Month: November 2011

The Schwartz lantern

Conan5 Comments

It’s thanksgiving, let’s have some fun in lantern-making!

This thing called Schwartz lantern initially came up in a talk some years ago, I vaguely remembered it as ‘a cool example where the refining triangle approximation of a smooth surface fail to converge in area’. Anyways, it came up again recently as I was talking to some German postdoc working in ‘discrete differential geometry’. As the example was mentioned, he took a napkin and started folding…and I just realized this lantern can actually be made with a piece of flat paper!

Given a compact, smooth surface (possibly with boundary) embedded in \mathbb{R}^3, we can approximate it by a PL surface with all vertices on the surface and having only triangular faces. (just like what’s done in many computer graphic softwares nowadays).

Question: When the vertices of the faces gets denser and denser and the diameter of triangles converge to 0, does the area of the PL surface converge to the area of the surface?

Ok, to explain why I found this being a quite curious little question, let’s prove a couple of trivial observations:

Trivial fact #1: The sequence of PL surfaces as described above does Hausdorff converge to the smooth surface.

Proof:Since our surface is smooth, as the diameter of the triangles converge to 0, the length of the geodesic between two vertices is roughly the length of the edge in the 1-skeleton of the PL surface. In particular, less than twice its length.

Now by compactness we have a positive injectivity radius, implying that for small enough length c, the diameter (on the surface) of region enclosed by any loop of length most c is controlled by c (say it’s \leq f(c) which converge to 0 as c \rightarrow 0). Now the \mathbb{R}^3 diameter of the region is of course even smaller than its surface diameter.

In conclusion, when all triangles have small diameter \varepsilon (hence all its sides have length at most \varepsilon), the geodesic triangle on the surface has parameter less than 6 \varepsilon . So \mathbb{R}^3 diameter of the geodesic triangle is no more than f(6 \varepsilon). Hence the surface is contained in the f(6\varepsilon)-neighbourhood of the vertex set.

Obviously the PL surface is also contained in this neighbourhood. Hence the Hausdorff distance is at most f(6\varepsilon), which converges to 0.

Trivial fact #2: For curves in \mathbb{R}^2 (in fact, or \mathbb{R}^n), the length converges.

Proof: Well…what can I say…see any undergrad calculus book? (well, all we need is that smooth curves are rectifiable. Of course they are…

So from the above observations, does it kinda smell like the area would converge? (If you know the answer, you should pretend you don’t and nod at this point :-P) Well, the fact is they don’t have to converge! (otherwise why are we making counter-examples here?) Furthermore, this is first discovered by a super-cool dude – Schwartz! He even wrote a paper about it back in 1880.

How can this be possible? You might have already observed that with some simple curvature bounding, we can push the argument for trivial fact #1 to show that area of the curved surface is controlled above by the straight surface, The point being (perhaps to one’s surprise) that the ‘straight surface’ can be a lot LARGER than the curved one!

So the example is a sequence of ‘lanterns’ converging to a standard cylinder (say of height and circumference both equal to 1), i.e. PL surfaces with triangular faces, vertices on the cylinder, with smaller and smaller ‘grids’, and the sum of areas of the triangle blows up to infinity.

As shown above, if we have put N^4 points on the cylinder, N points along the circumference and N^3 in the vertical direction (picture is not to scale); connected to form triangles in the above way.

Now all triangles are isosceles and identical. Doing some middle-school geometry shows that they have base length \geq \frac{1}{2N} and height \geq \frac{1}{4N}\tan^{-1}(\frac{\pi}{2N}) (this calculates the distance between the midpoint of the base to the cylinder surface).

Having 2N^4 triangles means the area A_N of the PL surface is at least N^4 \times \frac{1}{2N} \times \frac{1}{4N}\tan^{-1}(\frac{\pi}{2N}) when N large, \tan^{-1}(\frac{\pi}{2N}) \sim \frac{\pi}{2N} \geq \frac{1}{N}. Hence the A_N \geq \frac{N}{8} blows up to infinity.

Is this pretty cool? This lantern also have an interesting feature that, if we define ‘curvature’ on vertices to be the sum of angles attached to that vertex, (and of course the curvature on the edge between two flat faces shall be 0), then all lanterns have curvature 0 everywhere, just as in the smooth cylinder! i.e. it can be made by folding a single piece of flat paper.

Let’s note that although the triangles are getting uniformly smaller, they do become ‘thinner and thinner’ in the example. In fact this is the only way it can go wrong, i.e. it can be shown that if we further require the triangles to have bounded eccentricity then the area does converge.

Add-on: I actually made the lantern! They are interesting to fold, aesthetically pleasing and even functional! (you’ll see light flaring out in an interesting way)

Trying it out while one thinks about problems is highly amusing and recommended~

All one needs to do is:

Tips on folding:

1. Be sure to make all diagonal lines positive fold and horizontal lines negative.

2. Make the diagonals cross an even number of horizontals or else after you finish all diagonals, you’ll end up with left and right-facing diagonals not crossing on the horizontal (i.e. you’ll need to double the number of horizontals to make it work again)

3. After finishing all lines, it might be hard at first to make the whole thing ‘fold up’. The trick being to make sure all ‘crosses’ are ‘poped-out’ on the whole surface. The final folding process does not work locally!

4. Although theoretically you can take an arbitrarily long strip of paper with unit width to make unit-sized lantern, but in order to not make a million folds and have super-sharp angles between the diagonal and horizontal; I recommend not being too aggressive on the length :-P (square-ish papers are good enough)

Have fun!~

A not-very-good picture of my lantern (larking light bulb…>.<)

Haken manifolds and virtual Haken conjecture

Conan1 Comment

Hi people~ My weekends have been unfortunately filled up with grading undergrad assignments for the last couple of weeks >.< I'll try to catch up on blogging by finding some other time slot during the week.

As a grand-student of Thurston's I feel obligated to end my ignorance regarding Haken manifolds. I guess it's a good idea to start by writing my usual kids-friendly exposition here.

In the rest of the post, M is a compact (so perhaps with boundary), orientable, irreducible (meaning each embedded 2-sphere bounds a ball) 3-manifold.

Definition: A properly embedded oriented surface S \subseteq M is incompressible if S is not the 2-sphere and any simple closed curve on S which bounds an embedded disc in M \backslash S also bounds one in S.

Figure 1

In other words, together with Dehn’s lemma this says the map \varphi: \pi_1(S) \rightarrow \pi_1 (M) induced by the inclusion map is injective.

Note that the surface S could have boundary, for example:

Figure 2

Definition: M is Haken if it contains an incompressible surface.

Okay, at this point you should be asking, what’s good about Haken manifolds? The beauty about it is that, roughly speaking, once you find one incompressible surface in the manifold, you can just keep finding them until the manifold is completely chopped up into balls by incompressible surfaces.

Theorem: (Haken) Any Haken 3-manifold M contains a hierarchy S_0 \subseteq S_1 \subseteq \cdots \subseteq S_n where

1.S_0 is an incompressible surface in M
2.S_i = S_{i-1} \cup S where S is an incompressible surface for the closure of some connected component K of $latex M \backslash S_{i-1}
3.M \backslash S_n is a union of 3-balls

Sketch of proof: This is much simpler than it might appear to be. The point is (at least in my opinion), except for trivial cases as long as a manifold has boundary it must be Haken.

Lemma: If \partial M has a component that’s not \mathbb{S}^2 then M is Haken.

The proof of the lemma is merely that any such M will have infinite H_2 hence by the sphere theorem it will contain an embedded surface with non-trivial homology, if such surface is compressible then we just cut along the boundary of the compressing disc and glue two copies of it. This does not change the homology. Hence at the end we will arrive at a non-trivial incompressible surface.

Figure 3

Now back to proving of the theorem, so we start by setting S_0 to be an incompressible surface given by M being Haken.

Now since M is irreducible, we cut along S_0, i.e. take the closure of each component (may have either one or two components) of M\backslash S_0. Those will have a non-spherical boundary component, hence by lemma containing homologically non-trivial incompressible surface.

This process continuous as long as some pieces has non-spherical boundary components. But since M is irreducible, any sphere bounds a 3-ball in M, hence all components with sphere boundary are 3-balls. (In particular, the case where a component have multiple sphere boundary components cannot occur since the first boundary component bounds a 3-ball hence it can’t have any non-trivial incompressible surfaces on both sides.)

Now the only remaining piece is to show that this process terminates. We apply a standard ‘normal surface argument’ for this. Essentially if we fix a triangulation of M,

A normal surface in M is one that intersects each 3-simplex in a disjoint union of following two shapes:

Figure 4

There can’t be infinitely many non-parallel disjoint normal surfaces in M (in fact there can be no more than 6 times the number of 3-simplexes since each complementry component need to contain at least one non-I-bundle part from one 3-simplex).

Figure 5

However, if the above process do not terminate, we would obtain a sequence of non-parallel non-spherical boundary components:

Figure 6

They represent different homology classes hence can be represented by disjoint normal which results in a contradiction.

In general, this gives a way to prove theorems about Haken manifolds by using inductionL i.e. one may hope to just show the property trivially holds for 3-balls and is invariant under gluing two pieced along an incompressible surface. Note that the gluing surface being incompressible is in fact quite strong hence making the induction step possible in many cases.

For example, by applying an incredible amount of brilliant techniques, Thurston was able to prove his revolutionary result:

Hyperbolization theorem for Haken manifolds: Any Haken manifold M with tori boundary components that does not contain incompressible tori admits a complete hyperbolic structure of finite volume in its interior.

In other words, this is saying that given a Haken manifold, we cut along any incompressible tori, the resulting manifold with tori boundary must have a complete hyperbolic structure with cusps near each boundary component,

This is the best we could hope for since manifolds with incompressible tori would have their fundamental group split over Z^2 which of course imply they can’t be hyperbolic.

Now the more manifolds being Haken means the better this theorem is. Many evidences show that in fact a lot of manifolds are indeed Haken, in perticular we have:

Virtual Haken Conjecture: M is finitely covered by a Haken manifold as long as \pi_1(M) is infinite.

We can see that together with Thurston’s hyperbolization theorem, this would give full solution to the geometrization conjecture for general 3-manifolds.

However, although now Perelman has proved the geometrization conjecture, the virtual Haken conjecture remains open. But in light of Perelman’s result now we are able to try to ‘back-solve’ the puzzle and only prove the virtual Haken conjecture for hyperbolic manifolds.

(to be continued)