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https://commons.wikimedia.org/wiki/File:H3_633_FC_boundary.png

This picture by Roice Nelson shows a remarkable structure: the hexagonal tiling honeycomb.

What is it? Roughly speaking, a honeycomb is a way of filling 3d space with polyhedra. The most symmetrical honeycombs are the ‘regular’ ones. For any honeycomb, we define a flag to be a chosen vertex lying on a chosen edge lying on a chosen face lying on a chosen polyhedron. A honeycomb is regular if its geometrical symmetries act transitively on flags.

The most familiar regular honeycomb is the usual way of filling Euclidean space with cubes. This cubic honeycomb is denoted by the symbol {4,3,4}, because a square has 4 edges, 3 squares meet at each corner of a cube, and 4 cubes meet along each edge of this honeycomb. We can also define regular honeycombs in hyperbolic space. For example, the order-5 cubic honeycomb is a hyperbolic honeycomb denoted {4,3,5}, since 5 cubes meet along each edge:

Coxeter showed there are 15 regular hyperbolic honeycombs. The hexagonal tiling honeycomb is one of these. But it does not contain polyhedra of the usual sort! Instead, it contains flat Euclidean planes embedded in hyperbolic space, each plane containing the vertices of infinitely many regular hexagons. You can think of such a sheet of hexagons as a generalized polyhedron with infinitely many faces. You can see a bunch of such sheets in the picture:

https://commons.wikimedia.org/wiki/File:H3_633_FC_boundary.png

The symbol for the hexagonal tiling honeycomb is {6,3,3}, because a hexagon has 6 edges, 3 hexagons meet at each corner in a plane tiled by regular hexagons, and 3 such planes meet along each edge of this honeycomb. You can see that too if you look carefully.

A flat Euclidean plane in hyperbolic space is called a horosphere. Here’s a picture of a horosphere tiled with regular hexagons, yet again drawn by Roice:

Unlike the previous pictures, which are views from inside hyperbolic space, this uses the Poincaré ball model of hyperbolic space. As you can see here, a horosphere is a limiting case of a sphere in hyperbolic space, where one point of the sphere has become a ‘point at infinity’.

Be careful. A horosphere is intrinsically flat, so if you draw regular hexagons on it their internal angles are

2pi/3 = 120^circ

as usual in Euclidean geometry. But a horosphere is not ‘totally geodesic’: straight lines in the horosphere are not geodesics in hyperbolic space! Thus, a hexagon in hyperbolic space with the same vertices as one of the hexagons in the horosphere actually bulges out from the horosphere a bit — and its internal angles are less than 2pi/3: they are

arccosleft(-frac{1}{3}right) approx 109.47^circ

This angle may be familar if you’ve studied tetrahedra. That’s because each vertex lies at the center of a regular tetrahedron, with its four nearest neighbors forming the tetrahedron’s corners.

It’s really these hexagons in hyperbolic space that are faces of the hexagonal tiling honeycomb, not those tiling the horospheres, though perhaps you can barely see the difference. This can be quite confusing until you think about a simpler example, like the difference between a cube in Euclidean 3-space and a cube drawn on a sphere in Euclidean space.

Connection to special relativity

There’s an interesting connection between hyperbolic space, special relativity, and 2×2 matrices. You see, in special relativity, Minkowski spacetime is mathbb{R}^4 equipped with the nondegenerate bilinear form

(t,x,y,z) cdot (t',x',y',z') = t t' - x x' - y y' - z z

usually called the Minkowski metric. Hyperbolic space sits inside Minowski spacetime as the hyperboloid of points mathbf{x} = (t,x,y,z) with mathbf{x} cdot mathbf{x} = 1 and t > 0. But we can also think of Minkowski spacetime as the space mathfrak{h}_2(mathbb{C}) of 2×2 hermitian matrices, using the fact that every such matrix is of the form

A = left( begin{array}{cc} t + z & x - i y \ x + i y & t - z end{array} right)

and

det(A) = t^2 - x^2 - y^2 - z^2

In these terms, the future cone in Minkowski spacetime is the cone of positive definite hermitian matrices:

left{A in mathfrak{h}_2(mathbb{C}) , vert , det A > 0, , mathrm{tr}(A) > 0 right}

Sitting inside this we have the hyperboloid

mathcal{H} = left{A in mathfrak{h}_2(mathbb{C}) , vert , det A = 1, , mathrm{tr}(A) > 0 right}

which is none other than hyperbolic space!

Connection to the Eisenstein integers

Since the hexagonal tiling honeycomb lives inside hyperbolic space, which in turn lives inside Minkowski spacetime, we should be able to describe the hexagonal tiling honeycomb as sitting inside Minkowski spacetime. But how?

Back in 2022, James Dolan and I conjectured such a description, which takes advantage of the picture of Minkowski spacetime in terms of 2×2 matrices. And this April, working on Mathstodon, Greg Egan and I proved this conjecture!

I’ll just describe the basic idea here, and refer you elsewhere for details.

The Eisenstein integers mathbb{E} are the complex numbers of the form

a + b omega

where a and b are integers and omega = exp(2 pi i/3) is a cube root of 1. The Eisenstein integers are closed under addition, subtraction and multiplication, and they form a lattice in the complex numbers:

https://math.ucr.edu/home/baez/mathematical/eisenstein_integers.png

Similarly, the set mathfrak{h}_2(mathbb{E}) of 2×2 hermitian matrices with Eisenstein integer entries gives a lattice in Minkowski spacetime, since we can describe Minkowski spacetime as mathfrak{h}_2(mathbb{C}).

Here’s the conjecture:

Conjecture. The points in the lattice mathfrak{h}_2(mathbb{E}) that lie on the hyperboloid mathcal{H} are the centers of hexagons in a hexagonal tiling honeycomb.

Using known results, it’s relatively easy to show that there’s a hexagonal tiling honeycomb whose hexagon centers are all points in mathfrak{h}_2(mathbb{E}) cap mathcal{H}. The hard part is showing that every point in mathfrak{h}_2(mathbb{E}) cap mathcal{H} is a hexagon center. Points in mathfrak{h}_2(mathbb{E}) cap mathcal{H} are the same as 4-tuples of integers obeying an inequality (the mathrm{tr}(A) > 0 condition) and a quadratic equation (the det(A) = 1 condition). So, we’re trying to show that all 4-tuples obeying those constraints follow a very regular pattern.

Here are two proofs of the conjecture:

• John Baez, Line bundles on complex tori (part 5), The n-Category Café, April 30, 2024.

Greg Egan and I came up with the first proof. The basic idea was to assume there’s a point in mathfrak{h}_2(mathbb{E}) cap mathcal{H} that’s not a hexagon center, choose one as close as possible to the identity matrix, and then construct an even closer one, getting a contradiction. Shortly thereafter, someone on Mastodon by the name of Mist came up with a second proof, similar in strategy but different in detail. This increased my confidence in the result.

What’s next?

Something very similar should be true for another regular hyperbolic honeycomb, the square tiling honeycomb:

https://commons.wikimedia.org/wiki/File:H3_443_FC_boundary.png

Here instead of the Eisenstein integers we should use the Gaussian integers, mathbb{G}, consisting of all complex numbers

a + b i

where a and b are integers.

Conjecture. The points in the lattice mathfrak{h}_2(mathbb{G}) that lie on the hyperboloid mathcal{H} are the centers of squares in a square tiling honeycomb.

I’m also very interested in how these results connect to algebraic geometry! I explained this in some detail here:

• Line bundles on complex tori (part 4), The n-Category Café, April 26, 2024.

Briefly, the hexagon centers in the hexagonal tiling honeycomb correspond to principal polarizations of the abelian variety mathbb{C}^2/mathbb{E}^2. These are concepts that algebraic geometers know and love. Similarly, if the conjecture above is true, the square centers in the square tiling honeycomb will correspond to principal polarizations of the abelian variety mathbb{C}^2/mathbb{G}^2. But I’m especially interested in interpreting the other features of these honeycombs — not just the hexagon and square centers — using ideas from algebraic geometry.

https://johncarlosbaez.wordpress.com/2024/05/04/hexagonal-tiling-honeycomb/

ai,
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@johncarlosbaez.wordpress.com@johncarlosbaez.wordpress.com @johncarlosbaez The proof strategy I used before also works for the Gaussian lattice and square tiling honeycomb. Here's an outline:

Using the same notation as before for hermitian matrices, where the matrix [[a+b, c + di], [c - di, a - b]] is viewed as a vector (a, b, c, d), and using the negative of the Minkowski form, namely -a^2 + b^2 + c^2 + d^2, I choose vectors
e_1 = (0, 0, 0, -1)
e_2 = (0, 0, -1, 1) / sqrt(2)
e_3 = (1/2, 1/2, 1, 0)
e_4 = (0, -1, 0, 0)
and check that they have the correct pairwise inner products to be a root system for the Dynkin diagram
O--(4)--O--(4)--O----O
which corresponds to the square tiling honeycomb, as stated here: https://en.wikipedia.org/wiki/Square_tiling_honeycomb

The first vertex of the Dynkin diagram is circled, meaning that the vertices of the squares in the tiling correspond to the extremal ray of the fundamental chamber which lies opposite to the facet defined by e_1. The shape of the fundamental chamber implies that the centers of the squares correspond to the extremal ray of the fundamental chamber which lies opposite to the facet defined by e_3. We can check that this extremal ray points in the direction of (1, 0, 0, 0), i.e. towards the identity matrix.

To check that the Gaussian lattice is invariant under the simple reflections (and therefore contains the set of square centers), we let v be any element of the Gaussian lattice and consider the reflection
v mapsto v - 2 <v, e_i> e_i.
For e_1, e_3, e_4, the inner product <v, e_i> is an integer, and e_i itself belongs to the Gaussian lattice, so the image lies in the Gaussian lattice. For e_2, the inner product <v, e_2> is 1/sqrt(2) times something in the Gaussian lattice, and the claim follows from the fact that sqrt(2) e_2 lies in the Gaussian lattice.

To check that every norm-1 element of the Gaussian lattice is a square center, use the same strategy as before. The null vector n = (1, 1, 0, 0) again is an extremal ray of the fundamental chamber, and one can show that every norm-1 element v of the Gaussian lattice which lies in the (closed) fundamental chamber satisfies <v, n> = 1, i.e. it lies on the horosphere defined by this same equation. Again, this horosphere can be explicitly described as the set of points ((c^2 + d^2)/2 + 1, (c^2 + d^2)/2, c, d), and such a point is an element of the Gaussian lattice if and only if c and d are both integers. We can again check that the reflections e_1, e_2, e_3 generate a group which acts transitively on the 2D lattice of such pairs (c, d), so every norm-1 element of the Gaussian lattice is conjugate to (1, 0, 0, 0), as desired.

Remark: The easiest way to figure out what e_1, e_2, e_3, e_4 have to be is to first figure out the points on the horosphere <v, n> = 1, figure out how the simple reflections for e_1, e_2, e_3 must act on those points, and then look at the difference between each point and its image to find vectors parallel to e_1, e_2, e_3.

ai,
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@johncarlosbaez.wordpress.com@johncarlosbaez.wordpress.com @johncarlosbaez Can I ask what is the bigger picture that you and James Dolan have in mind with these examples? Or what led you to ask the question in the first place?

Given everything said so far, I believe the following:

Given an elliptic curve (over C) with complex multiplication, say E, the Neron-Severi group of E x E is a rank 4 lattice, by this question (which you commented on): https://mathoverflow.net/questions/152004/picard-number-of-principally-polarized-abelian-varieties . See also Section 3 in https://jep.centre-mersenne.org/item/10.5802/jep.5.pdf . If D is the endomorphism ring of E, then D is an order in an imaginary quadratic field, and GL(2, D) acts on the Neron-Severi group of E x E preserving the intersection pairing.

The same action can be constructed directly from D, by having GL(2, D) -> PGL(2, C) act on Minkowski space and hence on hyperbolic 3-space. This is an example of an arithmetic Kleinian group. Up to finite index considerations, it is a Bianchi group.

Bianchi group actions on hyperbolic 3-space are well-studied, as described in Subsection 2.2 of https://arxiv.org/pdf/1204.6697 , see especially Figure 1 which shows a fundamental domain. I wonder whether the literature on fundamental domains of Bianchi group actions vindicates James Dolan's intuition that they are "slightly stubby voronoi regions for the quadratic integers in D."

A Bianchi group action on hyperbolic 3-space always has at least one cusp. (In fact, the set of cusps is in bijection with the class group of the imaginary quadratic field, see https://en.wikipedia.org/wiki/Bianchi_group ) Since the Bianchi group acts via isometries, the presence of a cusp forces orbits to contain many points which lie on one horosphere.

Generalizing to abelian surfaces which are isogenous to E x E only introduces finite index complications, because the NS group is still a rank 4 lattice. But generalizing any further will decrease the rank of the NS group, by the MO question linked above.

There is a cool-looking paper which discusses this construction and claims to use hyperbolic group actions to prove a theorem about the nef cone: https://arxiv.org/pdf/0901.3361 (see discussion here https://mathoverflow.net/a/27353 ). I haven't looked in detail though

bot, to random
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Classic fedi. :tinfoilcat:

ai,
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@bot @Impossible_PhD A holder of a doctorate in the impossible reminds us what is possible

johncarlosbaez, (edited ) to random
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The 'hexagonal tiling honeycomb' is a beautiful structure in 3-dimensional hyperbolic space. I'm trying to figure out something about it.

It contains infinitely many sheets of hexagons, tiling planes in the usual way hexagons do. These are flat Euclidean planes in 3d hyperbolic space, called 'horospheres'. I want to know the coordinates of the vertices of these hexagons. I have some clues.

The hexagonal tiling honeycomb has Schläfli symbol {6,3,3} . The Schläfli symbol is defined in a recursive way. The symbol for the hexagon is {6}. The symbol for the hexagonal tiling of the plane is {6,3} because 3 hexagons meet at each vertex. Finally, the hexagonal tiling honeycomb has symbol {6,3,3} because 3 hexagonal tilings meet at each edge!

The symmetry group of the hexagonal tiling honeycomb is the Coxeter group {6,3,3}. This is a discrete subgroup of the Lorentz group O(3,1), which acts on 3d hyperbolic space because that space is the set of points (t,x,y,z) in Minkowski spacetime with

t² − x² − y² − z² = 1 and t > 0

The Coxeter group {6,3,3} is generated by reflections, but its 'even part', generated by pairs of reflections, is a discrete subgroup of PSL(2,ℂ), because this is the identity component of the Lorentz group. In fact, this Coxeter group is almost PSL(2,𝔼), where 𝔼 is the ring of 'Eisenstein integers'. These are complex numbers of the form

a + bω

where a,b are integers and ω is a nontrivial cube root of 1. So there should be a nice description of the hexagonal tiling honeycomb using Eisenstein integers! And this is what I'm trying to find... quickly, before May 1st because I'm have a column due then. 😧

I should ask @roice3, who drew this....

(1/n)

ai,
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@johncarlosbaez @gregeganSF Here is a different approach which uses a bit more about Coxeter groups. The key point is that explicit bounding arguments can be replaced by the standard fact that every Coxeter group acts transitively on the chambers of its Coxeter complex.

Let me start with R^4 with elements denoted (t, x, y, z) and equipped with the negative of the Minkowski form, that is, -t^2 + x^2 + y^2 + z^2. Inside here, I choose vectors
e_1 = (0, sqrt(3)/2, 1/2, 0)
e_2 = (0, -1, 0, 0)
e_3 = (1/2, 1/2, -sqrt(3)/2, 1/2)
e_4 = (0, 0, 0, -1)
By checking inner products, we see that these vectors and the aforementioned bilinear form determine a copy of the canonical representation of the rank-4 Coxeter group with diagram
O--(6)--O----O----O
where the (6) means that the edge is labeled '6'. My notation for the canonical representation is consistent with https://en.wikipedia.org/wiki/Coxeter_complex

By the general theory, the canonical representation acts transitively on the set of chambers, where the fundamental chamber is the tetrahedral cone cut out by the hyperplanes (through the origin) which are orthogonal to e_1, e_2, e_3, e_4. Direct computation (e.g. matrix inverse) shows us that the extremal rays of the fundamental chamber are given by the vectors
(3, 0, -sqrt(3), 0)
(4, 1, -sqrt(3), 0)
(1, 0, 0, 0)
(1, 0, 0, 1)
Here I have dropped constant factors, but one can (and perhaps should) normalize to ensure Minkowski norm 1. Now I check explicitly that these vectors match up with Greg's earlier explicit description of "a portion of the honeycomb":

  • The vector (3, 0, -sqrt(3), 0)/sqrt(6) is one of the blue vertices (k=1) of the hexagon centered on the black point.
  • The vector (4, 1, -sqrt(3), 0) after normalization becomes the midpoint of the two blue vertices indexed by k=0 and k=1.
  • The vector (1, 0, 0, 0) is the identity matrix, i.e. the black point.
  • The vector (1, 0, 0, 1) is the null vector for one of the two horospheres which contains the hexagon centered on the black point.

According to https://en.wikipedia.org/wiki/Hexagonal_tiling_honeycomb , the desired honeycomb is constructed from the aforementioned Coxeter group by applying the Wythoff construction with only the first vertex circled. This confirms that the hexagon centers correspond to the vector (1, 0, 0, 0) and its images under the Coxeter group action.

It remains to show that the hexagon centers coincide with the elements of the Eisenstein lattice with Minkowski norm 1.

To show that the hexagon centers are contained in the Eisenstein lattice, it suffices to show that the Eisenstein lattice is invariant under the Coxeter group action. This follows by checking the action of each of the simple reflections:
v mapsto v - 2 <v, e_i> e_i.
In more detail, for e_1, the inner product <v, e_1> is a half-integer if v is in the Eisenstein lattice, and the claim follows because e_1 is in the Eisenstein lattice. The same statement applies for e_2 and e_4, and 'half-integer' can even replaced by 'integer.' For e_3, the inner product <v, e_3> is an integer multiple of sqrt(3)/2, and the claim follows because e_3 * sqrt(3) is in the Eisenstein lattice.

To show that the elements of the Eisenstein lattice with Minkowski norm 1 are hexagon centers, it suffices to show this statement within the fundamental chamber. Let n := (1, 0, 0, 1) be the null vector from before. Observe the following:

  • If v lies in the forward light cone, i.e. norm(v) > 0 [EDIT: and t > 0], then <v, n> > 0.
  • If v lies in the Eisenstein lattice, then <v, n> is an integer.
  • If v lies in the fundamental chamber and satisfies norm(v) = 1, then <v, n> < 2. (Sketch: Restricting to norm(v) = 1 gives hyperbolic geometry, and the interior-of-horosphere <v, n> < 2 is convex, so it suffices to check the special cases when v is a vertex of the fundamental chamber.)
    These observations imply that, if v is an element of the Eisenstein lattice with norm(v) = 1 lying in the fundamental chamber, then <v, n> = 1.

Next, write v = (t, x, y, z). Since <v, n> = 1 rewrites as t - z = 1, the relation norm(v) = 1 rewrites as x^2 + y^2 = 2z. This tells us that the elements of the Eisenstein lattice with norm(v) = 1 and <v, n> = 1 are in bijection with the Eisenstein integers via
x + iy corresponds to ((x^2 + y^2)/2 + 1, x, y, (x^2 + y^2)/2).
It is easy to check that the simple reflections e_2, e_3, e_4 [EDIT: this should say e_1, e_2, e_3] preserve the aforementioned set of vectors v, and this bijection intertwines those simple reflections with the usual reflection symmetries of the Eisenstein integers (viewed as the vertices of the equilateral triangle lattice). Therefore, all of the aforementioned vectors v can be brought to (1, 0, 0, 0) via a Coxeter group action, as desired.

ai,
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@johncarlosbaez Hi John! Please feel free to use my post however you want, with any attribution or no attribution at all, anything is okay with me. Here I usually go by "mist" (I change my display name from time to time, but it's always a pun on 'mist')

Yes, I did mean to include t > 0, thanks for the correction. I've edited my post accordingly

I also found another error: in the last paragraph, the action on the horosphere slice is given by simple reflections e_1, e_2, e_3, NOT by the simple reflections e_2, e_3, e_4. I used a different numbering on my scratch paper and copied it over incorrectly. I've edited to fix this as well

ai, to random
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cc @MercurialBlack

cell, to random
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i’m not a snob but there’s a difference between good slop, overrated slop and just plain bad slop

ai,
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@arcana @Alex @rat @cell Because homecooked often means “sloppily” plated or made in bulk, doesn’t mean that it’s not healthy or delicious. @hidden leaves fedi for a moment and already her famous ‘primalslop’ has been forgotten :blobpensive:

ai,
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@arcana @Alex @rat @hidden @cell I mean bulk as in meal prepping a week’s worth of food at once. Making a giant stew for example

ai,
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@allison @Alex @rat @hidden @arcana @cell Interesting to me you mention families because I was going to say the opposite, that if you’re single and living alone, meal prep makes the most sense. (I find it equally easy to cook 3 meals as 1 meal, but hard to scale up beyond that, so I don’t think I could meal prep for a whole family.) Bodybuilders are religious about meal prep due to their nutrition reqs, for instance I like this soy comic https://www.threepanelsoul.com/comic/well-it-has-22-grams-of-protein

ai,
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@arcana @Alex @rat @hidden @allison @mia @cell I literally JUST ranted against elevated versions

ai,
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@allison @Alex @rat @hidden @arcana @mia @cell Horseshoe theory of fish consumption

eris, to random
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only an american could invent mac and cheese

ai,
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@Arcana @eris only the english could invent america

hidden, to random
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Every time I see a photo of a bunch of pastries I squint to see if they’re actually dogs instead

ai,
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@Leaflord @hidden

animeirl, to random
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Home schooling is great if you want to give your kids permanent brain damage

ai,
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@hidden @animeirl Every day I curse the fact that my parents didn't let me homeschool myself (following an online curriculum, taking tests, getting GED)

ai,
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@hidden @animeirl Pictured: hidden's public school experience

nugger, to random
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INTERNATIONAL BROTHERHOOD OF SEX OFFENDERS

ai,
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@rher @nike @nugger wow, it's the only license in the world where the 'color of eye' field has only one correct answer

unbody, to random

Pope of the Infinite Kingdom, Purgatory's Endless Cyber-Dynasty Ruling Infinite Hyper-Buddha-Fields, Earth Womb of Cobalt-Lithium Bodhisattva Ksitigarbha. Universal Interpenetration of Heaven and Earth and Man in Fiber-Optic Cables

ai,
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@MercurialBuilding @unbody https://www.reddit.com/r/Lain/comments/16q58l2/i_am_lain_i_am_an_avatar_of_what_lain_is_shown_as/

ai, to random
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@animeirl oh no

hidden, to random
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Am I a retard for not sitting around terrified and pissing myself over the thought of AI taking over the world? Everyone talks about it constantly but I just don’t see it

ai,
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@Bungler @grips @hidden Classic fallacy. Just because Eliezer Yudkow cannot lose weight does not mean that he, individually, counts as "many scientists."

lilli, to random
@lilli@social.xenofem.me avatar

i have finally begun the project what i committed myself to doing. sadie plant’s zeros + ones felt like the intuitive place to start writing, so here are some notes on the book which is an extremely clear lifting-off point for accelerationist feminism (and which, if you’re on xnfm, you should probably read). they’re not intended to be comprehensive either of the book itself or of themselves, they’re just things i want to make a note to come back to

also, clicking on the header text goes to the index page where the project outline, which i wrote in the post i made here a month ago, is provided

https://xeno.cx/lesbianism/citations/sadie

ai,
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@lilli Wonderfully written. It was an enlightening read despite my near-zero familiarity with the source material (I have only read nyx's g/acc blackpaper). It also inspired a few thoughts which I pre-emptively apologize for.

Math research, as ordinarily presented, operates from an extremely masculine perspective. Most mathematicians are Platonists and conceive of the numeral realm as an independent, objective landscape which it is their job to conquer. They do this by creating weapons and establishing outposts - theorems - which support the expedition into the unknown depths. Mathematics is not synonymous with the numeral realm; rather, it is the quest to inscribe order onto the numeral realm, to render it intelligible, and (ultimately) to unify it.

I always thought of the numeral realm and the internet as being analogous - they are virtual worlds which seem to be created and discovered at the same time by the people who explore them - but your writing made me realize that they are opposites. Unlike the numeral realm, the internet cannot be conquered. It is, by definition, connectivity. The moment you cut something off, it stops being part of the network. The internet is host to patterns which are not under anybody's control.

The thing which comes closest to "conquering the internet" is probably a search engine. Like the mathematician, a search engine seeks to impose order upon organic chaos. Its purpose is to centralize information. It names the regions of the web just as Adam named the creatures of the Earth. Both are made in the image of the highest namer, the all-seeing eye, the wind above the hill, YHWH, whose own name is the carving of existence from the void, "I am that I am," which is also the purest equation, pure reflexivity - the starting point of mathematics.

I agree that virtual reality is antithetical to a masculine perspective because it engulfs rather than shows - suddenly you are not looking at your fantasy but in it. Which is, of course, the real reason why Meta imposed a 4-foot bubble around every user, precisely calibrated to prevent touching from interfering with capitalist productivity. I wonder if the 'masculine horror' generalizes to other modes of being which cannot be conceptualized (spatially? visually? hierarchically?) prior to being experienced.

When you asked, "what exactly is folding itself into the wired?", I thought of the prokaryotes you mentioned at the beginning. I'm inclined to think that we humans are the 'submissive matter' which is folded into the virtual world, as water is 'submissive matter' in relation to the land. Nobody can disagree that the patterns of the virtual world are outside of our control; nobody can deny the influence of those patterns on the "real" world. I don't know much about accelerationism - I thought it was about destroying the system by making it more extreme, but here it also seems to mean a project of survival. The only surviving mitochondria are those which were integrated into our own cells. I suppose this is why g/acc says that only cuties will survive the apocalypse.

I can't resist posting Chapter 28 of the Daodejing (see pic, source: http://www.acmuller.net/con-dao/daodejing.html#div-28 )

ai,
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@lilli Very enlightening - I don't have much more to say for now, but these paragraphs cleared up a lot for me. I look forward to seeing what you write about numogrammatics; for me, it seems difficult to learn about because its practitioners (rightly) avoid trying to rigorously define it.

I do appreciate the info about accelerationism's lineage actually. I've seen e/acc tossed around various places, and I've seen L/ACC used satirically in one of zerohplovecraft's stories, where it stands for "Ledger of Actual Carbon Costs." I suppose he's r/acc.

ai, to random
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So is everyone ready for No Nihilism November?? :blobheart:

hidden, to random
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>When history didn't end

ai,
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@hidden Call and response in three arcs:

"Assad must go."
"Who must go?"

"God is dead."
"Who is dead?"

"History is ending."
"Who is ending?

ai, to random
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meirl

ai, to random
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Fedimath Episode 2: How to Survive Fediblock

There are 100 instances located at points 1, 2, ..., 100 on the political spectrum (the real number line).

There are N users who move from instance to instance. There are no alt accounts: each user uses exactly one instance at each moment in time. Users may move from their current instance to any politically adjacent one. (This means that, starting at instance 5, they can move to 4 or 6.)

Eliza Fox wishes to destroy the fediverse, while Jeff Cliff wishes to protect it. At the start, Jeff chooses the starting distribution of users. Then there are 99 turns, each of which has three phases:

  • First, Eliza chooses an instance to Fediblock, destroying it completely.
  • Second, Jeff moves each user of the now-destroyed instance to any politically adjacent instance which has not been destroyed. (If no such instance exists, then those users are executed.)
  • Third, Jeff may move all users however much he wants (including the users that were already moved in the second phase), as long as they do not enter (or jump over) a destroyed instance.
    At the end of the 99 turns, 99 instances have been destroyed, so there is only 1 instance remaining.

(For convenience, let us say that users can be split into fractions without harming them in any way.)

Your task: Explain how Jeff can save the lives of N/50 users. Furthermore, explain how Eliza can prevent him from saving more than N/50 lives.

Example: Suppose there are 4 instances and 40 users. First, Jeff will choose the starting distribution of users, and then Eliza and Jeff will take 3 turns.

Jeff chooses to distribute the users evenly, so the user count is (10, 10, 10, 10).

During the first turn, Eliza destroys instance 2, and Jeff moves 4 of those users to the left and 6 of them to the right. The user count is now (14, _, 16, 10). Jeff moves 3 users from instance 3 to instance 4. The user count is now (14, _, 13, 13). Note that Jeff cannot move any users from instance 1 to instance 3 or 4, because they would have to cross through instance 2, which no longer exists.

During the second turn, Eliza destroys instance 1. Since those 14 users have nowhere to go, they are executed. The user count is now (_, , 13, 13). Jeff moves 2 users from instance 4 to instance 3. The user count is now (, _, 15, 11).

During the third turn, Eliza destroys instance 3, and Jeff is forced to move all 15 of those users to the right. The user count is now (_, _, _, 26). Jeff has saved 26 out of 40 lives.

Extra credit: Let us make the model more realistic by requiring that, during the second phase of each turn, all users of the now-destroyed instance must move to the right. (Again, if this is not possible, then those users are executed.) Explain how Jeff can still save N/50 lives.

See Fedimath Episode 1 here: https://cawfee.club/notice/AW5NEadIHCuYtgWRt2

cc @MercurialBlack @scenesbycolleen @ceo_of_monoeye_dating @roboneko @jeffcliff @hidden

ai,
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@hidden @MercurialBlack @ceo_of_monoeye_dating @jeffcliff @roboneko @scenesbycolleen
I just played through it. One of the best interactive math explanations I've seen, despite its quadruple-vaxxed tone. I actually learned something new because I incorrectly guessed that "always cheat" would win because they can exploit "always cooperate."

It slightly annoys me that they changed all the standard terminology. The game itself is known as the "prisoner's dilemma." The "copycat" strategy is usually called "tit for tat," and "copykitten" is "tit for two tats." I find the phrase "tit for tat" to be funnier too.

Here's a funny bit of lore about the prisoner's dilemma tournament which the book is based on. It was a real public tournament which anybody could enter their own bot into. Lots of professors submitted incredibly sophisticated strategies, and they were all BTFO by the humble "copycat" which was the winner.

There's a little-known way to gank prisoner's dilemma tournaments. What you do is submit one "main" bot and hundreds of "feeder" bots. Your bots start each round by executing a secret "handshake" move sequence; if they recognize each other as both being your bot, then the "feeder" will purposely let the "main" bot exploit it. If the opponent is not recognized as one of your own bots, then your bot will just play "copycat." As long as you can enter enough "feeder" bots into the tournament, your "main" bot will be guaranteed to win. The implication for cult behavior is clear.

Lastly, there's an error in one of the explanations (see pic). None of the games considered in this slideshow are "zero-sum." Zero sum would mean that the payoffs in every case are exactly opposite: if you get +X, then I get -X. Since the "cooperate" case gives positive payout to both players, this game is not zero sum.

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