The Solomon–Yau least-volume theorem

Proof. For k+l=m, the product has volume

Define, for real j>0,

The ratio of the product volume to \sigma_m is d_kd_l/d_{k+l}. The usual series for the logarithmic derivatives of \Gamma gives

Indeed, strict convexity of t\mapsto(t+j/2)^{-2} implies

Thus, with k>0 fixed, the derivative in l of \log(d_kd_l/d_{k+l}) is negative. Increase the second entry of a minimizing pair for m by one to obtain a strictly smaller competitor for \vartheta_{m+1}. This proves the strict decrease. The stated values follow from k=l=1 and k=l=2, respectively.□

Proof. Push forward the parametrized Euclidean cone

with measure r^m\,dr\,d\mathrm{vol}_M. This gives an integral (m+1)-varifold \mathcal C_F, extended with zero mass at the origin. Minimality of F makes it stationary away from the origin. The boundary term for first variation after truncation at r=\epsilon is O(\epsilon^m); it tends to zero. Thus \mathcal C_F is stationary in all of Euclidean space. The corresponding calculation for general spherical varifolds is given in Lemma 6.1.

Its centered ball masses are exactly

and hence

Every fiber of a compact immersion is finite: it is compact and its points are isolated by local embedding neighborhoods. At a nonzero cone point over an image point with q preimages, these neighborhoods give q smooth local sheets, each of density one. Therefore the density there is q, even when tangent planes agree or sheets coincide.

For a fixed Euclidean center z, the inclusions

show that the large-radius mass ratio has limit \Theta(\mathcal C_F,0). Stationary monotonicity centered at the chosen nonzero point consequently gives

This proves the inequality. If F is not injective, take q\ge2 and use Lemma 2.1. Finally, an injective immersion from a compact manifold is an embedding.□

Proof. A closed embedded hypersurface in a sphere separates it into two regions and has a global unit normal. If its image were contained in a linear hyperplane, it would be an open and closed subset of the corresponding great sphere, by equality of dimensions. Connectedness and compactness would make the image that great sphere. This proves fullness.

The coordinate equation \Delta x=-mx gives the integral identity. In analytic ambient coordinates, a minimal hypersurface is locally an analytic graph by analytic regularity for the minimal graph equation [4, Theorem 1.1]. Thus its coordinate functions are analytic. Fullness makes each \langle a,x\rangle, a\ne0, nontrivial, and a nontrivial analytic function on a connected manifold has a zero set of measure zero [18, Proposition 0].□

Proof. We first describe the possible contacts. Let \gamma(t)=\exp_x(tv), where v\in T_x\Sigma is a unit vector, and regard \gamma as a curve in \mathbb R^{m+2}. The ambient differentiation formulas give

For example, \gamma''(0)=-x+A_x(v,v)N(x), while \langle x,\gamma'''(0)\rangle=0. Compactness makes these expansions uniform on the unit tangent bundle for sufficiently small |t|. Consequently, adjoining unit tangent directions at the diagonal extends the quotient continuously, with values |A_x(v,v)|.

For completeness, this extension also gives continuity when the base point varies. If x_j\to x and nearly maximizing points y_j stay away from x, a convergent subsequence gives an ordinary distinct-point contact. If y_j\to x, write y_j=\exp_{x_j}(t_jv_j) and take a convergent subsequence of unit tangent directions. The uniform expansions show that its limiting quotient is at most \mu(x). This proves upper semicontinuity. Holding fixed any y\ne x whose quotient is within \varepsilon of \mu(x) proves lower semicontinuity, after letting \varepsilon\downarrow0. The same compactness argument proves finiteness and shows that a maximum is realized either by distinct points or by a tangent direction. If \mu(x)=0, the entire hypersurface lies in N(x)^\perp, contrary to fullness from Lemma 2.4. Thus \mu>0.

Fix an upper test \phi at x. At a realizing contact, choose the sign of N to make the signed quotient positive, and change the sign of A at the same time. This choice is held fixed in a neighborhood. It requires no continuous choice of maximizing sign: for either fixed sign and for every nearby x' and every y'\ne x',

Choose a principal orthonormal basis e_1,\ldots,e_m at x, and write Ae_i=\lambda_i e_i. The diagonal limits give

We shall prove the stronger inequality

where \phi_i=d\phi(e_i) and every component with \lambda_i=\phi will be shown to vanish.

Distinct-point contact. Suppose y\ne x realizes the maximum with the chosen sign. On the product of a neighborhood of x with \Sigma, set

By (3.6), Z\ge0 and Z(x,y)=0. Write c=\langle x,y\rangle, s=1-c>0, and d_i=\langle e_i,y\rangle. The first derivatives at the minimum give

Reflection across the hyperplane perpendicular to x-y sends x to y, and sends N(x) to

Indeed \langle N(x),y\rangle=\phi s, so this vector is perpendicular to y and has length one. The y first-derivative equation for Z shows that n_y is perpendicular to T_y\Sigma. Therefore the reflected vectors

form an orthonormal basis of T_y\Sigma, and

The vector n_y may be either orientation of the normal at y; only minimality at y enters the following trace computation.

Use the product connection to compute the Hessian of Z. Denote its mixed entries in the indicated bases by Z_{x_i y_i}. From (3.1),

In particular, the nonnegative sum of Hessians along (e_i,\widetilde e_i) is

where we used \tr A=0 and then (3.9). If \lambda_i=\phi, (3.9) gives \phi_i=0. Dividing by s and substituting that equation in all other directions gives the first inequality of (3.8).

Diagonal contact, including repeated eigenvalues. Suppose instead that a unit tangent direction realizes the maximum. With our choice of normal sign, the top eigenvalue is \phi(x). Let E\subset T_x\Sigma be its full eigenspace. For every unit v\in E and every local smooth unit extension V of v,

The first derivative at this contact, in any direction w\in T_x\Sigma, is

The field-derivative term vanishes because A(v,\nabla_wV)=\phi\langle v,\nabla_wV\rangle=0. For the geodesic with initial direction v, formulas (3.4)–(3.5) give the signed quotient

It is at most the fixed value \mu(x)=\phi(x) for both signs of t. Hence (\nabla_vA)(v,v)=0, and (3.10) implies

This holds for every vector in E, not just for one principal vector. Moreover, polarizing (3.10) on E gives (\nabla_wA)|_{E\times E}=d\phi(w)g|_{E\times E}. These are pointwise statements; no smooth extension of the eigenspace is assumed.

Fix e_1\in E and choose a smooth unit field V with V(x)=e_1. Put

The first derivatives \nabla_{e_j}V may be prescribed arbitrarily in e_1^\perp: prescribe the corresponding jets of any local field and normalize it. Unit length determines precisely the second-derivative contraction that occurs in the calculation,

Thus, in a geodesic orthonormal frame at x,

No other second derivative of V contributes, since Ae_1=\phi e_1. Choose the free first jets by

The minimum of \phi-A(V,V) then yields

In the last step we used Simons’ identity and Codazzi: T_{1i}=(\nabla_{e_i}A)(e_1,e_1)=\phi_i. Equation (3.11) accounts for every omitted top-eigenspace gradient component. This proves the first inequality of (3.8) also at a repeated diagonal contact.

In both cases, (3.7) gives 0<\phi-\lambda_i\le2\phi in each retained direction. The vanished components in the other directions therefore give the second inequality of (3.8). Dividing by \phi and subtracting |\nabla\phi|^2/\phi^2 proves (3.3).□

Proof. If the inequality failed, then

Since m-P-c has mean zero, the Poisson equation on the closed connected manifold has a smooth solution. Indeed, the kernel of \Delta consists of constants by integration by parts; the compact elliptic Fredholm alternative and regularity [28, §10, equations (10.4)–(10.6)] give

Let K=\max_\Sigma(\log\mu-h) and let x attain this maximum. The smooth positive function \phi=\exp(h+K) is an upper test for \mu at x. Lemma 3.1 and (3.14) imply

a contradiction. This argument uses only continuity and positivity of \mu; it does not differentiate \mu or assume any weak regularity for its gradient.□

Proof. Suppose a\ne0 and \langle a,N\rangle=0 everywhere. Set f=\langle a,x\rangle. The ambient Gauss formula gives

This function is nonconstant: a constant solution would be zero, whereas f\equiv0 contradicts fullness. Obata’s sphere-rigidity theorem [21, Theorem A] applies to the complete connected manifold (\Sigma,g) and shows that it is isometric to the unit round S^m. The scalar Gauss equation for a minimal hypersurface, \operatorname{Scal}_\Sigma=m(m-1)-P, then gives P=0. This contradicts the hypothesis that \Sigma is not totally geodesic.□

Proof. Parametrize the transformations by z=(\tanh\tau)a in the open unit ball of \mathbb R^{m+2}, with the identity at z=0. Consider

As z\to a_*\in S^{m+1}, the integrand tends to -a_* at every point with x\ne a_*. This remains true for a moving direction a\to a_*. The measure u\,dV assigns zero mass to a point, and the integrands have length one. Dominated convergence extends F continuously to the closed ball, with F(a_*)=-a_*. If F had no zero, its normalization F/|F| would extend the antipodal map from the boundary sphere to the ball. The antipodal map has nonzero degree, so such an extension is impossible. A zero lies in the open ball and proves (3.17).

To prove (3.18), fix a and abbreviate f=\langle a,x\rangle, \eta=\langle a,N\rangle, and \beta=\beta_{\tau,a}(x). Orthogonal decomposition of a into position, tangent, and normal components gives

Integrating a divergence on the closed hypersurface yields

Differentiation in \tau therefore gives the exact formula

By Lemma 3.3, \eta is not identically zero. The derivative is strictly negative for every \tau>0. Starting from \beta_{0,a}=1 proves (3.18) and its strictness. Finally, Hölder’s inequality, including the identity case m=2, gives

The second inequality is strict for \tau>0, proving (3.19).□

Proof. By (3.1), the m+2 normal coordinate functions satisfy LN=mN; they are linearly independent by Lemma 3.3. The largest eigenvalue \lambda_1 of L is simple and has a strictly positive eigenfunction u. Apply the first-eigenfunction theorem to the shifted Schrödinger operator K-L, with K>1+\max_\Sigma(m+P); its positivity uses connectedness and the strong maximum principle [29, Proposition 2.9]. Testing its Rayleigh quotient with the constant function gives

These eigenfunctions prove \ind(\Sigma)\ge m+3.

Suppose that equality holds. The positive spectrum consists of the simple eigenvalue \lambda_1>m and the eigenvalue m with multiplicity m+2. The spectral theorem therefore gives

Choose the conformal map supplied by Lemma 3.4 for the weight u, and write v=T_{\tau,a}(x). Each coordinate v_\alpha satisfies the orthogonality in (3.22). Since |v|=1 and \sum_\alpha|\nabla v_\alpha|^2=m\beta^2, summing those inequalities and integrating by parts gives

Theorem 3.2 forces equality throughout. The strict part of (3.19) implies \tau=0, so v=x.

We now use equality in the individual spectral inequalities. For each \alpha, the defect

is nonnegative by (3.22), and (3.23) shows \sum_\alpha d_\alpha=0. Thus every defect vanishes. In the spectral decomposition on u^\perp, all eigenvalues are at most m; equality in its quadratic form implies Lx_\alpha=mx_\alpha for every \alpha. On the other hand, (3.1) gives Lx=Px. Taking the inner product of (P-m)x=0 with x yields

The pointwise conclusion uses spectral equality as well as the integrated curvature equality.

The last identity in (3.1) now gives \nabla A=0. We finish by deriving the product classification. Parallelness makes the eigenvalues constant and their eigendistributions parallel. If U and W are unit vectors in distinct eigendistributions, the curvature tensor preserves each of those distributions, so \langle R(U,W)W,U\rangle=0. The Gauss equation gives

for any two distinct eigenvalues \lambda,\kappa. There are at most two distinct eigenvalues: a third would have to equal -1/\lambda as well. Since A is nonzero and trace-free, there are exactly two, of opposite signs. After changing N if necessary, write them as

where k and \ell are their multiplicities.

Let E_\lambda and E_\kappa denote the parallel tangent eigendistributions. Consider the ambient subspaces, initially depending on x,

They are orthogonal and their dimensions sum to m+2, because \langle\lambda N-x,\kappa N-x\rangle=1+\lambda\kappa=0. For any tangent vector w and any local section V of E_\lambda, ambient differentiation gives

Both expressions lie in \mathcal E(x): the second vanishes on E_\kappa and maps E_\lambda to itself. Hence \mathcal E is a constant subspace of \mathbb R^{m+2} along the connected hypersurface; the same holds for its orthogonal complement \mathcal F. The projections of x to these fixed spaces have lengths

Thus the image of \Sigma lies in the displayed Clifford product. The inclusion is a local diffeomorphism onto this product, since both manifolds have dimension m. Its image is open, and compactness makes it closed. The product is connected because k,\ell\ge1, so the image is the entire product. The final assertion follows from the definition of a_m.□

Proof. We use the small-filling theorem in the following precise form: there are constants \nu>0 and c\ge1, depending on the sphere, such that if \mathcal F(T-T')<\nu, then there is an (m+1)-chain R in the sphere with

This is the isoperimetric choice described in [17, §3.1, p. 11]; all locators for this source refer to arXiv:1311.6501v2. Applied to a sufficiently fine partition of a flat-continuous path from zero to T, these fillings sum to a top-dimensional chain with boundary T. The preceding representation therefore writes T=\partial[E] with E of finite perimeter. Conversely, intersecting any such E with the increasing spherical caps \{y_{m+2}<t\}, -1\le t\le1, gives sets of finite perimeter continuous in volume, from the empty set to E. Thus every \partial[E] belongs to this path component.

If \partial[E]=\partial[F], then \partial[E\triangle F]=0. By (4.1), the characteristic function of E\triangle F has zero distributional gradient. Connectedness of the sphere makes it constant almost everywhere. Hence F=E or F=E^c.

For the covering topology, fix T=\partial[E] and choose \delta>0 so that 2\delta<\nu and 4c\delta<\sigma_{m+1}. For T' in the flat \delta-ball about T, choose the small filling R(T,T') from (4.2). This choice is unique: two such choices differ by zero or the whole sphere, and their total mass is smaller than \sigma_{m+1}. If T' and T'' are in the ball, the same argument applied to

shows that this cycle is zero: its mass is at most 4c\delta. Consequently the filling branch determined by

satisfies d(E_{T'},E_{T''})\le c\mathcal F(T'-T''). It and its complementary branch are continuous and exhaust the preimage of the ball. They are disjoint open branches in that preimage, distinguished by volume distance from E less than, or greater than, \sigma_{m+1}/2. Together with (4.1), this proves the assertion.□

Proof. Radial projection identifies the centrally symmetric convex body Q with the closed p-ball, respecting antipodes. Its boundary-antipodal quotient is therefore \mathbb{RP}^{p}. Complementary fillings have the same mod-two boundary, so (4.1) and the quotient property give a flat-continuous descended map \Phi. The fillings themselves specify local branches of its pulled-back cover, not a single filling on the quotient. At a boundary seam one uses E_q in one collar and the complementary choice in the collar of -q; these choices agree along the identified boundary and are continuous in volume.

The segment b\mapsto(0,b) becomes a closed loop in this quotient. Its displayed lift starts at the empty set and ends at the whole sphere. Hence \Phi^*\lambda is the nonzero degree-one class \alpha of \mathbb{RP}^{p}. The projective cohomology ring is

[14, Theorem 3.19, p. 220]. In particular (\Phi^*\lambda)^p=\alpha^p\ne0.□

Proof. First consider the constrained class of finite-perimeter sets

It is nonempty: the union of two antipodal caps, with their common radius chosen to give half the spherical volume, belongs to it. Define

This number is finite. A minimizing sequence has bounded perimeter; compactness for sets of finite perimeter on the compact sphere yields a subsequence converging in L^1 to a set E of finite perimeter. The volume and all centroid coordinates pass to the limit, while perimeter is lower semicontinuous. Thus E\in\mathcal A_m realizes (4.4). These are the usual BV compactness and lower-semicontinuity statements in local charts [25, Chapter 2, Theorem 2.6, pp. 54–55]; a finite chart cover gives the compact-sphere assertion used here.

Spherical isoperimetry says that a half-volume set has perimeter at least \sigma_m, with equality only for a hemisphere modulo null sets [5, §2.3, equation (2.2), p. 5]. The centroid of the hemisphere \{y\cdot a>0\}, |a|=1, is

The minimizer in \mathcal A_m therefore cannot attain \sigma_m, proving \gamma_m>\sigma_m.

Now pull the filling cover back by \Phi. On this cover the function

is continuous and changes sign under E\mapsto E^c. If it never vanished, normalization would give an equivariant map to S^{p-1} and hence a quotient map f:X\to\mathbb{RP}^{p-1}. Fiberwise this identifies the filling cover with the pullback of S^{p-1}\to\mathbb{RP}^{p-1}: each of its two points is sent to one of the two antipodal unit vectors. It follows that \Phi^*\lambda=f^*\alpha, so (\Phi^*\lambda)^p=f^*(\alpha^p)=0, contrary to detection. At a zero of (4.5) the filling belongs to \mathcal A_m, and its cycle has mass \operatorname{Per}(E)\ge\gamma_m by (4.1).□

Proof. Embed \mathbb{RP}^{p} in the Euclidean space of real symmetric (p+1)\times(p+1) matrices by [u]\mapsto uu^t, |u|=1. Each such matrix has a simple top eigenvalue 1, separated from the remaining eigenvalues 0. On a neighborhood U of this compact image the top eigenvalue remains simple, and its eigenline gives a continuous retraction to \mathbb{RP}^{p}. Choose a sufficiently fine finite cubical grid in a larger cube, and let K be the union of the closed cubes meeting the image, together with their faces. Then K\subset U. After an affine rescaling it is a finite cubical subcomplex of a subdivided unit cube.

Let i:\mathbb{RP}^{p}\to K be the inclusion and r:K\to\mathbb{RP}^{p} the restricted retraction. Since r\circ i=\mathrm{id}, the map r^* is injective on cohomology. Thus \Phi\circ r detects whenever \Phi does. Surjectivity of r makes the two cycle images identical, proving both remaining assertions. The dimension of K may exceed p. Only its finiteness and its detecting cohomology class are needed here.□

Proof. We specify both the imported results and the width comparison. For a discrete map \phi on the vertices of a cubical subdivision, its fineness is

An (X,\mathbf M)-homotopy sequence consists of uniformly mass-bounded discrete maps whose successive maps are joined, after subdivision, by discrete homotopies of fineness tending to zero. Two sequences represent the same class if their corresponding maps are joined by such homotopies with fineness tending to zero. These are the conventions of [17, §§2.2–2.6, pp. 6–7].

The discretization theorem [17, Theorem 3.9, p. 13], applied using (4.3), provides such a sequence \mathcal S=\{\phi_i\} with \mathbf f(\phi_i)\to0 and

For any sufficiently fine discrete map, the Almgren extension \widehat\phi:X\to\mathcal Z_m^0(S^{m+1};\mathbb Z_2) is continuous in the mass norm and satisfies

where C_X is fixed [17, Theorem 3.10, p. 14]. Moreover, the extensions of \phi_i are flat homotopic to \Phi for all sufficiently large i [17, Corollary 3.12(ii), p. 15].

Let \Pi be the class of \mathcal S, and write

Certainly \mathbf L(\Pi)\le D_0. Fix one competitor \mathcal S'\in\Pi. Its homotopies to \mathcal S have fineness tending to zero. Therefore their Almgren extensions are flat homotopic for all sufficiently large indices, by [17, Corollary 3.12(i), p. 15]. Each \widehat\phi_i' is then detecting. Lemma 4.4 and (4.7) give

Taking the upper limit proves \gamma_m\le\mathbf L(\mathcal S'). Since the competitor was arbitrary,

The index beyond which the extensions detect may depend on \mathcal S'; no threshold uniform over all competitors is used. This is the continuous/discrete detection compatibility also recorded in [17, Lemma 4.6, p. 16].

The Almgren–Pitts existence theorem in [17, Theorem 2.14, p. 10] now gives a stationary integral varifold at mass \mathbf L(\Pi), almost minimizing in annuli. That theorem has no dimension restriction. Its low-dimensional regularity consequence is [17, Theorem 2.11, pp. 8–9], which gives the stated smooth embedded support for 2\le m\le6. These results are the mod-two form of the Almgren–Pitts theory; the integrality input is [23, Theorem 3.13], and the regularity extension is [24, Theorem 4]. This proves (4.6). For projective parameters, first apply Lemma 4.5.□

Proof. Define H(z)=\max_{x\in\Sigma}\langle z,x\rangle. Lemma 2.4 gives \int_\Sigma x\,d\mu=0 and fullness. Thus H(z)>0 for z\ne0: a nonpositive coordinate function of mean zero would vanish identically. Compactness gives c=\min_{|z|=1}H(z)>0. Minimizing first along each ray of the cone gives

This proves (5.5) with K=\sqrt{1-c^2} and excludes the vertex as a closest point to nonzero z.

A closest ray is determined by a maximum x of \langle z,\cdot\rangle, and its radial coordinate is H(z). At this maximum the tangential component of z vanishes, so z=H(z)x+\widetilde hN(x), where |\widetilde h|=|h|. The segment from z to its closest point meets no earlier cone point. Its side near contact therefore agrees with the side of z, proving \widetilde h=h with our signed-distance convention. Finally, at this height maximum the spherical immersion formula gives

These arguments apply to each maximum separately and therefore also apply when there are several closest points.□

Proof. For fixed v,y, consider

The Lipschitz property of \varphi gives, when b_2>b_1,

The same estimate implies limits -\infty and +\infty at the two ends of the real line. Hence there is exactly one zero, and the local positive lower bound for C-S proves continuous dependence of that zero on finite parameters and y.

At a graph point with |y|=1, (5.5) and Lorentz invariance give

Equality |b|=1 would force z=h=0, contrary to |y|=1 and invertibility of the boost. Thus |b|<1. Since b+\langle v,y\rangle=h/C, we also obtain

This tends to zero as |v|\to1, uniformly in y and in the boost direction, proving the claimed extension.□

Proof. Take y\in\Lambda_{v,b} and transform (y,b) to (z,h) by (5.7). Lemma 5.2 supplies a closest point r_zx. Lorentz invariance gives r_z^2=|z|^2-h^2=1-b^2, so r_z=r. The inverse formulas then show that h=h_{v,b}(x), y=Y_{v,b}(x), and x\in T_{v,b}. Therefore

The nonnegative eigenvalues of rI-h_{v,b}A have sum mr, because \tr A=0. Their arithmetic–geometric mean inequality proves (5.15). The area formula [9, §3.2.3] [10, §3, pp. 299–300], with multiplicities counted in its source integral, now yields the second inequality in (5.14). The condition defining T_{v,b} is only necessary for a closest-point contact. Additional points in this source set, and multiple preimages of a level point, increase the upper bound and cause no difficulty.

In particular \mathcal H^m(\Lambda_{v,b})<\infty. Since E_{v,b} is open, its essential boundary is contained in its topological boundary, which is contained in \Lambda_{v,b}. The finite-essential-boundary criterion [10, §4, p. 302], applied in coordinate charts, gives finite perimeter. The intrinsic reduced-boundary representation [5, §2.2] then gives

This proves the first inequality with constant one.

For |v|<1 the endpoint sublevels are empty and full, respectively. For |v|=1, formula (5.8) makes them round caps, whose boundary masses are \sigma_m(1-b^2)^{m/2}, with zero mass at b=\pm1. These cases also have finite perimeter.□

Proof. We prove the sequential form of the assertion. Consider u_j=(v_j,b_j)\to(v_*,b_*), y_j\to y_*, and \rho_j\downarrow0. The case |b_*|=1 follows directly from (5.1). Parameters on |v|=1 are harmless as well: an m-sphere of Euclidean radius r in any affine (m+1)-plane has at most c_m\rho^m area in a ball of radius \rho, uniformly in r and the plane. For r\le2\rho use its total area; for r>2\rho the intersection lies in a spherical cap with angular radius at most \arcsin(\rho/r), and the same bound follows by the cap area formula. Spherical balls in the ambient unit sphere lie in Euclidean balls of the same radius. We may therefore pass to a subsequence with |v_j|<1 and |b_*|<1.

Put Y_j=Y_{v_j,b_j} and T_j=T_{v_j,b_j}. The local perimeter bound reduces the desired assertion to

The restricted bound gives 0\le I_j\le1 on the fixed compact source.

If |v_*|<1, the maps Y_j converge smoothly on \Sigma to Y_*=Y_{v_*,b_*}. Away from the fiber Y_*^{-1}(y_*), the ball indicator in (5.19) eventually vanishes. The area formula applied to the fiber gives

because a singleton has zero \mathcal H^m measure. Thus J_mY_*=0 almost everywhere on that fiber, and smooth convergence gives I_j\to0 there almost everywhere as well. Dominated convergence completes this case.

It remains to consider v_*=a with |a|=1. Write a_j=v_j/|v_j|, t_j=S_j/C_j=|v_j|, and f_j=\langle a_j,x\rangle, g_j=\langle a_j,N\rangle. The cancellation identity

gives

On compact subsets of \Sigma\setminus Z_a, where Z_a=\{x:N(x)=a\}, the denominators stay uniformly positive. Also

The maps and all their source derivatives therefore converge locally to

This computation allows the directions a_j to vary.

The exceptional normal fiber Z_a is source-null. Indeed Z_a\subset\{x:\langle a,x\rangle=0\}, whereas \langle a,x\rangle is a nontrivial analytic function by Lemma 2.4. Its zero set has zero m-dimensional measure. No rank assumption on the Gauss map is involved.

The point fiber F=Y_\infty^{-1}(y_*), defined off Z_a, may instead have positive source measure. On a countable cover by relatively compact coordinate patches in \Sigma\setminus Z_a, the area formula gives

for each such patch K. Hence J_mY_\infty=0 almost everywhere on F. Off F\cup Z_a the shrinking-ball indicator is eventually zero; on F, local smooth convergence gives J_mY_j\to0 almost everywhere. We conclude that I_j\to0 almost everywhere on \Sigma, and 0\le I_j\le1 again permits dominated convergence. In particular, convergence of the indicators \mathbf1_{T_j} is unnecessary. Only the restricted Jacobians are bounded; no bound for the unrestricted J_mY_j near Z_a has been used.

Finally, if (5.18) failed, compactness of Q and S^{m+1} would supply sequences of the above form whose ball masses stay bounded below by a positive constant. The sequential result excludes them.□

Proof. Suppose otherwise. There are u_j=(v_j,b_j)\in Q\setminus U with \mathbf M(\widehat\Phi(u_j))\to|\Sigma|. The mass bound forces b_j\to0. Parameters on the boost boundary have mass at most \sigma_m, which is strictly smaller by (5.22); discard those terms. After taking a subsequence, v_j\to v_*\ne0. Set t_*=|v_*|\in(0,1] and a=v_*/|v_*|. Discard finitely many terms so that v_j\ne0. For each j, set t_j=|v_j|, a_j=v_j/|v_j|, C_j=(1-t_j^2)^{-1/2}, r_j=\sqrt{1-b_j^2}, f_j=\langle a_j,x\rangle, and g_j=\langle a_j,N\rangle.

The open set \{A\ne0\} is nonempty. The analytic function f=\langle a,x\rangle cannot vanish on that entire open set, by fullness and connectedness. Choose a compact source patch K_0 of positive measure on which |A| and |f| are bounded below by positive constants. It is also disjoint from \{N=a\}, because N=a implies f=0. On this fixed patch, the formula for h_j=h_{v_j,b_j} becomes

uniformly. Its limit is finite and nonzero on K_0, including when t_*=1.

For a real symmetric matrix M, define its truncated determinant by

This is continuous: every boundary point of the positive-semidefinite cone is singular, so its determinant vanishes. At every x\in K_0, the matrix I-h_*(x)A(x) has trace m and is not the identity. If it is positive semidefinite, strictness in the arithmetic–geometric mean inequality gives determinant less than one; if it is not, \mathcal D is zero. Compactness of K_0 and continuity therefore give \delta>0 with

For all large j this yields \mathcal D(r_jI-h_jA)\le r_j^m-\delta on K_0. On all of \Sigma, the restricted determinant is at most r_j^m. Thus the area formula estimate gives

contradicting convergence to |\Sigma|. The proof supplies a uniform loss on a fixed portion of the source; it makes no upper-semicontinuity assumption about area in the flat topology.□

Proof. Use a tubular neighborhood of the smooth cone on a compact annulus containing \{sx:x\in\Sigma,\ 1/2\le s\le3/2\}, constructed inside a slightly larger annulus. Compact embeddedness supplies a uniform \epsilon>0 so that, for these central radii and |h|<\epsilon, the point sx+hN(x) has unique closest cone point sx, and signed distance h. This is a tube for the whole cone: after shrinking \epsilon, points at radii outside the larger annulus cannot be closest, and uniqueness within that annulus follows from its tubular neighborhood.

For any actual level point y\in\Lambda_{v,b}, the boost gives

Every closest cone point has radius r=\sqrt{1-b^2}\to1. Thus all actual contacts lie in this one fixed tube, so each is unique. Conversely, for every source point x, the inverse formula gives

uniformly on \Sigma. Consequently z=rx+h_{v,b}(x)N(x) lies in the same tube and really has signed distance h_{v,b}(x). Its transformed point Y_{v,b}(x) is therefore an actual level point. This proves equality with the entire level and uniqueness of its source representation, not merely existence of one nearby component.

For these parameters the matrix rI-h_{v,b}A is uniformly positive definite, so (5.13) gives an immersion. Uniqueness of the contact and invertibility of the boost give injectivity; compactness then gives an embedding. All the formulas are smooth in (v,b,x) near the center.

In the tube d\varphi=N\cdot dz. Differentiating the graph relation at one of the level points yields

The derivative with respect to b is D>0, so the graph is smooth near that point and its Euclidean gradient is N'. Projecting onto the tangent space of the unit sphere and using (5.12) proves (5.23). Each level point therefore has a neighborhood in which E_{v,b} is one side of a regular smooth hypersurface. This proves the claimed topological and reduced boundary identities, with multiplicity one.□

Proof. At a spherical point x with tangent plane P\subset x^\perp, assign to rx the plane \mathbb Rx\oplus P and the measure r^d\,dr\,dV(x,P). This defines the cone and gives (6.1). The polar map is a diffeomorphism away from the origin; rectifiability and integer multiplicity there follow from the area formula. The origin has zero mass.

For stationarity, first take a smooth ambient field X supported away from the origin and write X(rx)=f(r,x)x+Y_r(x), where Y_r(x)\perp x. The trace of its derivative on the cone plane is

After multiplication by r^d and integration, the first two terms integrate to the endpoint values of r^df, and the last term vanishes by spherical stationarity for each r. Thus the cone is stationary off the origin. For a general compactly supported X, insert a radial cutoff that is zero on B_\varepsilon and one outside B_{2\varepsilon}, with derivative bounded by c/\varepsilon. Formula (6.1) bounds the omitted first-variation terms by c_X(\varepsilon^{d+1}+\varepsilon^d), which tends to zero.

For the last assertion, [\mathcal C(V)] has zero boundary on every annulus: an odd component multiplicity gives a smooth mod-two product cylinder, and an even multiplicity gives zero. Truncate this chain inside radius \varepsilon. Its difference from the full chain has mass at most \mathbf M(V)\varepsilon^{d+1}/(d+1), whereas its inner boundary has mass at most \mathbf M(V)\varepsilon^d. The truncated chains therefore converge locally flat to [\mathcal C(V)] and their boundaries converge locally flat to zero. Continuity of the boundary operator proves \partial[\mathcal C(V)]=0 also at the vertex.□

Proof. Integrality gives density at least one at \|V\|-almost every point. Every neighborhood of z has positive mass, so choose such points x_i\to z. If \overline{B_r(z)}\subset U and d_i=|x_i-z|<r, monotonicity at x_i and ball inclusion give

Let i\to\infty, then r\downarrow0. This proves the asserted lower bound also at support points that are exceptional for rectifiability.

For a cone, its homogeneous mass measure gives \|\mathcal C\|(B_R(0))=\omega_k\Theta(\mathcal C,0)R^k. For R>|z|, the inclusions

identify the limit of the mass ratios centered at z as R\to\infty with \Theta(\mathcal C,0). Monotonicity proves the upper bound in (6.3). Tangent-cone existence and the monotonicity formula give the stated tangent density, which is positive by the first part.□

Proof. Write T_i=(\eta_{z,\rho_i})_\#\mathcal C\to T, where \eta_{z,\rho_i}(y)=(y-z)/\rho_i and \rho_i\downarrow0. For fixed b\in\mathbb R, dilation invariance of \mathcal C by the positive factors 1+b\rho_i implies invariance of T_i under

Passing to the limit shows that T is invariant under translation by bz. Its weight consequently disintegrates as \|T\|=\mathcal L^1\otimes\nu on \mathbb Re\oplus e^\perp.

We check that this is an integral varifold product. At almost every rectifiable point of T, a measure blow-up is an integer multiple of the measure on its approximate tangent plane. These blow-ups retain invariance under translations in the e-direction, so that plane contains e. The tangential Jacobian of the coordinate \pi(te+w)=t is therefore one almost everywhere. Rectifiable coarea gives, for almost every t, a (k-1)-rectifiable slice with the original integer multiplicities; its tangent plane is the intersection of the tangent plane of T with e^\perp. Product disintegration identifies the weight of this slice, after translation to e^\perp, with \nu. Thus \nu defines an integral varifold W and the tangent-plane identification gives T=\mathbf v(\mathbb Re)\times W as varifolds. Here we use rectifiable coarea [10, §3, p. 300].

For a compactly supported field Y on e^\perp and a compactly supported smooth function \chi on \mathbb R with \int\chi\ne0, test stationarity of T with X(te+w)=\chi(t)Y(w). Since Y is perpendicular to e,

Hence W is stationary. Dilation invariance of the tangent cone T and the product formula imply dilation invariance of W. Writing \theta=\Theta(W,0), integration over a ball gives

This proves (6.4); Lemma 6.2 makes W nonzero.

If \partial[\mathcal C]=0, every T_i is cyclic, and (6.2) implies \partial[T]=0. The mod-two slicing identity for the boundary [32, §3, equation (1) and Proposition 3.1] gives

where signs disappear modulo two. For the locally flat chain, apply the identity on a countable exhaustion by relatively compact regions; restriction compatibility gives one full-measure set of levels where it holds locally everywhere. Choose one such level where the preceding rectifiable slice description also holds. The slice there is the translate of [W] to te+e^\perp, because T is the product just established. Translation back proves \partial[W]=0. No assertion about whether zero is a good slicing level is needed.□

Proof. Apply Allard’s interior regularity theorem [1, §8], in the formulation of [25, Chapter 5, hypotheses (5.1) and Theorem 5.2], on a sufficiently small rescaled ball from the sequence giving that tangent. Its hypotheses hold as follows. Integrality supplies the almost-everywhere density lower bound one. Stationarity supplies zero generalized mean curvature and zero first variation, hence the required mean-curvature bound for any exponent greater than k. Convergence to the multiplicity-one plane supplies normalized mass arbitrarily close to \omega_k and arbitrarily small height and tangent-plane excess on a fixed smaller ball. The center remains in the support. Allard’s theorem therefore gives a single C^{1,\alpha} graph near the center. On that connected graph the stationary integral multiplicity is constant; its tangent multiplicity is one, so the constant is one. The higher-regularity conclusion of [25, Chapter 5, Theorem 6.1], applied with the smoothly prescribed mean curvature identically zero, makes the graph smooth; that statement permits arbitrary codimension. This theorem has no stability hypothesis and no upper bound on k.□

Proof. Hold m fixed throughout. We prove (i) by induction in k and then obtain (ii) from the final induction step.

An integral one-dimensional cone is a finite union of rays with positive integer weights: radial slicing of its conical rectifiable measure gives a zero-dimensional integral measure on the unit circle, and local finiteness makes its total weight finite. Write the distinct unit ray directions as u_j, their weights as q_j, and N=\sum_jq_j. Stationarity at the origin says

If m\geq4, Lemma 2.1 gives \vartheta_m\leq3/2, so the strict bound (6.5) gives N<3. Nonzero stationarity excludes N=1, leaving two opposite rays of weight one. If m=2 or 3, the same lemma gives \vartheta_m<2, hence N<4; cyclicity says that the boundary coefficient N at the origin is even. Again N=2 and balance gives a multiplicity-one line. This proves the base case in both branches.

Suppose next that 2\leq k\leq m+1 and that (i) has been proved in dimension k-1. At any nonzero support point z, Lemma 6.3 expresses a tangent as \mathbf v(\mathbb R)\times W. Its factor is nonzero, stationary and integral, is cyclic when required, and has density at most \Theta(\mathcal C,0)<\vartheta_m by Lemma 6.2. The induction hypothesis makes W a multiplicity-one plane. Thus this tangent of \mathcal C is a multiplicity-one plane, and Lemma 6.4 proves smooth embedded multiplicity-one structure near z.

It follows that \Lambda=\spt\|\mathcal C\|\cap S^k is a smooth closed embedded (k-1)-manifold. The radial direction lies in the cone tangent plane and is normal to the sphere, so this intersection is transverse. Polar coarea gives \mathcal C=\mathcal C(\mathbf v(\Lambda)) with multiplicity one. Reversing the angular first-variation calculation in Lemma 6.1 shows that \Lambda is minimal in S^k.

Each connected component \Lambda_j is compact and minimal, and its cone is stationary. Lemma 6.2 applied to that cone, together with (6.1), yields

Since the total cone density is less than \vartheta_m<2, there is only one component.

If k=2, a closed connected embedded minimal curve in S^2 is a great circle by the geodesic equation, so its cone is a plane. If 3\leq k\leq m and \Lambda were not totally geodesic, the outer induction in dimension k-1 would give

contrary to (6.5). Thus \Lambda is a great sphere and \mathcal C a multiplicity-one plane, completing (i). For k=m+1, the already proved plane assertion in dimension m gives exactly the smoothness and connectedness asserted in (ii). We do not apply the dimension-m volume statement to this last link.□

Proof. Lemma 6.1 and the mass bound put \mathcal C(V) under part (ii) of Proposition 6.5. Its smooth multiplicity-one link identifies V by the polar cone formula. An equator has area \sigma_m.□

Proof. For every member of \mathcal E_m, the family of Proposition 5.1 is detecting and has maximum mass at most its area. Lemma 4.4 therefore gives \Vol(\Sigma)\geq\gamma_m>\sigma_m. Nonemptiness gives H<a_m, proving (6.7).

Choose \Sigma_i\in\mathcal E_m with areas H_i\to H and set \mathcal C_i=\mathcal C(\mathbf v(\Sigma_i)). These cones are stationary, integral, and cyclic by Lemma 6.1. Their exact masses satisfy

Consequently their masses are uniformly bounded on every compact set. Allard’s integral-varifold compactness theorem and a diagonal subsequence give a stationary integral limit \mathcal C in all of \mathbb R^{m+2}. Every fixed dilation leaves each \mathcal C_i invariant, so by continuity of pushforward it leaves \mathcal C invariant as well. White’s theorem in the form (6.2) gives \partial[\mathcal C]=0.

We verify that the limiting mass is exactly H at the spherical level. The weight of the conical limit is homogeneous of degree m+1. It has zero mass on every sphere of positive radius: otherwise dilation would produce infinitely many disjoint spheres with uniformly positive mass in a fixed compact annulus. Weak convergence can thus be evaluated on every centered ball. Passing to the limit in (6.8) yields

In particular \mathcal C is nonzero. This exact identity retains the infimum value; a lower-semicontinuity bound would not give the same conclusion by itself.

Proposition 6.5(ii) now makes the link \Sigma_*=\spt\|\mathcal C\|\cap S^{m+1} a smooth closed connected embedded minimal hypersurface, with multiplicity one. Its area is exactly H. The bounds \sigma_m<H<a_m exclude an equator and put \Sigma_* in \mathcal E_m, as required.□

Proof of Theorem 1.1. We induct on m\ge2. Fix m, assuming the theorem in dimensions 2,\ldots,m-1; this hypothesis is empty when m=2. By Proposition 2.2, every counterexample in dimension m would be a smooth closed connected embedded minimal hypersurface.

Suppose such counterexamples exist. Proposition 6.7 gives one of least area H, denoted by \Sigma, with

Set p=m+3. Proposition 3.5 gives \ind(\Sigma)\ge p. Equality would make \Sigma a minimal Clifford product and force H\ge a_m. Therefore \ind(\Sigma)>p.

By Proposition 5.1, there is a detecting family without mass concentration whose supremum mass D_1 satisfies D_1<H. Theorem 4.6 yields a stationary integral m-varifold V on S^{m+1} with

When m=2,3, choose the smooth embedded min–max output provided by that theorem, with its integer component multiplicities. Its Euclidean cone is cyclic modulo two by Lemma 6.1. In every higher dimension the cone is stationary and integral, which are the required hypotheses.

Corollary 6.6, using the outer induction hypothesis, now shows that V is a smooth connected embedded minimal hypersurface of multiplicity one. Its mass exceeds \sigma_m, so it is non-equatorial. It is therefore a counterexample with area strictly smaller than H, a contradiction.

This proves the embedded assertion in dimension m, and Proposition 2.2 gives the full immersion assertion. The induction proves the theorem for every m\ge2.□