Colin Defant's General Exam Committee: Peter Sarnak (Chair), Chris Skinner, and Noga Alon Special Topics: Algebraic Number Theory and Ergodic Number Theory April 30, 2018. Sarnak's office at 3:30 PM. Below, I have written the dialogue as I remember. I have put the names of the professors in all capital letters when they ask questions. ================================================================== COMPLEX ANALYSIS ================================================================== We started with the professors trying to decide whether or not Alon should ask the first question since this was his first general exam. Alon said "I can try! I think I can do it." He started with a reasonable question. ALON: Tell us Liouville's theorem. Me: A bounded entire function is constant. ALON: How do you prove it? Me: Choose a point z in \mathbb C, and use Cauchy's integral formula to write f'(z)=\frac{1}{2\pi i}\int_{|\zeta-z|=R}\frac{f(\zeta)}{(\zeta-z)^2}d\zeta. A basic estimate shows that this is O(1/R), so let R tend to infinity and conclude f' vanishes identically. SARNAK: So suppose I have a bounded harmonic function in the entire complex plane. What can you say about it? Me: It's constant. SARNAK: Proof? Me: The harmonic function is the real part of an entire function f. Then |e^{f(z)}|=e^{\Re(f(z))} is bounded, so e^{f(z)} is constant. This forces f (and hence also its real part) to be constant. SARNAK: What if I'm in three dimensions? Me: So the proof is basically that you use the Mean Value Property for harmonic functions along with the fact that a sequence of balls centered at the origin in \mathbb R^n with radii tending to infinity forms a Folner sequence. Sarnak: Woah! Just be careful. You opened that can of worms. Me: Okay, I won't open any more cans of worms. Anyway, suppose our bounded harmonic function (in \mathbb R^n) is u, and pick a point x_0 in \mathbb R^n. Use the Mean Value Property for harmonic functions to say that u(x_0) is equal to the average of u on B_R(x_0), the ball of radius R centered at x_0. Also, u(0) is the average value of u on B_R(0). If |u(x)|\leq M for all x\in\mathbb R^n, then a simple estimate shows that |u(x_0)-u(0)|\leq M\mu(B_R(x_0)\Delta B_R(0))/\mu(B_R(0)), where \mu is Lebesgue measure. This tends to 0 as R tends to infinity. SARNAK: Okay good. What's a subharmonic function? Me: So a subharmonic function is an upper semicontinuous function that--- SARNAK: (Laughing) Woah you're going to have to remind me what that means. What does that mean? Me: A function f is upper semicontinuous at a point x_0 if \limsup_{x\to x_0}f(x)}\leq f(x_0). Then a function defined on a region is upper semicontinuous if it is upper semicontinuous at every point in the region. Then a function u defined on a region is called subharmonic if it is upper semicontinuous and has the property that whenever you have a ball B contained in the region and you have a harmonic function \phi defined on a region containing the closure of B such that u(z)\leq\phi(z) for all z\in \partial B, it follows that u(z)\leq\phi(z) for all z\in B. SARNAK: Okay, so if I have a subharmonic function in the plane, and it's bounded above by 3, what can I say about it? Me: It's constant. SARNAK: And can you prove THAT? Me: Yes, so let's say our function is u. Then u(z)\leq 3 for all z\in\mathbb C. Let M=\max_{z\in \mathbb D}u(z) (\mathbb D is the unit disc). By the maximum principal for subharmonic functions, M is the maximum of u over the boundary of the unit disc, also known as the unit circle. So M=\max_{z\in\partial \mathbb D}u(z). Sarnak: I'm impressed! We used the put this on the quals at Stanford, and people usually wouldn't get it! (He's also asked it in previous general exams, which is how I knew how to solve it ....) Me: Now let v(z)=u(z)-\epsilon\log|z|. Sarnak: He knows how to construct a barrier! Now I'm REALLY impressed! Again, he's asked this before .... Me: Now \log|z| is harmonic away from the origin, so v(z) is subharmonic away from the origin. Notice that v(z)\to -\infty as |z|\to\infty because u is bounded above. Also, M=\max_{z\in\partial\mathbb D}v(z) because u and v agree on the unit circle. By the maximum principal for subharmonic functions, v(z)\leq M for all z\in\mathbb C with |z|\geq 1. Let \epsilon\to 0 to see that u(z)\leq M for all z\in\mathbb C with |z|\geq 1. This means that u(z)\leq M for ALL z. However, u attains this maximum value M on the unit circle, which contradicts the maximum principal for subharmonic functions. ALON: Do you know the Picard theorem? Me: Well there's the Little Picard theorem, which states that if f is entire and there are two distinct complex numbers that are not in the image if f, then f is constant. Also, there's the Big Picard theorem, which says that if a function has an essential singularity, then on any punctured neighborhood of that essential singularity, the function attains every complex value, with the exception of one value, infinitely often. Sarnak: Draw a semicircle. I drew a semicircle. I knew what was coming, so I also marked a point in the interior of the semicircle and drew the angle subtended at that point by the diameter. Me: Should I also draw this? Sarnak: What? Do I ask this every time? Me: Yeah. Sarnak: (Laughing) All right well he already knows what I'm going to ask! I better ask something else. Alon was curious to see what the question was, so Sarnak asked the question. SARNAK: I have a holomorphic function that is bounded by 2 on the circular part of the semicircle and is bounded by 1 on the diameter. Give an upper bound for the function in the interior of the circle. Me: You want to find a harmonic function u that has value \log 2 on the circular part and has value 0 on the diameter. If f is the holomorphic function in question, then \log|f(z)| is subharmonic and is bounded above by u(z) on the semicircular boundary. Hence, \log|f(z)|\leq u(z) for all z in the interior. You can find u by taking v(z) to be the angle subtended at z by the diameter. This function is harmonic, has value \pi/2 on the circular part of the semicircle, and has value \pi on the diameter. Then u(z)=\frac{2\log 2}{\pi}(\pi-v(z)). SARNAK: So you know what harmonic measure is? Me: Yes, it's a measure, dependent on a point in a domain, which is defined on the boundary of a domain. It has the property that if you integrate boundary values against the harmonic measure, then you obtain the value at the specified point of the harmonic function in the domain that has those given boundary values. Sarnak: All right, you seem to know everything. I smiled and thought "No, I just know the things you've asked in previous exams." Sarnak: Draw a triply connected region! In my head I was thinking "Why is he starting with triply connected? Why not simply connected and doubly connected?" I had learned five proofs that annuli with different radial ratios are not conformally equivalent because I knew he would ask about that and I wanted to give multiple proofs. Now he was skipping over annuli and going for triply connected regions. I drew a blob with two circular holes missing. Sarnak: Draw another triply connected region. I drew a blob with a circular hole missing and a slit missing (just for fun). Sarnak: Ahh, I see he knows the answer! I'm not sure why he said this. This time I didn't know what he was planning on asking, so I actually didn't know the answer yet. SARNAK: Can I map the first region to the second biholomorphically? Me: Not necessarily, since these are just arbitrary triply connected regions. There are restrictions for when you can. SARNAK: Okay, so can I map these regions to some standard types of regions? Me: Oh yeah! You could map them to annuli with circular slits removed. Sarnak: Draw that. I drew an annulus with one slit removed. SARNAK: Okay, so how do I map the first region to this new region? I didn't know what he meant by the word "how." I knew you could do this theoretically by solving the Dirichlet problem on the first region, but I thought at the time that he was asking me for an explicit map. Looking back on it now, it should have been clear to me what he wanted. Me: I'm not sure. SARNAK: Okay, suppose I just have a doubly connected region and I want to map it to an annulus. How can I do that? In my head: "Yes! My plan is back on track!" Me: I can use the Uniformization theorem to--- SARNAK: What is the Uniformization theorem? Me: Every simply connected Riemann surface is conformally equivalent to the Riemann sphere, the complex plane, or the upper half-plane. Sarnak: Okay good, so go on. Me: I can use the Uniformization theorem to say that my simply connected region is a quotient of its universal cover by some group of Deck transformations, where the universal cover is the Riemann sphere, the complex plane, or the upper half-plane. The group of Deck transformations is isomorphic to \mathbb Z since that's the fundamental group of my region. Now, any nontrivial Deck transformation has no fixed points. Since every automorphism of the Riemann sphere has a fixed point, the universal cover can't be the Riemann sphere. So let's suppose it's the complex plane. In this case, you can use the fact that the Deck transformation group acts properly discontinuously on \mathbb C, along with the fact that nontrivial Deck transformations have no fixed points, to show that the only possible Deck transformation groups for \mathbb C are \{1\}, \mathbb Z\cdot\omega (with \omega\in\mathbb C\setminus\{0\}), and \mathbb Z\cdot\omega_1\times\mathbb Z\cdot\omega_2 (with \omega_1, \omega_2 linearly independent over \mathbb R). Again, our group is isomorphic to \mathbb Z, so it must be \mathbb Z\cdot\omega. Now the map z\mapsto e^{2\pi iz/\omega} gives a biholomorphic map from \mathbb C/(\mathbb Z\cdot\omega) to \mathbb C\setminus\{0\}, showing that in this case our region is conformally equivalent to the punctured plane. Next, suppose the universal cover is the upper half-plane \mathbb H. The automorphism group of \mathbb H is PSL_2(\mathbb R). Take a generator for the Deck transformation group. This generator is not elliptic since elliptic elements have fixed points in \mathbb H, so it is either parabolic or hyperbolic. If it is parabolic, the Deck group is conjugate to the group generated by the map T given by T(z)=z+1. In this case, the map z\mapsto e^{2\pi iz} gives an explicit biholomorphic map from \mathbb H/ to \mathbb D\setminus\{0\}. Finally, if the generator of the Deck group is hyperbolic, then it is conjugate to the map V_\lambda given by V_\lambda(z)=\lambda^2 z, where \lambda>1 is some real number. In this case, the map z\mapsto e^{-\pi i\log z/\log\lambda} is a biholomorphic map from \mathbb H/ to A(1,e^{\pi^2/\log\lambda})=\{z\in\mathbb C:1<|z|1. By inverting if necessary, I can assume f maps the inner circle to the inner circle and maps the outer circle to the outer circle. Expand f as a Laurent series, say f(z)=\sum_{n\in\mathbb Z}c_nz^n. It's an easy consequence of Green's theorem that the area bounded by a Jordan curve \gamma is Area(\gamma)=\frac{1}{2i}\int_\gamma\overline z dz.\] Let C(t)=\{z\in\mathbb C:|z|=t\}. We have Area(f(C(t)))=\frac{1}{2i}\int_{f(C(t))}\overline z dz=\frac{1}{2i}\int_0^{2\pi}\overline{f(te^{i\theta})}f'(te^{i\theta})ti e^{i\theta}d\theta. You can then just plug in the Laurent series and its derivative for f and f'. After you rearrange terms, you get Area(f(C(t)))=\pi\sum_{n\in\mathbb Z}n|c_n|^2t^{2n}. This holds for 11+3\pi/2. I said I thought this was what Weierstrass originally did. SARNAK: Uhh okay I don't know how to check that .... Do you know how to give a conceptual proof? I said something about taking a sequence of functions that have lots of triangular spikes that get really small so you get a uniform limit of continuous functions. Then the limit is continuous while the spikes keep it from being differentiable anywhere. They agreed that the a^n in the expression I wrote down was giving the uniform convergence while the b^n was giving the oscillation. SARNAK: What's the dual of L^1? Me: L^\infty. SARNAK: What's the dual of L^\infty? Me: Something that properly contains L^1. I can use the Hahn-Banach theorem to construct an element of the dual of L^\infty that isn't in L^1. SARNAK: How about the dual of L^p? Me: L^q, where $1/p+1/q=1. SARNAK: So is the dual of the dual the space? Me: It is when for L^p when 1> d^{-\epsilon}, but then he said we would talk about L(1,\chi) in a minute. He asked what \Delta_K is. I said it's d if d is 1 mod 4 and it's 4d otherwise. He said to ignore the constants and just say that the expression I had written was roughly h_K/\sqrt d. He then told me to write down L(1,\chi) as a product, so I wrote L(1,\chi)=\prod_p \frac{1}{1-\chi(p)p^{-1}}. SARNAK: Now suppose I want the class number to be 1. Then 1/\sqrt d is small, so what signs should you choose for the terms \chi(p)? I said I'd need them to be negative for lots of small primes p. Sarnak then told me to observe that this meant lots of small primes had to be inert. He then started talking about Siegel's theorem and its ineffectivity. Sarnak: So do you know how to prove Sieg--- Wait .... What are we testing him on? Skinner: We're doing algebraic number theory .... Sarnak: Right, so maybe we should go back to that. This led to one of my favorite parts of the exam. Skinner: So I feel obliged to ask you about class field theory. ALON: Do you know the Cauchy-Davenport theorem? Me: (???) Uhh yeah ... it says that if A,B \subseteq \mathbb F_q, then either A+B=\mathbb F_q or |A+B|\geq |A|+|B|-1. ALON: Do you know how to prove it? I smiled and said that you can use the Combinatorial Nullstellensatz. I think Skinner and Sarnak got a kick out of that. ALON: So do you know why it is called the Cauchy-Davenport theorem? They couldn't have had a joint paper. Me: I don't know. Alon: Well Davenport proved it several years after Cauchy died, but then he later found out that Cauchy had proved it. So you see, it's never too late to prove a theorem. We laughed a bit. Alon: So does this count as algebraic number theory? Sarnak: Not in THIS world! Skinner decided to get us back on track. SKINNER: So I feel obliged to ask you about class field theory. State your favorite version of the main theorems of class field theory. I struggled with choosing a favorite (mainly because I was trying and failing to anticipate what would come afterward), but eventually decided to go with the statements of global class field theory in terms of ideals. Me: Should I define the Artin map? Skinner: Yes. I defined the Artin map and stated the Reciprocity Law in terms of ideals. I think Sarnak made some comment about some famous mathematician, but I don't remember what it was. Eventually, he turned the floor over to Skinner again. SKINNER: Do you know the Kronecker-Weber theorem? Me: Every finite abelian extension of \mathbb Q is contained in a cyclotomic extension. SKINNER: Can you prove it from what you've written? I think I said that we just needed to show that the ray class fields of moduli of the form (m)\infty were cyclotomic extensions. I tried to indicate that I actually needed the Existence theorem, which I hadn't stated yet, but they cut me off (I never got back to stating the Existence theorem or Classification theorem). SARNAK: Can you deduce quadratic reciprocity from what you've written? (This might have been asked by Skinner, but I'm not sure. It seemed like they were often trying to ask things at the same time.) Me: Let p and q be distinct odd primes. Let L=\mathbb Q(\zeta_p). The map (\mathbb Z/p\mathbb Z)^\times \to Gal(L/\mathbb Q) given by a\mapsto \sigma_a is an isomorphism, where \sigma_a(\zeta_p)=\zeta_p^a. Let H be the image of ((\mathbb Z/p\mathbb Z)^\times)^2 under this isomorphism. Then H is the unique subgroup of Gal(L/\mathbb Q) of index 2, so L^H is the unique quadratic extension of \mathbb Q contained in L. This extension is \mathbb Q(\sqrt{p*}), where p* is p if p is 1 mod 4 and p* is -p if p is 3 mod 4. Now, (q/p)=1 if and only if \sigma_q is in H, and this occurs if and only if \sigma_q fixes \mathbb Q(\sqrt{p*}). It follows from the Reciprocity Law that \sigma_q is the Frobenius element of q in Gal(L/\mathbb Q). Choose a prime Q in L lying over the ideal (q), and form the decomposition group D(Q|(q)) (the group doesn't actually depend on the choice of the prime Q since the extension is abelian). The Frobenius element \sigma_q generates D(Q|(q)), so we find that \sigma_q fixes \mathbb Q(\sqrt{p*}) if and only if \mathbb Q(\sqrt{p*}) is contained in L^{D(Q|(q))}. This fixed field is the maximal subfield of L in which the prime ideal (q) splits completely, so \mathbb Q(\sqrt{p*}) is contained in L^{D(Q|(q))} if and only if q splits in \mathbb Q(\sqrt{p*}). This happens if and only if (p*/q)=1, so (q/p)=(p*/q). The night before the exam, I had studied the history of the development of Artin L-functions (Sarnak seems to like asking historical questions). I was almost sure this subject would arise since Sarnak and Skinner were on my committee. Strangely enough, it never did. I tried to get them to ask me something about this, but they decided we should move on to Ergodic Number Theory. ================================================================== ERGODIC NUMBER THEORY ================================================================== We started with Sarnak making fun of me for making up the phrase "ergodic number theory." I didn't really know what else to call it. I could have called it "applications of ergodic theory in number theory," but that doesn't really roll off the tongue as nicely as "ergodic number theory." Sarnak asked what I had read. I said that I had read a book of Einsiedler and Ward and that I had read Furstenberg's monograph. I didn't mention this at the time, but I had also reviewed a very recent paper of Frantzikinakis and Host concerning Sarnak's Mobius Disjointness conjecture. This turned out to be useful when Sarnak asked me about the theorem of Host, Kra, and Ziegler. Sarnak: What's in Einsiedler and Ward? Me: Some standard ergodic theory (such as the basic ergodic theorems), Weyl's polynomial equidistribution theorem (proven ergodically), Furstenberg's proof of Szemeredi's theorem, and some homogeneous dynamics (it also covers some diophantine approximation and the theory of continued fractions, but I chose not to mention that because I hadn't reviewed it). SARNAK: How did Furstenberg proved Szemeredi's theorem (Alon was also involved in deciding to ask this question)? I essentially prepared for this question by pretending I was going to give a seminar talk about Szemeredi's theorem. Below is the outline of the proof in the order that I had prepared. I think this is the most natural order in which to organize the proof. During the actual exam, I had to readjust the order because Sarnak told me to start in the middle, go back to the beginning, and then go to the end. Szemeredi's theorem states that every set of integers with positive upper Banach density contains arbitrarily long arithmetic progressions. In order to prove this, Furstenberg first proves the following. Multiple Recurrence Theorem: If T_1,\ldots,T_l are commuting measure-preserving transformations of a measure space (X,\mathcal B,\mu) and A \in \mathcal B is a set with \mu(A)>0, then there exists an integer b\geq 1 such that \mu(T_1^{-b}(A)\cap...\cap T_l^{-b}(A))>0. From the Multiple Recurrence Theorem, we can actually prove the multiple-dimensional analogue of Szemeredi's theorem quite easily. Multiple-Dimensional Szemeredi's Theorem: If S \subseteq \mathbb Z^r is a set of positive upper Banach density and F=\{u_1,...,u_l\} \subseteq \mathbb Z^r, then there exist a \in \mathbb Z^r and b\geq 1 such that a+bF \subseteq S. To deduce this last theorem from the Multiple Recurrence theorem, start by putting X=\{0,1\}^{\mathbb Z^r}. This has the structure of a compact metric space, so we can let \mathcal B denote the Borel \sigma-algebra. For u \in \mathbb Z^r, let T_u: X \to X be the "shift by u" map. Because S has positive upper Banach density, we can find a sequence (B_n) of blocks with widths tending to infinity such that |B_n\cap S|/|B_n|>\eta for all n, where \eta>0 is some fixed constant. Let \mu_n=\frac{1}{|B_n|}\sum_{u\in B_n}\delta_{T_u(1_S)}, where \delta_x denotes the Dirac measure at a point x and 1_S is the indicator function of S, which we can view as an element of X. By the Banach-Alaoglu theorem, the sequence (\mu_n) has a weak* subsequential limit \mu. If we let A=\{\omega\in X : \omega(0)=1\}, then it follows from our definition of \mu_n that \mu_n(A)>\eta. Thus, \mu(A)\geq \eta>0. This is why we need S to have positive upper Banach density. If we now apply the Multiple Recurrence theorem with the commuting transformations T_{u_1},...,T_{u_l}, then we find that there exists b\geq 1 such that \mu(T_{u_1}^{-b}(A)\cap...\cap T_{u_l}^{-b}(A))>0. Note that A is an open set and that the shift maps T_{u_i} are all continuous. Thus, T_{u_1}^{-b}(A)\cap...\cap T_{u_l}^{-b}(A) is open. The measure \mu is supported on the closure of the set of translates of 1_S, so it follows that T_a(1_S) \in T_{u_1}^{-b}(A)\cap...\cap T_{u_\ell}^{-b}(A) for some a \in \mathbb Z^r$. If we unwind the definitions, we find that this is saying precisely that a+bF \subseteq S. Now how does Furstenberg go about proving the Multiple Recurrence theorem? He starts by making the following definition. Definition: Say a system (X,\mathcal B,\mu,\Gamma) has the SZ property if \liminf_{N\to\infty}\frac{1}{N}\sum_{n=1}^N \mu(T_1^{-n}(A)\cap...\cap T_l^{-n}(A)) > 0 for all T_1,...,T_l \in \Gamma and A \in \mathcal B with \mu(A)>0. Here, \Gamma is a free abelian group of finite rank acting via measure-preserving transformations on (X,\mathcal B,\mu). Furstenberg actually proves that every measure-preserving system (with some very mild regularity conditions) has the SZ property. This is much stronger than the Multiple Recurrence theorem, but he decides to prove the stronger statement because he wants to use transfinite induction. In order to make the inductive argument work, he needs this stronger inductive hypothesis. Morally speaking, why should we expect every system to have the SZ property? Well, there are two opposite extreme types of systems. There are systems that are very rigid and predictable. A canonical example of this would be a rotation on a compact abelian group. In this type of system, the sets T_1^{-n}(A),...,T_l^{-n}(A) move around as n increases in a predictable fashion. They overlap significantly with each other at very predictable times. This leads to a significant positive contribution to the average at very regular time intervals, which leads to the positivity of the liminf. The other extreme type of system is a chaotic system that mixes everything together. In this case, the sets T_1^{-n}(A),...,T_l^{-n}(A)$become almost independent, so \mu(T_1^{-n}(A)\cap...\cap T_l^{-n}(A)) is roughly \mu(A)^l for most positive integers n. This again leads to positivity of the liminf. Now, Furstenberg's idea is to show that every system can be built up from rigid parts and chaotic parts. These parts have the SZ property for different reasons, and together they form a system that still has the SZ property. More precisely, Furstenberg defines two types of extensions of systems. The first is a compact extension. This is an extension of systems in which the extended system is very rigid relative to the base system. The other type of extension is a weak-mixing extension. This is an extension in which the extended system is very chaotic relative to the base system. He also defines a primitive extension to be an extension that is, in a precise sense, formed by combining a compact extension with a weak-mixing extension (I offered to define these terms formally, but Sarnak said I didn't need to do that). Furstenberg proves that weak-mixing extensions and compact extensions both preserve the SZ property, and he deduces that primitive extensions preserve the SZ property. He also defines how to take limits of systems and shows that a limit of systems with the SZ property has the SZ property. He is then able to show that every system (with very mild regularity conditions) can be obtained from the trivial system by a (possibly transfinite) sequence of primitive extensions and limits of extensions. The trivial system certainly has the SZ property, so he deduces that every system has the SZ property. SARNAK: What does it mean for a system to be ergodic? It's not often that you define ergodicity AFTER sketching an ergodic-theoretic proof of Szemeredi's theorem. I gave the standard definition (for a system with a single measure-preserving transformation). Sarnak said some things about how he thought Furstenberg's method was so incredible. Somehow this led him to ask the following. SARNAK: Do you know the theorem of Host, Kra, and Ziegler? Me: Yes. Suppose (X,\mathcal B,\mu,T) is an ergodic measure-preserving system. There is a factor (Z_\infty,\mathcal C,\mu_\infty,T) of (X,\mathcal B,\mu,T), called the infinite-step nilfactor, with the following properties. First, (Z_\infty,\mathcal C,\mu_\infty,T) is isomorphic to an infinite-step nilsystem. Second, if we are given any f_1,...,f_l \in L^\infty(X), then \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N \prod_{j=1}^l f_j \circ T^{nj}=\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N\prod_{j=1}^l E(f_j | Z_{\infty}) \circ T^{nj}, where the equality is in L^2(X). Here, E(f_j | Z_\infty) denotes the conditional expectation. There is actually an explicit description of the infinite-step nilfactor, but Sarnak didn't expect me to go into that (which is good because I didn't remember it). SARNAK: Can you prove that the horocycle flow is uniquely ergodic? Let's say we have a quotient of SL_2(\mathbb R) by some discrete subgroup. I asked if I could assume the quotient is compact. Then I realized that I HAD to assume the quotient is compact because the statement is false otherwise. SARNAK: So we have a lattice \Gamma in SL_2(\mathbb R), and the quotient is compact. Can you give an example? Me SL_2(\mathbb Z)! Sarnak: That's not compact! Me: Oh! Compact! Right. Well, you can take a hyperbolic quadrilateral on the upper half-plane in which the interior angles are all \pi/3 and--- Sarnak: Oh! Then you take the reflection group? Me: Yes. Sarnak: All right then! Why don't we use the Uniformization theorem? I said "okay" and started trying to show how to construct a compact Riemann surface of genus 2. Sarnak quickly stopped me. Sarnak: No, we can assume those exist! Me: Oh okay. Then just take a compact Riemann surface of genus at least 2. By the Uniformization theorem, it will be a quotient of the upper half-plane by some group of Deck transformations. Then that group of Deck transformations is the cocompact subgroup we want. Sarnak made some comment about moduli that I don't remember. Then we got back to proving the unique ergodicity. I first went to a separate board to write down all the notation I needed. I wrote X=\Gamma \ PSL_2(\mathbb R), T=\{lower triangular matrices in PSL_2(\mathbb R)\}, a_t={{e^{-t/2}, 0}, {0, e^{t/2}}, u^-(s)={{1, s}, {0, 1}}. I also let m_X be the Haar measure on X (that is, the push-forward of the Haar measure on PSL_2(\mathbb R) under the quotient map). Finally, I defined R_g: X \to X by R_g(x)=xg^{-1} for each g \in PSL_2(\mathbb R). While preparing for the exam, I didn't know how much detail I would need to know if Sarnak asked me this question. To compensate, I memorized more than I probably needed to. A couple nights before the exam, I wrote the whole proof out on a blackboard four times. When Sarnak asked me to go through it, he seemed impressed that I knew so many details. I'll probably forget many of them within the next week. Below is the sketch of the proof that I gave (although Sarnak cut me off just before the end because it was getting late). I took this from Chapter 11 of Einsiedler and Ward. Let B_r^T denote the ball in T of radius r centered at the identity. Choose \eta>0 such that the map from u^-(-[0,\eta])B_\eta^T to X given by g\mapsto yg is injective for every y in X (this is possible because X is compact). The necessity of choosing this \eta is mostly a technicality that I won't discuss. Choose f \in C(X) and x_0 \in X. Fix \epsilon>0. Since f is uniformly continuous, there exists \delta \in (0,\eta) such that |f(x)-f(y)|<\epsilon whenever d_X(x,y)<\delta. Here, d_X is a metric on X induced by a left-invariant metric on PSL_2(\mathbb R)$. We will consider x_0 u^-(-[0,\eta e^t]), the stretch of the orbit of x_0 under the horocycle flow of length \eta e^t. We want to find the average of f along this stretch. Instead, we will form a thin "tube" along this stretch and use the uniform continuity of f to say that the average of f on that tube is close to the average of f on the stretch of the horocycle orbit. Let Q_\delta=u^-(-[0,\eta]) B_\delta^T. Let B_t=R_{a_t}^{-1}(R_{a_t}(x_0)Q_\delta). The set B_t is our tube. Indeed, one can show that B_t \subseteq x_0 u^-(-[0,\eta e^t]) B_\delta^T. In other words, every element of B_t can be written in the form x_0u^-(-s)h, where s \in [0,\eta e^t] and h \in B_\delta^T. For such s and h, we have d_X(x_0u^-(-s)h, x_0u^-(-s)) \leq d_{PSL_2(\mathbb R)}(x_0u^-(-s)h, x_0u^-(-s))=d_{PSL_2(\mathbb R)}(h,I) < \delta. It follows from the choice of \delta that |f(x_0u^-(-s)h)-f(x_0u^-(-s))| < \epsilon. There is a way (discussed in Eisiedler and Ward) to decompose the Haar measure on PSL_2(\mathbb R) into two "pieces." One piece is a left Haar measure on \{u^-(s) : s \in \mathbb R\}, which is essentially the Lebesgue measure ds. The other piece is a right-invariant Haar-measure m_T^r on T. Using this decomposition, we can write \frac{1}{m_X(B_t)} \int_{B_t} f dm_X = \frac{1}{\eta e^t} \int_0^{\eta e^t} \frac{1}{m_T^r(a_t^{-1} B_\delta^T a_t)} \int_{a_t^{-1} B_\delta^T a_t} f(x_0u^-(-s)h) dm_T^r(h) ds (imagine that we are decomposing the integral over the tube into an integral in the "s direction" of an integral in the "h direction"). Using our above estimate, we find that this last integral is within \epsilon of \frac{1}{\eta e^t} \int_0^{\eta e^t} f(x_0u^-(-s)) ds. Our whole goal here is to show that this last integral approximates \frac{1}{m_X(X)} \int_X f dm_X when t is large. To do this, we argue that \frac{1}{m_X(B_t)} \int_{B_t} f dm_X approximates \frac{1}{m_X(X)} \int_X f dm_X when t is large. This is essentially because the set B_t is defined as a preimage of a set under the geodesic flow map R_{a_t} and because the action of the geodesic flow is mixing. In fact, this argument would complete the proof immediately if B_t were the preimage of a FIXED set under the geodesic flow (this is essentially what it means for an action to be mixing). However, the set R_{a_t}(x_0)Q_\delta is not fixed; it is dependent on t. We can get around this by using the fact that X is compact. Roughly speaking, the sets of the form R_{a_t}(x_0)Q_\delta all have the same "shape"; it is really the position that varies with t. With a compactness argument, one can show that all of these sets can be approximated by finitely many FIXED sets. We can then apply the mixing argument with each of these fixed sets and finish the proof with a final approximation argument. Sarnak ended by saying some things about Ratner's theorem, but I don't remember anything he said. ================================================================== AFTERMATH ================================================================== They kicked me out. I went into the hall and jumped around because I was so happy to have this behind me. It felt like they kept me waiting for ten minutes, but maybe it wasn't that long. They eventually opened the door and told me I passed. The whole exam lasted about 2.5 hours. It was actually really fun. I don't have too much advice that differs significantly from the advice given in the other past general exams. My strategy for studying for the standard topics was to go through every standard topic question on this website. I did the same for algebraic number theory. For ergodic number theory, I read Einsiedler and Ward along with Furstenberg's monograph. I had to guess what I thought Sarnak would ask in this area. This wasn't too hard since I knew roughly what he found interesting within ergodic theory (such as homogeneous dynamics). Good luck to all the students who have yet to pass their generals. Don't worry too much about it. Have fun with it!