Liouville’s Theorem for Projective Varieties?

Wow. I hate looking at the dates on old posts. I think that maybe a few days have gone by, and I’m horrified to find that 11 or 12 days have passed. It is hard to keep track of time in grad school.

The goal of this post is to prove the theorem: If V is an irreducible projective variety over an algebraically closed field k, then every regular function on V is constant. Note this says that \mathcal{O}(V)\cong k. Also, an exercise is to think about how this relates to Liouville’s Theorem if our field is \mathbb{C}.

Proof: Let V be an irreducible projective variety in \mathbb{P}_k^n. WLOG V is not contained in a hyperplane, since then we could just eliminate a variable and work in \mathbb{P}_k^{n-1} and repeat this until it was not in any hyperplane.

Let f\in\mathcal{O}(V). Consider the affine covering from last time V=V_0\cup\cdots\cup V_n. Note that f\big|_{V_i} is regular as an affine morphism on V_i. So we can write this f\big|_{V_i} as a polynomial in x_j/x_i where 1\leq j\neq i\leq n. i.e. we can factor out the homogeneous part of the denominator variable to get \displaystyle f\big|_{V_i}=\frac{g_i}{x_i^{N_i}} where g_i\in S(V)=k[x_0, \ldots, x_n]/I(V) is homogeneous of degree N_i.

But we assumed V irreducible, so I(V) is prime and hence S(V) is an integral domain. Let’s take the field of fractions then, L=Frac(S(V)). Then \mathcal{O}(V), k(V) and S(V) are all embedded in L. So in L we can multiply by that denominator we had before to get x_i^{N_i}f\in S_{N_i}(V).

Recall that S(V) is a graded ring, so I just am denoting S_{N_i}(V) to be the N_i-graded part. Thus if we take any integer N\geq\sum N_i, then S_N(V) is a finite-dimensional k-vector space. Moreover, the monomials of degree N span the space.

Let m\in S_N(V) be a monomial. Then it is divisible by x_i^{N_i} for some i, so mf\in S_N(V). Thus S_N(V)f\subset S_N(V).

So we have a chain: S_N(V)f^q\subset S_N(V)f^{q-1}\subset \cdots \subset S_N(V)f\subset S_N(V). So x_0^Nf^q\in S_N(V) for any q\geq 1. Thus S(V)[f]\subset x_0^{-N}S(V)\subset L.

But x_0^{-N}S(V) is Noetherian, since it is finitely generated as a S(V)-module, so S(V)[f] is also finitely generated over S(V). Thus f is integral over S(V).

i.e. there are a_i\in S(V) such that f^m+a_{m-1}+\ldots + a_0=0. But this shows that f is homogeneous of degree 0. i.e. f\in S_0(V)\cong k. So f is constant.


Are Projective Varieties Secretly Affine Varieties?

Last time I recoiled from my statement that projective spaces are manifolds, but let’s explore this a little. We don’t want manifolds, but an idea comes up in showing projective space is one that we can use. The fact that we have natural local “coordinate patches” on projective space. i.e. \mathbb{P}_k^n=U_0\cup \cdots \cup U_n, where U_i=\{(x_0 : \ldots : x_n)\in \mathbb{P}_k^n : x_i\neq 0. For a manifold, we would show these are diffeomorphic to copies of \mathbb{R}^n, but this isn’t the right topology (and has far too restricting of structure), so we really want to show that each is homeomorphic to affine space p_i: U_i\to \mathbb{A}_k^n by (x_0 : \ldots : x_i : \ldots : x_n)\mapsto \left(\frac{x_0}{x_i}, \ldots , \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \ldots , \frac{x_n}{x_i}\right).

So to reiterate, we have the Zariski topology on projective space by closed sets defined as zero sets of homogeneous polynomials, and the Zariski topology on affine space by zero sets of polynomials. What we want to show is that we can cover projective space with affine spaces, so that locally projective space looks affine.

Let’s just work with \pi_0, since there is nothing special about any of the coverings. As far as sets are concerned, \pi_0 is certainly bijective, so we just need to check that \pi_0 and \pi_0^{-1} are continuous. So we’ll check that they are closed maps.

The polynomials that we are concerned with defining closed sets on U_0 are from the set S=\{f\in k[x_0, \ldots , x_n] : f ~ homogeneous\} and the polynomials defining closed sets on \mathbb{A}_k^n are from k[x_1, \ldots , x_n].

So now let \alpha : S\to A by f\mapsto f(1, x_1, \ldots , x_n), and let \beta: A\to S by g\mapsto x_0^{deg(g)}g(\frac{x_1}{x_0}, \ldots , \frac{x_n}{x_0}).

It is easy to check that \alpha\circ \beta=Id.

Now let X be any closed set of U_0. Then X=E\cap U_0 where E is a closed set in \mathbb{P}_k^n. i.e. E is a projective variety, so there is a set T\subset S such that E=V(T) and \pi_0(X)=V(\alpha(T)).

Any closed set of \mathbb{A}_k^n has the form W=V(T') where T'\subset A and \pi_0^{-1}(W)=V(\beta(T'))\cap U_0.

But now the image of closed sets under \pi_0 is precisely \pi_0(V(T)\cap U_0)=V(\alpha(T)). So \pi_0 is a closed map. Also, the image of a closed set \pi_0^{-1}(V(T'))=V(\beta(T')))\cap U_0, so both are closed and hence continuous, so \pi_0 is a homeomorphism.

But this proof actually gives us more. Namely that irreducible projective varieties X\subset \mathbb{P}_k^n such that X\cap \{(x_0: \ldots : x_n) : x_0=0\}=\emptyset is in 1-1 correspondence with irreducible affine varieties X_0\subset \mathbb{A}_k^n.

Intro to Projective Varieties

I will assume a basic familiarity with projective space from now on (I don’t think Ive covered it in any previous posts). For a quick recap and a guide to my notation, we can define projective space on a finite dimensional vector space V over k by defining a n equivalence relation uRv iff there is a non-zero scalar \lambda\in k such that u=\lambda v. Then \mathbb{P}(V)=\frac{\left( V\setminus\{0\}\right)}{ R}.

Review/facts: dim\mathbb{P}(V)=dim V -1. Also, the way we usually think is \mathbb{P}(\mathbb{C})\cong S^2. The complex projective space (line) is just the compactification of the plane and hence the Riemann sphere. The fact that this is a compact manifold is no coincidence, in fact, all projective spaces can be given the structure of a compact manifold…but maybe this should not be mentioned, since I want to put a different topology on it and talk about varieties.

So our field and the dimension of the vector space are in some sense more important than the vector space itself, so I’ll notate from now on \mathbb{P}(V) as \mathbb{P}_k^n. If we take \pi: V\setminus\{0\}\to \mathbb{P}(V) as the standard projection onto the quotient, we use the notation (x_0 : \ldots : x_n) for \pi((x_0, \ldots , x_n)). These are called homogeneous coordinates. Note that these are only well-defined up to scalar multiple.

I haven’t developed on this blog any good motivation to now switch to projective space, but there are some good reasons. At first, it seems to just make things more complicated, but really it simplifies things in the long run. Also, there are some nice properties that you should check. In the affine case, lines can either be parallel and never intersect, or they intersect somewhere. In the projective case, all lines intersect.

So now we just extend the same definitions from the affine case to the projective case, but we are careful to make sure everything is well-defined.

Since homogeneous coordinates are determined up to multiplication by a scalar, we need to make sure our polynomials can deal with this. So we call a polynomial homogeneous of degree d, if every monomial has degree d. i.e. f(x_0, \ldots, x_n)=\sum a_{b_0\cdots b_n}x_0^{b_0}\cdots x_n^{b_n} where b_0+\cdots b_n=d. So, we have well-defined zero sets of polynomials since we can pull scalar multiples out: f(\lambda x_0, \ldots , \lambda x_n)=\lambda^d f(x_0, \ldots, x_n), i.e. the variety V(f)=\{ (x_0: \ldots : x_n)\in \mathbb{P}_k^n : f(x_0, \ldots, x_n)=0\} is well-defined. Note: If d=2, then we call these quadratic forms.

We call a subset V\subset \mathbb{P}_k^n a “projective variety” if there is a set of homogeneous polynomials T\subset k[x_0, \ldots , x_n] such that V=V(T)=\{P\in \mathbb{P}_k^n : f(P)=0 ~ \forall f\in T\}.

I’ll let you digest that, and learn a little more about projective space if this was your first experience dabbling with it. Our future plans are to try to figure out which parts of the affine theory we developed carries over unscathed, which parts break down, and which parts get simplified.

Albums I’d Love to Hate

I’m not sure if I’ve talked about this before, but I have something I call a “five-listens rule”. Which means that I require myself to listen to an album five times all the way through no matter how much I hate it the first time, and not skipping any tracks even if they annoy the crap out of me. This is an excellent rule if you were wondering. It has saved me from making a fool of myself with things that I wasn’t too impressed with at first, but have since become some of my favorite albums. Sometimes it just takes time for something to grow on you. After the first five listens, I allow myself to actually make some judgements and skip around or never listen again if I want.

Now that that has been explained, I can move on to what I really wanted to discuss. There is so much music that comes out all the time that it is just impossible to listen to even a small fraction of it (unfortunately). This is where albums that I’d love to hate come in. I wish there were more things that I listen to once through and say, I hate this and then have more time to move on. What usually happens, though, is that after five listens I start to understand things a little better and don’t actually give up on things I should.

My first example is Beck. I find nothing overly amazing about Beck. I’ve never really fallen in love with any of his albums. I really wish they weren’t on that border, though. When he comes out with something new, I think to myself, “Well, I kind of liked the last one, so maybe I’ll love this one.” If I just hated it, then I could move on, not get anymore, and not keep returning. I’d love to hate Sea Change. I think it is his best. I’ve been listening to it recently, and am borderline loving it. Due to this, it is eating up time I could be listening to current music.

Next example, M. Ward. I was pretty bored with his new one Hold Time. Nothing particularly good lyrically or musically. Just pretty standard sounding stuff. I was on my fourth listen, and I still thought I wouldn’t listen after the fifth. But on the fifth something started to change. For some reason I started to like it. It is probably still my least favorite thing that has come out this year that I got, but it isn’t bad enough for me to stop listening. I’d love to hate it. (Actually, this album is the source of this post).

Luckily, after last year my wish for wanting to hate Conor Oberst finally came true. He releases stuff so frequently, and I loved the early stuff so much, that I just really wanted a good reason to stop getting his albums. Last year it finally happened with his self-titled solo album. I see that a new one came out, but I will successfully avoid getting it and listening to it.

Lastly, I should talk about someone I’d hate to love. That would definitely have to be The Mountain Goats. With 16 albums and 27 singles/EP’s, talk about time and money killing. People that love them must never get to listen to anything else.