diff --git a/scv.tex b/scv.tex index 85f290b..9c6b38b 100644 --- a/scv.tex +++ b/scv.tex @@ -489,7 +489,7 @@ \chapter*{Introduction} \label{ch:intro} can help or even be required to solve later exercises. The prerequisites are a decent knowledge of vector calculus, basic -real-analysis, and a working knowledge of complex analysis in one variable. +real analysis, and a working knowledge of complex analysis in one variable. Measure theory (Lebesgue integral and its convergence theorems) is useful, but it is not essential except in a couple of places later in the book. The first two chapters and most of the third @@ -1446,7 +1446,7 @@ \section{Onto several variables} \label{sec:ontosevvar} \beta_j \, d \bar{z}_j , \end{equation*} where $\alpha_j$ and $\beta_j$ are functions (of $z$). -Also recall the wedge product $\omega \wedge \eta$ is anti-commutative on +Also recall the wedge product $\omega \wedge \eta$ is anticommutative on one-forms, that is, for one-forms $\omega$ and $\eta$, $\omega \wedge \eta = - \eta \wedge \omega$. A wedge product of two @@ -1713,7 +1713,7 @@ \section{Power series representation} Fortunately power series converge absolutely, and so the ordering does not matter. So if you want to write down the limit in terms of partial sums, you pick some ordering of the -multiindices: $\alpha(1), \alpha(2), \ldots$, and then +multi-indices: $\alpha(1), \alpha(2), \ldots$, and then \begin{equation*} \sum_{\alpha} c_\alpha {z}^\alpha @@ -2304,7 +2304,7 @@ \section{Power series representation} \item Prove that if an entire holomorphic function in $\C^2$ is bounded on countably many distinct -complex lines through the origin then it is constant. +complex lines through the origin, then it is constant. \item Find a nonconstant entire holomorphic function in $\C^3$ that is bounded on @@ -2443,7 +2443,7 @@ \section{Derivatives} \text{for all $j=1,\ldots,n$.} \end{equation*} -For functions that are neither holomorphic or anti-holomorphic we +For functions that are neither holomorphic or antiholomorphic we pretend they depend on both $z$ and $\bar{z}$. Since we want to write functions in terms of $z$ and $\bar{z}$, let us figure out how the chain rule works for Wirtinger derivatives, @@ -2627,7 +2627,7 @@ \section{Derivatives} Because of the proposition, when we deal with an arbitrary, possibly -non-holomorphic, function $f$, we often write $f(z,\bar{z})$ and treat $f$ as +nonholomorphic, function $f$, we often write $f(z,\bar{z})$ and treat $f$ as a function of $z$ and $\bar{z}$. \begin{remark} @@ -2765,7 +2765,7 @@ \section{Derivatives} \end{equation*} \end{prop} -In short-hand we often simply write $D(g \circ f) = Dg Df$. +In shorthand we often simply write $D(g \circ f) = Dg Df$. \begin{exbox} \begin{exercise} @@ -3001,7 +3001,7 @@ \section{Inequivalence of ball and polydisc} prominently appears in complex geometry. \begin{defn} -A non-constant holomorphic mapping +A nonconstant holomorphic mapping $\varphi \colon \D \to \C^n$ is called an \emph{\myindex{analytic disc}}. If the mapping $\varphi$ extends continuously to the closed unit disc $\overline{\D}$, then the mapping @@ -3208,7 +3208,7 @@ \section{Inequivalence of ball and polydisc} \end{exercise} \end{exbox} -The key take-away from this section is that +The key takeaway from this section is that in several variables, to see if two domains are equivalent, the geometry of the boundaries makes a difference, not just the topology @@ -3393,7 +3393,7 @@ \section{Cartan's uniqueness theorem} \begin{proof} The map $g(z) = f^{-1}\bigl(e^{-i\theta}f(e^{i\theta} z)\bigr)$ is an automorphism of $U$ and via the -chain-rule, $g'(0) = I$. Therefore, +chain rule, $g'(0) = I$. Therefore, $f^{-1}\bigl(e^{-i\theta}f(e^{i\theta} z)\bigr) = z$, or in other words \begin{equation*} f(e^{i\theta} z) = e^{i\theta}f(z) . @@ -3483,10 +3483,10 @@ \section{Cartan's uniqueness theorem} \end{exercise} \begin{exercise} -Suppose $U \subset \C^n$ is an open set, $a \in U$. -Suppose $f \colon U \to U$ is a holomorphic mapping, +Suppose $U \subset \C^n$ is an open set, $a \in U$, +$f \colon U \to U$ is a holomorphic mapping, $f(a) = a$, and suppose that $\sabs{\lambda} < 1$ -for any eigenvalue $\lambda$ of +for every eigenvalue $\lambda$ of $D f(a)$. Prove that there exists a neighborhood $W$ of $a$, such that $\lim_{\ell \to \infty} f^{\ell}(z) = a$ for all $z \in W$. \end{exercise} @@ -3680,7 +3680,7 @@ \section{Riemann extension theorem, zero sets, and injective maps} \end{example} \begin{example} -The theorem is not true in the non-holomorphic setting. For example, $x_1^2 +The theorem is not true in the nonholomorphic setting. For example, $x_1^2 + x_2^2 = 0$ in $\R^2$ is only the origin, clearly not a graph of any function of one variable. The first part of the theorem works, but the $h$ you find is either $2x_1$ or $2x_2$, and its zero set is too big. @@ -3737,7 +3737,7 @@ \section{Riemann extension theorem, zero sets, and injective maps} $f(z) = c + {\bigl( g(z) \bigr)}^d$. Such a simple result does not hold in several variables in general, but if the mapping is -locally one-to-one then the present theorem says that such a mapping can be +locally one-to-one, then the present theorem says that such a mapping can be locally written as the identity. \begin{proof}[Proof of the theorem] @@ -3762,7 +3762,7 @@ \section{Riemann extension theorem, zero sets, and injective maps} that $J_f = 0$ on the set given by $z_n=0$. We may also assume that $f(0) = 0$. -If $J_f$ vanishes identically then there is no need to do anything other +If $J_f$ vanishes identically, then there is no need to do anything other than a translation. In either case, we may assume that $0 \in U$, $f(0)=0$, and $J_f = 0$ when $z_n=0$. @@ -3860,7 +3860,7 @@ \section{Riemann extension theorem, zero sets, and injective maps} \end{exercise} \begin{exercise} -Prove that if a holomorphic $f \colon \C \to \C$ is injective then it is +Prove that if a holomorphic $f \colon \C \to \C$ is injective, then it is onto, and therefore $f(z) = az + b$ for $a \not= 0$. \end{exercise} \end{exbox} @@ -3888,7 +3888,7 @@ \section{Domains of holomorphy and holomorphic extensions} \begin{defn} \label{defn:domainofhol} Let $U \subset \C^n$ be a domain\footnote{\emph{Domain of -holomorphy} can make sense for disconnected sets (non-domains), and some authors +holomorphy} can make sense for disconnected sets (not domains), and some authors do define it so.} (connected open set). The set $U$ is a \emph{\myindex{domain of holomorphy}} if there do not exist @@ -4003,8 +4003,8 @@ \section{Domains of holomorphy and holomorphic extensions} \begin{thm} \label{thm:extensionhartogsfigure} Let $(z,w) = (z_1,\ldots,z_m,w_{1},\ldots,w_{k}) \in \C^m \times \C^k$ be the coordinates. For two numbers -$0 < a,b < 1$, let the set $H \subset \D^{m+k}$ -be defined by +$0 < a,b < 1$, define the set $H \subset \D^{m+k}$ +by \begin{equation*} H = \bigl\{ (z,w) \in \D^{m+k} : \sabs{z_j} > a ~\text{for $j=1,\ldots,m$} \bigr\} \cup @@ -4265,7 +4265,7 @@ \section{Tangent vectors, the Hessian, and convexity} In particular, the exercise says that given any domain $U \subset \C$ and any domain $V \subset \C$, the domain $U \times V$ is a domain of holomorphy in $\C^2$. The domains -$U$ and $V$, and therefore $U \times V$ can be spectacularly non-convex. +$U$ and $V$, and therefore $U \times V$ can be spectacularly nonconvex. But we should not discard convexity completely, there is a notion of \emph{pseudoconvexity}, which vaguely means ``convexity in the complex directions,'' that is the correct notion to classify which @@ -4699,13 +4699,13 @@ \section{Tangent vectors, the Hessian, and convexity} In the following, \glsadd{not:Ol}% we use the \emph{\myindex{big-oh notation}}, -although we use a perhaps less standard short hand\footnote{% +although we use a perhaps less standard shorthand\footnote{% The standard notation for $O(\ell)$ is $O(\snorm{x}^{\ell})$ and is usually defined to mean that $\abs{\frac{f(x)}{\snorm{x}^\ell}}$ is bounded as $x \to p$.}. A smooth function is $O(\ell)$ at a point $p$ (usually the origin), if all its derivatives of order $0, 1, \ldots, \ell-1$ vanish at $p$. -For example, if $f$ is $O(3)$ at the origin +For example, if $f$ is $O(3)$ at the origin, then $f(0)=0$, and its first and second derivatives vanish at the origin. For computations it is often useful to use a more convenient @@ -4815,7 +4815,7 @@ \section{Tangent vectors, the Hessian, and convexity} We say $M$ is convex from both sides at $p$ if both the set given by $r > 0$ and the set given by $r < 0$ are convex at $p$. Prove that if a hypersurface $M \subset \R^n$ is convex from both sides at all -points then it is locally just a hyperplane (the zero set of a real affine +points, then it is locally just a hyperplane (the zero set of a real affine function). \end{exercise} \end{exbox} @@ -5358,13 +5358,14 @@ \section{Holomorphic vectors, the Levi form, and pseudoconvexity} 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 -\propref{prop:graphcoordinatesCn}, that is, if our boundary (the -hypersurface) is given by +\propref{prop:graphcoordinatesCn}, that is, when our boundary (the +hypersurface) is given near the origin by \begin{equation*} -\Im w = \varphi(z,\bar{z},\Re w) . +\Im w = \varphi(z,\bar{z},\Re w) , \end{equation*} +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. Then the complex Hessian of $r$ is of the form +defining function. The complex Hessian of $r$ has the form \begin{equation*} \begin{bmatrix} L & 0 \\ @@ -5376,20 +5377,21 @@ \section{Holomorphic vectors, the Levi form, and pseudoconvexity} \right]_{j \ell} . \end{equation*} Note that $L$ is an $(n-1) \times (n-1)$ matrix. -The vectors in $T_0^{(1,0)} \partial U$ are in the span +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 $(a,0) \in \C^n$ for an -arbitrary $a \in \C^{n-1}$. The Levi form is then $a^* L a$, +a vector in $T_0^{(1,0)} \partial U$ is represented by $(a,0) \in \C^n$ for +some +$a \in \C^{n-1}$. 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 $0$. \begin{example} -Let us change variables to show how we write the ball $\bB_n$ in different +Let us change variables to write the ball $\bB_n$ in different local holomorphic coordinates where the Levi form is displayed nicely. Suppose the sphere $\partial \bB_n$ is defined in the variables $Z = (Z_1,\ldots,Z_n) \in \C^n$ by $\snorm{Z} = 1$. @@ -5406,7 +5408,7 @@ \section{Holomorphic vectors, the Levi form, and pseudoconvexity} $(0,\ldots,0,-1)$, 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). +the disc to the upper half-plane). We claim that the mapping takes the unit sphere given by $\snorm{Z} = 1$ (without the point $(0,\ldots,0,1)$), to @@ -5465,13 +5467,13 @@ \section{Holomorphic vectors, the Levi form, and pseudoconvexity} 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. -Of course not the entire sphere gets transformed, the points where $Z_n=1$ +Not the entire sphere gets transformed, the points where $Z_n=1$ get ``sent to infinity.'' The hypersurface $\Im w = \sabs{z_1}^2 + \cdots + \sabs{z_{n-1}}^2$ is sometimes called the \emph{\myindex{Lewy hypersurface}}, and in the literature some even say it \emph{is} the sphere\footnote{That is not, in fact, completely incorrect. -If you think of the sphere in the complex projective space, +If we think of the sphere in the complex projective space, we are simply looking at the sphere in a different coordinate patch.}. Pretending $z$ is just one real direction, it looks like this: @@ -5500,11 +5502,11 @@ \section{Holomorphic vectors, the Levi form, and pseudoconvexity} between two domains in $\C^n$, and let $r \colon V' \to \R$ be a smooth function with nonvanishing derivative. Let us compute the Hessian of $r \circ f \colon V \to \R$. -We first compute what happens to the non-mixed derivatives. +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 the derivatives as functions. Also for clarity, let $z$ be the coordinates in $V$ -and $\zeta$ 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 $\bar{z}$. @@ -6356,7 +6358,7 @@ \section{Harmonic, subharmonic, and plurisubharmonic functions} \begin{exercise} Let $U \subset \C$ be open. -Show that if $f \colon U \to \R \cup\{- \infty \}$ is subharmonic +Show that if $f \colon U \to \R \cup\{- \infty \}$ is subharmonic, then \begin{equation*} \limsup_{w \to z} f(w) = f(z) @@ -7815,7 +7817,7 @@ \section{Holomorphic convexity} \begin{exercise} Let $U \subset \C^n$ be a domain and $f \in \sO(U)$. Show that if -$U$ is holomorphically convex then +$U$ is holomorphically convex, then $\widetilde{U} = \bigl\{ z \in U : f(z) \not= 0 \bigr\}$ is holomorphically convex. Hint: First see \exerciseref{exercise:connectedcomplement}. @@ -7845,7 +7847,7 @@ \section{Holomorphic convexity} \begin{exercise} Show that if domains $U_1 \subset \C^n$ and $U_2 \subset \C^n$ are -holomorphically convex +holomorphically convex, then so are all the topological components of $U_1 \cap U_2$. \end{exercise} @@ -7999,8 +8001,8 @@ \section{Holomorphic convexity} \begin{exercise} Extend the proof to show that if $U \subset \C^n$ -is holomorphically convex then there -exists a single function $f \in \sO(U)$, that does not extend through any +is holomorphically convex, then there +exists a single function $f \in \sO(U)$ that does not extend through any point $p \in \partial U$. \end{exercise} @@ -8278,7 +8280,7 @@ \section{Real-analytic functions and complexification} \end{equation*} Notice that a holomorphic function -is real-analytic, but not vice-versa. A holomorphic function +is real-analytic, but not vice versa. A holomorphic function is a real-analytic function that does not depend on $\bar{z}$. Before we discuss complexification in terms of $z$ and $\bar{z}$ we need @@ -10832,7 +10834,7 @@ \section{The Bochner--Martinelli kernel} We will, again, apply Stokes formula, but we need to apply it to forms of higher degree. As before, we split the derivatives into the holomorphic and -antiholomorphic parts. We work with multiindices. For $\alpha$ and $\beta$ +antiholomorphic parts. We work with multi-indices. For $\alpha$ and $\beta$ with $\abs{\alpha}=p$ and $\abs{\beta}=q$, a differential form \begin{equation*} @@ -11453,7 +11455,7 @@ \section{The Bergman kernel} \begin{exercise} \begin{exparts} \item -Show that if $U \subset \C^n$ is bounded then for all $z \in U$ +Show that if $U \subset \C^n$ is bounded, then for all $z \in U$ we find $K_U(z,\bar{z}) > 0$. \item Why can this fail if $U$ is unbounded? @@ -12718,7 +12720,7 @@ \section{Properties of the ring of germs} \label{sec:propertiesofringofgerms} $\widetilde{P}_j$. We obtain $f = u \widetilde{P}_1 \widetilde{P}_2 \cdots \widetilde{P}_m$ for a unit $u$. By uniqueness part of the preparation theorem we obtain -$P = \widetilde{P}_1 \widetilde{P}_2 \cdots \widetilde{P}_m$. Conclusion is then +$P = \widetilde{P}_1 \widetilde{P}_2 \cdots \widetilde{P}_m$. Conclusion is obtained by noting that ${}_{n-1}\sO_0[z_n]$ is a UFD\@. \end{proof} @@ -12874,7 +12876,7 @@ \section{Varieties} \label{sec:varieties} The subvariety $X = \{ 0 \} \subset \C^2$ can be given by $\sF = \{ z_1^2, z_2^2 \}$. If $I = (z_1^2,z_2^2) \subset \sO_0$ is the ideal of germs -generated by these two functions then $I_0(X) \not= I$. We have +generated by these two functions, then $I_0(X) \not= I$. We have seen that the ideal $I_0(X)$ is the maximal ideal $\mathfrak{m}_0 = (z_1,z_2)$. If we prove that all the nonconstant monomials are in $\sqrt{I}$, then @@ -13374,7 +13376,7 @@ \section{Irreducibility, local parametrization, and Puiseux theorem} For complex subvarieties, a subvariety is irreducible if and only if the set of regular points is connected. We omit the proof in the general case, and for hypervarieties it is an exercise above. -It makes sense then, that we can split a subvariety into its irreducible parts. +It then makes sense that we can split a subvariety into its irreducible parts. \begin{prop} Let $(X,p) \subset (\C^n,p)$ be a germ of a subvariety. Then there exist @@ -13502,8 +13504,9 @@ \section{Irreducibility, local parametrization, and Puiseux theorem} becomes $\alpha_\ell(z)$ (and by continuity it is the same root along the entire $N$). So there is a permutation $\sigma$ on $m$ elements such that as $z$ moves -counter-clockwise around the origin from the upper half plane across $N$ to the lower half -plane, $\alpha_j(z)$ is continued as $\alpha_{\sigma(j)}(z)$. +counter-clockwise around the origin from the upper half-plane across $N$ to the +lower half-plane, +$\alpha_j(z)$ is continued as $\alpha_{\sigma(j)}(z)$. There exists some number $k$ (e.g.\ $k=m!$) such that $\sigma^k$ is the identity. As $\xi$ goes around a circle around the origin, $\xi^k$ goes around the origin $k$ times. @@ -14511,7 +14514,7 @@ \chapter{Results from one complex variable} \label{ap:onevarresults} An isolated singularity that is neither removable nor a pole is said to be \emph{essential}\index{essential singularity}. -At non-essential isolated singularities +At nonessential isolated singularities the function blows up to a finite integral order. The first part of the following proposition is usually called the \emph{\myindex{Riemann extension theorem}}. @@ -14702,7 +14705,7 @@ \chapter{Results from one complex variable} \label{ap:onevarresults} Since we stated the big theorem, let us also state the little theorem. \begin{thm}[Picard's little theorem]\index{Picard's little theorem} -If $f \colon \C \to \C$ is holomorphic +If $f \colon \C \to \C$ is holomorphic, then $f(\C)$ is either $\C$ or $\C$ minus a point. \end{thm} @@ -15083,7 +15086,7 @@ \chapter{Differential forms and Stokes' theorem} \label{ap:diffforms} like a curve needs to be oriented in order to define a line integral. However, unlike for curves, not every submanifold of dimension more than one is \emph{orientable}, that is, admits an orientation. -A classical non-orientable example is the M\"obius strip. +A classical nonorientable example is the M\"obius strip. \medskip @@ -15168,7 +15171,7 @@ \chapter{Differential forms and Stokes' theorem} \label{ap:diffforms} \end{equation*} The definition makes sense only if this sum actually exists. For example, if $\omega$ is -compactly supported then this sum is only finite, and so it exists. +compactly supported, then this sum is only finite, and so it exists. Higher degree forms are constructed out of $1$-forms and $0$-forms by the so-called wedge product. Given a $k$-form $\omega$ @@ -15195,9 +15198,8 @@ \chapter{Differential forms and Stokes' theorem} \label{ap:diffforms} \begin{equation*} \omega \wedge \eta = - \eta \wedge \omega . \end{equation*} -The negative keeps track of orientation for us. -When $\omega$ is a $k$-form and $\eta$ is an $\ell$-form then -we get +The negative keeps track of orientation. +When $\omega$ is a $k$-form and $\eta$ is an $\ell$-form, \begin{equation*} \omega \wedge \eta = {(-1)}^{k\ell} \eta \wedge \omega . \end{equation*} @@ -15219,7 +15221,7 @@ \chapter{Differential forms and Stokes' theorem} \label{ap:diffforms} dx_{j_k} , \end{equation*} where $g_{j_1,\ldots,j_k}$ are smooth functions. -But we can simplify even more. Since the wedge +We can simplify even more. Since the wedge is anticommutative on $1$-forms, \begin{equation*} dx_j \wedge dx_m = @@ -15735,15 +15737,15 @@ \chapter{Basic terminology and results from algebra} \label{ap:algebra} additive group we just mean we use $+$ for the operation and $0$ for the respective identity), with multiplication thrown in. If the multiplication -also defines a group on the set of nonzero elements then the ring is a -field. A ring that is not commutative, is one that doesn't satisfy -commutativity of multiplication. Some authors also define ring -without asking for an existence of $1$. +also defines a group on the set of nonzero elements, then the ring is a +field. A ring that is not commutative is one that does not satisfy +commutativity of multiplication. Some authors define ring +without asking for the existence of $1$. A ring that often comes up in this book is the ring of holomorphic functions. Let $\sO(U)$ be the set of holomorphic functions defined -on an open set $U$. Then pointwise addition and multiplication give -a ring structure on $\sO(U)$. The set of units are the set of +on an open set $U$. Pointwise addition and multiplication give +a ring structure on $\sO(U)$. The set of units is the set of functions that never vanish in $U$. The set of units is a multiplicative group.