A Mind for Madness

Musings on art, philosophy, mathematics, and physics


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Classical (Lagrangian) Mechanics

It turns out that because I work with Calabi-Yau varieties I often encounter various ideas and terms from physics. In particular, quantum field theory is a something that comes up a lot. I took a lot of physics as an undergrad, and I’ve pieced together a tiny bit about what is meant by “quantum field theory.” In order to record this somewhere before I forget it, I’m going to blog some stuff. This should be a very short series, because I don’t want to get hung up on it.

The main point is to try to express the idea of quantum field theory in a way a mathematician would understand it. Before we can do that I need to spend a post on classical mechanics. This post is going to present what is done over the course of a semester long undergrad class, so it will go fast. I’ll give you the take away up front. In a mathematically rigorous way one can prove that the “Lagrangian formalism” we’ll look at soon is exactly equivalent to Newton’s law {F=ma}.

Suppose we have some particle in space. If it is moving, that motion has something called kinetic energy. For simplicity, we’ll call this a function of time {K(t)=\frac{1}{2}mv(t)^2}. The formula isn’t important here. Usually you also have something called potential energy. For example, a ball is on a table. There is the potential energy of falling to the ground. Technically you can figure out the potential the same way you’d find the potential of any vector field (this took me awhile to connect as an undergrad).

Suppose your particle is moving in {\mathbb{R}^n}, then we can describe it by a function {q: \mathbb{R}\rightarrow\mathbb{R}^n}. There could be some ambient force (gravity, electromagnetic field, etc doesn’t matter). This is a vector field {F: \mathbb{R}^n\rightarrow \mathbb{R}^n}. The potential then is just a function {V: \mathbb{R}^n\rightarrow \mathbb{R}} such that {\nabla F=V}. This is what we tell our calculus students, so it shouldn’t be surprising. Of course, we must assume our force is conservative for a potential to exist, so we do that.

Now we define {L=K(t)-V(q(t))}. This is called the Lagrangian. We define the action over some path {q:[t_0, t_1]\rightarrow \mathbb{R}^n} to be the integral {S(q)=\int_{t_0}^{t_1}L(t)dt}. Now we get to the point. If we let our paths vary, then we get a bunch of real numbers by evaluating {S(q)}. From standard calculus we could find the minimum. This is the path of least action, and our particle will follow that path if and only if in our system Newton’s law {F=ma} holds.

We could go off and try to describe physically why one would think of this weird formalism. For example, integrating force over distance is the work needed to move the particle from point {a} to point {b}. We would expect that the particle will naturally follow the path that requires the least work. This has roughly the same flavor, but takes into account some extra stuff. Whatever the physical reason, it shouldn’t really matter to us, because it is exactly equivalent to the law we all know ought to be true.

In a classical mechanics class you’d probably take many weeks now just being handed various scenarios where you figure out the Lagrangian, and then given some inital starting point figure out the path by taking the derivative {\delta S(q)} and setting it to {0} and solving. Note: for practical purposes this is a little tricky, because the so-called variation of the action involves differentiating with respect to paths. Since we aren’t computing these, we won’t go through this, but the idea of how to do this is to just parametrize your paths in some nice way (think about a smooth {1}-parameter homotopy connecting them for the picture).

Now we must generalize a bit. Suppose we have some physical system (maybe a double pendulum for sufficient complicatedness). There’s more than just one particle, and there are constraints for how things can move in relation to eachother. What we do now is consider the space of all configurations. Think of this as the moduli space of all positions the system could ever take. A point in this space {Q} is one particular configuration. Now a path {q:[t_0, t_1]\rightarrow Q} is just a description of how the system evolves over that time period. This configuration space we assume is a smooth manifold.

This means the velocity, which is the time derivative {\dot{q}(t)\in T_{q(t)}Q} is actually a tangent vector now (it was before, but we just made the canonical identification). Let’s pick a starting point and ending point {a, b\in Q}. Then we can formalize what we did last time as follows. Define {\Gamma=\{q:[t_0, t_1]\rightarrow Q : q(t_0)=a, \ q(t_1)=b\}} to be the path space (of smooth paths) with those endpoints. Let {L: TQ\rightarrow \mathbb{R}} be a smooth function called the Lagrangian of the system.

Now the action is {S:\Gamma\rightarrow \mathbb{R}} defined by {S(q)=\int_{t_0}^{t_1}L(q, \dot{q})dt}. The path that our system will take in the configuration space will be a minimum of {S}. Thus to find it we just solve {\delta S=0}.

In order to test whether you follow this, a really quick (if you get it, but painful if you don’t) and wonderful exercise is to figure out the equation of motion of a single free particle in {\mathbb{R}^3}. What does this mean? Well, there is no force in the system at all, so the potential is {0}, and hence {L=K(t)=\frac{1}{2}mv(t)\cdot v(t)}. We already know the answer. No force means no acceleration. Thus from basic calculus the answer is that velocity is a constant {v_0} and the path is {q(t)=a+v_0t} where {a} is the initial starting point. Try to get that using the Lagrangian!


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Mori’s Bend and Break

I noticed that recently people were clicking a lot of the links I had on my blogroll. Since many of the blogs were defunct, or I didn’t read them anymore I chopped a lot off. I also added a few that I found myself returning to frequently (including no longer active ones). So that has been updated for the first time in years.

The last little bit we’ll do that is related to moduli spaces and deformation theory is something called Mori’s bend and break argument. It says that if {X} is a nonsingular projective variety of dimension {n} over an algebraically closed field, {k}, of positive characteristic {p} and if there is an irreducible curve {C\subset X} with {C. K_X<0}, then {X} contains a rational curve. In this context a rational curve is an integral curve whose normalization is {\mathbb{P}^1}. The condition on {K_X} is sometimes stated as being not numerically effective (not nef).

Suppose {C_0} is an integral curve such that {C_0. K_X<0}. If the normalization {C_1\rightarrow C_0} has genus {0}, then we are done. Let {g:=g(C_1)}. Choose {r} large enough so that {-p^r(C_0. K_X)\geq ng+1}. Define {q=p^r}. Let {F:C\rightarrow C_1} be the {q}-th power, {k}-linear Frobenius map and denote {f:C\rightarrow C_0} the composition. We only changed the structure sheaf and not the topological space, so {C} still has genus {g}.

Fix some point {P\in C}. Let {Hom_P(C,X)} be the quasi-projective scheme that represents the functor of families of maps from {C} to {X} that keep the image of {P} fixed (it's a subscheme of the usual Hom scheme). Standard deformation theory tells us that the tangent space is {H^0(C, f^*T_X(-P))} and the obstruction space is {H^1(C, f^*T_X(-P))}.

We didn't do this, but it is all part of the package that we've talked about. When the functor is representable we get the natural estimate that {\dim Hom_P(C, X)\geq h^0(f^*T_X(-P))-h^1(f^*T_X(-P))}. This just comes from the fact that if every possible obstruction is realized, then each one will cut the dimension down, but often despite an obstruction space being non-zero the obstruction itself might vanish. This will only make the dimension bigger.

Now Riemann-Roch gives us that {\dim Hom_P(C, X)\geq -q(C_0. K_X)-n+n(1-g)\geq 1} by our choice of {q}. In particular, we can find a nonsingular curve {D} and a morphism {g:C\times D\rightarrow X}, thought of as a nonconstant family of maps all sending {P} to the same point {P_0}. You can argue here that {D} cannot be complete, otherwise the family would have to be constant.

So let {D\subset \overline{D}} be a completion where {\overline{D}} is a nonsingular projective curve. Let {G:C\times \overline{D} \dashrightarrow X} be the rational map. Blow up a finite number of points to resolve the undefined points to get {Y\rightarrow C\times \overline{D}} whose composition given by {\pi: Y\rightarrow X} is an honest morphism. Let {E\subset Y} be the exceptional curve of the last blow up needed. Since it was actually needed, it can't be collapsed to a point, and hence {\pi(E)} is our desired curve.

This is one of those interesting things where it is easier in positive characteristic than characteristic {0} because you have the Frobenius at your disposal. It allowed us to jack up the tangent space without affecting the obstruction space to produce our curve. Mori actually does relate this back to varieties in characteristic {0} to prove Hartshorne's conjecture which says that a nonsingular, projective variety with ample tange bundle is isomorphic to {\mathbb{P}^n}.


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Moduli of Vector Bundles on Elliptic Curves

We’ve been talking about moduli problems, and one notoriously hard type of moduli problem is to “classify” vector bundles on some variety. Even when you restrict yourself to some special case like a specific surface and try to classify only vector bundles of certain rank or Chern class you run into trouble. To this day, these types of problems are a very active area of research.

It is fairly well-known (due to Grothendieck, but a problem in Hartshorne as well) that any finite rank vector bundle over {\mathbb{P}^1} is just a finite direct sum {\oplus_i\mathcal{O}(n_i)}. The next interesting case would be to move up to genus {1} curves. It turns out that Atiyah in the 1957 paper, Vector Bundles over an Elliptic Curve, worked out a classification. I just learned about this a month or two ago, and it is pretty cool so I’d like to briefly describe the idea.

Fix an algebraically closed field {k} of characteristic {0}, and let {E/k} be an elliptic curve. Define {V(r,d)} to be the set of indecomposable vector bundles (up to isomorphism) on {E} of rank {r} and degree {d}, where degree just means the degree of the determinant. It is well known that {V(1,0)} can be identified with {E}, because {V(1,0)=Pic^0(E)} is the set of degree {0} divisors. In particular, this says that the moduli space of degree {0} divisors (line bundles) on {E} is fine and representable by {E}.

Let’s continue with this idea of classifying degree {0} vector bundles. It turns out there is a unique vector bundle {T_r\in V(r,0)} with the property that {\Gamma (E, T_r)\neq 0}, i.e. there are non-trivial global sections. Now, recall how {V(1,0)} works. Let {0\in E} be the origin. Then our isomorphism {E\rightarrow V(1,0)} is given by {P\mapsto \mathcal{O}(-P)\otimes \mathcal{O}([0])}. Our {T_r} is going to play the role of {\mathcal{O}([0])} here. In complete analogy we get a bijection {V(1,0)\rightarrow V(r,0)} by {L\mapsto L\otimes T_r}.

This finishes off the case of degree {0} vector bundles, because we get that {E\simeq V(1,0)\simeq V(r,0)} (roughly speaking, of course there’s a lot more structure we need to know about to say something about the moduli spaces).

In more general situations we can use the same sort of trick. Note that we always have a bijection {V(r,d)\rightarrow V(r, d+nr)} given by {V\mapsto V\otimes \mathcal{O}(n[0])}. Thus one reduction Atiyah makes right away is that (by the Euclidean algorithm) we only need to consider {V(r,d)} for {0\leq d < r}.

Now suppose the rank and degree are non-zero, and {n=\text{gcd}(r,d)}. We can establish a bijection {E\rightarrow V(r,d)}, so that the composed map {E\rightarrow V(r,d)\stackrel{det}{\rightarrow} V(1, d)\rightarrow E} is the map multiplication by {n}. The idea again is that one can find a certain vector bundle {T_{r,d}\in V(r,d)} that is unique up to isomorphism from which all the others can be produced. Most people call this the Atiyah bundle. As a corollary, we see that if {(r,d)=1}, then the moduli space of indecomposable rank {r} and degree {d} vector bundles on {E} is fine and again representable by {E}.

In modern language, we could work out that stable or semi-stable sheaves are the appropriate things to classify if we want a hope for the moduli space to be fine. It turns out that this condition of certain numerical invariants being coprime often implies that the sheaves are stable. For example, on an elliptic curve, a K3 surface, or even when dealing with relative moduli of sheaves on a K3 fibration. We may get to this another day.


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Basic Properties of Moduli Spaces

I realized I left off in really strange place last time. Sorry about that. There should be a burning question in everyone’s mind. To recap, we’ve seen that sometimes coarse moduli spaces don’t exist. When they do, sometimes there is no universal family. When there is, sometimes it is not a fine moduli space. Despite all these examples I’ve shown, I forgot to show you an example where a fine moduli space exists! So the question should be: is fine too strong a condition to ever exist?

Of course not. We can sort of cheat to make an example by working backwards. Our moduli functor {\mathcal{F}} is fine if it is naturally isomorphic to {h_M} for some scheme {M/k}. Thus take any scheme {M/k} and make the moduli functor {h_M}. This is the moduli space of points on {M}, so it is not surprising that {M} represents this functor. Still, this is a good example to understand as sort of the easiest of all possible moduli problems.

Lots of other schemes are defined to be the scheme whose functor of points is some moduli problem. The Hom scheme, the Quot scheme, the Hilbert scheme, and the Picard scheme are all examples of schemes (when they are schemes) that are fine moduli spaces for less trivial moduli problems.

Let’s look at some standard properties of moduli spaces. The moduli functor is called bounded if there is some finite type scheme {S/k} and a family {X\in \mathcal{F}(S)} such that for any object {Z\in \mathcal{F}(k)} there is some fiber {X_s} such that {Z\simeq X_s}. This is just saying that the objects you are trying to make into a moduli space fit into some finite type scheme.

The moduli functor is called separated if for any nonsingular curve {S/k} and a fixed {s_0\in S} if {X, X'\in \mathcal{F}(S)} with all fibers {X_s, X_s'} isomorphic for all {s\neq s_0} then {X_{s_0}\simeq X_{s_0}'}. This is essentially the functor of points translation of the valuative criterion for separatedness, so the term makes sense. Intuitively this is just saying that if you have a family of objects over a punctured curve, there is at most one way to fill in an object over that point to make a family over the whole curve.

The moduli functor is called complete if you can always fill in a family over a punctured curve. If there is a fine moduli space {M} associated to {\mathcal{F}}, then these properties translate exactly to the corresponding properties for the scheme {M}. Namely, if {M/k} is of finite type, then {\mathcal{F}} is bounded. Also, {M} is separated if and only if {\mathcal{F}} is separated, and {M} is proper if and only if {\mathcal{F}} is complete. The proofs are that these are exactly the respective valuative criteria.

Fix a positive integer {N} and a Hilbert polynomial {p}. Let’s do a different type of example for today. The Hilbert functor assigns to {S\in Sch_k} the set of subschemes {Y\subset \mathbb{P}^N_S} flat over {S} whose fibers all have Hilbert polynomial {p}. We won’t prove that there is a fine moduli space for this problem, since this is a fairly long proof due to Grothendieck.

One way to prove that the functor is bounded is to convert the moduli problem to one that involves the ideal sheaves of the closed subsets. Once this is done, bounded becomes equivalent to finding a single {m_0} such that all coherent sheaves in the problem are {m_0}-regular. This can be found using Castelnuovo-Mumford regularity. Thus the Hilbert scheme {Hilb_p(n)} is of finite type.

The other two conditions we talked about hold for more trivial reasons. A family over a punctured curve {S_0} is a closed subscheme of {Y\subset\mathbb{P}^N_{S_0}}, flat over {S_0}. Thus we can always fill in the punctured point by taking the scheme theoretic closure of {S} in {\mathbb{P}^N_S} where {S} is the filled in curve. This is the unique way of doing it. This shows that {Hilb_p(n)} is proper. Actually, it is projective, but this takes more work.


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Fine Moduli Spaces

To make things easy, recall we are assuming we have a fixed algebraically closed field {k}. Last time we talked about what it means for a moduli functor {\mathcal{F}: Sch_k\rightarrow Set} to have a coarse moduli space associated to it. Even though this notion is fairly weak (recall that it essentially only requires the closed points to match up with the objects {\mathcal{F}(k)} we are classifying) it can still sometimes fail to exist.

One example is to make the moduli problem of vector bundles of rank {2} and degree {0} over {\mathbb{P}^1}. You can form a family over {Spec(k[t])} that is {\mathcal{O}(-1)\oplus \mathcal{O}(1)} over {\mathbb{A}^1\setminus \{0\}} and becomes {\mathcal{O}\oplus \mathcal{O}} over {0}. The existence of such a family that jumps at this closed isolated point rules out the possibility of a coarse moduli space. This is why stable vector bundles turn up in the theory of moduli of vector bundles. Maybe later we’ll elaborate on this example, but it isn’t worth it right now.

Today we’ll discuss a stronger notion than a coarse moduli space. If our functor {\mathcal{F}} is naturally isomorphic to {h_M} for some {M\in Sch_k}, then {M} represents the moduli functor and we call {M} a fine moduli space for {\mathcal{F}}.

The reason this is a strictly stronger condition is that we automatically have the universal property since they are isomorphic, and we have a bijection {\mathcal{F}(S)=h_M(S)} for all {S}. In particular, if {S=Spec(k)} we have a bijection. This shows that every fine moduli space is a coarse one. Moreover, even coarse moduli spaces may not have a universal family, but a fine moduli space always does since we can just take the object corresponding to {1_M\in h_M(M)=Hom(M,M)}.

Let {X\in \mathcal{F}(S)} be a family of objects. We say the family is trivial if it is the pullback of some object over {Spec(k)}, i.e. {X\rightarrow S} is obtained by the base change:

\displaystyle \begin{matrix} X & \rightarrow & P \\ \downarrow & & \downarrow \\ S & \rightarrow & Spec(k) \end{matrix}

A family {X/S} is called fiberwise trivial if all fibers {X_s} are isomorphic. One very important property of fine moduli spaces is that any fiberwise trivial family must be trivial. Suppose {M/T} is the universal family. By definition, every family must be obtained by some map {S\rightarrow T}. Thus our fiberwise trivial family is the pullback

\displaystyle \begin{matrix} X & \rightarrow & M \\ \downarrow & & \downarrow \\ S & \rightarrow & T \end{matrix}

But every fiber {X_s} is isomorphic to all others, so the image of {S\rightarrow T} is a single point. Thus the family {X/S} is the pullback over a point and hence trivial.

Let’s now use this fact to show that the example we did last time that everyone probably thought was way too simple and easy to illustrate anything actually has no fine moduli space.

Let {\mathcal{F}} be the functor that assigns to {S} a smooth family of genus {0} curves over {S}. Since {\mathcal{F}(k)=\mathbb{P}^1} we have a nice coarse moduli space {M=Spec(k)} for the functor. It even has a universal family {\mathbb{P}^1\rightarrow Spec(k)}. From the chapter on surfaces in Hartshorne it is a standard fact that if you have a ruled surface {X\rightarrow C} with a section, then there is a rank {2} vector bundle {\mathcal{E}} on {C} such that {X\simeq \mathbb{P}(\mathcal{E})}. But every fiber is isomorphic to {\mathbb{P}^1} and hence we can make fiberwise trivial families that are not trivial ({X} is not just a product {C\times\mathbb{P}^1}).

A high-level explanation of this is that if {\mathcal{F}} were represented by some scheme, then the functor would have to be a sheaf, but sheaves are determined completely by local data. Thus if you are locally trivial, then you must be globally trivial.


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The Coarse Moduli Space

On this blog we’ve extensively looked at lots of things from deformation theory over the past few years. Deformation theory is in some sense a local examination of a more global object called a moduli space. Today we’ll start a brief series on moduli spaces.

Roughly speaking a moduli space is a “space” whose “points” are objects you are trying to classify. You could make the moduli space of elliptic curves in which the points are elliptic curves. You could make the moduli space of rank {3} vector bundles over some {X}. Each point of this space would correspond to a vector bundle of rank {3} on {X} (up to isomorphism). You could make a moduli space of morphisms between {X} and {Y}.

In theory, any type of mathematical thing you think up you could try to make a space of them. Algebraic geometers do this a lot, but I see no reason why you couldn’t try to study Borel measures on some metric space by trying to make a space whose points correspond to Borel measures (maybe up to mutual absolute continuity or something).

We’ve talked about the deformation functor of an object. Roughly speaking you should expect something along the following lines. Fix an object {P}. This corresponds to a point on your moduli space {M}. The tangent space of {M} at {P} should correlated to the first order infinitesimal deformation of {P}. Nearby points to {P} on {M} should correspond to more similar objects (whatever that means) and far away objects should correspond to quite different objects.

The reason I want to keep this series brief is that the subject turns incredibly technical quickly because there are lots of conditions that people impose on their objects to get the moduli space to be small enough and nice enough to study. We’ll restrict ourselves to some fairly straightforward examples.

The general idea behind constructing a moduli space is that by specifying what the type of object you want to make a space out of, you’ve told me what the functor of points of the space is. In order for it to be some sort of “space” all that we need to do is figure out what space represents this functor.

Let’s start with making some of this more precise. Unfortunately, even the easiest cases have some strange technical points that can’t be avoided. Fix an algebraically closed field (unnecessary in general), {k}. The moduli functor will be a functor {\mathcal{F}: Sch_k\rightarrow Set} from schemes over {k} to sets. The set {\mathcal{F}(S)} is the set of equivalence classes of our objects “over” {S} (which will have a meaning depending on the type of object).

The coarse moduli space (if it exists) for this functor is a scheme {M} (a highly restrictive condition we’ll remove if this series goes very far) over {k} with the property that there is a natural transformation {\mathcal{F}\rightarrow h_M} such that {\mathcal{F}(k)\rightarrow h_M(k)} is bijective and satisfies a universal property: given any other natural transformation {\mathcal{F}\rightarrow h_N} where {N\in Sch_k} there is a unique map of schemes {M\rightarrow N} so that the original map factors {\mathcal{F}\rightarrow h_N\rightarrow h_M}.

If our moduli functor has a coarse moduli space {M}, then we define a universal family for the moduli problem to be a family of objects {X} over {M} (i.e. an element of {\mathcal{F}(M)}) with the property that for each closed point {m\in h_M(k)}, the object {X_m} over {M} is the one corresponding to {m} under the bijection {\mathcal{F}(k)\rightarrow h_M(k)}.

How should we think of this? Well, our coarse moduli space is just a scheme {M} whose closed points are the objects we are considering. This was our motivating definition. (Un)fortunately, schemes have a ton more structure than their closed points. This is what is meant by “coarse”. Other than being the universal space in some sense and actually having as its points the objects we want the rest of the scheme structure is basically irrelevant for a coarse moduli space. The universal family is essentially geometrically designating to each point of {M(k)} the object that it corresponds to.

We’ll end on an extremely simple example that will become somewhat annoying next time when we want a better notion of moduli space. Let {\mathcal{F}} be the functor that classifies smooth, projective, genus {0} curves over {k} up to isomorphism. We need to make precise our notion of a relative curve over some {S\in Sch_k} if we want a well-defined functor on all of {Sch_k}.

Define {\mathcal{F}(S)} to be the set of {X\rightarrow S} which are smooth and projective with geometric fibers curves of genus {0}. This is exactly what one would expect a family of genus {0} curves to be. The functor’s value on {Spec(k)} is just a single point {\mathcal{F}(k)=\{\mathbb{P}^1\}} because all genus {0} curves (over an algebraically closed field) are isomorphic. This easily shows that {M=Spec(k)} is the coarse moduli space and there is a universal family which is to just put the one object over {M}, i.e. {\mathbb{P}^1\rightarrow Spec(k)} is the universal family.

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