On the connectedness of the boundary of 𝑞-complete domains (2024)

Definition 1 ([MR3700709]*Definition 3.1.3).

Let $X$ be a complex manifold. We say that a domain $\Omega\subset X$ is $q$-complete, if there exists a $\mathcal{C}^{2}$-smooth exhaustion function $\varphi\colon\Omega\to[0,+\infty)$ that is $q$-convex, i.e.if its Levi form has at most $q-1$ negative or zero eigenvalues at each point of $\Omega$.

Theorem 2.

Let $X$ be a complex manifold with $\dim_{\mathbb{C}}X=n$ and let $\Omega\subset X$ be a relatively compact $q$-complete domain with $n>q$. Then the boundary $\partial\Omega$ is connected.

Corollary 3.

Let $X$ be a complex manifold with $\dim_{\mathbb{C}}X=n>1$ and let $\Omega\subset X$ be a relatively compact Stein domain. Then the boundary $\partial\Omega$ is connected.

Corollary 4.

Let $X$ be a Stein manifold with $\dim_{\mathbb{C}}X=n>1$ and let $\Omega\subset X$ be a relatively compact domain of holomorphy. Then the boundary $\partial\Omega$ is connected.

Theorem 5.

Let $(X,J)$ be an almost complex manifold with $\dim_{\mathbb{R}}X\geq 4$, and let $\Omega\subset X$ be a relatively compact domain with a $\mathcal{C}^{2}$-smooth strictly $J$-plurisubharmonic exhaustion function. Then the boundary $\partial\Omega$ is connected.

Remark 6.

Remark 7.

For symplectic manifolds with contact type boundaries, the boundary does not need to be connected, see McDuff [MR1091622] and Geiges [MR1328705].

Definition 8.

Remark 9 ([MR1410261]*Example3(i)).

Let $X$ be a topological space with a proper map $\varphi\colon X\to[0,+\infty)$ which is onto, and such that the inverse images $U_{t}=\varphi^{-1}(t,+\infty)\subseteq X$, $t\geq 1$ are connected. Then $X$ has one end.

Lemma 10.

Let $X$ be a locally compact Hausdorff space with a countable basis of topology. Then $X$ has one end if and only if there exists an exhaustion by compacts $(K_{j})_{j\in\mathbb{N}}$ of $X$ such that $X\setminus K_{j}$ is connected for every $j\in\mathbb{N}$.

Proof.

The “if” part is straightforward, see also Remark 9. We only need to provide a proof for the “only if” part: Assume to get a contradiction that no exhaustion by compacts $(K_{j})_{j}$ of $X$ exists such that $X\setminus K_{j}$ is connected for every $j\in\mathbb{N}$, but that $X$ has only one end. We can always pass to a subsequence, and thus we may assume that $X\setminus K_{j}$ always has at least two connected components. Moreover, we can assume that none of these components is contained in a compact $K_{\ell(j)},\ell>j,$ for otherwise we could have included this component already in the compact $K_{j}$.Now we pick a connected component $U_{1}$ of $X\setminus K_{1}$. For every $j\geq 2$ we a pick a connected component of $X\setminus K_{j}$ that is contained in $U_{j-1}$. This is always possible, since we eliminated superficial connected components by our choice of compacts $K_{j}$. This sequence of neighborhoods $(U_{j})_{j}$ defines an end of $X$.Since $X\setminus K_{1}$ has at least two connected components, we choose another sequence of neighborhoods $(V_{j})_{j}$ such that $U_{j}$ and $V_{j}$ are different connected components of $X\setminus K_{j}$ for every $j\in\mathbb{N}$, and hence define two different ends.∎

Definition 11.

A continuum is a non-empty, compact, connected metric space.

Lemma 12 ([MR1192552]*Theorem 1.8).

The intersection of a decreasing sequence of continua is a continuum.

Proposition 13.

Let $X$ be a manifold with countable basis of topology and let $\Omega\subset X$ be a relatively compact domain with one end. Then $\Omega$ has connected boundary.

Proof.

We apply Lemma 10 to obtain an exhaustion by compacts $K_{j}$ such that $\Omega\setminus K_{j}$ is connected for every $j\in\mathbb{N}$. Then $\Omega\setminus(K_{j})^{\circ}$ are a decreasing sequence of continua. Thus its limit, which is the boundary $\partial\Omega$, is also a continuum by Lemma 12.∎

Remark 14.

If the boundary $\partial\Omega$ of the relatively compact domain $\Omega$ admits a collar, then we have a one-by-one correspondence between ends and boundary components.

Proof of Theorem 2.

Let $\dim_{\mathbb{C}}X=n$. Since $\Omega$ is relatively compact, by a small perturbation we may assume that the $\mathcal{C}^{2}$-smooth exhaustion function $\varphi\colon\Omega\to[0,+\infty)$ is a Morse function and still $q$-complete. The Morse index at a critical point is at most $n+q-1$ (see the monograph of Forstnerič [MR3700709]*Sections 3.10 and 3.11 for more details). Note that $2n-1>n+q-1\Longleftrightarrow n>q$ is satisfied by assumption. Since we glue only handles of dimension $\leq n+q-1$, and $\Omega$ has real dimension $2n$, the complement of any sublevel set of $\varphi$ is connected. By Lemma 10 the domain $\Omega$ has only one end. Proposition 13 now gives the conclusion.∎

Proof of Theorem 5.

Let $\dim_{\mathbb{R}}X=2n$. Let $\varphi\colon\Omega\to\mathbb{R}$ be a strictly $J$-plurisubharmonic exhaustion function. Since the domain $\Omega$ is relatively compact, we may slightly disturb $\varphi$ if necessary to obtain a strictly $J$-plurisubharmonic Morse exhaustion function. By [MR3012475]*Corollary 3.4 the Morse index of $\varphi$ at a critical point in an almost complex manifold is at most $n$. Now the conclusion is the same as in the proof of Theorem 2 above.∎

Remark 15.

The proofs of these two theorems can also be given without using the theory of ends, by instead considering the exhaustion functions and their connected superlevel sets, and applying Lemma 12 directly to them. However, conceptually, it seems more natural to use Proposition 13 which reflects that the “reason” comes from the fact that these relatively compact domains have one end.