From fd804da5942e87d32e2b152898ccf4eb529a828b Mon Sep 17 00:00:00 2001 From: user202729 <25191436+user202729@users.noreply.github.com> Date: Fri, 12 Apr 2024 05:58:27 +0700 Subject: [PATCH] Add proof of Cauchy-Goursat theorem, and various small changes --- tex/alg-NT/classgrp.tex | 3 +- tex/alg-NT/dedekind.tex | 12 +- tex/alg-NT/norm-trace.tex | 3 - tex/complex-ana/holomorphic.tex | 218 ++++++++++++++++++++++++++++++-- tex/diffgeo/multivar.tex | 2 +- 5 files changed, 219 insertions(+), 19 deletions(-) diff --git a/tex/alg-NT/classgrp.tex b/tex/alg-NT/classgrp.tex index 8735d5f1..3149785c 100644 --- a/tex/alg-NT/classgrp.tex +++ b/tex/alg-NT/classgrp.tex @@ -643,7 +643,7 @@ \section{Computation of class numbers} each of the three branches has only one (large) cherry on it. That means any time we put together an integral ideal with norm $\le M_K$, it is actually principal. - In fact, these guys have norm $4$, $9$, $25$ respectively\dots + In fact, these guys have norm $4$, $9$, $25$ respectively\dots\ so we can't even touch $(3)$ and $(5)$, and the only ideals we can get are $(1)$ and $(2)$ (with norms $1$ and $4$). @@ -701,6 +701,7 @@ \section{Computation of class numbers} First, we use a lemma that will help us with narrowing down the work in our cherry tree. \begin{lemma}[Ideals divide their norms] + \label{lemma:ideal_divides_norm} Let $\kb$ be an integral ideal with $\Norm(\kb) = n$. Then $\kb$ divides the ideal $(n)$. \end{lemma} diff --git a/tex/alg-NT/dedekind.tex b/tex/alg-NT/dedekind.tex index cb958afa..35d7c225 100644 --- a/tex/alg-NT/dedekind.tex +++ b/tex/alg-NT/dedekind.tex @@ -169,6 +169,12 @@ \section{Dedekind domains} \emph{every nonzero prime ideal of $\mathcal A$ is in fact maximal}. (The last condition is the important one.) \end{definition} + +\begin{remark} + Note that $\mathcal A$ is a Dedekind domain if and only if $\mathcal A = \OO_K$ for some field + $K$, as we will prove below. We're just defining this term for historical reasons\dots +\end{remark} + Here there's one new word I have to define for you, but we won't make much use of it. \begin{definition} Let $R$ be an integral domain and let $K$ be its field of fractions. @@ -205,7 +211,8 @@ \section{Dedekind domains} \begin{proof} Boring, but here it is anyways for completeness. - Since $\OO_K \cong \ZZ^{\oplus n}$, we get that it's Noetherian. + Since $\OO_K \cong \ZZ^{\oplus n}$,\footnote{By \Cref{thm:OK_free_Z_module}.} + we get that it's Noetherian. Now we show that $\OO_K$ is integrally closed. Suppose that $\eta \in K$ is the root of some polynomial with coefficients in $\OO_K$. @@ -214,13 +221,14 @@ \section{Dedekind domains} + \dots + \alpha_0 \] where $\alpha_i \in \OO_K$. We want to show that $\eta \in \OO_K$ as well. - Well, from the above, $\OO_K[\eta]$ is finitely generated\dots + Well, from the above, $\OO_K[\eta]$ is finitely generated\dots\ thus $\ZZ[\eta] \subseteq \OO_K[\eta]$ is finitely generated. So $\eta \in \ol\ZZ$, and hence $\eta \in K \cap \ol\ZZ = \OO_K$. \end{proof} Now let's do the fun part. We'll prove a stronger result, which will re-appear repeatedly. \begin{theorem}[Important: prime ideals divide rational primes] + \label{thm:prime_ideals_over_rational} Let $\OO_K$ be a ring of integers and $\kp$ a nonzero prime ideal inside it. Then $\kp$ contains a rational prime $p$. diff --git a/tex/alg-NT/norm-trace.tex b/tex/alg-NT/norm-trace.tex index afe18c85..5a0c3c41 100644 --- a/tex/alg-NT/norm-trace.tex +++ b/tex/alg-NT/norm-trace.tex @@ -114,9 +114,6 @@ \section{Norms and traces} Then, $\Tr_{K/\QQ}(\alpha) = \sum_{i=1}^d \sigma_i(\alpha)$ and $\Norm_{K/\QQ}(\alpha) = \prod_{i=1}^d \sigma_i(\alpha)$. - - There is exactly $d$ embeddings, regardless of the number of Galois - conjugates of $\alpha$. \end{remark} \begin{theorem}[Morally correct definition of norm and trace] diff --git a/tex/complex-ana/holomorphic.tex b/tex/complex-ana/holomorphic.tex index f6e2d1f7..b3d1fda3 100644 --- a/tex/complex-ana/holomorphic.tex +++ b/tex/complex-ana/holomorphic.tex @@ -358,16 +358,6 @@ \section{Cauchy-Goursat theorem} and then applies the so-called Green's theorem. But it was Goursat who successfully proved the fully general theorem we've stated above, which assumed only that $f$ was holomorphic. -I'll only outline the proof, and very briefly. -You can show that if $f \colon \Omega \to \CC$ has an antiderivative $F \colon \Omega \to \CC$ which is also holomorphic, -and moreover $\Omega$ is simply connected, then you get a ``fundamental theorem of calculus'', a la -\[ \oint_\alpha f(z) \; dz = F(\alpha(b)) - F(\alpha(a)) \] -where $\alpha \colon [a,b] \to \CC$ is some path. -So to prove Cauchy-Goursat, you only have to show this antiderivative $F$ exists. -Goursat works very hard to prove the result in the special case that $\gamma$ is a triangle, -and hence by induction for any polygon. -Once he has the result for a rectangle, he uses this special case to construct the function $F$ explicitly. -Goursat then shows that $F$ is holomorphic, completing the proof. Anyways, the theorem implies that $\oint_\gamma z^m \; dz = 0$ when $m \ge 0$. So much for all our hard work earlier. @@ -479,7 +469,7 @@ \section{Cauchy's integral theorem} + 2\pi i f(a). \end{align*} - where we've used \Cref{thm:central_cauchy_computation} + where we've used \Cref{thm:central_cauchy_computation}. Thus, all we have to do is show that \[ \oint_{\gamma_\eps} \frac{f(z)-f(a)}{z-a} \; dz = 0. \] For this we can basically use the weakest bound possible, the so-called $ML$ lemma @@ -541,7 +531,7 @@ \section{Holomorphic functions are analytic} bounded by a circle $\gamma$. Suppose $D$ is contained inside $U$. Then $f$ is given everywhere in $D$ by a Taylor series \[ - f(z) = c_0 + c_1(z-p) + c_2(z-p)^2 + \dots + f(z) = c_0 + c_1(z-p) + c_2(z-p)^2 + \cdots \] where \[ @@ -561,6 +551,33 @@ \section{Holomorphic functions are analytic} that a function being complex differentiable once means it is not only infinitely differentiable, but in fact equal to its Taylor series. +\begin{remark} + If you're willing to assume this, you can see why Cauchy-Goursat theorem should be true: + assuming + \[ + f(z) = c_0 + c_1 z + c_2 z^2 + \cdots + \] + then, with $\gamma$ the unit circle, + \begin{align*} + \oint_\gamma f(z) \; dz + &= \oint_\gamma c_0 + c_1 z + c_2 z^2 + \dots \; dz \\ + &= \left( \oint_\gamma c_0 \; dz \right) + + \left( \oint_\gamma c_1 z \; dz \right) + + \left( \oint_\gamma c_2 z^2 \; dz \right) + \cdots + \end{align*} + We have already proven that each $\oint_\gamma z^m \; dz = 0$, so the sum ought to be $0$ as + well. + + Of course the argument is not completely rigorous, it exchanges the integration and the + infinite sum without justification. +\end{remark} + +\begin{remark} + You can see where the term $\frac{f(w-p)}{(w-p)^{k+1}}$ comes from in + \Cref{remark:cauchy_integral_intuition}. It is very intuitive that even if you forget it, + you can derive it yourself as well! +\end{remark} + I should maybe emphasize a small subtlety of the result: the Taylor series centered at $p$ is only valid in a disk centered at $p$ which lies entirely in the domain $U$. If $U = \CC$ this is no issue, since you can make the disk big enough to accommodate any point you want. @@ -572,6 +589,183 @@ \section{Holomorphic functions are analytic} Indeed, as you'll see in the problems, the existence of a Taylor series is incredibly powerful. +\section{Optional: Proof that holomorphic functions are analytic} + +It is recommended to read the next chapter first to understand the origin of the term +$\frac{f(w-p)}{(w-p)^{k+1}}$ in Cauchy's differentiation formula above. + +Each step of the proof is quite intuitive, if not a bit long. +The outline is: +\begin{itemize} + \ii We pretend that the function $f$ is analytic. (Yes, this is not circular reasoning!) + \ii We use Cauchy's differentiation formula to write down a power series:\footnote{Assume $0 \in + U$.} + \[ c_0 + c_1 z + c_2 z^2 + \cdots \] + \ii We prove that the power series coincide with $f$ using Cauchy-Goursat theorem. + \ii Note that the statement ``$f$ is analytic'' literally means ``for every $k \geq 0$, then + $f^{(k)}$ is differentiable''. So, we write down a power series for $f^{(k)}$, and show that it + is differentiable. + (We already did this for the real case in \Cref{prop:taylor_series_are_analytic}.) +\end{itemize} + +\subsection{Proof of Cauchy-Goursat theorem} + +Suppose $f$ is holomorphic i.e. differentiable. We wish to prove $\oint_\gamma f \; dz = 0$. + +How may we attack this problem? Looking at the conclusion, we may want to stare at some function +where $\oint_\gamma f \; dz \neq 0$. + +We readily got an example from the previous chapter: $f(z) = \frac{1}{z}$. +\begin{ques} + What part of the hypothesis does not hold? +\end{ques} + +In any case, you see the problem is it's because $f$ has a singularity at $0$ (even though we +haven't formally defined what a singularity is yet). So, we try to prove the contrapositive: +\begin{theorem} + Suppose $\oint_\gamma f \; dz \neq 0$. + Then something weird happens to $f$ somewhere inside $\gamma$. +\end{theorem} +(For arbitrary loops, it gets a bit more difficult, however. What does ``inside $\gamma$'' mean?) + +Phrasing like this, it isn't that difficult. You may want to look at $f(z) = \frac{1}{z}$ a bit and +try to figure out how the proof follows before continue reading. + +For simplicity, I will prove the statement for $\gamma$ being a rectangle, leaving the case e.g. +$\gamma$ is a circle to the reader. The case of fully general $\gamma$ will be handled later on. + +As you may figured out, for $f(z) = \frac{1}{z-w}$, you can try to locate where the singularity $w$ +is by ``binary search'': compute $\oint_\gamma f \; dz$, if it is $2\pi i$, we know $w$ is inside +$\gamma$. We're going to do just that. + +What should we search for? Let's see: +\begin{exercise} + Suppose $\oint_\gamma f \; dz \neq 0$. + Must there be a point where $f$ blows up to infinity, like the point $z = 0$ in $\frac{1}{z}$? +\end{exercise} +Answer: no, unfortunately. You can certainly take the function $f$ above, and ``smooth out'' the +singularity. +\begin{center} + \begin{asy} + import graph; + draw(graph(new real(real x){ return 1/x; }, -3, -1/3), red); + draw(graph(new real(real x){ return 1/x; }, 1/3, 3), red); + draw((-1/2, -2){(1, -2)} .. tension 5 .. {(1, -2)}(1/2, 2), blue+dashed); + graph.xaxis(); + graph.yaxis(); + \end{asy} +\end{center} +(Only real part depicted. You can imagine the imaginary part.) + +The best we can hope for, then, is to find a point where $f$ is not holomorphic (complex +differentiable). + +Construct $4$ paths $\gamma_a$, $\gamma_b$, $\gamma_c$ and $\gamma_d$ as follows. The margin is only +for illustration purpose, in reality the edges directly overlap on each other. +\begin{center} + \begin{asy} + void drawrect(real x1, real y1, real x2, real y2){ + draw((x1, y1)--(x2, y1), MidArrow); + draw((x2, y1)--(x2, y2), MidArrow); + draw((x2, y2)--(x1, y2), MidArrow); + draw((x1, y2)--(x1, y1), MidArrow); + } + var margin=0.2, halfmargin=margin/2; + drawrect(0, 0, 6, 4); + drawrect(0+margin, 0+margin, 3-halfmargin, 2-halfmargin); + drawrect(3+halfmargin, 0+margin, 6-margin, 2-halfmargin); + drawrect(0+margin, 2+halfmargin, 3-halfmargin, 4-margin); + drawrect(3+halfmargin, 2+halfmargin, 6-margin, 4-margin); + label("$\gamma$", (6, 4), dir(45)); + label("$\gamma_a$", (3-halfmargin, 4-margin), dir(-135)); + label("$\gamma_b$", (6-margin, 4-margin), dir(-135)); + label("$\gamma_c$", (3-halfmargin, 2-halfmargin), dir(-135)); + label("$\gamma_d$", (6-margin, 2-halfmargin), dir(-135)); + \end{asy} +\end{center} + +Notice that, because all the inner edges cancel out, +\[ + \oint_\gamma f \; dz = + \oint_{\gamma_a} f \; dz + \oint_{\gamma_b} f \; dz + + \oint_{\gamma_c} f \; dz + \oint_{\gamma_d} f \; dz. +\] +Which means $\oint_{\gamma_i} f \; dz \neq 0$ for some $i \in \{ a, b, c, d \}$. +(Idea: we have more accurately located the singularity, now we know it is inside $\gamma_i$. +Of course it's also possible that there are multiple singularities.) + +We also have $|\oint_{\gamma_i} f \; dz| \geq \frac{1}{4} \cdot |\oint_\gamma f \; dz|$ +for some $i$. +The reason why we must carefully keep track of the magnitude (instead of just saying it's $\neq 0$) +will become apparent later. + +So, we keep doing that, and get a decreasing sequence of rectangles $\{ \gamma_j \}$. Because the +edge length gets halved each time, the rectangles converge to a single point $p$. + +How would the rectangle perimeter decrease? Perhaps something like the following: +\begin{center} + \begin{tabular}{ccc} + $j$ & Perimeter of $\gamma_j$ & $|\oint_{\gamma_j} f \; dz|$ \\ \hline + $0$ & $1$ & $1$ \\ + $1$ & $\frac{1}{2}$ & $\geq \frac{1}{4}$ \\ + $2$ & $\frac{1}{4}$ & $\geq \frac{1}{16}$ \\ + $3$ & $\frac{1}{8}$ & $\geq \frac{1}{64}$ + \end{tabular} +\end{center} +$|\oint_{\gamma_j} f \; dz|$ decreases quite quickly compared to the perimeter --- as expected, we +cannot hope for $f$ to blow up at $p$, but this is sufficient to show $f$ is not holomorphic. + +For the sake of contradiction, assume otherwise. Then, by definition, +\[ \lim_{h \to 0} \frac{f(p+h)-f(p)}{h} = f'(p) \] +where $p$ is the point that the rectangles $\{ \gamma_j \}$ converges to as defined above, +and $f'(p) \in \CC$ is the derivative. In other words, for $h \in \CC$ close enough to $0$, +\[ f(p+h) = f(p) + f'(p) \cdot h + \eps(h) \cdot h\text{ for }\eps(h) \in o(1). \] +Why is this a problem? Notice that $f(p)$ and $f'(p) \cdot h$ are both polynomials, so +\[ \oint_{\gamma_j} f(p) + f'(p) \cdot (z-p) \; dz = 0, \] +which means +\[ \oint_{\gamma_j} f(z)\; dz = \oint_{\gamma_j} \eps(h) \cdot (z-p) \; dz. \] +We know the left hand side decreases as $4^{-j}$, but the integral on the right hand side is over a +curve with length decreasing as $2^{-j}$. +\begin{exercise} + Finish the proof. (Use the $ML$ estimation lemma.) +\end{exercise} + +Finally, what to do with arbitrary curve (which may not even have an interior\footnote{A +space-filling curve is an example.})? + +We construct the antiderivative $F \colon \Omega \to \CC$ by integrating $f$ across the side of a +rectangle, prove $F' = f$, +and get a ``fundamental theorem of calculus'', that is +\[ \oint_\alpha f(z) \; dz = F(\alpha(b)) - F(\alpha(a)) \] +where $\alpha \colon [a,b] \to \CC$ is some path. +Considering $\alpha = \gamma$, +because the starting and ending point for a loop $\gamma$ is the same, of course the integral would +be $0$. + +\subsection{The rest} + +Next step, we should show the power series coincide with $f$, that is +\[ f(z) = +\oint_\gamma \frac{f(t)}{t} \; dt + +\oint_\gamma \frac{f(t)}{t^2} \; dt \cdot z + +\oint_\gamma \frac{f(t)}{t^3} \; dt \cdot z^2 + \cdots \] +Here we assume $\gamma$ is the unit circle, the power series is centered at $0$, and $t$ is inside +the unit disk. +\begin{exercise} + Prove it. (You only need to know that you can interchange the infinite sum and the integral in + this situation,\footnote{Look at \Cref{ex:failure_interchange_lim_int} for some horror stories + where you cannot interchange a limit and an integral.} + how to sum a geometric series, and Cauchy's integral formula) +\end{exercise} + +\begin{remark} +\emph{Wait, where was Cauchy-Goursat theorem used?} If you forgot, it is used in the proof of +Cauchy's integral formula. +\end{remark} + +After we have proven that $f$ is a power series, then using \Cref{prop:taylor_series_are_analytic} +(suitably adapted for the case of complex holomorphic functions), the result follows. + \section\problemhead These aren't olympiad problems, but I think they're especially nice! In the next complex analysis chapter we'll see some more nice applications. diff --git a/tex/diffgeo/multivar.tex b/tex/diffgeo/multivar.tex index 8fdddcc1..69d34c7c 100644 --- a/tex/diffgeo/multivar.tex +++ b/tex/diffgeo/multivar.tex @@ -270,7 +270,7 @@ \section{Total and partial derivatives} \[ Df = \sum_{i=1}^n \fpartial{f}{\ee_i} \cdot \ee_i^\vee. \] \end{theorem} \begin{proof} - Not going to write out the details, but\dots + Not going to write out the details, but\dots\ given $v = t_1e_1 + \dots + t_ne_n$, the idea is to just walk from $p$ to $p+t_1e_1$, $p+t_1e_1+t_2e_2$, \dots, up to $p+t_1e_1+t_2e_2+\dots+t_ne_n = p+v$,