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Single-variable holomorphic function

[definition] Let $U \subseteq \CC$ be open. A function $f:U \to \CC$ is called holomorphic if for any point $z_0 \in U$, there exists a ball $B_\eps(z_0) \subseteq U$ of radius $\eps > 0$ around $z_0$ such that $f$ on $B_\eps(z_0)$ can be written as a convergent power series, i.e.

[/definition]

Cauchy-Riemann equations

[theorem] Write $z = x + iy$ and $f:U \to \CC$ as $f(x, y) = u(x, y) + iv(x, y)$. Then $f$ is holomorphic if and only if $u$ and $v$ are $C^1$ and

on $U$. [/theorem]

Wirtinger operators

Recall that the Wirtinger operators are the differential operators

Then the Cauchy-Riemann equations are equivalent to $\pdv{f}{\overline{z}} = 0$.

Relationship to real Jacobian

Let $U \subseteq \CC = \RR^2$ be an open subset. Then we can regard the (real) differentiable map $f:U \to \CC$ as a map $U \to \RR^2$. Then its differential

can be represented in the standard coordinates $z = x + iy$ (at $z \in U$) and $w = r + is$ (at $f(z)$) as the real Jacobian

We can extend $\dd{f(z)}$ to a $\CC$-linear map by extension of scalars:

At this point, we choose different bases: namely $\pdv{z}$ and $\pdv{\overline z}$. Thus, the differential is represented by the matrix

Using the fact that $\pdv{\overline f}{\overline z} = \overline{\left(\pdv{f}{z}\right)}$, $f$ being holomorphic implies that above matrix is

Cauchy integral formula

[theorem] $f:U \to \CC$ is holomorphic if and only if $f$ is continuously differentiable and for any $B_\eps(z_0) \subseteq U$

[/theorem]

Other standard single-variable theorems

Maximum (modulus) principle

[theorem] Let $U \subseteq \CC$ be open and connected. If $f:U \to \CC$ is holomorphic and non-constant, then $|f|$ has no local maximum in $U$. If $U$ is bounded and $f$ can be extended to a continuous function $f:\overline{U} \to \CC$, then $|f|$ attains its maximum on $\partial U$. [/theorem]

Identity theorem

[theorem] If $f, g:U \to \CC$ are two holomorphic function son a connected open subset $U\subseteq \CC$ such that $f(z) = g(z)$ for all $z$ in a nonempty open subset $V \subseteq U$, then $f = g$ on $U$. [/theorem]

Riemann extension theorem

[theorem] Let $f:B_\eps(z_0) \setminus z_0 \to \CC$ be a bounded holomorphic function. Then $f$ can be extended to a holomorphic function $f:B_\eps(0) \to \CC$. [/theorem]

Riemann mapping theorem

[definition] Let $U$ and $V$ be open subsets of $\CC$. We say that $U$ and $V$ are biholomorphic if there is a bijective holomorphic map $f:U \to V$ with holomorphic inverse $f^{-1}:V \to U$. [/definition]

[theorem] Let $U \subseteq \CC$ be a simply-connected proper open subset. Then $U$ is biholomorphic to $B_1(0)$. [/theorem]

Liouville

[theorem=Liouville’s theorem] Every bounded holomorphic function $f:\CC \to \CC$ is constant. [/theorem]

[corollary] There is no biholomorphism of $\CC$ and $B_\eps(0)$ with $\eps < \infty$. [/corollary]

[proof] Any holomorphic map $f:\CC \to B_{\eps}(0) \subseteq \CC$ were such a map is bounded and Liouville’s theorem implies that $f$ must be constant. So certainly $f$ is not a bijection, let alone a biholomorphism. [/proof]

Definition (polydisc)

[definition] Let $\eps \coloneqq (\eps_1, \ldots, e_n)$. We define the polydisc $B_\eps(w)$ to be the set

That is, a polydisc is a product of $\CC$-discs. [/definition]

Definition 1.1.1 (holomorphic function)

[definition] Let $U \subseteq \CC^n$ be an open subset and let $f:U \to \CC$ be continuously differentiable. Then $f$ is holomorphic if the Cauchy-Riemann equations holds for all coordinates $z_j = x_i + iy_j$, i.e.

where $j = 1, \ldots, n$. [/definition]

Thus, by definition, a continuously differentiable function $f$ is holomorphic if the induced functions

are holomorphic for all fixed $z_1, \ldots, z_{i-1}, z_{i+1}, \ldots, z_n \in \CC$.

Wirtinger formulation

If we define

then the multivariable Cauchy-Riemann equations can be restated as simply just that

for all $j = 1, \ldots, n$. For simplicity, we will write $\overline{\partial} f = 0$.

Proposition 1.1.2 (multivariable Cauchy integral formula)

[proposition] Let $f:\overline{B_\eps(w)} \to \CC$ be a continuous function such that $f$ is holomorphic with respect to every single component $z_i$ in any point of $B_\eps(w)$. Then, for any $z \in B_\eps(w)$,

[/proposition]

[proof] Just apply the single-variable Cauchy integral formula multiple times. Turning the iterated integral into the multiple integral follows by Fubini’s theorem since the integrand is continuous on the boundary of $B_\eps(w)$. [/proof]

Osgood’s lemma

Notice that if $f$ is holomorphic with respect to every single coordinate, then $f$ is necessarily holomorphic itself by just simply doing the usual “differentiation under the integral” trick that we do in single-variable complex analysis (recall that this is how we derive the Cauchy integral formulae for derivatives).

[corollary] If $f:U \to \CC$ is continuous and holomorphic with respect to every coordinate, then $f$ is holomorphic. [/corollary]

Power series expansion

Just like in single-variable complex analysis, the Cauchy integral formulae give formulas for the coefficients of the power series representation of a holomorphic function. In particular,

where

Single-variable theorems that generalize

The maximum modulus principle, identity theorem, and Liouville’s theorem generalize to the multivariable case. The Riemann extension theorem holds but is not trivial to prove, the Riemann mapping theorem fails badly.

Lemma 1.1.3

The holomorphicity of a function of serveral complex variables is typically shown by rrealizing it as an integral of a function known to be holomorphic.

[lemma] Let $U \subseteq \CC^n$ be an open subset and let $V \subseteq \CC$ be an open neighborhood of the booundary of $B_\eps(0) \subseteq \CC$. Assume that $f:V \times U \to \CC$ is a holomorphic function. Then

is a holomorphic function on $U$. [/lemma]

Proposition 1.1.4 (Hartogs’ theorem)

[theorem] Let $\eps = (\eps_1, \ldots, \eps_n)$ and $\eps’ = (\eps_1’, \ldots, \eps_n’)$ are given such that for all $i$, one has $\eps_i’ < \eps_i$. If $n > 1$, then any holomorphic map $f:B_\eps(0) \setminus \overline{B_{\eps’}(0)} \to \CC$ can be uniquely extended to a holomorphic map $f: B_\eps(0) \to \CC$. [/theorem]

Definition 1.1.5 (Weierstrass polynomial)

[definition] A Weierstrass polynomial is a polynomial in $z_1$ of the form

where the coefficients are holomorphic functions on some small disc in $\CC^{n-1}$ vanishing at the origin. [/definition]

Motivation

The idea behind Weierstass polynomials comes from the single-variable case. Any holomorphic function $f(z)$ in one variable with a zero of order $d$ at the origin can be written as

If we let this decomposition depend on extra parameters, then we can degenerate $z^d$ into an arbitrary polynomial of degree $d$.

More generally, a zero of order $d$ of $f_0(z_1)$ can deform to a collection of zeros of $f_w(z_1)$ whose orders sum to $d$.

Proposition 1.1.6 (Weierstrass preparation theorem)

[proposition] Let $f:B_\eps(0) \to \CC$ be a holomorphic function on the polydisc $B_\eps(0)$. Assume $f(0) = 0$ and $f_0(z_1) \not\equiv 0$. Then there exists a Weierstrass polynomial $g(z_1, w) = g_w(z_1)$ and a holomorphic function $h$ on some smaller polydisc $B_{\eps’}(0) \subseteq B_\eps(0)$ such that $f = gh$ and $h(0) \neq 0$. The Weierstrass polynomial $g$ is unique. [/proposition]

Definition (Zero set)

[definition] Let $f$ be a holomorphic function with domain $U\subseteq \CC^n$. Then

[/definition]

Proposition 1.1.7 (Riemann extension theorem)

[proposition] Let $f$ be a holomorphic function on an open subset $U \subseteq \CC^n$. If $g:U\setminus Z(f) \to \CC$ is holomorphic and locally bounded near $Z(f)$, then $g$ can uniquelly be extended to a holomorphic function $\tilde{g}:U \to \CC$. [/proposition]

Definition 1.1.8 (Vector-valued holomorphic functions)

[definition] Let $U \subseteq \CC^m$ be an open subset. A function $f:U \to \CC^n$ is holomorphic if all coordinate functions $f_1, \ldots, f_n$ are holomorphic functions $U \to \CC^n$.

In addition, a map $f:U \to V$ between two open subsets $U,V \subseteq \CC^n$ is biholomorphic if and only if $f$ is bijective and holomorphic along with holomorphic inverse. [/definition]

Definition 1.1.9 (Jacobian and regularity)

[definition] Let $U \subseteq \CC^m$ be an open subset and let $f:U \to \CC^n$ be a holomorphic map. The (complex) Jacobian of $f$ at a point $z \in U$ is the matrix

A point $z \in U$ is called regular if $J(f)(z)$ is surjective. If every point $z \in f^{-1}(w)$ is regular, then $w$ is called a regular value. [/definition]

Real and complex Jacobians

As usual, the real differentiable map

induces an $\RR$-linear map by making the usual identification $\CC^m = \RR^{2m}$ and $\CC^n = \RR^{2n}$. In particular, we obtain the map

by choosing the bases

for $T_z\RR^{2m}$ and

Then the linear map $df(z)$ is given by the real Jacobian

And then, of course, the $\CC$-linear extension

with respect to the bases

and

is

And then, of course, if $f$ is actually holomorphic then the Jacobian just becomes

Notice that, from this perspective, we recover the good ol’ result from single-variable complex analysis:

Proposition 1.1.10 (Inverse function theorem)

[proposition] Let $f:U \to V$ be a holomorphic map between two open subsets $U, V \subseteq \CC^n$. If $z \in U$ is regular, then there exists open subsets $z \in U’ \subseteq U$ and $f(z) \in V’ \subseteq V$ such that $f$ induces a biholomorphic map $f:U’ \to V’$. [/proposition]

To-do: This is a slightly stronger statement that what we do for real manifolds, right? Usually we do not say that $z$ must be regular.

Proposition 1.1.11 (Implicit function theorem)

[proposition] Let $U \subseteq \CC^m$ be an open subset and let $f:U \to \CC^n$ be a holomorphic map, where $m \ge n$. Suppose $z_0 \in U$ is a point such that

Then there exist open subsets $U_1 \subseteq \CC^{m - n}$, $U_2 \subseteq \CC^n$, and a holomorphic $g:U_1 \to U_2$ such that $U_1 \times U_2 \subseteq U$ and $f(z) = f(z_0)$ if and only if

[/proposition]

Corollary 1.1.12

[corollary] Let $U \subseteq \CC^m$ be an open subset and let $f:U \to \CC^n$ be a holomorphic map. Assume $z_0 \in U$ such that $\operatorname{rk}(J(f)(z_0))$ is maximal.

i) If $m \ge n$ then there exists a biholomorphic map $h:V \to U’$, where $U’$ is an open subset of $U$ containing $z_0$ such that

for all $(z_1, \ldots, z_m) \in V$.

ii) If $m \le n$ then there exists a biholomorphic map $g:V \to V’$ where $V$ is an open subset of $\CC^n$ containing $f(z_0)$, such that

[/corollary]

Proposition 1.1.13 (Biholomorphic is equivalent to bijective + holomorphic)

[proposition] Let $f:U \to V$ be a bijective holomorphic map between two open subsets $U, V \subseteq \CC^n$. Then for all $z \in U$ one has $\det J(f)(z) \neq 0$. In particular, $f$ is biholomorphic. [/proposition]

Definition 1.1.14 (Sheaf of holomorphic functions)

[definition] By $\mathcal{O}_{\CC^n}$ we denote the sheaf of holomorphic functions on $\CC^n$. As usual, the stalk $\mathcal{O}_{\CC^n,z}$ of $\mathcal{O}_{\CC^n}$ at a point $z \in \CC^n$ is the direct limit over all the germs containing $z$. [/definition]

Stalks at different points are isomorphic as $\CC$-algebras

Indeed, we consider $\mathcal{O}_{\CC^n, z}$ and $\mathcal{O}_{\CC^n, w}$. Then precomposing with the translation map does the trick.

Partial derivatives are $\CC$-linear operators on stalks

Yes, of course.

$\mathcal{O}_{\CC^n,0}$ is a commutative local ring

There is a natural set that happens to be a maximal ideal

[proposition] $\mathfrak{m}$ is a maximal ideal of $\mathcal{O}_{\CC^n, 0}$. [/proposition]

[proof] Consider the map $F:\mathcal{O}_{\CC^n,0} \to \CC$ given by sending $f\in\mathcal{O}_{\CC^n,0}$ to $f(0)$.

(Well-definedness) Suppose $\overline{(U, f)}$ and $\overline{(V, g)}$. Thus, there exists some $W \subseteq U\cap V$ that is an open neighborhood of $0\in\CC^n$ for which $f|_{W} = g|_{W}$. Since $0 \in W$, $f(0) = g(0)$.

($\CC$-algebra homomorphism) This is obvious.

($F$ is surjective) Constants functions are trivially holomorphic.

($\mathfrak{m}$ is a maximal ideal) By definition, $\mathfrak{m} = \ker F$. Then apply the first isomorphism theorem. [/proof]

It’s also the case that $\mathfrak{m}$ is the unique maximal ideal of the stalk at zero.

[proposition] Let $J$ be the Jacobson radical of the stalk at zero. We claim that $\mathfrak{m} = J$. Thus, in particular, $\mathcal{O}_{\CC^n,0}$ is local with maximal ideal $\mathfrak{m}$. [/proposition]

[proof] Since $J$ is the intersection of all the maximal ideals of $\mathcal{O}_{\CC^n,0}$, it follows that $J \subseteq \mathfrak{m}$. So it suffices to prove $\mathfrak{m} \subseteq J$. Let $f \in \mathfrak{m}$ and $g \in \mathcal{O}_{\CC^n,0}$. Then we claim that $1 + fg$ is a unit $\mathfrak{m}$. Of course, this just amounts to noting that pointwise $1 + fg \neq 0$ in a neighborhood of zero. But this is clear, because $1 + f(0)g(0) = 1$. Thus, $f \in \mathfrak{m}$. [/proof]

Units of $\mathcal{O}_{\CC^n,0}$

Of course, $\mathcal{O}_{\CC^n, 0}^*$ is the set of $f$ with $f(0) \neq 0$.

Proposition 1.1.15 ($\mathcal{O}_{\CC^n, 0}$ is a UFD)

[proposition] The local ring is a $\mathcal{O}_{\CC^n,0}$. [/proposition]

By the WPT, by making an appropriate choice of coordinates, any $f \in \mathcal{O}_{\CC^n,0}$ can be uniquely factored as $f = gh$ where $h$ is a unit in $\mathcal{O}_{\CC^n, 0}$ and $g \in \mathcal{O}_{\CC^{n-1},0}[z_1]$ is a Weierstrass polynomial. Continuing this way eventually gives a unique factorization of $f$.

Proposition 1.1.17 (Weierstrass division theorem)

[proposition] Let $f \in \mathcal{O}_{\CC^n, 0}$ and $g \in \mathcal{O}_{\CC^{n-1}, 0}[z_1]$ be a Weierstrass polynomial of degree $d$. Then there exist $r\in\mathcal{O}_{\CC^{n-1}, 0}[z_1]$ of degree $< d$ and $h \in \mathcal{O}_{\CC^n,0}$ such that $f = gh + r$ with $h$ and $r$ uniquely determined. [/proposition]

Proposition 1.1.18 ($\mathcal{O}_{\CC^n,0}$ is noetherian)

[proposition] The local UFD $\mathcal{O}_{\CC^n,0}$ is noetherian. [/proposition]

Corollary 1.1.19

[corollary] Let $g\in \mathcal{O}_{\CC^n,0}$ be an irreducible function. If $f \in \mathcal{O}_{\CC^n, 0}$ vanishes on $Z(g)$, then $g$ divides $f$. [/corollary]

Definition 1.1.21

[definition] The germ of a set in the origin $0\in \CC^n$ is given by a subset $X \subseteq \CC^n$ define the same germ if there exists an open neighborhood $0 \in U \subseteq \CC^n$ with $U \cap X = U \cap Y$. [/definition]

Zero set

Sometimes one writes $(X, 0)$ for a germ of a set in the origin. Let $f \in \mathcal{O}_{\CC^n, 0}$. Then we define

Note that $Z(f)$ doesn’t depend on the choice of representative $f$ because if $(U, f) \sim (V, g)$ (as germs of functions, not sets) and we define $S_f$ to be the zero set of $f$ and $S_g$ similarly, the germ equivalence (as functions) tells us that there is open $W\subseteq U \cap V$ so that $f|_W = g|_W$. So then $S_f$ and $S_g$ define the same germ via $W$ (as sets).

On a related note, notice that if $f$ is a unit, then $Z(f)$ is the empty set since we can just take a very small neighborhood around $0$.

In addition, we define

and

Definition 1.1.22 (analytic germ)

[definition] A germ $X \subseteq \CC^n$ in $0$ is called analytic if there exists elements $f_1, \ldots, f_k \in \mathcal{O}_{\CC^n, 0}$, such that $X$ and $Z(f_1, \ldots, f_k)$ define the germ. [/definition]

Definition 1.1.23 (analytic subset)

[definition] Let $U \subseteq \CC^n$ be an open subset. An analytic subset of $U$ is a closed subset $X \subseteq U$ such that for every $x \in $, there is an open neighborhood $V\subseteq U$ of $x$ and holomoprhic functions $f_1, \ldots, f_k: V \to \CC$ such that

[/definition]

Definition 1.1.24 (ideals)

[definition] Let $X \subseteq \CC^n$ be a germ in the origin. Then $I(X)$ denotes the set of all elements $f \in \mathcal{O}_{\CC^n,0}$ with $X \subseteq Z(f)$. [/definition]

Lemma 1.1.25

[lemma] For any germ $X \subseteq \CC^n$ the set $I(X) \subseteq \mathcal{O}_{\CC^n, 0}$ is an ideal. If $(A) \subseteq \mathcal{O}_{\CC^n, 0}$ denotes the ideal generated by a subset $A \subseteq \mathcal{O}_{\CC^, 0}$, then $Z(A) = Z((A))$ and $Z(A)$ is analytic. [/lemma]