# Surviving Upper Division Math

It’s that time of the year. Classes are starting up. You’re nervous and excited to be taking some of your first “real” math classes called things like “Abstract Algebra” or “Real Anaylsis” or “Topology.”

It goes well for the first few weeks as the professor reviews some stuff and gets everyone on the same page. You do the homework and seem to be understanding.

Then, all of a sudden, you find yourself sitting there, watching an hour-long proof of a theorem you can’t even remember the statement of, using techniques you’ve never heard of.

You panic. Is this going to be on the test?

We’ve all been there.

I’ve been that teacher, I’m sad to say, where it’s perfectly clear in my head that the students are not supposed to regurgitate any of this. The proof is merely there for rigor and exposure to some ideas. It’s clear in my head which ideas are the key ones, though I maybe forgot to point it out carefully.

It’s a daunting situation for the best students in the class and a downright nightmare for the weaker ones.

Then it gets worse. Once your eyes glaze over that first time, it seems the class gets more and more abstract as the weeks go by, filled with more and more of these insanely long proofs and no examples to illuminate the ideas.

Here’s some advice for surviving these upper division math classes. I’m sure people told me this dozens of times, but I tended to ignore it. I only learned how effective it was when I got to grad school.

Disclaimer: Everyone is different. Do what works for you. This worked for me and may only end up frustrating someone with a different learning style.

Tip Summary: Examples, examples, examples!

I used to think examples were something given in a textbook to help me work the problems. They gave me a model of how to do things.

What I didn’t realize was that examples are how you’re going to remember everything: proofs, theorems, concepts, problems, and so on.

Every time you come to a major theorem, write out the converse, inverse, switch some quantifiers, remove hypotheses, weaken hyphotheses, strengthen conclusions, and whatever you can think of to mess it up.

When you do this you’ll produce a bunch of propositions that are false! Now come up with examples to show they’re false (and get away from that textbook when you do this!). Maybe some rearrangement of the theorem turns out to be true, and so you can’t figure out a counterexample.

This is good, too! I cannot overstate how much you will drill into your memory by merely trying unsuccessfully to find a counterexample to a true statement. You’ll start to understand and see why it’s probably true, which will help you follow along to the proof.

As someone who has taught these classes, I assure you that a huge amount of problems students have on a test would be solved by doing this. Students try to memorize too much, and then when they get to a test, they start to question: was that a “for every” or “there exists?” Does the theorem go this way or that?

You must make up your own examples, so when you have a question like that, the answer comes immediately. It’s so easy to forget the tiniest little hypothesis under pressure.

It’s astounding the number of times I’ve seen someone get to a point in a proof where it looks like everything is in place, but it’s not. Say you’re at a step where $f: X\to Y$ is a continuous map of topological spaces, and $X$ is connected. You realize you can finish the proof if $Y$ is connected.

You “remember” this is a theorem from the book! You’re done!

Woops. It turns out that $f$ has to be surjective to make that true.

But now imagine, before the test, you read that theorem and you thought: what’s a counterexample if I remove the surjective hypothesis?

The example you came up with was so easy and took no time at all. It’s $f: [0,1] \to \{0\} \cup \{1\}$ given by $f(x) = 1$. This example being in your head saves you from bombing that question.

If you just try to memorize the examples in the book or that the professor gives you, that’s just more memorization, and you could run into trouble. By going through the effort of making your own examples, you’ll have the confidence and understanding to do it again in a difficult situation.

A lesser talked about benefit is that having a bunch of examples that you understand gives you something concrete to think about when watching these proofs. So when the epsilons and deltas and neighborhoods of functions and uniform convergence and on and on start to make your eyes glaze over, you can picture the examples you’ve already constructed.

Instead of thinking in abstract generality, you can think: why does that step of the proof work or not work if $f_n(x) = x^n$?

Lastly, half the problems on undergraduate exams are going to be examples. So, if you already know them, you can spend all your time on the “harder” problems.

Other Tip: Partial credit is riskier than in lower division classes.

There’s this thing that a professor will never tell you, but it’s true: saying wrong things on a test is worse than saying nothing at all.

Let me disclaimer again. Being wrong and confused is soooo important to the process of learning math. You have to be unafraid to try things out on homework and quizzes and tests and office hours and on your own.

Then you have to learn why you were wrong. When you’re wrong, make more examples!

Knowing a bunch of examples will make it almost impossible for you to say something wrong.

Here’s the thing. There comes a point every semester where the professor has to make a judgment call on how much you understand. If they know what they’re doing, they’ll wait until the final exam.

The student that spews out a bunch of stuff in the hopes of partial credit is likely to say something wrong. When we’re grading and see something wrong (like misremembering that theorem above), a red flag goes off: this student doesn’t understand that concept.

A student that writes nothing on a problem or only a very small amount that is totally correct will be seen as superior. This is because it’s okay to not be able to do a problem if you understand you didn’t know how to do it. That’s a way to demonstrate you’re understanding. In other words: know what you don’t know.

Now, you shouldn’t be afraid to try, and this is why the first tip is so much more important than this other tip (and will often depend on the instructor/class).

And the best way to avoid using a “theorem” that’s “obviously wrong” is to test any theorem you quote against your arsenal of examples. As you practice this, it will become second-nature and make all of these classes far, far easier.

# The Stack of Pitch Class Sets

I know it’s been a while since I’ve talked about either of these topics, but I’ve always been meaning to point something funny out. I thought I might formally work it out and write it up to submit to a music theory journal, but no one would probably accept it anyway. So I’ll sketch the idea now. Back here I talked about stacks as a useful way to generalize what we mean by a “space.” Back here I talked about the math behind the idea of pitch class sets.

I know Mazzola wrote a whole book on using topos theory in music, but I’ve never dug into it very deeply. I fully admit this is probably just a special case of something from that book. But it’s always useful to work out special cases.

Recall that a pitch set (or chord) is just converting notes to numbers: 0 is C, 1 is C#, 2 is D, etc. A given collection of pitches can be expressed in a more useful notation when there isn’t a key we’re working in. For example, a C major chord is (047).

A pitch class set is then saying that there are collections of these we want to consider to be the same. For one, our choice of 0 is completely arbitrary. We could have set 0 is A, and we should get the same theory. This amounts to identifying all pitch sets that are the same after translation.

We also want to identify sets that are the same after inversion. In the previous post on this topic, I showed that if we label the vertices of a dodecagon, this amounts to a reflection symmetry. The reflections together with the translations generate the dihedral group ${D_{12}}$, so we are secretly letting ${D_{12}}$ act on the set of all tuples of numbers 0 to 11, where each number only appears once and without loss of generality we can assume they are in increasing order.

Thus a pitch class set is just an equivalence class of a chord under this group action. It is not the direction I want this post to go, but given such a class, there is always a unique representative that is usually called the “prime form” (basically the most “compact” representative starting with 0).

Here’s where we get to the part I never really worked out. The set of all “chords” should have some sort of useful topology on it. For example, (0123) should be related to (0124), because they are the same chord except for one note. I don’t think doing something obvious like defining a distance based on the coordinates works. If you try to construct the lattice of open sets by hand based on your intuition, a definition might become more obvious. Call this space of chords ${X}$.

Now we have a space with a group action on it. One might want to merely form the quotient space ${X \rightarrow X/G}$. This will be 24 to 1 at most points, but it will also forget which chords were fixed by elements of the group. Part of the “theory” in music theory is to remember that information. This is why I propose making the quotient stack ${[X/G]}$. It seems like an overly complicated thing to do, but here’s what you gain.

You now have a “space” whose points are the pitch class sets. If that class contains 24 distinct chords, then the point is an “honest” point with no extra information. The fiber of the quotient map contains the 24 chords, and you get to each of them by acting by the elements of ${D_{12}}$ (i.e. it is a torsor under ${D_{12}}$). Now consider something like the pitch class set [0,2,4,6,8,10]. The fiber of the quotient map only contains ${2}$ elements: (02468T) and (13579E). The stack will tag these points with ${D_6}$, which is the subgroup of symmetries which send this chord to itself.

Now that I’ve drawn this, I can see that many of you will be skeptical about the simplicity. Think of it this way. The bottom thing is the space I’m describing. Each point in the space is tagged with the prime form representative together with the subgroup of symmetries that preserve the class. That’s pretty simple. Yet it remembers all of the complicated music theory of the top thing! If the topology was defined well, then studying this space may even lead to insights on how symmetries of classes are related to each other. Let me know if anyone has seen anything like this before.

# An Application of p-adic Volume to Minimal Models

Today I’ll sketch a proof of Ito that birational smooth minimal models have all of their Hodge numbers exactly the same. It uses the ${p}$-adic integration from last time plus one piece of heavy machinery.

First, the piece of heavy machinery: If ${X, Y}$ are finite type schemes over the ring of integers ${\mathcal{O}_K}$ of a number field whose generic fibers are smooth and proper, then if ${|X(\mathcal{O}_K/\mathfrak{p})|=|Y(\mathcal{O}_K/\mathfrak{p})|}$ for all but finitely many prime ideals, ${\mathfrak{p}}$, then the generic fibers ${X_\eta}$ and ${Y_\eta}$ have the same Hodge numbers.

If you’ve seen these types of hypotheses before, then there’s an obvious set of theorems that will probably be used to prove this (Chebotarev + Hodge-Tate decomposition + Weil conjectures). Let’s first restrict our attention to a single prime. Since we will be able to throw out bad primes, suppose we have ${X, Y}$ smooth, proper varieties over ${\mathbb{F}_q}$ of characteristic ${p}$.

Proposition: If ${|X(\mathbb{F}_{q^r})|=|Y(\mathbb{F}_{q^r})|}$ for all ${r}$, then ${X}$ and ${Y}$ have the same ${\ell}$-adic Betti numbers.

This is a basic exercise in using the Weil conjectures. First, ${X}$ and ${Y}$ clearly have the same Zeta functions, because the Zeta function is defined entirely by the number of points over ${\mathbb{F}_{q^r}}$. But the Zeta function decomposes

$\displaystyle Z(X,t)=\frac{P_1(t)\cdots P_{2n-1}(t)}{P_0(t)\cdots P_{2n}(t)}$

where ${P_i}$ is the characteristic polynomial of Frobenius acting on ${H^i(X_{\overline{\mathbb{F}_q}}, \mathbb{Q}_\ell)}$. The Weil conjectures tell us we can recover the ${P_i(t)}$ if we know the Zeta function. But now

$\displaystyle \dim H^i(X_{\overline{\mathbb{F}_q}}, \mathbb{Q}_\ell)=\deg P_i(t)=H^i(Y_{\overline{\mathbb{F}_q}}, \mathbb{Q}_\ell)$

and hence the Betti numbers are the same. Now let’s go back and notice the magic of ${\ell}$-adic cohomology. If ${X}$ and ${Y}$ are as before over the ring of integers of a number field. Our assumption about the number of points over finite fields being the same for all but finitely many primes implies that we can pick a prime of good reduction and get that the ${\ell}$-adic Betti numbers of the reductions are the same ${b_i(X_p)=b_i(Y_p)}$.

One of the main purposes of ${\ell}$-adic cohomology is that it is “topological.” By smooth, proper base change we get that the ${\ell}$-adic Betti numbers of the geometric generic fibers are the same

$\displaystyle b_i(X_{\overline{\eta}})=b_i(X_p)=b_i(Y_p)=b_i(Y_{\overline{\eta}}).$

By the standard characteristic ${0}$ comparison theorem we then get that the singular cohomology is the same when base changing to ${\mathbb{C}}$, i.e.

$\displaystyle \dim H^i(X_\eta\otimes \mathbb{C}, \mathbb{Q})=\dim H^i(Y_\eta \otimes \mathbb{C}, \mathbb{Q}).$

Now we use the Chebotarev density theorem. The Galois representations on each cohomology have the same traces of Frobenius for all but finitely many primes by assumption and hence the semisimplifications of these Galois representations are the same everywhere! Lastly, these Galois representations are coming from smooth, proper varieties and hence the representations are Hodge-Tate. You can now read the Hodge numbers off of the Hodge-Tate decomposition of the semisimplification and hence the two generic fibers have the same Hodge numbers.

Alright, in some sense that was the “uninteresting” part, because it just uses a bunch of machines and is a known fact (there’s also a lot of stuff to fill in to the above sketch to finish the argument). Here’s the application of ${p}$-adic integration.

Suppose ${X}$ and ${Y}$ are smooth birational minimal models over ${\mathbb{C}}$ (for simplicity we’ll assume they are Calabi-Yau, Ito shows how to get around not necessarily having a non-vanishing top form). I’ll just sketch this part as well, since there are some subtleties with making sure you don’t mess up too much in the process. We can “spread out” our varieties to get our setup in the beginning. Namely, there are proper models over some ${\mathcal{O}_K}$ (of course they aren’t smooth anymore), where the base change of the generic fibers are isomorphic to our original varieties.

By standard birational geometry arguments, there is some big open locus (the complement has codimension greater than ${2}$) where these are isomorphic and this descends to our model as well. Now we are almost there. We have an etale isomorphism ${U\rightarrow V}$ over all but finitely many primes. If we choose nowhere vanishing top forms on the models, then the restrictions to the fibers are ${p}$-adic volume forms.

But our standard trick works again here. The isomorphism ${U\rightarrow V}$ pulls back the volume form on ${Y}$ to a volume form on ${X}$ over all but finitely primes and hence they differ by a function which has ${p}$-adic valuation ${1}$ everywhere. Thus the two models have the same volume over all but finitely many primes, and as was pointed out last time the two must have the same number of ${\mathbb{F}_{q^r}}$-valued points over these primes since we can read this off from knowing the volume.

The machinery says that we can now conclude the two smooth birational minimal models have the same Hodge numbers. I thought that was a pretty cool and unexpected application of this idea of ${p}$-adic volume. It is the only one I know of. I’d be interested if anyone knows of any other.

# Volumes of p-adic Schemes

I came across this idea a long time ago, but I needed the result that uses it in its proof again, so I was curious about figuring out what in the world is going on. It turns out that you can make “${p}$-adic measures” to integrate against on algebraic varieties. This is a pretty cool idea that I never would have guessed possible. I mean, maybe complex varieties or something, but over ${p}$-adic fields?

Let’s start with a pretty standard setup in ${p}$-adic geometry. Let ${K/\mathbb{Q}_p}$ be a finite extension and ${R}$ the ring of integers of ${K}$. Let ${\mathbb{F}_q=R_K/\mathfrak{m}}$ be the residue field. If this scares you, then just take ${K=\mathbb{Q}_p}$ and ${R=\mathbb{Z}_p}$.

Now let ${X\rightarrow Spec(R)}$ be a smooth scheme of relative dimension ${n}$. The picture to have in mind here is some smooth ${n}$-dimensional variety over a finite field ${X_0}$ as the closed fiber and a smooth characteristic ${0}$ version of this variety, ${X_\eta}$, as the generic fiber. This scheme is just interpolating between the two.

Now suppose we have an ${n}$-form ${\omega\in H^0(X, \Omega_{X/R}^n)}$. We want to say what it means to integrate against this form. Let ${|\cdot |_p}$ be the normalized ${p}$-adic valuation on ${K}$. We want to consider the ${p}$-adic topology on the set of ${R}$-valued points ${X(R)}$. This can be a little weird if you haven’t done it before. It is a totally disconnected, compact space.

The idea for the definition is the exact naive way of converting the definition from a manifold to this setting. Consider some point ${s\in X(R)}$. Locally in the ${p}$-adic topology we can find a “disk” containing ${s}$. This means there is some open ${U}$ about ${s}$ together with a ${p}$-adic analytic isomorphism ${U\rightarrow V\subset R^n}$ to some open.

In the usual way, we now have a choice of local coordinates ${x=(x_i)}$. This means we can write ${\omega|_U=fdx_1\wedge\cdots \wedge dx_n}$ where ${f}$ is a ${p}$-adic analytic on ${V}$. Now we just define

$\displaystyle \int_U \omega= \int_V |f(x)|_p dx_1 \cdots dx_n.$

Now maybe it looks like we’ve converted this to another weird ${p}$-adic integration problem that we don’t know how to do, but we the right hand side makes sense because ${R^n}$ is a compact topological group so we integrate with respect to the normalized Haar measure. Now we’re done, because modulo standard arguments that everything patches together we can define ${\int_X \omega}$ in terms of these local patches (the reason for being able to patch without bump functions will be clear in a moment, but roughly on overlaps the form will differ by a unit with valuation ${1}$).

This allows us to define a “volume form” for smooth ${p}$-adic schemes. We will call an ${n}$-form a volume form if it is nowhere vanishing (i.e. it trivializes ${\Omega^n}$). You might be scared that the volume you get by integrating isn’t well-defined. After all, on a real manifold you can just scale a non-vanishing ${n}$-form to get another one, but the integral will be scaled by that constant.

We’re in luck here, because if ${\omega}$ and ${\omega'}$ are both volume forms, then there is some non-vanishing function such that ${\omega=f\omega'}$. Since ${f}$ is never ${0}$, it is invertible, and hence is a unit. This means ${|f(x)|_p=1}$, so since we can only get other volume forms by scaling by a function with ${p}$-adic valuation ${1}$ everywhere the volume is a well-defined notion under this definition! (A priori, there could be a bunch of “different” forms, though).

It turns out to actually be a really useful notion as well. If we want to compute the volume of ${X/R}$, then there is a natural way to do it with our set-up. Consider the reduction mod ${\mathfrak{m}}$ map ${\phi: X(R)\rightarrow X(\mathbb{F}_q)}$. The fiber over any point is a ${p}$-adic open set, and they partition ${X(R)}$ into a disjoint union of ${|X(\mathbb{F}_q)|}$ mutually isomorphic sets (recall the reduction map is surjective here by the relevant variant on Hensel’s lemma). Fix one point ${x_0\in X(\mathbb{F}_q)}$, and define ${U:=\phi^{-1}(x_0)}$. Then by the above analysis we get

$\displaystyle Vol(X)=\int_X \omega=|X(\mathbb{F}_q)|\int_{U}\omega$

All we have to do is compute this integral over one open now. By our smoothness hypothesis, we can find a regular system of parameters ${x_1, \ldots, x_n\in \mathcal{O}_{X, x_0}}$. This is a legitimate choice of coordinates because they define a ${p}$-adic analytic isomorphism with ${\mathfrak{m}^n\subset R^n}$.

Now we use the same silly trick as before. Suppose ${\omega=fdx_1\wedge \cdots \wedge dx_n}$, then since ${\omega}$ is a volume form, ${f}$ can’t vanish and hence ${|f(x)|_p=1}$ on ${U}$. Thus

$\displaystyle \int_{U}\omega=\int_{\mathfrak{m}^n}dx_1\cdots dx_n=\frac{1}{q^n}$

This tells us that no matter what ${X/R}$ is, if there is a volume form (which often there isn’t), then the volume

$\displaystyle Vol(X)=\frac{|X(\mathbb{F}_q)|}{q^n}$

just suitably multiplies the number of ${\mathbb{F}_q}$-rational points there are by a factor dependent on the size of the residue field and the dimension of ${X}$. Next time we’ll talk about the one place I know of that this has been a really useful idea.

# Topological Modular Forms

This will be my first and last post on this topic, since it will take us too far from the theme for this year which is arithmetic geometry. It took awhile for me to write this because something feels wrong in the last post and I wanted to correct it before doing this one. Unfortunately, I can only make a guess at what is happening. I’ll explain it when it comes up. Thus as a warning everything in the last post and in this post should be taken as approximately true (of course, this is a blog, so this warning should probably always be in place).

Recall briefly that we now have a description of (weight ${2}$) modular forms just as a global ${1}$-form on a certain moduli space of elliptic curves with level ${N}$ structure. To get the weight ${2k}$ modular forms we just take tensor powers, so ${H^0(X_0(N), \Omega^{\otimes k})\simeq M_{2k}(\Gamma_0(N))}$. It is funny to notice that a priori it is completely unclear that the collection of all modular forms of a fixed level should form a graded commutative ring, but with this description it falls right out. We define the graded ring of modular forms of level ${N}$ to be ${\displaystyle M(\Gamma_0(N))=\bigoplus_{k=1}^\infty H^0(X_0(N), \Omega^{\otimes k})=\bigoplus_{k=1}^\infty M_{2k}(\Gamma_0(N))}$.

Now notice that if we take ${N=1}$ in our moduli problem we are just taking an elliptic curve plus a cyclic subgroup of order ${1}$, i.e. we are marking the identity. Thus what ought to be the case is that ${\overline{\mathcal{M}_{1,1}}=X_0(1)}$. This is the part that confuses me. Last time I said that ${X_0(N)}$ was a smooth Riemann surface, but ${\overline{\mathcal{M}_{1,1}}}$ is a DM stack. My guess at what is going on is that since we only defined the moduli functor for ${X_0(N)}$ for elliptic curves over ${\mathbb{C}}$, we are maybe just taking the ${\mathbb{C}}$-valued points. Thus maybe ${\overline{\mathcal{M}_{1,1}}(\mathbb{C})\simeq X_0(1)}$. In any case, there is certainly some relation between the two so it isn’t unreasonable to try to figure out what happens when we replace ${X_0(1)}$ with ${\overline{\mathcal{M}_{1,1}}}$.

Now we’ll start the crazy generalizations. There is something called a derived DM stack. Since it would take a lot to define, we’ll just say that it is one of these things where “${\infty}$-blah” gets thrown around. The important idea here is that we can take ${\pi_0}$ and get back an honest DM stack. The big theorem of Hopkins, Miller, and Lurie is that there exists a derived DM stack ${(\mathcal{M}, \mathcal{O})}$ whose underlying DM stack is ${\overline{\mathcal{M}_{1,1}}}$ such that ${\pi_{2k}\mathcal{O}\simeq \omega^k}$ and ${\pi_{2k+1}\mathcal{O}=0}$.

Now “tmf” is something called a commutative ring spectrum and it is formed by taking the derived global sections of ${\mathcal{O}}$. Generalities give us a descent spectral sequence ${H^s(\mathcal{M}, \omega^t)\Rightarrow \pi_{2t+s}\mathbf{tmf}}$. An open and interesting question is to determine which modular forms give homotopy classes in ${\mathbf{tmf}}$ since the classical modular forms form the ${0}$-row of this spectral sequence. I’ll just end by pointing out how mind boggling this is. Modular forms have had such great success in number theory. Now they are successfully being used to understand homotopy groups of spheres and other extremely topological questions. The reverse has been done as well. Topological methods can transfer information back to modular forms and give number theoretic theorems such as congruence relations between ${p}$-adic modular forms. How amazing!

# Classical Local Systems

I lied to you a little. I may not get into the arithmetic stuff quite yet. I’m going to talk about some “classical” things in modern language. In the things I’ve been reading lately, these ideas seem to be implicit in everything said. I can’t find this explained thoroughly anywhere. Eventually I want to understand how monodromy relates to bad reduction in the ${p}$-adic setting. So we’ll start today with the different viewpoints of a local system in the classical sense that are constantly switched between without ever being explained.

You may need to briefly recall the old posts on connections. The goal for the day is to relate the three equivalent notions of a local system, a vector bundle plus flat connection on it, and a representation of the fundamental group. There may be some inaccuracies in this post, because I can’t really find this written anywhere and I don’t fully understand it (that’s why I’m making this post!).

Since I said we’d work in the “classical” setting, let’s just suppose we have a nice smooth variety over the complex numbers, ${X}$. In this sense, we can actually think about it as a smooth manifold, or complex analytic space. If you want, you can have the picture of a Riemann surface in your head, since the next post will reduce us to that situation.

Suppose we have a vector bundle on ${X}$, say ${E}$, together with a connection ${\nabla : E\rightarrow E\otimes \Omega^1}$. We’ll fix a basepoint ${p\in X}$ that will always secretly be lurking in the background. Let’s try to relate this this connection to a representation of the fundamental group. Well, if we look at some old posts we’ll recall that a choice of connection is exactly the same data as telling you “parallel transport”. So what this means is that if I have some path on ${X}$ it tells me how a vector in the fiber of the vector bundle moves from the starting point to the ending point.

Remember, that we fixed some basepoint ${p}$ already. So if I take some loop based at ${p}$ say ${\sigma}$, then a vector ${V\in E_p}$ can be transported around that loop to give me another vector ${\sigma(V)\in E_p}$. If my vector bundle is rank ${n}$, then ${E_p}$ is just an ${n}$-dimensional vector space and I’ve now told you an action of the loop space based at ${p}$ on this vector space.

Visualization of a vector being transported around a loop on a torus (yes, I’m horrible at graphics, and I couldn’t even figure out how to label the other vector at p as $\sigma (V)$):

This doesn’t quite give me a representation of the fundamental group (based at ${p}$), since we can’t pass to the quotient, i.e. the transport of the vector around a loop that is homotopic to ${0}$ might be non-trivial. We are saved if we started with a flat connection. It can be checked that the flatness assumption gives a trivial action around nullhomotopic loops. Thus the parallel transport only depends on homotopy classes of loops, and we get a group homomorphism ${\pi_1(X, p)\rightarrow GL_n(E_p)}$.

Modulo a few details, the above process can essentially be reversed, and hence given a representation you can produce a unique pair ${(E,\nabla)}$, a vector bundle plus flat connection associated to it. This relates the latter two ideas I started with. The one that gave me the most trouble was how local systems fit into the picture. A local system is just a locally constant sheaf of ${n}$-dimensional vector spaces. At first it didn’t seem likely that the data of a local system should be equivalent to these other two things, since the sheaf is locally constant. This seems like no data at all to work with rather than an entire vector bundle plus flat connection.

Here is why algebraically there is good motivation to believe this. Recall that one can think of a connection as essentially a generalization of a derivative. It is just something that satisfies the Leibniz rule on sections. Recall that we call a section, ${s}$, horizontal for the connection if ${\nabla (s)=0}$. But if this is the derivative, this just means that the section should be constant. In this analogy, we see that if we pick a vector bundle plus flat connection, we can form a local system, namely the horizontal sections (which are the locally constant functions). If you want an exercise to see that the analogy is actually a special case, take the vector bundle to be the globally trivial line bundle ${\mathcal{O}_X}$ and the connection to be the honest exterior derivative ${d:\mathcal{O}_X\rightarrow \Omega^1}$.

The process can be reversed again, and given any locally constant sheaf of vector spaces, you can cook up a vector bundle and flat connection whose horizontal sections are precisely the sections of the sheaf. Thus our three seemingly different notions are actually all equivalent. I should point out that part of my oversight on the local system side was thinking that a locally constant sheaf somehow doesn’t contain much information. Recall that it is still a sheaf, so we can be associating lots of information on large open sets and we still have restriction homomorphisms giving data as well. Next time we’ll talk about some classical theorems in differential equation theory that are most easily proved and stated in this framework.

# Gerbes 3: Another Example and Some Caution

This might be my last post on gerbes (explicitly for gerbe’s sake), so as in my last ‘stacks for stack’s sake’ post I’ll try to clarify some things with more examples and then give some cautions. Last time I mentioned the classifying stack ${BA}$. Let’s first actually construct it better than the quick idea I gave.

Let ${B}$ be a topological space, and ${A}$ a sheaf of abelian groups on ${B}$ (note that I’ll use ${A}$ instead of ${\mathcal{A}}$ to avoid typing the script, but it is a ${\mathit{sheaf}}$ and not just a group, otherwise we’ll just recover the classifying space).

Define a functor ${BA: \text{Top}(B)\rightarrow \text{Grpds}}$, where ${\text{Grpds}}$ is the category of groupoids, by ${BA(U)=}$ groupoid of ${A_U}$-torsors over ${U}$. This is a sheaf, say ${T}$, on ${U}$ with an action ${A_U\times T\rightarrow T}$ such that if ${T(V)\neq \emptyset}$, then ${A_U(V)}$ acts simply transitively on ${T(V)}$.

Again, this is just fancy language for something that is probably familiar to you. Since we have a sheaf of groups, just think an open set at a time. ${A_U(V)}$ is a group, call it ${G}$. Then ${G\times T\rightarrow T}$ is really just an honest group action, and “acting simply transitively” means that if we pick out some ${t\in T}$, then we have a way to identify ${G}$ with ${T}$, namely ${G\stackrel{\sim}{\rightarrow} T}$ (as sets) via the action ${g\mapsto g\cdot t}$.

You could also think of this as a “relative” principal bundle. The group that it is a principal bundle of gets to change locally, but if it is a constant sheaf and hence not changing, then we really do just get the classifying space.

I told you ${BA}$ as a functor to Grpds and not as a functor to ${\text{Top}(B)}$ which is how stacks were defined, but recall if we have a sheaf on ${B}$, then we can convert it to that form by taking our category to have objects the pairs ${\{(s, U)\}}$ where ${s\in BA(U)}$, and the maps in the category are inclusions and restricting to the right thing. If we were doing all the details we’d have to check all of this and then check it is actually a gerbe and that it is actually an ${A}$-gerbe, etc, but we’d be stuck here forever and these are all straightforward enough that it would make a great exercise if you don’t see it right away.

Recall last time that an ${A}$-gerbe, ${G}$, is isomorphic to ${BA}$ if and only if it has a global object. Recall that ${\text{Vect}^1}$, the stack of rank one vector bundles, was a ${\mathbb{G}_m}$-gerbe, and it has the trivial bundle as a global object, so ${B\mathbb{G}_m\simeq \text{Vect}^1}$.

Let’s actually prove it now. If ${G\simeq BA}$, then ${A(B) \in BA(B)}$ and hence ${G(B)\neq \emptyset}$. For the reverse direction, suppose there is some ${s\in G(B)}$, then we get a map ${G\rightarrow BA}$ which we’ll denote ${t\mapsto \text{Isom}(t,s)}$. One can check that this induces the isomorphism. In fact, one can check that whenever you have a map ${G_1\rightarrow G_2}$ in the category of ${A}$-gerbes, it will be an isomorphism.

This is why I wanted to bring this example up. Here are some of the cautions that jump to my mind. Something might feel fishy to you right now. That’s because I haven’t really told you the proper way to think about these things. When I say “isomorphism” what does that mean? Well, it really means as ${2}$-categories.

Also, suppose you are an algebraic geometer and you say you have a gerbe on the étale site of ${X}$. This isn’t precise enough, since funny differences can happen whether or not you’re on the big or small site. I guess because of all that I’ve left out in an attempt to bring the concept out, my main caution is to consult the literature and not any of these blog posts if you want to know if something is true.