Ianagol

@gregeganSF@mathstodon.xyz Vasudha has claimed that the Gassner representation is faithful, which is maybe a less well-known problem, but probably a bigger deal (it implies linearity of the braid groups with a much simpler to describe representation than the Lawrence-Krammer representation proved to be faithful by Bigelow & Krammer). https://arxiv.org/abs/2208.12378

@johncarlosbaez@mathstodon.xyz Quasicrystals were first predicted by Alan Mackay (he knew Penrose and may have been inspired by Penrose tilings) https://en.wikipedia.org/wiki/Alan_Lindsay_Mackay I met his son Bob at a climbing gym in Berkeley. http://bobmackay.com/ Bob met Penrose as a teenager through his father and made some of the first computer designs of Penrose tilings.

@mathowie @mekkaokereke Now it’s not even showing up in the list of charities on Fidelity Charitable.

@johncarlosbaez@mathstodon.xyz This is a rational multiple of the cusp density of the hyperbolic structure on the figure eight knot complement.

@johncarlosbaez@mathstodon.xyz Looks like MacKay’s paper appeared in 1981, and Levine-Steinhardt 1984. The three jointly received the Buckley prize for this in 2010.

@johncarlosbaez@mathstodon.xyz I hadn’t seen this before. Presumably these are the orbits of Z_10 x Z_10 subgroups of the 14400 - order symmetry group. The orientation preserving subgroup is generated by the product of binary icosahedral groups acting on the right and left on the unit quaternions (with center acting trivially). This is then extended by an orientation-reversing element. In the icosahedral group (which is A_5), there is a cyclic element of order 5. The preimage of this element in the binary icosahedral group is order 10. So one gets (Z_10xZ_10)/(-1,-1). Maybe one can adjoin an orientation-reversing element to get the full Z_10 x Z_10? Or maybe it’s some sort of Z/2-extension, like a Klein bottle group? In any case, this suggests that on the Clifford torus, the tetrahedra come in two mirror image groups of 50. This seems to be the case looking at the visualization.

@robinhouston Tesseractivists?

@georgetakei@universeodon.com So he admits it’s a war now?

@hallasurvivor@sunny.garden @henryseg@mathstodon.xyz Now do the Monster group.

@johncarlosbaez@mathstodon.xyz This is coming from Humbert’s volume formula for the quotient of hyperbolic space by a Bianchi group. https://en.wikipedia.org/wiki/Bianchi_group

@johncarlosbaez@mathstodon.xyz The figure eight knot group Γ is generated by the Mobius transformations z-> z+ω, z-> 1/(1+z), acting on the upper half-space model of hyperbolic space, with coordinates (z,t), z in C, t>0, and where ω is a third root of unity (-1+sqrt(-3))/2. There is a cusp H (or horoball) centered at infinity, which is the set of points t> 1. Looking at the image of H under the group Γ, one gets balls tangent to the boundary t=0 and of height 1, centered at the lattice Z+Zω, as well as horoballs of various heights tangent to Q(ω). This is what is shown in the image. They are tangent to H, and so are the largest horoball whose interior is disjoint from all of its images. The volume of the image of H in the quotient is sqrt(3), and the volume of the quotient is 2.0298. This ratio is a rational multiple of their value (and I think I had it flipped, 1/ cusp density, so volume / cusp volume).

@johndcook@mathstodon.xyz should be a sinh to prove.

@standupmaths Next do pi in the sky

@gregeganSF@mathstodon.xyz Hint: the answer is invariant under affine transformation.

@standupmaths Is the New Scientist quote meant to be sarcastic?

@DrewKadel@social.coop These LLMs have made some progress in the art of bullshitting, which is a nontrivial art to master.