## Large Countable Ordinals (Part 1)

29 June, 2016

I love the infinite.

It may not exist in the physical world, but we can set up rules to think about it in consistent ways, and then it’s a helpful concept. The reason is that infinity can be simpler to think about than a very large finite number.

Finding rules to work with the infinite is one of the great triumphs of mathematics. Cantor’s realization that there are different sizes of infinity is truly wondrous—and by now, it’s part of the everyday bread and butter of mathematics.

Trying to create a notation for these different infinities is very challenging. It’s not a fair challenge, because there are more infinities than expressions we can write down in any given alphabet! But if we seek a notation for countable ordinals, the challenge becomes more fair.

It’s still incredibly frustrating. No matter what notation we use it fizzles out too soon… making us wish we’d invented a more general notation. But this process of ‘fizzling out’ is fascinating to me. There’s something profound about it. So, I would like to tell you about this.

Today I’ll start with a warmup. Cantor invented a notation for ordinals that works great for ordinals less than a certain ordinal called ε0. Next time I’ll go further, and bring in the Veblen hierarchy! The single-variable Veblen functions let us describe all ordinals below a big guy called the ‘Feferman–Schütte ordinal’.

If I’m still feeling energetic after that, I will write another post that introduces the multi-variable Veblen functions. These get us all the ordinals below the ‘small Veblen ordinal’. And if I’m feeling ultra-energetic, I’ll talk about Veblen functions with infinitely many variables. These let us describe ordinals below the ‘large Veblen ordinal’.

But all this is really just the beginning of a longer story. That’s how infinity works: the story never ends!

To describe countable ordinals beyond the large Veblen ordinal, most people switch to an entirely different set of ideas, called ‘ordinal collapsing functions’. I doubt I’ll have the energy to tackle these. Maybe I will after a year-long break. My interest in the infinite doesn’t seem to be waning. It’s a decadent hobby, but I figure: some middle-aged men buy fancy red sports cars and drive them really fast; studying notions of infinity is even more intense, but it’s environmentally friendly.

I can even imagine writing a book about the infinite. Maybe these posts will become part of that book. But one step at a time…

### Cardinals versus ordinals

Cantor invented two kinds of infinities: cardinals and ordinals. Cardinals say how big sets are. Two sets can be put into 1-1 correspondence iff they have the same number of elements—where this kind of ‘number’ is a cardinal. You may have heard about cardinals like aleph-nought (the number of integers), 2 to power aleph-nought (the number of real numbers), and so on. You may have even heard rumors of much bigger cardinals, like ‘inaccessible cardinals’ or ‘super-huge cardinals’. All this is tremendously fun, and I recommend starting here:

• Frank R. Drake, Set Theory, an Introduction to Large Cardinals, North-Holland, 1974.

There are other books that go much further, but as a beginner, I found this to be the most fun.

But I don’t want to talk about cardinals! I want to talk about ordinals.

Ordinals say how big ‘well-ordered’ sets are. A set is well-ordered if it comes with a relation ≤ obeying the usual rules:

Transitivity: if x ≤ y and y ≤ z then x ≤ z

Reflexivity: x ≤ x

Antisymmetry: if x ≤ y and y ≤ x then x = y

and one more rule: every nonempty subset has a smallest element!

For example, the empty set

$\{\}$

is well-ordered in a trivial sort of way, and the corresponding ordinal is called

$0$

Similarly, any set with just one element, like this:

$\{0\}$

is well-ordered in a trivial sort of way, and the corresponding ordinal is called

$1$

Similarly, any set with two elements, like this:

$\{0,1\}$

becomes well-ordered as soon as we decree which element is bigger; the obvious choice is to say 0 < 1. The corresponding ordinal is called

$2$

Similarly, any set with three elements, like this:

$\{0,1,2\}$

becomes well-ordered as soon as we linearly order it; the obvious choice here is to say 0 < 1 < 2. The corresponding ordinal is called

$3$

Perhaps you’re getting the pattern — you’ve probably seen these particular ordinals before, maybe sometime in grade school. They’re called finite ordinals, or "natural numbers".

But there’s a cute trick they probably didn’t teach you then: we can define each ordinal to be the set of all ordinals less than it:

$0 = \{\}$ (since no ordinal is less than 0)
$1 = \{0\}$ (since only 0 is less than 1)
$2 = \{0,1\}$ (since 0 and 1 are less than 2)
$3 = \{0,1,2\}$ (since 0, 1 and 2 are less than 3)

and so on. It’s nice because now each ordinal is a well-ordered set of the size that ordinal stands for. And, we can define one ordinal to be "less than or equal" to another precisely when its a subset of the other.

### Infinite ordinals

What comes after all the finite ordinals? Well, the set of all finite ordinals is itself well-ordered:

$\{0,1,2,3,\dots \}$

So, there’s an ordinal corresponding to this — and it’s the first infinite ordinal. It’s usually called $\omega,$ pronounced ‘omega’. Using the cute trick I mentioned, we can actually define

$\omega = \{0,1,2,3,\dots\}$

What comes after this? Well, it turns out there’s a well-ordered set

$\{0,1,2,3,\dots,\omega\}$

containing the finite ordinals together with $\omega,$ with the obvious notion of "less than": $\omega$ is bigger than the rest. Corresponding to this set there’s an ordinal called

$\omega+1$

As usual, we can simply define

$\omega+1 = \{0,1,2,3,\dots,\omega\}$

At this point you could be confused if you know about cardinals, so let me throw in a word of reassurance. The sets $\omega$ and $\omega+1$ have the same cardinality: they are both countable. In other words, you can find a 1-1 and onto function between these sets. But $\omega$ and $\omega+1$ are different as ordinals, since you can’t find a 1-1 and onto function between them that preserves the ordering. This is easy to see, since $\omega+1$ has a biggest element while $\omega$ does not.

Indeed, all the ordinals in this series of posts will be countable! So for the infinite ones, you can imagine that all I’m doing is taking your favorite countable set and well-ordering it in ever more sneaky ways.

Okay, so we got to $\omega + 1.$ What comes next? Well, not surprisingly, it’s

$\omega+2 = \{0,1,2,3,\dots,\omega,\omega+1\}$

Then comes

$\omega+3, \omega+4, \omega+5,\dots$

and so on. You get the idea.

I haven’t really defined ordinal addition in general. I’m trying to keep things fun, not like a textbook. But you can read about it here:

• Wikipedia, Ordinal arithmetic: addition.

The main surprise is that ordinal addition is not commutative. We’ve seen that $\omega + 1 \ne \omega,$ since

$\omega + 1 = \{1, 2, 3, \dots, \omega \}$

is an infinite list of things… and then one more thing that comes after all those!. But $1 + \omega = \omega,$ because one thing followed by a list of infinitely many more is just a list of infinitely many things.

With ordinals, it’s not just about quantity: the order matters!

### ω+ω and beyond

Okay, so we’ve seen these ordinals:

$1, 2, 3, \dots, \omega, \omega + 1, \omega + 2, \omega+3, \dots$

What next?

Well, the ordinal after all these is called $\omega+\omega.$ People often call it "omega times 2" or $\omega 2$ for short. So,

$\omega 2 = \{0,1,2,3,\dots,\omega,\omega+1,\omega+2,\omega+3,\dots.\}$

It would be fun to have a book with $\omega$ pages, each page half as thick as the previous page. You can tell a nice long story with an $\omega$-sized book. I think you can imagine this. And if you put one such book next to another, that’s a nice picture of $\omega 2.$

It’s worth noting that $\omega 2$ is not the same as $2 \omega.$ We have

$\omega 2 = \omega + \omega$

while

$2 \omega = 2 + 2 + 2 + \cdots$

where we add $\omega$ of these terms. But

$2 + 2 + 2 + \cdots = (1 + 1) + (1 + 1) + (1 + 1) \dots = \omega$

so

$2 \omega = \omega$

This is not a proof, because I haven’t given you the official definition of how to add ordinals. You can find the definition here:

• Wikipedia, Ordinal arithmetic: multiplication.

Using this definition you can prove that what I’m saying is true. Nonetheless, I hope you see why what I’m saying might make sense. Like ordinal addition, ordinal multiplication is not commutative! If you don’t like this, you should study cardinals instead.

What next? Well, then comes

$\omega 2 + 1, \omega 2 + 2,\dots$

and so on. But you probably have the hang of this already, so we can skip right ahead to $\omega 3.$

In fact, you’re probably ready to skip right ahead to $\omega 4,$ and $\omega 5,$ and so on.

In fact, I bet now you’re ready to skip all the way to "omega times omega", or $\omega^2$ for short:

$\omega^2 = \{0,1,2\dots\omega,\omega+1,\omega+2,\dots ,\omega 2,\omega 2+1,\omega 2+2,\dots\}$

Suppose you had an encyclopedia with $\omega$ volumes, each one being a book with $\omega$ pages. If each book is twice as thin as one before, you’ll have $\omega^2$ pages — and it can still fit in one bookshelf! Here’s the idea:

What comes next? Well, we have

$\omega^2+1, \omega^2+2, \dots$

and so on, and after all these come

$\omega^2+\omega, \omega^2+\omega+1, \omega^2+\omega+2, \dots$

and so on — and eventually

$\omega^2 + \omega^2 = \omega^2 2$

and then a bunch more, and then

$\omega^2 3$

and then a bunch more, and then

$\omega^2 4$

and then a bunch more, and more, and eventually

$\omega^2 \omega = \omega^3$

You can probably imagine a bookcase containing $\omega$ encyclopedias, each with $\omega$ volumes, each with $\omega$ pages, for a total of $\omega^3$ pages. That’s $\omega^3.$

### ωω

I’ve been skipping more and more steps to keep you from getting bored. I know you have plenty to do and can’t spend an infinite amount of time reading this, even if the subject is infinity.

So if you don’t mind me just mentioning some of the high points, there are guys like $\omega^4$ and $\omega^5$ and so on, and after all these comes

$\omega^\omega$

Let’s try to we imagine this! First, imagine a book with $\omega$ pages. Then imagine an encyclopedia of books like this, with $\omega$ volumes. Then imagine a bookcase containing $\omega$ encyclopedias like this. Then imagine a room containing $\omega$ bookcases like this. Then imagine a floor with library with $\omega$ rooms like this. Then imagine a library with $\omega$ floors like this. Then imagine a city with $\omega$ libraries like this. And so on, ad infinitum.

You have to be a bit careful here, or you’ll be imagining an uncountable number of pages. To name a particular page in this universe, you have to say something like this:

the 23rd page of the 107th book of the 20th encyclopedia in the 7th bookcase in 0th room on the 1000th floor of the 973rd library in the 6th city on the 0th continent on the 0th planet in the 0th solar system in the…

But it’s crucial that after some finite point you keep saying “the 0th”. Without that restriction, there would be uncountably many pages! This is just one of the rules for how ordinal exponentiation works. For the details, read:

• Wikipedia, Ordinal arithmetic: exponentiation.

As they say,

But for infinite exponents, the definition may not be obvious.

### Ordinals up to ε0

Okay, so we’ve reached $\omega^\omega.$ Now what?

Well, then comes $\omega^\omega + 1,$ and so on, but I’m sure that’s boring by now. And then come ordinals like

$\omega^\omega 2,\dots, \omega^\omega 3, \dots, \omega^\omega 4, \dots$

$\omega^\omega \omega = \omega^{\omega + 1}$

Then eventually come ordinals like

$\omega^\omega \omega^2 , \dots, \omega^\omega \omega^3, \dots, \omega^\omega \omega^4, \dots$

and so on, leading up to

$\omega^\omega \omega^\omega = \omega^{\omega + \omega} = \omega^{\omega 2}$

This actually reminds me of something that happened driving across South Dakota one summer with a friend of mine. We were in college, so we had the summer off, so we drive across the country. We drove across South Dakota all the way from the eastern border to the west on Interstate 90.

This state is huge — about 600 kilometers across, and most of it is really flat, so the drive was really boring. We kept seeing signs for a bunch of tourist attractions on the western edge of the state, like the Badlands and Mt. Rushmore — a mountain that they carved to look like faces of presidents, just to give people some reason to keep driving.

Anyway, I’ll tell you the rest of the story later — I see some more ordinals coming up:

$\omega^{\omega 3},\dots \omega^{\omega 4},\dots \omega^{\omega 5},\dots$

We’re really whizzing along now just to keep from getting bored — just like my friend and I did in South Dakota. You might fondly imagine that we had fun trading stories and jokes, like they do in road movies. But we were driving all the way from Princeton to my friend Chip’s cabin in California. By the time we got to South Dakota, we were all out of stories and jokes.

Hey, look! It’s

$\omega^{\omega \omega}= \omega^{\omega^2}$

That was cool. Then comes

$\omega^{\omega^3}, \dots \omega^{\omega^4}, \dots \omega^{\omega^5}, \dots$

and so on.

Anyway, back to my story. For the first half of our half of our trip across the state, we kept seeing signs for something called the South Dakota Tractor Museum.

Oh, wait, here’s an interesting ordinal — let’s slow down and take a look:

$\omega^{\omega^\omega}$

I like that! Okay, let’s keep driving. Here comes

$\omega^{\omega^\omega} + 1, \omega^{\omega^\omega} + 2, \dots$

and then

$\omega^{\omega^\omega} + \omega, \dots, \omega^{\omega^\omega} + \omega 2, \dots, \omega^{\omega^\omega} + \omega 3, \dots$

and then

$\omega^{\omega^\omega} + \omega^2, \dots, \omega^{\omega^\omega} + \omega^3, \dots$

and eventually

$\omega^{\omega^\omega} + \omega^\omega$

and eventually

$\omega^{\omega^\omega} + \omega^{\omega^\omega} = \omega^{\omega^\omega} 2$

and then

$\omega^{\omega^\omega} 3, \dots, \omega^{\omega^\omega} 4, \dots, \omega^{\omega^\omega} 5, \dots$

and eventually

$\omega^{\omega^\omega} \omega = \omega^{\omega^{\omega + 1}}$

and then

$\omega^{\omega^{\omega + 2}}, \dots, \omega^{\omega^{\omega + 3}}, \dots, \omega^{\omega^{\omega + 4}}, \dots$

This is pretty boring; we’re already going infinitely fast, but we’re still just picking up speed, and it’ll take a while before we reach something interesting.

Anyway, we started getting really curious about this South Dakota Tractor Museum — it sounded sort of funny. It took 250 kilometers of driving before we passed it. We wouldn’t normally care about a tractor museum, but there was really nothing else to think about while we were driving. The only thing to see were fields of grain, and these signs, which kept building up the suspense, saying things like

ONLY 100 MILES TO THE SOUTH DAKOTA TRACTOR MUSEUM!

We’re zipping along really fast now:

$\omega^{\omega^{\omega^\omega}}, \dots, \omega^{\omega^{\omega^{\omega^\omega}}},\dots , \omega^{\omega^{\omega^{\omega^{\omega^{\omega}}}}},\dots$

What comes after all these?

At this point we need to stop for gas. Our notation for ordinals just ran out!

The ordinals don’t stop; it’s just our notation that fizzled out. The set of all ordinals listed up to now — including all the ones we zipped past — is a well-ordered set called

$\epsilon_0$

or "epsilon-nought". This has the amazing property that

$\epsilon_0 = \omega^{\epsilon_0}$

And it’s the smallest ordinal with this property!

### Cantor normal form

I’ll tell you the rest of my road story later. For now let me conclude with a bit of math.

There’s a nice notation for all ordinals less than $\epsilon_0,$ called ‘Cantor normal form’. We’ve been seeing lots of examples. Here is a typical ordinal in Cantor normal form:

$\omega^{\omega^{\omega^{\omega+\omega+1}}} \; + \; \omega^{\omega^\omega+\omega^\omega} \; + \; \omega^\omega \;+\; \omega + \omega + 1 + 1 + 1$

The idea is that you write it out using just + and exponentials and 1 and $\omega.$

Here is the theorem that justifies Cantor normal form:

Theorem. Every ordinal $\alpha$ can be uniquely written as

$\alpha = \omega^{\beta_1} c_1 + \omega^{\beta_2}c_2 + \cdots + \omega^{\beta_k}c_k$

where $k$ is a natural number, $c_1, c_2, \ldots, c_k$ are positive integers, and $\beta_1 > \beta_2 > \cdots > \beta_k \geq 0$ are ordinals.

It’s like writing ordinals in base $\omega.$

Note that every ordinal can be written this way! So why did I say that Cantor normal form is nice notation for ordinals less than $\epsilon_0$? Here’s the problem: the Cantor normal form of $\epsilon_0$ is

$\epsilon_0 = \omega^{\epsilon_0}$

So, when we hit $\epsilon_0,$ the exponents $\beta_1 ,\beta_2, \dots, \beta_k$ can be as big as the ordinal $\alpha$ we’re trying to describe! So, while the Cantor normal form still exists for ordinals $\geq \epsilon_0,$ it doesn’t give a good notation for them unless we already have some notation for ordinals this big!

This is what I mean by a notation ‘fizzling out’. We’ll keep seeing this problem in the posts to come.

But for an ordinal $\alpha$ less than $\epsilon_0,$ something nice happens. In this case, when we write

$\alpha = \omega^{\beta_1} c_1 + \omega^{\beta_2}c_2 + \cdots + \omega^{\beta_k}c_k$

all the exponents $\beta_1, \beta_2, \dots, \beta_k$ are less than $\alpha.$ So we can go ahead and write them in Cantor normal form, and so on… and because ordinals are well-ordered, this process ends after finitely many steps.

So, Cantor normal form gives a nice way to write any ordinal less than $\epsilon_0$ using finitely many symbols! If we abbreviate $\omega^0$ as $1,$ and write multiplication by positive integers in terms of addition, we get expressions like this:

$\omega^{\omega^{\omega^{\omega^{1 + 1} +\omega+1}}} \; + \; \omega^{\omega^\omega+\omega^\omega} \; + \; \omega^{\omega+1+1} \;+\; \omega + 1$

They look like trees. Even better, you can write a computer program that does ordinal arithmetic for ordinals of this form: you can add, multiply, and exponentiate them, and tell when one is less than another.

So, there’s really no reason to be scared of $\epsilon_0.$ Remember, each ordinal is just the set of all smaller ordinals. So you can think of $\epsilon_0$ as the set of tree-shaped expressions like the one above, with a particular rule for saying when one is less than another. It’s a perfectly reasonable entity. For some real excitement, we’ll need to move on to larger ordinals. We’ll do that next time.

For more, see:

• Wikipedia, Cantor normal form.

## Azimuth News (Part 5)

11 June, 2016

I’ve been rather quiet about Azimuth projects lately, because I’ve been too busy actually working on them. Here’s some of what’s happening:

Jason Erbele is finishing his thesis, entitled Categories in Control: Applied PROPs. He successfully gave his thesis defense on Wednesday June 8th, but he needs to polish it up some more. Building on the material in our paper “Categories in control”, he’s defined a category where the morphisms are signal flow diagrams. But interestingly, not all the diagrams you can draw are actually considered useful in control theory! So he’s also found a subcategory where the morphisms are the ‘good’ signal flow diagrams, the ones control theorists like. For these he studies familiar concepts like controllability and observability. When his thesis is done I’ll announce it here.

Brendan Fong is also finishing his thesis, called The Algebra of Open and Interconnected Systems. Brendan has already created a powerful formalism for studying open systems: the decorated cospan formalism. We’ve applied it to two examples: electrical circuits and Markov processes. Lately he’s been developing the formalism further, and this will appear in his thesis. Again, I’ll talk about it when he’s done!

Blake Pollard and I are writing a paper called “A compositional framework for open chemical reaction networks”. Here we take our work on Markov processes and throw in two new ingredients: dynamics and nonlinearity. Of course Markov processes have a dynamics, but in our previous paper when we ‘black-boxed’ them to study their external behaviour, we got a relation between flows and populations in equilibrium. Now we explain how to handle nonequilibrium situations as well.

Brandon Coya, Franciscus Rebro and I are writing a paper that might be called “The algebra of networks”. I’m not completely sure of the title, nor who the authors will be: Brendan Fong may also be a coauthor. But the paper explores the technology of PROPs as a tool for describing networks. As an application, we’ll give a new shorter proof of the functoriality of black-boxing for electrical circuits. This new proof also applies to nonlinear circuits. I’m really excited about how the theory of PROPs, first introduced in algebraic topology, is catching fire with all the new applications to network theory.

I expect all these projects to be done by the end of the summer. Near the end of June I’ll go to the Centre for Quantum Technologies, in Singapore. This will be my last summer there. My main job will be to finish up the two papers that I’m supposed to be writing.

There’s another paper that’s already done:

Kenny Courser has written a paper “A bicategory of decorated cospans“, pushing Brendan’s framework from categories to bicategories. I’ll explain this very soon here on this blog! One goal is to understand things like the coarse-graining of open systems: that is, the process of replacing a detailed description by a less detailed description. Since we treat open systems as morphisms, coarse-graining is something that goes from one morphism to another, so it’s naturally treated as a 2-morphism in a bicategory.

So, I’ve got a lot of new ideas to explain here, and I’ll start soon! I also want to get deeper into systems biology.

In the fall I’ve got a couple of short trips lined up:

• Monday November 14 – Friday November 18, 2016 – I’ve been invited by Yoav Kallus to visit the Santa Fe Institute. From the 16th to 18th I’ll attend a workshop on Statistical Physics, Information Processing and Biology.

• Monday December 5 – Friday December 9 – I’ve been invited to Berkeley for a workshop on Compositionality at the Simons Institute for the Theory of Computing, organized by Samson Abramsky, Lucien Hardy, and Michael Mislove. ‘Compositionality’ is a name for how you describe the behavior of a big complicated system in terms of the behaviors of its parts, so this is closely connected to my dream of studying open systems by treating them as morphisms that can be composed to form bigger open systems.

Here’s the announcement:

The compositional description of complex objects is a fundamental feature of the logical structure of computation. The use of logical languages in database theory and in algorithmic and finite model theory provides a basic level of compositionality, but establishing systematic relationships between compositional descriptions and complexity remains elusive. Compositional models of probabilistic systems and languages have been developed, but inferring probabilistic properties of systems in a compositional fashion is an important challenge. In quantum computation, the phenomenon of entanglement poses a challenge at a fundamental level to the scope of compositional descriptions. At the same time, compositionally has been proposed as a fundamental principle for the development of physical theories. This workshop will focus on the common structures and methods centered on compositionality that run through all these areas.

I’ll say more about both these workshops when they take place.

## Very Long Proofs

28 May, 2016

In the 1980s, the mathematician Ronald Graham asked if it’s possible to color each positive integer either red or blue, so that no triple of integers $a,b,c$ obeying Pythagoras’ famous equation:

$a^2 + b^2 = c^2$

all have the same color. He offered a prize of \$100.

Now it’s been solved! The answer is no. You can do it for numbers up to 7824, and a solution is shown in this picture. But you can’t do it for numbers up to 7825.

To prove this, you could try all the ways of coloring these numbers and show that nothing works. Unfortunately that would require trying

3 628 407 622 680 653 855 043 364 707 128 616 108 257 615 873 380 491 654 672 530 751 098 578 199 115 261 452 571 373 352 277 580 182 512 704 196 704 700 964 418 214 007 274 963 650 268 320 833 348 358 055 727 804 748 748 967 798 143 944 388 089 113 386 055 677 702 185 975 201 206 538 492 976 737 189 116 792 750 750 283 863 541 981 894 609 646 155 018 176 099 812 920 819 928 564 304 241 881 419 294 737 371 051 103 347 331 571 936 595 489 437 811 657 956 513 586 177 418 898 046 973 204 724 260 409 472 142 274 035 658 308 994 441 030 207 341 876 595 402 406 132 471 499 889 421 272 469 466 743 202 089 120 267 254 720 539 682 163 304 267 299 158 378 822 985 523 936 240 090 542 261 895 398 063 218 866 065 556 920 106 107 895 261 677 168 544 299 103 259 221 237 129 781 775 846 127 529 160 382 322 984 799 874 720 389 723 262 131 960 763 480 055 015 082 441 821 085 319 372 482 391 253 730 679 304 024 117 656 777 104 250 811 316 994 036 885 016 048 251 200 639 797 871 184 847 323 365 327 890 924 193 402 500 160 273 667 451 747 479 728 733 677 070 215 164 678 820 411 258 921 014 893 185 210 250 670 250 411 512 184 115 164 962 089 724 089 514 186 480 233 860 912 060 039 568 930 065 326 456 428 286 693 446 250 498 886 166 303 662 106 974 996 363 841 314 102 740 092 468 317 856 149 533 746 611 128 406 657 663 556 901 416 145 644 927 496 655 933 158 468 143 482 484 006 372 447 906 612 292 829 541 260 496 970 290 197 465 492 579 693 769 880 105 128 657 628 937 735 039 288 299 048 235 836 690 797 324 513 502 829 134 531 163 352 342 497 313 541 253 617 660 116 325 236 428 177 219 201 276 485 618 928 152 536 082 354 773 892 775 152 956 930 865 700 141 446 169 861 011 718 781 238 307 958 494 122 828 500 438 409 758 341 331 326 359 243 206 743 136 842 911 727 359 310 997 123 441 791 745 020 539 221 575 643 687 646 417 117 456 946 996 365 628 976 457 655 208 423 130 822 936 961 822 716 117 367 694 165 267 852 307 626 092 080 279 836 122 376 918 659 101 107 919 099 514 855 113 769 846 184 593 342 248 535 927 407 152 514 690 465 246 338 232 121 308 958 440 135 194 441 048 499 639 516 303 692 332 532 864 631 075 547 542 841 539 848 320 583 307 785 982 596 093 517 564 724 398 774 449 380 877 817 714 717 298 596 139 689 573 570 820 356 836 562 548 742 103 826 628 952 649 445 195 215 299 968 571 218 175 989 131 452 226 726 280 771 962 970 811 426 993 797 429 280 745 007 389 078 784 134 703 325 573 686 508 850 839 302 112 856 558 329 106 490 855 990 906 295 808 952 377 118 908 425 653 871 786 066 073 831 252 442 345 238 678 271 662 351 535 236 004 206 289 778 489 301 259 384 752 840 495 042 455 478 916 057 156 112 873 606 371 350 264 102 687 648 074 992 121 706 972 612 854 704 154 657 041 404 145 923 642 777 084 367 960 280 878 796 437 947 008 894 044 010 821 287 362 106 232 574 741 311 032 906 880 293 520 619 953 280 544 651 789 897 413 312 253 724 012 410 831 696 803 510 617 000 147 747 294 278 502 175 823 823 024 255 652 077 422 574 922 776 413 427 073 317 197 412 284 579 070 292 042 084 295 513 948 442 461 828 389 757 279 712 121 164 692 705 105 851 647 684 562 196 098 398 773 163 469 604 125 793 092 370 432

possibilities. But recently, three mathematicians cleverly figured out how to eliminate most of the options. That left fewer than a trillion to check!

So they spent 2 days on a supercomputer, running 800 processors in parallel, and checked all the options. None worked. They verified their solution on another computer.

This is one of the world’s biggest proofs: it’s 200 terabytes long! That’s about equal to all the digitized text held by the US Library of Congress. There’s also a 68-gigabyte digital signature—sort of a proof that a proof exists—if you want to skim it.

It’s interesting that these 200 terabytes were used to solve a yes-or-no question, whose answer takes a single bit to state: no.

I’m not sure breaking the world’s record for the longest proof is something to be proud of. Mathematicians prize short, insightful proofs. I bet a shorter proof of this result will eventually be found.

Still, it’s fun that we can do such things. Here’s a story about the proof:

• Evelyn Lamb, Two-hundred-terabyte maths proof is largest ever, Nature, May 26, 2016.

and here’s the actual paper:

• Marijn J. H. Heule, Oliver Kullmann and Victor W. Marek, Solving and verifying the Boolean Pythagorean triples problem via cube-and-conquer.

The ‘cube-and-conquer’ paradigm is a “hybrid SAT method for hard problems, employing both look-ahead and CDCL solvers”. The actual benefit of this huge proof is developing such methods for solving big problems! In the comments to my G+ post, Roberto Bayardo explained:

CDCL == “conflict-directed clause learning”: when you hit a dead end in backtrack search, this is the process of recording a (hopefully small) clause that prevents you from making the same mistake again…a type of memorization, essentially.

Look-ahead: in backtrack search, you repeat the process of picking an unassigned variable and then picking an assignment for that variable until you reach a “dead end” (upon deriving a contradiction). Look-ahead involves doing some amount of processing on the remaining formula after each assignment in order to simplify it. This includes identifying variables for which one of its assignments can be quickly eliminated. “Unit propagation” is a type of look-ahead, though I suspect in this case they mean something quite a bit more sophisticated.﻿

Arnaud Spiwack has a series of blog posts introducting SAT solvers here:

### A much longer proof

By the way, despite the title of the Nature article, in the comments to my G+ post about this, Michael Nielsen pointed out a much longer proof by Chris Jefferson, who wrote:

Darn, I had no idea one could get into the media with this kind of stuff.

I had a much larger “proof”, where we didn’t bother storing all the details, in which we enumerated 718,981,858,383,872 semigroups, towards counting the semigroups of size 10.

Uncompressed, it would have been about 63,000 terabytes just for the semigroups, and about a thousand times that to store the “proof”, which is just the tree of the search.

Of course, it would have compressed extremely well, but also I’m not sure it would have had any value, you could rebuild the search tree much faster than you could read it from a disc, and if anyone re-verified our calculation I would prefer they did it by a slightly different search, which would give us much better guarantees of correctness.

His team found a total of 12,418,001,077,381,302,684 semigroups of size 10. They only had to find 718,981,858,383,872 by a brute force search, which is 0.006% of the total:

• Andreas Distler, Chris Jefferson, Tom Kelsey, and Lars Kottho, The semigroups of order 10, in Principles and Practice of Constraint Programming, Springer Lecture Notes in Computer Science 7514, Springer, Berlin, pp. 883–899.

### Insanely long proofs

All the proofs mentioned so far are downright laconic compared to those discussed here:

• John Baez, Insanely long proofs, 19 October 2012.

For example, if you read this post you’ll learn about a fairly short theorem whose shortest proof using Peano arithmetic contains at least

$\displaystyle{ 10^{10^{1000}} }$

symbols. This is so many that if you tried to write down the number of symbols in this proof—not the symbols themselves, but just the number of symbols—in ordinary decimal notation, you couldn’t do it if you wrote one digit on each proton, neutron and electron in the observable Universe!

## The Busy Beaver Game

21 May, 2016

This month, a bunch of ‘logic hackers’ have been seeking to determine the precise boundary between the knowable and the unknowable. The challenge has been around for a long time. But only now have people taken it up with the kind of world-wide teamwork that the internet enables.

A Turing machine is a simple model of a computer. Imagine a machine that has some finite number of states, say N states. It’s attached to a tape, an infinitely long tape with lots of squares, with either a 0 or 1 written on each square. At each step the machine reads the number where it is. Then, based on its state and what it reads, it either halts, or it writes a number, changes to a new state, and moves either left or right.

The tape starts out with only 0’s on it. The machine starts in a particular ‘start’ state. It halts if it winds up in a special ‘halt’ state.

The Busy Beaver Game is to find the Turing machine with N states that runs as long as possible and then halts.

The number BB(N) is the number of steps that the winning machine takes before it halts.

In 1961, Tibor Radó introduced the Busy Beaver Game and proved that the sequence BB(N) is uncomputable. It grows faster than any computable function!

A few values of BB(N) can be computed, but there’s no way to figure out all of them.

As we increase N, the number of Turing machines we need to check increases faster than exponentially: it’s

$(4(n+1))^{2n}$

Of course, many could be ruled out as potential winners by simple arguments. But the real problem is this: it becomes ever more complicated to determine which Turing machines with N states never halt, and which merely take a huge time to halt.

Indeed, matter what axiom system you use for math, as long as it has finitely many axioms and is consistent, you can never use it to correctly determine BB(N) for more than some finite number of cases.

So what do people know about BB(N)?

For starters, BB(0) = 0. At this point I should admit that people don’t count the halt state as one of our N states. This is just a convention. So, when we consider BB(0), we’re considering machines that only have a halt state. They instantly halt.

Next, BB(1) = 1.

Next, BB(2) = 6.

Next, BB(3) = 21. This was proved in 1965 by Tibor Radó and Shen Lin.

Next, BB(4) = 107. This was proved in 1983 by Allan Brady.

Next, BB(5). Nobody knows what BB(5) equals!

The current 5-state busy beaver champion was discovered by Heiner Marxen and Jürgen Buntrock in 1989. It takes 47,176,870 steps before it halts. So, we know

BB(5) ≥ 47,176,870.

People have looked at all the other 5-state Turing machines to see if any does better. But there are 43 machines that do very complicated things that nobody understands. It’s believed they never halt, but nobody has been able to prove this yet.

We may have hit the wall of ignorance here… but we don’t know.

That’s the spooky thing: the precise boundary between the knowable and the unknowable is unknown. It may even be unknowable… but I’m not sure we know that.

Next, BB(6). In 1996, Marxen and Buntrock showed it’s at least 8,690,333,381,690,951. In June 2010, Pavel Kropitz proved that

$\displaystyle{ \mathrm{BB}(6) \ge 7.412 \cdot 10^{36,534} }$

You may wonder how he proved this. Simple! He found a 6-state machine that runs for

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13052411058118309311147944260478608


steps and then halts!

Of course, I’m just kidding when I say this was simple. The machine is easy enough to describe, but proving it takes exactly this long to run takes real work! You can read about such proofs here:

• Pascal Michel, The Busy Beaver Competition: a historical survey.

I don’t understand them very well. All I can say at this point is that many of the record-holding machines known so far are similar to the famous Collatz conjecture. The idea there is that you can start with any positive integer and keep doing two things:

• if it’s even, divide it by 2;

• if it’s odd, triple it and add 1.

The conjecture is that this process will always eventually reach the number 1. Here’s a graph of how many steps it takes, as a function of the number you start with:

Nice pattern! But this image shows how it works for numbers up to 10 million, and you’ll see it doesn’t usually take very long for them to reach 1. Usually less than 600 steps is enough!

So, to get a Turing machine that takes a long time to halt, you have to take this kind of behavior and make it much more long and drawn-out. Conversely, to analyze one of the potential winners of the Busy Beaver Game, people must take that long and drawn-out behavior and figure out a way to predict much more quickly when it will halt.

Next, BB(7). In 2014, someone who goes by the name Wythagoras showed that

$\displaystyle{ \textrm{BB}(7) > 10^{10^{10^{10^{10^7}}}} }$

It’s fun to prove lower bounds on BB(N). For example, in 1964 Milton Green constructed a sequence of Turing machines that implies

$\textrm{BB}(2N) \ge 3 \uparrow^{N-2} 3$

Here I’m using Knuth’s up-arrow notation, which is a recursively defined generalization of exponentiation, so for example

$\textrm{BB}(10) \ge 3 \uparrow^{3} 3 = 3 \uparrow^2 3^{3^3} = 3^{3^{3^{3^{\cdot^{\cdot^\cdot}}}}}$

where there are $3^{3^3}$ threes in that tower.

But it’s also fun to seek the smallest N for which we can prove BB(N) is unknowable! And that’s what people are making lots of progress on right now.

Sometime in April 2016, Adam Yedidia and Scott Aaronson showed that BB(7910) cannot be determined using the widely accepted axioms for math called ZFC: that is, Zermelo—Fraenkel set theory together with the axiom of choice. It’s a great story, and you can read it here:

• Scott Aaronson, The 8000th Busy Beaver number eludes ZF set theory: new paper by Adam Yedidia and me, Shtetl-Optimized, 3 May 2016.

• Adam Yedidia and Scott Aaronson, A relatively small Turing machine whose behavior is independent of set theory, 13 May 2016.

Briefly, Yedidia created a new programming language, called Laconic, which lets you write programs that compile down to small Turing machines. They took an arithmetic statement created by Harvey Friedman that’s equivalent to the consistency of the usual axioms of ZFC together with a large cardinal axiom called the ‘stationary Ramsey property’, or SRP. And they created a Turing machine with 7910 states that seeks a proof of this arithmetic statement using the axioms of ZFC.

Since ZFC can’t prove its own consistency, much less its consistency when supplemented with SRP, their machine will only halt if ZFC+SRP is inconsistent.

Since most set theorists believe ZFC+SRP is consistent, this machine probably doesn’t halt. But we can’t prove this using ZFC.

In short: if the usual axioms of set theory are consistent, we can never use them to determine the value of BB(7910).

The basic idea is nothing new: what’s new is the explicit and rather low value of the number 7910. Poetically speaking, we know the unknowable starts here… if not sooner.

However, this discovery set off a wave of improvements! On the Metamath newsgroup, Mario Carneiro and others started ‘logic hacking’, looking for smaller and smaller Turing machines that would only halt if ZF—that is, Zermelo–Fraenkel set theory, without the axiom of choice—is inconsistent.

By just May 15th, Stefan O’Rear seems to have brought the number down to 1919. He found a Turing machine with just 1919 states that searches for an inconsistency in the ZF axioms. Interestingly, this turned out to work better than using Harvey Friedman’s clever trick.

Thus, if O’Rear’s work is correct, we can only determine BB(1919) if we can determine whether ZF set theory is consistent. However, we cannot do this using ZF set theory—unless we find an inconsistency in ZF set theory.

For details, see:

• Stefan O’Rear, A Turing machine Metamath verifier, 15 May 2016.

I haven’t checked his work, but it’s available on GitHub.

What’s the point of all this? At present, it’s mainly just a game. However, it should have some interesting implications. It should, for example, help us better locate the ‘complexity barrier’.

I explained that idea here:

• John Baez, The complexity barrier, Azimuth, 28 October 2011.

Briefly, while there’s no limit on how much information a string of bits—or any finite structure—can have, there’s a limit on how much information we can prove it has!

This amount of information is pretty low, perhaps a few kilobytes. And I believe the new work on logic hacking can be used to estimate it more accurately!

## Shelves and the Infinite

6 May, 2016

Infinity is a very strange concept. Like alien spores floating down from the sky, large infinities can come down and contaminate the study of questions about ordinary finite numbers!﻿ Here’s an example.

A shelf is a set with a binary operation $\rhd$ that distributes over itself:

$a \rhd (b \rhd c) = (a \rhd b) \rhd (a \rhd c)$

There are lots of examples, the simplest being any group, where we define

$g \rhd h = g h g^{-1}$

They have a nice connection to knot theory, which you can see here if you think hard:

My former student Alissa Crans, who invented the term ‘shelf’, has written a lot about them, starting here:

• Alissa Crans, Lie 2-Algebras, Chapter 3.1: Shelves, Racks, Spindles and Quandles, Ph.D. thesis, U.C. Riverside, 2004.

I could tell you a long and entertaining story about this, including the tale of how shelves got their name. But instead I want to talk about something far more peculiar, which I understand much less well. There’s a strange connection between shelves, extremely large infinities, and extremely large finite numbers! It was first noticed by a logician named Richard Laver in the late 1980s, and it’s been developed further by Randall Dougherty.

It goes like this. For each $n,$ there’s a unique shelf structure on the numbers $\{1,2, \dots ,2^n\}$ such that

$a \rhd 1 = a + 1 \bmod 2^n$

So, the elements of our shelf are

$1$

$1 \rhd 1 = 2$

$2 \rhd 1 = 3$

and so on, until we get to

$2^n \rhd 1 = 1$

However, we can now calculate

$1 \rhd 1$

$1 \rhd 2$

$1 \rhd 3$

and so on. You should try it yourself for a simple example! You’ll need to use the self-distributive law. It’s quite an experience.

You’ll get a list of $2^n$ numbers, but this list will not contain all the numbers $\{1, 2, \dots, 2^n\}.$ Instead, it will repeat with some period $P(n).$

Here is where things get weird. The numbers $P(n)$ form this sequence:

1, 1, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, …

It may not look like it, but the numbers in this sequence approach infinity!

if we assume an extra axiom, which goes beyond the usual axioms of set theory, but so far seems consistent!

This axiom asserts the existence of an absurdly large cardinal, called an I3 rank-into-rank cardinal.

I’ll say more about this kind of cardinal later. But, this is not the only case where a ‘large cardinal axiom’ has consequences for down-to-earth math, like the behavior of some sequence that you can define using simple rules.

On the other hand, Randall Dougherty has proved a lower bound on how far you have to go out in this sequence to reach the number 32.

And, it’s an incomprehensibly large number!

The third Ackermann function $A_3(n)$ is roughly 2 to the $n$th power. The fourth Ackermann function $A_4(n)$ is roughly 2 raised to itself $n$ times:

$2^{2^{2^{2^{\cdot^{\cdot^\cdot}}}}}$

And so on: each Ackermann function is defined by iterating the previous one.

Dougherty showed that for the sequence $P(n)$ to reach 32, you have to go at least

$n = A(9,A(8,A(8,255)))$

This is an insanely large number!

I should emphasize that if we use just the ordinary axioms of set theory, the ZFC axioms, nobody has proved that the sequence $P(n)$ ever reaches 32. Neither is it known that this is unprovable if we only use ZFC.

So, what we’ve got here is a very slowly growing sequence… which is easy to define but grows so slowly that (so far) mathematicians need new axioms of set theory to prove it goes to infinity, or even reaches 32.

I should admit that my definition of the Ackermann function is rough. In reality it’s defined like this:

$A(m, n) = \begin{cases} n+1 & \mbox{if } m = 0 \\ A(m-1, 1) & \mbox{if } m > 0 \mbox{ and } n = 0 \\ A(m-1, A(m, n-1)) & \mbox{if } m > 0 \mbox{ and } n > 0. \end{cases}$

And if you work this out, you’ll find it’s a bit annoying. Somehow the number 3 sneaks in:

$A(2,n) = 2 + (n+3) - 3$

$A(3,n) = 2 \cdot (n+3) - 3$

$A(4,n) = 2^{n+3} - 3$

$A(5,n) = 2\uparrow\uparrow(n+3) - 3$

where $a \uparrow\uparrow b$ means $a$ raised to itself $b$ times,

$A(6,n) = 2 \uparrow\uparrow\uparrow(n+3) - 3$

where $a \uparrow\uparrow\uparrow b$ means $a \uparrow\uparrow (a \uparrow\uparrow (a \uparrow\uparrow \cdots ))$ with the number $a$ repeated $b$ times, and so on.

However, these irritating 3’s scarcely matter, since Dougherty’s number is so large… and I believe he could have gotten an even larger upper bound if he wanted.

Perhaps I’ll wrap up by saying very roughly what an I3 rank-into-rank cardinal is.

In set theory the universe of all sets is built up in stages. These stages are called the von Neumann hierarchy. The lowest stage has nothing in it:

$V_0 = \emptyset$

Each successive stage is defined like this:

$V_{\lambda + 1} = P(V_\lambda)$

where $P(S)$ is the the power set of $S,$ that is, the set of all subsets of $S.$ For ‘limit ordinals’, that is, ordinals that aren’t of the form $\lambda + 1,$ we define

$\displaystyle{ V_\lambda = \bigcup_{\alpha < \lambda} V_\alpha }$

An I3 rank-into-rank cardinal is an ordinal $\lambda$ such that $V_\lambda$ admits a nontrivial elementary embedding into itself.

Very roughly, this means the infinity $\lambda$ is so huge that the collection of sets that can be built by this stage can mapped into itself, in a one-to-one but not onto way, into a smaller collection that’s indistinguishable from the original one when it comes to the validity of anything you can say about sets!

More precisely, a nontrivial elementary embedding of $V_\lambda$ into itself is a one-to-one but not onto function

$f: V_\lambda \to V_\lambda$

that preserves and reflects the validity of all statements in the language of set theory. That is: for any sentence $\phi(a_1, \dots, a_n)$ in the language of set theory, this statement holds for sets $a_1, \dots, a_n \in V_\lambda$ if and only if $\phi(f(a_1), \dots, f(a_n))$ holds.

I don’t know why, but an I3 rank-into-rank cardinal, if it’s even consistent to assume one exists, is known to be extraordinarily big. What I mean by this is that it automatically has a lot of other properties known to characterize large cardinals. It’s inaccessible (which is big) and ineffable (which is bigger), and measurable (which is bigger), and huge (which is even bigger), and so on.

How in the world is this related to shelves?

The point is that if

$f, g : V_\lambda \to V_\lambda$

are elementary embeddings, we can apply $f$ to any set in $V_\lambda.$ But in set theory, functions are sets too: sets of ordered pairs So, $g$ is a set. It’s not an element of $V_\lambda,$ but all its subsets $g \cap V_\alpha$ are, where $\alpha < \lambda.$ So, we can define

$f \rhd g = \bigcup_{\alpha < \lambda} f (g \cap V_\alpha)$

Laver showed that this operation distributes over itself:

$f \rhd (g \rhd h) = (f \rhd g) \rhd (f \rhd h)$

And, he showed that if we take one elementary embedding and let it generate a shelf by this this operation, we get the free shelf on one generator!

The shelf I started out describing, the numbers $\{1, \dots, 2^n \}$ with

$a \rhd 1 = a + 1 \bmod 2^n$

also has one generator namely the number 1. So, it’s a quotient of the free shelf on one generator by one relation, namely the above equation.

That’s about all I understand. I don’t understand how the existence of a nontrivial elementary embedding of $V_\lambda$ into itself implies that the function $P(n)$ goes to infinity, and I don’t understand Randall Dougherty’s lower bound on how far you need to go to reach $P(n) = 32.$ For more, read these:

• Richard Laver, The left distributive law and the freeness of an algebra of elementary embeddings, Adv. Math. 91 (1992), 209–231.

• Richard Laver, On the algebra of elementary embeddings of a rank into itself, Adv. Math. 110 (1995), 334–346.

• Randall Dougherty and Thomas Jech, Finite left distributive algebras and embedding algebras, Adv. Math. 130 (1997), 201–241.

• Randall Dougherty, Critical points in an algebra of elementary embeddings, Ann. Pure Appl. Logic 65 (1993), 211–241.

• Randall Dougherty, Critical points in an algebra of elementary embeddings, II.

## Statistical Laws of Darwinian Evolution

18 April, 2016

guest post by Matteo Smerlak

Biologists like Steven J. Gould like to emphasize that evolution is unpredictable. They have a point: there is absolutely no way an alien visiting the Earth 400 million years ago could have said:

Hey, I know what’s gonna happen here. Some descendants of those ugly fish will grow wings and start flying in the air. Others will walk the surface of the Earth for a few million years, but they’ll get bored and they’ll eventually go back to the oceans; when they do, they’ll be able to chat across thousands of kilometers using ultrasound. Yet others will grow arms, legs, fur, they’ll climb trees and invent BBQ, and, sooner or later, they’ll start wondering “why all this?”.

Nor can we tell if, a week from now, the flu virus will mutate, become highly pathogenic and forever remove the furry creatures from the surface of the Earth.

Evolution isn’t gravity—we can’t tell in which directions things will fall down.

One reason we can’t predict the outcomes of evolution is that genomes evolve in a super-high dimensional combinatorial space, which a ginormous number of possible turns at every step. Another is that living organisms interact with one another in a massively non-linear way, with, feedback loops, tipping points and all that jazz.

Life’s a mess, if you want my physicist’s opinion.

But that doesn’t mean that nothing can be predicted. Think of statistics. Nobody can predict who I’ll vote for in the next election, but it’s easy to tell what the distribution of votes in the country will be like. Thus, for continuous variables which arise as sums of large numbers of independent components, the central limit theorem tells us that the distribution will always be approximately normal. Or take extreme events: the max of $N$ independent random variables is distributed according to a member of a one-parameter family of so-called “extreme value distributions”: this is the content of the famous Fisher–Tippett–Gnedenko theorem.

So this is the problem I want to think about in this blog post: is evolution ruled by statistical laws? Or, in physics terms: does it exhibit some form of universality?

### Fitness distributions are the thing

One lesson from statistical physics is that, to uncover universality, you need to focus on relevant variables. In the case of evolution, it was Darwin’s main contribution to figure out the main relevant variable: the average number of viable offspring, aka fitness, of an organism. Other features—physical strength, metabolic efficiency, you name it—matter only insofar as they are correlated with fitness. If we further assume that fitness is (approximately) heritable, meaning that descendants have the same fitness as their ancestors, we get a simple yet powerful dynamical principle called natural selection: in a given population, the lineage with the highest fitness eventually dominates, i.e. its fraction goes to one over time. This principle is very general: it applies to genes and species, but also to non-living entities such as algorithms, firms or language. The general relevance of natural selection as a evolutionary force is sometimes referred to as “Universal Darwinism”.

The general idea of natural selection is pictured below (reproduced from this paper):

It’s not hard to write down an equation which expresses natural selection in general terms. Consider an infinite population in which each lineage grows with some rate $x$. (This rate is called the log-fitness or Malthusian fitness to contrast it with the number of viable offspring $w=e^{x\Delta t}$ with $\Delta t$ the lifetime of a generation. It’s more convenient to use $x$ than $w$ in what follows, so we’ll just call $x$ “fitness”). Then the distribution of fitness at time $t$ satisfies the equation

$\displaystyle{ \frac{\partial p_t(x)}{\partial t} =\left(x-\int d y\, y\, p_t(y)\right)p_t(x) }$

whose explicit solution in terms of the initial fitness distribution $p_0(x):$

$\displaystyle{ p_t(x)=\frac{e^{x t}p_0(x)}{\int d y\, e^{y t}p_0(y)} }$

is called the Cramér transform of $p_0(x)$ in large deviations theory. That is, viewed as a flow in the space of probability distributions, natural selection is nothing but a time-dependent exponential tilt. (These equations and the results below can be generalized to include the effect of mutations, which are critical to maintain variation in the population, but we’ll skip this here to focus on pure natural selection. See my paper referenced below for more information.)

An immediate consequence of these equations is that the mean fitness $\mu_t=\int dx\, x\, p_t(x)$ grows monotonically in time, with a rate of growth given by the variance $\sigma_t^2=\int dx\, (x-\mu_t)^2\, p_t(x)$:

$\displaystyle{ \frac{d\mu_t}{dt}=\sigma_t^2\geq 0 }$

The great geneticist Ronald Fisher (yes, the one in the extreme value theorem!) was very impressed with this relationship. He thought it amounted to an biological version of the second law of thermodynamics, writing in his 1930 monograph

Professor Eddington has recently remarked that “The law that entropy always increases—the second law of thermodynamics—holds, I think, the supreme position among the laws of nature”. It is not a little instructive that so similar a law should hold the supreme position among the biological sciences.

Unfortunately, this excitement hasn’t been shared by the biological community, notably because this Fisher “fundamental theorem of natural selection” isn’t predictive: the mean fitness $\mu_t$ grows according to the fitness variance $\sigma_t^2$, but what determines the evolution of $\sigma_t^2$? I can’t use the identity above to predict the speed of evolution in any sense. Geneticists say it’s “dynamically insufficient”.

### Two limit theorems

But the situation isn’t as bad as it looks. The evolution of $p_t(x)$ may be decomposed into the evolution of its mean $\mu_t$, of its variance $\sigma_t^2$, and of its shape or type

$\overline{p}_t(x)=\sigma_t p_t(\sigma_t x+\mu_t)$.

(We also call $\overline{p}_t(x)$ the “standardized fitness distribution”.) With Ahmed Youssef we showed that:

• If $p_0(x)$ is supported on the whole real line and decays at infinity as

$-\ln\int_x^{\infty}p_0(y)d y\underset{x\to\infty}{\sim} x^{\alpha}$

for some $\alpha > 1$, then $\mu_t\sim t^{\overline{\alpha}-1}$, $\sigma_t^2\sim t^{\overline{\alpha}-2}$ and $\overline{p}_t(x)$ converges to the standard normal distribution as $t\to\infty$. Here $\overline{\alpha}$ is the conjugate exponent to $\alpha$, i.e. $1/\overline{\alpha}+1/\alpha=1$.

• If $p_0(x)$ has a finite right-end point $x_+$ with

$p(x)\underset{x\to x_+}{\sim} (x_+-x)^\beta$

for some $\beta\geq0$, then $x_+-\mu_t\sim t^{-1}$, $\sigma_t^2\sim t^{-2}$ and $\overline{p}_t(x)$ converges to the flipped gamma distribution

$\displaystyle{ p^*_\beta(x)= \frac{(1+\beta)^{(1+\beta)/2}}{\Gamma(1+\beta)} \Theta[x-(1+\beta)^{1/2}] }$

$\displaystyle { e^{-(1+\beta)^{1/2}[(1+\beta)^{1/2}-x]}\Big[(1+\beta)^{1/2}-x\Big]^\beta }$

Here and below the symbol $\sim$ means “asymptotically equivalent up to a positive multiplicative constant”; $\Theta(x)$ is the Heaviside step function. Note that $p^*_\beta(x)$ becomes Gaussian in the limit $\beta\to\infty$, i.e. the attractors of cases 1 and 2 form a continuous line in the space of probability distributions; the other extreme case, $\beta\to0$, corresponds to a flipped exponential distribution.

The one-parameter family of attractors $p_\beta^*(x)$ is plotted below:

These results achieve two things. First, they resolve the dynamical insufficiency of Fisher’s fundamental theorem by giving estimates of the speed of evolution in terms of the tail behavior of the initial fitness distribution. Second, they show that natural selection is indeed subject to a form of universality, whereby the relevant statistical structure turns out to be finite dimensional, with only a handful of “conserved quantities” (the $\alpha$ and $\beta$ exponents) controlling the late-time behavior of natural selection. This amounts to a large reduction in complexity and, concomitantly, an enhancement of predictive power.

(For the mathematically-oriented reader, the proof of the theorems above involves two steps: first, translate the selection equation into a equation for (cumulant) generating functions; second, use a suitable Tauberian theorem—the Kasahara theorem—to relate the behavior of generating functions at large values of their arguments to the tail behavior of $p_0(x)$. Details in our paper.)

It’s useful to consider the convergence of fitness distributions to the attractors $p_\beta^*(x)$ for $0\leq\beta\leq \infty$ in the skewness-kurtosis plane, i.e. in terms of the third and fourth cumulants of $p_t(x)$.

The red curve is the family of attractors, with the normal at the bottom right and the flipped exponential at the top left, and the dots correspond to numerical simulations performed with the classical Wright–Fisher model and with a simple genetic algorithm solving a linear programming problem. The attractors attract!

### Conclusion and a question

Statistics is useful because limit theorems (the central limit theorem, the extreme value theorem) exist. Without them, we wouldn’t be able to make any population-level prediction. Same with statistical physics: it only because matter consists of large numbers of atoms, and limit theorems hold (the H-theorem, the second law), that macroscopic physics is possible in the first place. I believe the same perspective is useful in evolutionary dynamics: it’s true that we can’t predict how many wings birds will have in ten million years, but we can tell what shape fitness distributions should have if natural selection is true.

I’ll close with an open question for you, the reader. In the central limit theorem as well as in the second law of thermodynamics, convergence is driven by a Lyapunov function, namely entropy. (In the case of the central limit theorem, it’s a relatively recent result by Arstein et al.: the entropy of the normalized sum of $n$ i.i.d. random variables, when it’s finite, is a monotonically increasing function of $n$.) In the case of natural selection for unbounded fitness, it’s clear that entropy will also be eventually monotonically increasing—the normal is the distribution with largest entropy at fixed variance and mean.

Yet it turns out that, in our case, entropy isn’t monotonic at all times; in fact, the closer the initial distribution $p_0(x)$ is to the normal distribution, the later the entropy of the standardized fitness distribution starts to increase. Or, equivalently, the closer the initial distribution $p_0(x)$ to the normal, the later its relative entropy with respect to the normal. Why is this? And what’s the actual Lyapunov function for this process (i.e., what functional of the standardized fitness distribution is monotonic at all times under natural selection)?

In the plots above the blue, orange and green lines correspond respectively to

$\displaystyle{ p_0(x)\propto e^{-x^2/2-x^4}, \quad p_0(x)\propto e^{-x^2/2-.01x^4}, \quad p_0(x)\propto e^{-x^2/2-.001x^4} }$

### References

• S. J. Gould, Wonderful Life: The Burgess Shale and the Nature of History, W. W. Norton & Co., New York, 1989.

• M. Smerlak and A. Youssef, Limiting fitness distributions in evolutionary dynamics, 2015.

• R. A. Fisher, The Genetical Theory of Natural Selection, Oxford University Press, Oxford, 1930.

• S. Artstein, K. Ball, F. Barthe and A. Naor, Solution of Shannon’s problem on the monotonicity of entropy, J. Am. Math. Soc. 17 (2004), 975–982.

## Diamonds and Triamonds

11 April, 2016

The structure of a diamond crystal is fascinating. But there’s an equally fascinating form of carbon, called the triamond, that’s theoretically possible but never yet seen in nature. Here it is:

In the triamond, each carbon atom is bonded to three others at 120° angles, with one double bond and two single bonds. Its bonds lie in a plane, so we get a plane for each atom.

But here’s the tricky part: for any two neighboring atoms, these planes are different. In fact, if we draw the bond planes for all the atoms in the triamond, they come in four kinds, parallel to the faces of a regular tetrahedron!

If we discount the difference between single and double bonds, the triamond is highly symmetrical. There’s a symmetry carrying any atom and any of its bonds to any other atom and any of its bonds. However, the triamond has an inherent handedness, or chirality. It comes in two mirror-image forms.

A rather surprising thing about the triamond is that the smallest rings of atoms are 10-sided. Each atom lies in 15 of these 10-sided rings.

Some chemists have argued that the triamond should be ‘metastable’ at room temperature and pressure: that is, it should last for a while but eventually turn to graphite. Diamonds are also considered metastable, though I’ve never seen anyone pull an old diamond ring from their jewelry cabinet and discover to their shock that it’s turned to graphite. The big difference is that diamonds are formed naturally under high pressure—while triamonds, it seems, are not.

Nonetheless, the mathematics behind the triamond does find its way into nature. A while back I told you about a minimal surface called the ‘gyroid’, which is found in many places:

It turns out that the pattern of a gyroid is closely connected to the triamond! So, if you’re looking for a triamond-like pattern in nature, certain butterfly wings are your best bet:

• Matthias Weber, The gyroids: algorithmic geometry III, The Inner Frame, 23 October 2015.

Instead of trying to explain it here, I’ll refer you to the wonderful pictures at Weber’s blog.

### Building the triamond

I want to tell you a way to build the triamond. I saw it here:

• Toshikazu Sunada, Crystals that nature might miss creating, Notices of the American Mathematical Society 55 (2008), 208–215.

This is the paper that got people excited about the triamond, though it was discovered much earlier by the crystallographer Fritz Laves back in 1932, and Coxeter named it the Laves graph.

To build the triamond, we can start with this graph:

It’s called $\mathrm{K}_4,$ since it’s the complete graph on four vertices, meaning there’s one edge between each pair of vertices. The vertices correspond to four different kinds of atoms in the triamond: let’s call them red, green, yellow and blue. The edges of this graph have arrows on them, labelled with certain vectors

$e_1, e_2, e_3, f_1, f_2, f_3 \in \mathbb{R}^3$

Let’s not worry yet about what these vectors are. What really matters is this: to move from any atom in the triamond to any of its neighbors, you move along the vector labeling the edge between them… or its negative, if you’re moving against the arrow.

For example, suppose you’re at any red atom. It has 3 nearest neighbors, which are blue, green and yellow. To move to the blue neighbor you add $f_1$ to your position. To move to the green one you subtract $e_2,$ since you’re moving against the arrow on the edge connecting blue and green. Similarly, to go to the yellow neighbor you subtract the vector $f_3$ from your position.

Thus, any path along the bonds of the triamond determines a path in the graph $\mathrm{K}_4.$

Conversely, if you pick an atom of some color in the triamond, any path in $\mathrm{K}_4$ starting from the vertex of that color determines a path in the triamond! However, going around a loop in $\mathrm{K}_4$ may not get you back to the atom you started with in the triamond.

Mathematicians summarize these facts by saying the triamond is a ‘covering space’ of the graph $\mathrm{K}_4.$

Now let’s see if you can figure out those vectors.

Puzzle 1. Find vectors $e_1, e_2, e_3, f_1, f_2, f_3 \in \mathbb{R}^3$ such that:

A) All these vectors have the same length.

B) The three vectors coming out of any vertex lie in a plane at 120° angles to each other:

For example, $f_1, -e_2$ and $-f_3$ lie in a plane at 120° angles to each other. We put in two minus signs because two arrows are pointing into the red vertex.

C) The four planes we get this way, one for each vertex, are parallel to the faces of a regular tetrahedron.

If you want, you can even add another constraint:

D) All the components of the vectors $e_1, e_2, e_3, f_1, f_2, f_3$ are integers.

### Diamonds and hyperdiamonds

That’s the triamond. Compare the diamond:

Here each atom of carbon is connected to four others. This pattern is found not just in carbon but also other elements in the same column of the periodic table: silicon, germanium, and tin. They all like to hook up with four neighbors.

The pattern of atoms in a diamond is called the diamond cubic. It’s elegant but a bit tricky. Look at it carefully!

To build it, we start by putting an atom at each corner of a cube. Then we put an atom in the middle of each face of the cube. If we stopped there, we would have a face-centered cubic. But there are also four more carbons inside the cube—one at the center of each tetrahedron we’ve created.

If you look really carefully, you can see that the full pattern consists of two interpenetrating face-centered cubic lattices, one offset relative to the other along the cube’s main diagonal.

The face-centered cubic is the 3-dimensional version of a pattern that exists in any dimension: the Dn lattice. To build this, take an n-dimensional checkerboard and alternately color the hypercubes red and black. Then, put a point in the center of each black hypercube!

You can also get the Dn lattice by taking all n-tuples of integers that sum to an even integer. Requiring that they sum to something even is a way to pick out the black hypercubes.

The diamond is also an example of a pattern that exists in any dimension! I’ll call this the hyperdiamond, but mathematicians call it Dn+, because it’s the union of two copies of the Dn lattice. To build it, first take all n-tuples of integers that sum to an even integer. Then take all those points shifted by the vector (1/2, …, 1/2).

In any dimension, the volume of the unit cell of the hyperdiamond is 1, so mathematicians say it’s unimodular. But only in even dimensions is the sum or difference of any two points in the hyperdiamond again a point in the hyperdiamond. Mathematicians call a discrete set of points with this property a lattice.

If even dimensions are better than odd ones, how about dimensions that are multiples of 4? Then the hyperdiamond is better still: it’s an integral lattice, meaning that the dot product of any two vectors in the lattice is again an integer.

And in dimensions that are multiples of 8, the hyperdiamond is even better. It’s even, meaning that the dot product of any vector with itself is even.

In fact, even unimodular lattices are only possible in Euclidean space when the dimension is a multiple of 8. In 8 dimensions, the only even unimodular lattice is the 8-dimensional hyperdiamond, which is usually called the E8 lattice. The E8 lattice is one of my favorite entities, and I’ve written a lot about it in this series:

To me, the glittering beauty of diamonds is just a tiny hint of the overwhelming beauty of E8.

But let’s go back down to 3 dimensions. I’d like to describe the diamond rather explicitly, so we can see how a slight change produces the triamond.

It will be less stressful if we double the size of our diamond. So, let’s start with a face-centered cubic consisting of points whose coordinates are even integers summing to a multiple of 4. That consists of these points:

(0,0,0)   (2,2,0)   (2,0,2)   (0,2,2)

and all points obtained from these by adding multiples of 4 to any of the coordinates. To get the diamond, we take all these together with another face-centered cubic that’s been shifted by (1,1,1). That consists of these points:

(1,1,1)   (3,3,1)   (3,1,3)   (1,3,3)

and all points obtained by adding multiples of 4 to any of the coordinates.

The triamond is similar! Now we start with these points

(0,0,0)   (1,2,3)   (2,3,1)   (3,1,2)

and all the points obtain from these by adding multiples of 4 to any of the coordinates. To get the triamond, we take all these together with another copy of these points that’s been shifted by (2,2,2). That other copy consists of these points:

(2,2,2)   (3,0,1)   (0,1,3)   (1,3,0)

and all points obtained by adding multiples of 4 to any of the coordinates.

Unlike the diamond, the triamond has an inherent handedness, or chirality. You’ll note how we used the point (1,2,3) and took cyclic permutations of its coordinates to get more points. If we’d started with (3,2,1) we would have gotten the other, mirror-image version of the triamond.

### Covering spaces

I mentioned that the triamond is a ‘covering space’ of the graph $\mathrm{K}_4.$ More precisely, there’s a graph $T$ whose vertices are the atoms of the triamond, and whose edges are the bonds of the triamond. There’s a map of graphs

$p: T \to \mathrm{K}_4$

This automatically means that every path in $T$ is mapped to a path in $\mathrm{K}_4.$ But what makes $T$ a covering space of $\mathrm{K}_4$ is that any path in $T$ comes from a path in $\mathrm{K}_4,$ which is unique after we choose its starting point.

If you’re a high-powered mathematician you might wonder if $T$ is the universal covering space of $\mathrm{K}_4.$ It’s not, but it’s the universal abelian covering space.

What does this mean? Any path in $\mathrm{K}_4$ gives a sequence of vectors $e_1, e_2, e_3, f_1, f_2, f_3$ and their negatives. If we pick a starting point in the triamond, this sequence describes a unique path in the triamond. When does this path get you back where you started? The answer, I believe, is this: if and only if you can take your sequence, rewrite it using the commutative law, and cancel like terms to get zero. This is related to how adding vectors in $\mathbb{R}^3$ is a commutative operation.

For example, there’s a loop in $\mathrm{K}_4$ that goes “red, blue, green, red”. This gives the sequence of vectors

$f_1, -e_3, e_2$

We can turn this into an expression

$f_1 - e_3 + e_2$

However, we can’t simplify this to zero using just the commutative law and cancelling like terms. So, if we start at some red atom in the triamond and take the unique path that goes “red, blue, green, red”, we do not get back where we started!

Note that in this simplification process, we’re not allowed to use what the vectors “really are”. It’s a purely formal manipulation.

Puzzle 2. Describe a loop of length 10 in the triamond using this method. Check that you can simplify the corresponding expression to zero using the rules I described.

A similar story works for the diamond, but starting with a different graph:

The graph formed by a diamond’s atoms and the edges between them is the universal abelian cover of this little graph! This graph has 2 vertices because there are 2 kinds of atom in the diamond. It has 4 edges because each atom has 4 nearest neighbors.

Puzzle 3. What vectors should we use to label the edges of this graph, so that the vectors coming out of any vertex describe how to move from that kind of atom in the diamond to its 4 nearest neighbors?

There’s also a similar story for graphene, which is hexagonal array of carbon atoms in a plane:

Puzzle 4. What graph with edges labelled by vectors in $\mathbb{R}^2$ should we use to describe graphene?

I don’t know much about how this universal abelian cover trick generalizes to higher dimensions, though it’s easy to handle the case of a cubical lattice in any dimension.

Puzzle 5. I described higher-dimensional analogues of diamonds: are there higher-dimensional triamonds?

### References

The Wikipedia article is good:

• Wikipedia, Laves graph.

They say this graph has many names: the K4 crystal, the (10,3)-a network, the srs net, the diamond twin, and of course the triamond. The name triamond is not very logical: while each carbon has 3 neighbors in the triamond, each carbon has not 2 but 4 neighbors in the diamond. So, perhaps the diamond should be called the ‘quadriamond’. In fact, the word ‘diamond’ has nothing to do with the prefix ‘di-‘ meaning ‘two’. It’s more closely related to the word ‘adamant’. Still, I like the word ‘triamond’.

This paper describes various attempts to find the Laves graph in chemistry:

• Stephen T. Hyde, Michael O’Keeffe, and Davide M. Proserpio, A short history of an elusive yet ubiquitous structure in chemistry, materials, and mathematics, Angew. Chem. Int. Ed. 47 (2008), 7996–8000.

This paper does some calculations arguing that the triamond is a metastable form of carbon:

• Masahiro Itoh et al, New metallic carbon crystal, Phys. Rev. Lett. 102 (2009), 055703.

Abstract. Recently, mathematical analysis clarified that sp2 hybridized carbon should have a three-dimensional crystal structure ($\mathrm{K}_4$) which can be regarded as a twin of the sp3 diamond crystal. In this study, various physical properties of the $\mathrm{K}_4$ carbon crystal, especially for the electronic properties, were evaluated by first principles calculations. Although the $\mathrm{K}_4$ crystal is in a metastable state, a possible pressure induced structural phase transition from graphite to $\mathrm{K}_4$ was suggested. Twisted π states across the Fermi level result in metallic properties in a new carbon crystal.

The picture of the $\mathrm{K}_4$ crystal was placed on Wikicommons by someone named ‘Workbit’, under a Creative Commons Attribution-Share Alike 4.0 International license. The picture of the tetrahedron was made using Robert Webb’s Stella software and placed on Wikicommons. The pictures of graphs come from Sunada’s paper, though I modified the picture of $\mathrm{K}_4.$ The moving image of the diamond cubic was created by H.K.D.H. Bhadeshia and put into the public domain on Wikicommons. The picture of graphene was drawn by Dr. Thomas Szkopek and put into the public domain on Wikicommons.