Curvature of the Bergman Metric

Abstract

We begin with a little introductory material on the scaling method. Then we use these ideas to discuss Klembeck’s theorem about the boundary asymptotics of the curvature of the Bergman metric on a strictly pseudoconvex domain.

APPENDIX: Scaling in Dimension One

7.1.1 The Scaling of the Unit Disc

Let D be the open unit disc in the complex plane $\mathbb{C}$. Choose a sequence a _j in D satisfying the conditions

$$\displaystyle{0 < a_{j} < a_{j+1} < \cdots < 1,\ \forall j = 1,2,\ldots,}$$

and

$$\displaystyle{\lim _{j\rightarrow \infty }a_{j} = 1.}$$

Consider the sequence of dilations

$$\displaystyle{L_{j}(z) = \frac{1} {1 - a_{j}}(z - 1).}$$

Let us write $\lambda _{j} = 1 - a_{j}$. Then one sees immediately that

$$\displaystyle\begin{array}{rcl} L_{j}(D)& =& \{\zeta \in \mathbb{C}\mid (1 +\lambda _{j}\zeta )(1 +\lambda _{j}\bar{\zeta }) < 1\} {}\\ & =& \{\zeta \in \mathbb{C}\mid 2\ Re\,\zeta < -\lambda _{j}\vert \zeta {\vert }^{2}\}\,. {}\\ \end{array}$$

It follows that the sequence of sets L _j (D) converges to the left half plane $H =\{\zeta \in \mathbb{C}\mid Re\,\zeta < 0\}$ in the sense that

$$\displaystyle{L_{j}(D) \subset L_{j+1}(D),\ \forall j = 1,2,\ldots,}$$

and

$$\displaystyle{\bigcup _{j=1}^{\infty }L_{ j}(D) = H.}$$

[Compare the concept of convergence in the Hausdorff metric on sets—see [FED].]

Now we combine this simple observation with the fact that there exists the sequence of maps

$$\displaystyle{\varphi _{j}(z) = \frac{z + a_{j}} {1 + a_{j}z}}$$

that are automorphisms of D satisfying $\varphi _{j}(0) = a_{j}$. Consider the sequence of composite maps

$$\displaystyle{\sigma _{j} \equiv L_{j} \circ \varphi _{j}: D \rightarrow \mathbb{C}.}$$

A direct computation yields that

$$\displaystyle\begin{array}{rcl} L_{j} \circ \varphi _{j}(z)& =& \frac{1} {1 - a_{j}}\left ( \frac{z + a_{j}} {1 + a_{j}z} - 1\right ) {}\\ & =& \frac{z - 1} {1 + a_{j}z}. {}\\ \end{array}$$

Hence, in fact we see that the sequence of holomorphic mappings $L_{j} \circ \varphi _{j}$ converges uniformly on compact subsets of D to the mapping

$$\displaystyle{\hat{\sigma }(z) = \frac{z - 1} {z + 1}}$$

that is a biholomorphic mapping from the open unit disc D onto the left half plane H. (We have in effect discovered here a means to see the Cayley map by way of scaling.)

The point is that we have exploited the automorphism of the disc to see that the disc is conformally equivalent to a certain canonical domain—namely, the half plane. This result is neither surprising nor insightful. But it is a toy version of the main results that we shall present below.

7.1.2 A Generalization

We now expand the simple observations of the preceding subsection to yield the statement and the proof of the following one-dimensional version of the Wong–Rosay theorem:

Proposition:

Let Ω be a domain in the complex plane $\mathbb{C}$ admitting a boundary point p such that

• There exists an open neighborhood U of p in $\mathbb{C}$ such that $U \cap \partial \Omega$ is a C ¹ curve.

• There exists a sequence $\varphi _{j}$ of automorphisms of Ω and a point q ∈ Ω such that

  $$\displaystyle{\lim _{j\rightarrow \infty }\varphi _{j}(q) = p.}$$

Then Ω is biholomorphic to the open unit disc.

See [KRA13] for this theorem. We use this simple result to illustrate the technique of scaling.

In order to be consistent with the remainder of this chapter, we change a bit the notation for the orbit accumulation point and the point whose orbit we are calculating. This will all make sense in context.

Sketch of the Proof: Let $q_{j} =\varphi _{j}(q)$ for each j. Choose the closest point in the boundary to q _j and call it p _j . If the closest boundary point p _j to q _j is not unique, then make a choice. As j tends to infinity, p _j converges to p because q _j converges to p. Then we select θ _j and apply the map $\rho _{j}(z) \equiv {\mathrm{e}}^{i\theta _{j}}(z - p_{j})$ so that

$$\displaystyle{\rho _{j}(p_{j}) = 0\ \ \mbox{ and }\ \ \rho _{j}(q_{j}) > 0}$$

for each j. Now consider the sequence of mappings

$$\displaystyle{\psi _{j}(z) = \frac{1} {\rho _{j}(q_{j})}\left (\rho _{j} \circ \varphi _{j}(z)\right ).}$$

Notice that $\psi _{j}(\Omega ) = \frac{1} {\rho (q_{j})}\ \rho _{j}(\Omega )$ for each j. Thus we expect that ψ _j (Ω) is almost the right half plane as j becomes very large. At least every ψ _j (Ω) is contained in $\mathbb{C}\setminus \ell$ for some line segment ℓ of positive length and for every j. (Note that ℓ can be chosen independently of j.) Therefore one can select a subsequence from $\{\psi _{j}\}$ that converges uniformly on compact subsets of Ω. Let $\hat{\psi }$ be the limit mapping. Then we expect $\hat{\psi }: \Omega \rightarrow \mathbb{C}$ to be an injective holomorphic mapping, and furthermore, $\hat{\psi }(\Omega )$ is equal to the right half plane. Thus we hope to conclude that Ω is biholomorphic to the right half plane, which in turn is biholomorphic to the open unit disc. See Fig. 7.1.

Fig. 7.1

The scaling process

This plan actually works, but it is evident that there are several points that need clarification. We shall now present the precise proof, which will show much of the essence of the scaling method.

Rigorous Proof of the Main Result: Keeping the “Plan of the Proof” in mind, we present the precise proof in several steps. Let p ∈ ∂ Ω be as in the hypothesis of the proposition. Write $D(p,r) =\{ z \in \mathbb{C}\mid \vert z - p\vert < r\}$. Transforming Ω by a conformal mapping z↦e^iα(z − p), we may assume the following with no loss of generality:

• (a) p = 0

• (b) $\Omega \cap D(p,r) =\{ z = x + \mathit{iy}\mid y >\psi (x),\vert z - p\vert < r\}$ and $\partial \Omega \cap D(p,r) =\{ z\mid y =\psi (x),\vert z - p\vert < r\}$ for a real-valued C ¹ function ψ in one real variable satisfying ψ(0) = 0 and ψ ′(0) = 0.

Step 1. The Scaling Map. Notice that the sequence $\varphi _{j}(q)$ now converges to 0 as j → ∞. For each j, we choose a point p _j  ∈ ∂ Ω that is the closest to $\varphi _{j}(q)$. Since p _j also converges to 0, replacing $\varphi _{j}$ by a subsequence if necessary, we may assume that every p _j  ∈ D(p, r ∕ 4). Now, for each j, set

$$\displaystyle{\alpha _{j}(z) = i\ \frac{\vert \varphi _{j}(q) - p_{j}\vert } {\varphi _{j}(q) - p_{j}} \ (z - p_{j}).}$$

Notice that $\varphi _{j}(q) - p_{j}$ is a positive scalar multiple of the inward unit normal vector to ∂ Ω at p _j . Thus $\frac{\varphi _{j}(q) - p_{j}} {\vert \varphi _{j}(q) - p_{j}\vert }$ converges to the inward unit normal vector to ∂ Ω at 0. This implies that α _j in fact converges to the identity map. Consequently, there exist positive constants r ₁, r ₂ independent of j such that, for each j, there exists a C ¹ function ψ _j (x) defined for | x |  < r ₁ satisfying

$$\displaystyle{\alpha _{j}(z) \cap \left ([-r_{1},r_{1}] \times [-r_{2},r_{2}]\right ) =\{ x + iy\mid \vert x\vert < r_{1},\vert y\vert < r_{2},y >\psi _{j}(x)\}.}$$

Furthermore, for each ε > 0, there exists δ > 0 such that

$$\displaystyle{\psi _{j}(x) <\epsilon \vert x\vert \mbox{ whenever }\vert x\vert <\delta }$$

regardless of j.

Next, let $\lambda _{j} = \vert \varphi _{j}(q) - p_{j}\vert$ for each j. Consider the dilation map

$$\displaystyle{L_{j}(z) = \frac{z} {\lambda _{j}}.}$$

Then the sequence of holomorphic mappings we want to construct is given by

$$\displaystyle{\psi _{j} \equiv L_{j} \circ \alpha _{j} \circ \varphi _{j}: \Omega \rightarrow \mathbb{C}.}$$

Before starting the next step, we make a few remarks. The automorphism $\varphi _{j}$ preserves the domain Ω but moves q to $\varphi _{j}(q)$ so that $\varphi _{j}(q)$ converges to the origin—recall that we made changes so that p became the origin at the beginning of the proof. Then the affine map α _j adjusts Ω so that the direction vector $\frac{\varphi _{j}(q) - p_{j}} {\vert \varphi _{j}(q) - p_{j}\vert }$ is transformed to a purely imaginary number. The final component L _j in the construction simply magnifies the domain α _j (Ω), while the map L _j itself diverges.

Step 2. Convergence of the ψ _j . We shall actually choose a subsequence from {ψ _j } that converges uniformly on compact subsets of Ω. Observe first that

$$\displaystyle{\psi _{j}(\Omega ) = L_{j} \circ \alpha _{j} \circ \varphi _{j}(\Omega ) = L_{j} \circ \alpha _{j}(\Omega )}$$

since $\varphi _{j}(\Omega ) = \Omega$. Choosing a subsequence of ψ _j , we may assume that λ _j  < 1 for every j. Then, since L _j is a simple dilation by a positive number, and since α _j (Ω) will miss a line segment

$$\displaystyle{E =\{ -iy\mid 0 \leq y \leq b\}}$$

for some constant b independent of j, we see immediately that

$$\displaystyle{\psi _{j}(\Omega ) \subset \mathbb{C} \setminus E}$$

for every j = 1, 2, …. Therefore Montel’s theorem implies that every subsequence of {ψ _j } admits a subsequence, which we again (by an abuse of notation) denote by ψ _j , that converges uniformly on compact subsets of Ω. Denote by $\hat{\psi }$ the limit of the sequence ψ _j .

Step 3. Analysis of $\hat{\psi }(\Omega )$. We want to establish that

$$\displaystyle{\hat{\psi }(\Omega ) = \mathcal{U},}$$

where $\mathcal{U}\equiv \{ z \in \mathbb{C}\mid \ Im\,z > 0\}$.

Let ε be a positive real number and let K an arbitrary compact subset of Ω. We will show that $\hat{\psi }(K) \subset C_{\epsilon }$, where $C_{\epsilon } \equiv \{ z \in \mathbb{C}: -\epsilon <\arg z <\pi +\epsilon \}$.

Choose R > 0 such that $\hat{\psi }(K)$ is contained in the disc D(0, R) of radius R centered at 0.

The sequence $\varphi _{j}: \Omega \rightarrow \Omega$ is a normal family since $\mathbb{C} \setminus \Omega$ contains a line segment with positive length. Every subsequence of $\varphi _{j}$ contains a subsequence that converges uniformly on compact subsets, since $\varphi _{j}(q)$ converges to p. Let $g: \Omega \rightarrow \overline{\Omega }$ be a subsequential limit map. Then g(q) = p. Recall that $p \in \partial \Omega$. Hence, the open mapping theorem yields that g(z) = p for every z ∈ Ω. Thus the sequence $\varphi _{j}$ itself converges uniformly on compact subsets to the constant map with value p. Therefore we may choose N > 0 such that $\varphi _{j}(K)$ is contained in a sufficiently small neighborhood of the origin for every j > N, and hence $\alpha _{j} \circ \varphi _{j}(K) \subset C_{\epsilon }$ for every j > N. Then it follows immediately that $\psi _{j}(K) \subset C_{\epsilon }$ for every j > N and consequently that

$$\displaystyle{\hat{\psi }(K) \subset C_{\epsilon }.}$$

Since K is an arbitrary compact subset of Ω, it follows that $\hat{\psi }(\Omega ) \subset \overline{\mathcal{U}}$. We also have $\hat{\psi }(q) = i$, since $\psi _{j}(q) = L_{j} \circ \alpha _{j} \circ \varphi _{j}(q) = i$ for every j = 1, 2, …. Therefore $\hat{\psi }(\Omega ) \subset \mathcal{U}$.

Step 4. Convergence of $\psi _{j}^{-1}$. Let $\tilde{K}$ be an arbitrary compact subset of the upper half plane $\mathcal{U}$. Then choose ε > 0 so that $\tilde{K} \subset C_{\epsilon }$. Choose then r > 0 such that

$$\displaystyle{D(0,r) \cap C_{\epsilon } \subset \Omega \cap D(0,r).}$$

Shrinking r > 0 if necessary, since α _j converges to the identity map uniformly on compact subsets of $\mathbb{C}$, there exists N > 0 such that

$$\displaystyle{D(0,r) \cap C_{\epsilon } \subset \alpha _{j}(\Omega ) \cap D(0,r)}$$

for every j > N. Hence, we see that $\psi _{j}^{-1}$ maps K into Ω. Since $\Omega \subset \mathbb{C} \setminus E$ as observed before, we may again choose a subsequence of ψ _j , which we again denote by ψ _j , so that $\psi _{j}^{-1}$ converges to a holomorphic map, say $\tau: \mathcal{U}\rightarrow \overline{\Omega }$. Since τ is holomorphic and τ(i) = q, we see that τ maps the upper half plane $\mathcal{U}$ into Ω.

Step 5. Synthesis. We are ready to complete the proof. By the Cauchy estimates, the derivatives dψ _j of ψ _j as well as the derivatives $d[\psi _{j}^{-1}]$ both converge. Therefore $d\hat{\psi }(q) \cdot d\hat{\tau }(i) = 1$. This means that $\hat{\psi }\circ \tau: \mathcal{U}\rightarrow \mathcal{U}$ is a holomorphic mapping satisfying $\hat{\psi }\circ \tau (i) = i$ and $(\hat{\psi }\circ \tau )^{\prime}(i) = 1$. Then, by the Schwarz’s lemma, one concludes that $\hat{\psi }\circ \tau = \mbox{ id}$, where id is the identity mapping. Likewise, the same reasoning applied to $\tau \circ \hat{\psi }: \Omega \rightarrow \Omega$ implies that $\tau \circ \hat{\psi } = \mbox{ id}$. So $\hat{\psi }: \Omega \rightarrow \mathcal{U}$ is a biholomorphic mapping. □ 

Remark: The sequence of mappings ψ _j constructed above is often called a scaling sequence. It is constructed from a composition of

• The automorphisms carrying one fixed interior point successively to a boundary point

• Certain affine adjustments

• The stretching dilation map

The proof given above is a good example of the scaling technique. The main thrust of the method is that the image of the limit mapping is determined solely by the affine adjustments and the dilations, while the scaling sequence converges to a conformal mapping. □ 

Remark: As observed earlier, the main result here can be proved in a much simpler way. Namely, one may conclude immediately from the argument on the shrinking of $\varphi _{j}(K)$ into a simply connected subset of Ω that Ω must be simply connected. Then the conclusion follows by the Riemann mapping theorem. But we are trying to skirt around the Riemann mapping theorem. The goal of this argument is to provide a basis for the scaling method which can be applied to the higher-dimensional cases. □