How to read this file. This is a study guide to one research paper: the proof that in a Cartan-Hadamard manifold (a complete, simply connected Riemannian manifold with sectional curvature at most 0), every region has boundary at least as large as a round ball of the same volume in the matching constant-curvature space, and that equality happens only for honest Euclidean balls. It is written for a patient high-school reader. Each block starts with "Exercise:", then a short question, then a self-contained answer. A line beginning "Proof:" inside an answer is a compressed but complete outline of the actual argument; skip it on a first read and return later. Named results carry a short nickname in parentheses, for example "Lemma (Distance Comparison)"; find the full statement by searching for that nickname. Notation is defined at first use. The symbols you meet first are: M^n (an n-dimensional manifold, n >= 2); a point p and the distance d(p,x); Sec (sectional curvature, a number attached to each 2-plane direction); kappa (a curvature bound, kappa <= 0); S^n (the round unit sphere in n+1 dimensions); omega_n (volume of the unit ball in flat n-space) and s_m (area of the unit m-sphere, where m = n-1; note s_m = n*omega_n); Vol and P (volume and boundary "perimeter" of a region); and I_b (the "isoperimetric profile" of the model space of curvature -b^2, i.e. the boundary a round ball needs for a given volume).

Exercise: What is a Cartan-Hadamard manifold, in plain words?
It is a space that (1) is smooth, so you can measure lengths, angles and curvatures; (2) is complete, so you can walk forever or reach any point you aim at without falling off an edge; (3) is simply connected, so there are no holes or handles that would let two paths between the same points fail to be deformed into one another; and (4) has sectional curvature Sec <= 0 everywhere, so it curves inward like a saddle, never outward like a sphere. The three examples to hold in your head are flat Euclidean space R^n (curvature 0), hyperbolic space H^n (curvature -1), and any smooth closed surface shaped like a saddle such as a paraboloid (curvature < 0). A key consequence is the Cartan-Hadamard theorem: for every point p, the exponential map exp_p (which sends a tangent vector v to the point reached by walking distance |v| along the geodesic leaving p in direction v) is a bijection from the tangent space T_p M onto all of M. So every point has "polar coordinates" around p, and geodesics never refocus: two geodesics leaving p spread apart forever. This is why these spaces are sometimes called "spaces of nonpositive curvature" or "Hadamard manifolds".

Exercise: What does the isoperimetric problem ask?
It asks: among all regions of a fixed volume, which has the least boundary, and how small can that boundary be? In flat R^n the answer is a round ball, and the optimal bound is the Euclidean isoperimetric inequality P(E) >= n*omega_n^{1/n} * Vol(E)^{(n-1)/n}, with equality exactly for round balls. Writing it as P(E) >= I_0(Vol E) with I_0(V) = n*omega_n^{1/n} V^{(n-1)/n}. The whole paper replaces the flat model by a curved one: in a space of curvature at most kappa <= 0, the least boundary for volume V should be that of a round ball of volume V in the simply connected space form of constant curvature kappa. That model space is R^n when kappa = 0 and hyperbolic n-space when kappa < 0. This is the "Cartan-Hadamard isoperimetric conjecture", proved here in every dimension, for every measurable set of finite volume and finite perimeter (not just for nice smooth regions).

Exercise: State the main theorem precisely.
Theorem (Generalized Comparison). Let n >= 2, let kappa <= 0, and let (M^n, g) be a complete, simply connected, smooth Riemannian manifold with Sec_g <= kappa. Let b = sqrt(-kappa) (so b >= 0, and b = 0 means kappa = 0). Let I_b be the model isoperimetric profile defined below. Then every measurable set E in M with Vol_g(E) < infinity and P_g(E) < infinity satisfies P_g(E) >= I_b( Vol_g(E) ). Two special readings: when kappa = 0 this is P(E) >= n*omega_n^{1/n} Vol(E)^{(n-1)/n}; when kappa = -1 and Vol(E) is written as the volume of a radius-r ball, i.e. Vol(E) = n*omega_n * integral_0^r sinh^{n-1}(t) dt, it is P(E) >= n*omega_n sinh^{n-1}(r). Round balls in the model spaces attain equality.

Exercise: What is "perimeter" for a rough set, not just a smooth region?
For a smooth region, perimeter is just boundary area. For an arbitrary measurable set E one uses a weak definition, the total variation of its indicator function chi_E. Concretely, P_g(E) = sup { integral over E of div_g(X) dVol_g : X is a smooth compactly supported vector field on M with |X|_g <= 1 }. This supremum is the "size of the boundary measured against every possible smooth test field". It equals boundary area for smooth regions, and it can be finite even when the topological boundary is wild (fractals, dust, and so on). Sets with P_g(E) < infinity are exactly the sets of finite perimeter; they are the natural class for the theorem, and they are closed under taking limits in a suitable weak sense, which is what makes existence proofs possible. All statements are understood up to sets of volume zero, since adding or removing a zero-volume set changes neither volume nor perimeter.

Exercise: Define the model functions sn_b, cs_b and the profile I_b.
For b >= 0 and r >= 0 set sn_b(r) = r when b = 0, and sn_b(r) = sinh(br)/b when b > 0; then set cs_b(r) = sn_b'(r) (the derivative). So cs_b(r) = 1 for b = 0 and cs_b(r) = cosh(br) for b > 0. Two identities matter constantly: sn_b' = cs_b and cs_b^2 - b^2 sn_b^2 = 1. The area of a radius-r geodesic sphere and the volume of a radius-r geodesic ball in the model space of curvature -b^2 are Area_b(r) = s_m * sn_b(r)^m and Vol_b(r) = s_m * integral_0^r sn_b(t)^m dt, where m = n-1. Both rise continuously from 0 to infinity and are strictly increasing, so Vol_b has an inverse. The model isoperimetric profile is I_b(V) = Area_b( Vol_b^{-1}(V) ): "the boundary of a round ball of volume V". It is continuous, strictly increasing, and I_b(0) = 0. These are exactly the functions appearing on the right side of the main theorem.

Exercise: State the equality case (Euclidean rigidity) precisely.
Theorem (Euclidean Equality Rigidity). Let (M^n, g) be complete, simply connected, smooth with n >= 2 and Sec_g <= 0. Let E be a bounded measurable set with 0 < Vol_g(E) < infinity and P_g(E) < infinity. Then P_g(E) = n*omega_n^{1/n} Vol_g(E)^{(n-1)/n} (equality in the FLAT bound) if and only if E agrees, up to a set of volume zero, with an open region whose own induced Riemannian metric is isometric to a round Euclidean ball. Said plainly: the set together with the metric it inherits must literally be a flat ball. The ambient manifold may be curved everywhere outside that ball; only the interior of the equality region is forced to be flat.

Exercise: Why can't we just use "geodesic balls grow at most as fast as model balls"?
Because that compares round balls to round balls, not arbitrary regions to round balls. Volume comparison (Bishop-Gromov) says a geodesic ball in M has volume at least that of a model ball of the same radius, and it controls the boundary of BALLS. But the isoperimetric problem is about ALL regions: a long thin blob, a dumbbell, a region with many wiggles. Knowing how balls behave does not tell you the best shape for arbitrary volume. In flat space the optimal shape is a ball, but that is a theorem (the isoperimetric inequality), not a consequence of volume comparison. The paper's whole point is to prove the analogous sharp inequality for arbitrary regions in curved Cartan-Hadamard spaces, which volume comparison alone cannot do. The proof therefore has to build a genuinely new analytic tool (a sharp Sobolev inequality on hypersurfaces) and a minimization argument, not merely quote volume comparison.

Exercise: Who proved what before, and what is new here?
The story is old and incremental. On surfaces, Weil (1926) and the subharmonic-function method of Beckenbach-Rado (1933), sharpened by Bol (1941), give the curved comparison; Chavel-Feldman (1980) gave a clean smooth-surface formulation. Kleiner (1992) proved the model comparison in dimension three, including its equality case. Croke (1984) proved the sharp flat comparison in dimension four. Kloeckner-Kuperberg (2019) added optical and chord-based methods, covering dimensions two and four again, but their four-dimensional negative-curvature statement carries an extra restriction on the domain's chord length and the model ball's radius. Two further threads: Ghomi-Spruck (2022) show that a total-curvature lower bound implies the flat inequality; Ghomi-Stavroulakis (2026) handle small perturbations of a ball metric. More recent 2026 preprints (Chen-Ghomi-Wang; Agnoletto-Da-Silva-Nardulli-Resende; Wheeler) settle various dimensions or add curvature hypotheses. The present paper is the first proof in ALL dimensions, for ALL kappa <= 0, for arbitrary finite-volume finite-perimeter sets, with a complete equality classification. The new ingredients are: (i) a sharp critical Sobolev inequality on a hypersurface whose constant depends on the mean curvature, (ii) a curved-space adaptation of Kleiner's confined-profile minimization, and (iii) a flexibility trick (an offset sphere and a spectral / flux argument) that pins down the equality set without assuming the equality boundary has any symmetry.

Exercise: What are the spectral and torsion corollaries, and why are they mentioned?
Two classical functionals are the first eigenvalue of the p-Laplacian and the p-torsional rigidity. For a smooth bounded domain D in M and 1 < p < infinity define lambda_{1,p}(D) = inf over nonzero u vanishing on the boundary of ( integral_D |grad u|^p ) / ( integral_D |u|^p ), and T_p(D) = sup over nonzero u of ( integral_D |u| )^p / ( integral_D |grad u|^p ). These measure "how hard it is to vibrate" and "how hard it is to stretch" the domain. Corollary (Model-Ball Functionals): if D is a compactly contained smooth domain in a Cartan-Hadamard manifold with Sec <= kappa, and B_kappa is the model ball of the same volume, then lambda_{1,p}(D) >= lambda_{1,p}(B_kappa) and T_p(D) <= T_p(B_kappa). These follow from the main theorem by "rearranging" any test function into an equimeasurable radial function on the model ball without increasing energy; the rearrangement theorem used is due to Nobili-Violo (2025). They are mentioned because they show the isoperimetric theorem has immediate, sharp functional-analytic payoffs, and because they were the original motivation for caring about the equal-volume ball comparison.

Exercise: Give the skeleton of the whole proof in one breath.
Plan (Skeleton): Six stages. (1) Prove a sharp "critical" Sobolev inequality on any hypersurface inside M, where the constant is corrected by the hypersurface's mean curvature (Section maps+sobolev). (2) Minimize boundary among regions of fixed volume that are trapped inside a large geodesic ball; call the least boundary the "confined profile" I(v). Show minimizers are smooth except on a small singular set, and their mean curvature h satisfies 0 <= h <= h0 = I'(v)/m (Section profile). (3) Feed the constant test function into the Sobolev inequality; the curvature bound turns it into one scalar differential inequality I'(v) >= m sqrt(b^2 + (s_m/I(v))^{2/m}) (Section comparison). (4) Integrate that inequality; its exact integral is the model comparison, giving the theorem for bounded sets in dimensions >= 4 (Section comparison). (5) Handle dimensions 2 and 3 by classical results plus a smooth approximation, then remove the "bounded" restriction with a cutoff, then rescale to reach every kappa (Section dimensions). (6) For equality, re-run the machinery on an equality set to learn that its boundary is smooth with constant mean curvature, then locate that boundary on a round sphere by three different dimension-dependent tricks, then recover the set and its flat metric (Sections equality-regularity and rigidity). Finally extract the spectral corollary by rearrangement (Section consequences). The rest of this guide walks through these stages.

Exercise: What is the single analytic inequality that drives everything?
Theorem (Hypersurface Sobolev). Normalize the curvature bound to -b^2 with b in {0,1}. Let M^n be complete, simply connected, with Sec <= -b^2, and let Sigma^m with m = n-1 >= 3 be a locally C^{1,1} immersed hypersurface in M (it need not be complete, connected, or without boundary). Let dsigma be the induced area, grad_Sigma the tangential gradient, and h the signed normalized mean curvature for a local unit normal (so the trace of the shape operator is m*h; the sign flips with the normal, but h^2 does not). Put q = 2m/(m-2), c = 4/(m-2) = q-2, and Y_m = m * s_m^{2/m}. Then for every Lipschitz function v compactly supported in the interior of Sigma, integral_Sigma ( c |grad_Sigma v|^2 + m(h^2 - b^2) v^2 ) dsigma >= Y_m ( integral_Sigma |v|^q dsigma )^{2/q}. The inequality extends to v in the Sobolev space H^1_0(U) for any open U compactly contained in the interior of Sigma. It is sharp: on a round geodesic sphere in the model space of curvature -b^2, the constant test function makes it an equality.

Exercise: Where do the exponents q and c come from, and why the term m(h^2 - b^2)?
The exponents are dictated by scaling and Sobolev theory. In n = m+1 dimensional space, the critical Sobolev exponent for one derivative is 2m/(m-2); it is the exponent for which the inequality is invariant under scaling and dilation. The coefficient c = 4/(m-2) is chosen so that c = q-2 and so that the Aubin-Talenti sharp constant, once multiplied by c, becomes exactly Y_m = m s_m^{2/m}. The potential m(h^2 - b^2) is the curvature correction: h is the normalized mean curvature of the hypersurface and b^2 is minus the ambient curvature bound. Geometrically, h^2 - b^2 measures how much the boundary curves relative to the model sphere, and it is exactly the coefficient that appears when you integrate the ambient Hessian comparison against a test function. On the model geodesic sphere of curvature -b^2 the mean curvature is h = cs_b(r)/sn_b(r), and h^2 - b^2 = (cs_b^2 - b^2 sn_b^2)/sn_b^2 = 1/sn_b^2, which produces exact equality. This kind of "mean-curvature-corrected Sobolev" is the engine: every later estimate is obtained by plugging a cleverly chosen test function into it.

Exercise: State the sharp Euclidean Sobolev inequality (Aubin-Talenti) and check the constant is the same Y_m.
Theorem (Aubin-Talenti). In flat R^m with m >= 3, for every smooth compactly supported function phi, c * integral_{R^m} |D phi|^2 dx >= Y_m ( integral_{R^m} |phi|^q dx )^{2/q}, with q = 2m/(m-2), c = 4/(m-2), and Y_m = m s_m^{2/m}. Equivalent form: before multiplying by c the sharp constant is m(m-2)|S^m|^{2/m}/4; multiplying by c = 4/(m-2) gives m|S^m|^{2/m} = m s_m^{2/m} = Y_m. These are the same constant, just shuffled between the two sides. So Y_m is not an artifact of the curved setting; it is THE sharp constant of the flat model, and the theorem above says the curved hypersurface inequality survives with exactly this constant, the curvature only appearing through the h^2 - b^2 potential. Proof (why the constant is sharp): test with the Euclidean "bubble" function phi(x) = (epsilon^2 + |x|^2)^{-(m-2)/2}, whose kinetic and potential energies are exactly balanced against the critical L^q mass at a fixed ratio; as epsilon tends to 0 this ratio tends to Y_m from above and no smaller constant works. This is the standard concentration argument and is the reason Y_m is the "critical mass" threshold that later appears in the Brezis-Lieb step.

Exercise: Why must the theorem hold for incomplete hypersurfaces and for H^1_0 functions, not just nice compact ones?
Because the equality boundary and the confined minimizers are exactly such hypersurfaces. A confined minimizer's regular boundary is a hypersurface inside a compact ball, but its singular set has been removed, so the remaining regular part is typically INCOMPLETE (it can run up to the singular points and stop) and its components need not be compact. The later strategy tests the Sobolev inequality on cutoff functions that go to 1 everywhere on the regular part and whose Dirichlet energy goes to 0 near the deleted singular set. For that to become a legitimate test on the whole regular part, the inequality must already be known for compactly supported Lipschitz tests, and then for H^1_0(U) limits with U a compact piece of the internal manifold topology. It must not assume the regular part is complete or has finitely many components, because the singular set can split it into infinitely many pieces in high dimensions. That is exactly why the theorem is stated for arbitrary locally C^{1,1} immersed hypersurfaces.

Exercise: What is the geometric idea of the proof of the hypersurface Sobolev inequality?
Idea (Square Completion). Build a "test map" Phi from the hypersurface into a round unit sphere, together with a positive weight f = e^u, arranged so that two inequalities hold at once: the differential of the map contracts against f (|dPhi| <= f |X|), and the Hessian of u has a one-sided bound with a completed square. Take the tangential trace of that Hessian bound, use the restriction formula for the Laplacian, and multiply by v^2; after integration by parts the gradient terms assemble into an exact square c |grad v - alpha v grad u|^2 plus the term v^2 |d(Phi restricted to Sigma)|^2. The positive potential in the Sobolev inequality is then bounded below by this square plus the map-energy. If the critical quotient were BELOW the Euclidean threshold Y_m, minimization plus Brezis-Lieb would produce a minimizer psi, and the sphere coordinates would give a first-order identity that forces psi to be a weighted multiple of a function with zero gradient; a zero extension then contradicts the Dirichlet condition. That contradiction forces the quotient to be at least Y_m. So the proof is a soft "no subcritical minimizer" argument, not a direct computation of the best function.

Exercise: How do we choose the map Phi and its center and scale?
Answer (Comparison Maps). Fix a center p, a scale a > 0, and put r = d(p,x). Define the radial coordinate z = a * sn_b(r/2)/cs_b(r/2) (with sn_b, cs_b the model functions of curvature -b^2; recall sn_b(0)=0, cs_b(0)=1). Then send x to the point Phi_{p,a}(x) = ( 2z/(1+z^2) * theta, (1-z^2)/(1+z^2) ) of the unit sphere in T_pM + R, where theta = exp_p^{-1}(x)/r is the unit direction. This is exactly inverse stereographic projection applied to a conformally rescaled polar metric. The weight is f_{p,a}(x) = a / ( cs_b(r/2)^2 + a^2 sn_b(r/2)^2 ), and u = log f. The scale a is a free knob used later to make the sphere-coordinates have zero mean with respect to a given probability measure, and the center p is chosen by a fixed-point (Brouwer) argument. On the flat model this specializes to z = a r/2 and f = a/(1 + a^2 r^2/4), the familiar stereographic factor.

Exercise: Prove the Distance Comparison (convexity of squared distance).
Lemma (Distance Comparison). Let (M^n,g) be complete and simply connected with Sec <= 0. For a point p write D_p(x) = d(p,x)^2 / 2 and r_p(x) = d(p,x) and log_p = exp_p^{-1}. Then Hess_g D_p >= g (as quadratic forms), (exp_p)^* g >= g_p (the pullback metric dominates the flat metric of the tangent space), and |d log_p(X)|_{g_p} <= |X|_g. Proof: The hard case is the Hessian of the distance r itself. For a vector X perpendicular to a unit-speed radial geodesic gamma of length r, deform the geodesic in the direction X and let J be the resulting Jacobi field with J(0)=0 and J(r)=X. The second variation of length gives Hess r (X,X) = integral_0^r ( |D_t J|^2 - <R(J, gamma')gamma', J> ) dt. Since Sec <= 0 the curvature term is nonnegative when moved to the other side, so Hess r(X,X) >= integral_0^r |D_t J|^2 dt. Among all Jacobi-type fields from 0 to X the energy integral is minimized by the straight-line-in-parallel-transport field J_0(t) = (t/r) X (in the flat model), which gives energy |X|^2 / r; and in general the energy is at least that, because writing J = J_0 + V with V vanishing at both ends makes the cross term vanish by integration by parts. Hence Hess r >= (1/r)(g - dr x dr) when curvature is 0, and the same comparison relative to the model function sn_b generalizes this to Sec <= -b^2 as Hess r - (cs_b(r)/sn_b(r))(g - dr x dr) >= 0. Now Hess D_p = Hess (r^2/2) = dr x dr + r Hess r >= dr x dr + (g - dr x dr) = g. The metric statement (exp_p)^* g >= g_p is just the integrated form of the same inequality along radial geodesics (the polar comparison below). The contraction statement follows by applying that differential inequality to the inverse map.

Exercise: Prove the polar metric comparison and the map differential bound.
Lemma (Polar Comparison). Under Sec <= -b^2, in geodesic polar coordinates about p the metric satisfies g >= dr^2 + sn_b(r)^2 g_{S_p^{n-1}}, where g_{S_p^{n-1}} is the round unit-sphere metric on the direction space. Proof: Write the metric in polar form g = dr^2 + A(r, theta)^2 g_{S}, and let J be an angular Jacobi field. From Hess r >= (cs_b/sn_b) |J_perp|^2 and the identity (d/dr)|J|^2 = 2 Hess r(J,J) for an angular field, we get (d/dr) log |J|^2 >= 2 cs_b(r)/sn_b(r), so |J|^2 / sn_b(r)^2 is nondecreasing in r, and its limit at r=0 is the initial angular squared norm. Hence A(r,theta) >= sn_b(r), which is exactly the claim. Now apply it to the map. The round stereographic metric on the sphere (in the coordinate z) is 4(dz^2 + z^2 g_S)/(1+z^2)^2. Substituting z = a sn_b(r/2)/cs_b(r/2) and using (2z'/(1+z^2)) = f and (2z/(1+z^2)) = f sn_b(r) gives Phi_{p,a}^* g_S = f^2 ( dr^2 + sn_b(r)^2 g_S ) <= f^2 g, so |d Phi_{p,a}(X)| <= f_{p,a} |X| for every tangent X. This is bound one.

Exercise: Prove the Hessian bound for u = log f.
Lemma (Hessian Bound). With u = log f_{p,a} one has, as quadratic forms, Hess u <= du x du - (1/2)( b^2 + f^2 + |grad u|^2 ) g. Proof: Introduce A = a^2 + b^2 and D = cs_b(r/2)^2 + a^2 sn_b(r/2)^2 = 1 + A sn_b(r/2)^2, so f = a/D. One computes D' = (A/2) sn_b(r) and D'' = (A/2) cs_b(r), and the double-angle identities give the key algebraic identity b^2 D^2 + a^2 + (A^2/4) sn_b(r)^2 = A cs_b(r) D. (Check it by substituting t = sn_b(r/2)^2: both sides become A + (A^2 + 2 b^2 A) t + 2 b^2 A^2 t^2.) From this one derives the radial identities u' = -(A sn_b(r))/(2D) <= 0, u'' = (u')^2 + u' cs_b(r)/sn_b(r), and u' cs_b(r)/sn_b(r) = -(1/2)( b^2 + f^2 + (u')^2 ). Combining these with the distance Hessian comparison, written as Hess r = (cs_b/sn_b)(g - dr x dr) + H with H >= 0, produces Hess u = du x du - (1/2)( b^2 + f^2 + |grad u|^2) g + u' * H. Since u' <= 0 and H >= 0, the last term is negative semidefinite, giving the bound. At the pole p the expansions dPhi_{p,a}(X) = (aX, 0) and u = log a - ((a^2+b^2)/4) r^2 + O(r^4) show both estimates also hold there. This is bound two.

Exercise: State the two bounds together, and note the smooth dependence.
Lemma (Comparison Maps). The map Phi_{p,a} and the weight f_{p,a} are smooth on all of M (including the pole, because in normal coordinates their components are smooth even functions of r^2). For every tangent vector X: |dPhi_{p,a}(X)| <= f_{p,a}|X| and Hess u_{p,a} <= du x du - (1/2)( b^2 + f_{p,a}^2 + |grad u_{p,a}|^2 ) g. The pair (Phi, f) depends smoothly on (p, a, x), and for a fixed base point o the parallel transport of the sphere-valued first component of Phi from p back to o also depends smoothly on (p, a, x). The smooth dependence is what makes the next fixed-point argument legitimate; it follows because the map (p, y) -> (p, exp_p y) is a bijective local diffeomorphism (hence a diffeomorphism), and because solutions of the parallel-transport ODE depend smoothly on parameters.

Exercise: Why can we always center and scale the map so that the sphere coordinates have zero mean?
Lemma (Balancing). Let mu be a compactly supported Borel probability measure on M with no atoms (mu({x}) = 0 for every x). If supp(mu) lies in a ball B_D(o), then there exist a center p in B_D(o) and a finite scale a > 0 such that integral_M Phi_{p,a}(x) dmu(x) = 0 in T_pM + R (after parallel-transporting the first component to a common origin). Proof: Parametrize the center by xi = exp_o^{-1}(p) and parallel-transport the sphere-valued part radially to o. Define F(xi,a) = integral (P_{p->o} + id) Phi_{p,a}(x) dmu(x). This is continuous on every cylinder {|xi| <= D} x [a0, a1]. Three boundary estimates hold. (Side) On |xi| = D, for x in the support, convexity of the distance to o along the geodesic from p to x gives <xi, P_{p->o} theta_p(x)> <= D (rho(x) - D)/d(p,x) < 0, because every support point is strictly closer to o than D; so the first component of F points strictly inward. (Bottom) Choose a0 small enough that a0 * sn_b(2D)/cs_b(2D) < 1; then the last coordinate of Phi is at least (1 - a0^2 t^2)/(1 + a0^2 t^2) > 0 on all of the cylinder, so the last component of F points strictly up. (Top) Because mu has no atoms and is supported in a compact ball, the supremum of mu(B_delta(p)) over p tends to 0 as delta tends to 0 (otherwise a limiting center would carry an atom). Pick delta small and a1 large so a1 * sn_b(delta)/cs_b(delta) >= sqrt(3); then the last coordinate of Phi is at most -1/2 outside B_delta(p) and at most 1 everywhere, so the last component of F is at most -1/8, pointing strictly down. A continuous map of the compact convex cylinder into itself that points strictly inward on the whole boundary has an interior zero by the Brouwer fixed-point theorem: the map zeta -> projection onto the cylinder of zeta + F(zeta) has a fixed point, the boundary estimates rule out every face, and taking both signs of every small displacement shows F = 0 at the interior point. Undoing the parallel transport gives the claimed p and a. This centered map is the one used in the next step.

Exercise: State and prove the Geometric Square estimate.
Lemma (Square Estimate). Fix p and a > 0, let Phi = Phi_{p,a} and f = e^u = f_{p,a} be the comparison map and weight, and let v be a compactly supported Lipschitz function on the interior of Sigma. Then Q_Sigma(v) := integral_Sigma ( c|grad_Sigma v|^2 + m(h^2 - b^2) v^2 ) dsigma is at least integral_Sigma ( m f^2 v^2 + m(h + Nu)^2 v^2 + c |grad_Sigma v - alpha v grad_Sigma u|^2 ) dsigma, which is at least integral_Sigma ( v^2 |d(Phi restricted to Sigma)|^2 + c |grad_Sigma v - alpha v grad_Sigma u|^2 ) dsigma. Here Nu is the normal derivative of u along the chosen unit normal N, and alpha = (m-2)/2, c alpha = 2, c alpha^2 = m-2. In particular Q_Sigma >= 0 on all such tests. Proof: Write grad = grad_Sigma and beta = Nu. The restriction formula for the Laplacian is Delta_Sigma u = trace_{T Sigma} Hess u - m h beta. Take the trace of the Hessian bound Hess u <= du x du - (1/2)(b^2 + f^2 + |grad u|^2)g over the m-dimensional tangent space, using the ambient identity |grad_g u|^2 = |grad u|^2 + beta^2. This yields 2 Delta_Sigma u <= -(m-2)|grad u|^2 - m(b^2 + f^2) - m beta^2 - 2m h beta. Rearranging, m(h^2 - b^2) >= m f^2 + m(h+beta)^2 + (m-2)|grad u|^2 + 2 Delta_Sigma u. Multiply by v^2 and integrate by parts (the Delta_Sigma u term becomes -2 v <grad v, grad u> up to boundary terms, which vanish for compact support). Since c alpha = 2 and c alpha^2 = m-2, the gradient terms assemble as c|grad v|^2 - 4v<grad v, grad u> + (m-2)v^2|grad u|^2 = c|grad v - alpha v grad u|^2, giving the first inequality. For the second, note |d(Phi restricted to Sigma)|^2 <= m f^2, because in an orthonormal tangent frame the squared norm of the differential of Phi is a sum of m terms each bounded by f^2. Reversing the local normal reverses both h and beta, so the whole expression is globally defined. The same inequalities pass to H^1_0(U) for any open U compactly contained in the interior, because all coefficients are bounded there.

Exercise: Prove the local Sobolev threshold and compactness.
Lemma (Local Threshold). For every open U compactly contained in the interior of Sigma and every epsilon > 0 there is a finite constant C_{epsilon,U} such that (Y_m - epsilon) ||v||_q^2 <= c integral_Sigma |grad_Sigma v|^2 dsigma + C_{epsilon,U} integral_Sigma v^2 dsigma for all v in H^1_0(U); moreover H^1_0(U) embeds compactly into L^2(U). Proof: Aubin-Talenti in flat space is c integral |D phi|^2 dx >= Y_m ( integral |phi|^q dx )^{2/q}. In a small coordinate chart the metric matrix G satisfies (1-delta) I <= G <= (1+delta) I, so comparing inverse metrics and volume densities gives c integral_Sigma |grad_Sigma phi|^2 dsigma >= Y_m ((1-delta)/(1+delta))^{m/2} ||phi||_q^2 for chart-supported phi: the gradient integral is at least (1-delta)^{m/2}(1+delta)^{-1} times the flat one, while the squared L^q norm is at most (1+delta)^{m/q} times the flat one, and m/q = (m-2)/2. Choose delta so the coefficient exceeds Y_m - epsilon. Cover the compact closure of U by finitely many such charts and choose chart-supported Lipschitz functions chi_i with sum chi_i^2 = 1 near that closure. The triangle inequality in L^{q/2} gives ||v||_q^2 <= sum_i ||chi_i v||_q^2, and the product rule gives sum_i |grad(chi_i v)|^2 = |grad v|^2 + v^2 sum_i |grad chi_i|^2. Summing the chart inequalities over i gives the threshold with C_{epsilon,U} = ||sum_i |grad chi_i|^2||_infty, first for compactly supported Lipschitz functions and then (by density) for all of H^1_0(U). For the compactness, extend a bounded H^1_0(U) sequence by zero. In each chart the localized functions chi_i v_j are bounded Euclidean H^1 sequences supported in a fixed compact set, so Rellich compactness gives an L^2-convergent subsequence in each of the finitely many charts; the identity v_j = sum_i chi_i (chi_i v_j) reassembles them into an L^2(U)-convergent subsequence. No boundary regularity of U is needed.

Exercise: Prove that a subcritical quotient is attained.
Lemma (Attainment Below Threshold). For nonempty open U compactly contained in the interior, put Y(U) = inf { Q_Sigma(v) : v in H^1_0(U), ||v||_q = 1 }. If Y(U) < Y_m, then the infimum is attained. Proof: Write Y = Y(U) >= 0 (nonnegativity is the Square Estimate). Take a normalized minimizing sequence v_j. Holder's inequality plus h^2 - b^2 >= -b^2 gives ||v_j||_2^2 <= sigma(U)^{2/m} and c ||grad v_j||_2^2 <= Q_Sigma(v_j) + m b^2 sigma(U)^{2/m}, so the sequence is bounded in H^1_0(U). By the compactness above, after a subsequence, v_j converges weakly in H^1_0(U) and strongly in L^2(U) and almost everywhere to a limit psi. Put t = integral_U |psi|^q dsigma, in [0,1], and z_j = v_j - psi. The Brezis-Lieb lemma gives integral_U |z_j|^q dsigma -> 1 - t. Weak convergence splits the quadratic form: Q_Sigma(v_j) = Q_Sigma(psi) + Q_Sigma(z_j) + o(1). The local threshold plus boundedness of the potential gives Q_Sigma(z_j) >= (Y_m - epsilon) ||z_j||_q^2 - (C + m b^2) ||z_j||_2^2; letting j then epsilon go gives Y >= Q_Sigma(psi) + Y_m (1-t)^{2/q} >= Y t^{2/q} + Y_m (1-t)^{2/q}. If t = 0 this contradicts Y < Y_m. If Y = 0 it forces t = 1. If Y > 0 and 0 < t < 1, strict concavity of s -> s^{2/q} gives t^{2/q} + (1-t)^{2/q} > 1, making the right side strictly bigger than Y, a contradiction. So t = 1 in every case, i.e. psi has full L^q mass and is the minimizer: lower semicontinuity of the gradient energy and strong L^2 convergence of the potential term give Q_Sigma(psi) <= Y, and normalization gives the reverse.

Exercise: Prove the Hypersurface Sobolev inequality.
Theorem (Hypersurface Sobolev). For every Lipschitz v compactly supported in the interior of Sigma, Q_Sigma(v) >= Y_m ( integral_Sigma |v|^q dsigma )^{2/q}; and the same holds for the zero extension of every v in H^1_0(U), U compactly contained in the interior. Proof (Balanced Variations): First let Sigma be connected and boundaryless, and fix nonempty open U compactly contained in the interior with Sigma minus the closure of U nonempty. Suppose, for contradiction, that Y(U) < Y_m, and take the normalized minimizer psi of the Attainment lemma; write Y = Q_Sigma(psi). Push the probability measure |psi|^q dsigma (extended by zero) to M by the immersion. Its support is compact and it has no atoms: an immersion is locally an embedding, so any fiber over a point meets each member of a finite embedding-chart cover in at most one point, and every such fiber has zero area measure. By the Balancing lemma there are fixed p, a such that the coordinate functions w_1,...,w_{n+1} of Phi_{p,a} restricted to Sigma satisfy integral_Sigma |psi|^q w_j dsigma = 0 for every j, with sum_j w_j^2 = 1. These coordinates and u = u_{p,a} have bounded first derivatives near the compact set, so multiplication by w_j or w_j^2 preserves H^1_0(U). First variation: let B_Q be the symmetric bilinear form of Q_Sigma. For a bounded Lipschitz multiplier zeta near the closure of U, differentiating the Rayleigh quotient along psi(1 + s zeta) at s = 0 gives B_Q(psi, psi zeta) = Y integral_Sigma |psi|^q zeta dsigma (the numerator has zero linear term because we are at a constrained minimum and the mass is fixed). Second variation: fix one coordinate w = w_j and set B_w = integral_Sigma |psi|^q w^2 dsigma. Since |w| <= 1 and the weighted mean of w is zero, ||psi(1 + s w)||_q^2 = 1 + (q-1) B_w s^2 + o(s^2) (the linear term vanishes and the quadratic uses the usual expansion of |1 + t|^q at t = 0). Because the quotient is minimized, its second variation is nonnegative, giving Q_Sigma(psi w) >= (q-1) Y B_w. The product rule and the first variation with zeta = w^2 give Q_Sigma(psi w) = B_Q(psi, psi w^2) + c integral_Sigma psi^2 |grad_Sigma w|^2 dsigma = Y B_w + c integral_Sigma psi^2 |grad_Sigma w|^2 dsigma. Subtract the two displays: c integral_Sigma psi^2 |grad_Sigma w|^2 dsigma >= (q-2) Y B_w = c Y B_w. Sum over all coordinates and use sum_j w_j^2 = 1: integral_Sigma psi^2 |d(Phi restricted to Sigma)|^2 dsigma >= Y. On the other hand, the Square Estimate gives Y = Q_Sigma(psi) >= integral_Sigma psi^2 |d(Phi restricted to Sigma)|^2 dsigma + c integral_Sigma |grad_Sigma psi - alpha psi grad_Sigma u|^2 dsigma. Combining the last two displays forces grad_Sigma psi = alpha psi grad_Sigma u almost everywhere on U. Now choose a compactly supported Lipschitz cutoff chi on Sigma that equals one near the closure of U, and consider F = (chi e^{-alpha u}) E_0 psi, where E_0 psi is the zero extension of psi off U. Then F lies in H^1(Sigma) and grad_Sigma F = 0 almost everywhere on all of Sigma, because the product e^{-alpha u} psi is constant by the gradient identity, and the factor chi is harmless near U while F vanishes off U. The local Poincare inequality on coordinate balls makes F constant almost everywhere on each ball; overlapping balls share the constant, so connectedness of Sigma makes F constant on all of Sigma. But F vanishes on the nonempty open set Sigma minus the closure of U, hence F is identically 0, contradicting ||psi||_q = 1. Therefore Y(U) >= Y_m.

Exercise: Finish the Sobolev theorem for an entire compact component and for several components.
Proof (Finish): Still with Sigma connected and boundaryless. Every proper compact support in Sigma is contained in such a set U: choose a coordinate ball outside the support, then finitely many relatively compact neighborhoods covering the support that avoid a smaller ball; this proves the inequality unless the support is the entire compact component. In that remaining case, let D_delta be the image of a Euclidean ball of radius delta in a fixed chart around a chosen point, and remove a point with a cutoff eta_delta equal to 0 in D_delta and 1 outside D_{2delta} with |grad_Sigma eta_delta| <= C/delta. For Lipschitz v, metric comparability gives integral_Sigma |grad_Sigma((1 - eta_delta) v)|^2 dsigma <= 2 integral_{D_{2delta}} |grad_Sigma v|^2 dsigma + C ||v||_infty^2 delta^{m-2}, which tends to 0 because m > 2. Also (1 - eta_delta) v tends to v in L^2 and L^q, and the potential is bounded on the compact component, so the inequality passes to v. This is exactly where the restriction m >= 3 (i.e. n >= 4) is used. Finally, pass to the interior Sigma^0 when Sigma has boundary. A compact support meets only finitely many connected components (the components are open). If a_j = integral over the j-th component of |v|^q dsigma, the connected result on each component gives Q_Sigma(v) >= Y_m sum_j a_j^{2/q} >= Y_m ( sum_j a_j )^{2/q} by concavity of s -> s^{2/q}. For v in H^1_0(U), take its defining approximating sequence: the Local Threshold gives convergence in L^q, and boundedness of the potential near the closure of U gives convergence of the quadratic form. This proves the final assertion. This completes the engine.

Exercise: What is the confined isoperimetric profile, and why confine?
Answer (Confined Profile). Fix an open geodesic ball B = B_{R_B}(o) of finite radius in a Cartan-Hadamard manifold M (only Sec <= 0 is needed here). For 0 < v < Vol(B), define I(v) = inf { P(E) : E is measurable, E differs from a subset of B by a null set, and Vol(E) = v }. Here P(E) is the full ambient perimeter of E, counting any part of its boundary that lies ON the confining sphere dS. Instead of hunting for the best shape freely, we hunt for it among sets trapped inside B. Two good things happen: existence becomes routine because B is compact, and regularity of minimizers is available from known obstacle (variational) theory. When B is very large compared to the set we eventually care about, the confinement is not felt, and the confined optimum I(v) equals the free optimum; this is how the confinement device ends up proving the free theorem. The contact with dS must be counted in P(E), otherwise a minimizer would happily smear itself along the boundary and the derivative identities would fail. The whole job of Section 4 is to show that at almost every volume the derivative I'(v) controls the mean curvature of any minimizer.

Exercise: Prove that a confined minimizer exists and describe its regularity.
Proposition (Confined Regularity). For each v in (0, Vol(B)) there is a minimizer E of I(v). It has a representative for which Gamma = dE (topological boundary) is the support of the perimeter measure and lies in the closure of B. There is a compact set S inside B such that: (1) the Hausdorff dimension of S is at most n-8, and S is empty when n < 8; (2) the regular part Sigma = Gamma minus S is a boundaryless locally C^{1,1} hypersurface, smooth on Sigma intersected with the interior of B; (3) Gamma is C^{1,1} near the confining sphere dS, and S avoids a collar of dS; (4) with dsigma the area on Sigma, the perimeter measure is |D chi_E|(A) = integral over (A intersected with Sigma) dsigma for every Borel A, and P(E) = sigma(Sigma) = I(v). The unit normal on Sigma is the outward normal of E. Proof: Existence: concentric balls supply competitors of every volume, so a minimizing sequence has uniformly bounded perimeter and is supported (up to null sets) in the compact closure of B. The BV compactness lemma below gives an L^1 limit of the same volume with no larger perimeter; it vanishes almost everywhere outside the closure of B and is admissible, so it attains I(v). Regularity: take a minimizing region from Ghomi-Spruck (their Lemma 7.2, items (i),(ii), which do not assume any total-curvature inequality). Those give interior smoothness off a singular set of Hausdorff dimension at most n-8, and C^{1,1} regularity in a collar of the confining sphere (the obstacle boundary), using Stredulinsky-Ziemer's Euclidean obstacle regularity lifted to the Riemannian graph setting; Morgan supplies the interior theory. The singular set has zero m-dimensional Hausdorff measure. Off it chi_E is locally constant almost everywhere, so replace E by the union of the components of M minus K on which chi_E = 1 almost everywhere (K = support of the perimeter measure); this changes only a null set. Every ball centered on K has positive perimeter and contains both phases in positive volume, hence the topological boundary equals K. The area-at-regular-points statement and zero measure of the singular set give the perimeter-area identity. Free patches in the collar are smooth by interior regularity for constant-mean-curvature hypersurfaces, so the actual interior singular set avoids the collar, sits in the compact set Gamma, and is a compact subset of the interior of B. This gives everything claimed. A useful consequence: P(E; B) > 0 for every 0 < |E| < |B|, because D chi_E cannot vanish on the connected ball B without chi_E being constant there, contradicting the volume condition; so a minimizer always has a nonempty smooth free patch.

Exercise: State the two variational/bounded-variation lemmas we use, and what they are for.
Lemma (BV Compactness). Let K be a compact subset of a smooth Riemannian manifold without boundary. If integrable u_j vanish outside K almost everywhere and sup_j ( ||u_j||_1 + |D u_j|(M) ) < infinity, then a subsequence converges in L^1(M). The total variation |D u|(M) is lower semicontinuous under this convergence, and if the u_j are characteristic functions then the limit is a characteristic function. Proof: On a coordinate chart, smooth positivity of the metric and volume density compares the Riemannian BV norm to the Euclidean one. In Euclidean space a BV function v is controlled by its difference quotients: ||v(.,+h) - v||_1 <= |h| |Dv|(R^n), and convolution satisfies ||D(v * rho_eps)||_1 <= |Dv|(R^n) with ||v * rho_eps - v||_1 <= eps C |Dv|(R^n). For fixed eps the convolutions have uniform compact support and bounded first derivatives, hence are precompact in L^1 by finite-mesh approximation; a diagonal subsequence then gives compactness, and total variation being a supremum of continuous divergence integrals is lower semicontinuous. Lemma (Perimeter under Smooth Flows). If F: M -> M is a smooth orientation-preserving diffeomorphism and E has finite perimeter, then P(FE) = integral over the reduced boundary of E of J_{n-1}(dF restricted to the tangent plane) dH^{n-1}, where J_{n-1} is the (n-1)-dimensional Jacobian of dF on that plane. For a smooth flow F_t with velocity Z one has d/dt at 0 of P(F_t E) = integral over the reduced boundary of div_{tangent plane} Z dH^{n-1}, and for a fixed finite family of flows over a fixed compact set there are common C, tau > 0 with |P(F_t E) - P(E)| <= C|t| P(E) for |t| < tau. Proof: The Piola pullback Q_X = J_F (dF)^{-1} X of a test field X satisfies div Q_X = J_F (div X) composed with F, so the vector measure -D chi_{FE} is the pushforward of the density J_F (dF)^{-*} N_E against |D chi_E|; its polar direction is that density and its total variation is the pushforward of its norm, which linear algebra identifies with the (n-1)-Jacobian. Differentiating the Gram determinant at t = 0 gives the tangential divergence; uniform boundedness of the Jacobians and their time-derivatives over all tangent planes above a fixed compact set justifies differentiating under the integral and yields the linear bound. These two lemmas are the workhorses: compactness gives minimizers and compact families of competitors; the flow formula shows smooth deformations change volume at a controlled speed while changing boundary at a controlled cost, which is what the Lipschitz and curvature arguments need.

Exercise: Prove that the confined profile is locally Lipschitz.
Lemma (Profile Lipschitz). The profile I is positive and locally Lipschitz on (0, Vol(B)). Proof: Positivity was noted (a minimizer has a free patch, so I(v) > 0). Fix a compact interval [v_-, v_+] inside (0, Vol(B)). Concentric balls give finite-perimeter competitors at every volume, and their boundary area varies continuously with radius; so there is a finite C_0 with I(v) <= C_0 for v in [v_-, v_+]. Consider the family K of characteristic functions chi_F with v_- <= |F| <= v_+, supported in B up to null sets, and P(F) <= C_0. By BV compactness on a compact neighborhood of the closure of B, together with lower semicontinuity and L^1 continuity of volume, K is compact in L^1. For each F in K there is a smooth field Z_F compactly supported in B with A_{Z_F}(F) := integral_F div Z_F dVol not equal to 0; otherwise D chi_F would vanish in B, contradicting v_- <= |F| <= v_+. Flip the sign so A_{Z_F}(F) > 0. The functional A_Z is L^1-continuous for fixed Z, so a finite subcover of K gives smooth fields Z_1,...,Z_k supported in B and a delta > 0 with max_j A_{Z_j}(F) >= 2 delta for every F in K. Let Psi_j^t be the flow of Z_j; the flows preserve B in both time directions. Writing G_j(t,F) = |Psi_j^t(F)| = integral_F J_j(t,x) dVol(x), we have d/dt G_j(0,F) = A_{Z_j}(F); smoothness of the finitely many flows gives a uniform bound on the second time-derivative on a fixed compact set, hence a common tau > 0 with |d/dt G_j(t,F) - A_{Z_j}(F)| <= delta for |t| <= tau and all F in K and all j. For the index where A_{Z_j}(F) >= 2 delta, the function t -> G_j(t,F) has derivative at least delta on [-tau,tau], so every w with |w - |F|| < delta tau is attained as |Psi_j^t(F)| for some t with |t| <= delta^{-1}|w - |F||. The flow Jacobian formula gives |P(Psi_j^t F) - P(F)| <= C |t| P(F) <= C C_0 |t| for |t| <= tau, with C uniform over the finitely many flows and all tangent planes above the closure of B. Hence changing the volume by Delta v costs at most C C_0 delta^{-1} |Delta v| in perimeter. Apply this first to a minimizer at v and then to a minimizer at w (both in K when v, w are in [v_-,v_+] and |v - w| < delta tau); the resulting competitors give |I(v) - I(w)| <= C C_0 delta^{-1} |v - w|. This is the claimed local Lipschitz bound; no uniform regularity of minimizers is required.

Exercise: Prove the mean-curvature bounds at differentiability volumes.
Lemma (Profile Curvature). Let v be a differentiability point of I in (0, Vol(B)), and let E be a minimizer as in the regularity proposition. With mh = div_Sigma N the trace curvature and h0 = I'(v)/m, one has I'(v) > 0, h = h0 on the free part Sigma intersected with the interior of B, and 0 <= h <= h0 almost everywhere on all of Sigma; in particular h^2 <= h0^2. Proof: For a smooth flow Psi^t with E_t = Psi^t(E), set P(t) = P(E_t) and V(t) = |E_t|. Whenever the flow is admissible (keeps the set inside B), P(t) >= I(V(t)) with P(0) = I(v) and V(0) = v. For a field supported on a free regular patch, admissibility holds for both signs of t; differentiating at the minimum gives P'(0) = I'(v) V'(0). The first variation formulas are V'(0) = integral_Sigma <Z,N> dsigma and P'(0) = m integral_Sigma h <Z,N> dsigma. Arbitrary smooth compactly supported normal speeds on the patch therefore give mh = I'(v), i.e. h = h0 on every free patch; such patches exist because P(E;B) > 0. For the sign of h0, put rho(x) = d(o,x) and F = rho^2/2, which is smooth on all of M including o, with Hess F = d rho x d rho + rho Hess rho >= d rho x d rho + (g - d rho x d rho) = g by the distance Hessian comparison. The flow of -grad F is Psi^t(x) = exp_o( e^{-t} exp_o^{-1}(x) ), which sends B into itself for t >= 0. The Jacobian formulas give V'(0) = -integral_E Delta F dVol <= -n v < 0, and P'(0) = -integral over the reduced boundary of trace_{tangent plane}(Hess F) dH <= -m P(E) < 0, since trace of Hess F restricted to any m-plane is at least m. These formulas act on the reduced boundary as a rectifiable set, so they include any contact portion and need no smoothness at singular points, and the integrands are bounded on a fixed compact neighborhood of the closure of B. The one-sided derivative of P(t) >= I(V(t)) at t = 0 now gives P'(0) - I'(v) V'(0) >= 0; dividing by V'(0) < 0 reverses the inequality, so I'(v) >= P'(0)/V'(0) > 0. It remains to control the contact set C = Gamma intersected with dS. At every point of C the C^1 boundary of E is tangent to the sphere and has the same outward normal N = grad rho (the region sits on the inner side). At almost every such point, in normal coordinates with the common outward normal pointing up, write the boundary of E and the sphere as graphs u and beta: they have equal values and first derivatives and u <= beta, hence D^2 u(x) <= D^2 beta(x). The second fundamental forms for the outward normal are -D^2 u(x) and -D^2 beta(x), so by Hessian comparison on the sphere, II_Gamma(x) >= II_{dS}(x) and h(x) >= 1/R_B > 0 for almost every x in C. Choose an open collar U of dS whose closure avoids S and o and in which Gamma is C^{1,1}. For phi >= 0 in C_c(U), the field Z = -phi grad rho extended by zero is admissible for t >= 0 (rho is nonincreasing along its trajectories). The one-sided variation inequality gives m integral_Sigma (h - h0) <Z,N> dsigma >= 0. The integrand vanishes on the free part (where h = h0), and <Z,N> = -phi on C, so integral_C (h - h0) phi dsigma <= 0 for all nonnegative phi; smooth nonnegative tests determine the sign of a Radon measure, hence h <= h0 almost everywhere on C, whether or not C has interior. Combining h = h0 on the free part, h >= 1/R_B > 0 on C, and h <= h0 on C with h0 > 0 gives 0 <= h <= h0 and h^2 <= h0^2 almost everywhere on Sigma.

Exercise: State and prove the deletion estimate for a geodesic ball.
Lemma (BV Deletion). Let E have finite volume and perimeter in a complete smooth Riemannian manifold, let x be a point, and let 0 < r < inj(x). Then P(E minus B_r(x)) <= P(E; M minus closure(B_r(x))) + H^{n-1}(d B_r(x)), where the last term is the (n-1)-area of the distance sphere. For all but countably many radii in any bounded interval one has P(E; d B_r(x)) = 0, so the perimeters on the two open sides add up to P(E). Proof: For a bounded Lipschitz scalar phi, distributional testing gives D(phi chi_E) = phi D chi_E + chi_E grad phi dVol. Prove it first for smooth phi by approximating on a compact neighborhood of the support of the test field, uniformly and in W^{1,1}; uniform convergence handles the measure term and W^{1,1} convergence the second. Now take phi_eps(y) = min{1, max{0, (d(x,y) - r)/eps}} with r + eps < inj(x); its derivative is supported in a thin annulus. The products phi_eps chi_E converge in L^1 to chi_{E minus B_r(x)} because the smooth sphere has ambient volume zero, so lower semicontinuity and the product formula give P(E minus B_r(x)) <= liminf_eps ( integral phi_eps d|D chi_E| + |E intersected with (B_{r+eps}(x) minus B_r(x))| / eps ) <= P(E; M minus closure(B_r(x))) + (d/ds at s=r) |B_s(x)|. Polar coordinates identify that last derivative with the area of the distance sphere. Finally, a finite measure can put positive mass on only countably many pairwise disjoint distance spheres.

Exercise: Prove the area growth bound near the singular set.
Lemma (Area Growth). Let 0 < v < Vol(B) and let E be a confined minimizing region of volume v. There are constants C < infinity and r_0 > 0 (depending on E and B) with sigma( Sigma intersected with B_r(x) ) <= C r^m for every x in the singular set S and every 0 < r < r_0. Proof: Assume S is nonempty. Because P(E;B) > 0 and the perimeter-area identity holds, there is a free smooth boundary point; choose a neighborhood W of it whose closure sits in B minus S and on which Gamma is smooth. Extend a nonnegative, nonzero normal variation of a smaller patch to a smooth field Z compactly supported in W; with N the outward normal, the number a := integral_{Sigma intersected with W} <Z,N> dsigma is positive. Let phi_t be the flow and G(t) = Vol(phi_t(E)) - Vol(E); the volume Jacobian formula gives G'(0) = a, and by continuity G'(t) >= a/2 for 0 <= t <= t_0 (some small t_0). The flow preserves W and B and fixes the complement of W. Whenever a finite-perimeter set F agrees with E almost everywhere on W, the volume Jacobian minus one is supported in W, so Vol(phi_t(F)) - Vol(F) = G(t); and the perimeter change P(phi_t F) - P(F) equals integral over the reduced boundary within W of (J_m(D phi_t restricted to the tangent plane) - 1), whose absolute value is at most L P(E;W) t uniformly over all tangent m-planes above the compact support of the flow and all 0 <= t <= t_0. Hence every volume deficit 0 <= delta <= a t_0 / 2 can be repaired by a flow time 0 <= t <= 2 delta / a at perimeter cost at most C_R delta, where C_R = 2 L P(E;W) / a is independent of F as long as F = E almost everywhere on W. Compactness of S and its being disjoint from the closure of W and from dS give rho_* > 0 such that B_{2 rho_*}(x) lies in B minus the closure of W for every x in S; decreasing rho_* and using smoothness of the metric on a compact neighborhood of S, we get uniform constants C_V, C_A with Vol(B_r(x)) <= C_V r^n and H^m(d B_r(x)) <= C_A r^m for x in S and 0 < r < rho_*; decrease rho_* once more so C_V rho_*^n <= a t_0 / 2. Fix x in S. For almost every r in (0, rho_*) we have P(E; d B_r(x)) = 0; at such an r set F = E minus B_r(x). The deletion lemma applied to E gives P(F) <= P(E; M minus closure(B_r(x))) + H^m(d B_r(x)). The volume deficit is delta = Vol(E intersected with B_r(x)) <= C_V r^n, and F agrees with E on W; repair the deficit by flow time t <= 2 delta / a. The repaired set stays in B and has volume v, so minimality, the perimeter-change bound, and the deletion bound give P(E;B_r(x)) <= H^m(d B_r(x)) + C_R delta <= C_A r^m + C_R C_V r^n <= C_0 r^m, where C_0 = C_A + C_R C_V rho_* (we used n = m+1 so r^n <= rho_* r^m), uniformly in x in S. For an arbitrary 0 < r < rho_* / 2 choose a good radius rho in (r, 2r); monotonicity of the perimeter measure gives P(E;B_r(x)) <= P(E;B_rho(x)) <= 2^m C_0 r^m. Combined with the perimeter-area identity this gives the claim with r_0 = rho_* / 2.

Exercise: Prove the cutoff construction around a set of zero capacity.
Lemma (Capacity Cutoffs). Let Gamma be a compact subset of a smooth Riemannian manifold, and S a closed subset of Gamma such that Sigma = Gamma minus S is an embedded, boundaryless, locally C^{1,1} hypersurface of dimension m >= 3 and finite area. Assume H^{m-2}(S) = 0 and that, for some C, r_0 > 0, the area bound sigma( Sigma intersected with B_r(x) ) <= C r^m holds for all x in S and 0 < r < r_0. Then there are Lipschitz functions eta_j compactly supported in Sigma with 0 <= eta_j <= 1, eta_j(x) -> 1 for every x in Sigma, and integral_Sigma |grad_Sigma eta_j|^2 dsigma -> 0. If S is empty then eta_j = 1 works in every dimension. Proof: If S is empty, Sigma = Gamma is compact and eta_j = 1 is admissible. Otherwise, use H^{m-2}(S) = 0 as follows. For each j there is a finite covering of S by balls B_{r_ji}(x_ji) with x_ji in S, with 0 < r_ji < min(2^{-j}, r_0/4), and with sum_i r_ji^{m-2} < 2^{-j}: Hausdorff nullity first gives countable covers with arbitrarily small radii and sums, recentering at points of S at the cost of a fixed enlargement, and compactness extracts a finite subcover. Put theta(t) = min{1, max{0, t - 1}} and eta_ji(x) = theta( d(x, x_ji)/r_ji ), then eta_j = min_i eta_ji restricted to Sigma. Each ambient cutoff vanishes on B_{r_ji}(x_ji), equals one outside B_{2 r_ji}(x_ji), and has Lipschitz constant at most 1/r_ji; a finite minimum is Lipschitz, and almost everywhere |grad_Sigma eta_j|^2 <= sum_i |grad_Sigma eta_ji|^2. There is no restriction on overlaps. Applying the area bound at the doubled radii gives integral_Sigma |grad_Sigma eta_j|^2 dsigma <= sum_i r_ji^{-2} sigma( Sigma intersected with B_{2 r_ji}(x_ji) ) <= 2^m C sum_i r_ji^{m-2} < 2^m C 2^{-j}, which tends to zero. The open balls cover a neighborhood of S on which eta_j vanishes, so its support lies in a compact subset of Gamma minus S; embedded graph charts identify the subspace topology with the manifold topology of Sigma, so that support is compact in Sigma. For fixed x in Sigma, closedness of S gives d(x,S) > 0, and as the maximum radius tends to zero the doubled balls miss x, so eta_j(x) = 1. No completeness of Sigma and no finiteness of its components is assumed.

Exercise: Prove the constant test.
Lemma (Constant Test). Let v in (0, Vol(B)) be a differentiability point of I and E a confined minimizer at volume v. Then m integral_Sigma (h^2 - b^2) dsigma >= Y_m I(v)^{(m-2)/m}. Proof: The curvature lemma gives 0 <= h <= h_0 := I'(v)/m < infinity almost everywhere on Sigma. If S is nonempty, the singular-dimension bound (from the regularity proposition) gives dim_H S <= n - 8 = m - 7 < m - 2, hence H^{m-2}(S) = 0. The area-growth lemma, compactness of Gamma, and the embedded C^{1,1} structure verify the hypotheses of the capacity-cutoff lemma, which supplies eta_j with 0 <= eta_j <= 1, eta_j -> 1, and integral |grad_Sigma eta_j|^2 -> 0; when S is empty take eta_j = 1. Apply the Hypersurface Sobolev inequality to these cutoffs: c integral_Sigma |grad_Sigma eta_j|^2 dsigma + m integral_Sigma (h^2 - b^2) eta_j^2 dsigma >= Y_m ( integral_Sigma eta_j^q dsigma )^{2/q}. The area is I(v) < infinity, and |h^2 - b^2| <= h_0^2 + b^2 almost everywhere, so the potential dominates by an integrable function; the Dirichlet term tends to zero, eta_j^2 -> 1 pointwise, and eta_j^q -> 1 pointwise. Dominated convergence gives m integral_Sigma (h^2 - b^2) dsigma >= Y_m ( integral_Sigma 1 dsigma )^{2/q} = Y_m I(v)^{2/q} = Y_m I(v)^{(m-2)/m}, since 2/q = (m-2)/m. The same construction covers every dimension under consideration: when 4 <= n <= 7 the singular set is empty, and when n = 8 its compact zero-dimensional structure still admits the covers.

Exercise: Derive the profile differential inequality.
Proposition (Profile ODE). For almost every v in (0, Vol(B)), I'(v) >= m sqrt( b^2 + ( s_m / I(v) )^{2/m} ). Proof: The profile is locally Lipschitz, hence differentiable almost everywhere. At a differentiability volume choose a minimizer; the curvature lemma gives h_0 = I'(v)/m > 0 and h^2 <= h_0^2 almost everywhere on its regular boundary. The constant test then gives m (h_0^2 - b^2) I(v) >= Y_m I(v)^{(m-2)/m}. Since I(v) > 0 and Y_m = m s_m^{2/m}, divide by m I(v): h_0^2 - b^2 >= (Y_m/m) I(v)^{(m-2)/m - 1} = s_m^{2/m} I(v)^{-2/m} = ( s_m / I(v) )^{2/m}. Hence h_0^2 >= b^2 + ( s_m / I(v) )^{2/m}, and taking the positive square root (h_0 > 0) proves the differential inequality.

Exercise: Define the inverse model functions and compute the derivative that turns the ODE into an exact integral.
Answer (Inverse Model). Recall A_b(r) = s_m sn_b(r)^m and V_b(r) = s_m integral_0^r sn_b(t)^m dt, both strictly increasing from zero to infinity. For A > 0 set R_b(A) = A_b^{-1}(A) (the radius whose sphere has area A) and W_b(A) = V_b(R_b(A)) (the ball volume at that radius). Then W_b is strictly increasing (a composition of two strictly increasing functions). Its derivative is W_b'(A) = V_b'(R_b(A)) * R_b'(A). Now V_b'(r) = s_m sn_b(r)^m, and differentiating A_b(R_b(A)) = A gives R_b'(A) = 1/(s_m * m * sn_b(r)^{m-1} cs_b(r)) with r = R_b(A). So W_b'(A) = sn_b(r)/(m cs_b(r)). To recognize this as the reciprocal of the profile derivative, use A = s_m sn_b(r)^m, so (s_m/A)^{2/m} = 1/sn_b(r)^2, and b^2 + (s_m/A)^{2/m} = (b^2 sn_b^2 + 1)/sn_b^2 = cs_b^2/sn_b^2 by the identity cs_b^2 - b^2 sn_b^2 = 1. Hence sqrt(b^2 + (s_m/A)^{2/m}) = cs_b/sn_b, and W_b'(A) = 1/( m sqrt( b^2 + (s_m/A)^{2/m} ) ). This is exactly the reciprocal of the right side of the Profile ODE.

Exercise: Integrate the ODE and prove the comparison for bounded sets in dimensions at least four.
Theorem (Normalized Comparison, n >= 4). Let n >= 4 and b in {0,1}. If M^n is complete, simply connected, smooth, with Sec_M <= -b^2, then every bounded measurable set E of finite perimeter and positive volume satisfies P(E) >= I_b(Vol(E)), where I_b is the model profile. Proof: First, integrate the ODE. On any compact subinterval [epsilon, v] of (0, Vol(B)), the positive Lipschitz function I takes values in a compact subset of (0, infinity) on which W_b is continuously differentiable; hence W_b composed with I is absolutely continuous there. The Profile ODE, the chain rule, and the derivative computation above give (W_b o I)'(s) = W_b'(I(s)) I'(s) >= 1 for almost every s in [epsilon, v]. Integrating, W_b(I(v)) >= W_b(I(epsilon)) + (v - epsilon) >= v - epsilon, and letting epsilon go to zero (using only nonnegativity of W_b(I(epsilon))) we get W_b(I(v)) >= v for every v in (0, Vol(B)). Now take a bounded measurable set E of finite perimeter and positive volume V = Vol(E). Choose radii 0 < R_0 < R with E contained in B_{R_0}(o) up to a null set, and put B = B_R(o); the annulus has positive volume, so Vol(B) > V. Then E is an admissible competitor in the confined profile at volume V (it lies in B up to a null set), so P(E) >= I(V); this step uses the full ambient perimeter of E and imposes no regularity on its boundary. Let r > 0 be determined by V_b(r) = V, so that A_b(r) = I_b(V) is the model boundary of the equal-volume ball. Then W_b(A_b(r)) = V_b(r) = V. Since W_b(I(V)) >= V = W_b(A_b(r)) and W_b is strictly increasing, I(V) >= A_b(r) = I_b(V). Combining P(E) >= I(V) >= I_b(V) proves the theorem for bounded sets when n >= 4. The dimension restriction appears only because the Hypersurface Sobolev inequality (hence the constant test and the ODE) was proved for m = n-1 >= 3; dimensions 2 and 3 are handled separately below.

Exercise: Show the model profile is concave and subadditive, and use it to handle disconnected domains and dimension three.
Answer (Concavity and Subadditivity). For b >= 0 and r = V_b^{-1}(V) > 0, differentiate the model profile: I_b'(V) = m cs_b(r)/sn_b(r) > 0, and d/dr( cs_b(r)/sn_b(r) ) = -1/sn_b(r)^2 < 0, so I_b is concave. Because I_b(0) = 0, concavity between 0 and u + v gives I_b(u) >= (u/(u+v)) I_b(u+v) and I_b(v) >= (v/(u+v)) I_b(u+v); adding, I_b(u) + I_b(v) >= I_b(u+v), which is subadditivity. Consequence: if a domain splits into finitely many connected pieces of volumes u_1, ..., u_k with perimeters p_1, ..., p_k, then subadditivity gives sum_i p_i >= sum_i I_b(u_i) >= I_b(u_1 + ... + u_k), so proving the comparison for each connected piece proves it for the whole disconnected domain at total volume. Now dimension three. Kleiner's Theorem 2 (1992, p.~38) states the model comparison for every upper sectional-curvature bound kappa <= 0 on a complete simply connected 3-manifold, for arbitrary compact smooth domains. Apply it to each connected component of a relatively compact smooth domain and use subadditivity (and continuity) to reach disconnected domains. This settles every compact smooth domain in dimension three; passing to general finite-perimeter sets is done below by approximation.

Exercise: Use the classical surface inequality to settle dimension two.
Answer (Weil-Bol, Dimension Two). In dimension two the classical Weil-Bol inequality, in the smooth-disk formulation of Chavel-Feldman (Isoperimetric Inequality (I), p.~83), says: for a smooth disk of area V and boundary length L in a surface with Gaussian curvature K <= kappa <= 0, one has L^2 >= 4 pi V - kappa V^2. Direct evaluation of the two-dimensional model functions gives the model profile I_{sqrt(-kappa)}(V) = sqrt( 4 pi V - kappa V^2 ): indeed a disk of radius r in curvature kappa = -b^2 has area V_b(r) = 2 pi integral_0^r sn_b(t) dt and perimeter 2 pi sn_b(r), and one checks that peri^2 = 4 pi Vol - kappa Vol^2. This disk statement suffices for arbitrary relatively compact smooth domains as follows. Take one connected component; fill in its holes while keeping its outer boundary. The resulting smooth disk has at least the original area (we added area) and no greater boundary length (we removed inner boundary loops). The disk inequality at the filled area, combined with monotonicity of I_{sqrt(-kappa)}, bounds the original component's boundary length below by the model profile at the ORIGINAL area. Since the surface is a Cartan-Hadamard surface, it is topologically the plane, so this filling is legitimate. Summing over components and using subadditivity settles the disconnected case; the filled disks may overlap, which is harmless because we apply the inequality to each component before summing.

Exercise: Prove that a bounded finite-perimeter set is a strict limit of smooth domains.
Lemma (Smooth BV Approximation). Let M be a complete connected smooth Riemannian manifold without boundary. Let E have finite ambient perimeter and suppose E is contained in a compact set K up to a null set. For every open neighborhood U of K with compact closure there are relatively compact open sets E_j compactly contained in U with smooth boundary such that |E_j symmetric-difference E| -> 0 and P(E_j) -> P(E). The E_j need not be connected. Proof: Guneyu-Pallara's strict smooth approximation (their Theorem 2.12, valid on complete smooth manifolds without curvature hypotheses) provides smooth compactly supported functions f_j with ||f_j - chi_E||_1 -> 0 and integral_M |grad f_j| dVol -> P(E). Taking real parts cannot increase the L^1 error or the gradient integral, and lower semicontinuity gives the reverse limiting bound, so we may assume the f_j real. Choose eta in C_c(U) with 0 <= eta <= 1 and eta = 1 near K, and put u_j = eta f_j. Then u_j -> chi_E in L^1, and chi_E grad eta = 0 almost everywhere, so integral_M |grad u_j| dVol <= integral_M |grad f_j| dVol + ||grad eta||_infty ||f_j - chi_E||_1; lower semicontinuity shows the left side also converges to P(E). All u_j now have support in one compact subset of U. Write e_j = ||u_j - chi_E||_1 and choose 0 < delta_j < 1/2 with delta_j -> 0 and e_j/delta_j -> 0. For delta_j < t < 1 - delta_j, pointwise comparison with chi_E gives |{u_j > t} symmetric-difference E| <= delta_j^{-1} e_j. The ordinary coarea formula for the smooth function u_j gives integral_{delta_j}^{1-delta_j} P({u_j > t}) dt <= integral_M |grad u_j| dVol. By Sard's theorem almost every t is a regular value, so we can pick such a t_j in (delta_j, 1-delta_j) with P({u_j > t_j}) <= (1/(1-2 delta_j)) integral_M |grad u_j| dVol + 1/j. Set E_j = {u_j > t_j}; its boundary is smooth and compact in U. The threshold estimate gives L^1 convergence of characteristic functions, and the perimeter bound with lower semicontinuity gives P(E_j) -> P(E).

Exercise: Show that a lower bound valid for bounded sets extends to all finite-volume sets.
Lemma (Finite-Volume Cutoff). Let M be a complete connected smooth Riemannian manifold without boundary. Let J: [0, infinity) -> [0, infinity) be continuous and increasing with J(0) = 0. Suppose P(F) >= J(|F|) for every bounded finite-perimeter set F. Then the same holds for every measurable E of finite volume and finite ambient perimeter. Proof: We need two BV identities. First, the product rule D(varphi u) = varphi Du + u grad varphi dVol for compactly supported Lipschitz varphi and bounded BV u; it follows from the smooth-multiplier case by uniform approximation of varphi with gradients converging in L^1. Second, the coarea identity |Du|(M) = integral_0^1 P({u > t}) dt for 0 <= u <= 1; for ">=", apply strict smooth approximation (Guneyu-Pallara) and take a subsequence whose level sets converge in L^1 for almost every t, using lower semicontinuity and Fatou; for "<=", write u = integral_0^1 chi_{{u>t}} dt, test against div X for |X| <= 1, and take the perimeter supremum. This also shows almost every level has finite perimeter. Fix o in M and, for R > 0, define varphi_R(x) = min{1, max{0, R + 1 - d(o,x)}}. By completeness its support is compact; it equals 1 on B_R(o), vanishes outside B_{R+1}(o), and is 1-Lipschitz. The product rule and coarea give integral_0^1 P({varphi_R chi_E > t}) dt = |D(varphi_R chi_E)|(M) <= integral_M varphi_R d|D chi_E| + integral_E |grad varphi_R| dVol <= P(E) + |E minus B_R(o)|. Almost every level is a bounded finite-perimeter set, and for every 0 < t < 1 its volume is at least v_R = |E intersected with B_R(o)|. The hypothesis and monotonicity of J bound the left side below by J(v_R). Since v_R -> |E| and |E minus B_R(o)| -> 0 while J is continuous, the inequality passes to E.

Exercise: Assemble the full comparison theorem for all dimensions and all curvature bounds.
Theorem (Generalized Comparison). Let n >= 2, kappa <= 0, and (M^n, g) complete, simply connected, smooth, with Sec_g <= kappa. Every measurable E with Vol_g(E) < infinity and P_g(E) < infinity satisfies P_g(E) >= I_{sqrt(-kappa)}( Vol_g(E) ). Proof: Null sets have zero perimeter and I_b(0) = 0, so ignore them. For b in {0,1}, the normalized comparison is known for BOUNDED finite-perimeter sets: for n >= 4 this is the Normalized Comparison theorem; for n = 2 and n = 3 combine the classical componentwise results, the Smooth BV approximation (which preserves perimeter and volume), and continuity of I_b to pass to arbitrary bounded finite-perimeter E. The Finite-Volume Cutoff with J = I_b then removes boundedness, giving the comparison for all finite-volume finite-perimeter E and b in {0,1}. For an arbitrary negative bound, put b = sqrt(-kappa) > 0 and rescale the metric to g-tilde = b^2 g. Then Sec_{g-tilde} = b^{-2} Sec_g <= -1; volumes and perimeters scale as |E|_{g-tilde} = b^n |E|_g and P_{g-tilde}(E) = b^{n-1} P_g(E). The last holds for ambient BV perimeter because constant metric rescaling leaves divergence unchanged while dVol_{g-tilde} = b^n dVol_g and |X|_{g-tilde} = b |X|_g, so substituting Y = bX in the dual definition multiplies the supremum by b^{n-1}. The model functions match: V_b(r) = b^{-n} V_1(br), A_b(r) = b^{-(n-1)} A_1(br), and consequently I_b(V) = b^{-(n-1)} I_1(b^n V). Apply the normalized (b = 1) result to g-tilde and divide by b^{n-1}: P_g(E) = b^{-(n-1)} P_{g-tilde}(E) >= b^{-(n-1)} I_1( b^n |E|_g ) = I_b(|E|_g). The flat case kappa = 0 was already covered directly, without a curvature limit. This proves the main comparison theorem.

Exercise: Prove that an equality set is a local perimeter almost-minimizer with a regular C^{1,1} boundary of constant mean curvature.
Proposition (Equality Boundary). Let M^n be complete, simply connected, smooth, n >= 2, Sec_M <= 0. Suppose a bounded measurable set E has positive volume V, finite ambient perimeter A, and equality A = n omega_n^{1/n} V^{(n-1)/n}. Put m = n-1, R = (V/omega_n)^{1/n}, and H_0 = m/R = (n-1)/R; note also that A = s_m R^m with s_m = n omega_n. Then the perimeter support Gamma (= support of |D chi_E|) is compact and splits as a disjoint union Gamma = Sigma union Z with: (1) Sigma is relatively open, boundaryless, embedded, locally C^{1,1}, and Z is compact with Hausdorff dimension at most n-8 (in particular Z is empty when n < 8); (2) the side occupied by E defines an outward unit normal N on Sigma, and the perimeter measure is dsigma restricted to Sigma, with sigma(Sigma) = A = s_m R^m and the closure of Sigma equal to Gamma; (3) there are C, r_0 with sigma(Sigma intersected with B_r(x)) <= C r^m for x in Gamma and 0 < r < r_0; (4) the outward trace mean curvature equals H_0 almost everywhere on Sigma. These statements concern the perimeter support and the measurable class of E, not the topological boundary of an arbitrary representative. Proof: Almost-minimality. Write theta = (n-1)/n and Lambda = A/V. For every bounded finite-perimeter set F, the Generalized Comparison applied to both E and F gives P(E) - P(F) <= A - n omega_n^{1/n} |F|^theta = A ( 1 - (|F|/V)^theta ) <= Lambda |E symmetric-difference F|, where the last inequality uses z^theta >= min{z,1} for z >= 0 and the fact that the difference of volumes is bounded by the volume of the symmetric difference. If the symmetric difference of E and F lies in a compact subset of a bounded open set O, locality of perimeter cancels the part outside O, so P(E;O) <= P(F;O) + Lambda |E symmetric-difference F|; and it suffices to consider competitors with P(F;O) < infinity (agreement with E near the complement of O then gives finite global perimeter). Local regularity. Apply Antonelli-Pasqualetto-Pozzetta's local smooth-manifold conclusion (their Corollary 1.6). To check its quasi-perimeter hypothesis, on measurable subsets of O define G(F) = Lambda |F symmetric-difference (E intersected with O)|; this is finite at the empty set and |G(F_1) - G(F_2)| <= Lambda |F_1 symmetric-difference F_2|, so its continuity exponent is 1 > 1 - 1/n, exactly Condition (1.3) of that reference. The localized inequality says E minimizes P + G under compact modifications, in particular under the volume-preserving modifications of their Definition 1.2. For a ball O centered on Gamma, the remaining hypothesis P(E;O) > 0 follows from the definition of support. The cited conclusion applies in every dimension n >= 2 and needs no global lower Ricci bound. It provides a locally C^{1,alpha} regular part of full perimeter, a relatively closed singular part of Hausdorff dimension at most n-8, and local upper area growth; decrease alpha to lie in (0,1) if needed. These are statements about the perimeter support: take the density-one representative, whose boundary equals that support (the perimeter measure sits on the reduced boundary, and a boundary point of this representative is an essential boundary point, so perimeter cannot vanish nearby). Since E is bounded, Gamma and its closed singular subset Z are compact. The local area estimates become a uniform bound on Gamma by a finite cover. The occupied side and normal. Near a regular point the whole support is a graph dividing a coordinate neighborhood into two connected sides; on either side D chi_E = 0, so chi_E is constant almost everywhere there (mollify in coordinate balls; overlapping balls agree). The two side constants must differ, else chi_E would be constant across the neighborhood; so the occupied side defines a consistent outward normal, and the graph divergence formula identifies perimeter with induced area. Together with the full-perimeter conclusion this gives the perimeter-area identity and shows Sigma is dense in Gamma. The graph equation. Choose graph coordinates (y,z) with z = u(y) on Sigma and E below the graph up to a null set. Let G(y,z) be the ambient metric matrix and W = sqrt(det G). The graph area integrand is L_g(y,z,p) = W(y,z) [ (-p,1)^T G(y,z)^{-1} (-p,1) ]^{1/2}. For a compactly supported smooth zeta, the graph u + t zeta gives admissible sets for both signs of small t. The deficit P(F) - n omega_n^{1/n} |F|^theta is nonnegative and vanishes at F = E; its volume derivative is H_0 = n omega_n^{1/n} theta V^{theta-1} = (n-1)/R. Define B(y,z,p) = d/dz L_g(y,z,p) - H_0 W(y,z). Differentiating area and volume between the two graphs gives the weak equation integral [ d/dp L_g(y,u,Du) . D zeta + B(y,u,Du) zeta ] dy = 0. The matrix d^2/dp^2 L_g is positive definite (the Hessian of a positive-definite norm is positive transverse to its radial kernel, and a nonzero slope variation is not radial at (-p,1)), so the equation is uniformly elliptic. The graph-regularity lemma proved next then gives u in C^{1,1}; this does not require an advance bound on second derivatives. The signed curvature. On a C^{1,1} graph the tangential divergence splits as div_{T Sigma} Y = div_Sigma(Y^T) + H<Y,N>; take Y = zeta d/dz, use that the area derivative is integral div_{T Sigma} Y dsigma and the volume derivative is its outward flux, and the graph equation becomes integral (H - H_0)<d/dz,N> zeta dsigma = 0. The factor <d/dz,N> is strictly positive because E is on the lower side, so arbitrary compactly supported zeta give H = H_0 almost everywhere. The same global deficit produced H_0 on every patch, so the signed outward curvature is the same on every component.

Exercise: Prove the local graph-regularity lemma (a stationary graph is C^{1,1}).
Lemma (Graph Regularity). Let Omega be open in R^m, m >= 1, and let u in C^{1,alpha}_loc(Omega) for 0 < alpha < 1. Let G(y,z) be the matrix of a smooth Riemannian metric on a neighborhood of the graph of u in R^m x R, and put W(y,z) = sqrt(det G(y,z)) and L_g(y,z,p) = W(y,z) [ (-p,1)^T G(y,z)^{-1} (-p,1) ]^{1/2}. Fix H_0 > 0 and set B(y,z,p) = d/dz L_g(y,z,p) - H_0 W(y,z). If integral_Omega [ d/dp L_g(y,u,Du) . D zeta + B(y,u,Du) zeta ] dy = 0 for all zeta in C_c^infty(Omega), then u in C^{1,1}_loc(Omega). Proof (Difference Quotients). Positivity of d^2/dp^2 L_g was checked after the graph equation. Shrink the chart so that its arguments and the segments between any two nearby arguments lie in one compact ellipticity region. Choose a Euclidean ball B_{4rho} compactly contained in the parameter domain. For a coordinate unit vector e and 0 < delta < rho, put v(y) = (u(y + delta e) - u(y))/delta on B_{3rho}. Each v is C^1 (no bound on Dv is yet known), and the identity v(y) = integral_0^1 d_e u(y + s delta e) ds gives a uniform C^{0,alpha} bound. Subtract the graph equations for the two points and divide by delta; setting T_s(y) = (1-s)(y,u(y),Du(y)) + s(y + delta e, u(y + delta e), Du(y + delta e)), the result is a weak equation integral [ (aDv + f) . D zeta + (b . Dv + d) zeta ] dy = 0 with a = integral_0^1 d^2/dp^2 L_g(T_s) ds, b = integral_0^1 d/dp B(T_s) ds, f = integral_0^1 ( d_e d/dp L_g(T_s) + d/dz d/dp L_g(T_s) v ) ds, and d = integral_0^1 ( d_e B(T_s) + d/dz B(T_s) v ) ds, where d_e differentiates only the explicit y argument. The known C^{1,alpha} norm of u bounds the C^{0,alpha} norms of all T_s uniformly in s and delta; smooth composition on the compact argument set plus the bound on v shows a and f have uniform C^{0,alpha} bounds and b and d are uniformly bounded, with a symmetric and uniformly elliptic. Test with chi^2 v, chi supported in B_{3rho} and one on B_{2rho}: ellipticity gives a multiple of integral chi^2 |Dv|^2 on the left, and Young's inequality together with the uniform bounds absorbs every term linear in Dv (in particular b . Dv) into that multiple, leaving integral_{B_{2rho}} |Dv|^2 dy <= C uniformly in the difference step. No second derivative of u has entered. Frozen comparison. Fix x in B_rho and 0 < r <= min(rho,1). On B_r(x) let w - v in H^1_0(B_r(x)) solve div(a(x) Dw) = 0 (existence by minimizing the positive quadratic Dirichlet energy). Put z = v - w; subtract the equations and test with z, noting that the frozen constant vector f(x) integrates to zero against Dz. With all norms on B_r(x), ellipticity and the coefficient bounds give lambda ||Dz||_2^2 <= C r^alpha ( ||Dv||_2 + |B_r|^{1/2} ) ||Dz||_2 + C ( ||Dv||_2 + |B_r|^{1/2} ) ||z||_2; the second line carries the first-order term b . Dv. Zero-boundary Poincare gives ||z||_2 <= C r ||Dz||_2, and r <= r^alpha for r <= 1, hence ( average_{B_r(x)} |Dv - Dw|^2 )^{1/2} <= C r^alpha ( 1 + ( average_{B_r(x)} |Dv|^2 )^{1/2} ). Energy minimization after subtracting an affine function gives, for every constant vector p, ( average_{B_r(x)} |Dw - p|^2 )^{1/2} <= C ( average_{B_r(x)} |Dv - p|^2 )^{1/2}. Oscillation. For the constant-coefficient equation there is an interior estimate: for 0 < gamma <= 1/4, ( average_{B_{gamma r}(x)} |Dw - (Dw)_{B_{gamma r}(x)}|^2 )^{1/2} <= C gamma ( average_{B_r(x)} |Dw - p|^2 )^{1/2}, where the subscript denotes an average. To justify it at the weak regularity in use, mollify Dw - p in an interior ball: its components solve the same constant-coefficient equation, a linear coordinate change controlled by ellipticity makes them harmonic, the mean-value formula reproduces each by convolution with a smooth radial kernel of radius comparable to r, and differentiating the kernel with Cauchy-Schwarz bounds its gradient on B_{r/2}(x) by C r^{-1} ( average_{B_r(x)} |Dw - p|^2 )^{1/2}; the kernel radius is chosen so its support stays interior, and letting the mollification scale tend to zero gives the estimate for the representative, whose oscillation on B_{gamma r}(x) is at most that bound times 2 gamma r. Iteration. Define p_r = average_{B_r(x)} Dv and J(r) = ( average_{B_r(x)} |Dv - p_r|^2 )^{1/2}. Using p = p_r above and comparing averages on the two balls gives J(gamma r) <= C gamma J(r) + C gamma^{-m/2} r^alpha ( 1 + |p_r| + J(r) ) and |p_{gamma r} - p_r| <= gamma^{-m/2} J(r). Fix gamma so C gamma <= 1/2, and a fixed ell >= 1 with ell/2 >= gamma^{-m/2}; then S(r) = 1 + |p_r| + ell J(r) satisfies S(gamma r) <= (1 + C' r^alpha) S(r). Choose one starting radius r_* <= min(rho,1); its starting averages are bounded uniformly for every x in B_rho by the energy bound. On the geometric scales r_j = gamma^j r_* the product of the factors (1 + C' r_j^alpha) is finite because sum_j r_j^alpha < infinity, so |p_{r_j}| is uniformly bounded. Each fixed-step quotient is C^1, so these averages tend to Dv(x) as j tends to infinity, giving a bound on Dv(x) uniform in x in B_rho, in e, and in the sufficiently small step delta. Finish. Since Dv(y) = (Du(y + delta e) - Du(y))/delta holds classically, the uniform bound bounds increments of Du along every coordinate direction. On a smaller coordinate cube, join two points by coordinate segments; their total length is at most sqrt(m) times the Euclidean distance. Therefore Du is locally Lipschitz, and u in C^{1,1}_loc(Omega).

Exercise: Extend first variation and the Sobolev inequality across the singular set of an equality boundary.
Answer (Cutoffs and Extension). The equality boundary from the Equality Boundary proposition has a compact singular set Z of Hausdorff dimension at most n-8, so for n >= 4 we have dim_H Z <= m-7 < m-2, giving H^{m-2}(Z) = 0. The regular part Sigma carries all the perimeter and has finite area s_m R^m, is embedded and locally C^{1,1}, and, by the same proof as the area-growth lemma (or by reading off the local upper area growth from the regularity theorem and taking a finite cover), has sigma(Sigma intersected with B_r(x)) <= C r^m uniformly. Hence the Capacity Cutoffs lemma applies with S = Z and supplies eta_j in Lip_c(Sigma) with 0 <= eta_j <= 1, eta_j -> 1 on Sigma, and integral_Sigma |grad_Sigma eta_j|^2 dsigma -> 0; when Z is empty take eta_j = 1. Global first variation. For every smooth ambient field Y defined near Gamma, the cutoffs and the tangential-divergence formula imply the global identity integral_Sigma div_{T Sigma} Y dsigma = H_0 integral_Sigma <Y,N> dsigma. Indeed, apply the compact-support identity to eta_j Y and expand: integral_Sigma eta_j div_{T Sigma} Y dsigma + integral_Sigma <grad_Sigma eta_j, Y^T> dsigma = H_0 integral_Sigma eta_j <Y,N> dsigma. The middle integral has absolute value at most ||Y||_infty A^{1/2} ||grad_Sigma eta_j||_2, which tends to zero, and all other terms converge by dominated convergence since the ambient data are bounded near the compact set Gamma. On any compact boundaryless component of Sigma the identity is immediate by integrating the local formula. Extension of Sobolev. When n >= 4, the same cutoffs extend the Hypersurface Sobolev inequality to every smooth ambient function v restricted to Sigma. Here the signed normalized mean curvature is h = H/m = 1/R and we use the case b = 0, giving c integral_Sigma |grad_Sigma v|^2 dsigma + (m/R^2) integral_Sigma v^2 dsigma >= m s_m^{2/m} ( integral_Sigma |v|^q dsigma )^{2/q}. To pass from eta_j v to v: boundedness of v and its gradient near Gamma gives convergence in L^2 and L^q; and grad_Sigma(eta_j v) - grad_Sigma v = (eta_j - 1) grad_Sigma v + v grad_Sigma eta_j tends to 0 in L^2 because the first term goes to zero by dominated convergence and the second is bounded by ||v||_infty ||grad_Sigma eta_j||_2 -> 0. So the whole Dirichlet energy, including its cross term, converges. In particular the constant function and fixed ambient perturbations of it are now legitimate tests on Sigma, with no completeness of the regular locus assumed. This is the crucial legitimization: the rigidity argument will test with the sphere-coordinate functions w_j and their squares, which are smooth ambient restrictions and therefore fall under this extension.

Exercise: Prove the barycenter existence and the flux inequality (lower second moment).
Answer (Barycenter and Flux). Barycenter. If mu is a nonzero finite positive measure of compact support in M, the function F(p) = integral_M D_p(x) dmu(x), with D_p(x) = d(p,x)^2/2, is continuous and proper: if the support lies in B_a(o) then d(p,x) >= d(p,o) - a, so F(p) tends to infinity as p leaves compact sets. Completeness and Hopf-Rinow give a minimizer, and differentiation on compact sets at that minimizer gives integral_M log_p x dmu(x) = 0 (uniqueness is not needed). Flux. Let T be either the whole regular part Sigma (to which the global first-variation identity applies) or a compact boundaryless component of Sigma, and put a = area(T). The distance comparison gives Hess D_p >= g, so div_{T Sigma} grad D_p = trace_{T Sigma} Hess D_p >= m. Apply the global first-variation identity (multiply all ambient fields by a cutoff equal to one near the compact set Gamma, so their lack of compact support on M is no obstacle) to Y = grad D_p: ma <= integral_T div_{T Sigma} grad D_p dsigma = (m/R) integral_T <grad D_p, N> dsigma <= (m/R) ( a integral_T r_p^2 dsigma )^{1/2}, the last step by Cauchy-Schwarz and |grad D_p| = r_p (because |grad r_p| = 1). Rearranging, integral_T r_p^2 dsigma >= R^2 a: this is the lower second-moment bound, valid for every p in M. When the matching upper bound also holds, every step in this chain is an equality. Let us see what that forces. Equality in Cauchy-Schwarz and in the curvature comparison gives grad D_p = R N almost everywhere on T (equivalently, the squared L^2 norm of grad D_p - R N vanishes: integral_T r_p^2 + R^2 a - 2R integral_T <grad D_p,N> = 0, using the equality values of both integrals). Since |N| = 1, this reads |grad D_p| = R; but |grad D_p| = r_p, so r_p = R almost everywhere on T, i.e. T lies on the distance sphere of radius R about p. Continuity upgrades this to every point of T (a nonempty open regular patch has positive area).

Exercise: Prove that in dimensions at least four the support lies on a round sphere.
Answer (Spectral Rigidity, n >= 4). Here m = n-1 >= 3, so the extension of the Hypersurface Sobolev inequality to smooth ambient tests on Sigma is available. Retain q = 2m/(m-2) and c = 4/(m-2) = q-2. The constant function v = 1 attains equality in c integral_Sigma |grad v|^2 + (m/R^2) integral_Sigma v^2 >= m s_m^{2/m} ( integral_Sigma |v|^q )^{2/q}, because A = s_m R^m and m s_m^{2/m} A^{2/q} = (m/R^2) A. Let phi be a smooth ambient restriction with integral_Sigma phi dsigma = 0. Expand v = 1 + tau phi: ( integral_Sigma |1 + tau phi|^q )^{2/q} = A^{2/q} [ 1 + (q-1)(tau^2/A) integral_Sigma phi^2 + o(tau^2) ] (the linear term vanishes by the zero mean). On the left side, the linear term in tau is c * 2 tau integral_Sigma grad 1 . grad phi = 0 plus (m/R^2) * 2 tau integral_Sigma phi = 0. The constant terms agree; comparing quadratic terms, using m s_m^{2/m} A^{2/q-1} = m/R^2, gives c integral_Sigma |grad phi|^2 + (m/R^2) integral_Sigma phi^2 >= (m/R^2)(q-1) integral_Sigma phi^2, hence c integral_Sigma |grad phi|^2 >= (m/R^2)(q-2) integral_Sigma phi^2, i.e. the spectral gap integral_Sigma |grad_Sigma phi|^2 dsigma >= (m/R^2) integral_Sigma phi^2 dsigma. (The cutoff limit has already been taken before the expansion; no mean-zero condition on cutoff approximations is needed.) Upper moment. Choose a barycenter p for the area measure dsigma on Sigma (which has no atoms, being a positive area measure on a C^{1,1} hypersurface) and an orthonormal basis e_1,...,e_n of T_pM. The smooth ambient functions phi_i(x) = <log_p x, e_i> have zero mean by the barycenter identity, so each satisfies the spectral gap. Since d log_p is a contraction, sum_i |grad_Sigma phi_i|^2 = sum over an orthonormal tangent frame v_1,...,v_m of Sigma of sum_i <d log_p(v_j), e_i>^2 = sum_j |d log_p(v_j)|^2 <= sum_j |v_j|^2 = m. Summing the spectral gap over i gives (m/R^2) integral_Sigma r_p^2 dsigma = (m/R^2) integral_Sigma sum_i phi_i^2 <= sum_i integral_Sigma |grad_Sigma phi_i|^2 <= m * A, i.e. integral_Sigma r_p^2 dsigma <= R^2 A. Together with the lower second-moment bound for this same p, both are equalities, so the flux argument puts Sigma on the sphere {x : d(p,x) = R}, and density of Sigma in Gamma puts all of Gamma there: Gamma is contained in the exponential image of a Euclidean sphere of radius R in T_pM centered at the tangent-space origin. Neither connectedness nor completeness of the regular part was used.

Exercise: Prove the support lies on a sphere in dimension two.
Answer (Wirtinger, n = 2). Now m = 1 and Gamma = Sigma is a compact embedded C^{1,1} curve without boundary, with total length A = 2 pi R (since s_1 = 2 pi). Choose a connected component T and let L be its length, L > 0. A connected compact one-dimensional manifold without boundary has a periodic arclength parametrization of period L: continue an oriented arclength chart; a traverse of arbitrarily large length cannot stay injective in a component of finite length, and local uniqueness gives a least positive period, whose image is open and closed in the component, so one least period traverses it once. The periodic Wirtinger inequality says integral_T |grad_T phi|^2 dsigma >= (2 pi / L)^2 integral_T phi^2 dsigma for zero-mean phi; in arclength coordinates it follows by expanding a trigonometric polynomial in Fourier modes (the zero mode is absent), and periodic convolution uniformly approximates a C^1 function and its derivative while preserving the zero mean. Take a barycenter p of the area measure on T, apply Wirtinger to its logarithm coordinates, and use the contraction of d log_p exactly as in the higher-dimensional case: the result is integral_T r_p^2 dsigma <= L^3 / (2 pi)^2. On the other hand the lower second-moment bound gives integral_T r_p^2 dsigma >= R^2 L. Combining them, R^2 L <= L^3 / (2 pi)^2, so L^2 >= (2 pi R)^2 = A^2, i.e. L >= A = 2 pi R. Since L <= A, this component carries ALL the length and every inequality just used is an equality; by the flux argument r_p = R on T. Another component would contain a regular open arc of positive length and could not exist. Hence T = Gamma and Gamma lies on the sphere of radius R about p, i.e. on exp_p of a Euclidean circle of radius R in the tangent plane.

Exercise: Prove the support lies on a (possibly offset) sphere in dimension three.
Answer (Punctured Flux, n = 3). Now m = 2 and Sigma = Gamma is a compact embedded C^{1,1} surface, with A = 4 pi R^2. Write h = 1/R, so its trace curvature is 2h. Fix p in Sigma and, away from p, put r = r_p, Y = r^{-2} grad D_p = grad r / r, and t = <Y, N>. Since Hess D_p >= g and |grad r| = 1, compute div_{T Sigma} Y = r^{-2} trace_{T Sigma} Hess D_p - 2 r^{-2} |grad_Sigma r|^2 >= r^{-2}(2 - 2|grad_Sigma r|^2) = 2 r^{-2}(1 - |grad_Sigma r|^2) = 2 t^2, where the last identity uses 1 - |grad_Sigma r|^2 = <grad r, N>^2 = (r t)^2. Choose a smooth nondecreasing chi: [0,infinity) -> [0,1] that is 0 on [0,1] and 1 on [2,infinity). The field chi(r/eps) Y extends smoothly by zero near p. Apply the global first-variation identity and differentiate the cutoff: the cutoff-derivative term is nonnegative, and the divergence bound gives integral_Sigma chi(r/eps) [ h^2/2 - 2(t - h/2)^2 ] dsigma >= J_eps, where J_eps = integral_Sigma ( chi'(r/eps) / (eps r) ) |grad_Sigma r|^2 dsigma; this is because the integrand before completing the square is 2h t - 2t^2 = h^2/2 - 2(t - h/2)^2. Next, J_eps tends to 2 pi. Use exponential coordinates at p with the first two coordinate axes in T_p Sigma: in a neighborhood of p the whole support is a single graph (y, b(y)) with b(0) = 0 and Db(0) = 0, and for small eps that one graph contains the whole annulus contributing to J_eps, so no other sheet enters. Set y = eps z; for fixed z not zero, r/eps -> |z| (here r = |(y,b(y))| exactly), dsigma/eps^2 -> dz, and |grad_Sigma r| -> 1. On the support of chi', 1 <= r/eps <= 2, so z stays in a fixed bounded annulus away from zero, the area density is uniformly bounded, |grad_Sigma r| <= 1, and |chi'(r/eps)/(r/eps)| <= ||chi'||_infty. Dominated convergence after extending by zero gives lim J_eps = integral_{R^2} chi'(|z|)/|z| dz = 2 pi integral_1^2 chi'(s) ds = 2 pi. Because h^2 A / 2 = (1/R^2)(4 pi R^2)/2 = 2 pi, the display implies 0 <= integral_Sigma chi(r/eps)(t - h/2)^2 dsigma <= (h^2/4) integral_Sigma chi(r/eps) dsigma - (1/2) J_eps -> 0. Fatou's lemma yields t = h/2 almost everywhere off p, with no prior integrability assumption on t^2 at the pole; continuity makes it hold at every point of Sigma minus p. This off-diagonal identity has been proved for each fixed p, so fix x_0 in Sigma and evaluate it at x_0 for every p not equal to x_0 (no intersection of uncountably many exceptional null sets). Since grad D_p(x_0) = -log_{x_0} p and r = |log_{x_0} p|, the identity t = h/2 = 1/(2R) reads -<log_{x_0} p, N(x_0)> = |log_{x_0} p|^2 / (2R). Completing the square, including p = x_0 by continuity, gives |log_{x_0} p + R N(x_0)|^2 = R^2, i.e. Gamma is contained in exp_{x_0}( { v in T_{x_0}M : |v + R N(x_0)| = R } ). This is a Euclidean sphere of radius R centered at -R N(x_0); it passes through the origin of T_{x_0}M, so the logarithm origin sits on its boundary. That offset is the only difference from the higher-dimensional case.

Exercise: Recover the set and prove its interior is a flat ball.
Answer (Recovering the Ball and its Metric). The three dimension regimes give the same conclusion: for some point o in M, the perimeter support Gamma is contained in exp_o(S), where S is a Euclidean sphere of radius R in T_oM (possibly not centered at the origin, as in dimension three). Let U_0 be the open ball bounded by S, and put U = exp_o(U_0). Using the global exponential diffeomorphism, the two connected components of T_oM minus S map onto the two connected components of M minus exp_o(S). The interior component U is bounded (its closure is a compact image); the exterior component is unbounded, because vectors escaping to infinity in the Euclidean exterior have image distances d(o, exp_o v) = |v| escaping to infinity, and exterior connectedness uses n >= 2. The smooth hypersurface exp_o(S) has zero ambient volume. Phase recovery. The distributional derivative of chi_E is supported on Gamma, so it vanishes on each complementary component; a function with all weak coordinate derivatives zero is locally constant almost everywhere (mollify in coordinate balls and overlap), so chi_E is almost everywhere constant on each connected component. The two constants lie in {0,1}. Boundedness of E forces the exterior constant to be 0, because the open exterior contains positive-volume balls arbitrarily far from E; and since V > 0 and the separating sphere has zero volume, the interior constant is 1. Hence E = U up to a null set; only the containment of the perimeter support in the sphere was used. Flatness of the interior. By the distance comparison, the pullback metric g-tilde = (exp_o)^* g dominates the constant Euclidean metric g_o on all of T_oM, hence also on the offset ball U_0. On that ball, phase recovery gives Vol_{g-tilde}(U_0) = V = omega_n R^n = Vol_{g_o}(U_0). The smooth relative density sqrt(det_{g_o} g-tilde) is at least one; equality of its integral with the Euclidean volume and continuity force it to be identically one on U_0. All eigenvalues of g-tilde relative to g_o are then at least one and their product is one, so each eigenvalue is one, and (exp_o)^* g = g_o on U_0. Thus exp_o: U_0 -> U is a Riemannian isometry, which is necessity in the equality theorem. Converse. Let F: B_R(0) in R^n -> U in M be a Riemannian isometry onto an open region with its induced metric. Put p = F(0) and L = dF_0. For |v| < R, the image of the straight segment s -> sv is an ambient geodesic with initial data (p, Lv), because the Levi-Civita connection of an open region is the restriction of the ambient connection; geodesic uniqueness gives F(v) = exp_p(Lv). Consequently U = B_R(p). This formula extends smoothly to the closed Euclidean ball through the ambient exponential map; its pullback metric is Euclidean in the interior and hence on the boundary by continuity. The boundary is a smooth sphere of area s_{n-1} R^{n-1} = s_m R^m, which equals its ambient perimeter by the divergence formula, and its volume is omega_n R^n, so equality holds. Perimeter and volume are unchanged by null modifications. No flatness outside the equality region is implied.

Exercise: Derive the model-ball spectral and torsion comparisons by rearrangement.
Answer (Rearrangement Consequences). Let n >= 2, kappa <= 0, 1 < p < infinity, and let M be complete, simply connected, smooth, without boundary, with Sec <= kappa. Let D be a nonempty smooth domain compactly contained in M, and let B_kappa(|D|) be a geodesic ball of the same volume |D| in the model space of curvature kappa. Claim: lambda_{1,p}(D) >= lambda_{1,p}(B_kappa(|D|)) and T_p(D) <= T_p(B_kappa(|D|)), where lambda_{1,p} is the first Dirichlet eigenvalue of the p-Laplacian and T_p is the p-torsional rigidity. Proof: Fix a nonzero smooth test v compactly supported in D and extend u = |v| by zero; this is a compactly supported Lipschitz function. Apply the Nobili-Violo Polya-Szego rearrangement theorem (their Theorem 1.1, on M with constant one): its metric BV perimeter is exactly the ambient Riemannian perimeter, so the Generalized Comparison supplies the required support-neighborhood bound at every finite volume, and the positive superlevels of u have finite volume. The theorem provides a decreasing equimeasurable rearrangement u* (for the interval density a(r) = A_b(r), b = sqrt(-kappa)) that is bounded and locally absolutely continuous, with weighted derivative energy at most integral_D |grad_g v|_g^p dVol_g. Because the support of u is strictly inside D, its volume is strictly below |D|, so u* vanishes on an interval before R = V_b^{-1}(|D|). Define the radial function u-tilde(x) = u*( d_kappa(o,x) ) on the model ball B = B_kappa(|D|). In normal coordinates it is bounded and locally W^{1,p} away from the origin with finite gradient p-energy; its gradient is also locally integrable across the origin because integrating by parts outside the radius-eps ball leaves an inner boundary term O(eps^{n-1}) (bounded u*) which tends to zero since n >= 2, so the weak gradient extends. Equivalence of Sobolev norms in smooth coordinates and vanishing near R then give u-tilde in W^{1,p}_0(B). Its gradient is radial, so its energy is at most that of v, and the L^1 and L^p integrals are equal (equimeasurability). Taking the Rayleigh infimum for lambda_{1,p} and the torsion-quotient supremum for T_p proves both comparisons; the finite volume of D and density of C_c^infty(D) allow the variational values to be computed on these tests. This is the promised functional-analytic payoff of the isoperimetric theorem.

Exercise: Bookkeeping. What does the proof actually use, and what does it deliberately avoid?
Answer (Correctness Bookkeeping). The comparison theorem rests on four independent inputs. (1) The Cartan-Hadamard theorem (exp_p is a global diffeomorphism) and the distance Hessian / polar comparisons, which are pure nonpositive-curvature facts and use only Sec <= kappa. (2) The Aubin-Talenti sharp Euclidean Sobolev inequality, used only through its constant Y_m = m s_m^{2/m}. (3) The Brcaronzis-Lieb lemma for the subcritical-minimizer step and the Rellich compactness for the local threshold. (4) The Ghomi-Spruck obstacle regularity (Lemma 7.2(i),(ii), which does not assume any total-curvature inequality), the Stredulinsky-Ziemer Euclidean obstacle theory, and Morgan's interior regularity, used only to make confined minimizers smooth off a small singular set. The equality theorem adds (5) Antonelli-Pasqualetto-Pozzetta's local smooth-manifold regularity (Corollary 1.6, no global lower Ricci bound) and the elementary graph-regularity lemma proved here by difference quotients. Notably, the proof of the main comparison never needs the equality classification, never needs any symmetry or convexity of the minimizing set, never needs the ambient manifold to have a lower curvature bound, and never needs the boundary to avoid the singular set for the ODE step (the capacity cutoffs handle that). The rigidity argument deliberately avoids assuming the perimeter support is a sphere or that the set is convex: it derives the sphere from the second-moment identity, and derives the flat metric from a relative-volume comparison, so the only remaining freedom is the tangent-space center, which is then pinned down by phase recovery. Dimension two uses periodic Wirtinger because the log coordinates have zero mean; dimension three uses the punctured flux because the critical Sobolev exponent degenerates (m = 2 gives q = infinity), so the spectral test is unavailable and is replaced by the explicit annular calculation. This is why the proof is split three ways: not because the conclusion differs, but because the available analytic tool differs.

Exercise: What is the single sentence to remember, and the single picture?
Answer (Summary). Main comparison and equality in one breath: in any Cartan-Hadamard manifold, among all regions of a fixed volume, the round model ball of that volume in the constant-curvature space has the least boundary, and equality holds exactly when the region is itself a flat ball (its induced metric is a round Euclidean ball metric), with no ambient curvature outside that flat ball obstructing equality. The picture: confine the problem to a large geodesic ball and find the least-boundary region; run the sharp hypersurface Sobolev inequality on its boundary to obtain one scalar differential inequality for the profile; integrate it exactly to get the model comparison; rerun it on an equality set to force every regular boundary point onto a round sphere about a single point; compare the enclosed ball volume with the ambient ball volume to force the inherited metric to be exactly Euclidean; and finally read off the spectral and torsion comparisons by rearrangement. That is the whole paper.