User ryan budney - MathOverflow

Answer by Ryan Budney for What is an interpretation of the relation in the cohomology of the pure braid groups?

2013-05-18T20:19:49Z

You know some kind of relation has to hold, since the configuration space of three points in the plane $C_3 \mathbb R^2$ has the homotopy-type of $(S^1 \vee S^1) \times S^1$, so $H^2$ only has rank $2$. The homotopy-equivalence comes from noticing the Faddell-Neuwirth fibration $C_1 (\mathbb R^2 \setminus \{0,1\}) \to C_3 \mathbb R^2 \to C_2 \mathbb R^2$ is trivial. But why a relation of that specific type? You can follow this line of reasoning to its conclusion and derive the relation from a close inspection of this model, say, using cellular cohomology. Because of the action of the symmetric group, there's basically no other relation possible (modulo a small sign issue).

Another way to go about it would be to think of cohomology as dual to homology, but for that you need a compact manifold. So you could compactify the manifold in the Fulton-Macpherson manner. In this model, $\omega_{i,j}$ is dual to the subspace of the compactified configuration space where $j$ is directly above $i$. Cup product corresponds to intersection product, so your relation boils down to saying that the homology class where $m$ is directly over $l$ and $l$ is directly over $k$ (or any cyclic permutation of that) is a boundary. You can write that as a boundary of a class -- the idea is to swing the bottom point of the configuration around to be the top point.

Those are two things that come to mind. I suspect someone like Dev Sinha or Fred Cohen have cuter ways of thinking of this.

Answer by Ryan Budney for Circle bundles over $CP^1$ and self-intersection number of $CP^1$ embeddings

2013-05-15T17:17:56Z

You can find $\mathbb CP^1$ in a wide variety of $4$-manifolds having any Euler class you like. One really simple way is to take the connect-sum of $k$ copies of $\mathbb CP^2$. The idea is to embed $\mathbb CP^1$ in the connect sum so that you are simultaneously breaking the $\mathbb CP^1$ up as a connect sum in all the factors.

If you want a negative Euler class, you use the opposite orientation on $\mathbb CP^2$.

I'm not sure I understand your question about a "deeper relation" with the Hopf bundle. One way to interpret this is that if you remove a point from $\mathbb CP^2$ you get a complex line bundle over $\mathbb CP^1$ whose associated sphere bundle is the Hopf bundle.

Answer by Ryan Budney for Questions about knot (link) of surface in four dimension

2013-05-07T15:02:09Z

Given two disjoint surfaces $\Sigma_1, \Sigma_2$ in $\mathbb R^4$ there are the linking invariants

$$ l_1 : H_1 \Sigma_1 \to \mathbb Z $$

and

$$ l_2 : H_1 \Sigma_2 \to \mathbb Z $$

$l_1$ of a cycle $z \in H_1 \Sigma_1$ is the degree of the map

$$z \times \Sigma_2 \to S^3$$

which associates to a point in $z$ and point in $\Sigma_2$ the unit displacement vectors between them. Similarly for $l_2$.

That's the most direct analogy to the linking number. There's similarly higher-order linking invariants much like in the $3$-dimensional case.

Answer by Ryan Budney for How to visulize surface link in four dimension?

2013-05-03T05:45:21Z

To answer your questions,

1) No individual phenomena characterises knottedness. There won't be a simple answer to this question.

2) Yes, there is a knot theory of surfaces in $\mathbb R^4$. Perhaps start by reading standard references, like Hillman's book on knot theory?

3) There is an Alexander polynomial. But no invariant is known to fully characterize knots -- the fundamental group of the complement is quite strong. The 2nd homotopy group as a module over $\pi_1$ is also fairly useful but sometimes difficult to compute. There are duality pairings and such.

Regarding your comment on the Jones polynomial -- the Jones polynomial for links in $\mathbb R^3$ is not known to characterize knots in any way. It's a fairly strong invariant in terms of knots and links in low-crossing censuses, but many things share the same Jones polynomial.

Another very basic invariant is the Whitney class of the normal bundle of your surface. This is an invariant that takes finitely many values.

Answer by Ryan Budney for What kinds of manifolds admit concave boundary?

2013-04-05T19:18:50Z

I think the answer is yes and in a strong way.

Precisely, let $M$ be a compact manifold, and put any Riemann metric on $\partial M$ Then I claim there is a Riemann metric on $M$, extending the Riemann metric on $\partial M$ with $\partial M$ convex in the sense above -- that the geodesic curvature of boundary curves point out. Moreover, you can do the same if you replace convex with concave.

The idea is fairly simple. Take a collar neighbourhood of $\partial M$ in $M$. The collar neighbourhood is diffeomorphic to $[0,1] \times \partial M$. The idea is to pull-back the metric on $\partial M$ to $[0,1] \times \partial M$, and then linearly re-scale the metric but only in the $\partial M$ direction. If you re-scale so that $\{t\} \times \partial M$ has its Riemann metric $(t+1)$ times the Riemann metric for $\{0\} \times \partial M$, this does the job. You then use a partition of unity to extend the metric to all $M$.

Answer by Ryan Budney for injectivity radius of hyperbolic surface

2013-04-03T06:42:21Z

Buser in 1992 gives a lower bound on the Bers constant for surfaces of genus $g$, and it goes to infinity as $g$ goes to infinity. So this means there's compact hyperbolic surfaces whose injectivity radius is arbitrarily large, but you have to go to high genus to realize them.

P. Buser. Geometry and spectra of compact Riemann surfaces Prog. Math. Vol 106. (1992)

The genus needs to grow like the square of the injectivity radius. Probably a good way to find this surface would be to use Fenchel-Nielsen coordinates, where the pants correspond to points on a planar square lattice and you want to connect the vertices in some (likely non-planar) tri-valent graph configuration, so that it represents pants. Then you want to connect the edges in a way so that there's no short closed loops in the resulting graph. So I'm describing a surface that's less the gluing-together of pants, but more the gluing together of quadratically-many "short shorts".

Answer by Ryan Budney for Understanding four manifolds (more details inside)

2013-03-14T18:35:48Z

It sounds like you want to be able to compute the 2nd homotopy group of a space, and know just enough about bundles to determine when two surfaces have "essentially the same" tubular neighbourhood or not.

So you need a grounding in the 1st and 2nd homotopy groups + covering spaces, as well as homology and cohomology. There are lots of textbooks that cover that material. As mentioned, there's the Hatcher text, and there's also Peter May's text. There's many more, perhaps Bredon's "Geometry and Topology" or Kirk and Davis's book is more suited for your project. Milnor and Stasheff's "Characteristic Classes" would also be helpful.

Answer by Ryan Budney for Probing a manifold with closed curves

2013-03-14T17:43:21Z

The conjecture is false. For example, in a Moebius band, the central circle self-intersects once with many transverse perturbations of itself.

If you want a conjecture like yours to be true, you'll have to assume that a tubular neighbourhood of one of the curves is trivial. That ensures a regular neighbourhood of the union is a punctured torus or Klein bottle.

Answer by Ryan Budney for Symmetry group for the frame bundle of a G-space

2013-03-12T23:54:39Z

The chain rule for manifolds states that if $f : M \to N$ and $g : N \to Q$ are smooth, then the derivative $Df : TM \to TN$ and $Dg : TN \to TQ$ satisfy $$D(g \circ f) = Dg \circ Df$$

So if $g \in G$, let the action of $g$ on $Q$ be denoted $L_g : Q \to Q$. So $L_g \circ L_h = L_{gh}$, therefore $D(L_{gh}) = D(L_g) \circ D(L_h)$. So the action of $G$ on $TQ$ is the map $G \times TQ \to TQ$ given by $(g,v) \longmapsto D(L_g)(v)$.

For Q2, just lift the action on the tangent bundle to the frame bundle, i.e. $(g,(v_1,\cdots,v_n)) \longmapsto (D(L_g)(v_1), \cdots, D(L_g)(v_n))$.

Answer by Ryan Budney for Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)?

2013-03-07T19:31:22Z

The answer is yes for finite groups -- you can even ensure your space is a hyperbolic surface, or hyperbolic 3-manifold. The result for 3-manifolds is Sadayoshi Kojima's. For hyperbolic surfaces I forget who its due to, but I think the reference is in Kojima's paper.

For compact Lie groups I think you can just make your space some variant of the Lie group itself (with a left-invariant metric). The idea would be to take the Lie group $G$ with its left-invariant metric. If that has a bigger isometry group than $G$ itself, take the product of $G$ with a ball (with some metric), and perturb the metric on $G \times B$, $G$-equivariantly. Some generic perturbation of the metric should kill all isometries other than the ones coming from the $G$-action.

Andre raises the question of whether or not this construction could be performed for $S^1 \equiv SO_2$. The issue being that with its left-invariant metric the isometry group is $O_2$. I don't see any reason why it shouldn't work. For example, take $S^1 \times D^2$. Put a metric on this space which is locally a product, but where the fibre is a disc whose isometry group is $\mathbb Z_3$, orientation-preserving isometries of a triangle. We make the holonomy of the bundle around the base circle the generator of this isometry group. So there can be no isometry of this bundle that reverses the direction of the base space, since that would mean the holonomy is equal to its inverse, which in $\mathbb Z_3$ can not happen. So the only symmetries of this bundle act as orientation-preserving isometries of the base, and orientation-preserving on the fibre, and this group is $SO_2$.

Answer by Ryan Budney for Differences between various categories of surface embeddings in 4-space

2013-03-07T19:18:37Z

Regarding (1), yes there's a unique (up to smooth isotopy) smoothing of a PL locally-flat embedding. You can similarly ensure that if your original surfaces are close in some PL compact-open sense, that their smoothings can be made to be close in the smooth C^1 compact-open sense.

Regarding (2), yes, there are knots that are topologically locally-flat slice, yet they are not smoothly slice. A knot with a trivial Alexander polynomial does the job. These examples aren't as "concrete" as your cone example as they depend on theorems of Freedman which involve infinite constructions.

Answer by Ryan Budney for Characterizing Hessians among symmetric bilinear tensors

2013-03-05T19:58:44Z

Given $f : N \to \mathbb R$, there is its derivative $Df : TN \to \mathbb R$, which has its dual gradient $\nabla f: N \to TN$, which you can covariantly differentiate $c \circ D\nabla f : TN \to TN$, and this map is dual to the Hessian $Hf : TN \oplus TN \to \mathbb R$. From this perspective there's two things to determine. Here $c : T^2 N \to TN$ is the connection, in the Ehresmann formalism.

1) Your Hessian is adjoint to a bundle map $TN \to TN$, is this the covariant derivative of a vector field on $N$?

2) If the answer to (1) is yes, among the solution vector fields from (1) is there a vector field which is the gradient of a real-valued function on $N$?

(2) Has the traditional cohomological answer so I'll focus on (1).

A bundle map $TN \to TN$ is the covariant derivative of a vector field $N \to TN$ means that you could write the vector field as a type of holonomy integral, by adding to your parallel transport an integral of the bundle map $TN \to TN$. So there will be a local triviality condition on this holonomy integral, as well as a global triviality condition.

Answer by Ryan Budney for The role of ANR in modern topology

2013-03-04T18:16:38Z

I think the answer has more to do with the psychology of mathematicans as a culture than with actual mathematical facts.

I was not alive during the period where ANRs were mentioned in the topology literature but I've read quite a few early topology papers and also noticed before the 60's people couldn't seem to not mention them, and afterwards they were almost never mentioned.

I think this is mostly due to the more formal side of algebraic topology, with model categories. With the terminology cofibration one could largely avoid talking about ANRs and regular neighborhoods. You of course could continue to talk about those things but if you're attempting to write something short and concise with as few confusing side-roads as possible, you would omit it.

So fairly quickly people realized they didn't need to talk about ANRs. I think this kind of thing happens fairly often in mathematics, especially when the definition of a concept maybe slightly misses the mark of what you're aiming for, or if it isn't quite as general as you really need. Terminology like this cycles in and out of mathematics fairly frequently.

You could frame this in terms of the long-term survivability of a mathematical concept -- math verbiage evolution. The flaw in ANRs is they did not anticipate that point-set foundations would become less of a focus of topology, that the field would move on and become more scaleable.

Answer by Ryan Budney for Vassilliev invariants of knots and their cables

2013-02-27T22:14:27Z

Let $K$ be a knot, and $\Delta_K$ be the Alexander polynomial of $K$, $\Delta_K \in \mathbb Z[t^\pm]$.

Let's let $K(p,q)$ be the $(p,q)$-cable of $K$. Then

$$ \Delta_{K(p,q)} = \Delta_K(t^{p}) \cdot \Delta_{T_{p,q}}$$

where $\Delta_{T_{p,q}}$ is the Alexander polynomial of the $(p,q)$-torus knot. I believe that's

$$ \Delta_{T_{p,q}} = \frac{ (t^{pq}-1)(t-1) }{(t^p-1)(t^q-1)} $$

The above formulas are fairly classical. It appears at least as early as in Eisenbud and Neumann's book, but it's likely known much earlier.

The type-2 invariant of a knot is given in terms of the Alexander polynomial. In the Conway form it's the coefficient of $z^2$, but in the above Alexander normalization, you'll get it as some kind of linear combination of the first few coefficients. So you just apply whatever that formula is. At present I forget it!

Answer by Ryan Budney for Repertory of the different sorts of operads

2013-02-26T19:56:17Z

One operad that's popped up in my work is the kind of operad that has the same formal properties as the endomorphism operad of a space $X$, provided $X$ is equipped with an action of a topological group $G$. Level $n$ of the endomorphism operad is just the space of maps $Map(X^n, X)$, but we think of this space as one with an action of the group

$$ G \times (\Sigma_n \ltimes G^n) \equiv G \times (\Sigma_n \wr G). $$

I sometimes denote this group $\Sigma^*_n \wr G$, and the family of groups $\Sigma^* \wr G$. I've been calling these operads $\Sigma^* \wr G$-operads as I thought the name was kind of harmless and more or less descriptive.

The free objects over this operad look like a disjoint union of rooted trees, where the vertices of the trees are decorated by points in the generating $\Sigma^*_k \wr G$-spaces. There is an equivalence relation on these decorated trees, generated by one relation for each edge of the trees -- corresponding to the endomorphism operad's equivariance.

These types of operads come up (and are useful) with $G$ being various orthogonal groups, or diffeomorphism groups of balls (or other automorphism groups of balls) for certain embedding spaces, like spaces of knots. For classical knots, like knots in the 3-sphere, there is an almost free operad of this type that describes the homotopy-type of the space of knots completely, up to the computation of certain (finite) symmetry groups of some hyperbolic links in $S^3$. The operads I'm talking about I call splicing operads. These are subspaces of $Map(X^n,X)$ but in this case, $X$ is a ball, and the maps are smooth maps with various restrictions on them (to get knots of various types).

Answer by Ryan Budney for What manifolds are boundaries of euclidian spaces ?

2013-02-21T19:33:34Z

$N$ has to be a homotopy-sphere. So as long as it's dimension isn't $4$, there's a proof that it has to be the standard $S^{n-1}$.

These arguments appear in the Kosinski book on smooth manifolds. The basic idea goes like this.

1) $N$ is simply connected. There's the inclusion map $N \to M$, but if $p \in int(M)$ then there's also a retraction map $M \setminus \{p\} \to N$. Provided $n \geq 3$ removal of a point does not affect the fundamental group.

2) $N$ is a homology sphere by Alexander/Poincare duality of the pair $(M,N)$. Part (1) technically gives us this as well but this argument works even if $M$ is not contractible.

So by the Whitehead theorem, $N$ is a homotopy sphere. Moreover, $M$ is a contractible manifold whose boundary is a homotopy sphere. So $M$ is a disc by the h-cobordism theorem, and the disc has a unique smooth structure (not known in dimension $5$ still).

So that resolves all cases except $dim(N)=4$.

Answer by Ryan Budney for How to specify a finite group up to inner automorphism?

2013-02-15T01:15:27Z

Wouldn't the 2-skeleton of a finite CW-complex satisfy both your first and second conditions? You could choose a finite 2-dimensional simplicial complex if you wanted everything to be finitary, or you could add some restrictions to the attaching maps of the CW-complex, that they're PL.

From a finite simplicial complex you have the fundamental groupoid. Similarly, given a finitely-presentable groupoid, its presentation complex can be taken to be a finite 2-dimensional CW-complex.

Answer by Ryan Budney for Interpretation of Riemann tensor antisymmetry

2013-02-10T11:12:46Z

This is a re-flavouring of Alexander's answer but in a language I prefer.

Take two vectors $v,w \in T_p N$, and consider the `rectangle' $exp(xv+yw)$ where $0 \leq x \leq a$ and $0 \leq y \leq b$. The holonomy around the boundary of this rectangle is an orthogonal transformation of $T_p N$, and it looks approximately like

$$ Hol \simeq Id_{T_p N} + ab R(v,w) $$

where the approximation indicates the 2nd order taylor expansion of the holonomy with respect to the variables $a$ and $b$. Here $R(v,w)$ is the Riemann curvature tensor.

So from this point of view, the reason why it's anti-symmetric in the variables $v,w$ is that if you switch $v$ and $w$ you are essentially reversing the orientation of the rectangle you're computing the holonomy over, so the corresponding linear transformation (holonomy) is the inverse, which corresponds to negating the linear and quadratic parts of the Taylor expansion.

This is also the reason why the Riemann curvature tensor is skew-symmetric:

$$\langle R(v,w)z, y \rangle + \langle z, R(v,w)y \rangle = 0 $$

since the tangent space to the orthogonal group is consists of all the skew-symmetric linear transformations.

Has anyone thought about creating a formal proof wiki with verifier?

2010-10-05T22:06:26Z

Mathematics has undergone some rather nice developments recently with the adoption of new techologies, things like on-line journals, the arXiv, this website, etc. I imagine there must be many further developments that could be quite useful.

What I'm thinking of is a webpage where anyone can contribute formal proofs of theorems. In particular there would be many proofs of the same theorem provided the proof is different -- like a constructive proof of Brouwer's fixed point theorem, and non-constructive proof, etc.

The idea would be to build up a large web of formal proofs, one building on another so that one could eventually do searches through this space of formal proofs to find out what the most efficient proofs are, in the sense of how many ASCII characters it would take to write-up the proof using Zermelo-Frankel set theory. One hope would be to have a big, active database of verified formal proofs. Another would be to have a webpage where you could hope to discover whether or not there are simpler proofs of theorems you know, that you may have not been be aware of.

Being a web-page there would be certain useful efficiencies -- the webpage could "compile" your proof and check to see it's valid. Being a wiki would make it relatively easy for people to contribute and build on an existing infrastructure. And you'd be free to use pre-existing proofs (provided they've been verified as valid) in any subsequent proofs. One could readily check what axioms a proof needs -- for example to what extent a proof needs the axiom of choice, and so on.

Is there any efforts towards such a development? Such a tool would hopefully function like the publishing arm of some sort of modern internet-era Bourbaki.

Answer by Ryan Budney for Seifert surfaces via Alexander duality

2013-01-10T20:17:31Z

This is an old argument that essentially predates much modern knot theory, and goes back to Serre.

The basic idea goes like this: let $C$ be the complement of a co-dimension two knot in $S^n$. Apply Poincare/Alexander duality to deduce that $C$ is a homology $S^1 \times D^{n-1}$. So $H^1 C \simeq \mathbb Z$, and $H^1 \partial C$ is either $\mathbb Z^2$ or $\mathbb Z$ according to whether or not $n=3$ or $n>3$. In either case the restriction map is an injection.

Serre's theorem that $H^1 X = [X,S^1]$ gives you a map $C \to S^1$ which you make transverse to a point. Because the boundary is a product $\partial C \simeq S^1 \times S^{n-2}$, you can ensure the map $C \to S^1$ is not just homotopic to but actually projection onto the first factor (when restricted to $\partial C$).

This ensures the preimage of a regular value of $C \to S^1$ is a Seifert surface for the knot. By Seifert-surface I mean an orientable co-dimension one submanifold of $S^n$ whose boundary is the knot. If the surface is disconnected, you throw away any components that do not touch $\partial C$. Because $\partial C \to S^1$ is the projection onto the factor, only one path-component touches $\partial C$.

Serre and Thom used these ideas repeatedly in their early attacks on the Steenrod realization problem. This was the version where you're trying to realize the homology classes by embedded submanifolds.

The above is a relative version of their arguments. There's an old Springer Lecture Notes in Mathematics that outlines the history of this argument and when it was first brought to knot theory. Right, SLN in Math 685. If it's not in the essay by Cameron Gordon, it's in the other big essay, maybe it was Milgram?

Answer by Ryan Budney for Smooth maps transverse to a foliation

2012-12-13T18:42:19Z

I originally misunderstood your question. Here is a fairly simple example.

Let $N = \mathbb R^2$, $M=S^2$ and $S$ be the unit circle in $\mathbb R^2$ considered as foliated by its points -- the leaves are $0$-dimensional.

The map $M \to N$ will be a linear projection map. Let's project $S^2$ onto $\mathbb R^2$ in a way so that the image of $S^2$ is a disc in the plane, which half covers the unit circle.

So this map is transverse to the foliation at a point. But no nearby map is transverse to the foliation on all $M$. This is because the image of $M$ is a compact subset of the plane which does not contain the unit circle. So for a sufficiently small perturbation, you similarly do not cover the unit circle. Wherever the boundary of the image intersects the circle, you will not be transverse to the foliation of the circle.

Answer by Ryan Budney for Examples of sphere bundles

2012-12-10T20:03:53Z

As far as I know the only explicitly-described such bundles are in Hatcher's paper:

Hatcher. Concordance spaces, higher simple-homotopy theory, and applications. Algebraic and geometric topology (Proc. Sympos. Pure Math., Stanford Univ, Stanford Calif 1976), Part 1, pp. 321.

I think in Igusa's Higher Franz Reidemeister Torsion book there might also be these examples, although I don't have the book at home with me so I can't check. But that seems likely as Igusa has also developed these examples.

Persistent homology of Gaussian Fields in Euclidean space

2012-12-04T20:40:12Z

If you generate points in $\mathbb R^n$ via a process that respects a Gaussian normal distribution, then compute the persistent homology / barcodes, to my eye something fairly regular seems to be happening, with the barcodes tending towards something like a "wing" shape, fat in lower dimensions, thinning out towards dimension $n$.

Has anyone proven any theorems that describe the asymptotic "shape" of the barcodes?

Ideally I'd like a test so that I can look at some barcodes and say "that's typical of a Gaussian normal distribution".

The closest thing I've been able to find is experiments and results on the expected Euler characteristic of the persistent homology, in the references below.

http://webee.technion.ac.il/people/adler/larry.pdf

http://arxiv.org/abs/1003.5175

edit:

I did a very rough computation to try and get some kind of guess as to what the distribution of barcodes should look like. So I made a very coarse estimate based on a distribution of points that is roughly `locally cubical' and approximately respecting a normal distribution.

The density is given by:

$$\mu = N e^{-r^2}$$

where $r$ is the distance from the origin. Then if $\epsilon$ is the parameter for persistent homology, it appears that $H_0$ is rank approximately

$$N \int_{\sqrt{\ln(N\epsilon^{1/n})}}^\infty r^{n-1}e^{-r^2} dr$$

and $H_k$ for $k \in \{1,2,\cdots,n-1\}$ has rank approximately

$$ {n \choose k+1}\frac{(\sqrt{\ln(N\epsilon^{1/n}/\sqrt{k}))}^{n-2}}{4\sqrt{k}\epsilon^{1/n}} $$

These are fairly coarse estimates, and in no way rigorous. But if something like this is actually true it seems to be saying that for $N$ large and $n \geq 3$, the $H_0$ betti number tends to some asymptote (dependent on $\epsilon$), and $H_1, \cdots, H_{n-1}$ are non-trivial but small. So most of the points in the distribution are in a giant homology `black hole' at the centre and persitent homology sees the thin crust around the outside.

I'd be curious if people have done other similar guestimates (or better) and if they had similar-looking results.

Some mid-sized ¿hyperbolic? manifolds and SnapPea

2012-11-14T23:47:19Z

I've recently come across some mid-sized 3-manifolds that I think are likely hyperbolic, but SnapPea has some trouble with them. This is related to my previous question

http://mathoverflow.net/questions/4918/can-you-fool-snappea

but in this case I'm dealing with closed, orientable 3-manifolds instead of knot and link complements.

The 3-manifolds I've come across have 11, 12 and 13 tetrahedra in their triangulations. One of them SnapPea finds a solution to the gluing equations but it has "negatively oriented tetrahedra". Does this mean what I think it means -- that once you've put the geometric structure on the tetrahedra, you have a tetrahedron folded-over? If you view the gluing equations from the upper half-space model, they say that the sum of a bunch of angles should be $2\pi$. Is this the case where one of those angles is negative?

The other two triangulations SnapPea finds geometric structures with degenerate tetrahedra. In the upper half-space model this is where one of the angles is zero, I believe. Is there a way to fix this, so that I could get a Dirichlet domain, drill and fill, etc? This is my primary question.

Here are the triangulations, both in SnapPea format and Regina format.

Regina file, all triangulations

11-tet, SnapPea

12-tet, SnapPea

13-tet, SnapPea

With the 12-tetrahedron example, SnapPea can generate a Dirichlet Domain. This one has what looks almost like bevelled-edges. Is there a way to find a better basepoint to grow the Dirichlet domain from?

edit: After working a bit with Nathan's answer, I've identified these three manifolds and got a little closer to understanding how to work SnapPea to maximal advantage.

tri11, as Nathan mentioned, is hyperbolic. It has a fairly pretty Dirichlet domain.

Another common name for this manifold would be the 0-surgery on the 2-component link $7a_6$ (in the Thistlethwaite table). Similarly tri12 can be identified as Nathan says.

After playing around with Regina a bit I found an incompressible torus in tri13 that splits tri13 into the union of an orientable $I$-bundle over the Klein bottle and a figure-8 complement. So this answers the core of my question.

Answer by Ryan Budney for twisted bundle definition

2012-11-03T22:45:05Z

$S^2$ bundles over $S^2$ (smooth, PL, topological) are in bijective correspondence with homotopy-classes of maps:

$$ S^2 \to BSO_3 $$

which up to homotopy is

$$\pi_2 BSO_3 \simeq \pi_1 SO_3 \simeq \mathbb Z_2 $$

So there's precisely two non-isomorphic $S^2$-bundles over $S^2$. You can view the non-trivial $S^2$-bundle as the fibrewise one-point compactification of the vector bundle over $S^2$ whose Euler class is $1$. That vector bundle has fairly standard constructions.

Answer by Ryan Budney for A "dual" universal coefficient theorem

2012-11-02T01:51:29Z

Yes, there is such a universal coefficient theorem.

$$0 \to Ext(H^{q+1}(X,R), G) \to H_q(X, G) \to Hom(H^q(X, R), G) \to 0$$

see Theorem 6.5.12 in Spanier's textbook "Algebraic Topology". It's on page 248.

Answer by Ryan Budney for How Many 4-Manifolds are Symplectic?

2012-10-30T02:54:42Z

In an attempt to get a sense for the answer to Dmitri's 1st question I checked the euler characteristics of the 4-manifolds in the census of triangulated closed 4-manifolds containing 6 or less 4-dimensional simplices. I like to call a 4-dimensional simplex a pentachoron.

Non-orientable 2-pentachoron triangulations: 2. All have Euler characteristic 0.

Orientable 2-pentachoron triangulations: 9. Three have euler characteristic equal to 0. 6 have Euler char equal to 2.

Non-orientable 4-pentachoron triangulations: 184. 119 have euler char 0, 65 have euler char 1.

Orientable 4-pentachoron triangulations: 785. 131 have euler char 0. 3 have euler char 1. 647 have euler char 2. 4 have euler char 3.

Non-orientable 6-pentachoron triangulations: 60229. 28831 have euler char 0. 20 have euler char -1. 30824 have euler char 1. 554 have euler char 2.

Orientable 6-pentachoron triangulations: 440496. 29294 have euler char 0. 1477 have euler char 1. 405282 have euler char 2. 4423 have euler char 3. 20 have euler char 4. There are no negative euler characteristics.

So from this point of view, positive euler characteristics tend to dominate, at least for small numbers of pentachora.

Answer by Ryan Budney for (Non)-exoticness of a diffeomorphism of a sphere

2012-10-29T19:25:29Z

What definition of torus are you using Marco? The usual definition is $T^k = (S^1)^k$ but from the the way you've structured things above it looks like you're using the convention $T^k = S^1 \times S^{k-1}$. Either way, the homotopy-groups of these diffeomorphism groups are generally not trivial as they generally contain plenty of torsion. $Diff(S^1 \times S^{k-1})$ contains $SO_2 \times SO_k \times \Omega SO_k$ for example.

Question 1 is a standard pseudo-isotopy type question. Have you looked up the literature on pseudo-isotopy diffeomorphisms of $S^1 \times S^{n-2}$ ? For example, there's a recent arXiv paper of Crowley and Schick which states that there's elements of $Diff(S^n)$ with large Gromoll Degree, meaning they can be put into positions like in your questions 1 and 2, yet they're non-trivial diffeomorphisms of the sphere.

http://arxiv.org/abs/1204.6474

edit: For some basic results on the homotopy-type of $Diff((S^1)^n)$ see:

Hatcher, A. E. Concordance spaces, higher simple-homotopy theory, and applications. Algebraic and geometric topology (Proc. Sympos. Pure Math., Stanford Univ., Stanford, Calif., 1976), Part 1, pp. 3--21, Proc. Sympos. Pure Math., XXXII, Amer. Math. Soc., Providence, R.I., 1978. (Reviewer: Gerald A. Anderson) 57R52

Answer by Ryan Budney for Fundamental groups and homology groups of closed subsets of the plane

2012-10-27T04:08:09Z

Eda, K. Fundamental group of subsets of the plane. Topology and its Applications Volume 84, Issues 1-3, 24 April 1998, Pages 283-306

This is more or less a duplicate of the math.se thread: http://math.stackexchange.com/questions/36279/the-fundamental-group-of-every-subset-of-mathbbr2-is-torsion-free

Answer by Ryan Budney for Minimal number of cells of a CW complex (up to homotopy)

2012-10-24T23:33:49Z

To expand on my comment, there's a very general tool to manipulate CW-complexes, due to Whitehead. It tells you when you can in effect remove a cell from a CW-decomposition via `elementary moves', usually called Whitehead Moves. In smooth manifold theory there are parallel constructions -- people talk about "handle slides" and "handle cancellations". This comes up in the proof of the h and s-cobordism theorems, which are the smooth-category analogue of the Whitehead moves. Technically these moves have to do with a slightly more refined notion of homotopy-equivalence, called simple homotopy equivalence. Provided the fundamental group of the CW-complex is trivial, simple homotopy-equivalences are in effect the same as homotopy-equivalances, but in general they're a little more fussy.

What are the Whitehead moves? On the 0-skeleton, it's the move where you collapse a maximal forest in the 1-skeleton. On the 1-skeleton these are moves where you can cancel a 1-cell using a 2-cell that's incident to it only once. This is explained in detail in Marshall Cohen's "A course in simple-homotopy theory". GTM 10 Springer-Verlag.

Comment by Ryan Budney

2013-05-22T13:41:28Z

@Wlodzimierz: It's unclear what you are referring to, both with your "not exactly" comment and the "number of open maps of an atlas" comment.

Comment by Ryan Budney

2013-05-22T13:13:05Z

The package "Singular" is also analogous to GAP, but for rings.

Comment by Ryan Budney

2013-05-21T19:54:28Z

Yes, that's exactly how it works. The triangulation need not be "suitable", any triangulation works. You have to use signs if the triangulation is not compatible with the manifold's orientation but that's the only issue.

Comment by Ryan Budney

2013-05-18T20:28:54Z

Your question is very close in spirit to Abrams and Ghrist's work on discretized configuration spaces. It's not quite the same but it's close.

Comment by Ryan Budney

2013-05-18T08:24:53Z

Ah, right. I got really off-track there.

Comment by Ryan Budney

2013-05-17T22:00:05Z

But the projective general linear group is always a Lie group acting on the sphere, and it always contains the conformal group as a proper subgroup. How can the conformal group be maximal? By projective general linear group I mean $GL_n \mathbb R$ acting on the rays out of the origin in $\mathbb R^n$.

Comment by Ryan Budney

2013-05-17T21:07:58Z

What kind of properties do you want your "vector fields" to have? There can't be anything close to a literal vector field, since the tangent (micro)bundle does not have the structure of a vector space. But for various purposes there are approximations to the idea.

Comment by Ryan Budney

2013-05-16T10:20:53Z

At this point I think "big-list" would be appropriate.

Comment by Ryan Budney

2013-05-15T21:04:29Z

Such a DE is not exact -- it does not satisfy Clairault's theorem. So there's no solution.

Comment by Ryan Budney

2013-05-14T11:42:54Z

But pooper is at least living up to his or her name!

Comment by Ryan Budney

2013-05-14T11:39:13Z

Your question seems suitable for a blog. Since Terry Tao already has a blog discussion on this topic, I've voted to close.

Comment by Ryan Budney

2013-05-09T19:08:28Z

Why do you prefer that over the suspension of the Hopf fibration?

Comment by Ryan Budney

2013-05-09T09:03:13Z

To whomever is downvoting, it's probably better to either just vote to close or to flag as spam or for moderator attention.

Comment by Ryan Budney

2013-05-07T15:45:50Z

Thanks! . . . . .

Comment by Ryan Budney

2013-05-06T13:45:52Z

I think there's statements to this effect in Kosinski's Differential Topology text, have you looked? The idea is to cancel handles of low and high dimensions, leaving only the middle-dimensional handles.