The Horowitz–Myers Conjecture

The Horowitz–Myers conjecture [HM98] is a positive mass statement for spaces with negative cosmological constant. The point of this section is to explain the Riemannian version proved by Brendle–Hung [BH24], and then to outline how their systolic inequality proves it.

From AdS energy to a Riemannian inequality

Recall first the analogy with the positive mass theorem. For asymptotically flat initial data, Euclidean space is the reference geometry and the expected lower bound for the ADM mass is \(0\). With a negative cosmological constant, the natural reference geometry is anti-de Sitter space. In spacetime language the Einstein equation is

For AdS geometry one takes \(\Lambda\lt{}0\); in our normalization,

On a time-symmetric spacelike slice, the second fundamental form vanishes and the Hamiltonian constraint becomes a scalar-curvature condition. In vacuum it is

and under the dominant energy condition it becomes

For asymptotically hyperbolic data with spherical conformal infinity, the ground state is pure AdS, whose time-symmetric slice is hyperbolic space. The Horowitz–Myers phenomenon begins when the conformal infinity is toroidal. In that case the product hyperbolic end is not the expected lowest-energy geometry. One circle direction at infinity can fill in smoothly in the interior, producing the AdS soliton on \(\mathbb R^2\times T^{n-2}\). This metric has negative mass relative to the product hyperbolic end. The conjecture says that, among metrics with the same toroidal asymptotics and \(R_g\geq -n(n-1)\), the AdS soliton has the least possible mass.

The Brendle–Hung mass inequality

Let \(\gamma\) be a flat metric on \(S^1\times T^{n-2}\). The model end is

on \((r_0,\infty)\times S^1\times T^{n-2}\). An asymptotically Horowitz–Myers end has an expansion

where \(Q\) is a symmetric \(2\)-tensor on \(S^1\times T^{n-2}\). The quantity

is the mass term in the normalization used here.

There is also a systolic quantity built into the toroidal end. Let \(\xi:S^1\times T^{n-2}\to S^1\) be the circle projection and let \(\Xi\) be the pullback of the volume form on \(S^1\). Define

Thus \(\sigma\) is the length of the shortest loop which winds nontrivially in the distinguished \(S^1\)-direction.

The left-hand side is the Horowitz–Myers mass term. The right-hand side is the mass of the corresponding AdS soliton. The rigidity statement, proved by Brendle–Hung in a subsequent work [BH25], says that equality forces the metric to be locally isometric to a Horowitz–Myers metric.

The systolic boundary inequality

The mass inequality is proved by cutting off the end and applying a sharp boundary inequality to the resulting compact manifold. We state that inequality in the compact form in which it is used.

Let \(M\) be a compact, connected, orientable \(n\)-manifold with nonempty boundary. Suppose

are smooth maps such that \((\xi,\theta):\partial M\to S^1\times T^{n-2}\) has nonzero degree. Let \(\Xi=\xi^*(d\theta_{S^1})\), where \(d\theta_{S^1}\) is the volume form of the circle. Similarly, let \(\Theta_j=\theta_j^*(d\theta_j)\) denote the pulled-back volume forms from the circle factors of \(T^{n-2}\). Define \(\sigma\) to be the shortest length of a closed curve \(\alpha\subset\partial M\) with \(\int_\alpha\Xi\neq0\). Let \(H_{\partial M}\) be the mean curvature of \(\partial M\) with respect to the outward unit normal \(\eta\).

For the mass theorem, one only needs the scalar-curvature consequence obtained by taking \(\varphi\equiv0\) and letting \(\beta\to n\).

The AdS soliton and sharpness

The sharp example is the Horowitz–Myers, or AdS soliton, metric. Geometrically one should picture the \(S^1\)-factor at infinity as a polar-angle direction which collapses smoothly in the interior. Thus the underlying manifold is \(\mathbb R^2\times T^{n-2}\), while the conformal infinity is \(S^1\times T^{n-2}\).

For simplicity, assume that \((S^1,g_{S^1})\) has length \(4\pi/n\). On \((1,\infty)\times S^1\times T^{n-2}\), set

With the substitution

this becomes

Near \(\tilde\rho=0\), the first two directions look like

The choice \(\operatorname{length}(S^1)=4\pi/n\) is exactly the no-cone-angle condition, so the metric extends smoothly across the collapsed circle. The resulting metric has scalar curvature

To compare it with the asymptotic model, define \(r\) by

Then

Thus the tensor \(Q\) is

and

Since \(\sigma=4\pi/n\), the right-hand side in Theorem 6.5.1 is exactly the mass of this metric. Hence the inequality is sharp.

How the Brendle–Hung proof works

We keep the proof in three parts. The first part converts the boundary inequality into the mass inequality. The second part proves the boundary inequality by dimension reduction. The final part is the two-dimensional endpoint, where the sharp constant is computed.

Step 1: from the boundary inequality to the mass inequality.

The goal is to prove Theorem 6.5.1:

The compact input is Corollary 6.5.3. Thus we need to cut off the end of \(N\) in such a way that the boundary mean curvature sees the mass aspect.

Choose \(u\) and a constant \(\mu\) on \(S^1\times T^{n-2}\) by

Equivalently,

For large \(\widehat r\), cut off the end by the graph

The reason for using this graph, rather than the coordinate torus \(\{r=\widehat r\}\), is that the graph makes the first nontrivial term in the mean curvature constant. The two estimates one needs are

and

Substituting these into the level-set formula

gives

Therefore Corollary 6.5.3 applied to \(\widehat N\) gives

where \(\widehat\sigma\) is the boundary systole on \(\partial\widehat N\). Since

we pass to the limit and recover the mass inequality. Equivalently, one may argue by contradiction: if the mass integral were too negative, then for some \(\varepsilon\gt{}0\),

whereas the boundary inequality gives

for large \(\widehat r\). Hence \(\widehat\sigma/\widehat r\leq (1-\varepsilon)\sigma\), contradicting \(\widehat\sigma/\widehat r\to\sigma\).

Step 2: dimensional reduction for the systolic inequality.

The goal is now Theorem 6.5.2. The proof reduces the \(n\)-dimensional boundary inequality to a two-dimensional one by repeatedly taking free-boundary hypersurfaces which preserve the relevant cohomological information.

The inductive object is a pair \((\Sigma^k,\varphi_k)\), where \(\Sigma^k\) has boundary and carries the nontrivial topological information coming from \(\Xi,\Theta_1,\ldots,\Theta_{k-2}\). The pair is required to satisfy

and the boundary term is

Starting with \(\Sigma^n=M\) and \(\varphi_n=\varphi\), the reduction step is:

Reduction proposition.

Assume \(E_k(\varphi_k,g_k)\geq0\) and \(B_k\geq0\). Then there is a compact free boundary hypersurface \(\Sigma^{k-1}\subset\Sigma^k\), stable for the weighted area functional

and a function \(\varphi_{k-1}\) on \(\Sigma^{k-1}\), such that

and

The hypersurface is chosen in the homology class detected by the remaining forms, so the relevant systole cannot decrease in the direction needed for the final inequality.

Let us indicate why the differential inequality is preserved. Write \(\bar\Sigma=\Sigma^k\), \(\widetilde\Sigma=\Sigma^{k-1}\), and \(\bar\varphi=\varphi_k|_{\widetilde\Sigma}\). All geometric quantities below are computed in the original metric \(g_k\). If \(\widetilde\nu\) is the unit normal of \(\widetilde\Sigma\subset\bar\Sigma\), then the second variation of \(\mathcal A_k\) gives, for every smooth test function \(\zeta\),

Here \(\widetilde A\) is the second fundamental form of \(\widetilde\Sigma\subset\bar\Sigma\). The usual Gauss equation and the free-boundary relation give

and

Thus the stability inequality may be weakened to

By the first-eigenfunction argument for this Robin problem, there is a positive function \(\omega\) on \(\widetilde\Sigma\) and a number \(\lambda\geq0\) such that

with boundary condition

Define

The Schoen–Yau rearrangement of the stability inequality gives

The only algebraic point is

Taking \(p=1/(\beta-k)\) gives

Therefore

This is exactly the inequality \(E_{k-1}\geq0\). The boundary equality follows directly from the Neumann condition for \(\omega\):

Iterating the reduction either stops early, in which case the boundary infimum is already nonpositive and the desired inequality is immediate, or reaches a surface \((\Sigma^2,\varphi_2)\). In the latter case the boundary terms are the same along the construction and, if \(\sigma_2\) is the corresponding systole on \(\partial\Sigma^2\), then \(\sigma\leq\sigma_2\). The two-dimensional estimate therefore gives

Step 3: the two-dimensional endpoint and the monotonicity.

The dimension reduction leaves a sharp inequality on a surface. This is the only place where the numerical constant \((4\pi/\beta)^\beta\) is produced explicitly.

The second case is proved by reducing further to a free-boundary geodesic. The main new estimate is the disk case.

For the disk case define the parallel domains

and let \(l=\sup\{s:\Omega_s\neq\emptyset\}\). Put

For almost every \(s\), the curve \(\gamma_s\) is piecewise smooth. If \(\Gamma(s)\) denotes its total curvature, then the comparison geometry gives

Since the connected components of \(\Omega_s\) are disks, Gauss–Bonnet gives

Now define

The fundamental monotonicity statement is

The computation is short enough to record the structure. Differentiate \(I\), use \(L'(s)\leq-\Gamma(s)\), use Gauss–Bonnet to replace \(\Gamma(s)\), and use the scalar inequality to control the integral over \(\gamma_s\). The last point is the elementary estimate

These inequalities combine to give

Since

the equivalent formulation is

This is the monotonicity one should remember.

Finally \(J(0)\leq J(l)\). Using Gauss–Bonnet on the original disk, this gives

Let \(\sigma=|\partial\Sigma|\) in the disk case. Write

The preceding inequality controls the boundary average, hence also the infimum:

Multiplying by \(2\sigma^\beta\) gives

Since \(0\leq F(l)\lt{}1\), we have

Therefore

Now set

The right-hand side becomes

This one-variable expression is maximized at \(x=4\pi/\beta\), and its maximum is

This proves Theorem 6.5.4. Step 2 gives Theorem 6.5.2, Corollary 6.5.3 follows by taking \(\varphi\equiv0\) and \(\beta\to n\), and Step 1 proves the Horowitz–Myers mass inequality.