Symmetry and the Monster (8, finale): Moonshine

The eighth episode of the series, and also the finale. In the last episode, the Monster was truly built by one pair of hands — it lives in 196883 dimensions, and it carries a “+1” that at the time no one could explain (196884). In this episode we go and look at that “+1”: it appears, to the letter, in a number-theory formula written down more than a hundred years earlier. This is not a coincidence; it has a name, Moonshine. Who saw it first, how it was proved, and — why to this day no one can really explain it.

“So the real task is not so much to see what no one has seen, as to think — facing what everyone can see, to think a thought that no one has thought.” (translated)

— Arthur Schopenhauer (1788–1860), Parerga und Paralipomena, 1851

Author’s note

In just these two months while I was making and publishing the eighth episode, quite a few big things happened in the AI world and the mathematical world: Claude used Lean to formalize the proof of Fermat’s Last Theorem in full, that is, it wrote the proof that humans had completed thirty years ago, step by step, into code a machine can check; OpenAI announced that it had proved that solutions of the NS equations can blow up, and although whether it is right has not been settled as of today, it is already shocking enough. There are also many, many proofs that were previously out of reach that AI is now attempting one by one.

The first thing I felt was not joy but anxiety. This anxiety comes from thinking about the uncertainty of the future: the mental activity I take pride in is gradually being replaced by AI, so where will I myself go? The feeling that followed was bewilderment: since I do not know where to go, how should I live from here on?

What I can do is treat AI as a partner who is both teacher and friend, not as an intelligent agent that thinks for me; and see myself clearly: I am not as capable as AI, but I am a living person. In the first episode I wrote that the person reading this article must be alive; having reached the last episode, I come back to that sentence again.

More importantly, AI cannot take responsibility for our lives.

Companion video (YouTube)

Prologue · A coincidence that should not exist

In mathematics one occasionally meets this kind of thing: two objects that look completely unrelated, separated by a hundred years, belonging to different branches, nevertheless match on a set of numbers. You cannot prove that they must match, yet they simply do.

In November 1978, a British mathematician living in Montreal was casually reading a paper at home to pass the time. His own work was group theory, but the paper in his hands was number theory — he had assumed the two had nothing to do with each other. As he read, a number jumped out, and this number looked very familiar. He realized that the number he was really familiar with was another one: it differed from this number by just 1.

In that year, the two things linked by these two numbers had neither yet fully landed: on one side was the Monster, which at the time had not yet been built — how big it should be and what it should look like, mathematicians had computed a “medical report” for, but no one had seen it in person (that is exactly what EP7 was about); on the other side was a number-theory formula written on paper more than a hundred years earlier. A colossus not yet born, and an old formula long since yellowed, had, between 196883 and 196884, some kind of connection that could not be explained or put into words.

What this episode is about is investigating their connection (§1), what exactly this old formula is (§2), why these two unrelated fields match up (§3), and how it was finally proved, and why it is still a mystery to this day (§4). This is the last episode of the series. Having come this far, we will also look back at this road, which started from “why are there exactly five regular polyhedra” and took two thousand five hundred years to reach “Moonshine”.

§1 · McKay’s coincidence

The mathematician who, as just mentioned, came upon the coincidence was John McKay (1939–2022), a group theorist at Concordia University. The paper he was reading dealt with a classical object of number theory called the j-function (according to Ronan’s book, it was one by Atkin and Swinnerton-Dyer). The j-function can be expanded into a string of numbers, the first few of which are:

McKay’s gaze stopped at the first “interesting” coefficient: 196884. Trained in group theory, he recognized at a glance that this number was almost another number he was familiar with — the dimension of the smallest representation of the Monster, 196883 (the central number of EP7, $=47\times59\times71$) — with one more 1.

In 1978 the Monster had not yet been built — if this is not cleared up first, the whole story loses its mystery. Griess (Robert Griess) constructed it by hand only in 1980 (paper 1982). When McKay came upon this coincidence, 196883 was still only a predicted character degree that Bernd Fischer, Donald Livingstone and Michael Thorne had computed before “the Monster was proved to exist”. So the accurate statement is not “the representation dimension of the Monster” but “the smallest representation dimension that the Monster (at the time still a prediction) ought to have”. This is where it is most anomalous: it was discovered between two objects that had neither fully landed.

“Too close to be a coincidence.” McKay wrote a letter to John Thompson, the grand master of the group theory world at the time. (It was a letter, not a postcard; according to Ronan, he did not use the post, but entrusted the letter to Fischer, who had come from Princeton to Montreal to give a talk, to carry back — Thompson was at the time at the Institute for Advanced Study in Princeton.)

Someone else might have let it go as a coincidence; Thompson did not, he got to work — in his own words: “I started fooling around with it, and tried the next coefficient.” With this try he realized — because this was not a coincidence of one number at all, but of a whole string. The Monster has a “fingerprint table” (the $194\times194$ character table, discussed in EP7 §2); take its first few dimensions and do simple additions, and they match the coefficients of the j-function term by term:

Three lines, all correct term by term. (These can be checked independently up to the third — computing j on one side and matching it with Monster dimensions on the other; from the fourth coefficient, 20245856256, onward, more dimensions and less obvious multiplicities are needed, and it can only be checked by citing ATLAS data; see Appendix B for details.)

Thompson also took one more crucial step: since the coefficients of j correspond to the numbers of the Monster’s identity element in each dimension, could other elements of the Monster each also hide such a sequence? — this idea led straight to the whole set of conjectures in §3.

(A small episode, according to McKay’s later account as recorded in Marcus du Sautoy’s Finding Moonshine: the further decomposition of the higher coefficients was published first by Thompson; McKay himself had in fact also worked it out, “I was a bit peeved really”. And this string of series is today called the McKay–Thompson series, the honor going to both.)

Figure · McKay’s “portrait”: photographs of McKay from his lifetime exist, but we could not find one that can be freely used. What he left the world is the coincidence itself that he recognized at a glance — $196884 = 196883 + 1$. (Made by this channel.)

§2 · That old formula: what the j-function is

To understand this coincidence, we must first get to know the j-function. It has been in number theory for more than a hundred years, long an old acquaintance that “everyone can see” — only no one imagined it would get mixed up with a monster.

And to get to know j, we first need to change our angle.

2.1 Starting from complex numbers

In mathematics there is a concept called the upper half-plane:

First, what the complex numbers mean: $z$ is a complex number, that is, a point in the plane; $x$ is its real part, defined in the complex plane by the horizontal coordinate, $y$ is its imaginary part, defined by the vertical coordinate, and $i$ is the imaginary unit ($i^2=-1$). $y>0$ restricts this point to above the horizontal axis — hence the name “upper half” plane. But the horizontal axis itself ($y=0$, called the real axis) is not included.

On the upper half-plane, length is not measured the Euclidean way, but is defined by the following formula:

In this formula, $dz$ is a small step moved starting from $z$; $|dz|$ is the absolute distance between two points in the plane; $\operatorname{Im} z$ is read “the imaginary part of $z$”, that is, the $y$ of the upper half-plane mentioned above. $ds$ is then the length recorded for this small step under this standard.

Plug in two points and look: at $z=1+10i$, taking a step $|dz|=1$ gives $ds=1/10=0.1$; at $z=1+0.001i$, taking the same step gives $ds=1/0.001=1000$. For the same Euclidean length, the smaller $y$ is, the larger the reading; as $y\to 0$, $ds\to\infty$, so under this metric the real axis lies at infinity.

Area changes along with it:

$dx\,dy$ is the area of a small rectangle measured the usual way (width $dx$, height $dy$); $dA$ is the area it is recorded as under this standard. Here we divide by $y^{2}$ and not by $y$, for a very plain reason: area is the product of lengths in two directions, each direction has to be divided by $y$ once, and twice makes $y^{2}$.

The reason for choosing this metric is: it stays unchanged under the following set of transformations. First look at this set of transformations:

$z$ is the point before the transformation, $w$ is where it lands after the transformation; $a,b,c,d$ are four real numbers, the four parameters of this transformation. $ad-bc$ is called the determinant of this transformation, and it governs three things, three different conditions:

• $ad-bc\neq 0$ — this transformation can go back the way it came (if it equals $0$, it cannot go back);
• $ad-bc>0$ — after the transformation it still lands in the upper half-plane;
• $ad-bc=1$ — this is a normalization: for any positive determinant, dividing all four numbers $a,b,c,d$ by $\sqrt{ad-bc}$ turns it into $1$, and this scaling does not change $\frac{az+b}{cz+d}$ itself (scaling top and bottom at the same time cancels out). So requiring it to equal $1$ is not adding one more restriction; it is choosing one standard way of writing from the infinitely many ways of writing the same transformation.

One example makes all three clear together. Take $(a,b,c,d)=(3,1,1,1)$; then $ad-bc=2$:

The determinant is nonzero, so it is invertible (the inverse transformation is $(1,-1,-1,3)$); the determinant is positive, so $\operatorname{Im}w$ has the same sign as $\operatorname{Im}z$, and a point in the upper half-plane is still in the upper half-plane after the transformation. But in the numerator there is an extra $2$, which is exactly $ad-bc$. Divide all four numbers by $\sqrt2$, and the determinant becomes $1$, while the expression for $w$, after multiplying top and bottom by $\sqrt2$, is identical to the original letter for letter — the map has not changed, only the way of writing it, and the extra $2$ disappears.

After choosing this standard way of writing, only one doubling in the way of writing remains: $(a,b,c,d)$ and $(-a,-b,-c,-d)$ give the same transformation — in the $\mathrm{PSL}$ discussed in 2.2, that $\mathrm P$ takes care of exactly this. The modular group discussed in 2.2 later is the group obtained by further restricting $a,b,c,d$ to integers.

For such transformations, we can also derive the identities directly:

The first formula says: the height after the transformation equals the original height divided by $|cz+d|^{2}$. The second says: the small step after the transformation equals the original small step divided by $(cz+d)^{2}$. (Both formulas use the $ad-bc=1$ chosen above — in the general case each of their right-hand sides also carries a factor $ad-bc$, which disappears only after the standard way of writing is taken.) What appears in both formulas is the same $cz+d$ — this is not a coincidence; it is exactly what makes the next step work. Take absolute values and divide, and $|cz+d|^{2}$ cancels exactly top and bottom:

This is an important conclusion: the left side of the equals sign is the length recorded after the transformation, the right side is the length recorded before the transformation, and the two are, remarkably, exactly the same. That is to say, every transformation in the modular group is length-preserving under this standard.

So, take a region in the upper half-plane, and with these transformations we can get many results of the same lengths; in 2.3 next we will explain further how it comes about, and here we first call them tiles:

Those tiles are each the same size. “The closer to the edge, the smaller” is only an illusion of the Euclidean coordinate system — just as on a map Greenland looks bigger than Africa.

Let us combine this with a concrete example, and take this piece:

$\operatorname{Re}z$ is the real part of $z$ (that is, the $x$ above), and $|\operatorname{Re}z|\le\frac12$ says “no more than half a cell to the left or right”; $|z|$ is the distance from $z$ to the origin, and $|z|\ge1$ says “outside the unit circle”. This region is called the fundamental domain. Add up the $dA$ above over it (the lower limit of integration comes from $|z|\ge1$, i.e. $x^{2}+y^{2}\ge1$, which solves to $y\ge\sqrt{1-x^{2}}$):

Four equals signs, each with its reason:

1. Split into an iterated integral. $F=\left\{-\tfrac12\le x\le\tfrac12,\ y\ge\sqrt{1-x^{2}}\right\}$, so the outer integral is over $x$ and the inner over $y$.
2. Integrate $y$ first. $\displaystyle\int_{a}^{\infty}\frac{dy}{y^{2}}=\left[-\frac1y\right]_{a}^{\infty}=\frac1a$; substitute $a=\sqrt{1-x^{2}}$.
3. Recognize the antiderivative. $\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^{2}}}$.
4. Plug in the limits. $\arcsin\tfrac12=\tfrac{\pi}{6}$ (because $\sin 30^\circ=\tfrac12$), $\tfrac{\pi}{6}-\left(-\tfrac{\pi}{6}\right)=\tfrac{\pi}{3}$.

The area of every tile is $\pi/3$, however small it is drawn. Infinitely many pieces, each with the same area.

2.2 What those two transformations actually do in the plane

If the four parameters $a,b,c,d$ of the set of transformations above are restricted to integers (still requiring $ad-bc=1$), what we get is the modular group, written $\mathrm{PSL}(2,\mathbb{Z})$. Read this notation piece by piece: $2$ means these four numbers are arranged as $2\times2$; $\mathbb{Z}$ is the standard notation for all the integers; $\mathrm{SL}$ means the determinant is $1$; the $\mathrm{P}$ in front means $(a,b,c,d)$ and $(-a,-b,-c,-d)$ are regarded as the same transformation — because after plugging into $\frac{az+b}{cz+d}$, changing the sign of top and bottom at the same time cancels out, and the landing point is exactly the same.

This whole group of transformations is generated by two transformations. Let us look at them one at a time, at where they send a point.

The translation $T:\ z\mapsto z+1$. The whole picture shifts one cell horizontally to the right. The imaginary part does not move; the height does not change. (Corresponding to $a=1,b=1,c=0,d=1$.)

The flip $S:\ z\mapsto -1/z$. (Corresponding to $a=0,b=-1,c=1,d=0$.) It is clearest written in polar coordinates $z=re^{i\theta}$ — $r$ is the distance from this point to the origin, $\theta$ is its angle with the positive real axis:

The distance $r$ becomes $1/r$ — this is inversion in the unit circle, swapping inside and outside the circle; the angle $\theta$ becomes $\pi-\theta$ ($\pi$ is the radian way of writing $180^\circ$) — this is reflection in the imaginary axis, swapping left and right.

So: the unit circle as a whole does not move (when $r=1$, $1/r$ is still $1$), and points on the circle are only swapped left and right; the point $i$ does not move (its $r=1$, $\theta=\frac{\pi}{2}$, and $\pi-\frac{\pi}{2}$ is still $\frac{\pi}{2}$), it is the pivot of this transformation; the inside and the outside of the circle are swapped entirely.

Looking on the imaginary axis:

A point at height $y$ is sent to height $1/y$. A point at height $\tfrac1{100}$ lands at height $100$; and $i$ (that is, $y=1$) is the fixed point. In general

— the height after the transformation equals the original height divided by “the square of the distance to the origin”. So points near the spot directly above the origin (for example points on the imaginary axis close to $0$: $\operatorname{Im}z$ is very small, and the square of $|z|$ is even smaller) get flung, after the division, to very high places.

$T$ and $S$, these two, are enough to generate the whole modular group. This is a conclusion; Appendix A uses exactly it, and the main text only gives a general reading.

2.3 What a tile is: one fundamental domain, and all its transformed images

We cite the example from 2.1:

These two conditions correspond exactly to the two generators: $|\operatorname{Re}z|\le\frac12$ — go any further left or right and $T$ translates you back ($T$ translates one whole cell at a time, so half a cell is the dividing line); $|z|\ge1$ — go any further in and $S$ flips you out ($S:z\mapsto-1/z$ does exactly inversion: $\left|-1/z\right|=1/|z|$, a point at distance $r$ from the origin is sent to $1/r$ — so points inside the unit circle run to outside the circle, and points on the circle stay on the circle (swapped left and right, only $i$ does not move)).

So $F$ is “one representative of each point’s equivalence class under the modular group”. Any $z$ in the upper half-plane can be sent into $F$ by some element of the modular group, and in essentially only one way (with a few gluings on the boundary). The full proof is in Appendix A.

What is called a “tile” is all the copies obtained by applying this set of transformations to $F$ everywhere: $T(F)$, $S(F)$, $TS(F)$, …… infinitely many pieces, pairwise congruent under the standard of 2.1, filling the whole upper half-plane.

The relation between tiles and coordinates: a point in the plane is an ordinary complex number $z=x+iy$; a tile is not a pattern drawn on the plane, but a grouping of “which points should be regarded as the same point”. $z$, $z+1$, $-1/z$ are, in the eyes of the modular group, the same point, so only one piece $F$ needs to be kept as the representative.

2.4 The j-function: the characteristic of an elliptic curve

The j-function is a function defined on the upper half-plane: feed it a point $z$, and it gives back a complex number $j(z)$. What is special about it on this infinite tiled picture is two things.

The first: on every tile it takes the whole complex plane completely once — every complex number is taken, and each is taken only once — and it stays unchanged under all the transformations of the modular group (called modular invariant). Its meaning is defined by two:

The first equation in the first definition: move the point one cell to the right ($T$), and the value of $j$ does not change; the second says: flip the point over by $S$, and the value of $j$ also does not change. And 2.2 said that $T$ and $S$ generate the whole modular group — both generating transformations leave it unchanged, so the whole modular group leaves it unchanged.

The second definition is: in this position, such a function is it alone (the technical term is hauptmodul, “principal modular function”). The meaning is: any modular function (modular invariant, meromorphic, and also meromorphic at the cusp) can be written as a rational function of j — a rational function being one polynomial divided by another, of the form $\frac{P(j)}{Q(j)}$; the field of modular functions is $\mathbb{C}(j)$ ($\mathbb{C}$ is the standard notation for all the complex numbers, $\mathbb{C}(j)$ means “all the rational functions generated by $j$”). That is to say, in this world, j alone is one coordinate: all other modular invariants can be computed from j. Both of these definitions (uniqueness, and how j maps the whole picture one-to-one onto the whole complex plane) are fully proved in Appendix A — including the $\mathbb{C}(j)$ statement, which uses only Appendix A’s own valence formula.

What was said above is what j satisfies. But what it actually is has a more accurate answer, and this answer will follow us into §3 and §4.

First take a complex number $z$ (with positive imaginary part); it gives

The meaning is: $\mathbb{Z}$ is all the integers, so $\mathbb{Z}+\mathbb{Z}z$ means all points of the form $m+nz$ ($m,n$ ranging over all integers) — take $1$ and $z$ as two directions and walk out in integer steps, and you get an infinitely extending grid in the plane. $\Lambda_z$ is the set of all the intersection points of this grid, and $\subset\mathbb{C}$ means they are all complex numbers.

Roll up the complex plane according to this lattice (regard all points in the grid that differ by integer steps as the same point), and you get a torus — that is, an elliptic curve $\mathbb{C}/\Lambda_z$.

The key step: two $z$‘s produce the same elliptic curve (in the sense of isomorphism) if and only if they are related by the modular group:

$z_1$, $z_2$ are two starting points; $\cong$ is read “isomorphic”, meaning these two curves are essentially the same one, only written differently; $\iff$ is “if and only if”, meaning the two sides are necessary and sufficient conditions for each other.

So that tiled picture is no longer a decorative painting. Each elliptic curve has exactly one representative point in $F$, no more and no less. The other tiles are copies of $F$, holding another way of writing the same batch of curves; the various ways of writing the same curve fall at the corresponding positions of the various tiles.

And the j-function is what finds a unique characteristic for each elliptic curve:

It is modular invariant precisely because all the ways of writing the same curve must be unique; when we take the whole complex plane, and take it only once, then each elliptic curve has exactly one and only one counterpart.

These two statements do not conflict; they are two sides of the same thing: the $\mathbb{C}(j)$ proved in Appendix A is exactly the algebraic way of writing “j is a complete invariant”. And the statement of uniqueness explains two things: why §3 suddenly starts talking about “supersingular elliptic curves”, because those are the same batch of objects over a different number field; and why in §4 the coefficients of j turn out to be “dimensions”, because the generating function of §4 is computed on a torus, and a torus is an elliptic curve — it can only depend on the curve itself.

2.5 Escher’s disk

If you feel that this picture, which gets smaller and smaller as it tiles and never repeats, looks a lot like the tessellations in the disks of M. C. Escher, that is not an illusion, and there is an exact one-line explanation of it. That transformation is called the Cayley transform:

$z$ is a point in the upper half-plane, $w$ is where it lands in the disk. Plug in two numbers and you see what it is doing: when $z=i$, the numerator is $0$, so $i\mapsto 0$ — the “center” of the upper half-plane becomes the center of the disk; when $z$ is taken as a point on the real axis, $z-i$ and $z+i$ are a pair of conjugate complex numbers of equal length, and the result of dividing them has modulus $1$, so the whole real axis is sent to the boundary circle of the disk. The tiles in the upper half-plane that “get smaller and smaller against the real axis” are, in the disk, the ring that “gets smaller and smaller against the boundary” — the same batch of tiles, the same standard, only drawn a different way. What Escher drew and what number theory uses really are the same thing.

2.6 That 744, and the 0 after subtracting it

We talked about the j-function earlier; when we expand j at infinity, we get the string of coefficients of §1. The variable used for the expansion is

This is derived from $T$, one of the two generators. $T:z\mapsto z+1$ gives $j(z+1)=j(z)$, that is, $j$ has period $1$; and every function with period $1$ can be written as a function of $e^{2\pi iz}$ — so $q$ is not picked out; it is the unique natural coordinate. ($S$ does not take part in this step.)

At the same time it “rolls” the upper half-plane toward infinity: writing $z=x+iy$, $|q|=e^{-2\pi y}$, so the larger $z$‘s imaginary part $y$, the closer $q$ is to $0$; and when $z$ moves one cell to the right, the value of $q$ does not change at all (because $e^{2\pi i}=1$). The expansion starts from $q^{-1}$, and the coefficients are all integers (the first few terms are computed term by term in Appendix A.3; that all of them are integers follows immediately from Appendix A’s assembly $j = E_4^3/\Delta$ — the leading coefficient of $\Delta$ is $1$).

The standard j-function carries a constant term 744:

And what we use from here on is the version after subtracting 744:

Since the $q^{-1}$ term and the $196884\,q$ term are exactly the same, the constant term changes from $744$ to $0$. In §4 we will see that the coefficients of $J$ are exactly the “layer-by-layer dimensions” of something.

Where 744 comes from can in fact also be made clear: j can be assembled from two parts —

First make clear what “weight” is — it is about to appear in the next sentence. Weight says “how much it differs by after the transformation”. A modular form of weight $k$ satisfies

Note that it is not invariant: after the change it differs by a factor $(cz+d)^k$, and $k$ is the exponent of this factor. Only weight $0$ is truly invariant — the factor becomes $1$.

Plug in numbers and look: the weight of $E_4$ is $4$, so the weight of $E_4^{3}$ is $4\times3=12$; the weight of $\Delta$ is also $12$. Divide two things of weight $12$, and that factor is $(cz+d)^{12}$ on both top and bottom and cancels:

⟹ That $j$ is written in the form “one thing divided by another” is not a contrivance: to build a function of weight $0$, the numerator and denominator must have the same weight, otherwise that factor cannot cancel.

⚠️ But what this requirement fixes is the shape, not this one function. $E_6^{2}$ also has weight $12$ and is also holomorphic on the upper half-plane (differentiable everywhere in that region, with neither poles nor breaks), so $E_6^{2}/\Delta$ also has weight $0$ — it equals $j-1728$ (the relation between the two is the identity $E_4^{3}-E_6^{2}=1728\,\Delta$). The accurate statement comes in three parts: (1) wanting weight $0$ ⟹ the numerator and denominator must have the same weight, and there is no other choice at this step; (2) further requiring holomorphy on the upper half-plane and only a pole of order one at the cusp ⟹ it can only be “weight $12$ divided by $\Delta$”; (3) but what satisfies the first two is a whole family $a\,j+b$ — requiring the leading term of the $q$-expansion to be $q^{-1}$ fixes $a=1$; the constant $b$ can still be chosen freely: the constant term of $j$ is what the way of writing it as $E_4^{3}/\Delta$ computes, $744$, while the $J$ used later takes the constant term to be $0$.

Both are standard modular forms. $E_4$ is the Eisenstein series of weight 4, whose $q$-expansion is $E_4=1+240\sum_{n\ge1}\sigma_3(n)q^{n}$ ($\sigma_3(n)$ is the sum of the cubes of all the positive divisors of $n$); $\Delta$ is the discriminant, $\Delta=q\prod_{n\ge1}(1-q^{n})^{24}$, with leading coefficient $1$. The complete definitions of both, and the properties used here, are in Appendix A. Divide the $q$-expansions of the two, and what falls out in the constant place is 744. It is computed directly from this ratio; Appendix A computes this ratio in full from the Eisenstein series. And for j’s identity as “unique”, adding or subtracting a number in the constant term does no harm. What really needs attention is, on the contrary, the 0 after the subtraction: in §4 we will see that it equals the dimension of layer 1 of the moonshine module — that layer is empty; and “layer 1 is empty” is precisely the characteristic, in §4, of “the symmetry group can only be finite, not continuous”.

§3 · “Moonshine”: one word, and a whole set of conjectures

At the end of that year, Thompson returned from Princeton to Cambridge and told this strange affair to John Conway (1937–2020) — the same Conway as in EP6. Conway had in hand a huge amount of data on the Monster, and he immediately plunged into the library, and sure enough, in nineteenth-century number-theory literature, he found the same sequence. Together with a young mathematician, Simon Norton (1952–2019), he checked it again and again: between the Monster and number theory there really was a connection, but no one could say why.

The one who gave a name to this affair that could not be explained was Conway. He called it Moonshine. The word is a double pun, and it is worth citing Conway’s own words as they are (as recorded in Ronan’s book; given here in translation):

The things we had, with no logical argument holding them up, felt like dim moonlight shining on a group of dancing Irish leprechauns. And Moonshine can also mean illicitly distilled liquor — doing this research really did have something of the flavor of illicit distilling that cannot stand the light of day.

(“moonshine” in English already has the meaning of “foolish dreaming, absurd nonsense” — this layer is the common sense; the “bootleg liquor” layer of the pun is Conway’s own joke.)

Conway and Norton wrote this set of conjectures down in 1979, under the title “Monstrous Moonshine” (Bull. London Math. Soc. 11, 308–339). What the conjecture says goes far beyond the one function j: every conjugacy class of the Monster corresponds to a modular function of its own, and each one is the principal modular function of some genus 0 group.

What the conjecture says: four words, plugging in numbers one by one

There are four words in this sentence that need taking apart. Below, each is given a definition, and then numbers are plugged in.

Conjugacy classes: why by “class” and not by “element”

$g$ and $g'$ being in the same class means that there exists some $h$ in the group such that $g'=hgh^{-1}$. Count it out with $S_3$ (all the permutations of three letters, $|S_3|=6$): the six elements fall into 3 classes:

The reason for going by class is that the trace, which the next step takes, is unchanged under conjugation — let $\rho$ be any representation;

Elements of the same class give the same trace, so the whole series depends only on the class, not on which element of the class was picked. The Monster has 194 conjugacy classes (the number of rows of that $194\times194$ fingerprint table of EP7 §2 is exactly this), so there are at most 194 series.

The McKay–Thompson series $T_g$: the definition, then plug in $g=e$

The conjecture here first takes something on credit. Suppose there exists such a stack of vector spaces laid out in layers — this is what later became the moonshine module, written $V^\natural$:

Read it symbol by symbol. $\bigoplus$ is the direct sum, meaning the layers are laid side by side without being mixed up; $V_n$ is layer $n$, a finite-dimensional space, and its dimension $\dim V_n$ is exactly that string of coefficients of §1 ($\dim V_0=1$, $\dim V_1=0$, $\dim V_2=196884$, ……). The $\natural$ at the upper right is read “natural”; it is the traditional notation for this object and does not denote any operation. $\mathbb{M}$ is the Monster, and the Monster acts on each layer — every element $g$ in the group gives an invertible linear map on layer $n$. What matters is the timeline: in 1979, when the conjecture was written down, this stack of spaces had not yet been built — its very existence was part of the conjecture; §4 will tell how it was later built.

So for each $g\in\mathbb{M}$ define

In the formula, the vertical bar in $\operatorname{tr}\left(g\mid V_n\right)$ is read “restricted to”: regard $g$ only as that linear map on layer $n$, then take its trace (the sum of the entries on the main diagonal of the matrix). The exponent is written $n-1$ rather than $n$ to shift the whole series by one place so that it starts from $q^{-1}$ — aligned with the $J$ of §2.6; otherwise the two sides would differ by a factor of $q$.

First plug in $g=e$. The identity element is the identity matrix in any representation, and its trace is the dimension, $\operatorname{tr}(e\mid V_n)=\dim V_n$, so

This step connects §1 back in. Those three identities of §1 are exactly the first few coefficients of this series split up according to the dimensions of the irreducible representations of the Monster:

The left side is $\dim V_n$, the right side is layer $n$ split into irreducible blocks. When $g=e$ the trace can see only the dimension; change to another $g$, and the trace is no longer the dimension, and what we get is another series — 194 classes, so there are 194 of them.

Change the element: the same layer, another number

The step just now used only the identity element. The conjecture is about other elements, so now let us change it.

Take an element of order two in the Monster, called 2A by its ATLAS class name. The layer is still the same layer — $V_2$, the dimension is still 196884, and the splitting is still the same splitting:

The only thing that changes is the element plugged in. The traces of $e$ on the two blocks are the respective dimensions of the two blocks; the traces of $2A$ on the two blocks are two other numbers:

($4371$ is the character value of the $196883$-dimensional block of the representation at $2A$, read off the character table of the Monster.)

So we have the first few terms of the second series; place it side by side with the first:

How these two lines look side by side is how the whole set of conjectures looks: the same stack of spaces, the same splitting, change the element, and the whole series changes to another set of coefficients. That string of numbers in §1 that matched the j-function turns out to be only the first row of this table.

Genus 0 and the principal modular function

A group $\Gamma$ acts on the upper half-plane, and after taking the quotient we get a surface; then add the cusps back — the upper half-plane itself does not contain the cusps, and only after they are added is the surface compact — written $X_\Gamma$. Genus 0 means that this compact surface is a sphere; the field of modular functions on the sphere is generated by a single function, and this generator is called the principal modular function (hauptmodul).

The first example has already been given in §2. Take $\Gamma=\mathrm{PSL}(2,\mathbb{Z})$; $X_\Gamma$ is the surface generated from the fundamental domain $F$ as in §2.3, with that one cusp added back, and $j$ takes each complex number exactly once on $F$ (§2.4); counting the cusp in together with $\infty$, “exactly once” is

that is, $X_\Gamma$ has genus $0$ and $j$ is its principal modular function (full proof in Appendix A; the boundary note for the general case is also there).

The second example is the one just now. The group that goes with $T_{2A}$ is $\Gamma_0(2)$ with a Fricke involution $W_2:\tau\mapsto-1/(2\tau)$ added, written $\Gamma_0(2)+$; in Conway–Norton’s original table it is written “2+”. Its compactified quotient $X_0(2)^{+}$ is likewise a surface of genus $0$, and $T_{2A}$ is likewise its principal modular function — the function field is generated by this one function.

The group changed; the matter of “genus 0 plus one principal modular function” did not change. This shape is exactly what the conjecture says.

These functions were not made for moonshine. $T_{2A}$ can be written down without mentioning the Monster at all, using only the $\eta$ function of number theory (OEIS files this sequence under Fricke’s 1922 textbook on elliptic functions, and links it to Ramanujan’s paper on approximating $\pi$; we did not get to these two original books, and the entries are in the sources). Moonshine did not create this batch of functions; it recognized the Monster on a batch of functions that had long lived in number theory.

A complete analysis of the conjecture

Now we can say it in full: every conjugacy class of the Monster gives a $T_g$, and every $T_g$ is the principal modular function of some genus 0 group.

The one for $g=e$ we already saw in §2 — $T_e=J$, and the group is $\mathrm{PSL}(2,\mathbb{Z})$ itself; the one for $2A$ we just saw — the group is $\Gamma_0(2)+$. Each of the remaining ones has its own group.

A small note on counting: the 194 classes do not give 194 different series, but 171. The smallest example can be seen already on $C_3=\{e,g,g^{2}\}$ — $C_3$ is a commutative group, each element is a class by itself; take its representation as rotations of the plane ($g$ rotating by $2\pi/3$), and $\operatorname{tr}(g)=2\cos\frac{2\pi}{3}=-1$ while $\operatorname{tr}(g^{2})=2\cos\frac{4\pi}{3}=-1$: two different classes give the same trace. The mechanism is $g^{2}=g^{-1}$, and $\operatorname{tr}(g^{-1})=\overline{\operatorname{tr}(g)}$; when the trace is real the two coincide. It is the same in the Monster, and it happens on every layer at once: $g$ and $g^{-1}$ always share the same series, and this step brings 194 down to 172. For the last step, from 172 to 171, this article did not find the mechanism — Conway and Norton name it separately: the only other equality apart from inverses is between two conjugacy classes of order 27, and that there are no other equalities (Conway–Norton 1979, p. 310). Even in this comparison table there hides a coincidence without an explanation. (A class-by-class count is in Appendix F.)

In other words — every conjugacy class of the Monster has a principal modular function of its own.

Ogg’s other half of the riddle

In fact, nearly four years before McKay, another person had come upon the other half of the same fog.

On 14 January 1975, Jacques Tits gave his inaugural lecture at the Collège de France, and spoke of Fischer’s huge symmetry group that had not yet been built — “the Monster”. In the audience sat the Berkeley number theorist Andrew Ogg (1934– ). Ogg at the time had in hand a list of primes he had computed himself. After hearing that lecture, he compared the two sides and found them, astonishingly, exactly the same.

How that list was computed

What Ogg asked was a purely number-theoretic question: take the modular curve $X_0(p)$, then take the quotient by the Atkin–Lehner involution (when $p$ is prime, this is the Fricke involution), getting $X_0^{+}(p)$; when does this curve have genus 0 (that is, the simplest shape, a sphere)?

The answer is a finite list. And the way Ogg himself wrote it down shows that it has structure better than listing does:

$p\le 31$, or $p=41,\;47,\;59,\;71$.

There are 11 primes with $p\le 31$, and adding the 4 after them, 15 in all:

This whole derivation, from beginning to end, uses only number theory — modular curves, genus, involutions. It can be computed before the Monster was constructed, and in fact that is just how it was: in 1975 the Monster was still only a conjectured object; it would be truly built by Griess only in 1980 (the paper published in 1982). What Ogg himself wrote in that note was “if it exists”.

Plugging in $p=71$:

First the few objects that appear here. $X_0(p)$ is a modular curve, built exactly the same way as the $X_\Gamma$ above — take the quotient of the upper half-plane by a group to get a surface — only the group is changed to $\Gamma_0(p)$ (requiring that the lower-left entry $c$ of the matrix be divisible by $p$). The genus is how many holes the surface has: a sphere has genus $0$, a life ring has genus $1$. The cusps are the few points that land at infinity after taking the quotient.

The genus of $X_0(p)$ has a closed formula ($p\ge5$ prime, two cusps):

where $\left(\frac{a}{p}\right)$ is the Legendre symbol: when $a$ is a square modulo $p$ it takes $+1$, and when it is not, $-1$; $\nu_2,\nu_3$ count the elliptic points of orders 2 and 3 on the surface (special points fixed by some element of the group). Plug in a number and look: $\left(\frac{-1}{71}\right)=-1$, because $-1$ is not a square modulo $71$ (this is so for every prime $p\equiv3\bmod4$), so $\nu_2=1+(-1)=0$; likewise $\left(\frac{-3}{71}\right)=-1$, $\nu_3=0$. Plug in $p=71$:

Then take the quotient by the Fricke involution $w_p$ — it is a symmetry of order two of $X_0(p)$ onto itself (do it twice and you return to where you were); take the quotient by it too, and the resulting surface is written $X_0^{+}(p)$. This step is a double cover, and the Riemann–Hurwitz formula gives the genus of the new surface ($f$ is the number of fixed points of $w_p$; $h(-D)$ is the class number of the discriminant $-D$, that is, the number of equivalence classes of primitive binary quadratic forms of that discriminant, for example $h(-71)=7$):

These two class numbers can be obtained by directly counting reduced binary quadratic forms: $h(-284)=7$, $h(-71)=7$, so $f=14$,

It lands exactly on 0. And the next prime no longer does: plugging in the same way, $g(X_0(73))=5$, $g^{+}(73)=2\neq0$. So the list ends at 71, and it is not “these are all that have been found so far”: Ogg proved that after 71 the genus is never zero (this article recomputed up to 200, see Appendix C) — this is also why “the largest one is 71” is not an open case.

And on the other side, factor the order of the Monster group into primes:

The primes that appear are exactly 15, and written down they are

Identical to the list above, to the letter.

Each side can be computed independently, neither uses the other, and then they are identical to the letter. That is what makes it worth being astonished at. And Ogg in 1975 was taking a list computed perfectly clearly and comparing it with the order of a group that did not yet exist at the time.

What exactly is “super” about “supersingular”

The primes on this list are called supersingular primes in number theory.

Go back to those elliptic curves of §2.4. They live over the complex numbers; consider the same batch of equations instead over a finite field of characteristic $p$, and most properties of the curves stay as before, but a very few behave anomalously:

• Ordinary: the $p$-torsion points on the curve number $p$ in all (including the identity).
• Supersingular: there is not a single torsion point of order $p$.

The endomorphism ring of a supersingular curve is especially large. In the usual case it is an order in an imaginary quadratic field; in the supersingular case it is an order in a quaternion algebra — non-commutative. “So much symmetry that it does not look like what a curve ought to have” — this is where the name comes from.

Given $p$, there are only finitely many supersingular $j$-invariants (about $p/12$ of them), and all of them fall in $\mathbb{F}_{p^2}$. The 15 primes on the list are exactly those for which these $j$ values all fall in the smaller $\mathbb{F}_p$ (Appendix C).

That other half of the riddle

Look once more at the end of the list: 47, 59, 71. The last episode said that these are exactly the three largest prime factors of the order of the Monster, and their product is exactly 196883 — the number McKay recognized had been lying all along at the tail of Ogg’s list.

Ogg simply wrote this coincidence into the paper he was writing at the time (in a Remarque, not a theorem), offering a bottle of whiskey as a reward for an explanation. _{(What he wrote in the original was Jack Daniels, without the apostrophe; the distillery’s official spelling is Jack Daniel’s. When quoting his original words we follow the former.)} When McKay made his observation nearly four years later, the reward of this bottle was still open — and to this day (2026), no one has cleanly claimed the bottle.

He told this strange affair to Tits and to Serre (Jean-Pierre Serre, the one who wrote A Course in Arithmetic), and Serre’s reaction was a single “Sans blague” (no kidding). No one had the slightest idea why it should be so.

This riddle and the riddle of moonshine itself are two different matters, and must be looked at separately. First, which half is settled: the list of supersingular primes itself — including “the largest one is 71”. It is a classification result that Ogg computed from the side of modular curves: once $p$ exceeds 71, the genus of $X_0^{+}(p)$ is no longer zero, and the list stops there. What hangs open is why the two lists are equal. Moonshine proved one direction of it — “a prime divides $|M|$ ⟹ it is a supersingular prime” (so a prime greater than 71 certainly does not divide $|M|$); but the other way round — why each one of the 15 primes on the list happens to divide $|M|$, and why the prime factors of the Monster happen to reach exactly up to 71 — to this day there is no conceptual explanation; this is an independent unsolved riddle. (See Appendix C for details.)

§4 · The proof: the moonshine module, and a string theory

The conjecture was sitting there, matching term by term, but why did it match? No one could say. And the road that can at present be called an explanation is understanding it with the help of physics.

The moonshine module V♮

First comes the first step, completed by three mathematicians, Frenkel, Lepowsky and Meurman (FLM). They built a space — first announced in 1984, written up in 1988 as the book Vertex Operator Algebras and the Monster — and this space brings two properties together: its layer-by-layer dimensions are exactly the coefficients of $J=j-744$; and its symmetry group is exactly the Monster. This space is called the moonshine module $V^\natural$, and its structure is called a vertex operator algebra (VOA). The stack of spaces taken on credit in §3 was only here truly built.

In $V^\natural$, that $\natural$ is the natural sign of musical notation, read “natural” — it takes part in no operation; it is only a proper-name marker. $V^\natural$ as a whole is read “the moonshine module”, and refers to exactly the one particular space below.

In physics there is a concept called conformal field theory. A field theory describes “what is vibrating at each point of space”; conformal is a requirement added on top of it: enlarge, shrink or bend the whole picture arbitrarily, and as long as the angles everywhere are unchanged, this theory cannot tell the difference between before and after. ⟹ It has no built-in unit of length; “how big” carries no information for it. Two dimensions is the setting where this requirement is most powerful: the local angle-preserving transformations in two dimensions are so many that they form an infinite-dimensional symmetry, strong enough to almost pin down the theory itself; whereas in $d\ge3$ dimensions, such transformations form only a finite-dimensional group — two dimensions is the only exception. The theory in this episode has two identities: mathematically, its axiomatization is the VOA to be described below; physically, it lives on the surface swept out by a string (see the “worldsheet” later in this section).

A vertex operator algebra (VOA), then, is the structure that writes “two-dimensional conformal field theory” as axioms. Concretely, it is a graded vector space

($\bigoplus$ is the direct sum, laying the layers side by side without mixing them up), plus one rule: each state $a\in V$ is paired with an operator

$Y(a,z)$ is called the vertex operator of $a$; $z$ is not a number but a formal variable, used only to number a string of operators $a_{(k)}$ ($a_{(k)}$ is read “$a$‘s $k$-th mode”; it sends one state to another state). In this structure there is a special element $\omega$, which gives an operator $L_0$ and a number $c$: $L_0$ is responsible for the grading (layer $n$ is those states for which the eigenvalue of $L_0$ is $n$), and $c$ is called the central charge. ⟹ A VOA is the rigorous version of the “operator product expansion” of string theory (the physics-side account is in Appendix E, the mathematics-side axioms in Appendix D.1; this section uses only the two points “grading + each layer is a representation of the Monster”).

The “+1” of §1 here acquires an identity: the layer-by-layer dimensions of $V^\natural$ are the coefficients of $J$, and the Monster acts on it — the two things are no longer a numerical coincidence across empty space, but two faces of the same object.

“Layer by layer” deserves a sentence of expansion. The dimension of layer $n$ is exactly the coefficient that sits, in $J$, in the $q^{\,n-1}$ place — layer 1 corresponds to the $0$ in the constant place, layer 2 to the $q$ place’s $196884$, and so on (that leading $q^{-1}$‘s coefficient $1$ also has an identity: layer 0, the vacuum, one-dimensional). And the Monster acts on each layer separately — so each layer is itself a representation of the Monster. Look at layer 2: it is 196884-dimensional, and split up it is exactly the one-dimensional trivial representation $\oplus$ the 196883-dimensional smallest representation — McKay’s “+1” is the one-dimensional trivial direction in this layer. The additions that Thompson did back then when he “tried the next coefficient” acquire a geometric meaning here: that was splitting the representations layer by layer (the three identities of §1 are exactly the splittings of layers 2, 3 and 4).

The key is where the words “vertex operator” come from — they come from physics, from string theory. In the preface to their 1988 book, FLM say it very plainly (translated from the original English preface; given here in translation):

Our main theorem can be read as a quantum-field-theoretic construction of the Monster; in fact, what it says is: the Monster is the symmetry group of a certain special string theory.

This is the part of this episode written for readers in theoretical physics (the conceptual development is in Appendix E; here only the picture is given).

Where that 24 comes from. The critical dimension of the bosonic string is $26$; after light-cone gauge fixes away one pair of coordinates, $24$ transverse oscillators remain — where $26=24+2$ comes from (several different roads point to the same $26$) is in Appendix E.1, and the main text does not repeat it. It is precisely these 24 directions that need to be compactified, so a 24-dimensional lattice is needed.

Take the Leech lattice $\Lambda$ (the lattice of EP6). What is special about it is that it has no “roots”, roots meaning lattice vectors of squared length 2:

Two words are needed here. Even says that the squared length of every vector in the lattice is an even number; unimodular says that the determinant of the lattice is $1$ — an equivalent statement is that this lattice and its dual lattice are the same one. These two conditions pin things down extremely tightly: there are only 24 even unimodular lattices in 24 dimensions in all (this list is called the Niemeier classification, settled in 1973), and the other 23 all have roots; Leech is the only one that does not.

How is it constructed? Compactify these 24 bosons on the torus built from $\Lambda$, and you get a vertex operator algebra $V_\Lambda$; then fold away one $\mathbb{Z}/2$ symmetry:

In the formula above: $\theta$ is the symmetry obtained by negating lattice vectors ($\lambda\mapsto-\lambda$) and lifting it up; the superscript $\theta$ is read “take the part invariant under $\theta$”, that is, those states that stay exactly as they are after being acted on by $\theta$; the superscript $T$ is read twisted, and $V_\Lambda^{T}$ is the unique “twisted sector” of this symmetry. ⚠️ This is not as simple as “divide by $\mathbb{Z}/2$”: one must both project onto the invariant part and attach the twisted sector — taking only $\left(V_\Lambda\right)^\theta$ does not give $V^\natural$ (Appendix D.2).

The coincidence 196884 is assembled

What is most worth noticing about this construction is here: $196884$ is not a parameter tuned to match $j$; it is the result of adding two parts. Count layer 2 —

The untwisted half: the $\theta$-invariant states are of two kinds: lattice vectors of squared length 4 ($\lambda$ and $-\lambda$ paired up, each pair giving one invariant combination), and the symmetric square of the 24 directions:

⚠️ The “weight” here and the “weight 4” of §2 are two different quantities. The weight of §2 is the exponent that appears extra after a transformation; the weight here is the layer number — that number of layer 1, layer 2; it says only which layer this state sits in. That both places happen to be called “weight” is a collision of historical naming; when reading, understand each according to its own sentence.

The twisted half: the ground state of the twisted sector is $2^{24/2}=2^{12}=4096$-dimensional, at weight $3/2$; add one more mode to rise to weight $2$, and there are 24 directions that can be added:

Adding the two halves:

Every number on the left comes from the lattice and the construction itself; the right side is $J$‘s coefficient in the $q$ place. This step uses no information whatsoever about the $j$-function from start to finish.

Why layer 1 must be empty

Layer 1 of a lattice VOA has only two sources — roots, and the dimension of the lattice:

And $\theta$ acts on these 24 directions exactly as $-1$, so not a single one is invariant; and the twisted sector starts only from weight $3/2$ and cannot reach weight $1$. So

This step is the core, and the whole chain of logic is:

($\operatorname{Aut}$ is read “automorphism group”, meaning all the symmetries that keep this structure unchanged.) Why the middle step holds: the states of weight 1 are exactly the generators of continuous symmetries — each one gives a direction of a Lie algebra, along which one can rotate continuously. When layer 1 is empty, there is not a single continuous direction, the dimension of the symmetry group is $0$, and so it can only be a finite group (strictly speaking one also needs: such automorphism groups are algebraic groups, and an algebraic group of dimension $0$ must be finite) — and this finite group is exactly the Monster.

Change the lattice and the conclusion no longer holds, and this can be verified. Take any of the other 23, for example $E_8^3$ — three copies of $E_8$ placed side by side. In $E_8$, the vectors of squared length equal to $2$ number $240$ (such vectors are called “roots”), so three copies have $720$; whereas Leech has $0$:

As said earlier, there are 24 even unimodular lattices in 24 dimensions in all, and only Leech among them has $0$ roots. Back to $E_8^3$: after folding, layer 1 is $720/2=360\neq0$, and the continuous symmetry comes back. And layer 2 is still $196884$ — $E_8^3$ has fewer vectors of squared length $4$ than Leech ($179280$ versus $196560$), $17280$ fewer, which after halving is $8640$ states, made up exactly by the $24\times720/2=8640$ states of “one root plus one oscillator”. Change the lattice, and not one of the other coefficients of $J$ changes; the only thing that changes is the constant place. (This is not a coincidence: modular invariance pins down all the coefficients other than the constant place; see Appendix E.3.) So what singles out Leech is not $196884$, it is that $0$.

The “vertex operators” of this conformal field theory are exactly the operators of string interactions. And the most crucial point is: the partition function of this conformal field theory on a torus (that is, the generating function of its layer-by-layer dimensions) is exactly $J = j-744$ (the rigorous statement requires Zhu’s theorem; see Appendix E). So “why the j-function should happen to be a modular function” gets a clear explanation on the physics side — the modular invariance of the worldsheet torus; this phrase needs to be taken apart. A point particle moving in spacetime draws a line, called a worldline; but a string is not a point, it is something with length, and what it sweeps out as it moves is no longer a line, but a two-dimensional surface — this is called the worldsheet. All the calculations in string theory finally come down to “what happens on this surface”.

Next look at the shape of the surface. A closed string is a loop, and what it sweeps out when it propagates freely is a cylinder; and when this process is joined end to end — the string goes around one whole trip and returns to the state it started from — the two ends of the cylinder are glued together, and only then is what we get a torus (the surface of a doughnut). What needs computing is exactly this.

And the word “torus” has already appeared in §2.4: there it was the torus obtained by rolling up the complex plane according to the lattice $\Lambda_z$, its shape described by one complex number $z$ (in the appendices the same quantity is written $\tau$; it is the same thing). That sentence of §2.4 now comes into use a second time: two $z$‘s make the same torus if and only if they are related by the modular group. In other words, the same worldsheet can be labeled with different $z$‘s, and what separates these different labelings is exactly the action of $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. Since physics recognizes only the surface itself, and not which $z$ you chose to label it with, the quantities computed on this surface must be invariant under $\mathrm{PSL}(2,\mathbb{Z})$ — that is to say, it must be a modular function. The symmetry of §2.4, which looked like a convention internal to number theory, here becomes a rule that physics must obey. That old formula of number theory, more than a hundred years old, is in fact related to a special kind of string theory.

⚠️ But we cannot stop the statement here: modular invariance is a threshold, not a sieve. It requires the answer to be a modular function, but does not tell you which one — there is more than one candidate that clears this threshold. What really picks $J$ out from among them is the independent fact of construction above: layer 1 is empty. The rigorous statement is in Appendix E.3.

Borcherds’s proof, and the Fields Medal

The one who pushed FLM’s construction on to “proving the whole conjecture” was Richard Borcherds (1959– ) — when the moonshine module was announced, he was still a doctoral student under Conway. He built a new class of infinite-dimensional Lie algebras (today called Borcherds algebras), and picked out from them one key special case — the “Monster Lie algebra”, making use of the no-ghost theorem of string theory (Goddard–Thorn, 1972). In 1992, in the paper “Monstrous moonshine and monstrous Lie superalgebras” (Invent. Math. 109, 405–444), he completely proved the Conway–Norton conjecture: every McKay–Thompson series is indeed the principal modular function of a genus 0 group.

The difficulty of the conjecture is: on one side is a group, on the other side are functions. On the Monster side there are 194 conjugacy classes, on the number-theory side there are 171 series, and the Conway–Norton conjecture assigned a series to each class — but there was no reason at all for “why they match”; you had nothing in hand that could carry the information from this side to the other side.

What Borcherds built was exactly that bridge. His approach was not to check those 171 series coefficient by coefficient (each one has infinitely many coefficients; that could never be finished), but:

First build a “Lie algebra” from $V^\natural$. A Lie algebra is something that appeared in EP4 — the set of “infinitesimal generators + commutation relations” that describes continuous symmetry. The one he built (he called it “the Monster Lie algebra”) has two properties, and neither can be missing:

• It remembers the Monster: the Monster acts on it, so its numbers everywhere carry the Monster’s information;
• It remembers $j$: this class of Lie algebras (generalized Kac–Moody algebras) each come with a “denominator identity” — its structure list can be written down completely in two ways: one written as a product, one written as a sum; the two ways of writing describe the same thing. And the denominator identity of this one is exactly that product formula for $j$ already written down long ago in number theory.

The link is this one identity. The same algebra, seen from the group side, is a representation of the Monster; seen from the number-theory side, it is the expansion of $j$. What remains is to “twist” this identity once — twisting once for each conjugacy class — and one gets all 171.

Therefore, he did not prove, one by one, what each of those 171 series equals; he built a thing that ties them together, whereby each series is pinned down by its first few coefficients, and what is left is only one finite check.

The skeleton has four steps, and each step is given a formula one can look at (Appendix D.3 has five steps, cut differently from here).

① Build a Lie algebra. Tensor $V^\natural$ with another vertex operator algebra and take its physical states:

$\mathfrak{m}$ is a German blackletter m, read “the Monster Lie algebra”; $\mathrm{II}_{1,1}$ is a two-dimensional even unimodular Lorentzian lattice, the central charge of whose vertex operator algebra is $c=2$, so together it is $24+2=26$ — the same 26 as the critical dimension (for why, see Appendix E). ⚠️ It is $V^\natural\otimes V_{\mathrm{II}_{1,1}}$, not the Leech route; the latter gives a different algebra (Appendix D.3).

② This step is what the no-ghost theorem refers to. Goddard–Thorn identify, in $\mathfrak{m}$, the physical states of momentum $(m,n)$ with a certain whole layer of $V^\natural$ — and the isomorphism is as representations of the Monster, not just equality of dimensions. So the “root multiplicities” of the Lie algebra can be expressed as:

($\operatorname{mult}$ is read “multiplicity”, meaning how many linearly independent elements there are in the direction of this root; $c(N)$ is, of $J=\sum_n c(n)q^n$, the $N$-th coefficient.) Plug in a concrete root $(m,n)=(1,1)$:

The same 196884; this is its third identity in this episode — the left side of McKay’s equation, $J$‘s coefficient in the $q$ place, and now the multiplicity of a root.

③ The denominator identity. The “denominator identity” of this class of Lie algebras is an identity that writes its structure as a product equal to a sum. The one for $\mathfrak{m}$ is exactly the Koike–Norton–Zagier product already known in number theory (the $p$ here is only shorthand for $e^{2\pi i\sigma}$ and has nothing to do with the $p$ denoting a prime):

The left side is two values of $j$ subtracted, the right side is an infinite product, and the exponents are exactly those multiplicities $c(mn)$ above. The structure of the algebra and the arithmetic of $j$ are, in this one line, forcibly pressed into the same thing.

④ Only by twisting do we get all 171. The one above is the untwisted version; it recovers only the one for $g=e$ (that is, $J$ itself). What really gives all the $T_g$ is the “twisted” denominator identities, one for each conjugacy class — replace the dimensions in the formula above by the traces $\operatorname{tr}(g\mid V^\natural_n)$. “Twisting” is the key, not a footnote: what it gives is exactly the replication identities proposed by Norton (1984).

Closing up: complete replicability determines each $T_g$ uniquely by the first 5 coefficients of it (together with the various powers of $g$), and what remains is one finite computer check — so for the 194 classes and 171 series, every one lands on a principal modular function of genus 0.

In 1998, Borcherds received the Fields Medal for this (and related work) (at the International Congress of Mathematicians in August, Berlin), and the citation named moonshine.

About how he felt on receiving the prize, in the year he received it he said something to Simon Singh, who interviewed him. In translation, the gist: he said he did not really feel anything special; before receiving the prize he had always thought it was enormously important, and now he understood that it did not mean anything. But — at the moment he proved the moonshine conjecture, he was so happy he flew to the moon (“I was over the moon”). The prize is cold; the proof is warm. And over the moon literally means “up on the moon” — the man who proved moonshine said of himself that he had gone up on the moon.

The vertex operator algebras, generalized Kac–Moody algebras and the no-ghost theorem of string theory used in the course of the proof are a true high peak, and a blog appendix certainly cannot cover all the details, so here we pass over it in a stroke. So this episode cites it honestly and does not rebuild it: Appendix D gives the skeleton on the mathematics side (VOA, $V^\natural$, Borcherds’s strategy), and Appendix E gives the conceptual picture on the physics side. We prove to the end what can be proved (Appendices A, B), and state plainly what we cannot prove.

Coda · Proved, yet still a mystery; and a road walked for two thousand five hundred years

Moonshine has been proved; Borcherds’s theorem holds.

But “proved” and “understood” are indeed two different things. Why the Monster, why modular functions, why precisely the Leech lattice, why every series of that family is genus 0 — string theory has given the best clue so far (the modular invariance of the worldsheet, which connects the number-theory side and the symmetry side), but that is not a complete explanation. Ogg’s half of the riddle, the reverse direction, still has no resolution to this day (that is another riddle — why the two lists are equal; there has never been a conceptual reason); and that bottle of Jack Daniel’s has still not been claimed by anyone. This, perhaps, is what is most mysterious about it.

So this series too has come to its end. Let us look back once at this road:

From the five regular polyhedra in the hands of the ancient Greeks, asking “why are there only five”, the starting point was a geometric fact — there are exactly five regular polyhedra; the ancient Greeks counted them, but did not know that what they were counting was in fact symmetry. More than two thousand years later, Galois, to answer “why the quintic equation has no root formula”, made symmetry into an object that can be operated on — the group. Once there was an object, there was the next question: break all finite symmetries down until they cannot be broken further, and count these “atoms” clearly. Counting from Galois, it took more than a hundred years, and the answer is: eighteen infinitely extending families (exactly what EP5 took stock of), plus 26 sporadic groups that no family can take in (the box EP6 opened) — the largest of them is the Monster. The last step was one nobody expected: this largest exception matches, term by term, an old formula of number theory more than a hundred years old; this correspondence was proved to hold by Borcherds in 1992, and why it holds, no one to this day can say. From “why exactly five” to “why exactly a match” — the starting point and the end point of this road are the same kind of question.

Humankind has chased this thread of “symmetry” for two thousand five hundred years, from the regular polyhedra the ancient Greeks drew in the sand, to a monster living in over a hundred thousand dimensions, gazing across space at the depths of the whole of number theory. It did not stop at a full stop of “at last we understand it all”, but stopped before a riddle that can be proved, yet still cannot be understood.

The twenty years after the book

Ronan’s book stops at 2006. In these twenty years the riddle has not been solved — on the contrary, it has grown bigger.

In 2010, three physicists — Tohru Eguchi, Hirosi Ooguri and Yuji Tachikawa — came upon the same kind of coincidence in a completely different place: computing a function on the K3 surface (a standard geometric stage in string theory) — the elliptic genus — they found that the expansion coefficients decompose term by term into the dimensions of representations of M24 — the largest of the five Mathieu groups of EP6. This correspondence itself was rigorously proved to hold by Terry Gannon in 2012; but it is even more “mysterious” than Monster moonshine: Monster moonshine at least has the moonshine module $V^\natural$ as a “home”, while Mathieu moonshine has not even found its “home” yet — it has been proved that no concrete K3 model really carries this M24 symmetry, yet the correspondence is sitting there plain as day. The paper that discovered it writes, in the last sentence of its own abstract: “The reason is yet a mystery.”