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\author{Ji{\v r}\'i Lebl}
\title{Several Complex Variables are Better than One}

\begin{document}

\large

\maketitle

\begin{abstract}
Complex analysis in one complex variable is the calculus student's
dream come true.  No annoying pathological cases, no annoying extra
hypotheses to theorems, things just work, sometimes even better than
one thinks they should.  And it gets better in several variables,
including even more surprising results.  In this talk I will talk
about the Hartogs Phenomenon, a theorem with a coolness factor at
least double that of the Maximum Principle. In particular it will tell
us something about how the zero sets of holomorphic (analytic)
functions behave when we have more than one complex variable.  It says
that holomorphic functions simply extend across compact sets (no more
isolated singularities).  I will define holomorphic functions (both in
one and several variables), give you enough to understand the theorem
and even give a sketch of the proof (the d-bar equation makes an
appearance).  Only very basic knowledge of analysis will be required
to enjoy the talk.
\end{abstract}

\section{Introduction}

The field of Several Complex Variables (SCV) lies at the intersection of several
other fields.  There are deep connections to
Partial Differential Equations,
Functional Analysis,
Complex Dynamics,
Algebraic Geometry (both real and complex),
Pluripotential Theory,
Differential Geometry,
and of course One Complex Variable.
There are even connections to Combinatorics, Commutative Algebra, or
Representation Theory.  Basically any time you talk about several variables,
you often find they can be treated as complex numbers, and we arrive at SCV.

Whereas in basic real analysis one sees things go wrong in almost every way
imaginable, complex analysis deals with incredibly well-behaved objects.
Often it is more likely that in complex analysis you are surprised that some
simple minded idea works all too well, rather than being surprised that an
entirely reasonable idea breaks spectacularly.  One can vaguely think of
complex analysis as the ``study of polynomials of infinite degree,'' and we
all know how wonderful polynomials are.

We will go over a very fundamental theorem that will illustrate how several
complex variables is a very different field from one complex variable.
Let us however first start with one complex variable from the point of view
of an SCV person\ldots

\section{Holomorphic Functions in One Variable}

In complex analysis, we study \emph{holomorphic} (or \emph{analytic}) 
functions.
We identify $\C$ with $\R^2$ as $z = x+iy$.  The complex
conjugate is then $\bar{z} = x-iy$.
We start by defining the following formal differential operators (the
so-called Wirtinger operators)
\begin{equation*}
\frac{\partial}{\partial z}
=
\frac{1}{2}
\left(
\frac{\partial}{\partial x} - i
\frac{\partial}{\partial y}
\right),
~ ~ ~ ~ ~
\frac{\partial}{\partial \bar{z}}
=
\frac{1}{2}
\left(
\frac{\partial}{\partial x} + i
\frac{\partial}{\partial y}
\right)
.
\end{equation*}

A function on $\C$ is really a function defined on
$\R^2$ as identified above and hence it is a function of $x$ and $y$.
Writing
$x = \frac{z+\bar{z}}{2}$ and
$y = \frac{z-\bar{z}}{2i}$, we can think of it as a function of two
complex variables $z$ and $\bar{z}$.  Here we pretend for a moment as if $\bar{z}$ did not
depend on $z$.
The formal differential operators
work as if $z$ and $\bar{z}$ really were independent variables.  For example:
\begin{equation*}
\frac{\partial}{\partial z}
\left[ z^2 \bar{z}^3 + z^{10} \right]
=
2z \bar{z}^3 + 10 z^{9}
\qquad
\text{and}
\qquad
\frac{\partial}{\partial \bar{z}}
\left[ z^2 \bar{z}^3 + z^{10} \right]
=
z^2 ( 3 \bar{z}^2 ) + 0 .
\end{equation*}

A holomorphic function is a function not depending on $\bar{z}$.

\begin{defn}
Let $U \subset \C$ be open and $f \colon U \to \C$ continuously differentiable.
We say $f$ is \emph{holomorphic} if it satisfies
the \emph{Cauchy-Riemann equations}:
\begin{equation*}
\frac{\partial f}{\partial \bar{z}} = 0 .
\end{equation*}
\end{defn}

Examples of holomorphic functions are polynomials in $z$ or
elementary functions such as $\sin$, $\cos$, or $\exp$.  Combining
holomorphic functions algebraically or composing holomorphic functions
yields again holomorphic functions.

Holomorphic functions have a power series representation in $z$
at each point $a$:
\begin{equation*}
f(z) = \sum_{j=0}^\infty c_j {(z-a)}^j .
\end{equation*}
Notice there is no $\bar{z}$ necessary there
since $\frac{\partial f}{\partial \bar{z}} = 0$.

A basic problem in complex analysis of one
variable is the classification of the \emph{isolated singularities}
of holomorphic functions.
That is, you can have a holomorphic function defined everywhere but a single
point, for example:
\begin{equation*}
f(z) = \frac{1}{z},
\qquad
\text{or even worse,} \qquad
g(z) = e^{1/z} .
\end{equation*}
A related problem is to study sets where $f(z)$ is zero.  A holomorphic
function is locally just 
\begin{equation*}
{(z-a)}^k h(z)
\end{equation*}
where $h$ is nonzero and holomorphic.  In other words in one dimension,
all zeros are isolated.

\section{Several Variables}

The questions of singularities and zeros get a lot more interesting in
several variables.
We first need to define what do we mean
by holomorphic functions.  The definition could not be simpler.
In $\C^n$, we use the notation $z = (z_1,z_2,\ldots,z_n)$.

\begin{defn}
Let $U \subset \C^n$ be open.  A function $f \colon U \to \C$ 
is 
\emph{holomorphic} if it is holomorphic in each variable separately.  That
is, if
\begin{equation*}
\zeta \mapsto f(z_1,\ldots,z_{k-1},\zeta,z_{k+1},\ldots,z_n)
\end{equation*}
is holomorphic wherever it makes sense.  Or in other words if for all $k$
\begin{equation*}
\frac{\partial f}{\partial \bar{z}_k} = 0  .
\end{equation*}
\end{defn}

Examples are similar (polynomials, etc\ldots)
except now we allow $n$ variables instead of just one.
We also have the
power series representation of functions.  For example in two variables,
a power series at the origin looks like
\begin{equation*}
\sum_{j,k=0}^\infty a_{jk} {z_1}^j {z_2}^k
\end{equation*}
where $a_{jk}$ are complex numbers.

An important property of holomorphic functions (one or several variables)
arising from the power series
representation is that if a holomorphic function $f \colon U \to \C$ is zero on an open
subset of $U$ then the function is identically zero
everywhere in $U$, provided $U$ is connected.

We say $f \colon U \to \C$ has a holomorphic extension $\widetilde{f}$ to a
larger domain $\widetilde{U}$ where $U \subset
\widetilde{U}$ if $\widetilde{f} \colon \widetilde{U} \to \C$ is holomorphic and is equal to
$f$ on $U$.
By the uniqueness property above, this extension is unique provided
$\widetilde{U}$ is connected.  To see this fact, note
that if we have another extension on $\widetilde{U}$, say $\widehat{f}$, then
$\widehat{f}-\widetilde{f}$ is holomorphic on $\widetilde{U}$ and identically zero on $U$.
This uniqueness will be important in a moment.

\section{Hartogs' Phenomenon}

The Hartogs' phenomenon theorem tells us something
even more fascinating about extending functions, and also it tells us a
fundamental thing about the domains we use to define holomorphic functions.

\begin{thm}
Let $n \geq 2$, $U \subset \C^n$ be an open connected set (a domain), and let
$K \subset U$ be a compact set (closed and bounded) such that
$U \setminus K$ is connected.  Every holomorphic $f \colon U \setminus K \to \C$
extends uniquely to a holomorphic function on $U$.
\end{thm}

\begin{center}
\input hartogs-fig.pdf_t

The setup for Hartogs Phenomenon.
\end{center}

The result is not true in one complex dimension: Let $U = \C$ and $K = \{ 0
\}$.  Then let $f(z) = \frac{1}{z}$.  Clearly $f$ does not extend through
the origin (it blows up).

The theorem tells us something about
the zero set of a holomorphic function: in several variables
the set $\{ z \in \C^n : f(z) = 0 \}$ cannot be compact.
If it were, we could look at the function $\frac{1}{f(z)}$ and try to extend
this function.  Therefore the zero set must ``go all the way to the boundary,''
or perhaps ``off to infinity.''
So in some sense zero sets of
holomorphic functions in more than one variable are in some sense ``large.''
Similarly singular sets cannot be compact.

A more fundamental, and perhaps surprising, consequence of the theorem is that not
every domain is a natural domain for a holomorphic function if $n \geq 2$.
In one variable, on any domain we can cook up a function which does not
extend to anything larger.  In two or more variables, there are domains (for
example domains with compact holes) where every holomorphic function extends
to a larger domain.

It is the geometry of the boundary that makes the
difference.  Intuitively what happens is that near the ``hole''
(think $K$ being a closed ball or similar) the domain must be ``concave'' somewhere.
Therefore convexity enters the picture, although in a weaker and more
complicated form, we call it \emph{pseudoconvexity}, let us not define that
concept here however.

Let us sketch the general outline of the proof.
We solve the differential equation
\begin{equation*}
\bar{\partial} \psi = g ,
\end{equation*}
and use the solution to construct the needed extension.
The equation is called the \emph{inhomogeneous $\bar{\partial}$} (pronounced d-bar) equation.
Well I am cheating, I did not define the notation.
The $\psi$ is some smooth function, and $g$ is a
differential form, but we can simply think of it being $n$ different
smooth functions: $g_1$ through $g_n$.  The equation above is really $n$
equations
\begin{equation*}
\frac{\partial \psi}{\partial \bar{z}_k} = g_k ,
\end{equation*}
where the functions $g_k$ satisfy the compatibility conditions
\begin{equation*}
\frac{\partial g_k}{\partial \bar{z}_\ell} =
\frac{\partial g_\ell}{\partial \bar{z}_k} .
\end{equation*}
The compatibility conditions simply mean that
partial derivatives are supposed to commute, and the equation cannot be
solved without them.  Note that where $g = 0$, $\psi$ is holomorphic.  So
for a general $g$, what we are doing is finding a function that is not holomorphic in a very
specific way.

For Hartogs' theorem,
let us assume that all the $g_k$ have compact support.
Then these equations always have a solution $\psi$ (many solutions in fact),
and if $n \geq 2$, then we can find $\psi$ that has compact support.
We obtain a solution explicitly in fact:
\begin{equation} \label{dbarsol}
\psi(z)
=
\frac{1}{2\pi i}
\int_\C
\frac{
 g_1(\zeta,z_2,\ldots,z_n)
}{\zeta - z_1}
d\zeta \wedge d\bar{\zeta}
=
\frac{1}{2\pi i}
\int_\C
\frac{
 g_1(\zeta+z_1,z_2,\ldots,z_n)
}{\zeta}
d\zeta \wedge d\bar{\zeta} .
\end{equation}
Here
$d\zeta \wedge d\bar{\zeta}$ is just a fancy way to write the area measure
on $\C$ up to a funny constant and we really think of $\C$ as $\R^2$ for the
integral.  Note the singularity, and think why the integral still makes
sense.  Hint: because we integrate in $\R^2$, use polar coordinates and note the $r$ in
$r\,dr\, d\theta$.

To check we have the solution we have to do a bit of work.
Let us just give some hints for the intrepid reader to fill in.
Using the compatibility condition we get
\begin{equation*} 
\begin{split}
\frac{\partial\psi}{\partial \bar{z}_j}(z)
& =
\frac{1}{2\pi i}
\int_\C
\frac{
 \frac{\partial g_1}{\bar{z}_j}(\zeta+z_1,z_2,\ldots,z_n)
}{\zeta}
d\zeta \wedge d\bar{\zeta} 
\\
& =
\frac{1}{2\pi i}
\int_\C
\frac{
 \frac{\partial g_j}{\bar{z}_1}(\zeta+z_1,z_2,\ldots,z_n)
}{\zeta}
d\zeta \wedge d\bar{\zeta} 
=
g_j(z) .
\end{split}
\end{equation*}
The last equality follows by the generalized Cauchy formula (itself a
version of the generalized Stokes theorem).  Let $\varphi$ be any smooth
function (not necessarily holomorphic), then
\begin{equation*}
\varphi(z) =
\frac{1}{2\pi i}
\int_{\abs{\zeta}=R}
\frac{\varphi(\zeta)}{\zeta-z}
d \zeta
+
\frac{1}{2\pi i}
\int_{\abs{\zeta} \leq R}
\frac{\frac{\partial \varphi}{\partial \bar{z}}(\zeta)}{\zeta-z}
d\zeta \wedge d\bar{\zeta} .
\end{equation*}
The computation above uses
$\varphi(\cdot) = g_j(\cdot,z_2,\ldots,z_n)$ and
a large enough $R$ so that $\varphi$ is zero 
when $\abs{\zeta}=R$.

That $\psi$ has compact support follows because $g_1$ has compact
support and analytic continuation.  In particular note that $\psi$ is
holomorphic for very large $z$ and it is identically zero when
$z_2,\ldots,z_n$ are large.

Let us move onto the actual proof of the theorem.
First find a smooth function $\varphi$ that is 1 in a neighbourhood of
$K$ and is compactly supported in $U$.  Let
$f_0 = (1-\varphi)f$ on $U \setminus K$ and $f_0 = 0$ on $K$.  The function $f_0$
is smooth on $U$ and it is holomorphic
and equal to $f$ near the boundary of $U$, where $\varphi$ is 0.
We let $g = \bar{\partial} f_0$, that is $g_k = \frac{\partial
f_0}{\partial \bar{z}_k}$.  Let's see why $g_k$ is compactly supported.  The
only place to check is on $U \setminus K$ as elsewhere we have 0
automatically.  Note that $f$ is holomorphic and compute
\begin{equation*}
\frac{\partial f_0}{\partial \bar{z}_k}
=
\frac{\partial }{\partial \bar{z}_k}
\bigl((1-\varphi)f\bigr)
=
\frac{\partial f}{\partial \bar{z}_k}
- \varphi \frac{\partial f}{\partial \bar{z}_k}
- \frac{\partial \varphi}{\partial \bar{z}_k} f
=
- \frac{\partial \varphi}{\partial \bar{z}_k} f .
\end{equation*}
And 
$\frac{\partial \varphi}{\partial \bar{z}_k}$ must be compactly supported in
$U$.
And now we apply the above solution \eqref{dbarsol} to find a
compactly supported function $\psi$ such that $\bar{\partial}\psi = g$.  We
set $\widetilde{f} = f_0 - \psi$.  Let's check that $\widetilde{f}$ is the desired
extension.  Let's check it's holomorphic:
\begin{equation*}
\frac{\partial \widetilde{f}}{\partial \bar{z}_k}
=
\frac{\partial f_0}{\partial \bar{z}_k}
-
\frac{\partial \psi}{\partial \bar{z}_k}
=
g_k
-
g_k
= 0 .
\end{equation*}
It almost feels like we're cheating doesn't it?
OK, but does it extend $f$.  A bit of thought and the fact that $U \setminus
K$ is connected reveals that $\psi$ must be compactly supported in $U$.
This means that $\widetilde{f}$ agrees with $f$ near the boundary (in particular
on an open set) and thus everywhere in $U$ since $U$ is connected.

And that is how using sheep's bladders can prevent
earthquakes!\cite{Python}

\dots any questions?

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\bibitem{Boggess}
Boggess, Albert.
\emph{CR manifolds and the tangential Cauchy-Riemann complex},
CRC Press,
Boca Raton, Florida, 1991.
\bibitem{Conway:complexI}
Conway, John~B.
\emph{Functions of One Complex Variable I}.
Springer-Verlag, New York, New York, 1978.
\bibitem{DAngelo}
D'Angelo, John~P.
\emph{Several complex variables and the geometry of real hypersurfaces},
CRC Press, Boca Raton, Florida, 1993.
\bibitem{Hormander:several}
H\"ormander, Lars.
\emph{An Introduction to Complex Analysis in Several
Variables},
North-Holland Publishing Company, New York, New York, 1973.
\bibitem{Krantz:several}
Krantz, Steven~G.
\emph{Function Theory of Several Complex Variables},
AMS Chelsea Publishing, Providence, Rhode Island, 1992.
\bibitem{Python}
Python, Monty.
\emph{Monty Python and the Holy Grail},
EMI Films, 1975.
\end{thebibliography}

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