An L² Einstein Gap for Nonnegatively Curved Four-Manifolds

Proof. Simple connectivity implies orientability and b_1=0. Poincaré duality therefore gives \chi(M)=2+b_2(M)\ge2. By Chern–Gauss–Bonnet,

On the space of algebraic curvature tensors, the maximum absolute sectional curvature is a norm: sectional curvatures determine the tensor. Equivalence of norms in this finite-dimensional space, together with 0\le\sec\le R/2, gives a dimensional constant C_0 such that

The average scalar curvature is positive, since otherwise R\equiv0 and (2.5) contradicts (2.4). Rescale so that its average is 12. The L^2 norm of E is unchanged.

For completeness, the almost-Schur argument [5, Section 2.1] in the form needed here is as follows. Solve \Delta f=R-12. The contracted Bianchi identity says \diver E=\tfrac14dR, so integration by parts gives

Since \Ric\ge0, the integrated Bochner formula yields \norm{\nabla^2f}_2\le\norm{\Delta f}_2. Cauchy–Schwarz proves the second assertion in (2.2). Orthogonality of the trace and trace-free parts gives

Moreover,

Combining this identity with (2.4) and (2.5) gives V\ge v_0>0 for all sufficiently large j.

It remains to control the diameter. Write D=\diam(M,g) and choose points realizing it. Let A_1,A_2 be the balls of radius D/4 about these points. Bishop–Gromov comparison gives \Vol(A_i)\ge8^{-4}V, by comparison with the ball of radius 2D. For any nonnegative continuous function H, the segment inequality is

where \gamma_{xy} is the minimizing geodesic, unique for almost every pair. This is the nonnegative-Ricci case of [3, Theorem 2.11]. To see the estimate directly, fix x and use polar coordinates y=\exp_x(r\theta) before cut time. On the half-segment r/2\le s\le r, Jacobian comparison gives

Changing the order of integration bounds the integral over this half, and over y, by CD\int_MH\dd. Integrating in x\in A_1 gives its contribution to (2.7). Reverse the geodesic and fix y to treat the other half. The cut locus has measure zero.

Apply (2.7) to H=|\Ric-3g|. Some minimizing geodesic between the two balls has length \ell\ge D/2 and satisfies

Use three parallel orthonormal normal fields along \gamma, each multiplied by \sin(\pi s/\ell), in the index form. Summing gives

Because D\le2\ell, we obtain (3/2-2\delta_j)\ell^2\le3\pi^2/2. Thus D is uniformly bounded. Bishop comparison now gives the uniform upper volume bound as well.□

Proof. For k\in\mathbb Z define

The assumed estimate implies

The gradients of f_k occupy disjoint layers up to null sets, and \sum_k f_k^2\le C_1f^2 pointwise by a geometric series. Consequently \sum_k\E(f_k)\le C_2\E(f), with numerical constants. Put b_k=2^{4k}m_k. The preceding estimate becomes

The sum B=\sum_kb_k is finite and comparable, with numerical constants, to \int_M|f|^4\dd: integrate the geometric series \sum_{2^k<|f|}2^{4k} pointwise. Summing over k and using b_k^{1/2}\le B^{1/2} gives B\le16CC_2B^{1/2}\E(f). Division by B^{1/2} when B>0 proves the result; the case B=0 is immediate.□

Proof. We establish (2.9) with a=P=1. Let \bar f=V^{-1}\int_Mf\dd. Polar integration of |f(x)-f(y)| along minimizing segments yields

Indeed, on a ray from x, the coefficient of |\nabla f| at distance s is bounded using

This proves (2.11); absolute continuity along almost every ray justifies it for Lipschitz functions.

For any measurable set A of volume m>0, comparison with Euclidean polar measure gives

To verify this, split the integral at radius \rho=m^{1/4}. The inner integral is at most C\rho because J_x(r,\theta)\le r^3; the outer integral is at most m\rho^{-3}. By symmetry the same bound holds with x and y interchanged. The integral operator with this kernel, restricted to A in both variables, therefore has L^2 norm at most Cm^{1/4} by the Schur estimate.

Take A=\{f\ne0\}. The gradient vanishes almost everywhere off A, and (2.3) bounds D^4/V. Taking L^2(A) norms in (2.11) gives

If m/V\le1/2, absorb the last term and square. If m/V>1/2, then m\ge v_0/2, and (2.9) follows directly from the \int f^2 term in the energy. Lemma 2.2 completes the proof.□

Proof. In dimension four, Perelman’s entropy is

where the infimum is over smooth positive f. Its monotonicity gives

[12, Section 3]. Here \tau+t is an entropy parameter, not a later flow time, so (3.3) imposes no additional lifespan requirement.

At time zero, Jensen’s inequality and (2.10) give

for \int f^2\dd=1. Since R\ge0,

The last bound follows by writing z=\tau(y+1): the expression in braces is 4z-2\log z-4\tau, uniformly bounded below in this range. Consequently (3.3) gives, at all times under consideration,

whenever \norm f_2=1 and 0<\tau\le1. Approximation of |f| by positive smooth functions extends this to Lipschitz f, with 0\log0=0.

Let m=\Vol\{f\ne0\}. For a normalized f, Jensen’s inequality on its support gives \int f^2\log f^2\dd\ge\log(1/m). Choose \tau=(1+\E_0(f))^{-1} in (3.4). It follows that

Exponentiation, followed by homogeneity for arbitrary f, yields

Apply Lemma 2.2 with a=4 and P=R+1.□

Proof. We first work on any interval where \norm w_2\le\eta for a small fixed \eta>0. Since R\le C|\Rm|\le C(K+w), the last term in (3.2) is bounded by a fixed multiple of \int f^2 plus

Choose \eta so that this term is absorbed into the left side. This gives (3.8) with a fixed S_2; decreasing \eta further does not change the chosen S_2.

On \{u>K\}, the equation for K removes the bounded-background reaction:

Integration by parts and (3.1) therefore give

There is no adverse volume term, since -\int Rw^2\dd\le0. The calculation is valid for the positive part: K>0, so u is smooth near \{u\ge K\}; the function (u-K)_+^2 is continuously differentiable, and Lipschitz truncation justifies integration by parts.

Hölder’s inequality and (3.8) give

Decrease \eta so that 2C_q\eta S_2\le1. Then Y(t)=\int w^2\dd satisfies Y'\le CY as long as Y^{1/2}\le\eta. By (3.7), for large j we have Y(0)e^{CT}<\eta^2. Gronwall’s inequality and continuity exclude a first time with Y^{1/2}=\eta before \min\{T,\mathcal T_j\}. This proves (3.8) and (3.9) on the entire existing interval. Finally, u\le K+w and \Vol(M,g_j(t))\le V_0 give (3.10).□

Proof. Write V_x(r)=\Vol B(x,r) on one smooth slice. Insert into (3.8) the distance cutoff that is 1 on B(x,r/2), vanishes outside B(x,r), and has gradient at most 2/r. Then

For a(r)=V_x(r)/r^4, this gives a(r)\ge c\,a(r/2)^{1/2}. Iteration yields

For this fixed smooth metric, a(s)\to\omega_4 as s\downarrow0, where \omega_4 is the volume of the Euclidean unit ball in \R^4. Letting k\to\infty gives a(r)\ge c^2. No uniform scale for the Euclidean asymptotic is required.□

Proof. Fix A>2 with A/T\ge1 and A/T\ge4\max_{[0,T]}K. Suppose that t\max_Mu(t) first reaches A at an existing time t<T. Set Q=\max_Mu(t)=A/t. On [t-Q^{-1},t] the curvature is at most 2Q, because t-Q^{-1}\ge t/2 and this is a first crossing. Parabolic rescaling gives a smooth closed flow on [0,1] whose curvature norm is at most 2. Shi’s first derivative estimate [14] gives

This estimate requires no initial derivative or injectivity-radius bound; one may use the local formulation in [10, Theorem 3.27] with the entire closed manifold as its domain.

Let x realize the curvature maximum. Choose a fixed c\in(0,1] small enough that on B_{g_j(t)}(x,cQ^{-1/2}) the derivative estimate implies u\ge3Q/4. Since Q\ge4K(t), we have w\ge Q/2 there. Also Q\ge1, so this radius is at most 1. Lemma 3.3 now gives

This contradicts the uniform convergence of the tail norm to zero in Proposition 3.2. Thus, for large j, no crossing occurs, and (3.13) holds throughout existing times below T.

If \mathcal T_j\le T, the same bound controls curvature on [\mathcal T_j/2,\mathcal T_j). Curvature is already bounded on the earlier compact interval by smoothness of the individual flow. The curvature continuation criterion contradicts maximality. Hence \mathcal T_j>T.□

Proof. Write L=\Ric-\lambda g. In moving orthonormal frames, define the curvature action on a symmetric tensor B by

The Ricci evolution in these frames is

The frame motion cancels the -2\Ric^2 term in the fixed-frame formula; the components of g in the moving frames are constant. Since \lambda'=2\lambda^2, subtraction yields

Using R\ge0, |\Rm|\le K+w, and boundedness of K and \lambda on [0,T], we obtain

By Hölder, (3.8), and Kato’s inequality,

The coefficient tends to zero uniformly in time, by (3.9). Absorb the gradient term for large j and apply Gronwall’s inequality. The initial norm tends to zero by (2.6), proving (4.1).□

Proof. The function F is nonnegative, convex, orthogonally invariant, and Lipschitz: it is a supremum of a bounded family of linear functionals and the zero function. Let F_\delta be its convolution with a nonnegative smooth radial mollifier supported in a ball of radius \delta in the algebraic-curvature tensor space. Then F_\delta is smooth, nonnegative, convex, and orthogonally invariant, with

Orthogonal invariance makes this construction independent of the orthonormal frame and compatible with the metric connection.

Apply the chain rule to (3.5) in moving orthonormal frames. Convexity gives a favorable Hessian term:

After integration, the volume variation contributes -\int R F_\delta(\Rm)\dd\le0. Closedness and (3.10) therefore imply

The last step uses F(\Rm(0))=0. Letting \delta\downarrow0 proves (4.4).□

Proof. Fix t_*=T/2. By (3.13), the flows have uniform global curvature bounds on every compact time interval in (0,T). At t_*, the local volume lower bound (3.12) and the total volume upper bound V_0 also give a diameter bound. Indeed, points spaced a fixed distance apart along a minimizing geodesic have pairwise disjoint balls of a fixed smaller radius; each such ball has uniformly positive volume. Their number is bounded by V_0. Write the resulting bound as \diam(M_j,g_j(t_*))\le D_*.

Absolute sectional curvature bounds and the ball-volume lower bound give a uniform positive injectivity radius at t_*, by the injectivity radius estimate of Cheeger–Gromov–Taylor [4, Theorem 4.7]. Hamilton’s compactness theorem [8] now gives, after passing to a subsequence, smooth pointed convergence to a connected Ricci flow (N,h(t)) on (0,T), with h(t_*) complete. One can apply the theorem on nested compact time intervals containing t_* and take a diagonal subsequence. The curvature bounds used here are global in space; hence the complete limit version of compactness applies, without the completeness issue that can arise from purely local bounds [16, Theorem 1.1 and subsequent discussion]. The convergence maps are time-independent embeddings defined on an exhaustion of N.

We explain why the limit is compact and the embeddings are eventually diffeomorphisms. If the complete connected slice (N,h(t_*)) were noncompact, it would contain a compact closed ball \overline B(p,L) with a point q at distance L, for any L>0. Choose L>2D_*. For large j, a convergence embedding is defined on a neighborhood of this ball, and its pulled-back metric bounds lengths on the ball from below by half their h(t_*) values. Any curve in M_j joining the images of p and q must have length at least L/2: if it leaves the image of the ball, its initial portion must first reach that image’s boundary and already has this length. If it stays inside, the same bound follows by pullback. This contradicts the diameter bound. Thus N is compact. Eventually the convergence embeddings are defined on all of N; their images are open, by the inverse function theorem, and closed, by compactness. Since M_j is connected, each is a diffeomorphism. It follows also that N is simply connected.

Smooth convergence on this compact manifold and (4.1) give

The Ricci-flow equation therefore integrates to

For each fixed t>0, (4.4) also passes to the limit. The function F is positively homogeneous of degree one. Under metric scaling by a positive constant c, its value on curvature scales by c^{-1} and the four-dimensional volume form scales by c^2. Consequently

Let t\downarrow0. The nonnegative continuous integrand must vanish, so \sec_h\ge0. This proves the proposition.□

Proof of Theorem 1.2. If no universal positive gap existed, then for every positive integer j there would be a connected simply connected closed four-manifold (M_j,g_j) with \sec_{g_j}\ge0 and \mathcal D(g_j)<1/j, whose smooth manifold is outside the three stated diffeomorphism types. Proposition 5.1 gives a subsequence diffeomorphic to a closed simply connected (N,h) with \Ric_h=3h and \sec_h\ge0. Classification Assumption 1.1 applies exactly to this metric. Since N is already simply connected, it is one of the three model manifolds, a contradiction.□