I’ve been fiddling around on here for a few weeks trying to figure out what my next major set of posts should be about. I’ve finally settled. It turns out that algebraic geometry requires knowledge of a ridiculously large amount of commutative algebra. Now I usually try to avoid repeat posting when I know that I’m doing it, but I don’t think I’m going to stick to that rule for this set of posts. For probably at least the next month I’m just going to try to vastly improve my commutative algebra knowledge.

The first topic will be the Hilbert polynomial. The motivation here is that we are looking for some invariants of projective algebraic sets.

Suppose is a graded ring. Then a graded R-module, M, is a module with an abelian group decomposition such that .

Let be a finitely generated graded -module (graded by degree of the polynomial). Then we define the Hilbert function of to be . The function takes as input something from and outputs the dimension of that graded part.

Here is where the Hilbert polynomial enters in. It turns out that actually agrees with a polynomial of degree less than or equal to for large . We will denote this polynomial .

Let’s prove a general fact first. Suppose is defined for all natural numbers. Then if agrees with a polynomial (with rational coefficients) of degree less than or equal to for all , then agrees with a polynomial (with rational coefficients) of degree less than or equal to for all .

Suppose is a polynomial that satisfies the hypothesis of the preceding statement, i.e. for .

Set for and for .

Now just note that for all integers. So we are done since then is a polynomial with rational coefficients of degree less than or equal to .

As you may have guessed, this little fact was to set up an induction for the actual theorem. Let’s induct on the number of variables . The base case just puts us in the case where our graded module is over a field and hence is a finite-dimensional vector space. Thus dimensions all have to be zero at some grading, so for large and we are done.

Suppose the theorem holds in variables. Now let be the kernel of the multiplication map by . This is a submodule of , and we get an exact sequence . Where the means the grading is shifted by .

The exactness tells us something about the dimensions. So look at the part of the grading: . In terms of the Hilbert function, this says precisely that .

Since and are f.g. graded modules over we can apply the inductive hypothesis to the right side. But since the right side is a polynomial for large , so is the left side. Now the fact we proved before this gives us the full result.

There is much to say about Hilbert polynomials, so I’ll probably keep posting about them for awhile.

I’ve sort of posted about Joanna Newsom before, but I really must come back and give her a full post. I really hate to use the album Ys as a perfect example of exactly what I think art should be, since it is such a polarizing album. I understand why, too. I completely understand where people are coming from that hate this album. There are people who say it is unlistenable (probably not a real word).

In any case, as I understand the back story, this is an album that recounts a full year of Newsom’s life. The lyrics are directly referencing actual, real, exact events of her life. But the lyrics are incredibly abstract. The concreteness of the events that they are based on gives the songs every bit of emotion and realness as if she were telling the events straight-up. The actual lyrical abstraction into story-metaphors allows the listeners to interpret into their own situation.

I’ve put serious listening into this album at three very different points in my life. All three times I have been 100% sure that I knew exactly what had happened in Newsom’s life that she was referring to. All three times my interpretations have been radically different. This is because I was identifying so well with the emotion and metaphor in the song. I am completely baffled at my current listening, but again it fits my situation perfectly.

To me this is exactly what art should be. It should be an abstracting of real life in such a way that the viewer feels as if it is exactly their own situation.

There are some interesting cases out there that I could bring up. The first that comes to mind is Connor Oberst (at least in the early Bright Eyes stuff). It is incredibly emotive and about some really intense things. Overall, Oberst is very specific lyrically. I think in this case that is alienating. As a listener, it is hard to change the details of these specific stories to really relate to them.

One thing I haven’t put a lot of thought into is whether this interpretation of great art translates well outside of the song/poem medium. My guess is it doesn’t. It seems like it would be hard to write a novel about a specific event, but keep everything really vague so that you don’t know what the event is.

There are many, many other aspects of Newsom’s music I could go on about, but I think what I just mentioned is the key element.

Maybe I should give some examples of her lyrics. I wish I could post the entire song Sawdust & Diamonds. It is so ridiculously abstract, but as I sit here reading it, it couldn’t be any more obvious what it is about exactly. Anyway, here is a part of it:

and the little white dove
made with love, made with love:
made with glue, and a glove, and some pliers

swings a low sickle arc
from its perch in the dark:
settle down
settle down my desire

The white dove is the relationship she is in. Although, she has created the relationship with care and love, it is also ad hoc patched together in places (the fact that glue and pliers had to be used). This doesn’t matter because she still desires the person and they’ll fight through it.

then the system of strings tugs on the tip of my wings
(cut from cardboard and old magazines)
makes me warble and rise like a sparrow
and in the place where I stood, there is a circle of wood
a cord or two, which you chop and you stack in your barrow

(First off, the “system of strings” is a recurring theme. Earlier it was mentioned: there’s a light in the wings, hits this system of strings/ from the side while they swing;/ see the wires, the wires, the wires)

The dove (relationship) is being held up artificially with wires. Again, the ad hoc construction of the relationship is mentioned since the dove is made of cardboard and magazines. She sees the wires. She is aware that it is artificial in some sense.

There is evidence of her resistance to falling in love with this person (“love, you ought not!/no you ought not!”). This is probably due to her being aware of all the faults and artificiality of the relationship. Perhaps she fears that her construction isn’t strong enough and the system of strings will collapse.

But she becomes aware that every relationship and person has faults. Resisting falling in love is not an option and it overtakes her at some point (“then the furthermost shake drove a murdering stake in/and cleft me right down through my center/and I shouldn’t say so, but I know that it was then, or never”)

In any event, I certainly have never interpreted the song this way before (I used to be convinced that it was about death, actually). And it baffles me, since this must be correct. But I had just as much evidence for my last interpretation. I’m not sure I will ever grow tired of this album.

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

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

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

We can naturally think of paths and singular 1-simplices as the same thing, since they are both just continuous maps to the space out of a closed interval. So after rescaling, a loop is actually also a 1-cycle since .

The overall idea of this proof is then to show that is a well-defined homomorphism with image all of and kernel the commutator subgroup. Almost all of these facts are fairly straightforward.

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

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

Recall the definition . The second fact, gives us that is well-defined since any other representative of the equivalence class will be homotopic to the original, and hence the outputs will be homologous.

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

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

Since the are integers, we can assume each is by just repeating the as many times as needed. But all the with -1 in front can be replaced by by the fourth property. This converts all the to 1. Thus .

But , so all the endpoints must cancel. So for any that is not a loop, in order to cancel , there must be a such that . i.e. there is some that we can concatenate with to form . In order to cancel the some other must exists with endpoint .

So we can concatenate, then rescale, and group all of these cycles into a collection of loops by the third property. So the only remaining thing we must do is get it to be a single loop. But is path-connected, so pick some basepoint . For any of these possibly disjoint loops floating around, we can pick a basepoint at each and connect with a path from to the basepoint of the i-th loop. By the third and fourth properties . So now all loops start and end at and we can combine into a single loop . Thus .

Now comes the hard part. We want . The one containment is easy. Since is abelian, by the universal property of the commutator subgroup, . The method to get the other direction is to show that for any , we must have that is trivial in the abelianization.

Suppose such that . Since is a cycle, there is some 2-chain such that . So as before, we can assume each . Now the goal is to associate a 2-dimensional -complex to by taking for each a and identifying pairs of edges which we’ll call .

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

Writing this in detail algebraically is much trickier. Given any , we have , where are singular 1-simplices. Thus .

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

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

Now restrict to the simplices to get a new chain with boundary and loops at .

Now we just need to check whether the class is trivial or not: where . But gives a nullhomotopy of and we are done.

Thus and by the First Iso Theorem we have .

Nothing makes blogging stop dead like the start of the quarter. Well, that combined with the fact that I sort of reached the logical end to the topic I was posting on. There are lots of neat things we could do from here, but they are all rather technical (like sliding handles around and cancelling them to make our manifold simpler) but have nice simple geometrical interpretation. So I don’t really feel like going into that, since the bulk of the powerhouse tools of Morse theory that the everyday person needs were covered (I guess that isn’t totally true since we’re missing a few quick things for Lefschetz).

There are a few brief corollaries and applications left over that I’m contemplating. The most interesting to me is the one that says if a manifolds admits a smooth function to with exactly two critical points which are both non-degenerate then it is homeomorphic to a sphere. Yes I said homeomorphic! That is what is so interesting. We only have a diffeo if it is dimension less than or equal to 6. There are “exotic ’s”.

I feel ready to move on in general. The next logical step is to develop some homology, but this is a hard question. I certainly do not have the motivation or patience to build this from the ground up starting with definitions. So I’m not sure how to proceed. I also may give up on Lefschetz for a few months and do things more related to things I’m doing for classes.

In other news, I went to a neat talk on harmonic measure theory. Normally I’m not very fond of measure theory, but this was pretty cool. There was even a result that I don’t remember now that had to do with the set on which the measure was full was an algebraic variety or something.

Anyway, figured I’d update at least once this week.

Today I’ll just prove that a Morse function will give a handle decomposition of a closed manifold. Let’s use all the notation already set up (meaning critical points, values, attaching maps, dimension, Morse function, gradient-like vector field, etc).

We just induct on the subscripts of critical points. We’ve already done the base case (it is a min and hence a 0-handle from here). So we just need to show that if is a handlebody for , then is a handlebody with the appropriate handle attached.

So we’ve assumed that we have some decomposition . We also know that we attach a handle of index when crossing , so we do have a diffeo to a manifold with a -handle attached with attaching map .

Note that contains no critical values, so by flowing along we get a diffeo . Let be this diffeo.

So by inductive hypothesis, , so we can assume actually maps from the handlebody to . Now by composing we get our actual attaching map (note that before now the handle was attached to and not the handlebody itself).

i.e. . So let , and we get that , so we are done.

So I sort of dragged on longer than probably necessary there, since there was essentially nothing new. It was just being pedantic about the diffeo of the manifold and the handlebody.

There are some subtleties that should be pointed out, though. The index of the critical point did determine the index of the handle, and we went in “ascending” order. The other much more important and also more subtle point is that the choice of gradient-like vector field was how we constructed the attaching map. So even the same Morse function with a different choice of gradient-like vector field could actually give a “different” handle decomposition when considering attaching maps as part of the data.

I keep naming my posts “handlebodies”, so I think it is officially time to define what one is. A handlebody is a manifold obtained from by attaching various -handles successively. Thus a general handlebody will look like .

If you’re familiar with how to construct a CW-complex, this is pretty similar. You just inductively attach the handles using smooth maps, and then smooth out the manifold so that at each step we have a legitimate smooth manifold. It may be useful to introduce a notation for this. The first attaching with attaching map will be denoted . So inductively denote the i-th attaching by .

After i steps, we will always have i attaching maps even if some are formally meaningless (attaching a 0-handle is a disjoint union, so there is no attaching).

If we express M as a handlebody we call that a handle decomposition of the manifold.

Next time I’ll prove the result that everyone that has been reading the posts will have already guessed. Given a Morse function on a closed manifold, determines a handle decomposition of . Moreover the handles of this handlebody correspond to the critical points of , and the indices of the handles coincide with the indices of the corresponding critical points.

I’m short on time today, so I’m going to put off proving it.

I’m not sure how much to say about my other news, since it is still sort of up in the air. I passed two of my qualifying exams (which was all that was necessary for now), and I may officially lock myself into the path of algebraic geometry as my field in the next couple of days. Before I say too much about this, I’ll just say that I should have more information on Monday about what I’m officially doing.

Let’s think back to our example to model our -handle (where is not a max or min). Well, it was a “saddle point”. So it consisted of a both a downward arc and upward arc. If you got close enough, it would probably look like .

Well, generally this will fit with our scheme. An n-handle looked like … or better yet , and a 0-handle looked like , so maybe it is the case that a -handle looks like . Let’s call the core of the handle, and the co-core.

By doing the same trick of writing out what our function looks like at a critical point of index in some small enough neighborhood using the Morse lemma, we could actually prove this, but we’re actually more interested now in how to figure out what happens with as crosses this point.

By that I mean, it is time to figure out what exactly it is to “attach a -handle” to the manifold.

Suppose as in the last post that is a critical value of index . Then I propose that is diffeomorphic to (sorry again, recall my manifold is actually m-dimensional with n critical values).

I wish I had a good way of making pictures to get some of the intuition behind this across. I’ll try in words. A 1-handle for a 3-manifold, will be , i.e. a solid cylinder. So we can think of this as literally a handle that we will bend the cylinder into, and attach those two ends to the existing manifold. This illustration is quite useful in bringing up a concern we should have. Attaching in this manner is going to create “corners” and we want a smooth manifold, so we need to make sure to smooth it out. But we won’t worry about that now, and we’ll just call the smoothed out , say .

Let’s use our gradient-like vector field again. Let’s choose small enough so that we are in a coordinate chart centered at such that is in standard Morse lemma form.

Let’s see what happens on the core . At the center, it takes the critical value and it decreases everywhere from there (as we move from 0, only the first coordinates change). This decreasing goes all the way to the boundary where it is . Thus it is the upside down bowl (of dimension ). Likewise, the co-core goes from the critical value and increases (as in the right side up bowl) to the boundary of a disk at a value (where ).

Let's carefully figure out the attaching procedure now. If we think of our 3-manifold for intuition, we want to attach to by pasting along .

So I haven't talked about attaching procedures in this blog, but basically we want a map and then forming the quotient space of the disjoint union under the relation of identifying with . Sometimes this is called an adjunction space.

So really is a smooth embedding of a thickened sphere , since . And the dimensions in which it was thickened is . Think about the "handle" in the 3-dimensional 1-handle case. We gave the two endpoints of line segment (two points = ) a 2-dimensional thickening by a disk.

Now it is the same old trick to get the diffeo. The gradient-like vector field, , flows from to , so just multiply by a smooth function that will make match after some time. This is our diffoemorphism and we are done.

We now come to the main point of all these Morse theory posts. We want to somehow figure out what a closed manifold looks like based a Morse function that it admits (who knows how long I’ll develop this theory, maybe we’ll even get to how Smale proved the Poincare Conjecture in dimensions greater than or equal to 5).

Suppose is closed and a Morse function. We’ll use the convenient notation . So again, with the height analogy, as t increases, we will be looking at the entire manifold up to that height. Since M is compact, there is some finite interval such that and .

Note that with essentially no modification, we have already proved the Theorem that if contains no critical values, then . So really, the point is to now figure out what happens as we pass through the critical values.

First off, there are only finitely many critical points, and we can assume that each of these has distinct critical values by raising and lowering critical values. So if are the critical points and , we can order the indices so that .

To be explicit, is the min, so for and for t greater than , since is the max (also, wordpress hates inequalities, or me, I haven't decided yet, but it always cuts out lots of stuff and I just have to write the inequality in words).

These two critical points would be a nice place to start our examination. By the Morse lemma and the fact that a min has index 0, we know that there exists a neighborhood of on which (Alright, I’m sorry about that, but I just realized I have n critical points, so the dimension of my manifold is now m).

More explicitly there is some such that . So if we are thinking of height (of a 2-dim manifold), we’ll want to visualize this as a “bowl” where you have the bottom of the bowl the min and then it slopes upward along a sphere, and then you have the boundary circle at height .

So note that the only thing we used about this critical point is that it had index 0. This shape is called a (m-dimensional) 0-handle.

The reverse happens at our max. We have , since the critical point has index m. This is an -handle and thinking in 2-d height, it is a downward facing bowl.

Again, there is nothing special about being the absolute max, any index m critical point will locally be an -handle.

Index k critical points where are more complicated so I’ll leave those for next time.

Now we have a nice overview of how this will work. We just need to figure out what a -handle looks like, then as t increases through a critical value with index k, will “attach a k-handle”. When we are not near a critical value, the will not change diffeomorphism-type. We just need to make this a little more precise next time (or maybe even the time after).

I officially have a new favorite search for which someone found this blog: How to write a Japanese satire.

Let’s introduce a new term. Two Morse functions are considered equivalent if they have the same critical points and same index at each critical point.

The hope here is that two equivalent Morse functions will give the same topological data about our manifold, and so we want to develop techniques of altering our Morse function to something extremely nice to work with, but having it be equivalent to the origin one.

Our first excursion into this technique is the following: If M is a compact manifold and is a Morse function on M, then we can find an equivalent Morse function such that all the critical values are distinct.

If we’re going back to the height intuition, this is the technique that corresponds to “raising” or “lowering” critical points. So if you have two strange things happening at the same height (two mountain peaks that have the same height), the idea is sort of that you can slightly move the manifold around so that one is now higher than the other. Of course, we won’t actually move the manifold in any real sense, we’re going to construct the function.

This is going to be really nice, because it says that we can always get a Morse function in which only a single “change” can happen at any given height.

We’ll do this by first proving a Lemma which does all the work for us. Let be our Morse function, and a critical point. Then there is some such that for all there is an equivalent Morse function that has the same critical values as , except for .

The arguments here are essentially the same as in previous posts, so I’ll be a little looser and only outline the proof.

Since the critical points are isolated we can take a small coordinate chart centered at that contains no other critical points. Now let be a bump function that is 1 on some small neighborhood of and dies to zero before getting to the edge of the chart.

Then we define . We definitely have that all the critical points of are still critical points of and since on a neighborhood of any of those points the functions either agree or differ by adding a constant, they have the same index. Also, , so we have constructed our desired function as long as we don’t have any extra critical points.

But in the same was as before, for all where , since we're only concerned with the compact set on which is decaying, has a positive min , and has a finite max . Thus we do not gain any critical points in that set and we are done.

To get to the whole theorem all we need to do is note that there are only finitely many critical points (since compact). So if any of the values are shared, we can use the lemma to give an equivalent Morse function with shifted critical value, where we shift by a small enough value that it can't make it to any other critical value. We only have to apply this a finite number of times.

Now we want to start building some technique that will allow us to figure out what our closed manifold looks like based on the Morse functions it admits.

We’ll call a vector field , a gradient-like vector field for f, if away from critical points, and if is a critical point of index , then there is a coordinate neighborhood about such that f has the standard form as in the Morse lemma, and (i.e. it is the gradient in this neighborhood).

Intuitively, if we think back to our example, we visualize Morse functions as “height functions”. So we are attempting to construct in some sense an everywhere “upward” pointing vector field. If we’re thinking of the entire manifold flowing along this, then the only places where it is allowed to get “stuck” is at the critical points of .

The theorem is that there always exists a gradient-like vector field for a Morse function on a compact manifold.

Proof: As before, let be a finite subcover of coordinate charts, and be a compact refinement. Since the critical points are isolated (immediate corollary to the Morse lemma), there can only be finitely many since our manifold is compact. So we can assume that each critical point has a neighborhood small enough so that it is entirely contained in exactly one of the , and that the were chosen so that has standard form in those coordinates.

Let be a bump function for supported in . Then we get a smooth function on the entire manifold by letting outside of .

Let be the gradient of on . Let . The claim is that this is our gradient-like vector field for .

Let’s check at non-critical points. If is not a critical point, and , then since is the gradient and . All other terms of the sum are 0 since for any such that . Thus .

The other condition we have set up to work since each critical point has a neighborhood that is contained in precisely one of the , thus on that neighborhood is in standard form, and which is of the correct form. Thus is gradient-like for .

As a preview of things to come, I’ll prove our first result about what our manifold looks like using Morse functions. This is often called the Regular Interval Theorem.

Suppose that has no critical value in , then is diffeomorphic to .

Let be gradient-like for . Define which is smooth off of the critical points of , but since contains no critical points it is a smooth vector field there (in fact, on an open set containing ).

Let be an integral curve for starting at . But now

.

Thus, the integral curve continues along at constant speed 1 for the entire time it is in . But it starts at at time 0, so it reaches at time .

Thus by is a diffeomorphism. But rescaling gives the diffeo to .

This basically says that between critical points of a Morse function, we must have the manifold looking like cylinder built off of a single slice of the function (if we’re thinking in terms of height, we can pick any height, and at anywhere between the two nearest critical heights, all the level sets will look the same).