Posted by: hilbertthm90 | November 9, 2009

Some Corollaries

Today will just be some quick results we get from this build up.

First, if we localize a polynomial ring at a maximal ideal, say k[x_1, \ldots, x_n] at \frak{m}=(x_1, \ldots, x_n), then \dim R_\frak{m}=n=\dim R. This is because G_m(R) has Poincare series (1-t)^{-n} so the order of the pole is n which is the dimension by the last post.

This one will be really useful later: \dim R\leq \dim_k(\frak{m}/\frak{m}^2). Let \{x_i\}_1^r \subset\frak{m} such that \overline{x_i}\in \frak{m}/\frak{m}^2 are a basis for the vector space. Then by Nakayama’s Lemma the x_i generate \frak{m}. Thus \dim_k(\frak{m}/\frak{m}^2)=s\geq \dim R.

This one is also useful in algebraic geometry. If R is Noetherian, and x_1, \ldots , x_r\in R, then every minimal ideal \frak{p} belonging to (x_1, \ldots, x_r) has height \leq r. Unfortunately, we cannot push this to equality. Geometrically the example is that if Y is the twisted cubic, then I(Y) has height 2, but cannot be generated by less than 3 elements.

Lastly, we’ll do the famous Principal Ideal Theorem. If R is Noetherian and x\in R is neither a zero-divisor nor a unit, then every minimal prime ideal \frak{p} of (x) has height 1. By the last paragraph we know that ht(p)\leq 1. If ht(\frak{p})=0 then it belongs to (0). Thus every element of \frak{p} is a zero-divisor which is a contradiction since x\in \frak{p}.

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Posted by: hilbertthm90 | November 8, 2009

Finishing up Dimensions

We are now on the last inequality: \dim R\geq \delta(R). Recall we’re supposing (R, \frak{m}) is Noetherian and local. Let \dim R=d, then the inequality is saying we can find an ideal, \frak{q} that is an \frak{m}-primary ideal and is generated by d elements: x_1, \ldots, x_d.

Let’s construct these elements inductively. The way we want to do it is so that at each step any prime ideal containing (x_1, \ldots, x_i) has height bigger than or equal to i to force the dimension of R to be big.

Suppose the x_1, \ldots, x_{i-1} have been constructed in the given way. Let \{p_j\} be the minimal prime ideals of (x_1, \ldots, x_{i-1}) with height exactly i-1. But we have that \frak{m}\neq p_j for any j since ht(\frak{m})=\dim R=d>i-1=ht(p_j). Thus \frak{m}\neq \cup p_j (there is a well-known fact that if any ideal \frak{a}\subset \cup p_j, then in fact \frak{a}\subset p_j for some j).

Now pick some element x_i\in \frak{m}\setminus (\cup p_j), and let \frak{q} be a prime ideal containing (x_1, \ldots, x_i). We definitely have that ht(\frak{q})\geq i, since \frak{q} contains a minimal prime ideal of (x_1, \ldots , x_{i-1}), say \frak{p}. If for some j, \frak{p}=p_j, then since x_i\in\frak{q}\setminus \frak{p}, we have strictly \frak{q}\supset \frak{p} increasing the height. If \frak{p}\neq p_j for any j, then since \{p_j\} are all minimal primes of height i-1, we have ht(\frak{q})>i-1.

All that is left is to show that \frak{q}=(x_1, \ldots, x_d) is \frak{m}-primary. Let \frak{p} be any prime ideal of \frak{q}. Then if \frak{p}\subsetneq \frak{m} we have that ht(\frak{p})< ht(\frak{m})=d. So this is impossible and we have \frak{p}=\frak{m}. Thus \frak{q} is \frak{m}-primary.

Over the last couple of posts we have finally completed the first goal. We have \delta(R)=d(R)=\dim R. In other words, for Noetherian local rings we have an equivalence between the maximum length of chains of prime ideals, the degree of the Hilbert polynomial, and the least number of generators of an \frak{m}-primary ideal.

Next time I'll derive some results directly from this including the Principal Ideal Theorem. Then we'll move on to something different (only for awhile, then we'll return).

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Posted by: hilbertthm90 | November 3, 2009

The Next Inequality

Considering it has been at least a post removed, I’ll bring us back to our situation. We have a local Noetherian ring (R, \frak{m}). Our notation is that \delta(R) is the least number of generators of an \frak{m}-primary ideal (which was shown to be independent of choice of ideal here). The goal for the day is to show that d(R)\geq \dim R.

Suppose \frak{q} is \frak{m}-primary. We’ll prove something more general. Let M be a finitely generated R-module, x\in R a non-zero divisor in M and M'=M/xM. Then the claim is that \deg\chi_q^{M'}\leq \deg\chi_q^M -1.

Since x is not a zero-divisor, we have an iso as R-modules: xM\cong M. Define N=xM. Now take N_n=N\cap \frak{q}^nM. Since \frak{q}^nM is a stable \frak{q}-filtration of M, by Artin-Rees we get that (N_n) is a stable \frak{q}-filtration of N.

For each n we have 0\to N/N_n \to M/\frak{q}^nM\to M'/\frak{q}^nM'\to 0 exact.

Thus we get l(N/N_n)-l(M/\frak{q}^nM)+l(M'/\frak{q}^nM')=0. So if we let g(n)=l(N/N_n), we get for large n: g(n)-\chi_q^M(n)+\chi_q^{M'}(n)=0.

But (N_n) is also a stable \frak{q}-filtration of M, since N\cong M. We already showed that the degree and leading coefficient of g(n) depends only on M and \frak{q} and not on the filtration. Thus g(n) and \chi_q^M(n) have the same degree and leading coefficient, so the highest powers kill eachother which gives \deg\chi_q^{M'}\leq \deg \chi_q^M-1.

In particular, we will need that R as an R-module gives us d(R/(x))\leq d(R)-1.

Now we prove the goal for today. For simplicity, let d=d(R). We will induct on d. The base case gives that l(R/\frak{m}^n) is constant for large n. In particular, there is some N such that \frak{m}^n=\frak{m}^{n+1} for all n>N. But we are local, so \frak{m}=J(R) and hence by Nakayama, \frak{m}^n=0. Thus for any prime ideal \frak{p}, we have \frak{m}^k\subset \frak{p} for some k, so take radicals to get \frak{m}=\frak{p}. Thus there is only one prime ideal and we actually have an Artinian ring and hence have \dim R=0.

Now suppose d>0 and the result holds for \leq d-1. Let p_0\subset p_1\subset \cdots \subset p_r be a chain of primes. Choose x\in p_1\setminus p_0. Define R'=R/p_0 and \overline{x} be the image of x in R'.

Note that since R' is an integral domain, and \overline{x} is not 0, it is not a zero-divisor. So we use our first proof from today to get that d(R'/(\overline{x}))\leq d(R')-1.

Let \frak{m}' be the maximal ideal of R'. Then R'/\frak{m}' is the image of R/\frak{m}, so l(R/\frak{m}^n)\geq l(R'/\frak{m}'^n) which is precisely d(R)\geq d(R'). Plugging this into the above inequality gives d(R'/(\overline{x}))\leq d(A)-1=d-1.

So by the inductive hypothesis, \dim(R'/\overline{x})\leq d-1. Take our original prime chain. The images form a chain \overline{p}_1, \ldots , \overline{p}_r in R'/(\overline{x}). Thus r-1\leq d-1\Rightarrow r\leq d. Since the chain was arbitrary, \dim R\leq d(R).

A nice corollary here is that the dimension of any Noetherian local ring is finite. Another similar corollary is that in any Noetherian ring (drop the local) the height of a prime ideal is finite (and hence primes satisfy the DCC), since ht(p)=\dim A_p which is local Noetherian.

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Posted by: hilbertthm90 | November 2, 2009

The Artin-Rees Lemma

We have a somewhat bumpy road to traverse today. I’ll start with the Artin-Rees lemma and see if we can get to a use of it to continue our set of inequalities we’re trying to prove.

First we’ll need some new ideas. Suppose R is any old ring (in particular, we are dropping graded and Noetherian assumptions). Then if \frak{a} is an ideal, we can form a new ring R^*=\bigoplus_{n=0}^\infty \frak{a}^n which by construction is graded. Now for any R-module, say M and an \frak{a}-filtration M_n we can form a graded R^*-module, M^*=\bigoplus M_n.

Note that if R is Noetherian in the situation above, then \frak{a}=(x_1, \ldots, x_r), so R^*=R[x_1, \ldots , x_r], so by Hilbert Basis Theorem, we get R^* is Noetherian.

We’ll need that in the situation above the following two statements are equivalent: M^* is finitely generated as an R^*-module, and that the filtration M_n is stable.

Proof: Each M_n is finitely generated, so Q_n=\bigoplus_{r=0}^n M_r is finitely generated for all n. Let’s form M_n^*=Q_n\oplus\left(\bigoplus_{k=1}^\infty \frak{a}^kM_n\right). We have that each Q_n is finitely generated as an R-module, so we get that M_n^* is finitely generated as an R^*-module.

Clearly, M_0^*\subset M_1^*\subset \cdots, so since R^* is Noetherian we get that M^* is finitely generated iff the ascending chain terminates iff M_{n_0+r}=\frak{a}^r M_{n_0} for some n_0 and for all r\geq 0 iff the filtration is stable.

Now we can prove the Artin-Rees Lemma which says that if R is a Noetherian ring, \frak{a} an ideal, M a finitely generated R-module, M_n a stable \frak{a}-filtration and M' a submodule of M, then M'\cap M_n is a stable \frak{a}-filtration of M'.

The situation is fairly simple from the previous fact. Note that \frak{a}(M'\cap M_n)\subset \frak{a}M'\cap \frak{a}M_n\subset M'\cap M_{n+1}. So we do indeed get a filtration. But M'^* is a graded A^*-submodule of M^*, so it is finitely generated. Now by the equivalence of finitely generated and stable we are done.

There are two important corollaries (both get referred to as the Artin-Rees Lemma as well). In the special case M_n=\frak{a}^nM we get that the stable filtration condition says that there is some integer N such that (\frak{a}^nM)\cap M'=\frak{a}^{n-N}((\frak{a}^NM)\cap M') for all n\geq N.

The other result uses the bounded difference result from last time. Since \frak{a}^nM' and (\frak{a}^nM)\cap M' are both stable \frak{a}-filtrations, they have bounded difference, so the \frak{a}-topology of M' coincides with induced topology from the \frak{a}-topology on M.

I think that is sufficient for today. Next time I’ll go ahead and knock off the next step of the inequalities: d(R)\geq \dim R.

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Posted by: hilbertthm90 | November 1, 2009

Beginning Dimension Theory

Recall the purpose of this development is to get some results on ring dimensions. All the hypothesis and notation from last time still hold (the important one to remember is that (R, \frak{m}) is a local ring).

Let’s introduce a new notation, which will disappear shortly. We call the characteristic polynomial of the \frak{m}-primary ideal \frak{q}, \chi_q^M(n)=l(M/\frak{q}^nM). An immediate corollary to the last post is that for large n, \chi_q(n)=l(R/\frak{q}^n) has degree \leq s where s is the least number of generators of \frak{q}.

Now we want to show that for our purposes the choice of \frak{m}-primary ideal doesn’t matter. The claim is that \deg \chi_q(n)=\deg \chi_m(n).

We know that there is some integer r such that \frak{q} contains \frak{m}^r. i.e. \frak{m}\supset \frak{q}\supset \frak{m}^r. Thus \frak{m}^n\supset \frak{q}^n \supset \frak{m}^{rn}. Thus for large n, we get \chi_m(n)\leq \chi_q(n)\leq \chi_m(rn). Since these are polynomials, we let n tend to \infty to get the claim.

Let’s denote the common degree d(R). Thus d(R) is the order of the pole at t=1 of the Hilbert function of G_\frak{m}(R).

Since this is short so far, we will very briefly start our first goal of showing that if \delta(R) is the least number of generators of an \frak{m}-primary ideal, and we impose Noetherian on R, then \delta(R)=d(R)=\dim R.

What we just showed above in this new notation is that \delta(R)\geq d(R). The way we will eventually show the equality is to get \delta(R)\geq d(R)\geq \dim R \geq \delta(R).

The next step is involved and needs the Artin-Rees Lemma, so I’ll hold off and do it next time.

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Posted by: hilbertthm90 | October 29, 2009

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.

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Posted by: hilbertthm90 | October 28, 2009

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.

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Posted by: hilbertthm90 | October 25, 2009

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.

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Posted by: hilbertthm90 | October 20, 2009

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 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.

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Posted by: hilbertthm90 | October 15, 2009

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}.

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