For any two points $p$, $q$ in $Y$, it is a general fact that a minimizing geodesic in $C(Y)$ connecting $p$ and $q$ must be of the form $(r(t),\gamma(t))$ where $\gamma(t)$ is a geodesic in $Y$, and $r$ is a universal function of $d_{Y}(p,q)$ and $t$ determined by elementary trigonometry. By recent result of Colding-Naber [7] we know $C(Y^{reg})$ is geodesically convex in $C(Y)$, so the lemma follows. ∎

By Lemma 5.2 for any $t\in[0,T]$, the minimizing geodesic $\gamma$ connecting $p_{1}(t)$ and $p_{2}(t)$ lies in $Y^{reg}$. So there is an $\epsilon>0$ so that the curve $\gamma_{s}=\exp(s\xi).\gamma$ is in $Y^{reg}$ for $s\in[0,\epsilon]$. Clearly the length of $\gamma_{s}$ is independent of $s$. Thus $f(t)$ is a decreasing function. Replace $\xi$ by $-\xi$ one sees that $f$ is also an increasing function. Thus $f$ is constant.
∎

Fix any point $p$ in $Y^{reg}$, choose a convex embedded ball $B(p,r)$ in $Y^{reg}$. We claim $B(p(t),r)\cap\Sigma=\emptyset$ for all $t$. For otherwise there is a $T>0$ such that $B(p(t),r)\cap\Sigma=\emptyset$ for $t\in[0,T]$ but $\partial B(p(T),r)\cap\Sigma$ is non-empty. Choose a point $q$ in this intersection. Let $\gamma:[0,1]$ be the radial geodesic connecting $p(T)$ and $q$, and let $p_{i}=\gamma(1-2^{-i})$. Then $B_{i}=B(p_{i},2^{-i}r)\subset B(p(T),r)$, and $d(p_{i},q)=2^{-i}$. Consider the pointed sequence $(Y,2^{i}d_{Y},q)$. By assumption we know as $i$ tends to infinity by passing to a subsequence this converges to a tangent cone $Y_{q}=\mathbb{R}\times\mathbb{C}^{2}/\Gamma$. Then the rescaled balls $2^{i}B_{i}$ converge to a ball $B(p_{\infty},r)$ in $Y_{q}$ and $d(p_{\infty},0)=r$. But $B_{i}$ is isometric to a ball in $B(p,r)$ so have uniformly bounded geometry and thus $2^{i}B_{i}$ converges to a flat ball $B_{\infty}$. Moreover by Lemma 5.3 the distance between any two points in $B_{\infty}$ is realized by the length of a geodesic within $B_{\infty}$. Clearly this can not happen on $Y_{q}$.

By the claim the isometric action $\exp(t\xi)$ is well defined on $Y^{reg}$ for all $t$. Then we can extend the action to an isometric action on $Y$: given $p\in\Sigma$ we pick a Cauchy sequence $p_{i}\in Y^{reg}$ converging to $q$; for any $t$, $p_{i}(t)=\exp(t\xi)$ is also a Cauchy sequence in $Y^{reg}$, so there is a unique limit $p(t)$. We define $exp(t\xi).p=p(t)$. Clearly $\exp(t\xi)$ is distance preserving. Moreover $\exp(t\xi)$ preserves both $Y^{reg}$ and $\Sigma$. ∎

If $q$ is a fixed point, then clearly $q\in\Sigma$. Choose a tangent cone $Y_{q}=\mathbb{R}\times\mathbb{C}^{2}/\Gamma$ at $q$. The action of $\psi$ induces a one parameter group of isometric actions on $Y_{q}$, which fixes the origin. On the other hand on the smooth part of $Y_{q}$ the corresponding infinitesimal action is given by a Killing field of constant length. Clearly such a Killing field can not have zeroes. Contradiction. ∎

Fix any $q\in\Sigma$. Since it is not a fixed point of $\psi$, we can choose a neighborhood $B_{r}(q)$ such that any path-connected component of the intersection of an orbit of $\psi$ with $\overline{B_{r}(q)}$ is compact. Let $O_{q}$ be one path-connected component of $\psi(q)$ in $B_{r}(q)$. We claim for $s>0$ sufficiently small, $\Sigma\cap B_{s}(q)=O_{q}\cap B_{s}(q)$. If not, then there is a sequence $p_{i}\in(B_{r}(q)\setminus O_{q})\cap\Sigma$ converging to $q$. We can choose $q_{i}$ on the path-connected component of the orbit of $p_{i}$ in $\overline{B_{r}(q)}$ which has least distance to $q$. Then $s_{i}=d(q,q_{i})>0$. For $i$ sufficiently large we have $d(q_{i},O_{q})=s_{i}$. Now consider the rescaled pointed sequence $(B_{r}(q),s_{i}^{-1}d_{Y},q)$. As $i\rightarrow\infty$, by passing to a subsequence, this converges to $\mathbb{R}\times\mathbb{C}^{2}/\Gamma$. Moreover, $O_{q}$ converges to $\mathbb{R}\times\{0\}$, and $q_{i}$ converges to $q_{\infty}$ which has distance $1$ to $\mathbb{R}\times\{0\}$. But $q_{i}$ is singular for all $i$, so $q_{\infty}$ is also singular. Contradiction. Then the Proposition follows from the claim and an obvious compactness argument. ∎

Assume the statement fails, then there is a constant $\tau>0$, a sequence $\epsilon_{i}\rightarrow 0$, and maps $\phi_{i}:\hat{A}(30,80)\rightarrow\mathbb{R}^{4}$ with $|\phi_{i}^{*}g_{0}-g_{0}|_{C^{4}(\hat{A}(30,80))}\leq\epsilon_{i}$, but for any isometry $P$ we have $|P\circ\phi_{i}-Id|_{C^{3}(\hat{A}(40,70))}\geq\tau$. Then $\phi_{i}$ converges to a map $\phi_{\infty}$ in $C^{3}(\hat{A}(40,70))$, such that $\phi_{\infty}^{*}g_{0}=g_{0}$. So $\phi_{\infty}$ is an isometry of $\mathbb{R}^{4}$. Since $|\phi_{\infty}^{-1}\circ\phi_{i}-Id|_{C^{3}(\hat{A}(40,70))}$ converges to zero as $i$ goes to infinity. We arrive at a contradiction. ∎

By the obvious scaling invariance we may assume $r=1$. Let $D=Im(f_{0})\cap Im(f_{1})$. Since $\epsilon$ is small, we may assume $\hat{A}(3,8)$ is contained in $f_{0}^{-1}(D)$. Then there is a constant $C_{1}$ independent of $\epsilon$ such that the map $\psi=10^{-1}f_{1}^{-1}\circ f_{0}:\hat{A}(3,8)\rightarrow\mathbb{R}^{4}$ satisfies $|\psi^{*}g_{0}-g_{0}|_{C^{4}}\leq C_{1}\epsilon$, and $(1-3\epsilon)|x|\leq|\psi(x)|\leq(1+3\epsilon)|x|$. By Lemma 5.8 there is an isometry $P$ of $\mathbb{R}^{4}$ such that $|P\circ\psi-Id|_{C^{3}}\leq K(C_{1}\epsilon)$ on $\hat{A}(4,7)$. We write $P(x)=R(x+\xi)$ for a rotation $R$ and a translation $\xi$. Then it is easy to see that $|R\circ\psi-Id|_{C^{3}}\leq C_{2}(\epsilon)$ with $\lim_{\epsilon\rightarrow 0}C_{2}(\epsilon)=0$, and $R(\hat{A}(1-\delta,100+\delta))$ contains $\hat{A}(1,100)$. Choose a cut-off function $\chi(x)$ on $\hat{A}(1,100)$ with $\chi(x)=1$ for $|x|\leq 5$ and $\chi(x)=0$ for $|x|\geq 6$. Using the map $f_{0}$ we get a corresponding cut-off function on $B(q,200)$, still denoted by $\chi$. Then $|\chi|_{C^{4}_{g}}\leq C_{3}$ for a constant $C_{3}$ independent of $\epsilon$. Clearly $\chi(p)=0$ when $p\notin Im(f_{1})$ and $\chi(p)=1$ when $p\notin Im(f_{0})$. Define $h:Imf_{0}\cup Imf_{1}\rightarrow\mathbb{R}^{4}$ sending $p$ to $10^{-1}\chi(p)R\circ f_{1}^{-1}(x)+(1-\chi(p))f_{0}^{-1}(x)$. Then for $\epsilon$ sufficiently small we have $h=f_{0}^{-1}$ on $A(8,100)$ and $h=10^{-1}R\circ f_{1}^{-1}$ on $A(10^{-1},3)$, and $|h^{*}g_{0}-g|_{C^{2}_{g}}\leq C(\epsilon)$ with $\lim_{\epsilon\rightarrow 0}C(\epsilon)=0$. Define $f=h^{-1}$. Then $f(x)$ meets the required properties. ∎

Since the problem is local, we may assume for all $r\leq 1$ that the above map $\phi_{r}$ exists and $\epsilon(1)$ is as small as we like. For simplicity we denote $\phi_{k}=\phi_{10^{-k}}$, and $\epsilon_{k}=\epsilon(10^{-k})$. Now we first define $F_{0}(x)=\phi_{0}(x)$ on $\hat{A}(1,100)$. Inductively suppose $F_{k}$ is defined on $\hat{A}(10^{-k},10^{-k+2})$ satisfying $F_{k}(x)=\phi_{k}\circ R_{k}^{-1}(10^{k}x)$ on $\hat{A}(10^{-k},20\cdot 10^{-k})$ for some rotation $R_{k}\in O(4)$, then we apply Lemma 5.9 to the two maps $\phi_{k}\circ R_{k}^{-1}$ and $\phi_{k+1}$ with $r=10^{-k}$ and $\epsilon=\max(\epsilon_{k-1},\epsilon_{k})$, and obtain a map $f_{k+1}$ defined on $\hat{A}(1/10,100)$ satisfying (2). Then we define $F_{k+1}(x)$ to be $f_{k+1}(10^{k}x)$ on $\hat{A}(10^{-k-1},10^{-k+1})$. By Lemma 5.9 we see that all the $F_{k}$’s match together to a map $F$ from $\hat{B}^{*}$ to $B(q,200)$, and we can modify $F$ slightly near $\partial\hat{B}$ so that the image is exactly $B^{*}$. It is easy to see that $|F^{*}g-g_{0}|_{L^{\infty}(\hat{A}(10^{-k},10^{-k+1}))}=|10^{2k}f_{k+1}^{*}g-g_{0}|_{L^{\infty}(\hat{A}(10,100))}\leq G(\max(\epsilon_{k-1},\epsilon_{k}))$, and $F^{*}g$ extends to a continuous metric tensor over $B$. ∎

Let $A_{\pm}$ be the connection induced by the Levi-Civita connection of $g$ on $\Lambda^{\pm}_{g}$. The Einstein condition implies $A_{+}$ is self-dual and $A_{-}$ anti-self-dual with respect to $g$. Thus

$$|Rm(g)|^{2}dvol_{g}=Tr(F_{A_{+}}\wedge F_{A_{+}}-F_{A_{-}}\wedge F_{A_{-}}).$$

By the tangent cone condition we can easily find a smooth family of spheres $S_{r}$ in $B^{*}$ with the property that as $r$ tends to zero, $(S_{r},r^{-2}g)$ converges smoothly to the round sphere in $\mathbb{R}^{4}$, and the restriction to $(S_{r},r^{-2}g)$ of the connection $A_{\pm}$ converges to the trivial flat connection. Then for any $s<r$

$$\int_{A(s,r)}TrF_{A_{+}}\wedge F_{A_{+}}=CS(A_{+},S_{r})-CS(A_{+},S_{s})(mod\ \mathbb{Z}),$$

where $CS(A,M)=\int_{M}dA\wedge A+\frac{2}{3}A\wedge A\wedge A$ is the Chern-Simons invariant of a connection $A$ over a three manifold $M$, defined modulo $\mathbb{Z}$. By assumption, $CS(A_{+},S_{r})=CS(A_{+},\frac{1}{r}S_{r})\rightarrow 0$ as $r\rightarrow 0$. So we choose $r$ small enough so that for any $s\leq r$ we have $|CS(A_{+},S_{r})|\leq 1/8$ modulo $\mathbb{Z}$. So $\int_{A(s,r)}TrF_{A_{+}}\wedge F_{A_{+}}$ is in $[-1/4,1/4]$ modulo $\mathbb{Z}$, and on the other hand it clearly depends continuously on $s$, so the integral is uniformly bounded for all $s<r$. One can similarly deal with $A_{-}$. Together this implies $\int_{B^{*}}|Rm(g)|^{2}dvol_{g}$ is finite.

∎

Since the metric $g$ is $C^{0}$ equivalent to the flat metric $g_{0}$, the Sobolev space $W^{1,p}$ is the same with respect to both metrics, and the Moser iteration works for the operator $\Delta=\Delta_{g}$. Here again we use the geometers’ convention for the sign. By Bochner formula there is a constant $C_{1}>0$ such that

$$\Delta|Rm|\leq C_{1}|Rm|^{2},$$

which is on the borderline of applying Moser iteration. Due to Bando-Kasue-Nakajima [2] (Corollary 4.10), there is an improved Kato’s inquality, namely, there are $C_{2}>0$ and $\delta\in(0,1)$, such that

$$\Delta|Rm|^{1-\delta}\leq C_{2}|Rm|^{2-\delta}.$$

Let $u=|Rm|^{1-\delta}$ and $f=|Rm|$. Then we can apply [19](Lemma 2.1) with $q=\frac{1}{1-\delta}$ and $q_{0}=\frac{1}{2(1-\delta)}$ to conclude that $|Rm|$ is in $W^{1,2}$. By Sobolev embedding we see $|Rm|\in L^{4}$. Also that $|\nabla Rm|\in L^{2}$ implies that the inequality $\Delta|Rm|\leq C_{1}|Rm|^{2}$ holds weakly on the whole ball $B$. Then we can apply the standard Moser iteration to conclude $|Rm|$ is uniformly bounded. Now consider $|\nabla Rm|$. For any $p\in B^{*}$ with $d(p,q)=r\leq 1/2$, the rescaled ball $r^{-1}B(p,r/2)$ has uniformly bounded geometry, so standard elliptic regularity for the Einstein equation then implies that $|\nabla Rm|\leq C_{3}r^{-1}.$ for some constant $C_{3}>0$. Thus $|\nabla Rm|\in L^{3}$. By Bochner formula again there is a constant $C_{4}>0$ such that

$$\Delta|\nabla Rm|\leq C_{4}|Rm||\nabla Rm|.$$

Let $u=|\nabla Rm|$, $f=C_{4}|Rm|$ and apply [19](Lemma 2.1) with $q=1$, and $q_{0}=3/4$, we get $|\nabla Rm|\in W^{1,2}$. Thus the inequality holds weakly on $B$ and by Moser iteration $|\nabla Rm|$ is uniformly bounded. Then similarly one can prove the bound for higher covariant derivatives of the curvature tensor. ∎

We define $F_{i}:S^{3}\times[i^{-1},1]$, sending $(x,t)$ to $\exp_{f_{i}(x)}((t-i^{-1})N(x))$, where $N(x)$ is the outward normal vector at $x$. Consider a Jacobi field $J(t)$ along a geodesic $\gamma_{x}(t)=F_{i}(x,t)$. Then since the curvature of $g$ is uniformly bounded, by Rauch comparison theorem there are constants $C_{1}>0$ and $\delta>0$ independent of $x$ and $i$ such that $C_{1}^{-1}|J(i^{-1})|_{g}\leq|J(t)|_{g}\leq C_{1}i|J(i^{-1})|_{g}$ for $t\in[i^{-1},\delta]$. For simplicity of notation we may assume $\delta=1$. So for $i$ large enough $F_{i}$ has no critical points in $[i^{-1},1]$. Indeed $F_{i}$ is a diffeomorphism. For otherwise there would be a geodesic loop $\sigma(s)(s\in[0,T])$ which is perpendicular to $S_{i}$ when $s=0$ and $s=T$. It is then easy to see this could not happen for sufficiently large $i$, by passing to a tangent cone.

Now we write $F_{i}^{*}g=dt^{2}+t^{2}h_{i}(t)$. First we notice that $|d_{g}(0,F_{i}(x,t))-t|\leq i^{-1}\epsilon_{i}$. Now we derive estimates for $g_{i}(t)$. Given a unit tangent vector $\xi$ at $x\in S^{3}$. Let $J(t)$ be the Jacobi field along $\gamma_{x}(t)$ with $J(i^{-1})=df_{i}(\xi)$ and $\dot{J}(i^{-1})=A_{S_{i}}(J(i^{-1}))$. Then $J(t)=d{F_{i}}_{(x,t)}(\xi)$. Clearly $||J(i^{-1})|_{g}-i^{-1}|\leq i^{-1}\epsilon_{i}$ and $|\dot{J}(i^{-1})-iJ(i^{-1})|_{g}\leq 2\epsilon_{i}$. Let $\{e_{1}(t),\cdots,e_{n}(t)=\dot{\gamma}_{x}(t)\}$ be an orthonormal frame of parallel vector fields along $\gamma_{x}(t)$, such that $J(i^{-1})=|J(i^{-1})|_{g}e_{1}$. Under the decomposition $J(t)=\sum_{\alpha}J_{\alpha}(t)e_{\alpha}(t)$ we have

$$\ddot{J}_{\alpha}(t)+\sum_{\beta}R_{\alpha n\beta n}(\gamma_{x}(t))J_{\beta}(t)=0,$$

where $R_{\alpha n\beta n}=R(e_{\alpha},e_{n},e_{\beta},e_{n})$. From the above discussion we have $|J(t)|\leq 2C_{1}$ for $t\in[i^{-1},1]$. So it is easy to see that there is a constant $C_{2}>0$ such that

$$||J(t)|_{g}-t|\leq C_{2}(i^{-1}+\epsilon_{i}t+t^{3}).$$

Thus

$$|h_{i}(t)-h_{0}|_{L^{\infty}_{h_{0}}}\leq C_{2}(i^{-1}t^{-1}+\epsilon_{i}+t^{2}).$$

Now take a unit tangent vector $X$ at $x$, we vary $J(i^{-1})$ so that $\nabla^{0}_{X}J(i^{-1})=0$ at $x$, and extend $X$ to a unit tangent vector field in a neighborhood $U$ of $x$ in $S^{3}$. We may also view $X$ as a tangent vector field on $U\times[i^{-1},1]$. Now we differentiate the Jacobi field equation, and similar arguments as above yield

$$|\nabla_{X}J(t)|_{g}\leq C_{3}(i^{-1}+\epsilon_{i}t+t^{3}),$$

for a constant $C_{3}>0$. This implies that there is a constant $C_{4}>0$ such that $t^{-1}|\nabla^{0}(h_{i}(t)-h_{0})|_{h_{0}}\leq C_{4}(i^{-1}t^{-2}+\epsilon_{i}t^{-1}+t).$ Similarly one can get bounds on higher derivatives of $h_{i}(t)-h_{0}$. The point is that for a fixed $\tau>0$ as $i$ goes to infinity we know $F_{i}(x,t)$ converges in $C^{3}$ to a limit $F^{\tau}_{\infty}(x,t)$ on $S^{3}\times[\tau,1]$. Then we can let $\tau\rightarrow 0$ and obtain a limit $F:S^{3}\times(0,1]$ with the property that $d_{g}(0,F(x,t))=t$, and

$$t^{-2}|F^{*}g-g_{0}|_{C^{0}_{g_{0}}}+t^{-1}|F^{*}g-g_{0}|_{C^{1}_{g_{0}}}+|F^{*}g-g_{0}|_{C^{2}_{g_{0}}}\leq C_{5},$$

for some constant $C_{5}>0$. This implies that $F^{*}g$ extends to a $C^{1,\alpha}$ metric on $B$. ∎