## July 13, 2009

### Categorifying Nicomachus of Gerasa’s Equation

#### Posted by John Baez

I’ve been having a fun conversation with Gavin Wraith, who allowed me to post parts of it here.

$1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$

I hope you know this one. Take the first $n$ natural numbers: the sum of their cubes is the square of their sum!

This formula is attributed to Nicomachus of Gerasa, a Pythagorean living in Syria from 60 to 120 AD, who is famous for his Manual of Harmonics and Introduction to Arithmetic.

As you’ll see if you click the link, the latter book was also deeply connected to Pythagorean music theory… a topic I had fun exploring in week266. It’s a digression from our main theme here, but I can’t resist mentioning that he thought a lot about this triangle I talked about:

1    2     4    8     16  ....
3    6    12    24  ....
9    18    36  ....
27   54  ....
81 ....


The first row consists of powers of two, each other number is the sum of the two above it… and the diagonal consists of powers of three! In fact, this triangle contains all the numbers of the form $2^n 3^m$. It shows up when you study the musical intervals made from octave and perfect fifths.

Anyway, Gavin wrote:

While on the topic of maths and antiquity, here is a bit of gossip. You know the nice formula that says

$1^3 + \cdots + n^3 = (1 + \cdots + n)^2,$

discovered by Nicomachus of Gerasa? Gerasa (‘cherry orchard’ in Greek) is the modern Jerash in Jordan. A friend of mine, Othman Malhas in the Physics department of Yarmouk University, has converted an old building there as a conference centre in honor of Nicomachus, with marble floors inlaid with geometrical constructions. So next time you are in that part of the world (Jerash has the best preserved Roman forum to be found anywhere) give the Nicomachus Centre a visit.

This reminded me how much I loved this formula for the sum of cubes, and it prompted in me the wish for a ‘bijective proof’. A bijective proof of an equation between natural numbers is one that actually constructs a one-to-one mapping between two sets having those numbers of elements. Since passing from natural numbers to finite sets is one of the most basic examples of ‘categorification’, and Gavin Wraith is fond of that, it seemed like a natural question.

And I hoped that this in this case, the bijective proof could be nicely geometrical:

Yes, I know and love that formula. It would be nice to exhibit it as a way of taking the balls in a 4-dimensional stack of cannonballs with cubical layers, and rearranging them to form a 4-dimensional shape that’s a product of two triangles.

And Gavin obliged me by coming up with such a proof!

Can you? Of course there could be many, in which case we want the ‘nicest’ one, whatever that means.

After some of you succeed, or you all give up, I’ll tell you Gavin’s proof.

Posted at July 13, 2009 2:38 PM UTC

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### Re: Categorifiying Nichomacus of Gerasa’s Equation

A proof by pictures is given on page 58 of Conway and Guy’s The Book of Numbers. If I have a little more time today, I might write out what the pictures say.

Posted by: Todd Trimble on July 13, 2009 4:33 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

Guess I might as well write out the Conway and Guy picture proof without pictures, but maybe without doing it complete visual justice.

Picture a square tile of side $1 + 2 + \ldots + n$ units long and one unit thick. Remove a square of side $1 + 2 + \ldots + (n-1)$ containing one of the corners, so that what remains is an L-shaped tile. The L-shaped area is of uniform width $n$, and can be decomposed into rectangular tiles; the total area of the tiles is

$1 \cdot n + 2 \cdot n + \ldots + (n-1) \cdot n + (n \cdot n) + n \cdot (n-1) + \ldots + n \cdot 2 + n \cdot 1$

Now stack these tiles, by putting the $n \cdot n$ tile at the bottom, and putting the $j \cdot n$ tile together with the $n \cdot (n-j)$ tile to form an $n \cdot n$ square, which you stack on successively for values $j = 1, \ldots, n-1$. In the end these stack up to a cube of dimension $n \cdot n \cdot n$. So each L-shaped tile can be chopped up and reassembled as a cube, so that the original large square tile has volume equal to a sum of cubes, $1^3 + 2^3 + \ldots + n^3$.

Posted by: Todd Trimble on July 13, 2009 8:09 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

There are several visual proofs linked as references from the Wikipedia article on this equation.

Posted by: D. Eppstein on July 13, 2009 4:47 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

D. Eppstein wrote: “There are several visual proofs linked as references from the Wikipedia article on this equation.”
————————————–

I thought this was an interesting article,

http://www.emis.de/journals/NNJ/Kappraff.html [algebraic proof at end]

“As we mentioned, if x = f the Nicomachus sequence collapses to a single line, that of the Fibonacci sequence known as the f-sequence, …

I have also shown that Farey series and the geometry of the Brunes star is suggested by the generalizations of Nicomachus sequences. Since Farey series is also fundamental to an understanding of a modern study of dynamical systems, we see that the voice of Nicomachus still echoes through the halls of mathematics.”

Posted by: Stephen Harris on July 13, 2009 11:19 PM | Permalink | Reply to this

### Neo-Pythagorean; Re: Categorifiying Nichomacus of Gerasa’s Equation

One of my favorite Theomathematicians.

http://www.britannica.com/EBchecked/topic/414433/Nicomachus-of-Gerasa

“Neo-Pythagorean philosopher and mathematician who wrote Arithmētikē eisagōgē (Introduction to Arithmetic), an influential treatise on number theory. Considered a standard authority for 1,000 years, the book sets out the elementary theory and properties of numbers and contains the earliest-known Greek multiplication table…. He was not interested, however, in abstract theorems on whole numbers and their proofs, as found in Books VII–IX of Euclid’s Elements; contrary to Euclid’s approach, Nicomachus would merely give specific numerical examples. A Latin translation of the Arithmētikē by Lucius Apuleius (c. 124–170) is lost, but a version by Ancius Boethius (c. 470–524) survived and was used in schools up to the Renaissance. Nicomachus also wrote Encheiridion Harmonikēs (‘Handbook of Harmony’) on the Pythagorean theory of music and the two-volume Theologoumena arithmetikēs (‘The Theology of Numbers’) on the mystic properties of numbers; only fragments of the latter survive.”

Posted by: Jonathan Vos Post on July 13, 2009 6:10 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

“Introduction to Arithmetic” was originally written using Arabic numbers, and the Arabs preserved and transformed Greek ideas during the Middle Ages.

This book is chock full of references to Plato, and so why does the Wikipedia entry on Nicomachus say that he was highly influenced by Aristotle?

Plato totally rules compared to Aristotle, and Plato would beat Aristotle on the football (soccer) field any day!

Posted by: Charlie Stromeyer on July 13, 2009 8:13 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

Plato totally rules compared to Aristotle

I prefer Aristotle, but that is mostly political rather than mathematical.

“Introduction to Arithmetic” was originally written using Arabic numbers

That's amazing if true, since the (so-called?) Hindu-Arabic numeral system is usually attributed to Indian mathematicians a few centuries later. (Their system was introduced to Arab-writing culture by al-Khwārizmī of Algebra fame; Leonardo Fibonacci spread it to Europe.)

Do you know a reference other than the MacTutor biography (which backs your claim but gives no clear reference itself)?

Posted by: Toby Bartels on July 13, 2009 9:22 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

We might be able to make this more of a mathematical than a political debate (but I don’t know whether Nicomachus would favor Plato over Aristotle).

For one example, two famous Princeton mathematicians named Conway and Kochen one day said let us define the concept of “free will” mathematically to mean the opposite of both randomness and physical determinism.

The Notices of the AMS just published their new “Strong Free Will Theorem”:

http://www.ams.org/notices/200902/rtx090200226p.pdf

I have already told Roger Penrose, Gerard t’Hooft and a few others why this theorem appears valid to me. However, I cannot figure out why via category theory.

Posted by: Charlie Stromeyer on July 14, 2009 1:18 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

Kenny Easwaran has commented on the Conway-Kochen paper on his blog, here.

Posted by: Todd Trimble on July 14, 2009 2:29 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

CS wrote:
I have already told Roger Penrose, Gerard t’Hooft and a few others why this theorem appears valid to me. However, I cannot figure out why via category theory.

Nor will you soundly, the reasoning for this theorem is at most valid. It makes the same category error that Penrose demonstrated in his attack on strong AI using Godel’s Incompleteness Theorem. It fails because one can’t fit a square peg into a round hole. It is beyond the scope of QM to generate some proof which applies to the realm of ideas which are abstract idealistic constructs of physical reality. QM’s mathematical formalisms are correct but they don’t ratchet any proven *interpretation* of physical reality, such as Many Minds or Worlds, nor the abstracted idea of free will. I mean the proof may be valid but that doesn’t mean that the proof exerts any pressure or influence on a concept such as free will, there is no relevance, because the argument contains hidden unsound premises, like QM has anything at all to do with free will. I’m really surprised that the Cafe was so skeptical of the proof of Goldbach’s Conjecture (GC), but let this “theorem” pass by so blithely. Maybe it’s all in the name … or maybe they felt the proof of GC was on topic but that this theorem is off topic. Since I don’t want to hijack even a dying thread, I’ll give my blog url so that any objectors to my pov can reply there, rather than wrongly employ the Cafe (unless approved). http://textonyx.blogspot.com/

Posted by: Stephen Harris on July 16, 2009 1:18 AM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

Maybe there is some confusion because the name of Aristotle’s father was Nichomachus.

Posted by: Stephen Harris on July 13, 2009 9:30 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

Dylan Thurston told me this one. I think it is due to Dylan, Walter Neumann, and David Bayer. The sum of cubes can be thought of as the cone on a cube. The square of triangles is that: a triangle times a triangle.

The cone on the cube can be decomposed as the union of two cones on two prisms. These cones on prisms can be re-glued as the triangle times a triangle.

These considerations provoked the works displayed here, and if that link is too slow to download then look here. Specifically, the pieces that have triangle in their names are relevant.

We were discussing this in Hanoi, and Dror Bar-Natan pointed out that in the continuous version you could conceive of the problem like this: A 3-cube consists of 3 un-ordered points in the interval. The cone on a cube is four points in the interval in which one is always to the left of the others. A triangle is an ordered pair of points in the interval, and the triangle cross the triangle is two ordered pairs of points in the interval in which the two pairs of points have no further relations. Let the two pairs of points be named (a,b) and (x,y). There are the following possible configurations: type A: (abxy), (axby), (axyb), and type B: (axyb) (xayb), (xyab). Those of type A form the cone on the prism b(xy) — with “a” being the cone point. Those of type B form the cone on the prism a(xy) with “b” as the cone point.

Meanwhile, the cone on the cube can be thought of as a{xyz} where the {xyz} go through the 6 permutations. Three of these (xyz, xzy, zxy) form one prism that is coned to a, and the other three (yxz,yzx,zyx) form the other prism.

The problem is related to categorification through John and Jeff’s musings about Falhauber and also with regards to some properadic considerations about brackets and parentheses.

Posted by: Scott Carter on July 13, 2009 8:49 PM | Permalink | Reply to this

### Re: Categorifiying Nichomacus of Gerasa’s Equation

The way I think about it (very close to Scott’s discussion, above) is to say the following. $f(n) = 1+2+\dots+n = \frac{n(n+1)}{2}$ is the area of a right-angle triangle, of sides, $n$ and $n+1$.

The product of two such triangles is a 4-dimensional figure, with volume $f(n)^2 = \frac{n^2 (n+1)^2}{4}$ Consider the difference in volume, when increasing $n$ by 1: $f(n)^2 - f(n-1)^2 = \frac{n^2[(n+1)^2-(n-1)^2]}{4} = n^3$

By induction on $n$, we obtain Nichomacus’s formula.

Posted by: Jacques Distler on July 13, 2009 9:56 PM | Permalink | PGP Sig | Reply to this

### Harmony of Nichomacus of Gerasa

What! no mention of the Pythagorean comma (musical)?

Posted by: jim stasheff on July 14, 2009 12:44 AM | Permalink | Reply to this

### Re: Harmony of Nichomacus of Gerasa

Jim wrote:

What! no mention of the Pythagorean comma (musical)?

Never fear! I referred back to week266, where I discussed the Pythagorean comma extensively.

Posted by: John Baez on July 14, 2009 8:23 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Very nice replies, all!

Here is Gavin’s solution. Remember, I’d said “It would be nice to exhibit Nicomachus of Gerasa’s formula as a way of taking the balls in a 4-dimensional stack of cannonballs with cubical layers, and rearranging them to form a 4-dimensional shape that’s a product of two triangles.”

Here’s how he did it:

Here is a bijection of which Nicomachus’s formula is the shadow. Let {$n$} denote the set of integers from $1$ to $n$ inclusive. Let us denote by $S(k,n)$ the coproduct of ${m}^k$ for $m$ running from $1$ to $n$. More concretely:

$S(3,n) = \{(t,x,y,z) | 1 \le t \le n, 1 \le x \le t, 1 \le y \le t, 1 \le z \le t\}$

$S(1,n) = \{(t,x) | 1 \le t \le n, 1 \le x \le t \}$

Define two functions $u,v : S(3,n) \to S(1,n)$ by:

$u(t,x,y,z) = (t,y)$ if $x \gt z$ else $(z,x)$

$v(t,x,y,z) = (t-z+1,t-x+1)$ if $x \gt z$ else $(t,y)$.

Then the induced function

$\langle u,v\rangle : S(3,n) \to S(1,n)^2$

is a bijection, I claim. This is about the simplest I can cook up. It has a simple geometrical meaning, more laborious to describe in words than with a picture. Are you acquainted with Dynnakov’s formula for braid ordering? I do not claim any connection but I am reminded by the little language: linear arithmetic extended by the .. if $\langle$cond$\rangle$ else .. operator.

I replied:

It looks like you’re slicing the ‘hyperpyramid’ $S(3,n)$ into two pieces along the hyperplane $x = z$, and then rearranging these to form $S(1,n)^2$. If so, I need to grok what these pieces look like.

He replied:

A ------ B -- C
|        |    |
|        |    |
D ------ E -- F
|        |    |
G ------ H -- J


$t$ = AC = AG

$z$ = AB

$t-z+1$ = FJ

Rectangles ABHG and DFJG fit together to make a square, the $z$th horizontal slice of the $t$th cube.

I think this idea of slicing a hyperpyramid into two pieces and fitting them together to form a product of two triangles is also the basis of Dror Bar-Natan’s idea as described by Scott Carter. But, I don’t know whether this solution is secretly ‘the same’ as Gavin’s.

Posted by: John Baez on July 14, 2009 7:50 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

This is, if I’m remembering correctly, more or less the solution given in “Proofs that Really Count” by Benjamin and Quinn.

Posted by: Qiaochu Yuan on July 14, 2009 7:30 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

I have a feeling a number of these solutions are secretly very close to one another, and differ just by a tweak or two.

For example, after seeing Gavin’s solution, it smelled similar to me to the Conway-Guy solution I described above, so I sat down to work out a possible explicit formula for the latter.

Following Gavin’s notation, let $T = S(1, n) = \{(x_0, x_1): 1 \leq x_1 \leq x_0 \leq n\}$ and write the typical element of $T \times T$ as $(x_0, x_1; y_0, y_1)$. Then, if I didn’t make a mistake, the Conway-Guy solution transcribes to a bijection $T^2 \to S(3, n)$, mapping

$(x_0, x_1; y_0, y_1) \mapsto (y_0, y_1, x_1, x_0 + 1) \qquad if y_0 \gt x_0$

$(x_0, x_1; y_0, y_1) \mapsto (x_0, x_1, x_0 - y_0 + y_1, x_0 - y_0 + 1) \qquad if x_0 \geq y_0$

which has a flavor somewhat similar to that of Gavin’s solution, even if his is “better” according to some reasonable set of aesthetic choices.

Also, I glanced at one of the references in the Wikipedia article that David Eppstein pointed to, which gave a q-analogue of the Nicomachus equation:

$\sum_{k=1}^n q^{2n - 2k} \frac{(1-q^k)^2 (1 - q^{2k})}{(1-q)^2 (1 - q^2)} = \binom{n+1}{2}_{q}^{2}$

However, this seems a bit formulaic, and it might be nice to give a categorified proof of this as well, in terms of combinatorial structures definable over $\mathbb{F}_q$, preferably one which would reproduce a Wraith-like solution when specialized to the “field with one element”.

Posted by: Todd Trimble on July 14, 2009 7:36 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Todd wrote:

However, this seems a bit formulaic, and it might be nice to give a categorified proof of this as well, in terms of combinatorial structures definable over $\mathbb{F}_q$, preferably one which would reproduce a Wraith-like solution when specialized to the “field with one element”.

Categorified $q$-deformed Nicomachus! Sounds great!

The right-hand side of the formula you write:

$\binom{n+1}{2}_{q}^{2}$

counts pairs of 2-dimensional subspaces of $\mathbb{F}_q^{n+1}$. The left-hand side expresses this as a sum… but unfortunately a sum over lots of terms, while there are just 3 basic incidence relations for a pair of 2d subspaces: they can intersect in a 0d subspace, or a 1d subspace, or a 2d subspace.

“Lots of terms.” But how many?

The left-hand side is a sum over $n$ terms. So, try duality: the right-hand side also counts pairs of $(n-1)$-dimensional subspaces of $\mathbb{F}_q^{n+1}$. And now there are $n$ basic incidence relations between these: they can intersect in a 0d subspace, a 1d subspace, … or an $(n-1)$d subspace.

So, I’ll wildly guess that the scary-looking summand in the formula you write:

$q^{2n - 2k} \frac{(1-q^k)^2 (1 - q^{2k})}{(1-q)^2 (1 - q^2)}$

counts the number of pairs of $(n-1)$-dimensional subspaces of $\mathbb{F}_q^{n+1}$ that intersect in a $k$-dimensional subspace. That would do the job.

Or an $(n+1-k)$-dimensional subspace — that would also do the job.

I prefer to quit here, feeling that I may have made progress, than check to see if I’m on the right track!

Posted by: John Baez on July 14, 2009 9:09 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

I’ll add my own hunch to the mix: if you look closely at these explicit categorified formulas for the case $q = 1$ ($\mathbb{F}_{un}$), that is to say bijective proofs of the classical Nicomachus formula, you find that there are these little braiding twists. For example, in Gavin’s formula you see $(t, y)$ appear in the first $u$-component under the condition $x \gt z$, and then appear in the second $v$-component under the condition $z \geq x$. Or, in the Conway-Guy picture as I described it first, there are moves where you stick a $j \times n$ rectangle together with an $n \times (n-j)$ rectangle to make an $n \times n$ rectangle, except that you have to interchange the dimensions of one to fit it with the other, hence a braiding twist. (Interestingly, Gavin also used the word ‘braiding’!)

The point, as you probably guess, is that I expect the Joyal-Street braiding on representations of the general linear groupoid

$GL(\mathbb{F}_q) = \sum_{n \geq 0} GL_n(\mathbb{F}_q) \to Vect$

also to make a similar but crucial appearance. You know about this, but I’ll recall anyway that the tensor product of two such representations is defined by the Green convolution formula

$(F \otimes G)(W) = \sum_{V \subseteq W} F(V) \otimes G(W/V)$

where $V$ ranges over subspaces of a finite-dimensional $\mathbb{F}_q$-vector space $W$, and the braiding is defined by taking into account all possible splittings of the exact sequence

$0 \to V \to W \to W/V \to 0,$

as described in

• A. Joyal and R. Street, The category of representations of the general linear groups over a finite field, J. Algebra 176 (1995), 908–946.

With luck I’ll find time to work through some of the details.

Posted by: Todd Trimble on July 15, 2009 1:42 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

The “derivative” of the continuous version is really easy to visualise.

Think of a hyperpyramid of dimension $L$ as a movie of a cube of side length $t$ that runs from $t=0$ to $t=L$. If you increase $L$ you don’t need to do anything except run the movie longer; the “derivative” wrt $L$ is just “the last frame”, the cube of side $L$.

Now, think of a product of triangles of dimension $L$ as a movie of a right triangular prism of height $t$ and a base that’s half of an $L\times L$ square, running from $t=0$ to $t=L$.

When you change $L$, there are now two distinct three-dimensional regions in the “derivative”. One is “the last frame”, the triangular prism that is half of an $L\times L\times L$ cube. The other is the three-dimensional region you get by following the top triangular face of the growing prism, as it grows from a height of $0$ to a height of $L$. But the second of these is identical to the first, so they fit together into a single $L\times L\times L$ cube.

Posted by: Greg Egan on July 15, 2009 2:35 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Sorry, I misidentified the second part of the “derivative” of the product of triangles.

It’s not the top triangular face of the growing triangular prism, it’s one rectangular face of dimensions $L\times t$, followed over the length of the movie. But again, this is itself a triangular prism identical to the other piece.

(Of course we’re free to make these 4-regions accrete with changing $L$ in all kinds of different ways, but these are choices that make things work out simply.)

Posted by: Greg Egan on July 15, 2009 3:12 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

This is a discrete version of what I was trying to describe in words above for the continuous case. The top row is the hyperpyramid, the bottom row is the product of triangles. The blue spheres are the ones you need to add to increment $n$; it’s clear that there are the same number of dark blue spheres in both cases, and easy to see that there are also the same number of light blue spheres.

Posted by: Greg Egan on July 15, 2009 6:40 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Doing the same thing for all $n$ gives the identification shown above.

Posted by: Greg Egan on July 15, 2009 1:34 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Beautiful! Now I see it.

Posted by: John Baez on July 16, 2009 8:28 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Posted by: steven e. landsburg on July 15, 2009 1:26 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Never mind! I posted before checking the wikipedia links; I see now that the same proof appears here including even a similar use of colors.

Posted by: steven e. landsburg on July 15, 2009 1:36 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

I know this discussion is getting old, but Greg’s pictures are really nice, and it reminded me of a couple of things. In a talk I gave in Tampa, I had a drawing of Dylan-Walter-Dave’s observation. This should give clues to Gavin’s and Greg’s version.

At the beginning of that talk, I asked, “If the integral of x to the third is one quarter, what is it a quarter of?” The answer is, of course, that it is a quarter of the 4d volume of a hyper-cube. Also, it is just as obvious that the integral of x to the nth takes up 1/(n+1) of the hyper volume of an n+1 cube, since the integral is measuring the hypervolume of the cone on an n-cube. Abhijit Champanerkar and I wrote down a proof here , but there were previous discoveries of this idea. Mathematica fans may enjoy playing with the code in the back of the paper — it has to be rewritten for the current version, but you get a nice rotating hyper-pyramid.

I still wonder what the Bernoulli numbers are measuring in these contexts.

Posted by: Scott Carter on July 15, 2009 5:05 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Scott wrote:

I know this discussion is getting old…

Getting old? Nicomachus proved his theorem around 100 AD… I posted a blog entry about it on Monday July 13th 2009 AD… and on Thursday you say the discussion is getting old?

For starters: is this theorem a special case of a general class of results? It boils down to saying that a cone on a 3-cube can be dissected into two copies of a cone on a 1-cube — and dissected in such a pretty way that the argument works ‘discretely’ as well as continuously. Could this be the only result of this type? It may be simplest and most beautiful… but if it’s the only one, that’s truly amazing.

Maybe we can dissect a cone on a 2-cube into a few parts, giving a nice proof of the less charismatic formula

$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$

(In fact I’m somewhat amazed that I know this formula: it’s only because it keeps coming up when I explain in calculus class how to compute the area under a parabola using Riemann sums. Which is not a very charming way to compute the area under a parabola. Scott’s way — reinterpreting it as the volume of a slice of a pyramid — is much prettier!)

Then there’s the problem of understanding the $q$-deformed version that Todd mentioned.

And so on. So, I hope people don’t give up on this discussion just because it’s a few days old.

Posted by: John Baez on July 16, 2009 8:46 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

If you multiply John’s formula through by 3, you get three pyramids with $n\times n$ bases being equal to a triangular prism of length $2n+1$. In the picture above, I’m adding successive layers to the structures for $n=1,2,3,4$; the three different hues are the three different pyramids, and the saturation of the colour represents the “level” within each pyramid.

Posted by: Greg Egan on July 16, 2009 11:26 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

In case the geometry isn’t clear in the picture above: the green and blue spheres comprise bona fide square-based pyramids, albeit drastically sheered. The red spheres comprise a triangle-based pyramid; each of the various layers of the notional red square-based pyramid – except the single-sphere layer (which here is the lowest red sphere) – have been split in two and stacked one on top of the other.

In the continuous case, I don’t know if there’s a nice isometric way to dissect a square-based pyramid and re-assemble it into a triangle-based pyramid twice as tall.

Posted by: Greg Egan on July 16, 2009 12:33 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

I guess the nicest way to rearrange three square-based pyramids into a double-length triangular prism is the one shown above. I figured out how to do this by looking at Scott’s tiling of the cube by three congruent pyramids here. But what I’m doing here is taking three congruent square pyramids, chopping just one of them in half (the red one), and rearranging the total of four pieces into a triangular prism.

Now I’ll try to figure out if there’s an exact discrete version of this that categorifies John’s discrete formula.

Posted by: Greg Egan on July 17, 2009 7:12 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

John wrote:

I’m not sure how it would work for the discrete case.

I guess I’m not 100% sure what the rules of the game are. I can construct a triangular pyramid by stacking $n$ layers, where the layer that is number $i$ starting from the vertex contains $i^2$ cannonballs, arranged as rows of length $1, 3, 5, ..., 2i-1$. That’s roughly the “right” shape and size, and the total number of cannonballs will, of course, match those in a square pyramid that also has $i^2$ in each layer. But however I tweaked the geometry of the stacking and the dissection, I’m not sure what would make it obvious that the total number of cannonballs in such a triangular pyramid is exactly:

$\frac{n(n+1)(2n+1)}{6}$

as opposed to some other very similar cubic expression. By this point, it seems like a losing proposition to me to expend any more effort trying to make the geometric process by which the cannonballs are classified into different subsets match the continuous version. At least the triangular prism construction where I multiplied everything by 3 made the formula apparent at a glance; even if I could find a way to rigorously align the geometry of the continuous and discrete triangular pyramid constructions, I’m not sure what the pay-off would be.

Posted by: Greg Egan on July 19, 2009 2:20 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

In the diagram above I was careless about chirality; when you slice Scott’s cube-tiling pyramids in half, the halves are chiral, but in the diagram above I’ve used two identical copies of one of the halves. A construction limited to dissections, rotations and translations is:

Posted by: Greg Egan on July 17, 2009 1:29 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Here’s a discrete version of the same construction, for $n=4$. When you “bisect” a discrete square pyramid it breaks into unequal parts, with one part getting the diagonal elements of the square layers and the other part missing out, which is where the length of $2n+1$ rather than $2n$ comes from.

Posted by: Greg Egan on July 17, 2009 2:16 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

I tried to find a way to carve a triangular pyramid out of the triangular prism assembled from red, green and blue square pyramids, in such a way that the union of the red, green and blue pieces lying within the triangular pyramid assembled into a single square pyramid. That would have given a nice categorification of the formula without the need to multiply through by 3, as a recipe for dissecting a single square pyramid and assembling it into a triangular pyramid twice as high. But this method never seems to give pieces that fit together, alas.

Posted by: Greg Egan on July 18, 2009 11:03 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Below is a particular square pyramid, seen from above; the apex is chosen to lie above a mid-point of one of the edges of the base.

By slicing down vertically through a plane that passes through the apex and the upper-right corner of the base, and then gluing the resulting tetrahedron on to the left-hand side of the remaining solid, we get a triangular pyramid with the same volume.

Posted by: Greg Egan on July 19, 2009 7:49 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Your argument is getting very nice. But in its last version I’m not sure how it would work for the discrete case. Maybe you can cut some cannonballs in half?

(Hmm, is there are known class of ‘almost bijective proofs’ where you show two sets of balls have the same cardinality not by giving an explicit bijection, but by chopping up the balls in one set and explicitly matching up these pieces with pieces in the other set of balls?)

Posted by: John Baez on July 19, 2009 12:43 PM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

The “square pyramid” discrete set above has a cannonball with centre $(k, j-i/2, -3i)$ for positive integers $i,j,k$ such that $1\leq i\leq n, 1\leq j\leq i, 1\leq k\leq i$; here I’ve used $n=5$. The cannonballs are coloured blue if $i-k\geq 2j-1$.

Here the blue cannonballs, those for which $i-k\geq 2j-1$, have been moved from $(k, j-i/2, -3i)$ to $(1-k, 1-j+i/2, -3i)$, i.e. subject to the translation and rotation $(x, y, z)\to (1-x, 1-y, z)$. This corresponds exactly to the “triangular pyramid” discrete set with centre $(i+1-k, j-i/2, -3i)$ for positive integers $i,j,k$ such that $1\leq i\leq n, 1\leq j\leq i, 1\leq k\leq 2j-1$, with the cannonballs coloured blue if $k\gt i$.

So you can construct an explicit bijection that doesn’t split any cannonballs … but I can’t see how this helps us prove that the total number of cannonballs is exactly $\frac{n(n+1)(2n+1)}{6}$.

Posted by: Greg Egan on July 20, 2009 12:56 AM | Permalink | Reply to this

### Re: Categorifiying Nicomacus of Gerasa’s Equation

Having converted square pyramids to triangular pyramids by the dissection described above, with a bit of shearing it’s possible to fit three of the triangular pyramids together into a triangular prism that makes the RHS of the counting formula manifest.

Posted by: Greg Egan on July 20, 2009 4:41 AM | Permalink | Reply to this

### Re: Categorifying Nicomacus of Gerasa’s Equation

Neat. I haven’t read all the comments so this might be redundant, but this reminds me of a binary tree:

• Heads, my wealth increases by a factor of 2.
• Tails, my wealth becomes the sum of my previous wealth and what my wealth would have been if we had flipped heads, which is (x+2*x) = 3x.

I wish I could play this game because my wealth would always increase! :)

Posted by: Eric on July 16, 2009 6:57 PM | Permalink | Reply to this

### Re: Categorifying Nicomachus of Gerasa’s Equation

Somehow in my N years of studying mathematics I’ve never come across this basic fact. Here’s my contribution to the discussion - a little proof-by-picture that popped into my head as I was lying in bed this morning - I came up with this without looking at any references:

This is not as cool as Greg Egan’s colored cannonballs, but is perhaps easier to understand. The idea is to decompose the N×N×N cube into N×N squares and pack them around the top and right of the existing square - the image shows this construction up through N=7. In the case N is odd, you neatly fit in N of these squares, when N is even one of the squares is split in two - the shaded regions in the image. (This reminds me a bit of the famous picture-proof of the Pythagorean theorem, where you have a square of size A, a square of size B, and two A×B rectangles).

Posted by: Charles Waldman on July 20, 2009 2:04 PM | Permalink | Reply to this

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