# 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.

# Lie groups have abelian fundamental group

Last year I wrote up how to prove that the fundamental group of a (connected) topological group was abelian. Since Lie groups are topological groups, they also have abelian fundamental groups, but I think there is a much neater way to prove this fact using smooth things. Here it is:

Lemma 1: A connected Lie group that acts smoothly on a discrete space is must be a trivial action.

Proof: Suppose our action is ${\theta: G\times M \rightarrow M}$ by ${\theta(g,x)=g\cdot x}$. Consider a point ${q}$ with non-trivial orbit. Then ${\text{im}\theta_{q}=\{q\}\cup A}$ where ${A}$ is non-empty. Thus ${G=\theta_{q}^{-1}(q)\cup \theta_q^{-1}(A)}$, a disconnection of ${G}$. Thus all orbits are trivial.

Lemma 2: A discrete normal subgroup of a Lie group (from now on always assumed connected) is central.

Proof: Denote the subgroup ${H}$. Then ${\theta(g,h)=ghg^{-1}}$ is a smooth action on ${H}$ (a discrete set). Thus by Lemma 1, ${ghg^{-1}=h}$ for all ${g\in G}$. Thus ${H}$ is central, and in particular abelian.

Now for the main theorem. Let ${G}$ be a connected Lie group. Let ${U}$ be the universal cover of ${G}$. Then the covering map ${p: U\rightarrow G}$ is a group homomorphism. Since ${U}$ is simply connected, the covering is normal and hence ${Aut_p(U)\simeq \pi_1(G, e)}$. By virtue of being normal, we also get that ${Aut_p(U)}$ acts transitively on the fibers of ${p}$. In particular, on the set ${p^{-1}(e)}$, which is discrete being the fiber of a discrete bundle. But this set is ${\ker p}$, which is a normal subgroup. I.e. a discrete normal subgroup, which by Lemma 2 is abelian.

Fix ${q\in \ker p}$. Then we get an ismorphism ${\ker p \simeq Aut_p(U)}$ by ${x\mapsto \phi_x}$ where ${\phi_x}$ is the unique covering automorphism that takes ${q}$ to ${x}$. Thus ${Aut_p(U)}$ is abelian which means ${\pi_1(G,e)}$ is abelian.

# Abelianization of the Fundamental Group

I guess I have no reason to offer explanation for lack of posting, but in general this has been one of the best weeks ever and at the same time one of the worst. The worst because I’ve been fairly ill and can’t seem to fully conquer it. It has been the best week for reasons I won’t mention, since I try to keep personal stuff out of this blog as much as possible (but if you know of my other blog which is purely my personal stuff, then you can read about it to your heart’s content, but I refuse to give any hints at all as to how to find that). Both of these factors has lead to a fairly unproductive week.

I may take a more algebraic topology approach for awhile. This is mainly since I’m doing a reading course on Hatcher (with two other students), and before I go present stuff to them and the prof I want to clarify my ideas.

Tomorrow I’m presenting the proof that $H_1(X)\cong \pi_1(X)^{ab}$ for path connected spaces. This is a pretty wonderful result if you think about it. We have exactly how first homology and the fundamental group relate. In fact, the first thing you’d think to do (granted, this might take a little while) is the thing that works.

We can naturally think of paths and singular 1-simplices as the same thing, since they are both just continuous maps to the space out of a closed interval. So after rescaling, a loop $f:[0,1]\to X$ is actually also a 1-cycle since $\partial f=f(1)-f(0)=0$.

The overall idea of this proof is then to show that $h: \pi_1(X, x_0)\to H_1(X)$ is a well-defined homomorphism with image all of $H_1(X)$ and kernel the commutator subgroup. Almost all of these facts are fairly straightforward.

First, we’ll need a few ways in which our different modes of thinking about loops versus 1-cycles correlate. If as a path $f\equiv c$, a constant, then $f\sim 0$ the cycle is homologous to 0. If two paths are homotopic (in the path homotopic and hence equivalence class of $\pi_1(X)$ sense), denoted $f\simeq g$ then they are homologous. Concatenation of paths (and hence the operation in the fundamental group) is homologous to addition of the cycles (the operation in first homology). Lastly, traversing a path backwards is homologous to negating the cycle: $\overline{f}\sim -f$.

So we’ll use these four facts without proof, since they are fairly standard and the proof is long enough as it is.

Recall the definition $h([f])=f$. The second fact, gives us that $h$ is well-defined since any other representative of the equivalence class will be homotopic to the original, and hence the outputs will be homologous.

The third fact gives us that $h$ is a homomorphism of groups.

Our first bit of effort comes from showing that $h$ is surjective. Here we will use the path-connected hypothesis (everything else so far is true without it). Let $\sum n_i\sigma_i$ be any 1-cycle. We must construct a loop that maps to it.

Since the $n_i$ are integers, we can assume each is $\pm 1$ by just repeating the $\sigma_i$ as many times as needed. But all the $\sigma_i$ with -1 in front can be replaced by $-\overline{\sigma_i}$ by the fourth property. This converts all the $n_i$ to 1. Thus $\sum n_i\sigma_i\sim \sum \sigma_k$.

But $\partial(\sum \sigma_k)=0$, so all the endpoints must cancel. So for any $\sigma_k$ that is not a loop, in order to cancel $\sigma_k(1)$, there must be a $\sigma_j$ such that $\sigma_j(0)=\sigma_k(1)$. i.e. there is some $\sigma_j$ that we can concatenate with to form $\sigma_k\cdot \sigma_j$. In order to cancel the $\sigma_k(0)$ some other $\sigma_j$ must exists with endpoint $\sigma_j(1)=\sigma_k(0)$.

So we can concatenate, then rescale, and group all of these cycles into a collection of loops by the third property. So the only remaining thing we must do is get it to be a single loop. But $X$ is path-connected, so pick some basepoint $x_0\in X$. For any of these possibly disjoint loops floating around, we can pick a basepoint at each and connect with a path $\gamma_i$ from $x_0$ to the basepoint of the i-th loop. By the third and fourth properties $\gamma_i\cdot \sigma_i\cdot \overline{\gamma_i}\sim \sigma_i$. So now all loops start and end at $x_0$ and we can combine into a single loop $\sigma$. Thus $h([\sigma])=\sum n_i\sigma_i$.

Now comes the hard part. We want $ker h=\pi_1(X, x_0)'$. The one containment is easy. Since $H_1(X)$ is abelian, by the universal property of the commutator subgroup, $\pi_1(X)'\subset ker h$. The method to get the other direction is to show that for any $h([f])=0$, we must have that $[f]$ is trivial in the abelianization.

Suppose $[f]\in\pi_1(X)$ such that $h([f])=0$. Since $f$ is a cycle, there is some 2-chain $\sum n_i\sigma_i$ such that $\partial (\sum n_i\sigma_i)=f$. So as before, we can assume each $n_i=\pm 1$. Now the goal is to associate a 2-dimensional $\Delta$-complex to $\sum n_i\sigma_i$ by taking for each $\sigma_i$ a $\Delta_i^2$ and identifying pairs of edges which we’ll call $K$.

So before writing this process down, we should examine what the process will be geometrically. It turns out that $K$ will be an orientable compact surface with boundary, since we are just fitting together a finite collection of disjoint 2-simplices (this is not meant to be obvious). The component containing the boundary is a closed orientable surface with an open disk removed. Since connected sums of tori can be expressed as a 2n-gon with pairs of edges identified in the manner $aba^{-1}b^{-1}, cdc^{-1}d^{-1}$ etc, we see that $f$ is homotopic to a product of commutators.

Writing this in detail algebraically is much trickier. Given any $\sigma_i$, we have $\partial \sigma_i=\tau_{i0}-\tau_{i1}+\tau_{i2}$, where $\tau_{ij}$ are singular 1-simplices. Thus $f=\partial(\sum n_i\sigma_i)=\sum (-1)^j n_i\tau_{ij}$.

Keep the picture of a triangle in your head. When we fit together the triangles we are getting pairs of edges. The signs on these pairs are opposite and so will cancel when we sum. The remaining (of the three sides) $\tau_{ij}$ is a copy of $f$. This forms our $\Delta$-complex $K$.

Now form $\sigma : K\to X$ by fitting together the $\sigma_i$ maps. Deform $\sigma$ relative the edges that correspond to $f$ by mapping each vertex to $x_0$. So we have a homotopy on the union of the 0-skeleton with edge $f$, so by the homotopy extension property we get a homotopy on all of $K$.

Now restrict $\sigma$ to the simplices $\Delta_i^2$ to get a new chain $\sum n_i\sigma_i$ with boundary $f$ and $\tau_{ij}$ loops at $x_0$.

Now we just need to check whether the class is trivial or not: $[f]=\sum (-1)n_i[\tau_{ij}]=\sum n_i [\partial \sigma_i]$ where $[\partial \sigma_i]=[\tau_{i0}]-[\tau_{i1}]+[\tau_{i2}]$. But $\sigma_i$ gives a nullhomotopy of $\tau_{i0}-\tau_{i1}+\tau_{i2}$ and we are done.

Thus $ker h=\pi_1(X, x_0)'$ and by the First Iso Theorem we have $H_1(X)\cong \pi_1(X, x_0)^{ab}$.

# The Standards

I’ve decided that these last couple of days I’ll post “standards” or as one professor used to say, “old standbys”. These are quick theorems that are in some sense standard in the literature, and so have at least some positive probability of showing up on a prelim exam.

Old standby 1: The fundamental group of every connected topological group is abelian. (Already a good thing I’m doing this one. I wrote “Lie group” thinking it was only true in this case, but ended up never using smooth structure).

Lemma: Let $X$ be a topological space. Let $F:I\times I\to X$ be continuous, and define the following paths in X: $f(s)=F(s,0)$, $g(s)=F(1,s)$, $h(s)=F(0,s)$, and $k(s)=F(s,1)$. Then $f\cdot g$ ~ $h\cdot k$.

This is just annoying to actually write for how little substance this actually has. But note that $f\cdot g$ is a path starting at $F(0,0)$ and ending at $F(1,1)$, as is $h\cdot k$. Thus it is possible to be path homotopic. Now the homotopy itself is just deforming the arguments of $F$ through $I\times I$, and since $F$ is defined and continuous through all that, it is just composing with a continuous function and hence is itself continuous.

Now for the actual problem. Fix $g\in G$, then $\pi_1(G, g)\cong \pi_1(G, e)$ by the standard trick of left multiplication being a homeomorphism, so WLOG we figure out whether or not $\pi_1(G, e)$ is abelian.

Let $f,g\in \pi_1(G, e)$. Define $F:I\times I\to G$ by $F(s,t)=f(s)g(t)$ (note that this is actual group multiplication whereas in the Lemma the $\cdot$ meant path concatenation). Since we are in a topological group, $F$ is continuous since it is multiplication. Now by the lemma, $F(s,0)\cdot F(1,s)$ ~ $F(0,s)\cdot F(s,1)$. Note where $f, g$ start and end and we get $f(s)g(s)$ ~ $g(s)f(s)$. Thus $\pi_1(G, g)$ is abelian.

This set of posts might not be as useful as I thought it would be considering I left out all the parts I didn’t want to fill in, and the point is to sort of force me to go through it before actually taking the test…

Next I think I’ll do $\mathbb{R}P^n$ is orientable if and only if n is odd.

P.S. Ack. WordPress weirdness strikes again! Who knows a hack to make a ~ without leaving the latex environment?