On Boundary points at which the squeezing function tends to one
J.E. Fornæss posed the question whether the boundary point of smoothly bounded pseudoconvex domain is strictly pseudoconvex, if the asymptotic limit of the squeezing function is 1. The purpose of this paper is to give an affirmative answer when the domain is in with smooth boundary of finite type in the sense of D’Angelo .
Key words and phrases:Holomorphic mappings, Automorphisms of
2010 Mathematics Subject Classification:32H02, 32M17
Let . For a domain in and , let . Then the squeezing function is defined to be
Note that for any .
This concept first appeared in [13, 14] in the context concerning the holomorphic homogeneous regular manifolds. But the name squeezing function comes from ; this concept is closely related to the concept of bounded geometry by Cheng and Yau , as one sees from . It is obvious from its construction that the squeezing function is a biholomorphic invariant.
Recent studies have shown:
If a bounded domain in has a boundary point, say , about which the boundary is strictly pseudoconvex, then .
J. E. Fornæss asked recently whether its converse is true (cf. , Sections 1 and 4), i.e., he posed the following question.
If is a bounded domain with smooth boundary, and if , then is the boundary of strictly pseudoconvex at ?
In a recent article, A. Zimmer has shown that the answer is affirmative if the bounded domain is also assumed to be convex .
The main purpose of this article is to present the following result.
Let be a bounded domain in with smooth pseudoconvex boundary. If is a boundary point of of finite type, in the sense of D’Angelo , and if , then is strictly pseudoconvex at .
We remark that our proof is only for complex dimension 2. It is mainly due to the limitation of current knowledge concerning the convergence of the scaling methods (cf. ). If the domain is convex and Kobayashi hyperbolic for instance, then there is no such restriction. Consequently in case the domain is bounded convex, as treated in , our proof-arguments also answer Question 1 affirmatively, which we shall put an explication of, in a remark at the end.
2. Construction of the scaling sequence
Let and the boundary point be as in the hypothesis of the theorem.
Let be the unit normal vector to at pointing outward. Then take a sequence such that for some satisfying
for every and
We also set for every . Since by assumption, there exists, for each , an injective holomorphic map such that and for all .
The next step is to use the dilation sequence introduced in , Section 2.2. (We remark that this is a mild but necessary modification of Pinchuk’s stretching sequence [1, 16]). If the type of is for some integer , then there is a neighborhood of and satisfying the following properties:
The map is the composition (in order) of a translation, a unitary map, a triangular map and a dilation map.
and for all .
The local defining function of at is represented by
is a nonzero real-valued homogeneous subharmonic polynomial of degree with no harmonic terms,
and are real-valued smooth functions satisfying the conditions on the vanishing: order and.
Note that the convergences and are uniform on compact subsets of , while is a nonzero real-valued subharmonic polynomial of degree . This is proved in detail in Lemma 2.4 of . Consequently, converges uniformly to on compact subsets of .
Define by . Note that for every .
3. Convergence of the “reverse” scaling sequence
Recall the concept of normal set-convergence introduced in , Section 9.2.2; it will give the necessary control for the convergence.
Let be domains in for each . The sequence is said to converge normally to a domain , if the following two conditions hold:
For any compact set contained in the interior of for some positive integer , .
For any compact subset of , there exists a constant such that .
If is a sequence of domains in that converges normally to the domain , then
If a sequence of holomorphic mappings from to another domain converges uniformly on compact subsets of , then its limit is a holomorphic mapping from into the closure of the domain .
If a sequence of holomorphic mappings converges uniformly on compact subsets of , if is pseudoconvex, and if there are a point and a constant so that the inequality holds for each , then is a holomorphic mapping from the domain into .
In our construction, the set-convergences and are in accordance with the sense of normal set-convergence with
Notice that is unbounded, so the convergence of the forward scaling sequence is not immediately obvious. On the other hand, one easily observes that the inverse sequence (i.e., the reverse scaling sequence) converges, choosing a subsequence when necessary, by Proposition 3.2 and Montel’s theorem. So we take a convergent subsequence and call it again, and denote the limit map by . Since , it holds that .
Now we show:
is a biholomorphic map.
The proof is almost the same as those for Propositions 2.8 and 2.10 of . However, the surjectivity part of requires a few, simple but perhaps subtle adjustments. Therefore, we choose to include the detail here.
Suppose that is not onto. Then there is a boundary point of in . Notice that converges to since is a peak point of . Denote by . Now we construct a new scaling sequence where is a stretching map (in the sense of Pinchuk) with respect to as above. In the same way, there is a subsequential limit map of , where is the limit domain of the sequence in the sense of normal set-convergence. We have already observed that is 1-1. Taking a subsequence if necessary, we may assume that the uniform convergence holds for and on compact subsets of and respectively. (See Lemma 2.9 in ).
Denote by . Then the map is well-defined. Actually, , and this, for each , is a polynomial automorphism of with degree less than or equal to . On the other hand, converges to uniformly on compact subsets of . By calculation, turns out to be a polynomial automorphism of of degree less than or equal to . Now Proposition 3.2 guarantees that the restriction is a 1-1 holomorphic map from into . In a similar way for the inverse sequence , the map is a 1-1 holomorphic map from into . Hence is a biholomorphism.
On the other hand, we see that a sequence of points in convergent to gives rise to a sequence in via that approaches the boundary and also to a sequence in via which, this time, converges to the interior point . This results in that the biholomorphism maps a sequence approaching the boundary to a sequence convergent to an interior point. So fails to be proper, and the surjectivity of follows by this contradiction. ∎
4. Proof of the Theorem 1.2
We are ready to complete the proof of Theorem 1.2. Recall that , where is a nonzero real-valued subharmonic polynomial of degree . Note that the unit ball is biholomorphic to the Siegel half space . So the theorem of Oeljeklaus in , which says that these two domains must be affinely biholomorphic in such a case, implies in particular that for some . Therefore, the origin must have been the boundary point of type 2 in the first place. Since the convergence is uniform on each jet-level, there exist positive contants and such that the smallest eigenvalue of the Levi form of each at is larger than whenever . Consequently, the Levi form of at is strictly positive-definite. This completes the proof of Theorem 1.2.
5. A remark for the convex case
Notice that the scaling sequence converges without difficulties in case the domain is bounded convex. See e.g., . If the boundary point under consideration satisfies the condition and if were convex of infinite type, then the scaled limit turns out to be a convex domain that contains a one-dimensional disc with a positive radius in its boundary. (cf. , Theorem 3.7).
On the other hand, the preceding arguments imply that this domain has to be biholomorphic to the ball. But this is impossible, for instance by . So the boundary point has to be of finite type. Notice that our arguments work in all dimensions in this case, as the scaling method for bounded convex domains converges regardless of dimension (cf., e.g., ). This reconfirms Zimmer’s affirmative answer  to Question 1 for the smoothly bounded convex domains in for all .
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