Levent’s generals
Committee: Bhargava, Sarnak, Dafermos (henceforth B, S, D, respectively)
Special topics: analytic number theory and representation theory
May 9th, 2016, Sarnak’s office
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unfortunately i have an extremely bad memory but i think i remembered most of the stuff below, anyway here we go!
S sits on Weyl’s ‘book’ (a volume in his collected works maybe?) and shows it to B: “should we start with Weyl stuff?”
me, trying to save myself: “am i allowed a preference? how about...complex analysis? :)”
Complex Analysis
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S: state the Riemann mapping theorem
S: prove it
i cheekily asked if they wanted Riemann’s proof or a rigorous (i.e. Koebe’s) proof
it bit me in the behind:
S: “sure, do Riemann, and then we’ll have you make it rigorous”
so i laughed and gave Koebe’s argument cause i didn’t want to start solving the Dirichlet problem... (reduce to a bounded domain, define the usual normal family, extract a subsequence whose derivatives at 0 converge to the sup, ...)
S: what are you using? (i wrote Montel, but i never remember it correctly, so i just did the precompactness directly by Arzela-Ascoli (which is how Montel goes))
S: alright let’s get back to the Dirichlet problem. draw a blob.
i wrote \Omega
S: draw something!!
me: (thinking im gonna have to write down an explicit Riemann map) can it be polygonal...? [Schwarz-Christoffel]
S: just draw a blob. (i drew a blob)
S: suppose i have a C^\infty function on the boundary (“can i assume the boundary is C^\infty?” “yeah sure”). what can you say?
me: by Riemann i reduce to the case of the disk and i can just solve the problem explicitly there
S: you’re using something --- what are you using?
me: Riemann mapping...??
i wrote down \Delta u = f, precomposed u with g, and soon enough they pointed out that i probably didn’t want to be solving that equation --- of course i meant \Delta u = 0, u\vert_{\partial \Omega} = f!
i replaced \Delta = \d \dbar, and used that \dbar g = 0, and then i got confused because the chain rule is a funny thing
S: what are you using? what is this function u you’ve written down called? (harmonic) what can you say about harmonic functions? (ok, they’re the real part of holomorphic functions and that property transfers) [i realize now the mistake that was confusing me is that \dbar does *not* satisfy the usual chain rule --- obviously the notion of harmonic function should behave well under a conformal isomorphism, but to show this i was imagining using the chain rule naively for \d and \dbar to show that u\circ f is harmonic if u is and f is holomorphic, and strangely that seemed to only require that f be holomorphic, hence my confusion... but yes you can show the claim formally and easily without using that harmonic functions are real parts of holomorphic functions by just computing \d\dbar(u\circ f). anyway]
S: so now do it on the disk. (just expand in a Fourier series on the circle and replace the positive powers by powers of z and the negative powers by powers of \zbar)
S: ok fine --- what are you really doing? (i wrote down the Poisson formula)
S: is the plane conformally equivalent to the disk? (no) why? (Liouville) prove it (write down the Cauchy formula for the derivative and let the radius go to \infty)
S: alright what about doubly connected regions? (they’re annuli/punctured disk)
S: prove it [i invoked uniformization, took the generator of the deck group (= \Z) and noted that it’s not conjugate inside SO(2) cause it is of infinite order and its powers act properly discontinuously, hence it’s conjugate to (1 x; 0 1) or (t 0; 0 t^{-1}), and then you can explicitly write down isomorphisms of the disk mod each of those guys to annuli. they yelled at me the whole time cause they didn’t like the proof and apparently i kept saying disk when i meant annulus. anyway i like the proof :P]
S: no!!! ok fine, maybe this proof works. whatever. when are two annuli conformally equivalent (i’d written A(1, r), so iff r = r’, note that r can be infinite in my notation (that’s the punctured disk)). prove it (i pointed to the deck group elements: their quotients are annuli with distinct radii, and if you have an isomorphism of the annuli it lifts to an element of PSL_2(\R) that then conjugates the two deck automorphisms to each other, QED)
S: no! gimme another proof (ok, solve the Dirichlet problem on the annulus where you specify boundary values 1 on the outer circle and 0 on the inner circle. the solution on one is log|z|/log r, the solution on the other is log|z|/log r’, and they have to go to each other (once you get the outer circles to go to each other rather than switch with the inner circles), QED).
S: fine but. no!!! the plane is the important player here. make a function on the plane. (use Schwarz reflection to extend the function by reflecting through each of the circles. doing this infinitely many times gives you an automorphism of \C cause it’s clearly injective and then clearly (since otherwise you’d have a bounded entire function) surjective. the automorphisms of \C are just az + b, but you arrange it to fix 0, so it’s scaling, QED)
S: what about multiply connected domains? what if i removed two circles? (i drew a disk minus a circle about 0 minus a disk about a positive real). that’s a deep theorem due to Koebe, who by the way was the first to prove Riemann mapping and used these sorts of methods. (i had seen a theorem of Schramm’s but i think that was new only when the boundary is extremely horrible maybe. anyway the general answer is circular domains in the disk, or parallel slit domains in the plane, which is i think the more classical result)
S: ok, you guys ask him something. it can still be complex analysis, something with more formulas maybe
B: how do you evaluate the sum of the reciprocals of the squares? (write down the Hadamard product for sin and take the z^3 coefficients)
D: let’s get back to that Dirichlet problem. how do you solve it? (i muttered something, i don’t remember what [how you solve it is you produce a weak solution and then use precompactness in the same way to get a strong one, i think --- there is a prettier way using Brownian motion but i didn't bring that up for obvious reasons])
D: what would you do in \R^n? (the key point is that the Poisson kernel gives the fundamental solution, i.e. \Delta_z P(z,\zeta) = \delta_\zeta. i know the Green's functions in \R^n but i dunno about in general!)
S: draw a semicircle. solve the Dirichlet problem where i specify boundary values to be 1 on the top part and 0 on the straight part. (use the isomorphism with the disk and the upper half plane to find that your region is the upper right quadrant, then just use the argument function. or draw lines to -1 and 1 from your point and measure the angle between the two lines). ok fine. (B and i laugh cause S asks this like every time) so why did i ask that question? (i try to read S’s mind and fail) suppose i have a holomorphic function bounded by 5 on the top and 2 on the bottom, how do i bound it? (2 + 3*that previous function, by writing down the Poisson formula)
Real Analysis
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D: ok let’s get to real analysis. this guy went to Harvard, he probably doesn’t even know what a measurable function is (the joke is that he and B also went to Harvard!) (“it’s a function that pulls back open sets to measurable sets”, not getting the joke)
S: what is Lebesgue measure? (defined it)
S: what is a measurable set? (Caratheodory def, told them i could also use Riesz)
S: what is a Borel set? (take the closure of the open sets under complements, countable unions and intersections)
S: are there more measurable sets than Borel sets? give an example of one (yes, the point is any subset of a measure zero set is automatically measurable. so just take a section of the projection of a bounded positive measure set A to it mod translation by rationals in [0,1] (via AC), claim: that’s a nonmeasurable set. proof: the union of that guy translated by rationals contains A, hence one, hence all of the translates have positive measure, hence the (disjoint) union has infinite measure, but it’s bounded, qed claim. now use x\mapsto x + the Cantor function(x) to pull back a nonmeasurable subset of the image of the Cantor set into the Cantor set: it’s a homeomorphism so preserves Borelness, so since the original guy wasn’t even measurable the pullback can’t be Borel, but it’s measurable since the Cantor set has measure 0)
S: you used AC. can you do it without AC? (no) that’s a theorem of Solovay! (Solovay!) no Solovay! (aand i’ve already forgotten how to properly pronounce it)
S: which L^p is a Hilbert space? (L^2, the others aren’t even their own duals) you mentioned Riesz before. which Riesz? (well there’s the easy one with (L^p)^* = L^q when 1/p + 1/q = 1 and 1\leq p < \infty, the one from before was the other one) what’s it say? (bounded linear functionals on continuous functions on a compact space are just integration against a signed measure, and the ones that take nonnegative values on squares are integration against a measure) (we argued about whether measures were signed or actual measures should be called ‘positive measures’)
B: what is a Hilbert space? (complete inner product space)
S: let’s focus on (L^p)^* for a moment. what about the case p=1? ((L^1)* = L^\infty). what about (L^\infty)^*? (oh man...it certainly contains L^1 but that isn’t the full thing) what can you say about a double dual? (it contains a copy of the space) is the dual of a Banach space Banach? (yes...) ok anyway what is an explicit element that isn’t in the dual? (well i could evaluate --- no, these are just L^\infty guys...uhhh)
S: what does Hahn-Banach say? (W\subseteq V, V Banach implies V^*\to W^* surjective)
S: no, no, write it how i know it (for \phi: W\to \R, \exists \tilde{\phi}: V\to \R such that \tilde{\phi}\vert_W = \phi, “and often one wants to know” (i think this is what he was looking for) ||\tilde{\phi}||\leq ||\phi|| [something seems fishy about what i wrote but they were happy]).
S then explained the history of the problem, with somebody whose name i forgot proving that there were finitely additive measures that weren’t countably additive, thus integrating against them wouldn’t come from L^1 but would be in (L^\infty)^*. [apparently this leads to the Ruziewicz problem, and Drinfeld resolved it for S^2 and S^3 using...Ramanujan!!!!]
D: prove completeness of L^p. “can i do it for L^1?” “yeah fine” (take a super quickly converging Cauchy sequence f_n with ||f_m - f_n||_1 < 2^{-\min(m,n)}, this converges pointwise to an L^1 function cause you can write down sets where it isn’t Cauchy and use Borel-Cantelli a few times to show it’s a.s. Cauchy hence a.s. convergent, i confused myself with how many parameters i needed, it worked in the end, then use dominated convergence. to use dominated convergence i wrote down \sum_{n=1}^N f_{n+1} - f_n, and then realized i could have just said from the start that that converged pointwise to my function and since the L^1 norms are summable the sum of the absolute values gives you the upper bound L^1 function for dominated convergence to apply, anyway QED x2)
D: you’re missing one step. you extracted a subsequence, why’s the full Cauchy sequence converge to your function? (take a limit in ||f_m - f_n||_1 < \eps)
S: wait wait wait --- what is a Cauchy sequence? (??!?: \forall \eps \exists N \forall m,n > N ||f_m - f_n|| < \eps). ok fine yeah
S: alright, let’s turn the tables on this guy. what courses did you take at Cambridge? (my life flashes before my eyes...suffice it to say i was a little lazy while there...) (“analytic number theory...functional analysis...percolation theory...combinatorics...brownian motion......” [do i remember anything from those? basically no]) alright alright alright (i live to fight another day)
Algebra
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B: what can you say if the automorphism group of a finite group is cyclic? (it’s immediately cyclic! no wait there can be a centre. ok well you get that G/Z(G) is cyclic) why? (it’s a subgroup of a cyclic group, via inner automorphisms). can you prove G is abelian? (uhhh...) write it out (let \phi map to a generator, then for every g\in G there is a k such that g\phi^{-k}\in Z(G)). no more explicitly! (oh duh, every element is of the form \phi^k z, z\in Z(G), and they clearly then all commute! amazing!)
B: ok, so G is abelian. now what? (by the classification G = \oplus_p \oplus_i \Z/p^{e_i}......i become afraid of automorphisms mixing everything up.)
B: what do you think follows about the i's? (there's almost certainly no multiplicity...maybe lemme write G = A\times B, A and B with coprime orders)
S: hold on, let’s do explicit examples. \Z/p (the automorphism group is noncanonically \Z/(p-1))
B: prove it (classification means it suffices to rule out a \Z/q x \Z/q, but those are all solutions of x^q - 1 = 0, and there are at most q of those cause \F_p is a field).
B: what about \Z/p x \Z/p? (the auts are \GL_2(\F_p), which is noncommutative)
B: what about in general? (i said i was still afraid of a \Z/p^e mixing with \Z/p^f). ok let’s do prime powers. (it’s noncommutative cause the two noncommuting elements of (\Z/p)^2, namely (1 1; 0 1) and (1 0; 1 1) generalize: i wrote down (x,y)\mapsto (x+y,y) and (x,y)\mapsto (x,p^{f-e} x + y), where let’s say f\geq e, and we’re in \Z/p^e x \Z/p^f)
B: ok, so there’s no multiplicity. now classify all groups with cyclic automorphism groups. (they’re cyclic by the above!) no classify them! (oh! ok this is a problem Gauss studied. (\Z/n)^\times is cyclic iff n = 4, p^a, 2p^a, p an odd prime, a\geq 0)
B: why do those have cyclic automorphism groups? (Chinese remainder theorem plus (\Z/p^e)^\times = (\Z/(p-1)) x (\Z/p^{e-1}), and those are coprime)
B: can you prove that theorem? (factor n and use the Chinese remainer theorem, and then observe that 2 divides \phi(p^e) so you have to have at most one odd prime in there, and also the Klein four group is inside (\Z/2^a)^\times once a\geq 3).
B: can you construct towers of automorphism groups where each guy is cyclic? (well they’re always finite cause the order is decreasing, and idk you probably can’t get arbitrary length. i mean the p-1 factor would have to be trivial...wait what about 5...now im confused)
B: what’s an example where the p-1 factor is trivial? (oh man! 3^k! ok so you can get arbitrary length) or 2*3^k! (yeah! those are just auts of the previous guys)
B: anything else? (uh... i’d need a lot of primes in very special positions...) well you could have primes of the form 2*3^k + 1 or 2p = 3^k + 1, for example. it’s an unpublished theorem of Conway that these are all the infinite families of examples, and there are a few sporadic examples! like here’s something interesting about 94: do it for \Z/94! (i get really scared cause i factor it as 2*47 and then remember the famous Grothendieck story... then 46 = 2*23, then 22 = 2*11, then 10 = 2*5, then 4 = 2*2, then 2, then 1!!!!! all cyclic!!!!!)
B: classify groups of order 94. (47-Sylow is normal) nah don’t use Sylow (ok, Cayley finds you a 47-cycle, whose subgroup is index 2, hence normal) why? (write down the cosets and check that the nontrivial coset rep preserves H cause left and right cosets are the same). (so you get a direct product or a dihedral group)
B: can you produce a noncommutative Galois group of order 20? (x^5 - 2) what about commutative? (\Q(\zeta_{44})
B: hm maybe we should do some rings... can you realize the automorphism group as units of some ring? (i freaked out cause i thought the group was nonabelian and wrote down \C[G] and stared at it and prayed they’d stop me... they shook their heads) no this is an abelian group! you had the idea before. (???) what about (\Z/p)^2? (GL_2(\F_p)). is that units in something? (oh duh! \End_\Z(G)^\times). ok what if G isn’t abelian but its automorphism group is? can you get them as units in a commutative ring? (uhhh...i guess the natural thing to do would be to take the units in Z(\End(G))...) ah yeah i guess [don’t think what i said was correct, phew], or you could just take the ring generated by the automorphisms! (oh man that’s much smarter)
D: define the trace, the Harvard way. (this was the best moment in the whole exam. B, D, and i started cracking up, S had no idea what’s going on, and B explained to him that D took Mochizuki’s math 55 class [probably the most legendary class ever taught in the math department] at Harvard while B was the TA at the beginning. what’s great is that this is how i first learned about the trace too...) (V^*\otimes V\to End(V) is an iso for V f.d., and there’s an evident map of the left-hand side to the base field k via evaluation on simple tensors, and that’s it)
S: when can i solve e^A = B? (in complex matrices?) yeah (always when B is invertible, write down the Jordan normal form) ok, ok. what about over the reals? (not always, e.g. (-1 1; 0 -1). you need an even number of negative eigenvalue Jordan blocks)
B: what’s the Galois group of a random polynomial (S_n). how do you prove it? (write down...what were they called? the guys who have a root in your base field iff the Galois group is in a subgroup, e.g. the discriminant being a square iff Gal is in A_n ... not resultants) resolvents! (yeah! ok i’d write down resolvents and observe that by Hilbert irreducibility they don’t have roots in the base field generically)
S: what’s Hilbert irreducibility? (stated it) what do you mean by generic? (i said in terms of height, i.e. counting, and then said probably in the alg. geom. sense too) no! it’s not true if you use the alg. geom. sense (i really don’t know this stuff very well so anyway)
B: what about more explicitly? what if e.g. the degree were prime? (well the general method is to reduce mod p and look for a p-cycle for various p, i.e. see if the polynomial stays irreducible mod p, or you can reduce mod primes dividing the discriminant and use the fact that inertia groups generate the Galois group cause \Q has no unramified extensions)
B: yeah, that works. should we start on the topics?
S: alright let’s take a 5 minute break.
i drank the tea i brought in with me and we stood around awkwardly for a few minutes while B and S discussed Rivin’s new result on the error term for the number of polynomials that don’t have S_n Galois group
Representation Theory [of compact Lie groups]
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B: how do you get from a Lie algebra to a Lie group? (embed the thing as a subalgebra of gl_n and then exponentiate)
B: what’s the exponential map? (g\to G, i start writing down the differential equation)
S: what’s a Lie group to you? what’s a Lie algebra? what’s the bracket of two tangent vectors? (extend to left-invariant guys, view them as differential operators, take the usual bracket)
D: what about more geometrically? (you can do it in charts but that’s not very geometric... integrate the first guy for time \eps, then go along the second for time \eps, and compare the values to the other way around, divide by \eps, take \eps\to 0. i got very afraid because i am afraid of Lie derivatives and anyway)
B: ok so what’s the exponential map? (wrote down the differential equation) can you do this in terms of 1-parameter subgroups? (wrote down what such a thing is) what’s exp(tX) in terms of it? (f(t)). ok what’s the exponential map from \R\to \R? (the identity) how do you see this in terms of 1-parameter subgroups? (well the subgroup is just t\mapsto a*t). what about on \R^\times? (actually exp, cause it’s also known as GL_1, and it’s exp on GL_n) that’s why it’s called the exponential map. what’s that got to do with S’s e^A = B question? (the image is the full connected component of the identity, GL_n(\C) is connected, GL_n(\R) isn’t, since e.g. the determinant takes on positive and negative values).
S: alright. give me a representation of a compact Lie group (canonical answer...: the trivial rep). ok nontrivial (G\to U(L^2(G)), write down an invariant convolution operator, use the spectral theorem to get f.d. eigenspaces, and there are your reps!). why’s the operator compact? (Arzela-Ascoli again!) (everyone laughed) (i stated the spectral theorem for compact operators.) do you know how to show the operator has spectrum? (i wrote down the resolvent kernel and used Liouville) you’re using something here, the operator has to be self-adjoint (no, it’s ok if it’s normal) yeah but you’re using something here. [this just shows it has some spectrum but it could all be at zero. i think the point is the spectral radius is the norm] you know, we don’t teach this stuff here anymore! you should use the analytic way of doing it for f.d.v.s (oh! via Rayleigh quotients, ok fine) and a variational argument.
S: what is L^2 here? (i say take a metric and average it, not realizing...) write it down explicitly! (\int h(g_* X, g_* Y) d---wait...crap. ok fine Haar measure exists) you need Haar measure! (somehow i always have the knee jerk reaction to average, cause that’s how you do it for finite groups, but of course you have to know how to average first...)
S: anyway, so does this give all irreps? how do you know an irrep is finite dimensional? (let’s say you have a unitary rep, let w be a unit vector, by Schur, the invariant operator v\mapsto \int dg gw ---) i like this guy! like the physicists! he writes his dg first!
B: and the brackets! (i hate the (--,--) notation)
(--- anyway that operator by Schur is scaling by some \lambda. maybe it’s 0! nah, it’s not 0. evaluate on w and pair with w to get \lambda = \int dg ||^2, the integrand is continuous and takes the value 1 at e, hence is positive in a nbhd of e, hence the integral is positive, so \lambda > 0. consequently 1/\lambda \int dg gw is the identity. but each of those operators is trace class! QED)
S: alright, good for you. now write down SL_n(\C). what about representations of those? how do you know they’re finite dimensional? (complex analytic reps are in bijection with reps of the maximal compact, SU(n). i complained about the name “unitarian”. it’s the unitarization trick!! not some religious sect!! S thought at first that i was talking about the term 'trick', but i like tricks)
S: so what are the finite dimensional reps? (Schur-Weyl duality --- apply Schur functors to \C^n w.r.t. partitions with at most n-1 parts (n-1 cause it’s SL_n not GL_n so \wedge^n \C^n is trivial) to get all irreps)
S: alright what about L^2? why do they all occur in L^2? (i got super confused. in the finite case you use...brain stopped working...oh yeah complete reducibility!) with what multiplicity? (the dimension) this is *the* fundamental object!
S: does SL_2(\R) have any finite dimensional unitary reps? (no, (1 x; 0 1) are all conjugate, hence all go to id by taking a limit, same for lower triangular guys, QED) there’s the trivial rep! (ah crap yeah that too)
S: i assume you did many examples. [just before the exam i completely worked out everything for su, sp, so, just so i’d know all of everything off the top of my head] alright you mentioned Weyl in there, tell us about the Weyl character formula. (G compact Lie, T\subseteq G a maximal torus) define it (defined, W = NT/T, roots, coroots, weight lattice, ordering, Weyl vector, dominant weights, det w, bam the Weyl character formula in all its glory). they laughed cause i remembered it all off the top of my head. lol no i studied like a madman
S: you mentioned maximal tori in there. prove that any two are conjugate. everyone thinks he or she has the best proof of this, this is where everyone’s ego lies. i assume you’ve studied this? (yes, and i have the best proof of this!!! use Lefschetz not on G/T, but on G/NT! now the rational cohomology is 1, but you don’t need to think so hard. Lefschetz says you can calculate the alternating sum w.r.t. any element, so choose a dense generator of the torus, it explicitly now has *one* fixed point, so the sum is \pm 1. indeed if ghNT = hNT, then h^{-1}gh\in NT, so h^{-1}Th\subseteq NT, but T is the connected component of the identity in NT (which is why W is finite), so h conjugates T to T, hence h\in NT! now apply this to a dense generator of the other torus, QED no calculating signs! [Snaith gave an amazing proof of Brauer induction like this too. if you have a map G\to U(n), it suffices to find a point in U(n)/NT fixed by the whole group. if there is one your rep was the sum of Ind(abelian chars), QED. if there isn’t one then triangulate G/NT in such a way as to have simplices mapped to simplices. by the Euler characteristic computation (i.e. using Lefschetz) since the alternating sum of the traces of action of any g on chain complexes (= the same thing on cohomology) is the same for any g it’s then 1 since we saw it was for a dense generator of T (you can get the sign right). hence as a virtual rep it’s the trivial rep. but the rep acting on i-chains is just a sum of Ind_H^G(1)’s for some subgroups H. if there were a simplex fixed by the full group (i.e. some H = G) then you could retriangulate and eventually you would get the simplex to not be fixed, else you’d get a convergent sequence of fixed simplices converging to a pt, whence a fixed pt, contra. hence all the H\neq G, and you’ve written 1 = alternating sum of Ind_{H_i}^G(1)’s, multiply by your character \chi and use Frobenius reciprocity to get Ind_{H_i}^G(\chi\vert_{H_i})’s, and you’ve reduced it to proper subgroups H_i so you’re done by induction!!])
[sorry for that long aside]
B: alright let’s do some examples. what about sl_3? (i wrote down the Lie alg., the Cartan, the roots, drew the picture of the \alpha_i at the 3rd roots of unity and the usual hexagon for the roots (it wasn’t so awful a hexagon, but it wasn’t pretty either)) what’s the Weyl group? (S_3) what’s Sym^2 of the standard rep? (it has highest weight \alpha_1 + \alpha_2) so what is it? is it irreducible? (so confused cause that highest weight suggests it’s the dual rep and i mixed up a factor of 2 in the dimension) (oh crap that’s the highest weight of \wedge^2 std! Sym^2 std has 2\alpha_1. it’s 6 dimensional, and ok). what about Sym^2 Sym^2 std? (i kept mixing up the dimension counts, but anyway that is 21 dimensional). what rep does it contain? (Sym^4, which is 15 dimensional). any guesses as to the rest? (probably Sym^2 std?) actually it’s Sym^2 std^*! what is the corresponding map? what can you do to a 3x3 symmetric matrix as a quadratic function of the matrix to get another 3x3 symm mat? (take \wedge^2!) more explicitly (take \wedge^2...?? ie take 2x2 minors...) yeah! that’s called the adjoint (oh god, of course. wedging with \wedge^2 gives you the det, same thing). yep
B: what’s the weight lattice? (\Z[\alpha_1, \alpha_2, \alpha_3]). draw the fundamental Weyl chamber (the angle has to be 60 degrees cause there are 6 elements of the Weyl group...so i drew two lines and by then i got scared my brain wasn’t working but it was ok). yep! what are the weights for the standard rep? (the \alpha_i, and the \alpha_i - \alpha_j are the ones for the adjoint rep).
Analytic Number Theory
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S: give me a lower bound for L(1,\chi). (i could give you the ‘trivial’ lower bound... (i.e. via h(d)\geq 1)) to B: this is how you can tell if someone knows their stuff
(ok fine, proved Siegel’s theorem)
S stopped me right before the crescendo: tell me what's really at the heart of the argument. (nonnegativity of a biquadratic Dedekind zeta function + repulsion, it's always nonnegativity that allows us to prove anything). why is it not effective? (you assume there are Siegel zeroes in one case) alright fine fine
S: let’s see if you actually know what’s in the books i told you to read. what else was in Davenport? (i laughed cause i’d told him earlier in the exam that i didn’t really like the other Davenport [Waring’s problem is kinda uninspiring, no?] and im very bad with books in general) tell me about twin primes. (wait twin primes? well there’s a parity obstruction) fine that but let’s do two primes --- tell him (ie Dafermos) about Goldbach! (i explain the circle method. what works for three primes is you can do L^\infty L^2 and you drop the fact that you’re on the minor arcs for L^2 and just need a nontrivial bound in L^\infty on the minor arcs, and now you can use Vaughan and Cauchy-Schwarz like a madman) what fails for two primes? (if you insert absolute values you only can use L^2 and that’s larger than the main term. the real reason is ---) well ok, what about trying to use the L^1 norm? somebody had a failed argument that tried to say the L^1 norm is bounded by a power of log once. what can you say? (the L^1 norm is lower bounded by N^{1/2}/(maybe some powers of logs)) oh man, you know that?? (...i saw it asked on previous generals...) so how do you prove it? (just calculate on the major arcs, QED) yep. (the real issue with Goldbach is that maybe you’ll next try (L^\infty)^\delta L^{2-\delta}. but the problem ends up being you need some power savings in L^\infty, and to get something like that you’d need Siegel-Walfisz up to a power of N, not a power of log! and then suddenly you need to get rid of Siegel zeroes, not gonna happen.) [nb: this isn’t the real reason --- i always get exponents (so, addition) mixed up in my head: i thought that if you did (L^\infty)^\delta L^{2-\delta}, used Holder on the latter and prayed for a power savings on the former you would be ok with less than what you need for L^\infty L^1 (namely square-root cancellation), but nope! Goldbach is a strange beast]
B: what about a more recent result? (like bounded gaps?) yeah sure, like that, or how about writing integers as sums of powers? how many integers \leq N do you expect to be able to write as x^d + y^d, for example? (N^{2/d}, let’s say x and y are positive). yeah, but something goes wrong for d=2, what goes wrong? (there are congruence obstructions)
S: let’s see about this guy’s intuition. how many numbers up to X are sums of two squares? (X/\sqrt{\log{X}}) nice! (i must have thought about this a million times in my life...but yeah let’s call it intuition)
S,B: yeah alright alright let’s stop
B: oh yeah, we have to deliberate (i kick myself out)
they sit around for a bit and then come out and wahoo!
D: tebrikler!(!!!)
time: 2-4:45
notes: to prep i literally did every single previous general topics/analytic nt/rep theory problem on this website, i can guarantee a) it works b) it’s far, far too much (but i learn better from doing problems than reading, so i enjoyed it) c) you’ll learn a lot!!!
good luck!!!