Easy cases of the volume conjecture?

 November 18, 2013 mathematics

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The volume conjecture relates the hyperbolic volume of a knot complement to quantum invariants of the knot. Specifically, the conjecture is that

\operatorname{Vol}(S^3 \setminus K) = \lim_{N \to \infty} \frac{2 \pi \log |J_N(K;\xi_N)|}{N}

where J_N computes the colored Jones polynomial and \xi_N = e^{2\pi i / N} . For some knots K , there are nice formulas for J_N(K;\xi_N) . For instance, if K is the figure eight knot, then J_N(K;q) can be written as

J_N(K;q) = \sum_{j=0}^{N-1} \prod_{\substack{k=-j \\ k \neq 0}}^j \left(q^{(N-k)/2} - q^{-(N-k)/2}\right).

When q = \xi_N and one takes the limit, this sum transforms quite nicely into the Riemann sum which computes 6 \,\Lambda(\pi/3) , which is the hyperbolic volume of the figure eight knot complement.

Are there other cases in which one can verify the volume conjecture directly by finding a nice form for the colored Jones polynomial? Yamazaki-Yokota in “On the limit of the colored Jones polynomial of a non-simple link” verified, in the same direct way, the volume conjecture for a certain link with volume 6\,\Lambda(\pi/3)+16\,\Lambda(\pi/4) . I hope one could find other examples by searching for links and knots with “nice” volumes in terms of the Lobachevsky function \,\Lambda .

From my search with SnapPy,

Of course, there are lots of nonhyperbolic knots and it is hard to prove the volume conjecture even for those, in spite of the fact that their hyperbolic volume is “really nice” (namely, zero!).

Many more Lights Out

 July 17, 2010 mathematics

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A very long while ago I posted some solutions to Lights Out; back then, I solved the n -by- n board by row-reducing an n^2 -by- n^2 matrix.

Since then, both Boris Okun and Brent Werness pointed out to me that I should’ve solved Lights Out by using a scanning algorithm: propagating the button presses down one row at a time, and exponentiating the propagation matrix to make sure that I don’t get stuck at the last row.

This is much faster.

With this method, here is a (scaled down, auto-leveled) 2000-by-2000 solution:

Solution to 2000x2000 Lights Out

And here is a (very much scaled-down, auto-leveled) 5000-by-5000 solution:

Solution to 5000x5000 Lights Out

Projector on Blackboard

 January 19, 2010 mathematics

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I recently gave a beamer talk, which gave me the chance to point the beamer at my blackboard.

Classifying manifolds is impossible.

 February 12, 2007 mathematics

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At a recent Pizza Seminar, Matt Day gave a lovely talk explaining why it isn’t possible to classify 4-manifolds.

An algorithm for deciding whether two closed 4-manifolds are homeomorphic gives an algorithm for deciding whether a closed 4-manifold is simply connected, and therefore (since every finitely presented group is the fundamental group of a 4-manifold), and algorithm for deciding when a group is trivial. Here’s the reduction: we are given a 4-manifold M , and we compute its signature \sigma(M) . By Freedman, there are no more than two closed simply connected 4-manifolds, M_1 and M_2 , having the same signature as M ; we construct M_1 and M_2 , and we use the homeomorphism decision procedure to test if M \cong M_1 or M \cong M_2 .

Since there is no algorithm for deciding when a group is trivial, there can not be an algorithm for deciding when two closed 4-manifolds are homeomorphic.

There is a paper [1] discussing some of these issues. In particular, that paper discusses Novikov’s proof that S^n cannot be recognized for n \geq 5 .

[1] A.V. Chernavsky, V.P. Leksine, Unrecognizability of manifolds, Ann. Pure Appl. Logic. 141 (2006) 325–335. https://doi.org/10.1016/j.apal.2005.12.011.

Subgroups of products versus products of subgroups.

 February 4, 2007 mathematics

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This is a question I wandered into accidentally years ago now, which I think other people might be amused to think about (or more likely, put on an abstract algebra exam).

Let G be a group, and H a subgroup of G \times G . Is H always isomorphic to G_1 \times G_2 , for some subgroups G_1, G_2 < G ? But beware!–I am not requiring (or expecting) any canonicity or naturality for the isomorphism: for instance G sits diagonally in G \times G , and it just so happens that G = \{ 1 \} \times G = G \times \{ 1 \} , so this is not a counterexample, in spite of the fact that the “horizontal” or “vertical” subgroup is not a canonical choice for the diagonal subgroup.

What is a good name for groups with this property? It’s not completely trivial: cyclic groups, for instance, have this property–not that I think this property is important, but names can be amusing…

I have examples of groups G and H < G \times G with H not (abstractly!) a product of subgroups of G . My challenge to you is to find some explicit examples of H < G \times G and prove that H doesn’t decompose.

In the end, I think this is a fun problem for a group theory final exam; I think it nicely highlights the difference between “being isomorphic” and “being equal,” though if one completes the challenge as stated, one probably already understands that distinction… So maybe the best reason for blogging about this is that chiastic title.

Constructing a Lie group from a Lie algebra.

 November 30, 2006 mathematics

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Theorem. For every Lie algebra \mathfrak{g} , there is a simply-connected, connected Lie group G having \mathfrak{g} as its Lie algebra.

First, if \mathfrak{g} \subset \mathfrak{gl}(V) , then the exponential map gives U = \exp \mathfrak{g} , and we define G = \bigcup_{k=1}^\infty U^k \subset GL(V) . It turns out G is a Lie group, and \mathfrak{g} is its Lie algebra.

If \mathfrak{g} has no center, then \rm{ad} : \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g}) is injective, so we have realized \mathfrak{g} as a Lie subalgebra of endomorphisms of a vector space, and by the above, there is a Lie group G \subset GL(\mathfrak{g}) with \mathfrak{g} as its Lie algebra. Taking its universal cover \tilde{G} proves the theorem in this case.

Now we induct on the dimension of the center Z(\mathfrak{g}) . Let Z \subset Z(\mathfrak{g}) be a one-dimensional central subspace of \mathfrak{g} , and construct a short exact sequence 0 \to Z \to \mathfrak{g} \to \mathfrak{g}&39; \to 0 . But this central extension of \mathfrak{g}&39; by Z = \mathbb{R} corresponds to a 2-cocycle \omega \in H^2(\mathfrak{g}; \mathbb{R}) .

Lemma. Let D : H^2(G;\mathbb{R}) \to H^2(\mathfrak{g}; \mathbb{R}) be the map which differentiates a (smooth!) 2 -cocycle of the group cohomology of G . The map D is injective.

Consequently, we can find f \in H^2(G;\mathbb{R}) with Df = \omega . Since \dim Z(\mathfrak{g}&39;) < \dim Z(\mathfrak{g}) , by induction there is a Lie group G&39; having \mathfrak{g}&39; as its Lie algebra. We build the central extension of G&39; by \mathbb{R} using the cocycle f , namely, 0 \to \mathbb{R}\to G \to G&39; \to 0 , where G \cong G&39; \times \mathbb{R} and the operation is (g_1, t_1) \cdot (g_2, t_2) = (g_1 g_2, t_1 + t_2 + f(g_1,g_2)) . Since Df = \omega , it turns out that the Lie algebra corresponding to G is \mathfrak{g} . We finish the proof by taking the universal cover \tilde{G} .

[1] V.V. Gorbatsevich, Construction of a simply connected group with a given lie algebra, Uspekhi Mat. Nauk. 41 (1986) 177–178.

Coxeter group visualization.

 November 28, 2006 mathematics

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Jenn is a fabulous program for visualizing the Cayley graphs of finite Coxeter groups. The pictures are absolutely beautiful (oh, symmetry!).

Efficient construction of the reals.

 October 20, 2006 mathematics

Today in Geometry/Topology seminar, quasihomomorphisms \mathbb{Z}\to \mathbb{Z} were discussed, i.e., the set of maps f : \mathbb{Z}\to \mathbb{Z} such that | f(a+b) - f(a) - f(b) | is uniformly bounded, modulo the relation of being a bounded distance apart. These come up when defining rotation and translation numbers, for instance.

Anyway, Uri Bader mentioned that these quasihomomorphisms form a field, isomorphic to \mathbb{R} , under pointwise addition and composition. I hadn’t realized that this is a general construction. Given a finitely generated group (with fixed generating set, so we have the word metric d on the group), I can define a quasihomomorphism f : G \to G by demanding d(f(ab),f(a)f(b)) be uniformly bounded, and where two quasihomomorphisms f, g are equivalent if d(f(a),g(a)) is uniformly bounded. Let’s call the resulting object \hat{G} for now.

What can be said about \hat{G} ? For instance, what is \hat{F_2} ?

Outer Space. And real aliens.

 October 10, 2006 mathematics

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There are some questions about outer space that I would like to be able to answer. Some nice survey articles look to be [1] and also [2].

Here is a ridiculously simple question I have wondered about: given A, B \subset F_n , say with [F_n : A] = [F_n : B] , how can I tell if A and B are conjugate? I suspect I’m being stupid here.

In light of my recent comments on LINCOS and communicating with extraterristrials, I found the article [3].

Putnam also makes use of the idea of mathematicians from other planets, to more philosophical ends.

[1] M. Bestvina, The topology of {\rm Out}(F_n) , in: Proceedings of the International Congress of Mathematicians, Vol. II (Beijing, 2002), Higher Ed. Press, Beijing, 2002: pp. 373–384.

[2] K. Vogtmann, Automorphisms of free groups and outer space, in: Proceedings of the Conference on Geometric and Combinatorial Group Theory, Part I (Haifa, 2000), 2002: pp. 1–31. https://doi.org/10.1023/A:1020973910646.

[3] D. Ruelle, Conversations on mathematics with a visitor from outer space, in: Mathematics: Frontiers and Perspectives, Amer. Math. Soc., Providence, RI, 2000: pp. 251–259.

Euler characteristic of closed hyperbolic 4-manifolds.

 September 22, 2006 mathematics

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By the Gauss-Bonnet theorem, the volume of a hyperbolic 4-manifold is proportional to its Euler characteristic. There are examples, constructed explicitly in [1] of hyperbolic 4-manifolds with every positive integer as their Euler characteristic. These examples are non-compact (with five or six cusps, I believe). But [2] observes that there are restrictions on the Euler characteristic that a closed hyperbolic 4-manifold may possess. In particular, it is shown in [3] that the Pontrjagin numbers of a hyperbolic manifold M vanish. But the signature \sigma(M) is a rational linear combination of those Pontrjagin numbers, so \sigma(M) = 0 . And by Poincare duality, \chi(M) \equiv \sigma(M) \pmod 2 , so \chi(M) is even. A natural question to ask is: does there exist a hyperbolic 4-manifold M with \chi(M) = 2 ? Now if such an M also had H_1(M) \neq 0 , we would know the volume spectrum of closed hyperbolic 4-manifolds.

This certainly seems to parallel the case for 2-manifolds: all negative integers are the Euler characteristic of a hyperbolic 2-manifold, and all even negative integers are the Euler characteristic of a closed hyperbolic 2-manifold.

The vanishing of Pontrjagin numbers for hyperbolic manifolds also holds for pinched negative curvature under some conditions [1].

It is also a fact that the Stiefel-Whitney numbers vanish for a closed hyperbolic manifold (and the vanishing of the top Stiefel-Whitney class is the same thing as having even Euler characteristic).

[1] J.G. Ratcliffe, S.T. Tschantz, The volume spectrum of hyperbolic 4-manifolds, Experiment. Math. 9 (2000) 101–125. http://projecteuclid.org/getRecord?id=euclid.em/1046889595.

[2] J.G. Ratcliffe, The geometry of hyperbolic manifolds of dimension at least 4, in: Non-Euclidean Geometries, Springer, New York, 2006: pp. 269–286.

[3] S.-s. Chern, On curvature and characteristic classes of a Riemann manifold, Abh. Math. Sem. Univ. Hamburg. 20 (1955) 117–126.

Non-arithmetic lattices.

 September 14, 2006 mathematics

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Margulis’ amazing arithmeticity theorem says that irreducible lattices in Lie groups of high ( >2 ) rank are arithmetic. But {\rm SO}(n,1) has rank 1, so a question is how to produce non-arithmetic lattices. For {\rm SO}(3,1) , there are non-arithmetic lattices coming from hyperbolic knot complements.

G–P-S [1] produces higher dimensional examples by taking two hyperbolic (arithmetic) manifolds, cutting along totally geodesic hypersurfaces, and gluing. Are there are examples of non-arithmetic hyperbolic manifolds without any totally geodesics hypersurfaces?

There are complements of T^2 ’s in S^4 which are hyperbolic, and maybe these would provide some examples.

[1] M. Gromov, I. Piatetski-Shapiro, Nonarithmetic groups in lobachevsky spaces, Inst. Hautes Études Sci. Publ. Math. (1988) 93–103.

Tannakian Philosophy

 August 16, 2006 mathematics

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From Recent Advances in the Langlands Program, quoted in This Week’s Finds:

First of all, it should be remarked that according to the Tannakian phylosophy, one can reconstruct a group from the category of its finite-dimensional representations, equipped with the structure of the tensor product.

I suppose one should think of this as the categorification of Pontrjagin duality?

For a long while, I had wondered how this goes; this Introduction to Tannaka Duality and Quantum Groups will probably answer my questions.

Finite subgroups of rotation groups.

 April 5, 2006 mathematics

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Here is a question that I haven’t been able to find very much about:

What are the finite subgroups of the rotation groups SO(n) ?

For examples, I can take a Coxeter group, and choose elements corresponding to rotations (e.g., the subgroup generated by products of generators), but that’s not going to produce very many examples.

Orientable 3-manifolds are parallelizable

 March 5, 2006 mathematics

Here’s a very easy theorem.

Theorem. All closed orientable 3-manifolds are parallelizable. All closed orientable 3-manifolds are the boundary of a 4-manifold.

Proof. Let M be an orientable 3 -manifold. Recall that the Wu class v is the unique cohomology class such that \langle v \cup x, [M] \rangle = \langle Sq(x), [M] \rangle , and Wu’s theorem says that w(M) = Sq(v) . The up-shot is that Stiefel-Whitney classes are homotopy invariants, even though they are defined using the tangent bundle.

Since M is orientable, we have w_1(M) = 0 . Since \dim M = 3 , the Steenrod squares Sq^2 and Sq^3 kill everything, so v_2 = 0 and v_3 = 0 . By Wu’s theorem, w_2(M) = Sq^1(v_1) + v_2 = 0 , and w_3(M) = Sq^1(v_2) + v_3 = 0 . In other words, all the Stiefel-Whitney classes vanish.

Orientability matters; after all, being orientable is the same thing as w_1 vanishing. For example, RP^2 \times S^1 is not parallelizable, since w_1(RP^2 \times S^1) = w_1(RP^2) + w_1(S^1) \neq 0 .