Let $S^{1}_{\beta}$ denote the circle of circumference $\beta<2\pi$. It has been understood since [ChCo1] that to prove Theorem 1.1, the key step is to show that the cone $\mathds{R}^{n-2}\times C(S^{1}_{\beta})$ does not occur as the (pointed) Gromov-Hausdorff limit of some sequence $M^{n}_{j}$ with $|{\rm Ric}_{M^{n}_{j}}|\to 0$. This was shown in [CCT02] assuming just a lower bound ${\rm Ric}_{M^{n}_{i}}\geq-(n-1)$, but with the additional assumption that the $L^{1}$ norm of the curvature is sufficiently small. In [Ch2], it was proved for the Kähler-Einstein case, which was also done by Tian. A common feature of both of the proofs is an argument by contradiction, implemented by the use of harmonic almost splitting maps $u:B_{2}(p)\to\mathds{R}^{n-2}$, see Lemma 1.7. In each case, it is shown that for most points $s\in\mathds{R}^{n-2}$ in the range, the slice $u^{-1}(s)$ has a certain good property which, when combined with the assumed curvature bounds, enables one to deduce a contradiction. In particular, in [CCT02] it is shown that most slices $u^{-1}(s)$ have integral bounds on the second fundamental form, which when combined with the assumed integral curvature bounds, enables one apply the Gauss-Bonnet formula for $2$-dimensional manifolds with boundary, to derive a contradiction.

However, prior to the present paper it was not known how, in the general case, to implement a version of the above strategy which would rule out the cones $\mathds{R}^{n-2}\times C(S^{1}_{\beta})$ without assuming the integral curvature estimates. In the remainder of this subsection we will state the main results which are used in the present implementation and allow us to prove Theorem 1.1.

Thus, we consider a sequence of Riemannain manifolds $(M^{n}_{j},d_{j},p_{j})$, with $|{\rm Ric}_{M^{n}_{j}}|\to 0$ and ${\rm Vol}(B_{1}(p_{j})>{\rm v}>0$, such that

We wish to see that $\beta=2\pi$. As above, we have harmonic almost splitting maps

see Lemma 1.7 below. The key ingredient will be Theorem 1.8 (the Slicing Theorem), which states that there exist $s_{j}\in\mathds{R}^{n-2}$ such that for all $x\in u^{-1}_{j}(s_{j})$ and for all $r<1$, the ball $B_{r}(x)$ is $\epsilon_{j}r$-close in the Gromov-Hausdorff sense to a ball in an isometric product $\mathds{R}^{n-2}\times S_{j,x,r}$, where $\epsilon_{j}\to 0$ as $j\to\infty$.

Granted this, we can apply a blow up argument in the spirit of [A90] to obtain a contradiction. Namely, it is easy to see that if $\beta<2\pi$ then the minimum of the harmonic radius $r_{h}$ at points of the slice $u^{-1}_{j}(s_{j})$ is obtained at some $x_{j}\in u^{-1}_{j}(s_{j})$ and is going to zero as $j\to\infty$. We rescale the metric by the inverse of the harmonic radius $r_{j}=r_{h}(x_{j})$ and find a subsequence converging in the pointed Gromov-Hausdorff sense a smooth noncompact Ricci flat manifold,

such that $X=\mathds{R}^{n-2}\times S$ splits off $\mathds{R}^{n-2}$ isometrically, with $S$ a smooth two dimensional surface. It follows that $S$ is Ricci flat, and hence flat. From the noncollapsing assumption, it follows that $X$ has Euclidean volume growth. Thus, $X=\mathds{R}^{n}$ is Euclidean space. However, the $2$-sided Ricci bound implies that the harmonic radius behaves continuously in the limit. Hence, the harmonic radius at $x$ is $r_{h}(x_{\infty})=1$; a contradiction. See Section 5.1 for more details on the blow up argument.

Clearly then, the key issue is to show the existence of the points $s_{j}\in\mathds{R}^{n-2}$, such that at all points $x\in u_{j}^{-1}(s_{j})$, we have the above mentioned splitting property on $B_{r}(x)$ for all $r<1$. To indicate the proof, we now recall some known connections between isometric splittings, the Gromov-Haudorff distance and harmonic maps to Euclidean spaces $\mathds{R}^{k}$. We begin with a definition.

The following lemma summarizes the basic facts about splitting maps²² 2 In [ChCo1], only a uniform bound $|\nabla u|<C(n)$ is proved. This would actually suffice for our present purposes. The improved bound, $|\nabla u|<1+\epsilon$, in (1) above, is derived in (3.30)–(3.34), in a context that passes over almost verbatim to the present one.

Let us return to the consideration of the maps $u_{j}$ from (1.11), which in our situation arise from (2) of Lemma 1.7. We can thus assume that the $u_{j}$ are $\delta_{j}$-splitting maps, with $\delta_{j}\to 0$. We wish to find slices $u^{-1}_{j}(s_{j})$ such that $B_{r}(x)$ continues to almost split for all $x\in u_{j}^{-1}(s_{j})$ and all $r\leq 1$. One might hope that there always exist $s_{j}$ such that by restricting the map $u_{j}$ to each such ball $B_{r}(x)$, one obtains an $\epsilon_{j}$-splitting map. However, it turns out that there are counterexamples to this statement; see Example 2.1.

The essential realization is that for our purposes, it actually suffices to show the existence of $s_{j}$ such that for all $x\in u_{j}^{-1}(s_{j})$ and all $0<r\leq 1$, there exists a matrix $A=A(x,r)\in GL(n-2)$, such that the harmonic map $A\circ u_{j}:B_{r}(x)\to\mathds{R}^{n-2}$ is our desired $\epsilon_{j}$-splitting map. Thus, while $u_{j}$ might not itself be a splitting map on $B_{r}(x)$, it might only differ from one by a linear transformation of the image.³³ 3 Note that if such a matrix $A$ exists, without essential loss of generality, it can be chosen to be lower triangular. Since this condition also plays a role in the proof of Theorem 3.2, we will incorporate it from now on. This turns out to hold. More precisely, we have the following result.

The proof of the Slicing Theorem is given in Section 4. We now describe main steps in the proof.

To begin with, by using Bochner’s formula and the improved Kato inequality, $|\nabla|\nabla u^{a}|\,|^{2}\leq\frac{n-1}{n}|\nabla^{2}u^{a}|^{2}$, we show in Section 3.1 the following estimates on the ball $B_{2}(p)$.

As will be clear from Theorem 1.11 below (the Transformation theorem) that the following definition is key.

Note that there is an invariance property for (1.16). Namely, if (1.16) holds for $u$ then it holds for $A\circ u$ for any lower triangular matrix $A\in GL(k)$. That is, the singular scale of $u$ and the singular scale of $A\circ u$ are equal. In view of (1.15), this means essentially that (1.16) is a necessary condition for the existence of $A$ as in the Slicing theorem. Our next result, which is by far the most technically difficult of the paper, provides a sort of converse. We will not attempt to summarize the proof except to say that it involves a contradiction argument, as well as an induction on $\ell$. It is proved in Section 3:

Granted the Transformation theorem, let us return to the outline of the proof of the Slicing theorem. So consider the singular radius $s_{x}^{\eta}$, where $\eta(n,\epsilon)$ such that for $\delta<\eta^{2}$ the conclusions of Theorem 1.11 hold for $\epsilon>0$. Let $u:B_{2}(p)\to\mathds{R}^{n-2}$ denote a harmonic $\delta$-splitting map, and put

Let $|u(B_{r}(x))|$ denote the $(n-2)$-dimensional measure of the image $u(B_{r}(x))$. In view of the Transformation theorem, to conclude the proof of the Slicing theorem it suffices to show

for $\delta^{\prime}<<\epsilon$. To this end, we record two perhaps non-obvious, but easily verified consequences of Theorem 1.11.

Denote by $\mu$, the measure such that for all open sets $U$

The first consequence (see Lemma 4.1) is that for each $x$ and $1/2\geq r\geq s^{\eta}_{x}$, we have the doubling condition

Let $|u(B_{r}(x)|$ denote the $(n-2)$-dimensional measure of the image $u(B_{r}(x))$.

The second consequence (see Lemma 4.2) is that if $1/2\geq r\geq s^{\eta}_{x}$, then we have the volume estimate

The proof of these results exploits the fact that $A\circ u:B_{r}(x)\to\mathds{R}^{n-2}$ is an $\epsilon$-splitting map for some lower triangular matrix $A$.

By a standard covering lemma, there exists a collection of mutually disjoint balls, $\{B_{s_{j}}(x_{j})\}$ with $s_{j}=s^{\eta}_{x_{j}}$, such that

Since the balls $B_{s_{j}}(x_{j})$ are mutually disjoint, we can apply Theorem 1.9 together with (1.20) and the doubling property (1.19) of $\mu$ to obtain

where by Theorem 1.9 the last term tends to zero as $\delta\to 0$, as claimed. See Section 4 for a complete proof of the Slicing Theorem.