User dylan thurston - MathOverflow

Intrinsic characterization of Soergel bimodules?

2012-03-26T02:34:05Z

A Soergel bimodule (for $S_n$) is a bimodule over $R = \mathbb{Q}[x_1,\dots,x_n]$ which appears as a summand/grading shift of tensor products of the basic bimodules $$B_{i,i+1} = R \otimes_{i,i+1} R$$ where $\otimes_{i,i+1}$ means the tensor product over the subring of polynomials invariant under permuting $i$ and $i+1$. It follows immediately that every Soergel bimodule $M$ has the following properties:

(1) $M$ is free as a left module or as a right module, although not necessarily as a bimodule.

(2) $M$ commutes with invariant polynomials, in the sense that for every invariant polynomial $p \in \mathbb{Q}[x_1,\dots,x_n]$ and $m \in M$, we have $$ pm = mp. $$

I think they also have the following property:

(3) There is an invariant vector, an element $m_0 \in M$ so that $$ x_i m_0 = m_0 x_i $$ for every $i=1,\dots,n$.

Do these properties characterize Soergel bimodules? Without the third condition, you could have, for instance, a bimodule that just permuted the $x_i$: a one-dimensional module with a single generator $a$ as a right module, so that $ x_i a = a x_{\sigma(i)} $ for some permutation $\sigma$.

Edit: The natural generalization for a general Weyl group $W$ would be to replace the invariant polynomials in (2) by the polynomials that are invariant under $W$. Clearly all Soergel bimodules would still satisfy this generalization of (2).

Any references are welcome. If it's not known, I'll try to prove it.

Edit: Ben Webster gave a counterexample below. More generally, I'm still interested in some sort of intrinsic, elementary characterization.

Answer by Dylan Thurston for Does this knot invariant distinguish trefoil chiralities?

2012-04-30T18:13:35Z

I'm very curious where this came up. In any case, the answer to the first question is yes, it does distinguish these trefoils; you found the minimal representatives.

Let $a_0,\dots,a_{N-1}$ be the roots of unity that are visited along the knot, in (cyclic) order. Suppose we have a minimal representative for some non-trivial knot. Then we cannot have $|a_k - a_{k+1}| = 1$ for any $k$, as otherwise we could replace this pair $a_k, a_{k+1}$ by a single root of unity (for $N-1$), adjusting the other roots of unity as appropriate. A little more subtly, we cannot have $|a_{k-1} - a_{k+1}| = 1$ either, as then we could again delete $a_k$ from the sequence to get a smaller representation. With these simple constraints, the smallest possible sequence for a non-trivial knot is the one you found for one of the trefoils with $N=7$. There are several possibilities for $N=8$, including the one you found for the other trefoil. I've included a very short Haskell program below that computes this. The possibilities for $N=8$ are $$ (2,7,5,3,1,6,4,0)\quad (2,5,7,3,1,6,4,0)\quad (3,6,1,4,7,2,5,0)\quad (2,6,4,1,7,3,5,0) $$ $$ (3,1,6,4,2,7,5,0)\quad (2,4,6,1,3,7,5,0)\quad (3,5,1,7,4,2,6,0)\quad (4,2,7,5,1,3,6,0) $$ $$ (3,1,5,7,2,4,6,0)\quad (5,3,1,6,4,2,7,0)\quad (2,4,6,1,3,5,7,0) $$

For the second question, I have never heard of this representation before.

Here is the code, for anyone interested.

 -- A (partial) circular stick representation is a list of integers,
 -- the order of the roots of unity to visit in order
 type CircStick = [Int]

 -- The next element ak after a partial representation a1, ..., a{k-1} 
 -- must satisfy
 --   (a) ak has not already been seen
 --   (b) |ak - a{k-1}| > 1
 --   (c) |ak - a{k-2}| > 1
 -- There are a few more "easy" constraint, eg the first and last entries
 -- cannot differ by one.  We do not impose those constraint here.
 nexts :: Int -> CircStick -> [Int]
 nexts n [] = [0]
 nexts n [a1] = filter (\a -> abs (a-a1) > 1) [0..n-1]
 nexts n (a1:a2:as) =
   filter (\a -> not (elem a as)) $
   filter (\a -> abs (a-a1) > 1) $
   filter (\a -> abs (a-a2) > 1) $
     [1..n-1]

 completions :: Int -> CircStick -> [CircStick]
 completions n as | length as >= n = [as]
 completions n as =
   concat [completions n (a:as) | a <- nexts n as]

 -- Impose final constraints:
 --    (a) Last entry cannot be 1
 --    (b) Take entry that is lexicographically less than its reverse
 --    (c) first and next-to-last entries cannot differ by one
 circSticks :: Int -> [CircStick]
 circSticks n = 
   filter (\as -> abs ((as!!0) - (as!!(n-2))) > 1) $
   filter (\as -> as < tail (reverse as)) $
   filter (\as -> head as /= 1) $
   (completions n [])

Edit: For those interested, here are the 108 possibilities for $N=9$. I hope there's some way of checking what these are more efficiently than just going through them by hand.

[[2,7,5,3,8,1,6,4,0],[2,7,5,3,1,8,6,4,0],[2,5,7,3,1,8,6,4,0],[2,7,5,1,3,8,6,4,0],[2,5,7,1,3,8,6,4,0],[2,6,8,3,5,1,7,4,0],[2,7,5,3,1,6,8,4,0],[2,5,7,3,1,6,8,4,0],[2,7,5,1,3,6,8,4,0],[2,5,7,1,3,6,8,4,0],[3,8,6,1,4,7,2,5,0],[3,6,8,1,4,7,2,5,0],[3,7,1,4,6,8,2,5,0],[2,8,6,4,1,7,3,5,0],[2,6,8,4,1,7,3,5,0],[2,7,4,1,6,8,3,5,0],[2,4,8,6,3,1,7,5,0],[2,4,6,8,3,1,7,5,0],[3,8,1,6,4,2,7,5,0],[3,1,8,6,4,2,7,5,0],[3,6,1,8,4,2,7,5,0],[3,1,6,8,4,2,7,5,0],[2,8,4,6,1,3,7,5,0],[2,4,8,6,1,3,7,5,0],[2,6,4,8,1,3,7,5,0],[2,4,6,8,1,3,7,5,0],[3,7,1,4,6,2,8,5,0],[2,7,4,1,6,3,8,5,0],[3,8,5,1,7,4,2,6,0],[3,7,5,1,8,4,2,6,0],[3,5,7,1,4,8,2,6,0],[4,7,1,3,5,8,2,6,0],[4,1,7,3,5,8,2,6,0],[4,8,2,5,7,1,3,6,0],[4,7,2,5,8,1,3,6,0],[4,2,7,5,1,8,3,6,0],[2,4,7,1,5,8,3,6,0],[2,8,5,3,7,1,4,6,0],[2,7,5,3,8,1,4,6,0],[3,8,1,5,7,2,4,6,0],[3,7,1,5,8,2,4,6,0],[2,7,5,3,1,8,4,6,0],[2,5,7,3,1,8,4,6,0],[3,1,7,5,2,8,4,6,0],[4,2,7,5,3,1,8,6,0],[3,5,1,7,4,2,8,6,0],[4,2,7,5,1,3,8,6,0],[2,4,7,1,5,3,8,6,0],[3,1,5,7,2,4,8,6,0],[5,8,3,1,6,4,2,7,0],[5,3,8,1,6,4,2,7,0],[3,5,8,1,6,4,2,7,0],[5,3,1,8,6,4,2,7,0],[3,5,1,8,6,4,2,7,0],[5,1,3,8,6,4,2,7,0],[5,3,1,6,8,4,2,7,0],[5,1,3,6,8,4,2,7,0],[4,6,1,3,8,5,2,7,0],[4,1,6,3,8,5,2,7,0],[4,6,2,8,5,1,3,7,0],[5,8,2,4,6,1,3,7,0],[5,2,8,4,6,1,3,7,0],[2,6,4,8,1,5,3,7,0],[4,2,6,8,1,5,3,7,0],[2,4,6,8,1,5,3,7,0],[2,6,4,1,8,5,3,7,0],[2,4,6,1,8,5,3,7,0],[2,5,8,3,6,1,4,7,0],[5,3,1,8,6,2,4,7,0],[3,5,1,8,6,2,4,7,0],[5,1,3,8,6,2,4,7,0],[5,3,1,6,8,2,4,7,0],[5,1,3,6,8,2,4,7,0],[4,2,8,6,3,1,5,7,0],[2,4,8,6,3,1,5,7,0],[4,2,6,8,3,1,5,7,0],[2,4,6,8,3,1,5,7,0],[3,6,1,4,8,2,5,7,0],[3,1,6,4,8,2,5,7,0],[2,8,4,6,1,3,5,7,0],[4,2,8,6,1,3,5,7,0],[2,4,8,6,1,3,5,7,0],[2,6,4,8,1,3,5,7,0],[4,2,6,8,1,3,5,7,0],[2,4,6,8,1,3,5,7,0],[2,6,4,1,8,3,5,7,0],[2,4,6,1,8,3,5,7,0],[5,7,3,1,6,4,2,8,0],[4,6,1,3,7,5,2,8,0],[5,3,7,1,4,6,2,8,0],[3,5,7,1,4,6,2,8,0],[6,4,2,7,5,1,3,8,0],[5,7,2,4,6,1,3,8,0],[6,2,4,7,1,5,3,8,0],[2,6,4,1,7,5,3,8,0],[5,2,7,4,1,6,3,8,0],[6,3,1,5,7,2,4,8,0],[6,1,3,5,7,2,4,8,0],[2,7,5,3,1,6,4,8,0],[2,5,7,3,1,6,4,8,0],[3,6,1,4,7,2,5,8,0],[6,2,4,7,1,3,5,8,0],[2,6,4,1,7,3,5,8,0],[4,2,7,5,3,1,6,8,0],[5,3,1,7,4,2,6,8,0],[3,5,1,7,4,2,6,8,0],[4,2,7,5,1,3,6,8,0],[3,1,5,7,2,4,6,8,0]]

Answer by Dylan Thurston for Wanted: differential coming from higher genus surface in Heegaard Floer Homology

2012-03-29T13:16:06Z

Yes, $S \to \Sigma$ can fail to be an immersion. The failure is a branch point of the map $S \to \Sigma$, since it is (close to) a holomorphic map. This is fairly common as soon as the multiplicity of some region in $\Sigma$ gets higher than $1$. Of course, in the simple examples you can actually compute, this tends not to happen.
A "slit" along the $\alpha$ or $\beta$ curves looks like a perfectly ordinary piece of the boundary upstairs in $S$, including at the end of the slit. It's best to think about the end of the slit as a boundary branch point, looking locally like the map $z \to z^2$ restricted to the upper half-plane in the domain. Slits cannot collide, due to boundary monotonicity: above each point on $\partial \mathbb{D}^2$, the $g$ different points map to distinct $\alpha$-curves. I don't know what you mean by "independent" degenerations, sorry....
Yes, higher genus images can happen, and it's not too hard to construct examples that are forced by gluing, although again in most cases where you're able to compute Heegaard Floer homology by directly counting curves it does not. The index formula in Corollary 4.3 of Lipshitz's paper "A cylindrical reformulation of Heegaard Floer homology" will let you easily construct examples of high genus surface with index 1, and a little more playing around should let you see that some of these must actually have representatives.

Answer by Dylan Thurston for Maslov index and heegard floer homology

2012-03-29T12:53:59Z

Robert Lipshitz has the nicest formula, described in Corollary 4.3 of this paper:

Robert Lipshitz, "A cylindrical reformulation of Heegaard Floer homology", Geometry & Topology 10 (2006) 955–1096, DOI: 10.2140/gt.2006.10.955, arXiv: math.SG/0502404

At this stage, I would not worry so much about the proof of the formula, although it's also not so bad.

Answer by Dylan Thurston for The growthrate of the homology of $H_*(M^{\otimes_A n})$ for a DG-bimodule $M$

2012-03-26T02:13:29Z

I don't know in general, but in Heegaard Floer theory, there are bimodules naturally associated to a mapping class of a surface self-homeomorphism. The rank of $H_*(M^{\otimes_A n})$ grows exponentially iff the underlying mapping class group element is pseudo-Anosov. See our paper at http://front.math.ucdavis.edu/1012.1032. In particular, it's easy to give examples where the growth rate is linear, although I'm sure there are also more elementary constructions.

(There are also other, earlier constructions for the braid group, by Khovanov-Seidel and Khovanov-Thomas. I don't know if they explicitly stated the fact about growth rates, but it follows directly from their results.)

Answer by Dylan Thurston for Why is a 2d TQFT formulated as a functor?

2011-06-02T02:30:27Z

There are interesting examples where the duality you're trying to bake in does not hold. In particular, the 3+1 dimensional (pseudo-)TQFTs coming from Seiberg-Witten theory in 4 dimensions do not have such duality maps. This is possible since the TQFT is only defined for a subset of the 4-d cobordism, for instance cobordisms where both the input and the output are connected 3-manifolds. (The actual TQFT allows one of the two sides to be disconnected, depending on which flavor you look at.)

Presumably this is related to some strangeness from the physics point of view.

Answer by Dylan Thurston for Area-preserving map between rectangles and fat polygons

2011-06-01T20:34:54Z

There are lots of ways to do this; without more constraints (or indication of what is desired), it's hard to pick a best one. But here's one. We'll exploit the fact that for any two triangles, there is a unique affine linear map that takes one to the other (with vertices in a specified order); if the two triangles have the same area, this will be area-preserving.

Start by chopping up $C$ into triangles $T_1,\dots,T_k$ by drawing lines connecting the vertices to the center.
Now take $R$ to be a square with the same area as $C$. Divide up the boundary of $R$ into $k$ intervals $I_1,\dots,I_k$ whose lengths are proportional to the area of the $T_i$. (Some of the intervals may make a turn at the corner.) Divide the square into regions $S_i$ by connecting the ends of $I_i$ to the center of the square. Most of the $S_i$ will be triangles, except for the four at the corners, which are quadrilaterals.

Now, the area of $S_i$ is equal to the area of $T_i$, since the area of $S_i$ is proportional to the length of its base (as all $S_i$ have the same height), which is proportional to the area of $T_i$. Now just map $T_i$ to $S_i$ in an area preserving way. For the $S_i$ that are not at the corners, just take the affine linear map from above. For the $S_i$ that do go around the corner, cut both $S_i$ and $T_i$ into two triangles in the same proportion of area, and map the corresponding triangles to each other.

Answer by Dylan Thurston for The preimage of the boundary morphism in the Mayer-Vietoris sequence

2011-05-31T03:15:31Z

Take a contraction of $x$ in $A$. That gives an $n$-chain $y_A$ living in $A$, whose boundary is $x$. Similarly, a contraction of $x$ in $B$ gives an $n$-chain $y_B$, again with boundary equal to $x$. The desired preimage is $y_A - y_B$.

Answer by Dylan Thurston for Is there a version of Seiberg-Witten-Floer or Heegard-Floer homology for 3-manifolds with boundary?

2011-05-27T01:44:02Z

Andy Manion has already plugged our answer for Heegaard Floer homology. On the Seiberg-Witten side, not as much is known, but Tim Nguyen's thesis starts to attack the problem. However, there isn't as yet a complete answer on the Seiberg-Witten side.

(Of course the two theories for closed 3-manifolds are now known to be isomorphic, but this doesn't actually help as much as you'd like. In particular, the isomorphism is not known to be natural, and not known to be compatible with the 4-manifold invariants.)

Answer by Dylan Thurston for Is the following two-dimensional graph likely to be globally rigid?

2011-05-02T02:58:54Z

jc's answer touches most of the bases. Let me just clear up a few points.

You can, in fact, give concrete bounds on the probability a graph with vertices chosen from a finite grid will fail to be generic for the purposes of rigidity, as jc alluded to. The relevant theorem is called the Schwartz-Zippel(-DeMillo-Lipton) Lemma. Namely, suppose you are given a non-zero polynomial $P(x_1,\dots,x_n)$ of degree $d$ over some field $F$, and a finite subset $S$ of $F$. Then the probability that $P$ is zero when evaluated on arguments chosen uniformly at random from $S$ is zero at most $d/|S|$. This gives an easy randomized algorithm for testing whether a polynomial is zero or not, given bounds on the degree. (Lipton wrote up a nice history.)

The test for local rigidity boils down to checking the rank of the rigidity matrix (the Jacobian of the length function), which can in turn be interpreted as checking whether a certain polynomial is zero. Concretely, in my paper with Healy and Gortler we go through the analysis and get specific bounds (in Section 5). We also go through some analysis for global rigidity, but it's for whether the check for generic global rigidity works, not whether the particular framework is actually globally rigid. For concrete bounds there, you'd have to do a little more work.

(While I was writing this, my co-author Steven Gortler posted his own answer, but we cover different points so I'm leaving this up.)

I think the question about global rigidity is well-answered by now. The question also asked about algorithms for finding the realization. It sounds like the graphs are pretty dense, dense enough that they're likely to be universally rigid: the edge lengths will determine the positions in any dimension, not just 2 dimensions. (For instance, the square of any 3-connected graph is 3-connected.) For such graphs, there is a good algorithm, namely semidefinite programming: Consider the Gram matrix of the configuration, the positive semi-definite matrix formed by dot products between the coordinates. The length constraints give linear constraints on such matrices. If the graph is universally rigid, there will be essentially one PSD matrix satisfying the constraints; this gives you the embedding. This kind of problem (a PSD matrix with linear constraints) can be solved quickly, at least in practice.

Answer by Dylan Thurston for Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold?

2011-04-29T23:35:08Z

I know of several different arguments. You can decide which one you think is most elegant...

Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3-manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4-manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Richard Kent's answer above.
Thom's argument, with lots of algebraic topology. This is probably not the most elegant route if you only want this piece, although of course Thom tells you much more.
Rourke's argument as sketched by Daniel Moskovich above. Indeed, any proof that the mapping class group is generated by Dehn twists also gives a proof that $\Omega_3 = 0$. Dehn and Lickorish also have proofs of this.
I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3-manifold and look at a generic map to $\mathbb{R}^2$. The preimage of a generic point is a disjoint union of circles, which bounds a convenient canonical surface (a union of disks). Take these disks as the start of your 4-manifold. In codimension one singularities, two of these circles can merge, and the preimage of a little transversal is a pair of pants, which can be filled in with a 3-sphere (together with the disks already attached). In codimension 2, there are only two different interesting local models, and both can be filled in canonically with a 4-ball.

The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3-manifold triangulated with $n$ tetrahedra, it gives a 4-manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mapping-class group arguments of (3) tend to give a 4-manifold of complexity at least exponential in $n$, and usually a tower of exponentials. (You can see this already in the inductive argument sketched out in Daniel Moskovich's answer.) Thom's proof (2) is completely non-explicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4-manifold with $O(n^4)$ simplices, although I never worked out all the details.

Answer by Dylan Thurston for central extensions of Diff(S^1) and of the semigroup of annuli

2011-04-17T13:00:49Z

Let me give a method for answering the problem. I haven't yet done the relevant integral to get an actual answer.

In the setting you originally laid out for $\mathcal{A}$, actual diffeomorphisms are not included; however, you can find an annulus close to a diffeomorphism $f$ by taking an annulus of the form $[0,\epsilon] \times S^1$ with $\epsilon$ very small, and parametrizing the two sides differently: parametrize the left boundary by $x \mapsto (0,f(x))$ and parametrize the right boundary by $x \mapsto (1,x)$. Call this parametrized annulus $A_0(f)$.

$A_0(f)$ is not of the type that you apply the Liouville functional to, since the left boundary is not constant-speed geodesic. So let's construct another metrized torus where both boundaries are constant-speed. Let $B_\epsilon:[0,\epsilon] \to [0,1]$ be a bump function which is $1$ in a neighborhood of $0$ and $0$ in a neighborhood of $\epsilon$. Let $A_1(f)$ be the annulus $A_0(f)$ with the metric rescaled as $$ g_1(s,t) = \exp\Bigl(-2B_\epsilon(s)\ln \bigl(f'(t)\bigr)\Bigr)g_0(s,t) = \exp\Bigl(-2B_\epsilon(s)\ln \bigl(f'(t)\bigr)\Bigr)(ds^2 + dt^2). $$ Then $A_1(f)$, with the same boundary parametrization as in $A_0(f)$, is now parametrized by constant-speed geodesics. Call $(A_1(f), 0) \in \widetilde{\mathcal{A}}$ the canonical annulus of the diffeomorphism $f$.

Now suppose we have two diffeomorphisms $f,g$. We can glue the associated cannonical annuli $A_1(f), A_1(g)$. The result will represent the diffeomorphism $g \circ f$; but the metric will not be the canonical metric on $A_1(g \circ f)$. We have to rescale the metric by an appropriate scaling factor $\phi$ to get to the canonical metric. (Actually, the conformal width of the annulus is also now $2\epsilon$ rather than $\epsilon$, but that shouldn't matter.) The integral in the problem statement for this rescaling presumably reduces to the Bott-Virasoro integral as $\epsilon$ gets sufficiently small.

I'd still like to understand this geometrically well enough to not have to do the integral.

Answer by Dylan Thurston for Residual Finiteness of Fundamental Group of Compact 3-Manifold

2011-04-14T10:27:46Z

For the first question, the answer is yes. Geometrization implies that the only non-Haken manifolds irreducible manifolds are compact hyperbolic manifolds (with no cusps), and there again $\pi_1$ is residually finite.

For the second part, I'm not aware of other arguments, and I think this is still the standard way to prove this. Geometrization just says that the argument works in all cases.

Answer by Dylan Thurston for central extensions of Diff(S^1) and of the semigroup of annuli

2011-04-13T23:28:33Z

There's a more geometrically natural description of a $\mathbb{Z}$-central extension in both cases.

For $\operatorname{Diff}(S^1)$ , the central extension are diffeomorphisms $f: \mathbb{R} \to \mathbb{R}$ which are equivariantly periodic, i.e. $f(x+1) = f(x) + 1$.

For $\mathcal{A}$, the central extension consists of holomorphic structures on the strip $I \times \mathbb{R}$ (where $I$ is an interval) which are invariant under translation by $1$ in the $\mathbb{R}$ direction, together with an equivariant parametrization of both boundaries by $\mathbb{R}$. These are considered up to equivariant isomorphism. (If the interval $I$ has length $> 0$, then you can assume the parametrization is by the identity map by shearing your holomorphic structures. But this way makes the compatibility with $\widetilde{\operatorname{Diff}}(S^1)$ more obvious.)

In both cases the composition is clear, and it's also obvious that $\widetilde{\operatorname{Diff}}(S^1)$ acts on $\widetilde{\mathcal{A}}$.

Presumably these two extensions embed in the $\mathbb{R}$-central extension you describe.

Answer by Dylan Thurston for Using higher-order Bring radicals to solve arbitrary polynomials

2011-04-13T14:12:55Z

The answer to the second question is "no". For a family of polynomials $p_t$ depending polynomially on a complex parameter, as in the polynomials satisfied by your $B_r(t)$, define its Galois group to be the group of permutations of the roots you see by moving around the branch points. (Assume that the roots of $p_t$ are distinct for generic values of $t$ to make this work well. This is the same as the Galois group of `$\mathcal{C}(t)(\text{roots of }p_t)$ over $\mathcal{C}(t)$.) Then there are families of polynomials exhibiting an arbitrary finite group as its Galois group. Any finite family of $B^j(t)$ will only exhibit finitely many groups $G_j$, and any tower of roots of the $B^j$ will give only groups whose composition factors are among the $G_j$. Since there are infinitely many finite simple groups, you cannot acheive all possible finite groups this way.

However, I suspect the question was stronger than you meant to ask, since even the original conjecture that all polynomials are solvable by radicals wouldn't fit in to the framework of your second question.

Answer by Dylan Thurston for Is it reasonable to define `poset homotopy' as a `natural transformation of posets'?

2011-04-11T20:03:17Z

This is an interesting line of questions, but I think it doesn't quite work as stated. First off, your notion of "homotopy" is not an equivalence relation (as far as I understand it), so it won't agree with a topological notion.

But there are also other issues; basically, any notion of "poset homotopy classes" along the lines you suggest will have finitely many classes of maps between two spaces, unlike real homotopy classes.

For instance, the poset $P = {x,y,z,w}$ with $x > z$, $x > w$, $y > z$, and $y > w$ corresponds to a finite topological space that is weakly homotopy equivalent to $S^1$, so $\pi_1(P) = \mathbb{Z}$. But there are only finitely many poset maps from $P$ to $P$, even before taking any homotopy equivalence.

Answer by Dylan Thurston for Usefulness of using TQFTs

2011-04-05T01:08:44Z

All the answers so far have focused on 3 dimensions, but the answer is much more striking in 4 dimensions. Freedman's theorem tells you that classical homology invariants give you complete information about topological, simply-connected 4-manifolds. These classical invariants cannot, however, distinguish between distinct smooth structures on the same topological 4-manifold, and essentially our only technique for distinguishing smooth 4-manifolds is Donaldson's invariant or the Seiberg-Witten invariant or their relatives. These do not quite form a TQFT, but are related to TQFTs.

Edit: On request, a little about how the 4-manifold invariants are related to a TQFT. This is all nicely explained in the beginning of Kronheimer and Mrowka's book Monopoles and 3-manifolds.

There are actually three different theories, denoted $\widehat{\mathit{HM}}$ ("HM-from"), $\check{\mathit{HM}}$ ("HM-to", unfortunately typeset badly here), and $\overline{\mathit{HM}}$. All are close to satisfying axioms for a TQFT assigning a vector space to a 3-manifold and maps to a 4-manifold, at least for connected manifolds. (The vector spaces are infinite dimensional, but finite in each graded piece.) Unfortunately, however you slice it, in each case the invariant associated to a closed 4-manifold in the usual TQFT way (when defined) is zero.

Instead, you use the fact that there is an exact triangle \[ \cdots \longrightarrow \widehat{\mathit{HM}} \longrightarrow \overline{\mathit{HM}} \longrightarrow \check{\mathit{HM}}\longrightarrow \cdots \] (with right mapping to left), and the map $\overline{\mathit{HM}}(W)$ is $0$ for $b_2^+(W) \ge 1$.

If you have a 4-manifold $W$ with $b_2^+(W) \ge 2$, you factor it as two cobordisms $W = W_1 \cup_Y W_2$ for some 3-manifold $Y$, with $b_2^+(W_i) \ge 1$. Then the properties above let you map from $\check{\mathit{HM}}(S^3)$, to $\check{\mathit{HM}}(Y)$, backwards in the exact triangle to $\widehat{\mathit{HM}}(Y)$, and then forwards to $\widehat{\mathit{HM}}(S^3)$. The resulting map (from $\check{\mathit{HM}}(S^3)$ to $\widehat{\mathit{HM}}(S^3)$) gives the interesting Seiberg-Witten invariants of $W$.

Answer by Dylan Thurston for Do the results of (1/n)-surgery determine the link?...

2011-04-02T20:04:11Z

If the orginal link $U_1 \cup U_2$ was hyperbolic, the answer is yes. For large enough $n$, $S^3 \setminus K(n)$ will also be hyperbolic, and will approach $S^3 \setminus (U_1 \cup U_2)$ in the Gromov-Hausdorff topology; so the sequence $K(n)$ determines the complement of $U_1 \cup U_2$. The complement doesn't determine the link in general (unlike for knots), but we also have the marking of the component $U_1$ by its meridian, which I believe is enough.

The answer in general is also almost certainly yes, but I haven't thought through all the cases. Note that this operation has a simple geometric description: arrange $U_1 \cup U_2$ so that $U_2$ sits as a flat unknot in a plane. Then to get $K(n)$, remove $U_2$ and twist the bundle of strands that passed through $U_2$ by $n$ full twists.

(This is all much easier than Lackenby's result mentioned above.)

Examples for non-naturality of universal coefficients theorem

2011-03-29T04:11:48Z

Does anyone have good examples of a space $X$ and a map $f: X \to X$ so that $f_*: H_*(X) \to H_*(X)$ is the identity but (e.g.) $f_*: H_*(X; \mathbb{F}_2) \to H_*(X; \mathbb{F}_2)$ is not the identity?

Edit: As mentioned in the comments, $f_*$ is an isomorphism on the mod-2 homology, but I don't see why it needs to be the identity. More precisely, the exact sequence \[ C(X; \mathbb{Z}) \overset{\times 2}{\longrightarrow} C(X; \mathbb{Z}) \longrightarrow C(X;\mathbb{Z}/2) \] gives an exact triangle in homology, which in turn induces a 2-step filtration on $H_*(X; \mathbb{Z}/2)$ (where one step is the image of the map $H_*(X;\mathbb{Z}) \to H_*(X;\mathbb{Z}/2)$ ). The assumption that $f_*$ is the identity on integral homology implies that it is the identity on the associated graded space to this filtration, but that still doesn't imply it is the identity.

I came across a similar phenomenon in the context of Heegaard Floer homology, with the rings $(\mathbb{Z}/2)[U]$ and $\mathbb{Z}/2$ playing the roles of $\mathbb{Z}$ and $\mathbb{Z}/2$.

When is a quasi-isomorphism necessarily a homotopy equivalence?

2011-03-24T05:46:49Z

Under what circumstances is a quasi-isomorphism between two complexes necessarily a homotopy equivalence? For instance, this is true for chain complexes over a field (which are all homotopy equivalent to their homology). It's also true in an $\mathcal{A}_\infty$ setting.

Is it true for chain complexes of free Abelian groups? The case I'm particularly interested in is chain complexes of free $(\mathbb{Z}/2\mathbb{Z})[U]$ modules or free $\mathbb{Z}[U]$ modules, but I'm also interested in general statements.

Spaces with same homotopy and homology groups that are not homotopy equivalent?

2011-01-26T20:32:47Z

A common caution about Whitehead's theorem is that you need the map between the spaces; it's easy to give examples of spaces with isomorphic homotopy groups that are not homotopy equivalent. (See http://mathoverflow.net/questions/3540/are-there-two-non-homotopy-equivalent-spaces-with-equal-homotopy-groups). It's surely also true that the pair (homotopy groups, homology groups) is not a complete invariant, but can anyone give examples? That is, I'm looking for spaces $X$ and $Y$ so that $\pi_n(X) \simeq \pi_n(Y)$ and $H_n(X;\mathbb{Z}) \simeq H_n(Y; \mathbb{Z})$ but $X$ and $Y$ are still not (weakly) homotopy equivalent.

(Easier examples are preferred, of course.)

Answer by Dylan Thurston for A special class of regular languages: "circular" languages. Is it known?

2011-01-12T08:00:05Z

For deciding whether a language is "circular", you can just take the normalized DFA for the language (where the states correspond to sets of possible different completions). In that normalized DFA, a language is circular iff the only accept state is the start state, pretty much by definition.

I don't know what you want by a characterization. A language L has this property iff it is M* for some other language M, but that's not useful..

Answer by Dylan Thurston for How many edge-disjoint paths go from upper left to lower right in a $4 \times N$ rectangular gridwork of streets?

2011-01-03T16:37:34Z

To amplify on Christian's answer, the problem on a $K \times N$ grid for fixed $K$ and varying $N$ admits a finite-state transition model, so in particular is given by a linear recurrence.

The key is to find the right set of states. If you take an edge-disjoint path on a $K \times N$ grid and slice it on a vertical line through the middle passing through a set of horizontal edges, you'll see the path crossing along some odd number of these edges ($\le K$). On both the left and right we'll see a collection of paths with these endpoints. There's another constraint, that we end up with a single, connected path without disjoint loops; to take that in to account, also record a matching: which endpoints are paired up on the right hand side. All but one of the endpoints are paired up in this way. (You could also choose the left, and end up with a slightly different matrix.)

For instance, in the $3 \times N$ case, there are $6$ states. If we record an occupied edge by $\times$ and an unoccupied edge by $\circ$ and turn everything on its side, the states are $$ \times\circ\circ\quad\circ\times\circ\quad\circ\circ\times\quad\times_1\times_1\times \quad\times_1\times\times_1\quad\times\times_1\times_1 $$ where the subscript indicates the matching. (In this case, there is at most one matched pair.)

Next consider the transitions. If you consider two adjacent vertical slicings of a path, you'll see two possibly different states. The set of edges that are occupied in the middle is determined by which edges are occupied in the two different states. There is sometimes a choice about how the strands are connected up. However, some of these choices will be ruled out by the constraints on the connectivity; usually you will end up with just $0$ or $1$ possibilities.

For instance, in the $3 \times N$ case, with the states in the order above, I get the following matrix of possibilities: $$ M =\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 & 1\\ 1 & 1 & 1 & 0 & 1 & 1\\ 0 & 1 & 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 1 & 1 & 0 & 0 & 0 & 1 \end{pmatrix} $$

For the ultimate answer, you want to look at paths that start at the upper-left and go to the lower-right. You can incorporate that nicely by adding an extra slice to the left of the entire diagram, with only the top slot occupied, and another to the right of the diagram, with only tho lower slot occupied.

Concretely, in the $3 \times N$ case, the number of paths is given by the $(1,3)$ entry of $M^N$.

For the $4 \times N$ case, you would get a $16 \times 16$ matrix, which is straightforward but somewhat tedious to work out. As a result, the answer will satisfy a linear recurrence of order $16$.

An interesting variation is to consider only crossingless paths. In this case, the matching must be crossingless, so we only get 5 states in the $3 \times N$ case and $12$ in the $4 \times N$ case.

Update Jan 7: The matrix above is wrong: it should be $$ M =\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 2 & 2 & 2\\ 1 & 1 & 1 & 0 & 1 & 1\\ 0 & 1 & 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 1 & 1 & 0 & 0 & 0 & 1 \end{pmatrix} $$

Update 2: And here's an image illustrating what is actually being counted: I permuted the entries slightly, but they're labelled along the sides. The dotted paths are there to help in the counting: the non-allowed configurations would form a closed loop.

Answer by Dylan Thurston for pseudo-Anosov maps on surfaces with boundary

2010-10-26T01:49:52Z

First off, the mapping class group of a surface with boundary is generally taken to mean the group of diffeomorphisms that fix the boundary, up to isotopies fixing the boundary. In this context, a Dehn twist around one boundary component is not isotopic to the identity. What you're describing is usually called the mapping class group of a surface with punctures.

Under many definitions (including Casson and Bleiler's) the identity automorphism is also reducible (as well as periodic). So I don't see a counterexample in what you write.

If you're asking, for which simple closed curves is the Dehn twist around that curve not the identity (and also not periodic), the answer is simple: it's those curves that do not surround a single puncture. These are the same as the curves that are homotopic to a geodesic in any hyperbolic representative with cusps at the punctures. (For most curves, you look at a shortest representative in its homotopy class and you automatically get a geodesic. For a curve that surrounds a puncture, there is no shortest representative: you can make the curve arbitrarily short by pushing it out the cusp.)

Answer by Dylan Thurston for Numeric problem when evaluating log of a pdf

2010-10-23T13:59:46Z

Work with the logs of probability directly, rather than trying to compute the probability and then compute the log. You can do arithmetic with the logs, as well; multiplication becomes addition, of course. Addition is somewhat more complicated, but it's not too hard to work out how to do it without taking the exponential back.

In response to JM's question, it's easy to come up with practical problems where the probability of something happening is indeed so low that it underflows a float or double.

Answer by Dylan Thurston for Computing the hopf invariant (without integration or homology, as in Milnor) of the hopf map

2010-10-17T23:52:38Z

If you have the Hopf link embedded in some standard way in $\mathbb{R}^3$, you can see the linking number as given by the degree of a map $S^1 \times S^1 \to S^2$ in a number of ways. For instance, the pre-image of the north pole in $S^2$ consists of pairs of points stacked vertically above each other, i.e., crossings between the two components in the knot diagram given by projection to the $xy$ plane. (Crossings will correspond to preimages of the north pole or south pole, depending on your conventions.) For the standard diagram for the Hopf link, there's only one crossing that counts. The hard part from this point of view is getting the orientation right (is the Hopf invariant $-1$ or $+1$?), but that can be done with care and attention.

Answer by Dylan Thurston for If the series Σ pᵃ⁽ʷ⁾·xᴵʷᴵ is rational, is Σ a(w)·xᴵʷᴵ also rational (summation over words w in a regular language)?

2010-10-17T05:15:28Z

If there are only finitely many values for the $a_i$, then I believe the answer to both questions is affirmative for somewhat silly reasons. If you have any rational power series $f(x)$ with only finitely many different coefficients, then I believe the power series $f_c(x)$ with a coefficient of 1 when $f(x)$ has $c$ (and 0 otherwise) is also rational. (Restivo and Reutenauer, "On cancellation properties of languages which are supports of rational power series", cite Sontag, "On some questions of rationality and decidability" for this fact. I couldn't find the precise statement, but the techniques are plausibly related.) Then you can easily rearrange the coefficients as in Question 1. (This would also work for a slightly weaker version of Question 2 that seems likely to be relevant for you: instead of power series in $x$, work in non-commutative power series in the alphabet of $L$ and do the rearranging there. For this, you need to know that $\sum p^{a_i} w$ is rational, but it sounds like you do.)

(The theorems above can be made easier by assuming that the power series like $\sum p^{a_i} x^i$ are the weighted generating series of a regular language with positive weights, which again sounds like the relevant case; then you can just use the usual NDFA -> DFA reduction.)

However, your $a_i$ are not bounded, it sounds like. The linear recurrence on the coefficients is going to give strong constraints on the $a_i$, but I don't yet see how to finish this to a proof.

Terminology: Algebras where long strings of products are 0?

2010-04-02T18:08:29Z

I'd like a name for an augmented algebra $A = \langle 1\rangle \oplus A_+$ for which there is an $N$ so that any product of more than $N$ elements in the augmentation ideal is $0$, i.e., $(A_+)^N = 0$. Is there a name? It seems related to nilpotence, and it implies that all elements in $A_+$ are nilpotent, but is stronger than that. There is a uniform bound on the degree of nilpotence, but that's not enough either, as the example of the exterior algebra in infinitely many variables over $\mathbb{Z}/2$ shows.

MathWorld defines a nilpotent algebra or nilalgebra to be one where every element is nilpotent. (They are therefore not considering unital algebras, contrary to an earlier discussion here.) Is this standard? Is there a better term?

Answer by Dylan Thurston for Elementary functions with zeros only at the positive integers

2010-05-08T18:45:39Z

$1/\Gamma(1-z)$.

Fundamental domain for symmetric group $S_n$ acting on $\mathop{Gr}(k,n)$?

2010-05-07T00:23:22Z

Is there a nice fundamental domain for the symmetric group $S_n$ acting on the Grassmannian of $k$-planes in $\mathbb{R}^n$? (The action of $S_n$ is by permuting the coordinates, of course.)

I'm looking for a way to efficiently test whether two subspaces of $\mathbb{R}^n$ are related by permuting and/or negating the coordinates. (So I really care not about $S_n$, but about $(\mathbb{Z}/2)^n \rtimes S_n$, the Weyl group of the B/C series.) I'm somewhat skeptical this is easy, but has anyone thought about it?

Comment by Dylan Thurston

2013-01-01T17:54:23Z

You can take a CW-complex with only a single 4-cell, for instance.

Comment by Dylan Thurston

2012-06-27T14:17:32Z

I'm not sure what you are asking. Barring intersections, you can follow the roots. Are you looking for a numerical method?

Comment by Dylan Thurston

2012-06-26T03:42:49Z

This is pretty good, though of course rather more roundabout than I was hoping for. Thank you.

Comment by Dylan Thurston

2012-06-25T15:55:12Z

It seems like you expect the answer to be negative, since you're looking for obstructions?

Comment by Dylan Thurston

2012-06-16T04:52:32Z

Are you looking for an invariant associated to a prime in a number field, or for an expression for the Jones polynomial of a knot in $S^3$? For the latter, the number-theoretic limits giving the Jones polynomial are interesting; see front.math.ucdavis.edu/0211.5044 . For the former, AFAIK no one has a convincing analogue of the Jones polynomial. Milnor invariants can be interpreted and the Alexander polynomial is supposed to be related to Iwasawa theory (arxiv.org/abs/0904.3399v1), but I believe there's no good analogue of finite-type invariants with more loops.

Comment by Dylan Thurston

2012-06-07T22:57:56Z

You need backquotes to escape math code in certain cricumstances, as indicated in the box that should be on the right or in the FAQ. I fixed it.

Comment by Dylan Thurston

2012-06-07T00:28:53Z

This is a good answer, but it's worth noting that your theorem relies on the Reidemeister I move being true locally. This relation does not hold for the R-matrices from quantum groups, so you don't actually get invariants of virtual knots in the way you sketch in your answer, but rather virtual knots up to isotopy. I'm not aware of a uniqueness or classification result in this context.

Comment by Dylan Thurston

2012-06-02T13:24:39Z

You need to escape the math in certain cases, or it gets interpreted as markdown (italics in this case). There's a box to the right under 'How to write math' that explains this, or see the FAQ.

Comment by Dylan Thurston

2012-05-12T15:14:03Z

DoubleJay, I don't think that can be right: there must be more than one embedding of a Klein bottle in $\mathbb{R}^4$. Maybe it's true that if there's only a single circle of self-intersections then it's a standard embedding?

Comment by Dylan Thurston

2012-05-11T12:27:25Z

Thank you for both pointing out the mistake in both my example and giving the general argument; I agree now.

Comment by Dylan Thurston

2012-05-06T02:57:38Z

It's worth commenting (for those that don't know) that PL topology is the same as smooth topology in low dimensions (up to 6).

Comment by Dylan Thurston

2012-05-04T02:50:56Z

The edge does not contract. Please think things through more.

Comment by Dylan Thurston

2012-05-03T15:36:05Z

Isn't the non-trivial knot with $N=7$ a right-handed trefoil, not a left-handed trefoil? Or am I getting some convention backwards?

Comment by Dylan Thurston

2012-05-03T15:33:31Z

To reduce, you really need $|a_k - a_{k-1}| = 1$, not just that it's equivalent $\textrm{mod} N$. For instance, on the list with $N=9$, I find that $(6,3,1,5,7,2,4,8,0)$ represents the figure 8 knot, which according to your calculations was not representable with $N=8$. One point that is possibly confusing is that I switched the indexing, so that the points run from $0$ to $N-1$ rather than $1$ to $N$.

Comment by Dylan Thurston

2012-05-02T11:39:18Z

I don't think adjacent 0-7 pairs reduce, do they? In any case, I added the $N=9$ possibilities.