User lucas culler - MathOverflow

Local properties of analytic elliptic surfaces

2010-11-12T16:14:28Z

Let $D \subset \mathbb{C}$ be the unit disc, and let $E$ be a complex surface with a regular holomorphic map $\pi: E \to D$, whose fibers are all curves of genus one (so $E$ is a nonsingular elliptic surface over $D$). Given a point $z \in D$, let $j(z)$ be the $j$-invariant of the fiber over $z$. Two questions:

1) Is $j$ a holomorphic function of $z$?

2) Is $E$ isomorphic to the quotient of $D \times \mathbb{C}$ by a $\mathbb{Z}^2$-action of the form

$(n,m) \cdot (z,w) = (z, w + n + m\tau(z))$

for some holomorphic map $\tau:D \to \mathbb{H}$?

I assume the answers are yes, but it's not obvious to me. This seems like a pretty basic question, so I assume there's a reference which talks about it, but I haven't found one yet.

Answer by Lucas Culler for Are there examples of non-orientable manifolds in nature?

2010-11-12T16:25:38Z

There was a study where they took thousands of digital pictures of "natural images", rendered them in grayscale, and looked at all the 3x3 pixel squares which arose in such pictures. Using topological data analysis they found that (after some normalizations) their data points actually clustered around a Klein bottle embedded in the 7 sphere! Here's a paper that talks about it, and tells you where to look for the Klein bottle:

http://www.math.uiuc.edu/~ghrist/preprints/barcodes.pdf

Answer by Lucas Culler for Ways to prove the fundamental theorem of algebra

2010-09-20T12:35:58Z

As in one of the previous posts, consider the projective space $CP^n$ of nonzero all polynomials

$$ c_nT^n + c_{n-1} T^{n-1} + \cdots + c_1 T + c_0 $$ considered up to nonzero scalar multiple. We'll show directly that any such polynomial admits a factorization into linear factors.

Consider the map $\phi: CP^1 \times \cdots \times CP^1 \to CP^n$ given by

$$ ([\alpha_1:\beta_1],\dots,[\alpha_n:\beta_n]) \mapsto \prod_{i=1}^n (\alpha_i T - \beta_i) $$

In other words, this map sends a set of roots to the polynomial which has precisely those roots. It suffices for us to show that $\phi$ is surjective.

If the points $[\alpha_i:\beta_i]$ are distinct, it is easy to check that the differential $d\phi$ is nonzero. Hence the polynomial $T^n - 1$ (for example) is a regular value of $\phi$ with exactly $n!$ preimages (here we've used the fact that polynomials factor uniquely into irreducibles). Thus the map $\phi$ has positive degree.

It is a fact that any map of positive degree between compact connected complex manifolds of the same dimension is surjective. (Proof: Since any such map is orientation preserving, the number of preimages of any regular value must be exactly equal to the degree - not just up to multiplicity. Hence the image contains the set of regular values, which is dense by Sard's theorem. But the image is also closed since it is the image of a compact set, hence the map is surjective.)

We conclude that $\phi$ is surjective. In other words, every polynomial of degree $n$ has a factorization into linear factors.

Answer by Lucas Culler for Möbius and projective 3-manifolds

2010-05-05T13:48:57Z

If true, such a statement would immediately imply the Poincare conjecture. Indeed, suppose X were a closed, simply connected 3-manifold with a Mobius structure. Near any point of $X$, there would then be a Mobius map from $X$ to $S^3$. Analytically continuing this map (see Thurston's notes, chapter 3) would give a local homeomorphism $\phi: X \to S^3$. Since both spaces are compact, $\phi$ would be a covering map. Since $S^3$ is simply connected it would be a homeomorphism. A similar argument applies to projective structures, with $S^3$ replaced by $\mathbb{RP}^3$.

Answer by Lucas Culler for Is it always possible to compute the Betti numbers of a nice space with a well-chosen Lefschetz zeta function?

2010-04-28T04:13:59Z

Having resolved my ignorance concerning surface groups I can now answer question 1 negatively (or at least some formulation thereof). It is impossible if $Y$ is an oriented surface of genus at least $2$.

Suppose that $f: Y \to Y$ is a self map of the surface such that the eigenvalues of $f^*$ acting on each $H^i(Y)$ are all nonzero (otherwise we can't "detect" the betti numbers), and such that $H^i(Y)$ and $H^j(Y)$ do not have eigenvalues of common magnitude for $i \neq j$. Then in particular $f^*$ acts on $H^2(Y)$ nontrivially, say by multiplication by some integer $d$. This integer cannot be $\pm 1$ since then $H^0(Y)$ and $H^2(Y)$ would contain eigenvectors with eigenvalues of equal magnitude.

Consider the subgroup $H = f_*(\pi_1(Y))$ inside $G = \pi_1(Y)$. If this had infinite index, then $f$ would lift to a map to some infinite covering of $Y$, so it would induce a trivial map of $H^2$. So $H$ has finite index in $G$. Let $X \to Y$ be the corresponding covering space. Then $\pi_1(X)$ is a quotient of $\pi_1(Y)$, hence its abelianization has rank $\leq 2g$ where $g$ is the genus of $Y$. This implies that $X$ is a closed surface of genus at most $g$. But its Euler characteristic is precisely $[G:H]$ times the Euler characteristic of $Y$, so $X = Y$. Thus $f$ induces a surjection on $\pi_1(Y)$. By the post cited above, $f$ actually induces an isomorphism on $\pi_1(Y)$, so it is a homotopy equivalence. In particular, $d = \pm 1$, contrary to assumption.

After writing this it occurs to me that you might object to me ruling out the case $d = -1$... At any rate, this shows that the eigenvalues can't ever look like they do in the case of the Riemann hypothesis, with magnitude $q^{i/2}$ on $H^i$ for some $q>1$.

Self-homomorphisms of surface groups

2010-04-28T03:42:35Z

Let $X$ be a closed, orientable surface of genus at least 2, and let $\phi: \pi_1(X) \to \pi_1(X)$ be a surjective homomorphism. Is $\phi$ necessarily injective?

Answer by Lucas Culler for Linear elliptic partial differential equation with analytic coefficients

2010-04-22T00:41:04Z

In general this will not have a solution. Note that your equation is equivalent to the following:

\[ \Delta_g (u) = \sum_{i} \frac{1}{a} \frac{\partial}{\partial \theta_i}( a \frac{\partial u}{\partial \theta_i} ) = e^{i\theta_1} \]

where by definition $a = e^b$ . This is the Laplace equation (see the Wikipedia article "Laplace-Beltrami operator") with respect to the metric

\[ g_{ij} = e^{2b} \delta_{ij} \]

Hence it only has a solution if the average value of $e^{i\theta_1}$ with respect to the volume form of $g$ is zero. The reason for this is the integration by parts formula

\[ \int_{[0,2\pi]^d} (\Delta_g u)\cdot v dvol_g = \int_{[0,2\pi]^d} u \cdot (\Delta_g v) dvol_g \]

which when you plug in $v = 1$ shows that the average value of $\Delta_g u $ must be zero. The formula can be proved in much the same way as for the standard Laplacian. Thus a necessary (and sufficient) condition for a solution to exist is:

\[ \int_{[0,2\pi]^d} e^{b+ i \theta_1} d\theta_1 \cdots d\theta_n = 0\]

In general, even if a solution exists I wouldn't expect a particularly nice description of the Fourier coefficients. You might be able to get estimates of them, particularly if the function $b$ is very small, in which case the equation will closely resemble the standard Laplacian on the torus. If $b$ has very few terms you might try taking the Fourier transform of your equation, in which case the second term will become a convolution and you may be able to solve inductively for the Fourier coefficients if you're lucky.

Answer by Lucas Culler for Newlander-Nirenberg for surfaces

2010-04-21T06:27:38Z

I'm not sure if the following is elementary enough, but it does only use standard PDE machinery (plus some basic Riemannian geometry). It's also nice in that it suggests an approach to proving the uniformization theorem (via metrics of constant curvature).

Say you have an almost-complex structure on the unit disk. Your goal is to find a conformal isomorphism of this disk (or at least some neighborhood of the origin) with an open subset of the complex plane. To do it, first choose a metric $g$ on the disk which is compatible with the given complex structure. Let $K$ be the curvature of this metric. If you can find a flat metric $\tilde{g}$ on on the disk which is conformally equivalent to $g$ then you'll be done, since the exponential map with respect to $\tilde{g}$ will be an isometry, hence also a conformal isomorphism.

So, multiply $g$ by an arbitrary positive function $e^f$ , and compute the curvature of the new metric. You'll find that it's given by the formula:

$\tilde{K} = e^{-2f}(K - \Delta f) $

where $\Delta$ is the Laplacian with respect to the metric $g$ . Setting the left hand side equal to zero, you have reduced to solving the Laplace equation, which can be done locally using standard PDE techniques.

As for a more "direct" approach...

The equation you wrote down should reduce to solving the Laplace equation as well, using the notion of conjugate harmonic functions. However, solving your equation will inevitably be a bit more subtle due to your requirement that the solution have nonvanishing differential at the origin. There is a proof along the lines you're suggesting in Taylor's PDE book, chapter 5, section 11, and I think there's a similar one in Jost's "Postmodern analysis". Basically the idea is to rescale your coordinate system so that the metric is nearly flat, in which case you should have a conformal map that is close to the identity map in a high enough sobolev space, and therefore has a nonvanishing derivative at the origin.

Comment by Lucas Culler

2011-12-06T03:30:14Z

You can also use an embedding in S^6 (a bit easier to construct). If nM = normal bundle, then H^2(S^6-nM) = Z by Mayer Vietoris. A geometric representative of the generating class will be a 4-manifold X with dX = M. More precisely, dX will be cobordant to a nonvanishing section of nM.

Comment by Lucas Culler

2010-11-12T17:51:27Z

I meant to assume smoothness so that's not an issue. I agree that it's true for algebraic families, so assuming the last fact you mentioned I would agree that it's true for analytic families. Where would I find a proof of that fact?

Comment by Lucas Culler

2010-11-12T17:30:07Z

Rather, I just don't see how the arguments in Katz-Mazur apply in this situation.

Comment by Lucas Culler

2010-11-12T17:19:00Z

I'm not assuming that the generic fiber has genus one. Why does this follow from my assumptions?

Comment by Lucas Culler

2010-09-20T12:05:15Z

The diffeomorphism group is not a subgroup of the gauge group, because a diffeomorphism f induces maps $T_x M \to T_{f(x)} M$, rather than from $T_x M$ to itself. In other words, Df is not a map of bundles over $X$.

Comment by Lucas Culler

2010-05-06T03:43:39Z

Yeah, sorry for the lack of a real answer. I was just pointing out that if there was a proof it wouldn't be easy.

Comment by Lucas Culler

2010-04-29T00:17:50Z

I think exactness just refers to the fact that you prove it by solving df = (Q - P')dy + Rydx. So it seems to still be exactness of a 1-form you're interested in, not a 2-form. Though I agree the alternating sum is suggestive...

Comment by Lucas Culler

2010-04-28T23:33:42Z

C does not act on the fibers of a line bundle by addition, so line bundles can't be C-torsors. One thing to observe is that line bundles come with a canonical choice of zero section, and are not necessarily trivial, unlike the situation with torsors where having a section automatically makes the bundle trivial. One thing you can say is that a line bundle is a "bundle of groups", but this is very different from being a torsor.

Comment by Lucas Culler

2010-04-28T05:52:22Z

Yeah, it doesn't rule much out. I guess what I'm saying is that manifolds don't necessarily come with natural (or even any) self-maps of positive degree (like the frobenius in the case of varieties over F_p), and this seems to be an obstruction to doing what you want.

Comment by Lucas Culler

2010-04-28T04:50:10Z

Thanks a lot! I'll take a look at the references next time I give up. I clearly have a lot to learn about group theory.

Comment by Lucas Culler

2010-04-28T03:54:50Z

I knew I could count on you. Should I just keep trying to prove it myself or is there a reference?