Let’s look at a simple mathematical procedure to generate such a pattern, called Koch’s curve. In a polygon, whenever we encounter a line let’s replace it according to the following rule:

The result is again a polygon and we may iterate this step again and again. Here are two examples:

While this does look similar to an actual snowflake, it only accounts for a single shape. Of course, in nature the entire process takes place in reverse. It starts with a microscopic piece of ice and gradually over time, molecules attach to it. The rules for this attaching are probabilistic and depend on the outside temperature and pressure. Since every snowflake undergoes a unique trajectory of outer circumstances, this accounts for their individuality.

Roughly, the formation in nature can be described as follows. The hydrogen atom has its positive and negative poles arranged like the vertices of a tetrahedron. Under cold weather conditions, this leads to a tendency of hydrogen atoms to form hexagonal ice molecules. The result is a structure of ice similar to that of a honey comb. Now the electromagnetic forces favor some of its faces to grow faster than others, which leads again to hexagonal shapes on larger scales. This is how the six-fold symmetry can be explained.

For a detailed discussion from the point of view of chemistry, the following link is highly recommended reading:

]]>Bamler, Kleiner: Ricci flow and diffeomorphism groups of 3-manifolds, https://arxiv.org/pdf/1712.06197.pdf

The Smale conjecture in its original form asserted that the diffeomorphism group of the 3-sphere deformation retracts onto O(3), the isometry group of its “round” constant curvature metric. It was proved by Hatcher in 1983. Also for hyperbolic 3-manifolds it was proven by Gabai in 2001 that the inclusion of the isometry group into the diffeomorphism group is a homotopy equivalence. Together with other known results for e.g. lens spaces, this gave evidence for the generalized Smale conjecture, asserting that the inclusion of the isometry group into the diffeomorphism group should be a homotopy equivalence for all 3-manifolds of non-zero constant curvature.

The natural action of the diffeomorphism group on the space of constant curvature metrics is transitive (because of rigidity of constant curvature metrics) and defines a fibration with the isometry groups as its fiber. Thus, in view of the long exact homotopy sequence, the weak homotopy equivalence of Isom(X)–>Diff(X) is equivalent to weak contractibility of the space of constant curvature metrics. Because all involved spaces have the homotopy type of CW-complexes, one can replace “weak homotopy equivalence” by “homotopy equivalence” and “weak contractibility” by “contractibility” in this equivalence.

Thus the generalized Smale conjecture reduces to showing that the space of constant curvature metrics has trivial homotopy groups, which this paper proves for all 3-manifolds except the 3-sphere and the real projective space. (And the case of the 3-sphere is of course known from Hatcher.)

The new preprint attacks this problem via Ricci flow. The naive idea to prove vanishing of homotopy groups would be to take a sphere in the space of constant curvature metrics, which (by contractibility of the space of all Riemannian metrics) will bound a ball in the space of Riemannian metrics, and then use Ricci flow to deform this ball into a ball in the space of constant curvature metrics. Of course, this does not work that easily because the Ricci flow will develop singularities along the flow. The main part of the paper consists in analyzing these singularities and showing that the constant curvature metric can be extended over neighborhoods of the singularities.

As the authors point out, their proof also provides an alternative argument for Moscow rigidity. Namely, it shows that the space of constant curvature metrics modulo Isom(X) is connected, which together with the known finiteness implies that it consists of one point only.

]]>Cremaschi: A locally hyperbolic 3-manifold that is not hyperbolic, https://arxiv.org/pdf/1711.11568

By the proofs of hyperbolization and tameness, one knows precisely which irreducible 3-manifolds with finitely generated fundamental groups admit hyperbolic metrics: they have to be atoroidal and have infinite fundamental group.

The study of hyperbolic 3-manifolds with infinitely generated fundamental group is more involved. For example there are Cantor sets in the 3-sphere such that the complement admits a complete hyperbolic metric. (J. South, M. Store: A Cantor set with hyperbolic complement, Conformal Geometry and Dynamics 17, 58-67, 2013.)

For irreducible 3-manifolds \(M\) with infinitely generated fundamental group \(\Gamma \) one has two necessary conditions for hyperbolization:

* \(\Gamma \) has to be divisible, i.e., for every \(\gamma\in \Gamma \) there are infinitely many (distinct) \(\alpha\in\Gamma \) such that \(\alpha^n=\gamma\) for some $n$, and

* \(M=\Gamma\backslash {\bf H}^3 \) has to be locally hyperbolic, i.e., every cover with finitely generated fundamental group has to be hyperbolic.

One may wonder whether these conditions are already sufficient or whether there exist 3-manifolds satisfying both conditions without being hyperbolic. (This question is attributed to Agol.)

The new paper shows that such a manifold indeed exists. The example \(M_\infty\) is the thickening of the 2-complex obtained from gluing to an infinite annulus \(A \) countably many copies \(\Sigma_i\) of a genus 2 surface \(\Sigma\) such that each \(\Sigma_i\) is glued to \(S^1\times\left\{i\right\}\). The surfaces \(\Sigma_{\pm i}\) cobound a hyperbolizable manifold, which can be used to prove that \(M_\infty\) is locally hyperbolic. On the other hand, the infinite annulus \(A \) is used to show that \(M_\infty\) cannot be hyperbolic.

]]>Relationships between homology maps of cobordant manifolds

Sullivan conjecture for compact Lie groups

Quasi-isometric groups without common virtual geometric model

Several conjectured identities for polylogarithms

Is there a mathematical explanation for this dice packing phenomenon?

]]>Main event of the meeting were the lectures of Oscar Randal-Williams from Oxford, who discussed work on the cohomology of the mapping class group beyond the stable range.

Some more impressions:

In any case, if you‘d like to see the ceremony, the math part starts at 1:22:30.

The breakthrough prize for 2018 was given to Christopher Hacon and James McKernan for their contributions to the birational classification of higher-dimensional algebraic varieties.

The New Horizons Prizes went to Aaron Naber, Maryna Viazovska, Zhiwei Yun and Wei Zhang.

Stark, Woodhouse: Quasi-isometric groups with no common model geometry,

https://arxiv.org/pdf/1711.05026.pdf

If a group \(\Gamma\) acts geometrically (i.e., properly and cocompactly) on a space $X$, then it is quasi-isometric to $X$ by the Milnor-Svarc theorem. Thus, a standard way of proving two groups to be quasi-isometric is to let them act geometrically on the same space. For example, hyperbolic space in dimensions \(\ge 3\) admits many geometric actions by non-commensurable groups which are thus quasi-isometric.

One may ask for the converse: do quasi-isometries between groups always arise from geometric action on a common model space. (Note that groups always act geometrically on some space, namely their Cayley graph. However, the quasi-isometry between the Cayley graphs of two quasi-isometric groups only serves to promote an action on one Cayley graph to a quasi-action on the other and it is not always the case that a quasi-action can be promoted to an actual action.)

Known examples of quasi-isometric groups not having a common model geometry arise as groups of the form

\(\Gamma\times{\mathbb Z}\mbox{ and }\overline{\Gamma},\)

where \(\Gamma\subset G\) is a uniform lattice in a simple Lie group \(G\) with \(\pi_1G={\mathbb Z}\).

If $G$ has property $T$, then \(\overline{\Gamma}\) also does, while

\(\Gamma\times{\mathbb Z}\) does not, thus the two groups can not be quasi-isometric.

This argument does not apply to \(G=SL(2,{\mathbb R})\), i.e., for surface groups \(\Gamma\), but in this case another argument applies. Two groups \(\Gamma, \Lambda\) are called measure equivalent if they have a (not necessarily cocompact) proper action with finite-volume fundamental domains \(X_\Gamma,X_\Lambda\) on some measure space. Such a measure equivalence defines a coccycle \(\alpha\colon\Gamma\times X_\Lambda\to\Lambda \) by figuring out in which copy of the fundamental domain \(X_\Lambda\) an element \(\gamma x\) with \(\gamma\in\Gamma, x\in X_\Lambda\) lies. In the case of \(\Gamma\times{\mathbb Z}\mbox{ and }\overline{\Gamma}\) with \($\Gamma\subset SL(2,{\mathbb R})\) it turns out that the groups are measure equivalent but the cocycle $\alpha$ is not \(L^p\)-integrable. This, however, would necessarily be the case if the quasi-isometry would come from a cocompact action on a common model space. (Das, Tessera: Integrable measure equivalence and the central extension of surface groups, https://arxiv.org/pdf/1405.2667)

The examples in the new construction are of a different type. They are fundamental groups of amalgams of surfaces, i.e., of spaces obtained from $k$ compact surfaces with one boundary component by identifying all the boundary components. These fundamental groups are known to be quasi-isometric to each other (Malone, Topics in Geometric Group Theory, PhD-thesis, University of Utah, 2011) and for k=4 there is a classification up to commensurability (Stark, Abstract commensurability and quasi-isometry classification of hyperbolic surface group amalgams, Geometriae Dedicata, 186, 39–74, 2017) which in particular implies that there are infinitely many commensurability classes. The new paper shows that the non-commensurable groups can not act geometrically on the same space.

Very roughly, the idea of the proof is the following. If two groups act on the same model space, then they have the same boundary at infinity. From the boundary at infinity one can reconstruct the JSJ-tree of the JSJ-decomposition. The JSJ-tree can be used to promote the model space to a CAT(0) cube complex, on which both groups act geometrically. While the groups need not be finite index in the isometry group of the cube complex, It turns out that there is a certain subgroup of this isometry group in which both groups happen to be finite index subgroups. This proves commensurability.

]]>Michael Rios and David Chester in two videos try to explain the essence of the new work and, for example, the compatibility with Garrett Lise’s E8 theory. The video is remarkable not only for the content but also for the form: not a panel lecture, but a campus walk.

Automorphisms of genus 6 surfaces

Finiteness aspects of Deligne cohomology

Tubular Neighborhood Theorem for C1 Submanifold

When is it easier to work projectively?

Why do we need model categories?

In what respect are univalent foundations “better” than set theory?

]]>Kronheimer. Mrowka: A deformation of instanton homology for webs, https://arxiv.org/pdf/1710.05002.pdf

The four color theorem says that every planar map can be colored by four colors such that adjacent countries correspond to different colors.

It is known from Tait‘s work in the 19th century that the four color theorem is equivalent to the existence of Tait colorings on bridgeless trivalent graphs in the plane. Here a Tait coloring is a coloring of the edges by 3 colors such that each vertex is adjacent to 3 edges of different colors. A bridgeless graph is a connected graph which can not be disconnected by removing only one edge.

In an earlier paper, the authors used a variant of the instanton homology for knots to define an invariant of trivalent graphs in \({\mathbf R}^3\) (which they call „webs“). They prove that it is non-zero for bridgeless graphs, and they conjecture that it is equal to the number of Tait colorings of the graph. (Kronheimer, Mrowka: „Tait colorings, and an instanton homology for webs and foams“, https://arxiv.org/abs/1508.07205)

In the new paper, the authors construct a deformation of their homology theory for which they can actually show that its dimension agrees with the number of Tait colorings. The dimension of the deformed homology is at most that of the original homology theory. Of course, to prove the four color theorem one would need the opposite inequality.

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