# A Mind for Madness

## Applying the Hilbert Polynomial

Let’s start applying to some specific situations now. Suppose $R$ is a Noetherian local ring with maximal ideal $\frak{m}$. Let $\frak{q}$ be an $\frak{m}$-primary ideal. Let $M$ be a finitely-generated $R$-module, and $(M_n)$ a stable $\frak{q}$-filtration of $M$.

Don’t panic from the set-up. I think I haven’t talked about filtrations. All the stable $\frak{q}$-filtration means is that we have a chain of submodules $M=M_0\supset M_1\supset \cdots \supset M_n\supset \cdots$ such that $\frak{q}M_n=M_{n+1}$ for large $n$.

The goal for the day is to prove three things.

1) $M/M_n$ has finite length for all $n\geq 0$.

Define $G(M)=\bigoplus \frak{q}^n/\frak{q}^{n+1}$ and $G(M)=\bigoplus M_n/M_{n+1}$. We have a natural way to make $G(M)$ into a finitely-generated graded $G(R)$-module. The multiplication in the ring comes from the following. If $x_n\in\frak{q}^n$, then let the image in $\frak{q}^n/\frak{q}^{n+1}$ be denoted $\overline{x_n}$. We take $\overline{x_n}\overline{x_m}=\overline{x_nx_m}$. This does not depend on representative.

We’ll say $G_n(M)$ is the n-th grade: $M_n/M_{n+1}$. Now $G_0(R)=R/q$ is an Artinian local ring and each $G_n(M)$ is a Noetherian $R$-module annihilated by $\frak{q}$. Thus they are all Noetherian $R/\frak{q}=G_0(R)$-modules. So by the Artinian condition we get that each $G_n(M)$ is of finite length. Thus $l_n=l(M/M_n)=\sum_{r=0}^{n-1} l(G_r(M))<\infty$.

2) For large $n$, $l(M/M_n)$ is a polynomial $g(n)$ of degree $\leq s$ where $s$ is the least number of generators of $\frak{q}$.

Suppose $x_1, \ldots, x_s$ generate $\frak{q}$. Then $\{\overline{x_i}\}$ in $\frak{q}/\frak{q}^2$ generate $G(R)$ as an $R/\frak{q}$-algebra. But $l$ is an additive function on the filtration, so by last time we saw thatfor large $n$ there is some polynomial such that $f(n)=l(G_n(M))=l(M_n/M_{n+1})$, and each $\overline{x_i}$ has degree 1, so the polynomial is of degree $\leq s-1$.

Thus we get that $l_{n+1}-l_n=l(G_n(M))=f(n)$. So from two posts ago, we get for large $n$ that $l_n$ is some polynomial $g(n)$ of degree $\leq s$.

3) Probably the most important part is that the degree and leading coefficient of $g(n)$ depends only on $M$ and $\frak{q}$ and not on the filtration.

Let $(\overline{M_n})$ be some other stable $\frak{q}$-filtration with polynomial $\overline{g}(n)=l(M/\overline{M_n})$. Since any two stable $\frak{q}$-filtrations have bounded difference, there is an integer $N$ such that $M_{n+N}\subset \overline{M_n}$ and $\overline{M_{n+N}}\subset M_n$ for all $n\geq 0$. But this condition on the polynomials says that $g(n+N)\geq \overline{g}(n)$ and $\overline{g}(n+N)\geq g(n)$, which means that $\lim_{n\to\infty}\frac{g(n)}{\overline{g}(n)}=1$. Thus they have the same degree and leading coefficient.

That seems to be enough for one day. Unfortunately, I haven't quite got to the right setting that I want yet.

## Hilbert Polynomial II

My overall goal has not changed, but I definitely have a much clearer picture of where my posts are headed for right now. I recently was working on what happens to dimension when you intersect varieties, and I needed a commutative algebra result that sort of surprised me. So that is my first benchmark on this front. Lucky for me, there is a nice clean way to prove it using the Hilbert polynomial, so I can just continue this course for now.

Let’s now reconstruct the Hilbert polynomial in a different way. As before let $M$ be a finitely generated graded $R$-module. Then $M_n$ is finitely generated as an $A_0$-module.

Let $\lambda$ be an additive funtion (in $\mathbb{Z}$) on the class of finitely generated $A_0$-modules. We define the Poincare series of $M$ to be the generating funciton of $\lambda(M_n)$. So we get a power series with coefficients in $\mathbb{Z}$: $P(M, t)=\sum \lambda(M_n)t^n$.

By a remarkably similar argument to the last post we can check by induction that $P(M, t)$ is a rational function in $t$ of the form $\displaystyle \frac{f(t)}{\prod_{t=1}^s (1-t^{k_i})}$ where $f(t)\in\mathbb{Z}[t]$.

Let’s suggestively call the order of the pole at $t=1$, $d(M)$.

We now simplify the situation by taking all $k_i=1$. Then the main idea for today is that $\lambda(M_n)$ is a polynomial of degree $d-1$. In fact, $\lambda(M_n)=H_M(n)$.

Our simplification gives that $P(M, t)=f(t)\cdot (1-t)^{-s}$. So $\lambda(M_n)$ is the coefficient of $t^n$. If we cancel factors of $(1-t)$ out of $f(t)$ we can assume $f(1)\neq 0$ and that $s=d$. Write $f(t)=\sum_{k=0}^N a_kt^k$. Then since $\displaystyle (1-t)^{-d}=\sum_{k=0}^\infty \left(\begin{matrix} d+k-1 \\ d-1 \end{matrix}\right) t^k$ we get that $\lambda(M_n)=\sum_{k=0}^N a_k \left(\begin{matrix} d+n-k-1 \\ d-1 \end{matrix}\right)$ for all $n\geq N$.

Thus we get a polynomial with non-zero leading term. Note the values at integers are integers, but the coefficients in general are only rationals.

Since $\lambda$ was any additive function, this is a bit more general. But taking $\lambda(M_n)=\dim M_n$ we get the Hilbert polynomial from last time.

Next time we’ll start using this to streamline some proofs about dimension.

## Hilbert Polynomial I

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 $R=\oplus R_i$ is a graded ring. Then a graded R-module, M, is a module with an abelian group decomposition $\displaystyle M=\oplus_{-\infty}^\infty M_i$ such that $R_iM_j\subset M_{i+j}$.

Let $M$ be a finitely generated graded $k[x_1,\ldots, x_r]$-module (graded by degree of the polynomial). Then we define the Hilbert function of $M$ to be $H_M(s)=\dim_k M_s$. The function takes as input something from $\mathbb{Z}$ and outputs the dimension of that graded part.

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

Let’s prove a general fact first. Suppose $f(s)\in\mathbb{Z}$ is defined for all natural numbers. Then if $g(s)=f(s)-f(s-1)$ agrees with a polynomial (with rational coefficients) of degree less than or equal to $n-1$ for all $s\geq s_0$, then $f(s)$ agrees with a polynomial (with rational coefficients) of degree less than or equal to $n$ for all $s\geq s_0$.

Suppose $Q(s)$ is a polynomial that satisfies the hypothesis of the preceding statement, i.e. $Q(s)=g(s)$ for $s\geq s_0$.

Set $P(s)=f(s)$ for $s\geq s_0$ and $\displaystyle P(s)=f(s_0)-\sum_{t=s+1}^{s_0} Q(t)$ for $s\leq s_0$.

Now just note that $P(s)-P(s-1)=Q(s)$ for all integers. So we are done since then $P(s)$ is a polynomial with rational coefficients of degree less than or equal to $n$.

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 $r$. 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 $H_M(s)=0$ for large $s$ and we are done.

Suppose the theorem holds in $r-1$ variables. Now let $K$ be the kernel of the multiplication map by $x_r$. This is a submodule of $M$, and we get an exact sequence $\displaystyle 0\to K(-1)\to M(-1)\stackrel{x_r}{\to} M\to M/(x_rM)\to 0$. Where the $(-1)$ means the grading is shifted by $-1$.

The exactness tells us something about the dimensions. So look at the $s$ part of the grading: $\dim_kK(-1)_s-\dim_k M(-1)_s+\dim_k M_s-\dim_k (M/x_rM)_s=0$. In terms of the Hilbert function, this says precisely that $\displaystyle H_M(s)-H_M(s-1)=H_{M/x_rM}(s)-H_K(s-1)$.

Since $K$ and $M/x_rM$ are f.g. graded modules over $k[x_1, \ldots, x_{r-1}]$ we can apply the inductive hypothesis to the right side. But since the right side is a polynomial for large $s$, 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.

## Art done right

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 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.

## Abelianization of the Fundamental Group

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Quick Update

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 $\mathbb{R}$ 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 $S^7$‘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.

## Handle Decomposition

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 $M_{t}$ is a handlebody for $t\in (c_{i-1}, c_i)$, then $M_{c_i+\varepsilon}$ is a handlebody with the appropriate handle attached.

So we’ve assumed that we have some decomposition $M_{c_{i-1}+\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1})$. We also know that we attach a handle of index $\lambda_i$ when crossing $c_i$, so we do have a diffeo to a manifold $M_{c_i-\varepsilon}$ with a $\lambda_i$-handle attached with attaching map $\phi: \partial D^{\lambda_i}\times D^{m-\lambda_i}\to \partial M_{c_i-\varepsilon}$.

Note that $[c_{i-1}+\varepsilon, c_i-\varepsilon]$ contains no critical values, so by flowing along $X$ we get a diffeo $M_{c_{i-1}+\varepsilon}\cong M_{c_i-\varepsilon}$. Let $\psi:M_{c_{i-1}+\varepsilon}\to M_{c_i-\varepsilon}$ be this diffeo.

So by inductive hypothesis, $M_{c_i-\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1})$, so we can assume $\psi$ actually maps from the handlebody to $M_{c_i-\varepsilon}$. Now by composing we get our actual attaching map (note that before now the handle was attached to $M_{c_i-\varepsilon}$ and not the handlebody itself).

i.e. $\psi^{-1}\circ \phi : \partial D^{\lambda_i}\times D^{m-\lambda_i}\to \partial (\mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1}))$. So let $\phi_i=\psi^{-1}\circ \phi$, and we get that $M_{c_i+\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1}, \phi_i)$, 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.