Posted by: hilbertthm90 | January 24, 2010

Spectral Sequence Arrows Revisited

I don’t want to do much today. In fact, we’ll probably not cover any new ground, but start to fill in what is going on a little better. Notice that in order to work with the spectral sequence last time, we didn’t actually have to figure out what the {i, j,} or {k} or {d} maps were. We only needed that they existed. The main thing of today is to work out all these sequences. Some are exact. Some are chain complexes. Some are commutative diagrams. Most of them fit together in a really organized way, other parts are just related. It is useful to see how these fit together, since sometimes you can get by without the maps as long as you know it is part of a commutative exact diagram.

Recall that our current situation is that given a filtered chain complex {\cdots K^{p-1}\subset K^p\subset \cdots \subset K} we get an exact couple {(D_{p,q}, E_{p,q})} where {D_{p,q}= H_{p+q}(K^p)} and {E_{p,q}=H_{p+q}(K^p/K^{p-1})} by taking the long exact sequence in homology associated to the short exact sequence {0\rightarrow K^{p-1}\rightarrow K^p\rightarrow K^p/K^{p-1}\rightarrow 0}.

Two times ago we saw exactly what the {i_{p,q}}, {j_{p,q}} and {k_{p,q}} maps were.

Let’s develop them again, but being more careful as to how they all fit together. Well {i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})} and it comes from the induced maps on homology from the injection {K^p\hookrightarrow K^{p+1}}. So we get infinite strings of {i_{p,q}} for fixed {q} that relate as injections {p} changes. i.e we’ll get as part of a large diagram columns that look like {\cdots \rightarrow H_{p+q}(K^{p-2})\rightarrow H_{p+q}(K^{p-1})\rightarrow H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})\rightarrow\cdots}.

The {j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})} are induced on homology from the projection map {K^p\rightarrow K^p/K^{p-1}}.

The {k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})} comes from the snake lemma. It is the boundary map in the long exact sequence.

Note that these last two will fit into rows: {\cdots \rightarrow H_{p+q}(K^p)\stackrel{j}{\rightarrow} H_{p+q}(K^p/K^{p-1})\stackrel{k}{\rightarrow} H_{p+q-1}(K^{p-1})\stackrel{j}{\rightarrow} H_{p+q-1}(K^{p-1}/K^{p-2})\rightarrow\cdots}.

Thus we can form a large commutative diagram from these long exact sequences, and figure out how to extrapolate the exact couple from it. Below is the diagram relating what was just said:

(Sorry about the image. I’ll figure this out eventually. For now you can click on it and zoom to get the full size).

The labelled arrows are the {i,j,k} maps of just a single exact couple. Note that there are infinitely many exact couples going on here, and they are all related. Note also that the exact couples don’t just go across or down.

Now let’s figure out the differentials and how to get the spectral sequence. Well, take the {H_{p+q}(K^p/K^{p-1})} spot. Recall that in the exact couple, this is the {E} term which is where our chain complex comes from. Then doing {j\circ k} is the {d} map. Note this is the really the {d^1} map. So our page of {E^1_{p,q}} comes from this row only.

Take homology with respect to this, and due to the nature of the exact couples, we’ll get another page of these things. Now we are on the {E^2_{p,q}} page of groups. The {d^2} map now is to go right, then up, then right. This is the composition {k\circ i^{-1} \circ j: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-2}/K^{p-3})}.

Take homology again. In general the {d^r} map is going right, then up {r-1} times then right one more time. This gives us exactly what we said last time, {d^r_{p,q}: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}.

Hopefully this helps clarify some of the inner workings of what is going on here. Next time we’ll talk about why the conditions from last time imply that the spectral sequence converges. We won’t prove it, but there are some nice concepts behind why it works that should be enlightening.

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Posted by: hilbertthm90 | January 21, 2010

Serre Spectral Sequence

Today we’re going to back up and not worry about the tediousness of the last post. Let’s try to form some motivation and see how these things work even if we have to just assume some of the theory works for now. I think it will be helpful. So why care about spectral sequences? Well, one thing is that sometimes it is really hard to compute say {H_n(C)}, but it isn’t so bad to do {H_n(C\otimes A)} for some {A}’s. There is no good way to convert the information you get out of {H_n(C\otimes A)} to information about {H_n(C)}, but this is exactly what the Bockstein spectral sequence does.

Today, I’ll talk about the Serre spectral sequence. This will help us calculate the homology of the loop space of an {n}-sphere. There is a nice proof that Tor is independent of which resolution you take (i.e. {Tor(A, B)} can be calculated by resolving {A} or {B}). There is a nice proof of the Snake Lemma using spectral sequences. There is a proof of the K\:{u}nneth formula. For the algebraic geometers, the Leray spectral sequence tells us information about sheaf cohomology. Anyway, there are tons of examples hitting many areas of math of uses of spectral sequences.

Right now we’ll not worry about the technical details of when and how a spectral sequence converges, but recall that the situation we ended on was that given a complex {K}, we could form a spectral sequence from a filtration {\cdots \subset K^p\subset K^{p+1}\subset \cdots \subset K}. We’ll just accept for now that if we have {K^p=0} for {p\max(p, q+1)} then {E^r_{p,q}\simeq E^\infty_{p,q}}. This {E^\infty} term is the associated graded group of a filtration of {H_*(K)}.

This should be exciting. It means that if we can formulate a statement that reads, “There is a spectral sequence with {E^2_{p,q}=} (fill in the blank), differential map {d^r: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}, that converges strongly to {H_{p+q}}(fill in again).” Then we can just chase around differentials and hope that everything collapses fast (lots of times these things only have a few terms or even {E^2=E^\infty}). If this happens then we can just read off what the known things are, and we’ll have figured out new information. Note that being able to say such a statement is not bad. We only need the mere existence to begin work, and taking those conditions above for granted, we actually can formulate lots of statements such as these.

Let {\Omega S^n} be the loop space of {S^n}, namely the space of loops based at the north pole. Let {PS^n} be the path space, the space of paths starting at the north pole. Note here that our standard tools of algebraic topology are not very useful in trying to calculate {H_q(\Omega S^n)}. But we know that {PS^n} is contractible using the obvious map of retracting along all the paths simultaneously. And we also know something very useful, that there is a spectral sequence with {E^2_{pq}=H_p(S^n)\otimes_\mathbb{Z} H_q(\Omega S^n)}, differentials {d^r : E^r_{pq}\rightarrow E^r_{p-r, q+r-1}} converging strongly to {H_{p+q}(PS^n)}.

So recall that we get an entire page of groups. The {E^2} groups ({r=2}). Since {H_p(S^n)} is {0} for all {p} except {p=0, n}, and is {\mathbb{Z}} in those spots. We also know that {H_0(\Omega S^n)=\mathbb{Z}}. This allows us to fill in the whole {q=0} row, and all {p} columns are completely {0} except in {p=0} and {p=n}.

The differential map {d^r} goes to the left {r} and up {r-1}.

Examine the bottom row. Due to all the {0}’s in columns not {p=0} or {n}, all the differential maps that don’t land in the {p=0} column must be 0. Thus there is only one possible non-zero differential coming out of there. Namely, the {d^n_{n,0}} map. It lands in the {E^2_{0, n-1}=H_{n-1}(\Omega S^n)} spot. Now the complex of {d^n} maps is {\cdots \rightarrow 0\rightarrow \mathbb{Z}\stackrel{d^n_{n,0}}{\rightarrow} H_{n-1}(\Omega S^n)\rightarrow 0\rightarrow \cdots }. Recall that to get to the different {r} values in {E^r_{p,q}} you take homology with respect to the {d^r} maps. Since {E^\infty_{p,q}=H_{p+q}(PS^n)=0} for {p+q\neq 0} we must have {d^n_{n, 0}} an isomorphism in order for it to vanish. This gives us a {\mathbb{Z}} in the 0, {n-1} spot and hence another in the {n}, {n-1} spot. We can keep repeating this argument giving a {\mathbb{Z}} in the {p=0, n} and {q=} multiples of {n-1} spots.

Since the differentials are all {0} before {r=n}, nothing changes for {E^2=E^3=\cdots E^n}. Then at {r=n} we get isomorphisms except at the {E^2_{0,0}} position, so everything vanishes except that position which stays a {\mathbb{Z}}. Thus {H_q(\Omega S^n)=\begin{cases} \mathbb{Z} \ if \ (n-1)|q \\ 0 \ else\end{cases}}.

This is a pretty typical situation. Hopefully that gives a better overview of the process so that going back to the general case will have some sort of reference. If there are questions, feel free to ask. Maybe I’ll do another example before going back to theory.

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Posted by: hilbertthm90 | January 18, 2010

Exact Couples

Today we begin a long set of posts on spectral sequences. I must first off say that I only learned what a spectral sequence was about 2 weeks ago, so I am far from knowledgable about them. This means if you have questions, I may not know the answers. I’m also going to go incredibly slowly. I want to carefully develop everything and give lots of motivating examples. Lastly, all of this is coming from the notes I’ve taken from John Palmieri’s class this quarter, so any great insights should be attributed to him and mistakes should be attributed to myself.

Like I’ve already said, I’m not familiar with the literature, so I’m not sure what the standard development is. For people who know some of this already, we’re going to define the spectral sequence associated to an exact couple. Then anytime we want a spectral sequence, we just need to figure out the exact couple it comes from. This is a different approach than say Weibel.

In general, this is going to work in any abelian category, but we’ll use the category of {A}-modules just to keep it less scary for those that haven’t worked with abelian categories.

Our first definition is that of an exact couple. This is a diagram of {A}-modules, {D} and {E} such that:

is exact in every spot. Then we immediately get a map {d: E\rightarrow E} by {d=j\circ k}. This may look strange, since we’re not just going around the diagram. We sort of jump, but this is fine since at this stage we aren’t identifying {i(D)\subset D} or anything. Now this map has a nice property, {d\circ d=j\circ k\circ j\circ k=j\circ (k\circ j)\circ k=j\circ (0)\circ k=0} by exactness that middle term is {0} making the whole thing {0}.

This means that {\cdots \stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\rightarrow \cdots} is actually a chain complex. Thus we can take homology: \displaystyle {H(E, d)=\frac{ker d}{im d}}.

Now take {D'= im i}. Define the map {i': D'\rightarrow D'} by {i'=i\big|_{D'}}. Take {E'=H(E, d)}. We will now think of {j': D'\rightarrow E'} as {j\circ i^{'-1}}, but to be precise we pick an element whose preimage is {x=i^{-1}(y)}, and take an equivalence class, so {j'(x)=[j(y)]}. Lastly define {k': E'\rightarrow D'} by {k'([z])=k(z)}. Now we have made lots of choices, but as usual in homological things it turns out by examination that everything is well-defined. It is also the case that with these new maps, the corresponding diagram is exact at every place. We call this the derived exact couple.

Since the derived exact couple is an exact couple, there is nothing stopping us from defining the derived exact couple of the derived exact couple. Then do it again, and again, … . Obviously, if we aren’t careful we’ll lose track of where we are, so choosing notation is essential. Suppose {(D, E)} form an exact couple. Then let {D^1=D} and {E^1=E}. Next let {D^2=D'} and {E^2=E'}. This inductively gives us that {D^{r+1}=(D^r)'} and {E^{r+1}=(E^r)'}. In each of these we get three maps which tell us that each stage is an exact couple {i^r: D^r\rightarrow D^r}, also {j^r: D^r\rightarrow E^r}, and {k^r: E^r\rightarrow D^r}.

Since the triangle is exact, defining (which technically already happened in the definition of {E^{r+1}}) {d^r: E^r\rightarrow E^r} by {d^r=j^r\circ k^r} we get a chain complex and can take homology. We say that {\{(E^r, d^r)\}_{\{r\geq 1\}}} is the spectral sequence associated to the exact couple. Note that {E^{r+1}=H(E^r, d^r)}.

This is probably incredibly useless for people wanting to know what a spectral sequence is and how it is used. But in general we are going to have a lot more information floating around than just this abstract exact couple. More specifically, we have to get the exact couple somehow, and the process of forming it will give us things to work with. The other thing is that we want to know that {\displaystyle \lim_{r\rightarrow\infty}(E^r, d^r)} converges in some sense to something, and hopefully we have information about what this “{E^\infty}” term is.

I don’t want to go through an entire example today, but we’ll set up the situation for the example (for those who have seen this, it will be the Serre Spectral Sequence).

Suppose we have a chain complex {K}. Then given a filtration {\cdots \subset K^n\subset K^{n+1}\subset \cdots \subset K} (not necessarily starting at the 0 complex or ending at the total complex, but the {j}th grade of any {K^n} should be a submodule of the {j}th grade of {K} and the submodules should respect the differentials). Then we will get a short exact sequence of chain complexes {0\rightarrow K^{P-1}\rightarrow K^P\rightarrow K^P/K^{P-1}\rightarrow 0}. This is just because it works trivially on each grade.

Well, our old standby says that a short exact sequence of chain complexes induces a long exact sequence in homology, so {\cdots \rightarrow H_n(K^{P-1})\stackrel{i}{\rightarrow} H_n(K^P)\stackrel{j}{\rightarrow} H_n(K^P/K^{P-1})\stackrel{k}{\rightarrow} H_{n-1}(K^{P-1})\rightarrow \cdots}.

From here we define our exact couple. Let {D_{p,q}=H_{p+q}(K^p)} and {E_{p,q}=H_{p+q}(K^p/K^{p-1})}.

Now use the maps given in the long exact sequence.

So {i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})} which is really {D_{p,q}\rightarrow D_{p+1, q-1}}.

Our {j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})} which maps {D_{p,q}\rightarrow E_{p,q}}.

Lastly the map {k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})} which maps {E_{p,q}\rightarrow D_{p-1, q}}.

Lastly, the differential {d_{p,q} = j_{p-1, q}\circ k_{p,q}: E_{p,q}\rightarrow E_{p-1, q}}.

Don’t worry if this seems horribly confusing. We’ll look at a very concrete example next time in which we’ll see that often everyting stays stationary or goes to 0 leaving us with very little. I’ll just try to summarize a little of this filtered chain complex example. We are getting infinitely many exact couples all mingled together somehow. If we think of a {(p,q)}-plane, then at each integer valued place we get a group {E_{p,q}}. The differential maps move us to the “left” on place. So we have rows of chain complexes for each fixed {q}. Remember that when we form the spectral sequence this is really only the {E^1_{p,q}} part. We can think of this term as an entire “sheet” or “page” of groups related to each other. We get a page {E^r_{p,q}} for each {r\geq 1}.

Some things that would be nice (and to get you thinking about why this might be useful) is if for each fixed pair {(p,q)} the groups stabilize, i.e. {E^N_{p,q}=E^{N+1}_{p,q}=\cdots}. Then we would have a notion of {E^\infty_{p,q}}. It would be nice to know how this limit term relates to the original chain complex, or filtration. It would be nice if the spectral sequence limit were somehow independent of the filtration, that way taking a limit of different filtrations would give us information we might not know. Like I said earlier, don’t worry if this post was confusing. It just needed to be done. After some examples, this will probably not seem so bad.

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Posted by: hilbertthm90 | January 11, 2010

What to do next

Awhile ago I was talking mainly about algebra. I was mostly going by what I was reading in Matsumura, but recently the chapters in Matsumura that I’ve been reading I’ve already covered in previous posts, so I didn’t want to reopen those topics. This has been one of the main causes of my hiatus. My guess is that in general this quarter will be a lot sparser in posts.

Two ideas of where to go while I wait for my algebra to get back to a point where I can start posting again is to start in on some of the basic constructions underlying modern algebraic geometry, like developing sheaves and schemes. This seems to have been covered in a few other blogs, so I’m hesitant.

The other thing that would be really neat, and I’m pretty sure has not been covered by any other blog is to talk about spectral sequences. Right now I’m in a homological algebra class that is essentially devoted to them. I think they are quite a fascinating and brilliant construction. The main thing I’d like to do on this blog is to see where they come up in algebraic geometry, since this particular application is sure to not come up in class.

The problem with this is that it would be an incredibly massive undertaking. Just motivating and explaining the construction would take a long time. I would also really need to figure out a good way of doing diagrams in wordpress.

In any case, just thought I’d check in and see if anyone had opinions one way or another.

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Posted by: hilbertthm90 | December 31, 2009

Middlesex by Jeffrey Eugenides

I’ve always had some sort of aversion to this book. I never understood why it was so universally acclaimed by everyone I spoke to. The main plot is about two siblings that get married and have a child, who then goes on to marry his first cousin, who then has a child that is a hermaphrodite due to the genetic mutations passed on from incest. The story just never seemed to be the kind that all people can relate to.

So as usual with a book post, I want to touch on lots of things I thought about during the reading of the book without going into much detail on any of them. The other aspect of this post is to make some sort of argument that will get people to read the book even if they are sort of put off by the content.

The narrator is the hermaphrodite. So let’s address this right off the bat. He is born and raised as a female, even though he is genetically essentially a male. So a transformation takes place over the course of the book from Callie (female) to Cal (male). To me the “hermaphrodite nature” of the book is to completely miss the point. It seems more of a device to indicate transformation from being confused about your identity and sexuality to becoming more comfortable with yourself and accepting yourself for who you are. It is a device to indicate coming of age. It exists for literary reasons, and to me, this seems to be the only reason it exists. I believe everyone will be able to identify with Cal in a deep way.

The structure of the book is really effective as well. Cal is an adult narrating his family history. The “current” timeline is usually about the first paragraph or so of the chapter and then the chapter flashes back. Over time you start to better understand Cal’s current motivations and life. Due to having both male and female sex organs, Cal has spent his life very closed off. He is afraid of telling people, so never really let’s anyone in. To me, this is the most profound aspect of the book. Again, instead of the hermaphrodite aspect of the book alienating readers, it actually creates a universal situation. Everyone has a past and some secrets they are afraid of. This “current” timeline gives the book this edge: a story about finding someone who will accept these aspects of you. It is risky and scary to let someone get close to you, but if you are honest about yourself and they accept you as this person, the pay-off is much greater than living closed off.

As with The Virgin Suicides, Jeffrey Eugenides writes fantastically beautiful prose. He seems to be much better in this book, though. He quickly jumps around in style to perfectly set up and mimic exactly what the content of the sentences are trying to say. Some paragraphs I think of as just heart-wrenching snippets of truth. Of course, I didn’t mark them, so it will be hard to find now. But here is one I randomly opened to:

It occured to me today that I’m not as far along as I thought. Writing my story isn’t the courageous act of liberation I had hoped it would be. Writing is solitary, furtive, and I know all about those things. I’m an expert in the underground life. Is it really my apolitical temperament that makes me keep my distance from the intersexual movement? Couldn’t it also be fear? Of standing up. Of becoming one of them.

Still, you can only do what you’re able. If this story is written only for myself, then so be it. But it doesn’t feel that way. I feel you out there, reader. This is the only kind of intimacy I’m comfortable with. Just the two of us, here in the dark.

Another aspect of the book is about chance and how much of our lives are out of our control. The book often reads like Pynchon. It is full of reference, statistics, and facts. This is a very effective way of addressing the question of whether our lives are just the lump sum of determined information, or is there something more. This aspect of the book is hard to describe. It is very implicit and never explicitly addressed. It always is lurking there in the hard factual data that seems to be competing with the flowery undetermined prose around it. From which does Cal’s life get its meaning? This seems to also be the reason we follow three generations of this family. The point seems to be setting up the fact that all these things happen that directly affect and determine Cal’s life before he is even born. There seems to be one part of the book that tries to give a definitive answer, but I won’t spoil that plot point.

I could go on forever, so I’ll try to wrap it up. I didn’t even touch on comparisons to and use of Greek mythology. Or any of the great world and American historical events going on in the background, always subtly affecting everyone.

Overall, I can’t recommend this book enough. It raises the big questions in a new way. It is funny. It forces you to examine your own life. It is about learning to accept yourself and others. It is about the struggle between succumbing to all the predetermined genetic, scientific, political and historical forces working against you and actually living your life and overcoming these forces. It is about societal norms. It is about falling in love. There are so many instances that I thought it was reading my mind. There were other moments I knew it wasn’t, because it helped me figure out something I was struggling with. It is about the messiness of life. It is universal.

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Posted by: hilbertthm90 | December 22, 2009

Music 2009

I may as well post this today since it is done. I’ve officially relistened to everything I got this year and ranked it. I’m going to do two new things this year. First off, I need to explain a little about how my taste in music (and art in general) has changed drastically. I’ll start with a quote from Franz Kafka

I think we ought to read only the kind of books that wound and stab us. We need the books that affect us like a disaster, that grieve us deeply, like the death of someone we loved more than ourselves, like being banished into forests far from everyone, like a suicide. A book must be the axe to the frozen sea inside us.

I’ve sort of lost interest in things that don’t really affect me. I still get and listen to lots of technically amazing musicians who are doing great original things. If this were last year Andrew Bird would probably be in the top 5 because there is no denying that he wrote some of the most complicated and original music this year. He is a phenomenal violin player and shows this off as well. But his songs sort of lack any real meaning.

Let’s face it. I’m a very moody person. I like to have music that hits a huge range of emotions, so that I can find something to sympathize with me no matter my mood.

So without furthur ado, the top 10:

1. The Antlers – Hospice
2. Imogen Heap – Ellipse
3. Land of Talk – Fun and Laughter
4. Loney, Dear – Dear John
5. Doves – Kingdom of Rust
6. The Swell Season – Strict Joy
7. Wilco – Wilco (The Album)
8. Regina Spektor – Far
9. Grizzly Bear – Veckatimest
10. The Decemberists – The Hazards of Love

The honorable mentions:
Dirty Projectors – Bitte Orcha
Phoenix – Wolfgang Amadeus Phoenix
M. Ward – Hold Time
Sunn O))) – Monoliths & Dimensions

Now the ordering here was made with some consideration, but it shouldn’t be taken too seriously. Everything above I thought was all around great. Essentially no bad songs. All original creative things. All technically great things with a good deal of emotional content.

The middle ground stuff:
Andrew Bird, Neko Case, Dan Deacon, Tortoise, Dodos, Animal Collective, F**k Buttons, Duncan Sheik, Other Lives

These had really good aspects and had some aspects I didn’t really like. All worth getting in my opinion, but not as well rounded as the top list. For instance, I think the songs on Neko Case or Animal Collective are the best of those groups when they are good. But when they are bad, they are some of the worst either have put out. This dichotomy could not be overlooked. Actually having to skip songs on an album is a major turn-off for me.

The subpar group is as follows:
Bon Iver, Yeah Yeah Yeahs, The Field, Mount Eerie

These were not worth getting in my opinion, but not totally completely horrible. I enjoy some songs and aspects, but not enough to make the “middle ground” list.

The bottom three I’ll go back to ranking:
Bad: Volcano Choir – Unmap
Worse: Holopaw – Oh Glory, Oh Wilderness
Worst: Heartless Bastards – The Mountain

Other than The Antlers, everything on the top 10 I expected to be there. They’ve all impressed me in the past. So mostly my surprises are in the realm of disappointments. I’ve already said this, but for the most part I was severely disappointed this year. The other surprise was sort of Loney, Dear. I only got this because he was touring with Andrew Bird. After a few listens I thought I had gotten everything to get to it. To my surprise 8 months later, I was still not bored of it.

My biggest disappointment has to be Bon Iver. He was number 1 last year and was very close to bottom three this year. Another big disappointment was Holopaw. This was bottom three. The first time I heard this group was at a party, and it was on in the background. Music people and non-music people alike were so drawn to its greatness that there were constant comments and asking who it was. I’ve never seen anything like that occur since.

I’ll leave this post here. I didn’t rank a top 10 list for individual songs, but I probably should have. If anyone would like me to elaborate more on specific placements, just comment, I’d be happy to. Also, if there are things you thought were great but I missed, I’d love to hear about it.

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Posted by: hilbertthm90 | December 22, 2009

Associated Primes III

Hopefully today we can finish this topic off. We’ll jump right in. Let R be Noetherian and M finite, then N\subset M is primary if and only if Ass(M/N) consists of a single element.

We’ll use the second formulation of primary. Suppose Ass(M/N)=\{P\}. Then by last time Supp(M/N)=V(P), so we have P=\sqrt{(ann(M/N)}. Suppose r\in R is a zero divisor for M/N. Then from AP I we get that r\in P\Rightarrow r\in\sqrt{(ann(M/N))}. Thus by the second formulation, N is a primary submodule.

For the reverse, suppose N is a primary submodule and P\in Ass(M/N). Then every element of P is a zero-divisor for M/N. So r\in P\Rightarrow r\in\sqrt{ann(M/N)}. Thus P\subset \sqrt{ann(M/N)}. By definition of associated prime we get ann(M/N)\subset P, and primes are radical so taking radicals of both sides get the other inclusion and so \sqrt{ann(M/N)}=P. i.e. the only element of Ass(M/N) is P.

We now note that I=ann(M/N) is actually a primary ideal. Let r,s\in R and suppose rs\in I but that s\notin I. Then (rs)(M/N)=0 but s(M/N)\neq 0. So r is a zero-divisor for M/N which gives r\in P=\sqrt{I}. Thus the ideal I is P-primary.

Thus we make the definition for modules that if Ass(M/N)=\{P\} then the submodule N is P-primary (sometimes called a primary submodule belonging to P).

Note that the intersection of any two P-primary submodules is again P-primary. This is seen by embedding M/(N\cap N')\hookrightarrow (M/N)\oplus (M/N').

We call a submodule reducible if it can be written as such an intersection and irreducible otherwise (this is a property on submodules, not to be confused with the notion of irreducible for modules).

Any submodule of a Noetherian module can be written as a finite intersection of irreducible submodules. This is seen by applying Zorn’s lemma to the set of submodules having no such representation.

We are about to the point where we can define a primary decomposition. Of course there is not going to be a unique way of doing it, but we’ll make some contrived definitions to get it as unique as possible.

We’ll call an intersection irredundant if none of the components of the intersection can be omitted (in particular this will prevent unnecessary repetition sort of like multiplying by a bunch of 1’s in a prime factoring).

A decomposition of a submodule is an expression of the submodule as an intersection of a finite number of submodules, and if each component is irreducible, then we say it is an irreducible decomposition. Likewise, we define primary decomposition if each component is primary.

Now suppose we write N=\cap N_i as an irredundant primary decomposition with Ass(M/N_i)=\{P_i\}. Since the intersection of any finite number of P_i-primary submodules is again P_i-primary we can group that intersection together and consider it as just a single submodule. In this way we get a decomposition in which P_i\neq P_j when i\neq j. This makes the decomposition as short as possible.

To wrap up we need to prove that everything behaves the way we want (note that we’ve been assuming Noetherian ring and finite module).

I) An irreducible submodule of M is a primary submodule. Suppose N\subset M is not primary. We can assume N=0 without loss of generality by replacing M with M/N. Then by the first theorem in this post we get that Ass(M) has at least two elements P_1, P_2. i.e. M contains submodules K_i isomorphic to R/P_i, so K_1\cap K_2=0 which means N=0 is reducible.

II) Now for an important one. We want to be able to read off the associated primes from the decomposition, so If we have an irredundant primary decomposition of a proper submodule N=\cap N_i, then Ass(M/N)=\{P_1, \ldots, P_r\} where Ass(M/N_i)=\{P_i\}.

We’ll again assume WLOG that N=0=\cap N_i. Thus M is isomorphic to a submodule of M/N_1\oplus \cdots \oplus M/N_r. i.e. Ass(M)\subset Ass\left(\bigoplus M/N_i\right)=\bigcup Ass(M/N_i)=\{P_1, \ldots, P_r\}.

Now by being irredundant N_2\cap \cdots \cap N_r\neq 0. So pick a non-zero element x\in N_2\cap \cdots \cap N_r. Then ann(x)=(0:x)=(N_1:x). But we have (N_1 :M) is primary belonging to P_1, so P_1^nM\subset N_1 for some n. Thus P_1^nx=0 which gives that for some i<n we have P_1^ix\neq 0 and P_1^{i+1}x=0. Choose a non-zero element y\in P_1^ix. Then P_1y=0.

But note that y\in N_2\cap \cdots \cap N_r, so we have y\notin N_1. Since the submodule is primary ann(y)\subset P_1 so P_1=ann(y) giving P_1\in Ass(M). If we do this for all the other P_i we get that \{P_1, \cdots , P_r\}\subset Ass(M) giving equality of sets.

III) Lastly we want the existence and uniqueness. Every proper submodule has a primary decomposition. This is just because we know that there is an irreducible decomposition, so apply (I) to each irreducible component.

Uniqueness is a little trickier. We must restrict our attention to minimal primes. Suppose P is a minimal associated prime of M/N, then the P-primary component of N is \phi_P^{-1}(N_P) where \phi_P : M\to M_P. Thus it is uniquely determined given the data M, N and P. Non-minimal are not unique: Take the ring \mathbb{C}[x,y], then (x^2, xy)=(x)\cap (x^2, y)=(x)\cap (x^2, xy, y^2).

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Posted by: hilbertthm90 | December 21, 2009

Associated Primes II

Today we’ll continue towards a primary decomposition for modules. First, I’ll list two facts without proof that may come up (they are quite straightforward to prove if you want to try). If R is any ring and 0\to M'\to M\to M''\to 0 is an exact sequence of R-modules, then Ass(M)\subset Ass(M')\cup Ass(M''). Secondly, if R is a Noetherian ring and M a non-zero finite R-module, then there is a chain 0=M_0\subset M_1\subset \cdots \subset M_n=M of submodules of M such that for each i we have M_i/M_{i-1}\simeq R/P_i with P_i\in Spec R.

I don’t remember, but I may have even proved that second one when talking about Artin-Rees. Now let R be Noetherian and M a finite R-module.

I) Ass(M) is finite. We induct on the length of the chain in the second fact. Suppose this is true for all M having a chain of the above form of length n-1. If N is a finite module with a chain of length n, then since R is Noetherian and N_{n-1} is a submodule it is also finite. So by the inductive hypothesis, Ass(N_{n-1}) is finite. Now consider the exact sequence 0\to N_{n-1} \to N \to N/N_{n-1}\to 0. By the first fact Ass(N)\subset Ass(N_{n-1})\cup Ass(N/N_{n-1}). But the chain has the condition that N/N_{n-1}\simeq R/P for some P\in Spec(R). Since Ass(R/P)=\{P\} we have that the cardinality of Ass(N) can increase by at most one from the cardinality of Ass(N_{n-1}) which was finite.

II) Ass(M)\subset Supp(M). Suppose P\in Ass(M). Then M contains a submodule isomorphic to R/P (it is just the image of the hom r\mapsto r\cdot x and apply first iso theorem). So 0\to R/P\to M is exact, so when we localize we still have an exact sequence 0\to R_P/PR_P\to M_P. Since R_P/PR_P\neq 0, M_P\neq 0 which means P\in Supp(M).

III) The set of minimal elements of Ass(M) coincides with the minimal elements of Supp(M). Well, (II) gave one inclusion so suppose P\in Supp(M) is a minimal element. Then since M_P\neq 0 by the last post we get that Ass(M_P)\neq \emptyset. But we also figured out a formula for this set Ass(M_P)=Ass(M)\cap Spec(R_P)\subset Supp(M)\cap Spec(R_P)=\{P\}. Thus by non-emptyness we must have P\in Ass(M).

Recall when we working in the Zariski topology on Spec(R). We have an operator on ideals V(I)=\{p\in Spec(R) : p\subset I\}, and the Zariski closed sets of Spec(R) are precisely those sets that are of the form V(I) for some I.

So by definition of this operator, if P_1, \ldots , P_r are the minimal elements of Supp(M), then Supp(M)=V(P_1)\cup \cdots \cup V(P_r). Another property of the topological space Spec(R) is that a subspace is irreducible if and only if it is V(\frak{p}) for some minimal prime \frak{p}. So if we think of Supp(M) as a closed subspace of Spec(R), then the irreducible components are precisely V(P_i). We call the P_i the isolated associated primes of M. The other associated primes are called embedded primes.

Due to the above geomtric interpretation of what isolated and embedded primes are, the terms make sense. An isolated prime gives you full irreducible component of Supp(M) whereas an embedded prime gives some embedded subspace of the component generated by the prime it lies over.

I’ll finish with the new definitions. Suppose N\subset M is a submodule. Then we call N a primary submodule if for all r\in R and x\in M we have the condition x\notin N and rx\in N\Rightarrow r^nM\subset N for some n.

The above condition is equivalent to the condition: if r\in R is a zero-divisor for M/N, then r\in \sqrt{(ann(M/N)}. Showing these are equivalent is immediate when you write out what the definitions of all these things are. This shows that the property of being primary is dependent only on the quotient module M/N.

Sorry to end on some definitions, but I think if I do another theorem this post will become too long.

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Posted by: hilbertthm90 | December 20, 2009

Associated Primes I

I’d like to go over associated primes in general rather than just the Noetherian ring form of primary decomposition of ideals. The natural generalization is to modules, since ideals are sub-modules over the ring treated as a module over itself. We’ll need to define a few things first.

Let M be an R-module. Let P be a prime ideal of R. Then we call this an associated prime of M if P=ann(x) for some x\in M. The set of associated primes is denoted Ass_R(M) (since R will be understood, we’ll just drop that from here on).

Now suppose I\subset R an ideal. Then the elements of Ass(A/I) are called the “prime divisors” of I.

Now we’ll get some basics out of the way. Suppose that R is Noetherian. First off, Ass(M)\neq \emptyset when M\neq 0. We show this by showing that any maximal element of the family \mathcal{F}=\{ann(x) : 0\neq x\in M\} is an associated prime. This is an important fact on its own.

Note that all we really are trying to show is that a maximal element of \mathcal{F} is prime as an ideal, since it will already be of the form ann(x). Let A\in\mathcal{F} be a maximal element. Suppose A=ann(x) and that ab\in A and that b\notin A. Then abx=0 but bx\neq 0. Thus ann(x)\subset ann(bx). But by maximality, ann(x)\subset ann(bx). Thus ann(x)=ann(bx) which means a\in A. Thus A is prime and hence an associated prime of M.

The other fact we need is that the set of zero-divisors for M is precisely the set \displaystyle \bigcup_{P\in Ass(M)}P.

If a\in \displaystyle \bigcup_{P\in Ass(M)}P, then this just says there is some x\in M such that a\in ann(x)\Rightarrow ax=0 and hence a is a zero-divisor. The reverse inclusion just uses the previous fact. Let a be such that ax=0 for some x. So we have a\in ann(x). Then take a maximal element P\in\mathcal{F} containing ann(x). By the last fact P\in Ass(M) and hence a\in \displaystyle \bigcup_{P\in Ass(M)}P.

For the theorem of the day, you may need a refresher on the spectrum of a ring, and on localization.

We no longer assume R is Noetherian. Let S\subset R be multiplicative, and N an R_S-module. Viewing Spec(R_S)\subset Spec(R), then we have Ass_R(N)=Ass_{R_S}(N). In general, if R is Noetherian, then for any R-module M, we have Ass(M_S)=Ass(M)\cap Spec(R_S).

Let x\in N. Then ann_R(x)=ann_{R_S}(x)\cap R. This is just because the elements of R that kill x, are just the fractions that kill x that are “actually” in R. This immediately gives us one inclusion, since if P\in Ass_{R_S}(N) then P\cap R\in Ass_R(N).

Now suppose Q\in Ass_R(N). Then there is some x\in N such that Q=ann_R(x). Thus x\neq 0 giving Q\cap S=\emptyset. Thus QR_S\in Spec(R_S) with QR_S=ann_{R_S}(x). This proves the first statement.

We now show the second statement about M. Suppose P\in Ass(M)\cap Spec(R_S). Thus we again get that P\cap S=\emptyset and P=ann_R(x) for some non-zero x\in M. Suppose that that (r/s)x=0 in M_S. Thus there is some t\in S such that trx=0 in M. But we’ve already noted that t\notin P and tr\in P, thus by primality r\in P. So PR_S=ann_{R_S}(x). Thus PR_S\in Ass(M_S) giving one inclusion.

For the reverse, suppose Q\in Ass(M_S). By clearing the denominator we can assume that for a non-zero x\in M we have Q=ann_{R_S}(x). Let Q^c=Q\cap R. Then Q=Q^cR_S. We have that Q^c is finitely generated since R is Noetherian, so there is some t\in S such that Q^c=ann_R(tx). Thus Q^c\in Ass_R(M) which gives the reverse inclusion.

A nice little corollary is that for Noetherian rings a prime ideal P\in Ass_R(M) if and only if PR_P\in Ass_{R_P}(M_P).

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Posted by: hilbertthm90 | December 17, 2009

Dune

Bear with me. I’m going to attempt to do my Dune post as a single post, but there probably two theses worth of ideas in here. Well, the plane trip across the country allowed me the time needed to read most of Dune. I’ll start out by saying that this book was very challenging for me at first. There is a very complicated system and hierarchy set up. There are families and alliances and sworn enemies. There are traitors. There are lots and lots of confusing names. The book essentially throws you in and although the first 100 pages or so are very slow in what I imagine is the author’s mercy at trying to catch you up, it makes for tough reading.

The other tough part is that it is incredibly dense. It is like reading Wittgenstein’s Philosophical Investigations. There are only 100 aphorisms because you are supposed to stop and think after each one. This kept happening to me, except that it was a novel and so I probably wasn’t supposed to do that. Anyway, I hit a critical point when I was on the plane and had these two ideas about what the book was actually about, but my history isn’t very good and so I had no way to check. I also couldn’t see if other people had come up with this or not. So I just started testing it against everything I read, and it fit way too well to not have at least crossed the author’s mind. This also made the book way more exciting to read.

Here goes. The first thing I thought was that Dune was a metaphor for the Middle East. I immediately checked when the book was published, and it said 1965. This was rather unhelpful because I couldn’t even be sure which countries existed in the Mid East in that year, yet alone what sort of political things were going on. I haven’t really researched it because I wanted to at least get this post out there. So I apologize if this is way off base.

First off, the book takes place on Arrakis…a desert. Does this sound like “Iraq” to anyone (for future reasons, I actually think that Arrakis is Saudi Arabia in this metaphor)? The natives are the Fremen and they have a tribal society. What are some common strange names that keep appearing. Muad’Dib or Kwisatz Haderach and even at one point Rhamadan. Do these have an Arabic flare to them or am I imagining it? Alright, so this one can be resolved with a quick google search. Kefitzat Haderech is Hebrew for “short cut”, and the made up word means “shortening of the way”. Well, those are all things that got me thinking about this, but they are sort of the fluff of this argument.

The real thing was that the universe seems completely dependent on the “spice”. So the spice is the metaphor for oil. All the conflict is essentially based around possession of the spice. Here comes a spoiler if you haven’t read it. But to solidify the metaphor beyond a doubt, the spice is created by some chemical underground process that happens to dead gigantic worms (maybe they don’t have to be dead, this was unclear to me). Um, what is oil (re: fossil fuels)? It is just decomposition of buried organisms in a particularly well-suited environment.

Some things I still haven’t figured out that will take a bit of researching of what was going on at the time are who the families/people are. I’m assuming that the Atreides are a country and the Harkonnens are another country. Who would have Herbert have thought of as the “good guys” at that time and who the “bad guys?” I don’t think the U.S. had a big involvement yet, so it probably wouldn’t be one of them. The emperor sort of presides over all the families (re: countries), but seems to have no power in controlling them. Is this the UN? Also, is this just a metaphorical retelling of events, or did he take it further and insert some sort of warning/message about the situation? This brings me to my next topic.

I thought the only really clear overriding message was one of religion. I guess I can pitch this at two different levels. The more dramatic level is that religion is invented by people in order to control people. Whether you believe this or not, there is no doubt this was an intention of Herbert. I wish I had quotations on hand, but it is repeated time and time again that the Bene Gesserit invented legends and myths to create the religion of the Fremen a long time ago so that when this time came it could be exploited for their protection. Paul is some sort of messiah figure that most religions have, but he also clearly exploits this to gain power.

The less dramatic religious message seems to be that religion and politics need to stay as separate as possible. When people believe they are doing things for a religious cause, then they will stop at nothing since the cause is far greater than their mere earthly bodies. In particular, wars waged with religious overtones break from any sense of a “just war” (whatever that is). The end of Dune talks about women, children, and elderly throwing themselves onto swords so that the men can get in actually kill the other side. The Fremen are often so feared as fighters because they have no reason to fear death with the assurance of an afterlife.

I’ll stop here. I’m sure there are holes and errors, and I purposely skipped details and quotes. But a first sketch of these arguments is now out there for criticism. Be gentle, remember this was my first time through the book. It is possible a second reading would make me embarrassed that I ever thought this.

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