel for the lifted functions. Thus the method does not claim the same bound in the broader nonorientable scope of [S1, Theorem 1.2].

The equality of distances in $4$ is also indispensable. If one only prescribed boundary length, the ambient distance functions could be strictly smaller than the shorter boundary arcs and their image need not be the parallelogram. Even preserving distances of one antipodal pair would not determine the four affine segments or their signed area.

For a simple quantitative illustration, consider a rotationally symmetric disk metric \[ dr^2+f(r)^2d\theta^2,\qquad0\le r\le1, \] with $f(r)=r$ near the center for smoothness, $f>0$ away from the center, and $f(1)=1$ for boundary length $2\pi$. For small $\epsilon$, choose a smooth profile with $f\le\epsilon$ on $[0,1-\epsilon]$ and $f\le1$ everywhere, joining to the value one in the outer collar. Such profiles exist by smooth interpolation after a small initial segment where $f=r$. Their areas satisfy \[ A=2\pi\int_0^1 f(r)\,dr\le 2\pi\left(\epsilon(1-\epsilon)+\epsilon\right), \] which tends to zero. But boundary antipodes can travel radially to $r=1-\epsilon$, follow a half-circle, and return, with length at most $2\epsilon+\pi\epsilon<\pi$ for small $\epsilon$. These are not admissible fillings. The explicit shortcut explains why thin-neck or collar-collapse pictures do not disprove the conjecture.

6. Cutting topology away changes the problem

A tempting approach is to cut along loops until a disk remains and invoke the disk filling theorem. Cutting preserves area, but it adds boundary components or duplicates arcs in the boundary. The new boundary is not the original length-$2\pi$ circle, and its intrinsic boundary distances are not prescribed by the catalogue hypothesis. Gluing caps to restore a single boundary may add area or create paths which shorten the original boundary distances.

The Stokes argument avoids this issue because it uses a function on the original surface and only its original boundary trace. However, that same robustness loses enough metric information to give $7$ rather than the sharp constant. Thus topology is not removed for free; it is bypassed by a weaker estimate.

Likewise, a combinatorial filling estimate requires a controlled passage between graph length and Riemannian length and between numbers of cells and area. Uniformly small triangles do not automatically make graph distance approximate geodesic distance with relative error tending to zero; directions and edge shapes matter. The discrete theorem in [S1] addresses a specific approximation scheme. We have not supplied a new scheme attaining the hemisphere constant.

7. Verification and the exact remaining gap

The saved exact diagnostic computes the four boundary vertices and the signed shoelace sum for rational values of $\delta/\pi$, obtaining $2\delta(\pi-\delta)$ and its maximum at one half. This checks both signs and the lack of double counting. The geometric proof separately verifies the Lipschitz norm bounds and the limiting Stokes step, which a polygon calculation alone would not establish.

The lower bound $7$ is valid under the original orientability hypothesis and is consistent with the hemisphere's area. It does not give equality rigidity, and it does not prove the target by taking finer boundary samples: an additional estimate relating all those distance coordinates would be needed. A precise next objective is an area-controlled differential or integral functional whose prescribed boundary value is $2\pi$ and whose pointwise bound is attained by the hemisphere. The present two-distance functional fails that equality test by $8$.

Thus arbitrary orientable fillings satisfy the proved partial bound $\pi^2/2$, but the first remaining step is the sharp improvement from this calibration to $2\pi$ without introducing restrictions on topology or curvature. The filling area conjecture remains UNRESOLVED by this attempt.

Sources

• S1 Joseph Briggs and Chris Wells, “A Discrete View of Gromov's Filling Area Conjecture,” arXiv:2602.17859 $2026$. https://arxiv.org/abs/2602.17859. Formulation source.