We will prove the result by contradiction. So let us assume it is false. Then there exists a sequence $(M^{n}_{j},g_{j},p_{j})$ of Riemannian manifolds satisfying $|{\rm Ric}_{M^{n}_{j}}|\to 0$, ${\rm Vol}(B_{1}(p_{j}))>{\rm v}>0$ and such that

with $\beta<2\pi$ and $p$ a vertex.

Note first that by the noncollapsing assumption we have $\beta\geq\beta_{0}(n,v)$.

Now by Lemma 1.7, there exists $\delta_{j}$-splitting maps $u_{j}:B_{2}(p_{j})\to\mathds{R}^{n-2}$ with $\delta_{j}\to 0$. Fix some sequence $\epsilon_{j}\to 0$ which is tending to zero so slowly compared to $\delta_{j}$, that Theorem 1.8 holds for $u_{j}:B_{2}(0)\to\mathds{R}^{n-2}$ with $\epsilon_{j}$. Let $G_{\epsilon_{j}}\subseteq B_{1}(0^{n-2})$ be the corresponding good values of $u_{j}$, and let $s_{j}\in G_{\epsilon_{j}}\cap B_{10^{-1}}(0^{n-2})$ be fixed regular values.

Note that $\mathds{R}^{n-2}\times C(S^{1}_{\beta})$ is smooth outside of the singular set $\mathcal{S}=\mathds{R}^{n-2}\times\{0\}\subseteq\mathds{R}^{n-2}\times C(S^{1}_{\beta})$. In particular on $\mathds{R}^{n-2}\times C(S^{1}_{\beta})$ we have $r_{h}(x)\approx 1/d(x,\mathcal{S})$, where $r_{h}$ is the harmonic radius as in Section 1 and $d$ denotes distance. By the standard $\epsilon$-regularity theorem, it follows that the convergence of $M^{n}_{j}$ is in $C^{1,\alpha}\cap W^{2,q}$ away from $\mathcal{S}$, for every $\alpha<1$ and $q<\infty$. Let $f_{j}:B_{\epsilon^{-1}_{j}}(p)\to B_{\epsilon^{-1}_{j}}(p_{j})$ be the $\epsilon_{j}$-Gromov Hausdorff maps, and let us denote $\mathcal{S}_{j}\equiv f_{j}(\mathcal{S})\subseteq M^{n}_{j}$. Then by the previous statements, for every $\tau>0$, all $j$ sufficiently large, and $x\in B_{1}(p_{j})\setminus T_{\tau}(\mathcal{S}_{j})$, we have $r_{h}(x)\geq\frac{\tau}{2}$.

Consider again the submanifold $u^{-1}_{j}(s_{j})\cap B_{1}(p_{j})$. Define the scale

By the considerations of the previous paragraph, this minimum is actually obtained at some $x_{j}\in u^{-1}_{j}(s_{j})\cap B_{1}(p_{j})$, with $x_{j}\to\mathcal{S}_{j}\cap B_{10^{-1}}(p_{j})$. Moreover, since $S^{1}_{\beta}$, the cross-section of the cone factor, satisfies $0<\beta<2\pi$, it follows that $r_{j}\to 0$. According to Theorem 1.8, there exists a lower triangular matrix $A_{j}\in GL(n-2)$ such that $v_{j}\equiv A_{j}\circ\big(u_{j}-s_{j}\big):B_{r_{j}}(x_{j})\to\mathds{R}^{n-2}$ is an $\epsilon_{j}$-splitting map. Note that we have renormalized so that each of our regular values is the zero level set.

Now let us consider the sequence $(M^{n}_{j},r_{j}^{-1}d_{j},x_{j})$. After passing to a subsequence if necessary, which we will continue to denote by $(M^{n}_{j},r_{j}^{-1}d_{j},x_{j})$, have

in the pointed Gromov-Hausdorff sense, where $X$ splits off $\mathds{R}^{n-2}$ isometrically.

We begin by observing that by our noncollapsing assumption we have ${\rm Vol}(B_{1}(x_{j}))>c(n){\rm v}>0$, and hence, in the rescaled spaces, we have ${\rm Vol}(B_{r}(x_{j}))>c{\rm v}r^{n}$ for all $r\leq R_{j}\to\infty$. In particular, $X$ has Euclidean volume growth at $\infty$ i.e. ${\rm Vol}(B_{r}(x^{\prime}))>c{\rm v}\,r^{n}$ for all $r>0$.

After possibly passing to another subsequence, we can limit the functions $v_{j}$ to a function $v:X\to\mathds{R}^{n-2}$. Note that by our normalization, we have $v_{j}:B_{2}(x_{j})\to\mathds{R}^{n-2}$ are $\epsilon_{j}$-splittings, and that by Theorem 1.11, we have for each $R>2$ that $v_{j}:B_{R}(x_{j})\to\mathds{R}^{n-2}$ are $C(n,R)\epsilon_{j}$-splittings. In particular, we can conclude that

where $v:X\to\mathds{R}^{n-2}$ is the projection map and $S=u^{-1}(0)$.

Now by construction, in the rescaled spaces we have for any $y\in u^{-1}_{j}(0)$ that $r_{h}(y)\geq 1$. Therefore, the limit $X$ is $C^{1,\alpha}\cap W^{2,q}$ in a neighborhood of $u^{-1}(0)$, and hence $S=u^{-1}(0)$ is a nonsingular surface. Thus, since $X=\mathds{R}^{n-2}\times S$ it follows that $X$ is at least a $C^{1,\alpha}\cap W^{2,q}$ manifold with $r_{h}\geq 1$. Since the Ricci curvature is uniformly bounded, in fact tending to zero, we have by the standard $\epsilon$-regularity theorem that the convergence $(M^{n}_{j},r_{j}^{-1}d_{j},x_{j})\to(X,d_{X},x)$ is in $C^{1,\alpha}\cap W^{2,q}$. Because the convergence is in $C^{1,\alpha}\cap W^{2,q}$ we have that $r_{h}$ converges continuously; [A90]. In particular, we have that $r_{h}(x^{\prime}_{j})\to r_{h}(x^{\prime})$ and so $r_{h}(x^{\prime})=1$.

On the other hand, since $|{\rm Ric}_{M^{n}_{j}}|\to 0$ and $X$ is $C^{1,\alpha}\cap W^{2,q}$ it follows that $X$ is a smooth Ricci flat manifold. This is easiest to see by writing directly in harmonic coordinates on $X$, see [A90] for the argument. Now since $X=\mathds{R}^{n-2}\times S$, we can conclude that $S$ is smooth and Ricci flat, hence flat. In particular, we have that $X$ is flat. Since we have already shown that $X$ has Euclidean volume growth, this implies that $X=\mathds{R}^{n}$. However, we have also already concluded that $r_{h}(x^{\prime})=1$, which gives us our desired contradiction. ∎

Recall the standard stratification of $X$. In particular, if we consider the subset $\mathcal{S}^{n-3}\subset X$ we have that $\dim\mathcal{S}^{n-3}\leq n-3$, and that for every point $x\not\in\mathcal{S}^{n-3}$ there exists some tangent cone at $x$ which is isometric to $\mathds{R}^{n-2}\times C(S^{1}_{\beta})$. That is, there exists $r_{a}\to 0$ such that

However by Theorem 5.1 we then have $\beta=2\pi$, which is to say that

Thus, for $a\in\mathds{N}$ sufficiently large, we can apply the standard $\epsilon$-regularity theorem, Theorem 2.3, to see that a neighborhood of $x$ is a $C^{1,\alpha}$ Riemannian manifold, which proves the corollary. ∎

Let us assume that this is not the case and study such a limit space $\mathds{R}^{n-3}\times C(Y)$. The first observation is that by Corollary 5.2, it follows that $Y$ is a smooth surface. Indeed, if there were a point $y\in Y$ such that $r_{h}(y)=0$. Then since $X=\mathds{R}^{n-3}\times C(Y)$ it would follow that there is a set of codimension at least $2$ such that $r_{h}\equiv 0$, which cannot happen by Corollary 5.2.

Since $Y$ a $C^{1,\alpha}\cap W^{2,q}$ manifold and $|{\rm Ric}_{M^{n}_{j}}|\to 0$, it follows that $Y$ is a smooth Einstein manifold satisfying ${\rm Ric}_{Y}=g$. Because $Y$ is a surface, this means in particular that $Y$ has constant sectional curvature $\equiv 1$. Thus, either $Y=\mathds{R}\mathds{P}^{2}$ or $Y=S^{2}$, the unit $2$-sphere, and in the latter case we are done.

So let us study the case $Y=\mathds{R}\mathds{P}^{2}$. For $\epsilon>0$ small, choose $u_{j}:B_{2}(p_{j})\to\mathds{R}^{n-3}$ to be an $\epsilon$-splitting as in Lemma 1.7. Note that away from the singular set $\mathcal{S}\equiv\mathds{R}^{n-3}\times\{0\}$ we have that the $M_{j}^{n}$ converge to $\mathds{R}^{n-3}\times C(Y)$ in $C^{1,\alpha}$. If $f_{j}:B_{2}(p)\to B_{2}(p_{j})$ denote the Gromov-Hausdorff maps, we put $\mathcal{S}_{j}=f_{j}(\mathcal{S})$. Then for $\tau>0$ small but fixed, we have for $j$ sufficiently large, that on $B_{1}(p)\setminus T_{\tau}(\mathcal{S}_{j})$, the estimates $|\nabla u_{j}|>\frac{1}{2}$ and $|\nabla^{2}u_{j}|\leq 1$ hold.

Consider Poisson approximation $h_{j}$ to the square of distance function $d^{2}(x,p_{j})$ on $B_{2}(p_{j})$. That is, $\Delta h_{j}=2n$ and $h_{j}=1$ on $\partial B_{2}(p_{j})$. We have (see for instance [ChCo1]) that $|h_{j}-d(\cdot,p_{j})|\to 0$ uniformly in $B_{2}(p_{j})$, and again because the convergence is in $C^{1,\alpha}$ we have for $j$ sufficiently large that $|\nabla h|>\delta$ and $|\nabla^{2}h|\leq 4n$ on $B_{1}(p)\setminus B_{\tau}(\mathcal{S}_{j})$. Once again, appealing to the $C^{1,\alpha}$ convergence, for all $j$ sufficiently large and all $s\in B_{1}(0^{n-3})$ we have that $u^{-1}(s)\cap h^{-1}(1)$ is diffeomorphic to $\mathds{R}\mathds{P}^{2}$. By Sard’s theorem, there exists a regular value $s_{j}\in B_{1}(0^{n-3})$. Then for $j$ sufficiently large, $u^{-1}_{j}(s_{j})\cap\{h\leq 1\}$ is a smooth $3$-manifold, whose boundary is diffeomorphic to $\mathds{R}\mathds{P}^{2}$. However, the second Stiefel-Whitney number of $\mathds{R}\mathds{P}^{2}$ is nonzero, and in particular, $\mathds{R}\mathds{P}^{2}$ does not bound a smooth $3$-manifold. This contradicts $Y=\mathds{R}\mathds{P}^{2}$.

∎