# 254A announcement: Analytic prime number theory

In the winter quarter (starting January 5) I will be teaching a graduate topics course entitled “An introduction to analytic prime number theory“. As the name suggests, this is a course covering many of the analytic number theory techniques used to study the distribution of the prime numbers $latex {{mathcal P} = {2,3,5,7,11,dots}}&fg=000000$. I will list the topics I intend to cover in this course below the fold. As with my previous courses, I will place lecture notes online on my blog in advance of the physical lectures.

The type of results about primes that one aspires to prove here is well captured by Landau’s classical list of problems:

1. Even Goldbach conjecture: every even number $latex {N}&fg=000000$ greater than two is expressible as the sum of two primes.
2. Twin prime conjecture: there are infinitely many pairs $latex {n,n+2}&fg=000000$ which are simultaneously prime.
3. Legendre’s conjecture:…

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# MA1505 Summary

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# Perron-Frobenius Operator

## Perron-Frobenius Operator

Consider a map $f$ which possibly has a finite (or countable) number of discontinuities or points where possibly the derivative does not exist. We assume that there are points

$\displaystyle q_{0} or $q_{0}

such that $f$ restricted to each open interval $A_{j}=(q_{j-1},q_{j})$ is $C^{2}$, with a bound on the first and the second derivatives. Assume that the interval $[q_{0},q_{k}]$ ( or $[q_{0},q_{\infty}]$ ) is positive invariant, so $f(x)\in [q_{0},q_{k}]$ for all $x\in [q_{0}, q_{k}]$ ( or $f(x)\in [q_{0},q_{\infty}]$  for all $x\in[q_{0},q_{\infty}]$ ).

For such a map, we want a construction of a sequence of density functions that converge to a density function of an invariant measure. Starting with $\rho_{0}(x)\equiv(q_{k}-q_{0})^{-1}$ ( or $\rho_{0}(x)\equiv(q_{\infty}-q_{0})^{-1}$ ),assume that we have defined densities up to $\rho_{n}(x)$, then define define $\rho_{n+1}(x)$ as follows

$\displaystyle \rho_{n+1}(x)=P(\rho_{n})(x)=\sum_{y\in f^{-1}(x)}\frac{\rho_{n}(y)}{|Df(y)|}.$

This operator $P$, which takes one density function to another function, is called the Perron-Frobenius operator. The limit of the first $n$ density functions converges to a density function $\rho^{*}(x)$,

$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x).$

The construction guarantees that $\rho^{*}(x)$ is the density function for an invariant measure $\mu_{\rho^{*}}$.

Example 1. Let

$\displaystyle f(x)= \begin{cases} x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2x &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

We construct the first few density functions by applying the Perron-Frobenius operator, which indicates the form of the invariant density function.
Take $\rho_{0}(x)\equiv1$ on $[0,1]$. From the definition of $f(x)$, the slope on $(0,\frac{1}{2})$ and $(\frac{1}{2},1)$ are 1 and 2, respectively. If $x\in (\frac{1}{2},1)$, then it has only one pre-image on $(\frac{1}{2},1)$; else if $x\in(0,\frac{1}{2})$, then it has two pre-images, one is $x^{'}$ in $(0,\frac{1}{2})$, the other one is $x^{''}$ in $(\frac{1}{2},1)$. Therefore,

$\rho_{1}(x)= \begin{cases} \frac{1}{1}+\frac{1}{2} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{1}{2} &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

By similar considerations,

$\displaystyle \rho_{2}(x)=\begin{cases}1+\frac{1}{2}+\frac{1}{2^{2}} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{1}{2^{2}} &\mbox{if } x\in(\frac{1}{2},1).\end{cases}$

By induction, we get

$\displaystyle \rho_{n}(x)=\begin{cases}1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2^{n}} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{1}{2^{n}} &\mbox{if } x\in(\frac{1}{2},1).\end{cases}$

Now, we begin to calculate the density function $\rho^{*}(x)$. If $x\in(0,\frac{1}{2})$, then
$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x) =\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1} \sum_{m=0}^{n}\frac{1}{2^{m}} =\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\left(2-\frac{1}{2^{n}}\right)=2.$
If $x\in(\frac{1}{2},1)$, then
$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x) =\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\frac{1}{2^{n}} =\lim_{k\rightarrow \infty}\frac{1}{k}\left(2-\frac{1}{2^{k}}\right)=0.$
i.e.

$\displaystyle \rho^{*}(x)= \begin{cases} 2 &\mbox{if } x\in(0,\frac{1}{2}), \\ 0 &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

Example 2. Let

$\displaystyle f(x)=\begin{cases} 2x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2x-1 &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

Take $\rho_{0}(x)\equiv1$ on $(0,1)$. By induction, $\rho_{n}(x)\equiv1$ on $(0,1)$ for all $n\geq 0$. Therefore, $\rho^{*}(x)\equiv1$ on $(0,1)$.

Example 3. Let

$\displaystyle f(x)=\begin{cases} x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2^{n+1}\cdot\left(x-\left(1-\frac{1}{2^{n}}\right)\right) &\mbox{if } x\in\left(1-\frac{1}{2^{n}},1-\frac{1}{2^{n+1}}\right) \text{ for all } n\geq 1.\end{cases}$

Take $\rho_{0}(x)\equiv1$ on $(0,1)$. Assume

$\displaystyle \rho_{n}(x)= \begin{cases} a_{n} &\mbox{if } x\in(0,\frac{1}{2}), \\ b_{n} &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$

for all $n\geq 0$. It is obviously that $a_{0}=b_{0}=1$. By similar considerations,
$\displaystyle \rho_{n+1}(x)= \begin{cases} \frac{a_{n}}{1}+\frac{b_{n}}{4}+\frac{b_{n}}{8}+\frac{b_{n}}{16}+\cdot\cdot\cdot= a_{n}+\frac{b_{n}}{2} &\mbox{if } x\in(0,\frac{1}{2}), \\ \frac{b_{n}}{4}+\frac{b_{n}}{8}+\frac{b_{n}}{16}+\cdot\cdot\cdot = \frac{b_{n}}{2} &\mbox{if } x\in(\frac{1}{2},1). \end{cases}$
That means

$\displaystyle \left( \begin{array}{ccc} a_{n+1} \\ b_{n+1} \end{array} \right) =\left( \begin{array}{ccc} a_{n}+\frac{1}{2}b_{n} \\ \frac{1}{2}b_{n} \end{array} \right) = \left( \begin{array}{ccc} 1 & \frac{1}{2} \\ 0 & 1 \end{array} \right) \left( \begin{array}{ccc} a_{n} \\ b_{n} \end{array} \right)$

for all $n\geq 0$. From direct calculation, $\displaystyle a_{n}=2-\frac{1}{2^{n}}$ and $\displaystyle b_{n}=\frac{1}{2^{n}}$ for all $n\geq 0$. Therefore,

$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x)=\begin{cases} 2 &\mbox{if } x\in (0,\frac{1}{2}), \\ 0 &\mbox{if } x\in (\frac{1}{2},1). \end{cases}$

Example 4. Let

$\displaystyle f(x)=\begin{cases} 1.5 x &\mbox{if } x\in(0,\frac{1}{2}), \\ 2^{n+1}\cdot\left(x-\left(1-\frac{1}{2^{n}}\right)\right) &\mbox{if } x\in\left(1-\frac{1}{2^{n}},1-\frac{1}{2^{n+1}}\right) \text{ for all } n\geq 1.\end{cases}$

Take $\rho_{0}(x)\equiv1$ on $(0,1)$. Assume

$\displaystyle \rho_{n}(x)= \begin{cases} a_{n} &\mbox{if } x\in(0,\frac{3}{4}), \\ b_{n} &\mbox{if } x\in(\frac{3}{4},1). \end{cases}$

for all $n\geq 0$. It is obviously that $a_{0}=b_{0}=1$. By similar considerations,

$\displaystyle \left( \begin{array}{ccc} a_{n+1} \\ b_{n+1} \end{array} \right) =\left( \begin{array}{ccc} \frac{11}{12}a_{n}+\frac{1}{4}b_{n} \\ \frac{1}{4}a_{n}+\frac{1}{4}b_{n} \end{array} \right) = \left( \begin{array}{ccc} \frac{11}{12} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} \end{array} \right) \left( \begin{array}{ccc} a_{n} \\ b_{n} \end{array} \right)$

for all $n\geq 0$. From matrix diagonalization , $\displaystyle a_{n}=\frac{6}{5}-\frac{1}{5}\cdot\frac{1}{6^{n}}$ and $\displaystyle b_{n}=\frac{2}{5}+\frac{3}{5}\cdot\frac{1}{6^{n}}$ for all $n\geq 0$.

Therefore,

$\displaystyle \rho^{*}(x)=\lim_{k\rightarrow \infty}\frac{1}{k}\sum_{n=0}^{k-1}\rho_{n}(x)=\begin{cases} \frac{6}{5} &\mbox{if } x\in (0,\frac{3}{4}), \\ \frac{2}{5} &\mbox{if } x\in (\frac{3}{4},1). \end{cases}$

## Perron-Frobenius Theory

Definition. Let $A=[a_{ij}]$ be a $k\times k$ matrix. We say $A$ is non-negative if $a_{ij}\geq 0$ for all $i,j$. Such a matrix is called irreducible if for any pair $i,j$ there exists some $n>0$ such that $a_{ij}^{(n)}>0$ where $a_{ij}^{(n)}$ is the $(i,j)-$th element of $A^{n}$. The matrix $A$ is irreducible and aperiodic if there exists $n>0$ such that $a_{ij}^{(n)}>0$ for all $i,j$.

Perron-Frobenius Theorem Let $A=[a_{ij}]$ be a non-negative $k\times k$ matrix.

(i) There is a non-negative eigenvalue $\lambda$ such that no eigenvalue of $A$ has absolute value greater than $\lambda$.

(ii) We have $\min_{i}(\sum_{j=1}^{k}a_{ij})\leq \lambda\leq \max_{i}(\sum_{j=1}^{k}a_{ij})$.

(iii) Corresponding to the eigenvalue $\lambda$ there is a non-negative left (row) eigenvector $u=(u_{1},\cdot\cdot\cdot, u_{k})$ and a non-negative right (column) eigenvector $v=(v_{1},\cdot\cdot\cdot, v_{k})^{T}$.

(iv) If $A$ is irreducible then $\lambda$ is a simple eigenvalue and the corresponding eigenvectors are strictly positive (i.e. $u_{i}>0$, $v_{i}>0$ all $i$).

(v) If $A$ is irreducible then $\lambda$ is the only eigenvalue of $A$ with a non-negative eigenvector.

Theorem.
Let $A$ be an irreducible and aperiodic non-negative matrix. Let $u=(u_{1},\cdot\cdot\cdot, u_{k})$ and $v=(v_{1},\cdot\cdot\cdot, v_{k})^{T}$ be the strictly positive eigenvectors corresponding to the largest eigenvalue $\lambda$ as in the previous theorem. Then for each pair $i,j$, $\lim_{n\rightarrow \infty} \lambda^{-n}a_{ij}^{(n)}=u_{j}v_{i}$.

Now, let us see previous examples, again. The matrix $A$ is irreducible and aperiodic non-negative matrix, and $\lambda=1$ has the largest absolute value in the set of all eigenvalues of $A$. From Perron-Frobenius Theorem, $u_{i}, v_{j}>0$ for all pairs $i,j$. Then for each pari $i,j$,
$\lim_{n\rightarrow \infty}a_{ij}^{(n)}=u_{j}v_{i}$. That means $\lim_{n\rightarrow \infty}A^{(n)}$ is a strictly positive $k\times k$ matrix.

## Markov Maps

Definition of Markov Maps. Let $N$ be a compact interval. A $C^{1}$ map $f:N\rightarrow N$ is called Markov if there exists a finite or countable family $I_{i}$ of disjoint open intervals in $N$ such that

(a) $N\setminus \cup_{i}I_{i}$ has Lebesgue measure zero and there exist $C>0$ and $\gamma>0$ such that for each $n\in \mathbb{N}$ and each interval $I$ such that $f^{j}(I)$ is contained in one of the intervals $I_{i}$ for each $j=0,1,...,n$ one has

$\displaystyle \left| \frac{Df^{n}(x)}{Df^{n}(y)}-1 \right| \leq C\cdot |f^{n}(x)-f^{n}(y)|^{\gamma} \text{ for all } x,y\in I;$

(b) if $f(I_{k})\cap I_{j}\neq \emptyset$, then $f(I_{k})\supseteq I_{j}$;

(c) there exists $r>0$ such that $|f(I_{i})|\geq r$ for each $i$.

As usual, let $\lambda$ be the Lebesgue measure on $N$. We may assume that $\lambda$ is a probability measure, i.e., $\lambda(N)=1$. Usually, we will denote the Lebesgue measure of a Borel set $A$ by $|A|$.

Theorem.  Let $f:N\rightarrow N$ be a Markov map and let $\cup_{i}I_{i}$ be corresponding partition. Then there exists a $f-$invariant probability measure $\mu$ on the Borel sets of $N$ which is absolutely continuous with respect to the Lebesgue measure $\lambda$. This measure satisfies the following properties:

(a) its density $\frac{d\mu}{d\lambda}$ is uniformly bounded and Holder continuous. Moreover, for each $i$ the density is either zero on $I_{i}$ or uniformly bounded away from zero.

If for every $i$ and $j$ one has $f^{n}(I_{j})\supseteq I_{i}$ for some $n\geq 1$ then

(b) the measure is unique and its density $\frac{d\mu}{d\lambda}$ is strictly positive;

(c) $f$ is exact with respect to $\mu$;

(d) $\lim_{n\rightarrow \infty} |f^{-n}(A)|=\mu(A)$ for every Borel set $A\subseteq N$.

If $f(I_{i})=N$ for each interval $I_{i}$, then

(e) the density of $\mu$ is also uniformly bounded from below.

# Shape of Inner Space

String Theory and the Geometry of the Universe’s Hidden Dimensions

Chapter 3: P.39

My personal involvement in this area began in 1969, during my first semester of graduate studies at Berkeley. I needed a book to read during Chrismas break. Rather than selecting Portnoy’s Complaint, The Godfather, The Love Machine, or The Andromeda Strain-four top-selling books of that year-I opted for a less popular title, Morse Theory, by the American mathematician John Milnor. I was especially intrigued by Milnor’s section on topology and curvature, which explored the notion that local curvature has a great influence on geometry and topology. This is a theme I’ve pursued ever since, because the local curvature of a surface is determined by taking the derivatives of that surface, which is another way of saying it is based on analysis. Studying how that curvature influences geometry, therefore, goes to the heart of geometric analysis.

Having no office, I practically lived in Berkeley’s math library in those days. Rumor has it that the first thing I did upon arriving in the United States was visiting that library, rather than, say, explore San Francisco as other might have done. While I can’t remember exactly what I did, forty years hence, I have no reason to doubt the veracity of that rumor. I wandered around the library, as was my habit, reading every journal I could get my hands on. In the course of rummaging through the reference section during winter break, I came across a 1968 article by Milnor, whose book I was still reading. That article, in turn, little else to do at the time (with most people away for the holiday), I tried to see if I could prove something related to Preissman’s theorem.

Chapter 4: P.80

From this sprang the work I’ve become most famous for. One might say it was my calling. No matter what our station, we’d all like to find our true calling in life-that special thing we were put on this earth to do. For an actor, it might be playing Stanley Kowalski in A Streetcar Named Desire. Or the lead role in Hamlet. For a firefighter, it could mean putting out a ten-alarm blaze. For a crime-fighter, it could mean capturing Public Enemy Number One. And in mathematics, it might come down to finding that one problem you’re destined to work on. Or maybe destiny has nothing to do with it. Maybe it’s just a question of finding a problem you can get lucky with.

To be perfectly honest, I never think about “destiny” when choosing a problem to work on, as I tend to be a bit more pragmatic. I try to seek out a new direction that could bring to light new mathematical problems, some of which might prove interesting in themselves. Or I might pick an existing problem that offers the hope that in the course of trying to understand it better, we will be led to a new horizon.

The Calabi conjecture, having been around a couple of decades, fell into the latter category. I latched on to this problem during my first year of graduate school, though sometimes it seemed as if the problem latched on to me. It caught my interest in a way that no other problem had before or has since, as I sensed that it could open a door to a new branch of mathematics. While the conjecture was vaguely related to Poincare’s classic problem, it struck me as more general because if Calabi’s hunch were true, it would lead to a large class of mathematical surfaces and spaces that we didn’t know anything about-and perhaps a new understanding of space-time. For me the conjecture was almost inescapable: Just about every road I pursued in my early investigations of curvature led to it.

Chapter 5: P.104

A mathematical proof is a bit like climbing a mountain. The first stage, of course, is discovering a mountain worth climbing. Imagine a remote wilderness area yet to be explored. It takes some wit just to find such an area, let alone to know whether something worthwhile might be found there. The mountaineer then devises a strategy for getting to the top-a plan that appears flawless, at least on paper. After acquiring the necessary tools and equipment, as well as mastering the necessary skills, the adventurer mounts an ascent, only to be stopped by unexpected difficulties. But others follow in their predecessor’s footsteps, using the successful strategies, while also pursuing different avenues-thereby reaching new heights in the process. Finally someone comes along who not only has a good plan of attack that avoids the pitfalls of the past but also has the fortitude and determination to reach the summit, perhaps planting a flag there to mark his or her presence. The risks to life and limb are not so great in math, and the adventure may not be so apparent to the outsider. And at the end of a long proof, the scholar does not plant a flag. He or she types in a period. Or a footnote. Or a technical appendix. Nevertheless, in our field there are thrill as well as perils to be had in the pursuit, and success still rewards those of us who’ve gained new views into nature’s hidden recesses.