On the topology of stable minimal hypersurfaces in a homeomorphic $S^{4}$

If $\tilde{\Omega}_{M}$ is compact, then the desired result follows from Poincaré duality. In the following, we assume that $\tilde{\Omega}_{M}$ is non-compact. Recall that $\tilde{g}$ denotes the pull-back metric on $\tilde{\Omega}_{M}$ via the covering map.

Let $x$ be a fixed point in $\operatorname{int}(\tilde{\Omega}_{M})$, and let $B_{\rho}(x)$ be the geodesic ball with radius $\rho$ centered at $x$. For generic $\rho$, we have $\partial B_{\rho}(x)$ is a properly embedded, compact, smooth submanifold of $\tilde{\Omega}_{M}$ with codimension $1$. Since every homology class is represented by finitely many singular simplices, for $\rho$ sufficiently large, the homology class $a$ is contained in the image of $H_{n}(B_{\rho}(x))\to H_{n}(\tilde{\Omega}_{M})$. From now, let $\rho$ be a fixed number that is both sufficiently large and generic so that the above properties hold.

Since $\partial B_{\rho}(x)$ has only finitely many connected components, the set $\tilde{\Omega}_{M}\setminus B_{\rho}(x)$ has at most finitely many connected components. Let $C_{1},\dots,C_{n}$ be the closures of the connected components of $\tilde{\Omega}_{M}\setminus B_{\rho}(x)$. Then each $C_{i}$ is a manifold with corners, where the codimension of each corner stratum is at most $2$.

Now we finish the proof of the lemma using the claim. Let $\hat{\Omega}\subset\tilde{\Omega}_{M}$ be the union of $B_{\rho}(X)$ and all the $C_{i}$’s which are compact. Then $\hat{\Omega}$ is a compact manifold with corners, and $a$ is in the image of $H_{3}(\hat{\Omega})\to H_{3}(\tilde{\Omega}_{M})$.

Let $\hat{a}\in H_{3}(\hat{\Omega})$ be a preimage of $a$. Since $a\neq 0$, we know that $\hat{a}\neq 0$. Note that $\hat{\Omega}$ is a compact smooth manifold with boundary after smoothing the corners. By Poincaé duality, there exists a properly embedded smooth 1-manifold $(\hat{\gamma},\partial\hat{\gamma})\subset(\hat{\Omega},\partial\hat{\Omega})$ such that the intersection number of $\hat{a}$ with $\hat{\gamma}$ is non-zero, and we may perturb $\hat{\gamma}$ such that $\partial\gamma$ does not intersect the corners of $\partial\hat{\Omega}$.

For every $q\in\partial\hat{\gamma}$ with $q\notin\partial\tilde{\Omega}_{M}$, there exists a unique $C_{i}$ such that $C_{i}$ is non-compact and $q\in\partial C_{i}$. By the above claim, there exists $q^{\prime}\in\partial C_{i}\cap\partial\tilde{\Omega}_{M}$. Let $\gamma_{q}$ be a properly embedded arc in $C_{i}$ that connects $q$ and $q^{\prime}$. After perturbing $\gamma$ and $\gamma_{q}$ near $q$, we may further assume that $\gamma\cup\gamma_{q}$ is smooth near $q$.

Let

where the union takes over all $q$ such that $q\in\partial\hat{\gamma},q\notin\partial\tilde{\Omega}_{M}$. If $n\geq 2$, then after a generic perturbation, $\gamma$ is a properly embedded $1$–manifold in $\tilde{\Omega}_{M}$. If $n=1$, then after a generic perturbation $\gamma$ is a properly immersed $1$–manifold with transverse self-intersections, and we can resolve the self-intersection of $\gamma$ to obtain an properly embedded $1$–manifold in $\tilde{\Omega}_{M}$ with the same homology class.

Since all $\gamma_{q}$ are disjoint from $\operatorname{int}(\hat{\Omega})$, a Thom form of $\gamma$ on $\tilde{\Omega}_{M}$ restricts to a Thom form of $\hat{\gamma}$ on $\hat{\Omega}$. So the intersection number of $a$ and $\gamma$ in $\tilde{\Omega}_{M}$ is equal to the intersection number of $\hat{a}$ and $\hat{\gamma}$ in $\hat{\Omega}$, which is non-zero by the definition of $\hat{\gamma}$. Hence the lemma is proved. ∎

Consider the minimization problem

Let $(\gamma_{0},\partial\gamma_{0})\subset(\tilde{\Omega}_{M},\partial\tilde{\Omega}_{M})$ be the compact embedded curve constructed in Lemma 2.5.

Let $\omega_{0}$ be the Thom form of $\gamma_{0}$ as in Definition 2.2. By definition, $\omega_{0}$ is compactly supported in a neighborhood of $\gamma_{0}$. Then any $n$-cycle $N\in[Y_{0}]$ satisfies that $\int_{N}\omega_{0}\neq 0$.

For $\rho\gg 1$ consider the $\tilde{g}$-geodesic ball $B_{\rho}(\tilde{y})$, and let $S_{\rho}(\tilde{y})$ be the $\tilde{g}$-geodesic sphere. Choose $\rho$ large enough such that $[Y_{0}]\neq 0\in H_{n}(B_{\rho}(\tilde{y}))$. Perturbing $\rho$ a bit if necessary, we assume that $S_{\rho}(\tilde{y})$ meets $\partial\tilde{\Omega}_{M}$ transversely. Deform the metric $\tilde{g}$ to $\tilde{g}^{\prime}$ in a small neighborhood of $S_{\rho}(\tilde{y})$, such that $S_{\rho}(\tilde{y})$ is $\tilde{g}^{\prime}$ strictly mean convex and meets $\partial\tilde{\Omega}_{M}$ orthogonally. We then may minimize the $\mathcal{H}^{n}$ volume (with respect to $\tilde{g}^{\prime}$) in the nontrivial homology $[Y_{0}]$ in $(B_{\rho}(\tilde{y}),\tilde{g}^{\prime})$, and obtain a possibly disconnected area minimizing hypersurface $N_{\rho}$. Since

each $N_{\rho}$ has a non-trivial intersection with the compact set $\operatorname{supp}\omega_{0}$. Consider $N_{\rho}^{0}$ the union of all connected components of $N_{\rho}$ that intersect $\operatorname{supp}\omega_{0}$. Since $N_{\rho}^{0}$ has uniformly bounded $\mathcal{H}^{n}$-volume, standard curvature estimates imply that they subsequentially (which we do not relabel) $C^{\infty}$ graphically converge to a limit $N^{0}$ (possibly with integer multiplicity at this moment). Note that $N^{0}$ is compact: since $\tilde{\Omega}_{M}$ has bounded geometry, by the monotonicity formula, there exist $r_{0}>0$ and $\varepsilon_{0}>0$ depending only on $\Omega_{M}$ such that if $x\in N^{0}$, then the intersection of $N^{0}$ and $B_{r_{0}}(x)$ has $\mathcal{H}^{n}$-volume at least $\varepsilon_{0}$.

Therefore we conclude that $\{N_{\rho}^{0}\}_{\rho}$ is also compactly supported, and the convergence to $N^{0}$ in fact holds as currents. In particular, $N_{\rho}^{0}$ is homologous to $N^{0}$ for sufficiently large $\rho$.

If $N^{0}$ is homologous to $Y_{0}$, we are done. Otherwise, consider the homology class $[Y_{0}]-[N_{0}]$. By Lemma 2.5, there exists a properly embedded compact curve $(\gamma_{1},\partial\gamma_{1})\subset(\tilde{\Omega}_{M},\partial\tilde{\Omega}_{M})$ that has a non-zero intersection number with $[Y_{0}]-[N_{0}]$. Denote by $\omega_{1}$ the Thom class of $\gamma_{1}$. Then for all sufficiently large $\rho$, the minimizing sequence $N_{\rho}-N_{\rho}^{0}$ satisfies that

Let $N_{\rho}^{1}$ be the union of the connected components of $N_{\rho}-N_{\rho}^{0}$ that intersect the support of $\omega_{1}$. By passing to a further subsequence (which again we do not relabel), the same argument as above finds a compact minimizing hypersurface $N_{1}$ as the limit of $\{N_{\rho}^{1}\}$.

Inductively, assuming that we have constructed $N_{0},\cdots,N_{k}$. If $N_{0}+\cdots+N_{k}$ (taking the sum as currents) is not homologous to $Y_{0}$, we apply the above argument and construct a compact minimizing hypersurface $N_{k+1}$ from the minimizing sequence $\{N_{\rho}-(N_{\rho}^{0}+\cdots+N_{\rho}^{k})\}$. Note that since $\tilde{\Omega}_{M}$ is the covering space of a compact manifold, any minimal hypersurface has a uniform lower bound on its $\mathcal{H}^{n}$-volume. This implies that this construction terminates in finitely many steps, since $\{N_{\rho}\}$ has a uniform $\mathcal{H}^{n}$-volume upper bound. We thus conclude that there exists a compact area minimizing hypersurface $N$ (possibly disconnected) in the homology class $[Y_{0}]$. Note that $N$ is not contained entirely in $\partial\tilde{\Omega}_{M}$, as the only $n$-currents entirely contained in $\partial\tilde{\Omega}_{M}$ that are homologous to $[Y_{0}]$ are in the form $Y_{0}+k\partial\tilde{\Omega}_{M}$ for some integer $k$ (if $\partial\tilde{\Omega}_{M}$ is non-compact, then we must have $k=0$). Since $Y_{0}$ is isometric to $Y$ and $\pi^{-1}(Y)$ has at least two connected components, the mass of a current of the form $Y_{0}+k\partial\tilde{\Omega}_{M}$ is at least $\mathcal{H}^{n}(Y_{0})$. We know $N\neq Y_{0}$ (as otherwise we would have picked $M=Y$), so $N$ cannot be contained entirely in $\partial\tilde{\Omega}_{M}$.

Let $\hat{N}=\pi(N)$. If $\hat{N}$ is embedded, then it is homologous to $Y$ in $\Omega_{M}$, and bounds a region with smaller volume than $\mathcal{H}^{n+1}(\Omega_{M})$, contradicting the choice of $M$. If $\hat{N}$ is immersed, then it still represents the homology class $[Y]$ in $H_{n}(\Omega_{M},\mathbb{Z})$. We may then minimize the $n$-dimensional volume in an open neighborhood of $Y$ in $\Omega_{M}$, among hypersurfaces homologous to $Y$ that does not intersect $\hat{N}$ (note that immersed minimal hypersurfaces are weakly mean convex as self-intersections form an angle that is strictly less than $\pi$). This produces another stable minimal hypersurface $M^{\prime}\in\mathcal{S}$ such that $\mathcal{H}^{n+1}(\Omega_{M^{\prime}})<\mathcal{H}^{n+1}(\Omega_{M})$, contradiction. ∎

Let $D_{1},\dots,D_{k}$ be the closures of the connected components of $S^{4}\setminus A$. Then $\partial A$ is the disjoint union of $\partial D_{1},\dots,\partial D_{k}$. We first show that every $\partial D_{i}$ is connected. Assume $\partial D_{i}$ is not connected, let $M_{1},M_{2}$ be two connected components of $\partial D_{i}$, let $p$ be a point in the interior of $D_{i}$, let $q$ be a point in the interior of $A$. Then there exist arcs $\gamma_{1},\gamma_{2}$ from $p,q$ such that $\gamma_{i}$ intersects $\partial A$ transversely at one point in each $M_{i}$ ($i=1,2$). As a consequence, the arcs $\gamma_{1}$ and $\gamma_{2}$ combine to define a loop in $S^{4}$ that intersects each of $M_{1}$ and $M_{2}$ transversely at one point. This contradicts the fact that $H_{1}(S^{4})=0$.

Let $\hat{D}_{i}=D_{i}/\partial D_{i}$. By definition, $\hat{D}_{i}$ is the quotient space of $D_{i}$ by collapsing $\partial D_{i}$ to a point. The space $\hat{D}_{i}$ may not be a manifold. Let $S\subset\{1,\dots,k\}$ be non-empty, let $A_{S}=A\cup(\cup_{i\in S}D_{i})$. We claim that

where the right-hand side is the free product.

We prove (1) by induction on the number of elements of $S$. If $S=\emptyset$, the statement is trivial. Suppose (1) holds for all sets $S$ with $l$ elements, and consider a set $S^{\prime}$ of the form $S^{\prime}=S\cup\{i\}$ where $S$ has $l$ elements and $i\notin S$. Then $A_{S^{\prime}}=A_{S}\cup D_{i}$. By the van Kampen theorem,

Since the inclusion-induced map $\pi_{1}(\partial D_{i})\to\pi_{1}(A_{S})$ factors through $\pi_{1}(A)$, it is the trivial map. Therefore, we have

where $\pi_{1}(D_{i})/\pi_{1}(\partial D_{i})$ denotes the quotient of $\pi_{1}(D_{i})$ by the normal subgroup generated by $\pi_{1}(\partial D_{i})$. Hence (1) is proved.

Now we can prove the lemma. Let $S=\{1,\dots,k\}$, we have $S^{4}=A_{S}$, and hence (1) shows that $\pi_{1}(\hat{D}_{i})=1$ for all $i$. Applying (1) again, we conclude that $\pi_{1}(A_{S})=1$ for all $S$. Since the set $A^{\prime}$ in the statement of the lemma has the form $\Omega_{S}$ where $S$ contains all but one element of $\{1,\dots,k\}$, we know that $A^{\prime}$ is simply connected. ∎

(1) We follow an argument from [12]. Consider the following commutative diagram

where the maps are induced by the inclusions of spaces. Since $H_{2}(S^{4})=0$, and since by excision, the map $H_{2}(A,M)\to H_{2}(S^{4},B)$ is an isomorphism, we conclude that the map $H_{2}(A)\to H_{2}(A,M)$ is zero. Now consider the homology long exact sequence

we have $H_{1}(M)\cong H_{2}(A,M)$. By Lefschetz duality, we have $H_{2}(A,M)\cong H^{2}(A)$. By the universal coefficient theorem, the torsion of $H^{2}(A)$ is isomorphic to the torsion of $H_{1}(A)$, which is zero. So $H_{1}(M)$ has no torsion.

(2) By the Mayer-Vietoris sequence

we have $H_{2}(M)\cong H_{2}(A)\oplus H_{2}(B)$, so the map $H_{2}(M)\to H_{2}(A)$ induced by the inclusion is surjective. On the other hand, by the exact sequence

and Lefschetz duality $H_{3}(A,M)\cong H^{1}(A)\cong 0$, we know that the map $H_{2}(M)\to H_{2}(A)$ is injective. Hence it is an isomorphism.

(3) By Lefschetz duality, $H_{3}(A)\cong H^{1}(A,M)$. By the cohomology long exact sequence for $(A,M)$, we have an exact sequence

so $H^{1}(A,M)\cong 0$. ∎

$\partial\hat{A}$ is obtained from $\partial{A}$ by removing a tubular neighborhood of each attaching sphere and gluing 2 copies of $D^{2}$ for each removed neighborhood. So $\partial\hat{A}$ is homeomorphic to $M_{1}$.

Now we show that $H_{1}(\hat{A})\cong H_{2}(\hat{A})\cong H_{3}(\hat{A})\cong 0$.

Attaching 3-handles does not modify $H_{1}$, so we have $H_{1}(\hat{A})\cong H_{1}(A)\cong 0$.

Consider the exact sequence