2.3: Holomorphic Vectors, the Levi Form, and Pseudoconvexity

As \(\mathbb{C}^n\) is identified with \(\mathbb{R}^{2n}\) using \(z=x+iy\), we have \(T_p\mathbb{C}^n = T_p\mathbb{R}^{2n}\). If we take the complex span instead of the real span we get the complexified tangent space \[\mathbb{C} \otimes T_p\mathbb{C}^n = \operatorname{span}_{\mathbb{C}} \left\{ \frac{\partial}{\partial x_1}\Big|_p, \frac{\partial}{\partial y_1}\Big|_p, \ldots, \frac{\partial}{\partial x_n}\Big|_p , \frac{\partial}{\partial y_n}\Big|_p \right\} .\]

We simply replace all the real coefficients with complex ones. The space \(\mathbb{C} \otimes T_p\mathbb{C}^n\) is a \(2n\)-dimensional complex vector space. Both \(\frac{\partial}{\partial z_j}\big|_p\) and \(\frac{\partial}{\partial \bar{z}_j}\big|_p\) are in \(\mathbb{C} \otimes T_p\mathbb{C}^n\), and

\[\mathbb{C} \otimes T_p\mathbb{C}^n = \operatorname{span}_{\mathbb{C}} \left\{ \frac{\partial}{\partial z_1}\Big|_p, \frac{\partial}{\partial \bar{z}_1}\Big|_p, \ldots, \frac{\partial}{\partial z_n}\Big|_p , \frac{\partial}{\partial \bar{z}_n}\Big|_p \right\} .\]

Define \[T_p^{(1,0)} \mathbb{C}^n \overset{\text{def}}{=} \operatorname{span}_{\mathbb{C}} \left\{ \frac{\partial}{\partial z_1}\Big|_p, \ldots, \frac{\partial}{\partial z_n}\Big|_p \right\} \quad \text{and} \quad T_p^{(0,1)} \mathbb{C}^n \overset{\text{def}}{=} \operatorname{span}_{\mathbb{C}} \left\{ \frac{\partial}{\partial \bar{z}_1}\Big|_p, \ldots, \frac{\partial}{\partial \bar{z}_n}\Big|_p \right\} .\]

The vectors in \(T_p^{(1,0)} \mathbb{C}^n\) are the holomorphic vectors and vectors in \(T_p^{(0,1)} \mathbb{C}^n\) are the antiholomorphic vectors. We decompose the full tangent space as the direct sum

\[\mathbb{C} \otimes T_p\mathbb{C}^n = T_p^{(1,0)} \mathbb{C}^n \oplus T_p^{(0,1)} \mathbb{C}^n .\]

A holomorphic function is one that vanishes on \(T_p^{(0,1)} \mathbb{C}^n\).

Let us see what holomorphic mappings do to these spaces. Given a smooth mapping \(f\) from \(\mathbb{C}^n\) to \(\mathbb{C}^m\), its derivative at \(p \in \mathbb{C}^n\) is a real-linear mapping \(D_\mathbb{R} f(p) \colon T_p\mathbb{C}^n \to T_{f(p)} \mathbb{C}^m\). Given the basis above, this mapping is represented by the standard real Jacobian matrix, that is, a real \(2m \times 2n\) matrix that we wrote before as \(D_\mathbb{R} f(p)\).

Proposition \(\PageIndex{1}\)

Let \(f \colon U \subset \mathbb{C}^n \to \mathbb{C}^m\) be a holomorphic mapping with \(p \in U\). Suppose \(D_\mathbb{R} f(p) \colon T_p\mathbb{C}^n \to T_{f(p)} \mathbb{C}^m\) is the real derivative of \(f\) at \(p\). We naturally extend the derivative to \(D_\mathbb{C} f(p) \colon \mathbb{C} \otimes T_p\mathbb{C}^n \to \mathbb{C} \otimes T_{f(p)} \mathbb{C}^m\). Then \[D_\mathbb{C} f(p)\Bigl(T_p^{(1,0)} \mathbb{C}^n\Bigr) \subset T_{f(p)}^{(1,0)} \mathbb{C}^m \qquad \text{and} \qquad D_\mathbb{C} f(p)\Bigl(T_p^{(0,1)} \mathbb{C}^n\Bigr) \subset T_{f(p)}^{(0,1)} \mathbb{C}^m .\]

If \(f\) is a biholomorphism, then \(D_\mathbb{C} f(p)\) restricted to \(T_p^{(1,0)} \mathbb{C}^n\) is a vector space isomorphism. Similarly for \(T_p^{(0,1)} \mathbb{C}^n\).

Exercise \(\PageIndex{1}\)

Prove the proposition. Hint: Start with \(D_\mathbb{R} f(p)\) as a real \(2m \times 2n\) matrix to show it extends (it is the same matrix if you think of it as a matrix and use the same basis vectors). Think of \(\mathbb{C}^n\) and \(\mathbb{C}^m\) in terms of the \(z\)s and the \(\bar{z}\)s and think of \(f\) as a mapping \[(z,\bar{z}) \mapsto \bigl( f(z) , \bar{f}(\bar{z}) \bigr) .\] Write the derivative as a matrix in terms of the \(z\)s and the \(\bar{z}\)s and \(f\)s and \(\bar{f}\)s and the result will follow. That is just changing the basis.

For holomorphic functions and holomorphic vectors, when we say derivative of \(f\), we mean the holomorphic part of the derivative, which we write as \[D f(p) \colon T_p^{(1,0)} \mathbb{C}^n \to T_{f(p)}^{(1,0)} \mathbb{C}^m .\] That is, \(Df(p)\) is the restriction of \(D_\mathbb{C} f(p)\) to \(T_p^{(1,0)} \mathbb{C}^n\). In other words, let \(z\) be the coordinates on \(\mathbb{C}^n\) and \(w\) be coordinates on \(\mathbb{C}^m\). Using the bases \(\left\{ \frac{\partial}{\partial z_1} \big|_p,\ldots, \frac{\partial}{\partial z_n} \big|_p \right\}\) on \(\mathbb{C}^n\) and \(\left\{ \frac{\partial}{\partial w_1} \big|_{f(p)},\ldots, \frac{\partial}{\partial w_m} \big|_{f(p)} \right\}\) on \(\mathbb{C}^m\), the holomorphic derivative of \(f \colon \mathbb{C}^{n} \to \mathbb{C}^m\) is represented as the \(m \times n\) Jacobian matrix \[\left[ \frac{\partial f_j}{\partial z_k} \Big|_p \right]_{jk} ,\] which we have seen before and for which we also used the notation \(Df(p)\).

As before, define the tangent bundles \[\mathbb{C} \otimes T\mathbb{C}^n, \quad T^{(1,0)} \mathbb{C}^n, \quad \text{and} \quad T^{(0,1)} \mathbb{C}^n ,\] by taking the disjoint unions. One can also define vector fields in these bundles.

Let us describe \(\mathbb{C} \otimes T_pM\) for a smooth real hypersurface \(M \subset \mathbb{C}^n\). Let \(r\) be a real-valued defining function of \(M\) at \(p\). A vector \(X_p \in \mathbb{C} \otimes T_p\mathbb{C}^n\) is in \(\mathbb{C} \otimes T_pM\) whenever \(X_p r = 0\). That is, \[X_p = \sum_{j=1}^n \left( a_j \frac{\partial}{\partial z_j} \Big|_p + b_j \frac{\partial}{\partial \bar{z}_j} \Big|_p \right) \in \mathbb{C} \otimes T_p M \quad \text{whenever} \quad \sum_{j=1}^n \left( a_j \frac{\partial r}{\partial z_j} \Big|_p + b_j \frac{\partial r}{\partial \bar{z}_j} \Big|_p \right) = 0 .\] Therefore, \(\mathbb{C} \otimes T_p M\) is a \((2n-1)\)-dimensional complex vector space. We decompose \(\mathbb{C} \otimes T_p M\) as \[\mathbb{C} \otimes T_pM = T_p^{(1,0)} M \oplus T_p^{(0,1)} M \oplus B_p ,\] where \[T_p^{(1,0)} M \overset{\text{def}}{=} \bigl( \mathbb{C} \otimes T_pM \bigr) \cap \bigl( T_p^{(1,0)} \mathbb{C}^n \bigr), \qquad \text{and} \qquad T_p^{(0,1)} M \overset{\text{def}}{=} \bigl( \mathbb{C} \otimes T_pM \bigr) \cap \bigl( T_p^{(0,1)} \mathbb{C}^n \bigr) .\] The \(B_p\) is just the “leftover” and must be included, otherwise the dimensions will not work out.\(^{1}\)

Make sure that you understand what all the objects are. The space \(T_pM\) is a real vector space; \(\mathbb{C} \otimes T_pM\), \(T_p^{(1,0)}M\), \(T_p^{(0,1)} M\), and \(B_p\) are complex vector spaces. To see that these give vector bundles, we must first show that their dimensions do not vary from point to point. The easiest way to see this fact is to write down convenient local coordinates. First, let us see what a biholomorphic map does to the holomorphic and antiholomorphic vectors. A biholomorphic map \(f\) is a diffeomorphism. And if a real hypersurface \(M\) is defined by a function \(r\) near \(p\), then the image \(f(M)\) is also a real hypersurface is given by the defining function \(r \circ f^{-1}\) near \(f(p)\).

Proposition \(\PageIndex{2}\)

Suppose \(M \subset \mathbb{C}^n\) is a smooth real hypersurface, \(p \in M\), and \(U \subset \mathbb{C}^n\) is an open set such that \(M \subset U\). Let \(f \colon U \to \mathbb{C}^n\) be a holomorphic mapping such that \(D f(p)\) is invertible (a biholomorphism near \(p\)). Let \(D_{\mathbb{C}} f(p)\) be the complexified real derivative as before. Then \[D_{\mathbb{C}} f(p)\Bigl(T_p^{(1,0)}M\Bigr) = T_{f(p)}^{(1,0)}f(M), \qquad D_{\mathbb{C}} f(p)\Bigl(T_p^{(0,1)}M\Bigr) = T_{f(p)}^{(0,1)}f(M).\] That is, the spaces are isomorphic as complex vector spaces.

The proposition is local, if \(U\) is only a neighborhood of \(p\), replace \(M\) with \(M \cap U\).

Proof

We apply Proposition \(\PageIndex{1}\). As \(f\) is a biholomorphism at \(p\), \[\begin{align}\begin{aligned} D_{\mathbb{C}} f(p)\Bigl(T_p^{(1,0)}\mathbb{C}^n\Bigr) = T_{f(p)}^{(1,0)}\mathbb{C}^n, \quad D_{\mathbb{C}} f(p)\Bigl(T_p^{(0,1)}\mathbb{C}^n\Bigr) = T_{f(p)}^{(0,1)}\mathbb{C}^n, \quad \text{and} \\ D_{\mathbb{C}} f(p)\bigl(\mathbb{C} \otimes T_pM\bigr) =\mathbb{C} \otimes T_{f(p)}f(M) .\end{aligned}\end{align}\] Then \(D_{\mathbb{C}} f(p)\) must take \(T_p^{(1,0)}M\) to \(T_{f(p)}^{(1,0)}f(M)\) and \(T_p^{(0,1)}M\) to \(T_{f(p)}^{(0,1)}f(M)\).

In the next proposition it is important to note that a translation and applying a unitary matrix are biholomorphic changes of coordinates.

Proposition \(\PageIndex{3}\)

Let \(M \subset \mathbb{C}^n\) be a smooth real hypersurface, \(p \in M\). After a translation and a rotation by a unitary matrix, \(p\) is the origin, and near the origin, \(M\) is written in variables \((z,w) \in \mathbb{C}^{n-1} \times \mathbb{C}\) as \[\Im w = \varphi(z,\bar{z},\Re w) ,\] with the \(\varphi(0)\) and \(d\varphi(0) = 0\). Consequently \[\begin{align}\begin{aligned} T_0^{(1,0)} M = \operatorname{span}_{\mathbb{C}} \left\{ \frac{\partial}{\partial z_1}\Big|_0, \ldots, \frac{\partial}{\partial z_{n-1}}\Big|_0 \right\} , & \qquad T_0^{(0,1)} M = \operatorname{span}_{\mathbb{C}} \left\{ \frac{\partial}{\partial \bar{z}_1}\Big|_0, \ldots, \frac{\partial}{\partial \bar{z}_{n-1}}\Big|_0 \right\} , \\ & B_0 = \operatorname{span}_{\mathbb{C}} \left\{ \frac{\partial}{\partial (\Re w)}\Big|_0 \right\} .\end{aligned}\end{align}\] In particular, \(\dim_{\mathbb{C}} T_p^{(1,0)} M = \dim_{\mathbb{C}} T_p^{(0,1)} M = n-1\) and \(\dim_{\mathbb{C}} B_p = 1\).

If \(M\) is the boundary of a open set \(U\) with smooth boundary, the rotation can be chosen so that \(\Im w > \varphi(z,\bar{z},\Re w)\) on \(U\).

Proof

We apply a translation to put \(p=0\) and in the same manner as Lemma 2.2.1 in apply a unitary matrix to make sure that \(\nabla r\) is in the direction \(-\frac{\partial}{\partial (\Im w)}\big|_0\). That \(\varphi(0) = 0\) and \(d\varphi(0) = 0\) follows as before. As a translation and a unitary matrix are holomorphic and in fact biholomorphic, then via Proposition \(\PageIndex{1}\) we obtain that the tangent spaces are all transformed correctly.

The rest of the proposition follows at once as \(\frac{\partial}{\partial (\Im w)}\big|_0\) is the normal vector to \(M\) at the origin.

Remark \(\PageIndex{1}\)

When \(M\) is of smaller dimension than \(2n-1\) (no longer a hypersurface, but a submanifold of higher codimension), then the proposition above does not hold. That is, we would still have \(\dim_{\mathbb{C}} T_p^{(1,0)} M = \dim_{\mathbb{C}} T_p^{(0,1)} M\), but this number need not be constant from point to point. Fortunately, when talking about domains with smooth boundaries, the boundaries are hypersurfaces, and this complication does not arise.

Definition: Pseudoconvex

Suppose \(U \subset \mathbb{C}^n\) is an open set with smooth boundary, and \(r\) is a defining function for \(\partial U\) at \(p \in \partial U\) such that \(r < 0\) on \(U\). If \[\sum_{j=1,\ell=1}^n \bar{a}_j a_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_p \geq 0 \qquad \text{for all} \qquad X_p = \sum_{j=1}^n a_j \frac{\partial}{\partial z_j}\Big|_p \quad \in \quad T^{(1,0)}_p \partial U,\] then \(U\) is pseudoconvex at \(p\) (or Levi pseudoconvex). If the inequality above is strict for all nonzero \(X_p \in T^{(1,0)}_p \partial U\), then \(U\) is strongly pseudoconvex at \(p\). If \(U\) is pseudoconvex, but not strongly pseudoconvex, at \(p\), then \(U\) is weakly pseudoconvex.

A domain \(U\) is pseudoconvex if it is pseudoconvex at all \(p \in \partial U\). For a bounded\(^{2}\) \(U\), we say \(U\) is strongly pseudoconvex if it is strongly pseudoconvex at all \(p \in \partial U\).

For \(X_p \in T^{(1,0)}_p\partial U\), the sesquilinear form \[\mathcal{L}(X_p,X_p) = \sum_{j=1,\ell=1}^n \bar{a}_j a_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_p\] is called the Levi form at \(p\). So \(U\) is pseudoconvex (resp. strongly pseudoconvex) at \(p \in \partial U\) if the Levi form is positive semidefinite (resp. positive definite) at \(p\). The Levi form can be defined for any real hypersurface \(M\), although one has to decide which side of \(M\) is “the inside.”

The matrix \[\left[ \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_p \right]_{j \ell}\] is called the complex Hessian of \(r\) at \(p\).\(^{3}\) So, \(U\) is pseudoconvex at \(p \in \partial U\) if the complex Hessian of \(r\) at \(p\) as a sesquilinear form is positive (semi)definite when restricted to tangent vectors in \(T^{(1,0)}_p \partial U\). For example, the unit ball \(\mathbb{B}_{n}\) is strongly pseudoconvex as can be seen by computing the Levi form directly from \(r(z,\bar{z}) = ||z||^2-1\), that is, the complex Hessian of \(r\) is the identity matrix.

We remark that the complex Hessian is not the full Hessian. Let us write down the full Hessian, using the basis of \(\frac{\partial}{\partial z}\)s and \(\frac{\partial}{\partial \bar{z}}\)s. It is the symmetric matrix \[\left[\begin{array}{cccccc}{\frac{\partial^{2}r}{\partial z_{1}\partial z{1}}}&{\cdots}&{\frac{\partial^{2}r}{\partial z_{1}\partial z_{n}}}&{\frac{\partial^{2}r}{\partial z_{1}\partial\overline{z}_{1}}}&{\cdots}&{\frac{\partial^{2}r}{\partial z_{1}\partial\overline{z}_{n}}}\\{\vdots}&{\ddots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{\frac{\partial^{2}r}{\partial z_{n}\partial z_{1}}}&{\cdots}&{\frac{\partial^{2}r}{\partial z_{n}\partial z_{n}}}&{\frac{\partial^{2}r}{\partial z_{n}\partial\overline{z}_{1}}}&{\cdots}&{\frac{\partial^{2}r}{\partial z_{n}\partial\overline{z}_{n}}}\\{\frac{\partial^{2}r}{\partial\overline{z}_{1}\partial z_{1}}}&{\cdots}&{\frac{\partial^{2}r}{\partial\overline{z}_{1}\partial z_{n}}}&{\frac{\partial^{2}r}{\partial\overline{z}_{1}\partial\overline{z}_{1}}}&{\cdots}&{\frac{\partial^{2}r}{\partial\overline{z}_{1}\partial\overline{z}_{n}}}\\{\vdots}&{\ddots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{\frac{\partial^{2}r}{\partial\overline{z}_{n}\partial z_{1}}}&{\cdots}&{\frac{\partial^{2}r}{\partial\overline{z}_{n}\partial z_{n}}}&{\frac{\partial^{2}r}{\partial\overline{z}_{n}\partial\overline{z}_{1}}}&{\ddots}&{\frac{\partial^{2}r}{\partial\overline{z}_{n}\partial\overline{z}_{n}}}\end{array}\right]\]

The complex Hessian is the lower left, or the transpose of the upper right, block. That is, if you write the full Hessian as \(\left[ \begin{smallmatrix} X & L^t \\ L & \overline{X} \end{smallmatrix} \right]\), then \(L\) is the complex Hessian. In particular, \(L\) is a smaller matrix. And we also apply it only to a subspace of the complexified tangent space.

We illustrate the change of basis on one dimension. The change of variables is left to student for higher dimensions. Let \(z = x+iy\) be in \(\mathbb{C}\), and denote by \(T\) the change of basis matrix: \[T = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{-i}{2} & \frac{i}{2} \end{bmatrix} , \qquad T^t \begin{bmatrix} \frac{\partial^2 r}{\partial x \partial x} && \frac{\partial^2 r}{\partial x \partial y} \\ \frac{\partial^2 r}{\partial y \partial x} && \frac{\partial^2 r}{\partial y \partial y} \end{bmatrix} T = \begin{bmatrix} \frac{\partial^2 r}{\partial z \partial z} && \frac{\partial^2 r}{\partial z \partial \bar{z}} \\ \frac{\partial^2 r}{\partial \bar{z} \partial z} && \frac{\partial^2 r}{\partial \bar{z} \partial \bar{z}} \end{bmatrix} .\] The relationship between the eigenvalues of the real Hessian and the complex Hessian is not perhaps as straightforward as may at first seem, but there is a relationship there nonetheless.

Exercise \(\PageIndex{2}\)

If \(r\) is real-valued, then the complex Hessian of \(r\) is Hermitian, that is, the matrix is equal to its conjugate transpose.

Exercise \(\PageIndex{3}\)

Consider one dimension and \(z = x+iy\). Consider \[H=\begin{bmatrix} \frac{\partial^2 r}{\partial x\partial x} && \frac{\partial^2 r}{\partial x \partial y} \\ \frac{\partial^2 r}{\partial y \partial x} && \frac{\partial^2 r}{\partial y\partial y} \end{bmatrix} ,\] the real Hessian in terms of \(x\) and \(y\). Prove that the complex Hessian \(L\) (just a number now) is \(\frac{1}{4}\) of the trace of \(H\). Thus, if \(H\) is positive definite, then \(L > 0\), and if \(H\) is negative definite, then \(L < 0\). However, show by example that if \(H\) has mixed eigenvalues (positive and negative), then \(L\) can be positive, negative, or zero.

Exercise \(\PageIndex{4}\)

For any dimension, find the change of variables \(T^t H T\) to go from the real Hessian in terms of \(x\) and \(y\) to the Hessian in terms of \(z\) and \(\bar{z}\). Hint: If you figure it out for \(n=2\), it will be easy to do in general.

Exercise \(\PageIndex{5}\)

Prove in any dimension that if the real Hessian (in terms of \(x\) and \(y\)) is positive (semi)definite, then the complex Hessian is positive (semi)definite. Hint: A Hermitian matrix \(L\) is positive definite if \(v^*Lv > 0\) for all nonzero vectors \(v\) and semidefinite if \(v*Lv \geq 0\) for all \(v\).

Let us also see how a complex linear change of variables acts on the Hessian matrix. A complex linear change of variables is not an arbitrary \(2n \times 2n\) matrix. If the Hessian is in the basis of \(\frac{\partial}{\partial z}\)s and \(\frac{\partial}{\partial \bar{z}}\)s, an \(n \times n\) complex linear matrix \(A\) acts on the Hessian as \(A \oplus \overline{A} = \left[ \begin{smallmatrix} A & 0 \\ 0 & \overline{A} \end{smallmatrix} \right]\). Write the full Hessian as \(\left[ \begin{smallmatrix} X & L^t \\ L & \overline{X} \end{smallmatrix} \right]\), where \(L\) is the complex Hessian. The complex linear change of variables \(A\) transforms the full Hessian as \[{\begin{bmatrix} A & 0 \\ 0 & \overline{A} \end{bmatrix}}^t \begin{bmatrix} X & L^t \\ L & \overline{X} \end{bmatrix} \begin{bmatrix} A & 0 \\ 0 & \overline{A} \end{bmatrix} = \begin{bmatrix} A^tXA & {(A^*LA)}^t \\ A^*LA & \overline{A^tXA} \end{bmatrix} ,\] where \(A^* = \overline{A}^t\) is the conjugate transpose of \(A\). So \(A\) transforms the complex Hessian \(L\) as \(A^* L A\), that is, by \(*\)-congruence. Star-congruence preserves the inertia (the number of positive, negative, and zero eigenvalues) of a Hermitian matrix by the Sylvester’s law of inertia from linear algebra.

The Levi form itself does depend on the defining function, but the signs of the eigenvalues do not. It is common to say “the Levi form” without mentioning a specific defining function even though that is not completely correct. The proof of the following proposition is left as an exercise.

Proposition \(\PageIndex{4}\)

Let \(U \subset \mathbb{C}^n\) be an open set with smooth boundary and \(p \in \partial U\). The inertia of the Levi form at \(p\) does not depend on the defining function at \(p\). In particular, the definition of pseudoconvexity and strong pseudoconvexity is independent of the defining function.

Exercise \(\PageIndex{6}\)

Prove Proposition \(\PageIndex{4}\).

Exercise \(\PageIndex{7}\)

Show that a convex domain with smooth boundary is pseudoconvex, and show that (a bounded) strongly convex domain with smooth boundary is strongly pseudoconvex.

Exercise \(\PageIndex{8}\)

Show that if an open set with smooth boundary is strongly pseudoconvex at a point, it is strongly pseudoconvex at all nearby points.

We are generally interested what happens under a holomorphic change of coordinates, that is, a biholomorphic mapping. And as far as pseudoconvexity is concerned we are interested in local changes of coordinates as pseudoconvexity is a local property. Before proving that pseudoconvexity is a biholomorphic invariant, let us note where the Levi form appears in the graph coordinates from Proposition \(\PageIndex{1}\), that is, when our boundary (the hypersurface) is given near the origin by \[\Im w = \varphi(z,\bar{z},\Re w) ,\] where \(\varphi\) is \(O(2)\). Let \(r(z,\bar{z},w,\bar{w}) = \varphi(z,\bar{z},\Re w) - \Im w\) be our defining function. The complex Hessian of \(r\) has the form \[\begin{bmatrix} L & 0 \\ 0 & 0 \end{bmatrix} \qquad \text{where} \quad L = \left[ \frac{\partial^2 \varphi}{\partial \bar{z}_j \partial z_{\ell}}\Big|_0 \right]_{j \ell} .\] Note that \(L\) is an \((n-1) \times (n-1)\) matrix. The vectors in \(T_0^{(1,0)} \partial U\) are the span of \(\left\{ \frac{\partial}{\partial z_1}\big|_0, \ldots, \frac{\partial}{\partial z_{n-1}}\big|_0 \right\}\). That is, as an \(n\)-vector, a vector in \(T_0^{(1,0)} \partial U\) is represented by \((a,0) \in \mathbb{C}^n\) for some \(a \in \mathbb{C}^{n-1}\). The Levi form at the origin is then \(a^* L a\), in other words, it is given by the \((n-1) \times (n-1)\) matrix \(L\). If this matrix \(L\) is positive semidefinite, then \(\partial U\) is pseudoconvex at \(0\).

Example \(\PageIndex{1}\)

Let us change variables to write the ball \(\mathbb{B}_{n}\) in different local holomorphic coordinates where the Levi form is displayed nicely. The sphere \(\partial \mathbb{B}_{n}\) is defined in the variables \(Z = (Z_1,\ldots,Z_n) \in \mathbb{C}^n\) by \(||Z|| = 1\).

Let us change variables to \((z_1,\ldots,z_{n-1},w)\) where \[z_j = \frac{Z_j}{1-Z_n} \quad \text{ for all $j=1,\ldots,n-1$}, \qquad w = i\frac{1+Z_n}{1-Z_n} .\] This change of variables is a biholomorphic mapping from the set where \(Z_n \not= 1\) to the set where \(w\not= -i\) (exercise). For us it is sufficient to notice that the map is invertible near \((0,\ldots,0,-1)\), which follows by simply computing the derivative. Notice that the last component is the inverse of the Cayley transform (that takes the disc to the upper half-plane).

We claim that the mapping takes the unit sphere given by \(|Z| = 1\) (without the point \((0,\ldots,0,1)\)), to the set defined by \[\Im w = |z_1|^2 + \cdots + |z_{n-1}|^2 ,\] and that it takes \((0,\ldots,0,-1)\) to the origin (this part is trivial). Let us check:

\[\begin{align}\begin{aligned}|z_{1}|^{2}+\cdots +|z_{n-1}|^{2}-\Im w&=\left|\frac{Z_{1}}{1-Z_{n}}\right|^{2}+\cdots +\left|\frac{Z_{n-1}}{1-Z_{n}}\right|^{2}-\frac{i\frac{1+Z_{n}}{1-Z_{n}}-\overline{i\frac{1+Z_{n}}{1-Z_{n}}}}{2i} \\ &=\frac{|Z_{1}|^{2}}{|1-Z_{n}|^{2}}+\cdots +\frac{|Z_{n-1}|^{2}}{|1-Z_{n}|^{2}}-\frac{1+Z_{n}}{2(1-Z_{n})}-\frac{1+\overline{Z}_{n}}{2(1-\overline{Z}_{n})} \\ &=\frac{|Z_{1}|^{2}+\cdots +|Z_{n-1}|^{2}+|Z_{n}|^{2}-1}{|1-Z_{n}|^{2}}.\end{aligned}\end{align}\]

Therefore, \(|Z_1|^2 + \cdots + |Z_n|^2 = 1\) if and only if \(\Im w = |z_1|^2 + \cdots + |z_{n-1}|^2\). As the map takes the point \((0,\ldots,0,-1)\) to the origin, we can think of the set given by \[\Im w = |z_1|^2 + \cdots + |z_{n-1}|^2\] as the sphere in local holomorphic coordinates at \((0,\ldots,0,-1)\) (by symmetry of the sphere we could have done this at any point by rotation). In the coordinates \((z,w)\), the ball (the inside of the sphere) is the set given by \[\Im w > |z_1|^2 + \cdots + |z_{n-1}|^2 .\]

In these new coordinates, the Levi form is just the identity matrix at the origin. In particular, the domain is strongly pseudoconvex. We have not yet proved that pseudoconvexity is a biholomorphic invariant, but when we do, it will also mean that the ball is strongly pseudoconvex.

Not the entire sphere gets transformed, the points where \(Z_n=1\) get “sent to infinity.” The hypersurface \(\Im w = |z_1|^2 + \cdots + |z_{n-1}|^2\) is sometimes called the Lewy hypersurface, and in the literature some even say it is the sphere\(^{4}\). Pretending \(z\) is just one real direction, see Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\)

As an aside, the hypersurface \(\Im w = |z_1|^2 + \cdots + |z_{n-1}|^2\) is also called the Heisenberg group. The group in this case is the group defined on the parameters \((z,\Re w)\) of this hypersurface with the group law defined by \((z,\Re w)(z',\Re w') = (z+z',\Re w + \Re w' + 2 \Im z \cdot z')\).

Exercise \(\PageIndex{9}\)

Prove the assertion in the example about the mapping being biholomorphic on the sets described above.

Let us see how the Hessian of \(r\) changes under a biholomorphic change of coordinates. That is, let \(f \colon V \to V'\) be a biholomorphic map between two domains in \(\mathbb{C}^n\), and let \(r \colon V' \to \mathbb{R}\) be a smooth function with nonvanishing derivative. Let us compute the Hessian of \(r \circ f \colon V \to \mathbb{R}\). We first compute what happens to the nonmixed derivatives. As we have to apply chain rule twice, to keep track better track of things, we write where the derivatives are being evaluated, as they are, after all, functions. For clarity, let \(z\) be the coordinates in \(V\) and \(\zeta\) the coordinates in \(V'\). That is, \(r\) is a function of \(\zeta\) and \(\bar{\zeta}\), \(f\) is a function of \(z\), and \(\bar{f}\) is a function of \(\bar{z}\). Therefore, \(r \circ f\) is a function of \(z\) and \(\bar{z}\).

\[ \begin{align}\begin{aligned} \frac{\partial^2 (r \circ f)}{\partial z_j \partial z_k} \bigg|_{(z,\bar{z})} & = \frac{\partial}{\partial z_j } \sum_{\ell=1}^n \biggl( \frac{\partial r}{\partial \zeta_\ell} \bigg|_{(f(z),\bar{f}(\bar{z}))} \frac{\partial f_\ell}{\partial z_k} \bigg|_{z} + \frac{\partial r}{\partial \bar{\zeta}_\ell} \bigg|_{(f(z),\bar{f}(\bar{z}))} \cancelto{0}{\frac{\partial \bar{f}_\ell}{\partial z_k} \bigg|_{\bar{z}}} \biggr) \\ & = \sum_{\ell,m=1}^n \biggl( \frac{\partial^2 r}{\partial \zeta_m \partial \zeta_\ell} \bigg|_{(f(z),\bar{f}(\bar{z}))} \frac{\partial f_m}{\partial z_j} \bigg|_z \frac{\partial f_\ell}{\partial z_k} \bigg|_z + \frac{\partial^2 r}{\partial \bar{\zeta}_m \partial \zeta_\ell} \bigg|_{(f(z),\bar{f}(\bar{z}))} \smash{\cancelto{0}{\frac{\partial \bar{f}_m}{\partial z_j} \bigg|_{\bar{z}} }} \frac{\partial f_\ell}{\partial z_k} \bigg|_z \biggr) \\ & \qquad + \sum_{\ell=1}^n \frac{\partial r}{\partial \zeta_\ell} \bigg|_{(f(z),\bar{f}(\bar{z}))} \frac{\partial^2 f_\ell}{\partial z_j \partial z_k} \bigg|_z \\ &= \sum_{\ell,m=1}^n \frac{\partial^2 r}{\partial \zeta_m \partial \zeta_\ell} \frac{\partial f_m}{\partial z_j} \frac{\partial f_\ell}{\partial z_k} + \sum_{\ell=1}^n \frac{\partial r}{\partial \zeta_\ell} \frac{\partial^2 f_\ell}{\partial z_j \partial z_k} . \end{aligned}\end{align}\]

The matrix \(\left[ \frac{\partial^2 (r \circ f)}{\partial z_j \partial z_k} \right]\) can have different eigenvalues than the matrix \(\left[ \frac{\partial^2 r}{\partial \zeta_j \partial \zeta_k} \right]\). If \(r\) has nonvanishing gradient, then using the second term, we can (locally) choose \(f\) in such a way as to make the matrix \(\left[ \frac{\partial^2 (r \circ f)}{\partial z_j \partial z_k} \right]\) be the zero matrix (or anything else) at a certain point by choosing the second derivatives of \(f\) arbitrarily at that point. See the exercise below. Nothing about the matrix \(\left[ \frac{\partial^2 r}{\partial \zeta_j \partial \zeta_k} \right]\) is preserved under a biholomorphic map. And that is precisely why it does not appear in the definition of pseudoconvexity. The story for \(\left[ \frac{\partial^2 (r \circ f)}{\partial \bar{z}_j \partial \bar{z}_k} \right]\) and \(\left[ \frac{\partial^2 r}{\partial \bar{\zeta}_j \partial \bar{\zeta}_k} \right]\) is exactly the same.

Exercise \(\PageIndex{10}\)

Given a real function \(r\) with nonvanishing gradient at \(p \in \mathbb{C}^n\). Find a local change of coordinates \(f\) at \(p\) (so \(f\) ought to be a holomorphic mapping with an invertible derivative at \(p\)) such that \(\left[ \frac{\partial^2 (r \circ f)}{\partial z_j \partial z_k} \Big|_p \right]\) and \(\left[ \frac{\partial^2 (r \circ f)}{\partial \bar{z}_j \partial \bar{z}_k} \Big|_p \right]\) are just the zero matrices.

Let us look at the mixed derivatives: \[\begin{align}\begin{aligned} \frac{\partial^2 (r \circ f)}{\partial \bar{z}_j \partial z_k} \bigg|_{(z,\bar{z})} & = \frac{\partial}{\partial \bar{z}_j } \sum_{\ell=1}^n \left( \frac{\partial r}{\partial \zeta_\ell} \bigg|_{(f(z),\bar{f}(\bar{z}))} \frac{\partial f_\ell}{\partial z_k} \bigg|_z \right) \\ & = \sum_{\ell,m=1}^n \frac{\partial^2 r}{\partial \bar{\zeta}_m \partial \zeta_\ell } \bigg|_{(f(z),\bar{f}(\bar{z}))} \frac{\partial \bar{f}_m}{\partial \bar{z}_j} \bigg|_{\bar{z}} \frac{\partial f_\ell}{\partial z_k} \bigg|_z + \sum_{\ell=1}^n \frac{\partial r}{\partial \zeta_\ell} \bigg|_{(f(z),\bar{f}(\bar{z}))} \smash{\cancelto{0}{\frac{\partial^2 f_\ell}{\partial \bar{z}_j \partial z_k} \bigg|_z}} \\ &= \sum_{\ell,m=1}^n \frac{\partial^2 r}{\partial \bar{\zeta}_m \partial \zeta_\ell} \frac{\partial \bar{f}_m}{\partial \bar{z}_j} \frac{\partial f_\ell}{\partial z_k} . \end{aligned}\end{align}\]

The complex Hessian of \(r \circ f\) is the complex Hessian \(H\) of \(r\) conjugated as \(D^*HD\), where \(D\) is the holomorphic derivative matrix of \(f\) at \(z\) and \(D^*\) is its conjugate transpose. Sylvester’s law of inertia says that the number of positive, negative, and zero eigenvalues of \(D^*HD\) is the same as that for \(H\). The eigenvalues may change, but their sign do not. We are only considering \(H\) and \(D^*HD\) on a subspace. In linear algebra language, consider an invertible \(D\), a subspace \(T\), and its image \(DT\). Then the inertia of \(H\) restricted to \(DT\) is the same as the inertia of \(D^*HD\) restricted to \(T\).

Let \(M\) be a smooth real hypersurface given by \(r=0\), then \(f^{-1}(M)\) is a smooth real hypersurface given by \(r \circ f = 0\). The holomorphic derivative \(D = Df(p)\) takes \(T_{p}^{(1,0)}f^{-1}(M)\) isomorphically to \(T_{f(p)}^{(1,0)}M\). So \(H\) is positive (semi)definite on \(T_{f(p)}^{(1,0)}M\) if and only if \(D^*HD\) is positive (semi)definite on \(T_{p}^{(1,0)} f^{-1}(M)\). We have almost proved the following theorem. In short, pseudoconvexity is a biholomorphic invariant.

Theorem \(\PageIndex{1}\)

Suppose \(U, U' \subset \mathbb{C}^n\) are open sets with smooth boundary, \(p \in \partial U\), \(V \subset \mathbb{C}^n\) a neighborhood of \(p\), \(q \in \partial U'\), \(V' \subset \mathbb{C}^n\) a neighborhood of \(q\), and \(f \colon V \to V'\) a biholomorphic map with \(f(p) = q\), such that \(f(U \cap V) = U' \cap V'\). See Figure \(\PageIndex{2}\).

Then the inertia of the Levi form of \(U\) at \(p\) is the same as the inertia of the Levi form of \(U'\) at \(q\). In particular, \(U\) is pseudoconvex at \(p\) if and only if \(U'\) is pseudoconvex at \(q\). Similarly, \(U\) is strongly pseudoconvex at \(p\) if and only if \(U'\) is strongly pseudoconvex at \(q\).

Figure \(\PageIndex{2}\)

To finish proving the theorem, the only thing left is to observe that if \(f(U \cap V) = U' \cap V'\), then \(f(\partial U \cap V) = \partial U' \cap V'\), and to note that if \(r\) is a defining function for \(U'\) at \(q\), then \(f \circ r\) is a defining function for \(U\) at \(p\).

Exercise \(\PageIndex{11}\)

Find an example of a bounded domain with smooth boundary that is not convex, but that is pseudoconvex.

While the Levi form is not invariant under holomorphic changes of coordinates, its inertia is. Putting this together with the other observations we made above, we will find the normal form for the quadratic part of the defining equation for a smooth real hypersurface under biholomorphic transformations. It is possible to do better than the following lemma, but it is not possible to always get rid of the dependence on \(\Re w\) in the higher order terms.

Lemma \(\PageIndex{1}\)

Let \(M\) be a smooth real hypersurface in \(\mathbb{C}^n\) and \(p \in M\). Then there exists a local biholomorphic change of coordinates taking \(p\) to the origin and \(M\) to the hypersurface given by \[\Im w = \sum_{j=1}^\alpha |z_j|^2 - \sum_{j=\alpha+1}^{\alpha+\beta} |z_j|^2 + E(z,\bar{z},\Re w) ,\] where \(E\) is \(O(3)\) at the origin. Here \(\alpha\) is the number of positive eigenvalues of the Levi form at \(p\), \(\beta\) is the number of negative eigenvalues, and \(\alpha+\beta \leq n-1\).

Recall that \(O(k)\) at the origin means a function that together with its derivatives up to order \(k-1\) vanish at the origin.

Proof

Change coordinates so that \(M\) is given by \(\Im w = \varphi(z,\bar{z},\Re w)\), where \(\varphi\) is \(O(2)\). Apply Taylor’s theorem to \(\varphi\) up to the second order: \[\varphi(z,\bar{z},\Re w) = q(z,\bar{z}) + (\Re w) (Lz + \overline{Lz}) + a {(\Re w)}^2 + O(3) ,\] where \(q\) is quadratic, \(L \colon \mathbb{C}^{n-1} \to \mathbb{C}\) is linear, and \(a \in \mathbb{R}\). If \(L \not= 0\), do a linear change of coordinates in the \(z\) only to make \(Lz = z_1\). So assume \(Lz = \epsilon z_1\) where \(\epsilon = 0\) or \(\epsilon = 1\).

Change coordinates by leaving \(z\) unchanged and letting \(w = w'+bw'^2+cw'z_1\). Ignore \(q(z,\bar{z})\) for a moment, as this change of coordinates does not affect it. Also, only work up to second order.

\[\begin{align}\begin{aligned} -\Im w+\epsilon (\Re w)(z_{1}+\overline{z}_{1})+a(\Re w)^{2}&=-\frac{w-\overline{w}}{2i}+\epsilon\frac{w+\overline{w}}{2}(z_{1}+\overline{z}_{1})+a\left(\frac{w+\overline{w}}{2}\right)^{2} \\ &=-\frac{w'+bw'^{2}+cw'z_{1}-\overline{w}'-\overline{b}\overline{w}'^{2}-\overline{c}\overline{w}'\overline{z}_{1}}{2i} \\ &\quad +\epsilon\frac{w'+bw'^{2}+cw'z_{1}+\overline{w}'+\overline{b}\overline{w}'^{2}+\overline{c}\overline{w}'\overline{z}_{1}}{2}(z_{1}+\overline{z}_{1}) \\ &\quad +a\frac{(w'+bw'^{2}+cw'z_{1}+\overline{w}'+\overline{b}\overline{w}'^{2}+\overline{c}\overline{w}'\overline{z}_{1})^{2}}{4} \\ &=-\frac{w'-\overline{w}'}{2i} \\ &\quad +\frac{((\epsilon i-c)w'+\epsilon i\overline{w}')z_{1}+((\epsilon i+\overline{c})\overline{w}'+\epsilon iw')\overline{z}_{1}}{2i} \\ &\quad +\frac{(ia-2b)w'^{2}+(ia+2\overline{b})\overline{w}'^{2}+2iaw'\overline{w}'}{4i}+O(3) \end{aligned}\end{align}\]

We cannot quite get rid of all the quadratic terms in \(\varphi\), but we choose \(b\) and \(c\) to make the second order terms not depend on \(\Re w'\). Set \(b=ia\) and \(c=2i\epsilon\), and add \(q(z,\bar{z}) + O(3)\) into the mix to get

\[\begin{align}\begin{aligned}-\Im w+\varphi (z,\overline{z}, \Re w)&=-\Im w+q(z,\overline{z})+\epsilon (\Re w)(z_{1}+\overline{z}_{1})+a(\Re w)^{2}+O(3) \\ &=-\frac{w'-\overline{w}'}{2i}+q(z,\overline{z})-\epsilon i\frac{w'-\overline{w}'}{2i}(z_{1}-\overline{z}_{1})+a\left(\frac{w'-\overline{w}'}{2i}\right)^{2}+O(3) \\ &=-\Im w'+q(z,\overline{z})-\epsilon i(\Im w')(z_{1}-\overline{z}_{1})+a(\Im w')^{2}+O(3).\end{aligned}\end{align}\]

The right-hand side is the defining equation in the \((z,w')\) coordinates. However, it is no longer written as a graph of \(\Im w'\) over the rest, so we apply the implicit function theorem to solve for \(\Im w'\) and write the hypersurface as a graph again. The expression for \(\Im w'\) is \(O(2)\), and therefore \(-i\epsilon (\Im w')(z_1-\bar{z}_1)+a{(\Im w')}^2\) is \(O(3)\). So if we write \(M\) as a graph, \[\Im w' = q(z,\bar{z}) + E(z,\bar{z},\Re w'),\] then \(E\) is \(O(3)\).

We write the quadratic polynomial \(q\) as \[\label{eq:1} q(z,\bar{z}) = \sum_{j,k=1}^{n-1} a_{jk} z_jz_k + b_{jk} \bar{z}_j\bar{z}_k + c_{jk} \bar{z}_jz_k .\] As \(q\) is real-valued, it is left as an exercise to show that \(a_{jk} = \overline{b_{jk}}\) and \(c_{jk} = \overline{c_{kj}}\). That is, the matrix \([b_{jk}]\) is the complex conjugate of \([a_{jk}]\) and \([c_{jk}]\) is Hermitian.

We make another change of coordinates. We fix the \(z\)s again, and we set \[\label{eq:2} w' = w'' + i \sum_{j,k=1}^{n-1} a_{jk} z_jz_k .\] In particular, \[\Im w' = \Im w'' + \Im \biggl( i \sum_{j,k=1}^{n-1} a_{jk} z_jz_k \biggr) = \Im w'' + \sum_{j,k=1}^{n-1} \bigl( a_{jk} z_jz_k + b_{jk} \bar{z}_j\bar{z}_k \bigr) ,\] as \(\overline{a_{jk}} = b_{jk}\). Plugging \(\eqref{eq:2}\) into \(\Im w' = q(z,\bar{z}) + E(z,\bar{z},\Re w')\) and solving for \(\Im w''\) cancels the holomorphic and antiholomorphic terms in \(q\), and leaves \(E\) as \(O(3)\). After this change of coordinates we may assume \[q(z,\bar{z}) = \sum_{j,k=1}^{n-1} c_{jk} z_j \bar{z}_k .\] That is, \(q\) is a sesquilinear form. Since \(q\) is real-valued, the matrix \(C = [ c_{jk} ]\) is Hermitian. In linear algebra notation, \(q(z,\bar{z}) = z^*Cz\), where we think of \(z\) as a column vector. If \(T\) is a linear transformation on the \(z\) variables, say \(z'=Tz\), we obtain \({z'}^*Cz' = {(Tz)}^*CTz = z^* ( T^*CT) z\). Thus, we normalize \(C\) up to \(*\)-congruence. A Hermitian matrix is \(*\)-congruent to a diagonal matrix with only \(1\)s, \(-1\)s, and \(0\)s on the diagonal, again by Sylvester’s law of inertia. Writing out what that means is precisely the conclusion of the proposition.

Exercise \(\PageIndex{12}\)

Prove the assertion in the proof, that is, if \(q\) is a quadratic as in \(\eqref{eq:1}\) that is real-valued, then \(a_{jk} = \overline{b_{jk}}\) and \(c_{jk} = \overline{c_{kj}}\).

Lemma \(\PageIndex{2}\)

Narasimhan's Lemma

Let \(U \subset \mathbb{C}^n\) be an open set with smooth boundary that is strongly pseudoconvex at \(p \in \partial U\). Then there exists a local biholomorphic change of coordinates fixing \(p\) such that in these new coordinates, \(U\) is strongly convex at \(p\) and hence strongly convex at all points near \(p\).

Exercise \(\PageIndex{13}\)

Prove Narasimhan’s lemma. Hint: See the proof of Lemma \(\PageIndex{1}\).

Exercise \(\PageIndex{14}\)

Prove that an open \(U \subset \mathbb{C}^n\) with smooth boundary is pseudoconvex at \(p\) if and only if there exist local holomorphic coordinates at \(p\) such that \(U\) is convex at \(p\)

Narasimhan’s lemma only works at points of strong pseudoconvexity. For weakly pseudoconvex points the situation is far more complicated. The difficulty is that we cannot make a \(U\) that is weakly pseudoconvex at all points near \(p\) to be convex at all points near \(p\).

Let us prove the easy direction of the famous Levi problem. The Levi problem was a long-standing problem\(^{5}\) in several complex variables to classify domains of holomorphy in \(\mathbb{C}^n\). The answer is that a domain is a domain of holomorphy if and only if it is pseudoconvex. Just as the problem of trying to show that the classical geometric convexity is the same as convexity as we have defined it, the Levi problem has an easier direction and a harder direction. The easier direction is to show that a domain of holomorphy is pseudoconvex, and the harder direction is to show that a pseudoconvex domain is a domain of holomorphy. See Hörmander’s book [H] for the proof of the hard direction.

Theorem \(\PageIndex{2}\)

Suppose \(U \subset \mathbb{C}^n\) is an open set with smooth boundary and at some point \(p \in \partial U\) the Levi form has a negative eigenvalue. Then every holomorphic function on \(U\) extends to a neighborhood of \(p\). In particular, \(U\) is not a domain of holomorphy.

Proof

We change variables so that \(p = 0\), and near \(p\), \(U\) is given by \[\Im w > -|z_1|^2 + \sum_{j=2}^{n-1} \epsilon_j |z_j|^2 + E(z_1,z',\bar{z}_1,\bar{z}',\Re w) ,\] where \(z' = (z_2,\ldots,z_{n-1})\), \(\epsilon_j = -1,0,1\), and \(E\) is \(O(3)\). We embed an analytic disc via the map \(\xi \overset{\varphi}{\mapsto} (\lambda \xi, 0, 0, \ldots, 0)\) for some small \(\lambda > 0\). Clearly \(\varphi(0) = 0 \in \partial U\). For \(\xi \not= 0\) near the origin \[-\lambda^2 |\xi|^2 + \sum_{j=2}^{n-1} \epsilon_j |0|^2 + E(\lambda \xi,0,\lambda \bar{\xi},0,0) = -\lambda^2 |\xi|^2 + E(\lambda \xi,0,\lambda \bar{\xi},0,0) < 0 .\] That is because the function above has a strict maximum at \(\xi = 0\) by the second derivative test. Therefore, for \(\xi \not= 0\) near the origin, \(\varphi(\xi) \in U\). By picking \(\lambda\) small enough, \(\varphi(\overline{\mathbb{D}}\setminus\{0\}) \subset U\).

As \(\varphi(\partial \mathbb{D})\) is compact we can “wiggle it a little” and find discs in \(U\). In particular, for all small enough \(s > 0\), the closed disc given by \[\xi \overset{\varphi_s}{\mapsto} (\lambda \xi, 0, 0, \ldots, 0, i s)\] is entirely inside \(U\) (that is, for slightly positive \(\Im w\)). Fix such a small \(s > 0\). Suppose \(\epsilon > 0\) is small and \(\epsilon < s\). Define the Hartogs figure \[\begin{align}\begin{aligned} H = & \bigl\{ (z,w) : \lambda - \epsilon < |z_1| < \lambda + \epsilon, |z_j| < \epsilon \text{ for } j=2,\ldots,n-1, \text{ and } |w-is| < s+\epsilon \bigr\} \\ & \cup \bigl\{ (z,w) : |z_1| < \lambda + \epsilon, |z_j| < \epsilon \text{ for } j=2,\ldots,n-1, \text{ and } |w-is| < \epsilon \bigr\} . \end{aligned}\end{align}\]

The set where \(|z_1| = \lambda\), \(z' = 0\) and \(|w| \leq s\), is inside \(U\) for all small enough \(s\), so an \(\epsilon\)-neighborhood of that is in \(U\). For \(w = is\) the whole disc where \(|z_1| \leq \lambda\) is in \(U\), so an \(\epsilon\)-neighborhood of that is in \(U\). Thus, for small enough \(\epsilon >0\), \(H \subset U\). We are really just taking a Hartogs figure in the \(z_1,w\) variables, and then “fattening it up” to the \(z'\) variables. In , we picture the Hartogs figure in the \(|z_1|\) and \(|w-is|\) variables. The boundary \(\partial U\) and \(U\) are only pictured diagrammatically. Also, we make a “picture” the analytic discs giving the “tomato can.” In the picture, the \(U\) is below its boundary \(\partial U\), unlike usually.

Figure \(\PageIndex{3}\)

The origin is in the hull of \(H\), and so every function holomorphic in \(U\), and so in \(H\), extends through the origin. Hence \(U\) is not a domain of holomorphy.

Pseudoconvex at \(p\) means that all eigenvalues of the Levi form are nonnegative. The theorem says that a domain of holomorphy must be pseudoconvex. The theorem’s name comes from the proof, and sometimes other theorems using a similar proof of a “tomato can” of analytic discs are called tomato can principles. The general statement of proof of the principle is that “an analytic function holomorphic in a neighborhood of the sides and the bottom of a tomato can extends to the inside.” And the theorem we named after the principle states that “if the Levi form at \(p\) has a negative eigenvalue, we can fit a tomato can from inside the domain over \(p\).”

Exercise \(\PageIndex{15}\)

For the following domains in \(U \subset \mathbb{C}^2\), find all the points in \(\partial U\) where \(U\) is weakly pseudoconvex, all the points where it is strongly pseudoconvex, and all the points where it is not pseudoconvex. Is \(U\) pseudoconvex?

1. \(\Im w > |z|^4\)
2. \(\Im w > |z|^2(\Re w)\)
3. \(\Im w > (\Re z)(\Re w)\)

Exercise \(\PageIndex{16}\)

Let \(U \subset \mathbb{C}^n\) be an open set with smooth boundary that is strongly pseudoconvex at \(p \in \partial U\). Show that \(p\) is a so-called peak point: There exists a neighborhood \(W\) of \(p\) and a holomorphic \(f \colon W \to \mathbb{C}\) such that \(f(p)=1\) and \(|f(z)| < 1\) for all \(z \in W \cap \overline{U} \setminus \{ p \}\).

Exercise \(\PageIndex{17}\)

Suppose \(U \subset \mathbb{C}^n\) is an open set with smooth boundary. Suppose for \(p \in \partial U\), there is a neighborhood \(W\) of \(p\) and a holomorphic function \(f \colon W \to \mathbb{C}\) such that \(df(p) \not= 0\), \(f(p) = 0\), but \(f\) is never zero on \(W \cap U\). Show that \(U\) is pseudoconvex at \(p\). Hint: You may need the holomorphic implicit function theorem, see Theorem 1.3.2. Note: The result does not require the derivative of \(f\) to not vanish, but is much harder to prove without that hypothesis.

A hyperplane is the “degenerate” case of normal convexity. There is also a flat case of pseudoconvexity. A smooth real hypersurface \(M \subset \mathbb{C}^n\) is Levi-flat if the Levi form vanishes at every point of \(M\). The zero matrix is positive semidefinite and negative semidefinite, so both sides of \(M\) are pseudoconvex. Conversely, the only hypersurface pseudoconvex from both sides is a Levi-flat one.

Exercise \(\PageIndex{18}\)

Suppose \(U = V \times \mathbb{C}^{n-1} \subset \mathbb{C}^n\), where \(V \subset \mathbb{C}\) is an open set with smooth boundary. Show that \(U\) is has a smooth Levi-flat boundary.

Exercise \(\PageIndex{19}\)

Prove that a real hyperplane is Levi-flat.

Exercise \(\PageIndex{20}\)

Let \(U \subset \mathbb{C}^n\) be open, \(f \in \mathcal{O}(U)\), and \(M = \bigl\{ z \in U : \Im f(z) = 0 \bigr\}\). Show that if \(df(p) \not=0\) for some \(p \in M\), then near \(p\), \(M\) is a Levi-flat hypersurface.

Exercise \(\PageIndex{21}\)

Suppose \(M \subset \mathbb{C}^n\) is a smooth Levi-flat hypersurface, \(p \in M\), and a complex line \(L\) is tangent to \(M\) at \(p\). Prove that \(p\) is not an isolated point of \(L \cap M\).

Exercise \(\PageIndex{22}\)

Suppose \(U \subset \mathbb{C}^n\) is an open set with smooth boundary and \(\partial U\) is Levi-flat. Show that \(U\) is unbounded. Hint: If \(U\) were bounded, consider the point on \(\partial U\) farthest from the origin.

Footnotes

[1] The \(B_p\) is sometimes colloquially called the “bad direction.”

[2] The definition for unbounded domains is not consistent in the literature. Sometimes is used.

[3] People sometimes call the complex Hessian the “Levi form of \(r\),” which is incorrect. The Levi form is something defined for a boundary or a submanifold, not for \(r\).

[4] That is not, in fact, completely incorrect. If we think of the sphere in the complex projective space, we are simply looking at the sphere in a different coordinate patch.

[5] E. E. Levi stated the problem in 1911, but it was not completely solved until the 1950s, by Oka and others.