A Mind for Madness

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Liouville’s Theorem for Projective Varieties?

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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.

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Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

2 thoughts on “Liouville’s Theorem for Projective Varieties?

  1. Therefore, projective varieties can’t secretly be affine varieties. Unless they’re reduced to a point.

    Also, couldn’t you prove it like this- P^n(k) is a proper scheme over k, therefore a morphism from P^n -> k must have a closed image, and that image must be proper. But that image is just an affine variety, so it can’t be proper.

    I believe you can do something like this for the regular Liouville theorem too. If f: C -> C is bounded an entire, then by Riemann’s theorem on removable singularities it extends to F: S^2 -> C which is holomorphic, and whose image must be compact. But this contradicts the open mapping theorem. (This is, I think, a much more natural proof of Liouville’s theorem than the standard one.)

  2. I was trying to avoid the word “scheme,” since I haven’t defined it yet. I also haven’t talked about proper-ness, although that wouldn’t take long. Not that this blog is self-contained, but I’d like to avoid major assumptions of knowledge that aren’t really considered “standard”.

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