GG.

An angel came down for a meeting of the American Philosophical Association. Greeting the assembled philosophers, the angel offered to answer a single question for them. Immediately the philosophers set to arguing about what they should ask. So the angel said, “Alright, you figure out what you want to ask. I’ll come back tomorrow.” And he left the philosophers to deliberate.

Some of the philosophers favored asking conjunctive questions, but others argued persuasively that the angel probably wouldn’t count this as a single question. One philosopher wanted to ask “What is the best question to ask?”, in the hope that some day another angel might make a similar offer, at which point they could then ask the best question. But this suggestion was rejected by those who feared that no such opportunity would arise and did not want to waste their only question.

Finally, the philosophers agreed on the following question: “What is the ordered pair whose first member is the best question to ask, and whose second member is the answer to that question?” Satisfied with their decision, the philosophers awaited the angel’s return the next day, whereupon they posed their question. And the angel replied: “It is the ordered pair whose first member is the question you just asked, and whose second member is the answer I am now giving.” And then he disappeared.

Real world

Today, I went to the real world. This took over an hour. Upon arriving there, superfluous information. I am overwhelmed. Confused. I wander aimlessly for an eternity. Entering a store, I manage to locate an original, aesthetically appealing black object. However, one who belonged to the real world was quick to inform me that the object was not for me. I found this dubious. Hopf algebras. Quantum groups. Symmetric functions. Representations of semisimple Lie algebras. Weight. Unbearable weight of being. My shoulders hurt. I replaced the object. Then, I left that store. Then, I left the real world. This took over an hour. Goodbye, real world. I do not know when I shall return.

Not moving

Never mind. I’m not moving; just not ready yet. I will no longer bother trying to move until I feel the time is absolutely correct.

I will write a post soon. Until then, here are some good Epik High songs. For a kid who grew up hating rap, their music sure does resonate with me a lot… I’m realizing more and more that all the things I thought I disliked just needed to be given a chance. I can’t even overstate how much better life becomes when you make an effort to find good qualities about everything, instead of dismissing large amounts of it. Don’t miss out. Good night.

Bounded gaps seminar

From Stanley Xiao:

I am looking to organize a series of learning seminars on the topic of bounded gaps between primes. In particular we will focus on the paper of Yitang Zhang and the subsequent paper by the polymath project headed by Terence Tao. We will also look at the preliminary work of Bombieri, Friedlander, and Iwaniec, as well as the results of Goldston, Pintz, and Yildirim.

I would like to hold an organizational meeting on Wednesday, January 15th at 2:30 PM (time is subject to change), at MC 5046.

At the meeting we will discuss:
- How often will talks be given, and on what subject; and
- Who will contribute talks (the former will depend on the latter).

For undergraduates, they should have read some books or took some courses on analytic number theory and sieve theory before attending. I also assume that they are familiar with rudimentary material like real analysis and undergraduate algebra.

Hell

is more or less how I would describe my life this term. I almost never stop doing assignments. I really need to at least write a statement of purpose so that I’m ready to apply to grad schools, but I haven’t even had the chance to give that any thought lately. Before I give my impressions of courses so far, I want to ask for your opinions. Next Winter is my last term as an undergraduate, and I’m wondering which courses I should take (of the ridiculous number that are listed on my “Plan” page). I feel mainly interested in algebraic geometry and functional analysis, and I probably have a decent enough analysis background already (for an undergrad), so I was thinking about focusing on geometry, with something like this:

• Riemann surfaces
• Index theorems (I will have all the prerequisites for this, except algebraic topology)
• Commutative algebra or algebraic topology (which one?)
• 2 CO courses: nonlinear and combinatorial designs (to finish CO major requirements)
• Some lame non-math course (since I need 0.25 more non-math units… sigh)

(If you want to see the official course codes/titles, again, check the “Plan” page). Thoughts?

Here’s a summary of how I’m finding my courses this term. I’m also sitting in on PMATH 955, but I won’t talk about that here. I ordered them by ascending difficulty/time commitment.

PMATH 445 (Representations of Finite Groups): So far, this course has been really easy. I never thought I would use that word to describe a 4th year PMATH course. The assignments are doable in a couple of hours tops, and the material seems pretty standard. Initially I thought it was just because I had experience with representation theory from when I took Lie Groups, but it seems like almost everyone is finding the course easy going. I can’t complain since I honestly don’t think I would have the time to throw at this course if it were much more demanding. It’s my first class at 9:30am, and notes are posted online, so admittedly I haven’t been attending very regularly (things became hectic in the past week or two).

PMATH 900 (Valued Fields): This course was also easy going in the beginning. It’s a bit more technical now, but still rather palatable since all the objects involved are just fields and certain kinds of rings, and other stuff we all know and love. There’s no highly sophisticated machinery to deal with (I’ll save that for last). The first assignment was very reasonable, but not something I would want to attempt doing in a single day. Algebraic cleverness comes in short bursts for me.

PMATH 465 (Riemannian Geometry/”Diff Geo 2″): Just like representation theory, the material here is pretty tame, or maybe it just seems that way since (having been through 753, 763, 441, 365, and just about every other course) I’m used to seeing linear algebra everywhere. The concepts seem natural enough to me, although I feel kind of uncomfortable when I have to get my hands dirty and think about solutions of ODEs. I don’t really know the first thing about differential equations. Lectures aren’t hard to follow for the most part, but I just zone out when he starts grinding through horrific tensor calculations, mainly because I have to focus so much on TeXing them, which is always a pain. Assignments are long and routinely absorb my weekends, including the current one. :(

CO 430 (Algebraic Enumeration): This is a serious course on enumeration. The operations on species are really cleverly chosen to make the generating functions behave as you’d expect, and they allow you to write down extremely concise formulas that capture some pretty nontrivial ideas. Even if I’m dealing with an equation involving 3 species, it can take me a few minutes to unwrap the definition and grok what it’s actually trying to say. Lectures aren’t the easiest to follow, and the assignments take me a while to solve and write up (he could probably make them significantly more difficult while staying within the margins of reason, though).

PMATH 822 (Operator Spaces): I don’t even know where to start. A lot of people who took Lie Groups with me last Winter would describe it as the hardest course they ever took. This course is at least 5 times more insane. In addition to the disfigured tensor products everywhere, there are now von Neumann algebras flying around and we never even defined them. The other students (who are mostly PhD students) seem way better than I am at analysis, and math in general. I’m not blaming myself for this, after all I’m just an undergrad; hopefully I will be able to reach a similar level in a couple of years. To be honest, it feels like the lectures are aimed at the PhD students in analysis, and my background in functional analysis just feels inadequate (I also have no measure theory background beyond PMATH 450). Keep in mind that this is coming from someone who generally wrote perfect assignments in Functional Analysis and meticulously checked every detail. It seems like we are also expected to have a lot of time to spend thinking about the material outside of class, which is understandable since grad students usually only take two courses. Unfortunately, I have to deal with five. I can’t even emphasize this enough; if you can’t do functional analysis upside-down blindfolded while reciting the alphabet backwards in your sleep UNDERWATER, you probably barely have a chance of grokking this material during the lecture. I started the first assignment fairly late, and proceeded to sink an enormous amount of time into it. I still didn’t completely finish. I should have known better; operator theory is notoriously subtle. I did feel like I could solve the remaining things with a couple more days, but even so, there was no room in my schedule to continue working on it past today. :(

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Fibonacci and gold

Most people have heard either about the golden ratio

$\varphi := \frac{1}{2}(1 + \sqrt{5}) \approx 1.618\ldots$

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ….

They are intimately related, and I could write several enormous posts enumerating all of their amazing properties.

The golden ratio $\varphi$ is an irrational number which satisfies the polynomial equation $x^2 - x - 1 = 0$, that is, we have $\varphi^2 - \varphi - 1 = 0$. In fact, since this is the lowest-degree (hence “simplest”) polynomial annihilating $\varphi$, we refer to it as the minimal polynomial of $\varphi$. Many surprising facts can be derived from this innocuous-looking relation. For example, $\varphi^2 = 1 +\varphi$ immediately yields that $\varphi = \sqrt{1 + \varphi}$, from which we get the “continued surd” expression

$\varphi = \sqrt{1 + \sqrt{1 + \sqrt{\ldots}}}$

Stated differently, $\varphi$ is a fixed point of the mapping $x \mapsto \sqrt{1 + x}$. When you were a kid, if you were bored with a calculator, maybe you had the idea of starting with some number, adding 1 to it and taking its square root, and then repeating this process ad nauseum. If you did that, you would have found that eventually the numbers on your calculator stop changing at exactly the value $\varphi \approx 1.618 \ldots$.

For something else, take our original equation $\varphi^2 = \varphi + 1$, and divide through by $\varphi$ to obtain $\varphi = 1 + \frac{1}{\varphi}$. This tells us $\varphi$ is also fixed by the map $x \mapsto 1 + \frac{1}{x}$. It follows immediately that the so-called “continued fraction expansion” of $\varphi$, which (in a precise sense) provides the data of the “best rational approximants” to $\varphi$, must look like:

With a lot of irrational numbers, we get much less pretty continued fraction expansions:

One thing to note is that the appearance of a large number in the continued fraction expansion, like 292 above, is telling us something about Diophantine approximation: that is, how well we’re able to approximate our number by rationals of a given denominator. The rationals you obtain by truncating a number’s continued fraction expansion are provably always the “best” in this sense. Thus, if we look at $\varphi$, whose continued fraction is just all 1’s, we can say that in this precise sense, $\varphi$ is the number for which this “approximability” is the worst. It is as hostile as can be towards rational numbers.

Anyway, this is just a small taste of $\varphi$. Now let’s do something seemingly random: take the polynomial $x^2 - x - 1$ and “flip” the sequence of coefficients around, to obtain $1 - x - x^2$ (alternatively, replace each exponent $k$ in the expression with the new exponent $2-k$). Now we’ll take a reciprocal:

$\displaystyle \frac{1}{1-x-x^2} = \frac{1}{1-(x+x^2)}.$

We’re going to look at the series expansion of this thing, which we get from the geometric series formula. For what it’s worth, I don’t care at all about convergence (it’s the summer break; my operator theory course doesn’t start until 2 weeks from now!), so just work completely formally.

$\displaystyle \frac{1}{1-x-x^2} = \sum_{n=0}^\infty (x+x^2)^n = \sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k} x^k (x^2)^{n-k} = \sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k} x^{2n-k}.$

Stare at this thing for a while and you notice that the coefficient $F_k$ of $x^k$ in the above is given by

$\displaystyle F_k = \sum_{n=0}^\infty \binom{n}{2n-k}$

which is (of course) actually a finite sum since we nonchalantly ditch any terms with $2n-k < 0$ or $2n - k > n$. The sequence $F_k$ is the Fibonacci sequence (okay, except for being offset by one or something). It is not hard to show that as $k \to \infty$, we have $F_{k+1}/F_k \to \varphi$. In fact these common ratios are nothing more than the convergents of the continued fraction expansion of $\varphi$.

One cool thing Wikipedia mentions is that you can tile the plane with a “spiral” of squares whose side lengths are given by the Fibonacci sequence:

In the next post I will discuss why all this number-theoretic information related to $\varphi$ (root of the polynomial $x^2-x-1$) shows up in the power series expansion of the reciprocal of the “reversed” polynomial $1-x-x^2$. I’ll also apply the same general procedure to $x^3 - x - 1$, which has the so-called “plastic constant” $\rho$ as its root. The sequence we’ll obtain is called the Padovan sequence, and you can perform a similar tiling of the plane using triangles of those side lengths. Once you look at $x^4 - x - 1$, the sequence you get from the series expansion is no longer nice and monotonic. This is odd, but in hindsight unsurprising since by analogy we would expect it to correspond to some kind of “tiling of the plane by a spiral of 2-sided polygons”, which is absurd.

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Serre’s affineness criterion

Since we’ve been discussing sheaf cohomology for the last few weeks of the algebraic geometry seminar, and I’m leaving Waterloo soon, I was thinking about possible topics for what will probably be the last seminar talk. I figured that having drudged through all this machinery, it would be nice to look at a cohomological characterisation of affine schemes: namely, the fact that a scheme $(X, \mathcal{O}_X)$ is affine if and only if all quasi-coherent sheaves $\mathscr{F}$ of $\mathcal{O}_X$-modules are acyclic, i.e. $H^i(X, \mathscr{F}) = 0$ for $i>0$. In this post I’ll go over the treatment in [Hartshorne III.3, "Cohomology of a Noetherian Affine Scheme"]. I will probably explain all this stuff more coherently in a video sometime down the road.

This is called Serre’s affineness criterion, and the key to the proof (or at least one direction of it) lies in the fact that if you start with an injective $A$-module $I$, and consider its associated sheaf of $\mathcal{O}_{\text{Spec } A}$-modules $\widetilde{I}$ (just defined by $X_f \mapsto I_f$), then in fact this is flasque. We saw before that flasque sheaves are acyclic for the global sections functor $\Gamma(X,-)$, so in particular we can use flasque resolutions to compute cohomology (this will be important later).

We also saw that injective sheaves are flasque, so one might be tempted to claim that the “key” we mentioned above is a mere triviality: indeed, why not just observe that (in view of the equivalence of categories) any injective $A$-module will give rise to an injective sheaf, and then finish? The problem with this argument is that the category of $A$-modules is equivalent to the category of quasicoherent sheaves, and not the full category of $\mathcal{O}_X$-modules. So yes, we will always have an injective of the former category, but we would need an injective of the latter category to conclude flasqueness — and in general this does not happen.

The starting point is a theorem of Krull from commutative algebra. The full statement concerns the $\mathfrak{a}$-adic topology on an $A$-module, and I don’t really know much (nor do I currently have time to read Atiyah-Macdonald) about completions. However, we only really need one containment:

Krull’s Theorem. Let $A$ be a Noetherian ring and $\mathfrak{a} \subseteq A$ be an ideal. If $M \subseteq N$ are finitely generated $A$-modules, then for any $n > 0$ there is $n' \geq n$ such that $\mathfrak{a}^n M \supseteq M \cap \mathfrak{a}^{n'} N$.

Now, define the following submodule of $M$:

$\Gamma_{\mathfrak{a}}(M) = \{ m \in M \mid \mathfrak{a}^n m = 0 \text{ for some } n > 0 \}.$

Before proceeding, let us mention a remark about injectives. We said an object $I$ of an abelian category $\mathfrak{A}$ was injective if the functor $\mathrm{Hom}(-,I)$ is exact. This (contravariant) functor is always left exact, so the important thing to take away is the following: “$I$ injective” means that if $M' \subseteq M$ is a submodule and $\varphi : M' \to I$ is a morphism, then $\varphi$ extends to a morphism $\widetilde{\varphi}: M \to I$.

Surprisingly, the above turns out to be equivalent to the following seemingly weaker condition (Baer’s criterion), namely: if $\mathfrak{b}$ is an ideal of $A$ and $\varphi : \mathfrak{b} \to I$ is a morphism, then $\varphi$ extends to a morphism $\widetilde{\varphi} : A \to I$. This equivalence is a basic result from commutative algebra.

This reminds me of a similar thing that came up when trying to formulate the universal property of the Stone-Cech compactification: in some sense the closed unit interval $[0,1]$ is a “good enough” representative of the class of all compact Hausdorff spaces (this is formalised in the fascinating notion of an injective cogenerator).

Lemma 1. Let $A$ be a Noetherian ring, let $\mathfrak{a} \subseteq A$ be an ideal. Then if $I$ is an injective $A$-module, then $J = \Gamma_{\mathfrak{a}}(I)$ is also an injective $A$-module.

To prove this, we only need to establish Baer’s criterion for $J$, and this is done by observing one can apply Krull’s theorem to the inclusion $\mathfrak{b} \subseteq A$, pulling back from $\mathfrak{b}/(\mathfrak{b} \cap \mathfrak{a}^{n'})$ to $A/\mathfrak{a}^{n'}$, and finally using the natural map $A \to A/\mathfrak{a}^{n'}$ to pull back to $A$ as required.

Lemma 2. Let $A$ be a Noetherian ring, and $I$ an injective $A$-module. Then for any $f \in A$, the natural map $I \to I_f$ to the localisation is surjective.

This lemma isn’t very difficult either. If $\mathfrak{b}_i$ is defined as the annihilator of $f^i$, then you get some ascending chain of ideals in $A$, but $A$ is Noetherian, so $\mathfrak{b}_r = \mathfrak{b}_{r+1} = \ldots$, yada yada. Then, letting $\theta : I \to I_f$ be the natural map, you take some $x \in I_f$, write $= \theta(y)/f^n$ for some $y \in I$ and $n \geq 0$ (you can do this by definition of localisation), and define a map $\varphi : (f^{n+r}) \to I$ by sending $f^{n+r} \mapsto f^r y$ (this turns out to be fine since $(f^{n+r}) \cong A/{\text{Ann } (f^{n+r})}$ as $A$-modules, and $\text{Ann } f^{n+r} = \mathfrak{b}_{n+r} = \mathfrak{b}_r$). Lift $\varphi$ to a map $\psi : A \to I$ by injectivity of $I$, and then let $z = \psi(1)$. Then $\theta(z) = x$. Magic.

Proposition. If $I$ is an injective $A$-module, then $\widetilde{I}$ is a flasque sheaf of $\mathcal{O}_X$-modules, where $X = \text{Spec } A$.

To establish this, we use Noetherian induction on the support of the sheaf $\widetilde{I}$ (call it $Y$). The basic idea is, for some open $U \subseteq X$, to choose some $f$ and consider some open of the form $X_f \subseteq U$. Noting that $\Gamma(X_f,\widetilde{I}) = I_f$, we can invoke the lemma above, and then the problem reduces to showing $\Gamma_Z(X,\widetilde{I}) \to \Gamma_Z(U,\widetilde{I})$ is surjective, where $Z = X \setminus X_f$. But this follows by induction (put $J = \Gamma_{(f)}(I) = \Gamma_Z(X,\widetilde{I})$ and note this is an injective $A$-module by the lemma, hence $\widetilde{J}$, whose support is strictly contained in $Y$, is flasque; at this point we win since $\Gamma_Z(U,\widetilde{J}) = \Gamma(U,\widetilde{I})$ for all opens $U$).

Theorem. Let $X = \text{Spec } A$ for some Noetherian ring $A$. Then if $i>0$, $H^i(X,\mathscr{F}) = 0$ for all quasicoherent sheaves $\mathscr{F}$ on $X$.

To see this, let $M = \Gamma(X,\mathscr{F})$. Take an injective resolution in the category of $A$-modules, apply the Serre functor $M \mapsto \widetilde{M}$ to get a flasque resolution of $\mathscr{F}$. Applying the global sections functor, we just get back the original resolution, so we’re done.

Theorem (Serre). Let $X$ be a Noetherian scheme. Then TFAE:

1. $X$ is affine.
2. $H^i(X,\mathscr{F}) = 0$ for any quasicoherent sheaf $\mathscr{F}$ on $X$ and $i>0$.
3. $H^1(X,\mathscr{I}) = 0$ for any coherent sheaf of ideals $\mathscr{I}$ on $X$.

We’ve already shown that (1) => (2), and (2) => (3) is easy. (3) => (1) can be proved using the following characterisation of affineness: $X$ is affine if and only if there are $f_1, \ldots, f_r \in A := \Gamma(X,\mathcal{O}_X)$ such that $\langle f_1, \ldots, f_r \rangle = A$, each set $X_f$ is affine, and $X$ is covered by $X_{f_1}, \ldots, X_{f_r}$.