Posted by: hilbertthm90 | January 1, 2012

2012 Blogging

It’s the start of a new year, so I’m going to start up something new here. My research interests have recently tended towards arithmetic geometry, so my plan for 2012 is to write some basics of algebraic number theory and arithmetic geometry that I use a lot. I’ll try to avoid redoing some of the stuff that has already been done at Climbing Mount Bourbaki.

I’d like to explain what modular forms are and what some of their basic properties are. I may detour a little into Galois representations at some point. I definitely want to talk about L-functions of varieties and what it means for a variety to be modular. This may lead to some discussion about Fermat’s Last Theorem and the Taniyama-Shimura conjecture. Scattered throughout I’ll probably have to cover some more classical algebraic number theory.

If you’re interested in related topics just post a comment and maybe I’ll get to it. Maybe in a few weeks I’ll scratch this whole idea and do something else. Who knows?

Posted by: hilbertthm90 | December 27, 2011

Music 2011

It’s that time of the year again. Here is my favorite music list from 2011. I’m embarrassed by the top 2 since they are the top 2 on lots and lots of lists out there. It seems rather uninspired for me to not find something else. I’ve divided the list into three sections. The top 10, then the pretty good but not great set (in order of how much I like them), followed by the bottom part which I found to be sub-par.

1. Bon Iver – Bon Iver
2. James Blake – James Blake
3. Chris Merrit – Songs from Brokeland
4. Bjork – Biophilia
5. Incubus – If Not Now, When?
6. O’Death – Outside
7. Matt Nathanson – Modern Love
8. Moonface – Organ Music Not Vibraphone Like I’d Hoped
9. Loney Dear – Hall Music
10. Wilco – The Whole Love

The Chris Merrit album is a placeholder. About half way through this year I was so fed up with how boring and unoriginal all the music that was coming out was that I decided to go hunting for someone I wasn’t familiar with. I ran into Chris Merrit somehow. I spent the next 2 months slowly going through everything he had ever put out. I’m not sure I found a single song I didn’t like. I listened to no new music during this time period. My favorite album if you decide to check him out is Pixie and the Bear. Please check it out. You won’t regret it. My description of him is “Ben Folds … but good.” The album listed did come out this year, but it is only a demo so I’d wait until the full mastered album comes out next year.

I’ve written about everyone on this list with the exception of O’Death and Incubus, so I’ll try to keep this short. I’ve hated Bon Iver for the past two years. He put out my favorite album of 2008, but everything since then has been blah for me and I was about to give up hope. Instead he puts out this self-titled album that really is deserving of the number one spot.

Incubus is well-known from the early to mid 2000 time period for their singles “Drive,” “Pardon Me,” “Dig,” and “Oil and Water”. They were known as a mildly experimental, but mostly mainstream alternative rock band. I’m no expert on their history, but it seems they broke up around 2008 with no intention of recording anything together ever again. Good thing they did, because this is by far their most mature effort. It is incredibly subtle and restrained. The songs have a lot of emotion and energy behind them and they didn’t just let it out in a burst of rock. It is carefully constructed and beautiful at times. I’m honestly surprised this isn’t on any of the lists I’ve looked at.

O’Death could be considered “alternative country”. It is a really fun, creative, and often dark side to folk/Americana style music. Maybe you could call it a darker sounding Mumford and Sons. They liberally use banjo and other instruments that got them the name “country” but it never sounds like country at all. They sound more like a rock band with folk influence. I highly recommend it. To me it is the most successful attempt at such a fusion I’ve come across (much better than M&S).

Now onto the list of artist I found pretty good and enjoyable, but there were too many faults to call any of them honorable mentions like I usually do.

Grand Hallway, Florence + the Machine, Fleet Foxes, Death Cab for Cutie, Dodos, Bright Eyes, Elbow, Decemberists, Son Lux, and Wye Oak

Some of my favorite songs came off of these albums, but some of my least favorite songs came from here as well. Grand Hallway is a fantastic band from Seattle and if you ever get to see them live I recommend it. They pack tons of musicians on stage including violins, piano, bass, guitar, vocalists, slide guitar, drummers, and more. They have a great powerhouse sound that only comes across properly live. The only other thing I’d like to say is that Florence + the Machine is Adele done right. See my midway rankings for my complaints about Adele. If you want to know what I was talking about then listen to F+tM to see someone who fixes all those mistakes.

Now on to the bottom. These albums fell short in a major way.

Iron & Wine, Adele, Radiohead, The Antlers, Cold War Kids, Coldplay

Iron and Wine, Radiohead, and The Antlers have all been at the very, very top of my lists in the past. It was sad to have such disappointment in them. The Radiohead album is pretty horrible in my mind. They take all the things I love about them and remove all of those aspects to leave you with a shell of boring. I’ve liked Coldplay in the past, but this was worse than radio pop nonsense earning it the lowest ranking spot. The Cold War Kids also have put out some fantastic things in the past, but this was like an attempt to mimic the Kings of Leon style and the fresh originality of their old stuff got snuffed out.

Lastly, I got the Kate Bush album, but couldn’t fit it anywhere because it was weird enough and I haven’t listened to it enough to conclude whether or not it is nonsense or amazing. She reminds me of Tierney Sutton on this album who I used to love listening to, so there is a bit of nostalgia stuck in there muddling things.

As a concluding remark, this year turned out OK. I definitely listened to stuff not from this year more than any year in the past as I got bored with the current stuff. The finds from the past that I ended up loving involve things as diverse as Iceburn, Arvo Part, and My Bloody Valentine.

Please comment with things you’d think I’d like that I missed (aren’t on the list). I’ve already been informed I should check out the M83 album and the Destroyer album.

Posted by: hilbertthm90 | December 4, 2011

Multiplication is Still Repeated Addition…yet again

Keith Devlin has posted yet another blog post on how multiplication is defined. I left a comment there, but it wasn’t as good as I hoped it would be. Now that I’ve gone back and read all the comments, I am completely convinced that I know exactly where the misunderstanding is occurring. I should maybe throw in that the comments have been turned off for that post. Maybe my blog doesn’t get enough traffic, but I can’t ever, ever see a reason to do that. You filter out a few comments that are off topic, but don’t shut down comments completely. It seems to go in blatant opposition to the idea of using a blog rather than just publishing somewhere. But that is a completely side issue.

Devlin wants to claim that the definition of multiplication is to define a function f:\mathbb{N}\times \mathbb{N}\to \mathbb{N} in Peano arithmetic using recursion. Since the “definition” as repeated addition does not yield a function on the whole natural numbers it cannot possibly be the definition. In this strictest sense, the definition is not repeated addition because it is this other recursively defined function.

I’m with him on all of this. Of course that is true. Repeated addition only works for a finite stage and you must relying on something else to get the full function. The place we disagree is in what we mean by “definition”. Devlin wants to say that because it is necessarily the case that to define an honest multiplication function axiomatically in Peano arithmetic we must make some part of the definition that is not repeated addition, then we must say that the definition is not repeated addition.

That is a completely absurd idea. Devlin is conflating the two mathematical notions of definition and existence. You can make a definition of anything you want in math, but that doesn’t mean it exists. You then should prove that such a thing exists. I claim that the above “not repeated addition” definition of a function is actually the proof of the existence of a multiplication function and not a definition.

Recall that the context of this conversation is all about whether or not we should teach children that multiplication is repeated addition. The whole point of teaching children multiplication is precisely so that they can apply it to real world situations. The only sense in which multiplication can be applied to real world situations is if we know that it produces the same answer as repeated addition. I talk more about that in the original blog post. Even these strange analogies about stretching being a continuous rather than a discrete concept only makes sense after interpreting it as repeated addition.

Now that we’ve established the difference between definition and existence I’ll decompose what is going on to show that multiplication is repeated addition (by definition and hence ought to be taught to children because it isn’t incorrect). Recall that definitions in mathematics are technically biconditional statements. Definition: A function m:\mathbb{N}\times\mathbb{N}\to \mathbb{N} is multiplication if and only if it satisfies m(n,k)=k+k+\cdots +k repeated n times for any choice of n,k in \mathbb{N}. Note that this is not only the definition that will tell us the multiplication function is useful, but a priori we don’t know that such a function exists.

Devlin’s post uses recursion to prove that such a function exists, but once these two notions are pulled apart we see that Devlin is actually incorrect. The (useful) definition is the one I give above. If we try to conflate the existence and definition as Devlin does, then we have no idea if the recursively defined f can be used to compute something in reality. As I point out in the comments at that blog post, for all we know that nice recursive definition could yield f(n,k)=2 for all inputs. Are we going to call that multiplication? I think Devlin wrote that off as a flippant comment to be ignored, but I still think it illustrates the necessity of separating definition from existence perfectly and is precisely our point of disagreement.

Now to try to summarize briefly, it seems that when Devlin wants to use the phrase “multiplication is not defined to be repeated addition” he seems to mean that in order to have a function defined on \mathbb{N}\times\mathbb{N} it must be defined recursively without reference to repeated addition. I agree completely. I just disagree that the definition of such a function should be consider the definition of multiplication (is the confusion that “definition” is used in reference to “defining a function” versus defining the term “multiplication”?). The actual definition should be the one I gave (otherwise we’ve made a useless function), and then the proof that such a function exists is what Devlin wants. I don’t see how Devlin could disagree with what I’ve written here, and hence I see no way to conclude anything other than multiplication is repeated addition…unless we are going to use some non-standard idea of what it means to “be” something.

Posted by: hilbertthm90 | November 26, 2011

Music 2008-2010

In preparation for the music 2011 list I thought I’d look back at my 2008-2010 lists and see what I still listen to and what has fallen by the wayside as impressive at first but not lasting. It is kind of funny to do this, because really these lists tend to be way off in terms of long-term listening. They reflect how excited I was about certain albums at first, but when I want something solid to fall back on when I’m bored with the current I tend to pick things in the middle or lower of the list.

What do I continually go back to in 2008? Definitely NIN’s The Slip. Honestly, I still listen to this on a regular basis. It is fantastic. I sometimes have cravings for it. Out of everything I post this is probably the thing I return to the most. Lightspeed Champion’s Crossing the Lavender Bridge has not faded as a fantastic album either. The third place by a significant gap is The Helio Sequence’s Keep Your Eyes Ahead. It is quite good and I return to it frequently, but not the same as the other two. Another really good one that I periodically come back to (but not that frequently, more for nostalgia’s sake) is Land of Talk’s Some Are Lakes. I recommend this above all other things they’ve put out. The others are good as well, but this one I come back to the most.

How about 2009? I return to almost nothing from this year. What happened guys? Sometimes I come back to Loney, Dear’s Dear John or Regina Spektor’s Far. I still think that Phoenix’s Wolfgang Amadeus Phoenix is probably the best thing to come out that year. The Dirty Projector’s Bitte Orcha was amazing, but honestly I rarely return to it.

What about last year? Has it been long enough? Well, Joanna Newsom’s Have One on Me is still so good…so, so good. Really amazing. It is my ring tone for goodness sake. OK, I have several ring tones depending on who calls me, but all of them are off this album. Very rarely do I come back to anything else. I don’t think I gave The Tallest Man on Earth enough credit last year. I continually come back to that. It is really good. Other than that I sometimes come back to Moonface, Interpol, or … I hate to admit it Kanye West (come on, admit it, that was a pretty awesome album). I unfortunately missed Lightspeed Champion’s new album last year Life Is Sweet! Nice to Meet You which has been consistently amazing for me this year.

That is my summary of what in retrospect was good in 2008-2010. In a few weeks I’ll post my new best of 2011 series.

Posted by: hilbertthm90 | November 16, 2011

Mirror Symmetry A-branes

I started writing this post this past weekend, but got stuck really quickly and then kept putting it off. I don’t want to leave anyone following this hanging with no idea what the A-model is. This is harder for me to describe than the A-model for some reason. Mostly I’m running into the problem of either just saying what the A-side is without explanation or I’m getting too bogged down in details. Both seem bad. In conclusion, I think I’ll err on the side of too few details, and then hopefully make sense of what is going on by completely describing mirror symmetry in the easiest case possible: the one dimensional case, i.e. for an elliptic curve.

I’m going to semi-cheat right off and refer to posts over a year old. Recall what a symplectic form is on a smooth manifold is. It is just a closed non-degenerate 2-form. A smooth manifold plus symplectic form is called a symplectic manifold. The cotangent bundle always has a canonical symplectic form on it. An example that may be less well-known is that any smooth complex projective variety is symplectic because the Fubini-Study Kähler form on {\mathbb{P}^n} restricts to a symplectic form.

If we just think about vector spaces for a second, then given a symplectic form, we say that a subspace {S} is isotropic if {S\subset S^\perp} and coisotropic if {S^\perp \subset S}. The subspace is Lagrangian if it is both isotropic and coisotropic. This extends to manifold language easily by saying an embedded submanifold {S\subset M} is Lagrangian if the tangent subspace {T_sS\subset T_sM} is Lagrangian for every point of {S}. If you want to get used to these definitions, a quick exercise would be to check that the zero section of the cotangent bundle is Lagrangian with respect to the canonical symplectic structure.

My second semi-cheat is to ask you to recall the definition of an almost complex structure from close to two years ago. The way to think about it is that it is a bundle map {J: TM \rightarrow TM} that behaves similarly to “multiplication by {i}“. The condition is that {J^2=-Id}, and indeed multiplication by {i} when identifying {\mathbb{R}^2\simeq \mathbb{C}} gives an example of an almost complex structure. In fact, since we’ll always work over {\mathbb{C}}, any complex manifold does have multiplication by {i} as a natural almost complex structure.

It is possible that all these things are related by the following. Suppose {(M, \omega)} is a symplectic manifold, {J} an almost complex structre, and {g} a Riemannian metric. These three structures are called compatible if {\omega(J(-), -)=\langle - , -\rangle_g}. I am far out of my depth here, but I’m pretty sure such a manifold is called Kähler if this happens, but maybe some slight more conditions are needed (e.g. does this automatically imply that {g} is Hermitian? If so, then this is definitely what people call Kähler).

Now for the definition of the A-model. Let {(M, \omega)} be a Kähler (in the sense of the previous paragraph) manifold. We define the Fukaya category {Fuk(M)} to have as objects the Lagrangian submanifolds. The morphisms require a bit of technicality to define, but essentially are a way to intersect the submanifolds. It involves all the structures above and is called Floer cohomology. Recall that we’re merely sketching an idea here! Somehow this should be an {A_\infty} or dg-category if you remember from last time, and this just comes from the fact that the morphisms have to do with cohomology classes of intersections.

If you’ve been following this at all, then you should be in utter amazement. We can state mirror symmetry now as an equivalence of {A_\infty} categories {D^b(X)\rightarrow Fuk(\widehat{X})} where {X} is a Calabi-Yau. Why is this amazing (for those not following along)? Look at the left side of this equivalence. The bounded derived category of coherent sheaves (in the Zariski topology!!) on {X} is something that has to do purely with the algebraic data of {X}. I mean, the Zariski topology is algebraic, the definition of coherent is very algebraic, the construction of the derived category is algebraic, etc.

The right hand side seems to have forgotten all of the algebraic data. You forget that it is a variety and instead think of it as a smooth manifold. You consider a bunch of structure that helps you study the smooth structure. You consider Lagrangian submanifolds. The Fukaya category is almost entirely analytic in nature. But now the conjecture of Kontsevich mirror symmetry is that the two are always equivalent. That’s it for today. There should be one more post in this series in which I try to sketch the conjecture in the case of an elliptic curve.

Posted by: hilbertthm90 | November 9, 2011

Side Note

When I told people I was typing up notes on Deligne’s proof (written by Illusie) that all K3 surfaces lift from positive characteristic to characteristic zero they were really excited and wanted a copy. I’m not sure if any of those people read my blog (in fact I’m pretty sure they don’t!), but just in case other people are interested I just put up a rough copy under Expository/Notes at my department webpage.

Posted by: hilbertthm90 | November 5, 2011

The B-model

Sorry for the delay, but I’ve been incredibly busy and this topic is basically lowest priority right now. Today we will finish describing the “B-side” of mirror symmetry. The typical way this is done is with {A_\infty}-categories. I’m going to do it using dg-categories, because I’m much more comfortable with the language. Since we are working over {\mathbb{C}} it turns out these two things are exactly the same. I won’t go into why that is the case (probably because I don’t have an understanding why they are). The reason it is called the “B-side” is that we are constructing what is called the B-model. The statement of mirror symmetry will be something along the following lines: the B-model on {X} is equivalent to the A-model on {\widehat{X}} where {\widehat{X}} is that mirror pair we looked at here.

The two different “models” have something to do with string theory that I definitely don’t understand. As I said above we will need the notion of a dg-category. Without being incredibly careful with this definition (we’ll see later that this essentially adds no information to our derived category) a dg-category, {T}, (over {\mathbb{C}}) is a collection of objects {Ob(T)} such that for any pair of objects and any integer we have {T(x,y)^n} a {\mathbb{C}}-module. This should be thought of as morphisms in degree {n}. There is a composition of morphisms, and it should act like multiplication in a graded ring, so {T(x,y)^n\times T(y,z)^m\rightarrow T(x,z)^{n+m}} is bilinear and associative.

There needs to be an identity map in degree zero for any object which we notate {e_x\in T(x,x)^0}. So far nothing should feel strange to you, but our final condition is that there is a differential {d: T(x,y)^n\rightarrow T(x,y)^{n+1}} with the property that {d^2=0} (i.e. {Hom(x,y)} is a complex with respect to the differential) satisfying a graded Leibniz rule {d(fg)=d(f)g+(-1)^mfd(g)}.

If you want something to help you wrap your head around this here is an analogy. A groupoid is a category in which every morphism is invertible, so you can think of any group {G} as a groupoid by taking the category to have only one object and {Hom(*,*)=G}. In a similar way, a dg-category or “differential graded” category can be formed by taking any differential graded algebra {A=\bigoplus A^n} and then forming the category with exactly one object and {T(*,*)^n=A^n}. All those definitions are just basically turning the definition of a differential graded algebra into a category in a consistent way.

Let {X} be a smooth projective variety over {\mathbb{C}}. Then we have already constructed a triangulated category {D^b_{Coh}(X)}, the derived category. This is not a dg-category, but we can ask if there is a dg-enhancement. It is a bit technical to describe what this means, but roughly an enhancement is asking for a dg-category {T} where the homotopy category {H^0(T)} is equivalent to {D^b(X)}. Here {H^0(T)} means to take the same objects and define the morphisms to just be the {0}-th cohomology {H^0(Hom(x,y))}.

There are varying levels of strength of what it means to be a unique enhancement mostly stemming from the fact that there is a choice of equivalence {H^0(T)\rightarrow D^b(X)} that may or may not be respected. Again, we’ll skip over these technicalities because the hope is to take away from these posts a general flavor of what mirror symmetry says rather than trying to describe all the technical details (the Clay Math book called “Mirror Symmetry” is 929 pages for goodness sake!). Thanks to Lunts and Orlov when our variety is projective we get that there always exists a dg-enhancement and it is unique in the strongest sense.

A priori a dg-category is big with lots of information and behaves nicely (is relatively easy to work with). A triangulated category has very little structure and behaves rather poorly (is difficult to work with). This enhancement theorem says that it doesn’t really matter which one we work with if we care about {D^b(X)} which should be surprising. Somehow {D^b(X)} is just the {0}-th part of the big dg-category so for instance having an equivalence of the triangulated categories doesn’t seem strong enough to be able to extend it to an equivalence of the whole dg-structure, but it is.

To conclude this half of the series, suppose we have a Calabi-Yau threefold {X}, then for our purpose we will call the B-model of {X} denoted by {B(X)} to be the unique dg-enhancement of {D^b(X)}. Mirror symmetry will eventually be an equivalence (as dg-categories) with {A(\widehat{X})}. This means we need to move on to the {A}-model.

Posted by: hilbertthm90 | October 22, 2011

The Derived Category 2

I thought we would be able to move on to dg-categories today, but I was wrong. There was too much we didn’t do last time, plus an example might be nice. Recall that last time we took an arbitrary abelian category {\mathcal{A}} and created the derived category {D(\mathcal{A})} by making the category whose objects are complexes and morphisms are equivalence classes up to homotopy equivalence and we “invert” all quasi-isomorphisms (they become isomorphisms).

This post should be taken as just a smattering of remarks on this construction plus focusing in on the actual category we want to consider. Since the first step of the construction was to take the category of complexes we can actually alter this part of the construction in several natural ways and complete the next two steps without any problem. We called the category of complexes {Kom(\mathcal{A})}. Instead of using this whole category we might be interested in just the complexes that are bounded below (meaning {A^i=0} for all {i<<0}) or bounded above or just plain bounded. These will be denoted {Kom^+(\mathcal{A})}, {Kom^-(\mathcal{A})}, or {Kom^b(\mathcal{A})} respectively. Finishing the construction of the derived category in the same way with these gives {D^+(\mathcal{A})}, {D^-(\mathcal{A})} and {D^b(\mathcal{A})}.

For the purposes of this post we will focus on the bounded derived category. The first remark is that if {\mathcal{A}} has enough injectives, then in {D^b(\mathcal{A})} for any {A^\bullet} there is a complex {I^\bullet} of injective objects such that they are isomorphic {A^\bullet\simeq I^\bullet}. I.e. any bounded complex is quasi-isomorphic to a complex of injectives and so if all we care about are objects up to isomorphism we may as well just work with complexes of injective objects.

In fact, something much stronger is true. Take {\mathcal{I}\subset \mathcal{A}} to be the full subcategory of injective objects, then we have an equivalence of categories {K^+(\mathcal{I})\simeq D^+(\mathcal{A})}. Now let's shift our focus to {D^b(Coh(X))} where {X} is a scheme. Since this is the category that will appear the most we will simply denote it {D^b(X)} and call it "the derived category of {X}" instead of the more accurate and cumbersome "the bounded derived category of coherent sheaves on X".

If you're paying attention you should be objecting to all of this post at this point. My remarks are useless here because {Coh(X)} might be a nice abelian category, but it certainly doesn't have enough injectives. The fix for this is that in a sense that we'll make precise it is often good enough to take your complex of injectives in a category that has enough injectives as long as when you take cohomology it lands you back in the right place.

The exact statement is that if {\mathcal{A}\subset \mathcal{B}} is a thick subcategory and any {A} can be embedded in some {I\in \mathcal{A}} which is injective as an object of {\mathcal{B}} then we get an equivalence of categories {D^b(A)\simeq D^b_\mathcal{A}(\mathcal{B})} where the notation {D^b_\mathcal{A}(\mathcal{B})} means the full subcategory of {D^b(\mathcal{B})} of complexes with cohomology in {\mathcal{A}}.

Now to not get in trouble at some point let's suppose our scheme {X} is always a smooth projective variety over a field {k} (this is the only case we'll consider in mirror symmetry so it's fine for these purposes). If we just take for a moment the category {QCoh(X)} of quasi-coherent sheaves on {X}, then it is a standard theorem in algebraic geometry that any quasi-coherent sheaf {\mathcal{F}} has an injective resolution {0\rightarrow \mathcal{F}\rightarrow \mathcal{I}^\bullet} by quasi-coherent sheaves that are injective as {\mathcal{O}_X}-modules. If you think of the definition of the sheaf cohomology {H^i(X, \mathcal{F})} in relation to the cohomology of the injective resolution complex it immediately shows that we can view {D^b(QCoh(X))} as the full subcategory {D^b_{QCoh}(Sh_{\mathcal{O}_X}(X))} of the bounded derived category of sheaves of {\mathcal{O}_X}-modules on {X}.

Even though this can't be done directly by some general theorem as in the quasi-coherent case we actually do get a theorem that parallels that case (again, warning: this cannot be done the same way since coherent things will not be injective). There is a natural functor {D^b(X)\rightarrow D^b(QCoh(X))} and it induces an equivalence of {D^b(X)} with the full subcategory {D^b_{Coh}(QCoh(X))}. This means that to work with the derived category of {X} we can always replace (up to isomorphism) a complex of coherent sheaves by a complex of injective quasi-coherent sheaves whose cohomology is coherent.

There are so many interesting things we could talk about at this point (a derived form of Serre duality, how derived functors relate to this, when can you recover a variety from its derived category, Fourier-Mukai transforms, and so on…), but my goal is to keep trudging towards a statement of mirror symmetry and stopping to talk about all these things would lead us too far astray. Although, talking about a few of them might help some of you who might be in unfamiliar territory here feel a little more comfortable.

The last thing we'll look at today is an example. This might be completely unenlightening because I won't take the time to properly define everything, but we'll do it anyway. What is {D^b(\mathbb{P}^n)} or even just {D^b(\mathbb{P}^1)}? We will say that the derived category {D^b(X)} is generated by a collection of objects {S} if the smallest full subcategory that contains {S} and is closed under taking shifts and cones (we haven't defined that, but recall that the derived category satisfies this condition called being triangulated and those conditions just make sure the generated category is still triangulated) is {D^b(X)} itself.

For some intuition, what this means is that any object of {D^b(X)} can be reached by doing the two main operations of {D^b(X)} a finite number of times using only the objects of {S}. In the case of projective space we get that {D^b(\mathbb{P}^n)} is generated by just {n+1} objects {S=\{\mathcal{O}(n), \mathcal{O}(n-1), \ldots, \mathcal{O}(1), \mathcal{O}\}}.

Posted by: hilbertthm90 | October 20, 2011

The Derived Category 1

I promised some sort of an explanation of Kontsevich mirror symmetry. Going in there are two things to know. First, it is a conjecture so it hasn’t been established yet. Second, it isn’t even really a well-formulated conjecture. There are cases where there is a firm conjecture, but there are other cases where part of the conjecture is to figure out what the conjecture should be.

The other thing I should point out is that this is sometimes called “homological mirror symmetry”. The general form is that a certain derived category should be equivalent to a certain other category. This means that for at least the first two posts of this series I will only be describing two different categories and it won’t be at all obvious why this should be related to the mirror symmetry I alluded to last time.

Today we’ll discuss what the derived category of an abelian category is. In your head you can just keep thinking that our category is the abelian category of coherent sheaves on some sufficiently nice scheme or variety, since this is what will come up in mirror symmetry. I see absolutely no reason why I shouldn’t do the construction in the more general case of an aribtrary abelian category since it doesn’t introduce any extra difficulties (and for algebraists you can just think of {R}-modules or something).

The notation in the literature varies greatly. I’m going to follow Huybrechts. Let {\mathcal{A}} be an abelian category. One basic related category is to form {Kom(\mathcal{A})} which is the category of complexes. The objects are chain complexes {A^\bullet} and morphisms {f: A^\bullet \rightarrow B^\bullet} are just maps {A^i\rightarrow B^i} for all {i} that commute with the morphisms in the complex. This is again an abelian category and there is an embedding {\mathcal{A}\rightarrow Kom(\mathcal{A})} as a full subcategory by {A\mapsto (\cdots \rightarrow 0 \rightarrow A \rightarrow 0 \rightarrow 0 \rightarrow \cdots)} where {A} sits in degree {0}.

Given any object {(A^\bullet, d^\bullet)} in {Kom(\mathcal{A})} (i.e. any complex of objects from {A}) we can take cohomology {H^i(A^\bullet)=ker(d^i)/im(d^{i-1})}. The cohomology lies in {\mathcal{A}}. Given a morphism of complexes {f:A^\bullet \rightarrow B^\bullet} there is an induced map on each cohomology {H^i(f): H^i(A^\bullet)\rightarrow H^i(B^\bullet)}. If {f} induces an isomorphism for all {i} we say that {f} is a quasi-isomorphism.

Loosely speaking the derived category {D(\mathcal{A})} is a category (not necessarily abelian) in which quasi-isomorphisms become isomorphisms. One way to construct it is by proving the existence of a functor {Q: Kom(\mathcal{A})\rightarrow D(\mathcal{A})} that takes quasi-isomorphisms to isomorphisms and {Q} satisfies a certain universal property. This isn’t so easy to work with so we’ll describe {D(\mathcal{A})} in a different way.

Recall that two morphisms {f,g: A^\bullet \rightarrow B^\bullet} are homotopy equivalent if there is a collection of maps {h^i: A^i\rightarrow B^{i-1}} such that {f^i-g^i=h^{i+1}\circ d_A^i+d_B^{i-1}\circ h^i}. We can form {K(\mathcal{A})} often called the homotopy category of {\mathcal{A}} where the objects are still just complexes, but a morphism of complexes is an equivalence class where the relation is being homotopy equivalent.

The functor {Q} factors through {K(\mathcal{A})}. To describe the derived category {D(\mathcal{A})} I’ll just tell you the objects and the morphisms. The objects are still just complexes, i.e. the objects of {K(\mathcal{A})}. The morphisms are trickier to describe. A morphism {A^\bullet \rightarrow B^\bullet} is a choice of (equivalence class of) quasi-isomorphism {C^\bullet \rightarrow A^\bullet} plus a map (in {K(\mathcal{A})}) {C^\bullet \rightarrow B^\bullet}.

This may look weird at first, but recall that all we are really trying to do is make all our quasi-isomorphisms actual isomorphisms. Thus when giving a morphism {A^\bullet \rightarrow B^\bullet} you may need to pick something isomorphic {C^\bullet} before defining the map, and then since {Q} factors through {K(\mathcal{A})} everything must be done only up to homotopy equivalences.

The derived category is no longer abelian, but it is what is known as a triangulated category. We will not talk about the general definition of one of these, since it is a sort of ad hoc, difficult, poorly behaved definition. Some people might tell you that the reason for the bad behavior is actually a result of forgetting too much stuff. You are actually decategorifying (an {\infty}-delooping, whatever that means) a stable {(\infty, 1)}-category which is a very nicely behaved structure and you may as well just work with the nicer thing.

Now whether or not you actually believe this (I’m on the fence myself) there is a much nicer, easier to understand way to say this for the derived category of coherent sheaves on a projective variety. We’ll talk about this next time, because it is actually a necessary part of the statement of Kontsevich Mirror Symmetry.

Posted by: hilbertthm90 | October 10, 2011

Quick Update

I will be doing the Mirror Symmetry thing. I had someone “like” the post which I’ll count as a vote for it and someone explicitly vote for it. That’s two whole people! For all I know that is half of my regular readers, so by majority rule I have to do it.

I haven’t gotten around to a post because I’ve been incredibly busy. I’ve been doing my usual research and I’m taking two classes this quarter. In addition to all that I’ve been teaching a “mini-course” in our AG club. So for the next several weeks when I have down time I’ll probably think about what I’m going to say there and/or I’ll be typing up notes for that. I should point out that mirror symmetry has two parts, the derived category side and the Fukaya category side. The derived side is something that is really close to what I do, so it is without a doubt worthwhile to blog about that half, and I plan to soon. I may be a bit sketchier on the side that isn’t as important to me.

I’ll just leave you with a quick taste of what some people mean by “Mirror Symmetry” which is a bit different than Kontsevitch Mirror Symmetry which is what I’ll be explaining at some point. Suppose you have a Calabi-Yau threefold. By this I’ll mean a smooth, projective variety of dimension three over \mathbb{C} with \omega_X\simeq \mathcal{O}_X and H^1(X,\mathcal{O}_X)=H^2(X,\mathcal{O}_X)=0. By general Hodge theory there is a symmetry in the Hodge numbers h^{pq}(X)=\dim_{\mathbb{C}}H^q(X, \Omega^p), namely that h^{pq}(X)=h^{qp}(X). Also, you can use the fact that it is Calabi-Yau to check that all Hodge numbers are completely determined (independently of X) as either 0 or 1 except h^{11} and h^{12}. Fun exercise in Serre duality!

Suppose X is a Calabi-Yau threefold. There is some specified h^{11} and h^{12} (the only two unknown Hodge numbers). A mirror pair for X is another Calabi-Yau threefold with h^{11} and h^{12} swapped. In my brief encounter with Kontsevich Mirror Symmetry (which says something about an equivalence of categories) this will follow as a special case.

Since I’m in the mood I may as well say some things that immediately pop into mind when seeing this as someone that has recently been thinking in the arithmetic world. If we are over an algebraically closed field of characteristic 0, then there is a result that says h^{11}>0. In particular, there cannot be a rigid CY 3-fold if mirror symmetry is true, since h^{12} gives the space of deformations of X. But in positive characteristic there are tons of rigid CY 3-folds! Interesting.

I’ll leave you with that little taste of what is to come.

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