Binomial Coefficient Coincidences

20 September, 2026

These seven equations between binomial coefficients are ‘coincidences’: they aren’t among the four known infinite families. De Weger conjectured that there are no more such coincidences:

• Benjamin M. M. de Weger, Equal binomial coefficients: some elementary considerations, Journal of Number Theory 63, no. 2 (1997), 373–386.

At that time, he and his collaborators checked there were no others involving binomial coefficients less than 1030. Later they checked that there are none involving binomial coefficients less than 1060:

• Aart Blokhuis, Andries Brouwer and Benne de Weger, Binomial collisions and near collisions.

So, De Weger’s conjecture stands open. The four infinite families, by the way, are these:

\displaystyle{ \binom{n}{k} = \binom{n}{\,n-k\,}, \qquad 0 \le k \le n}

\displaystyle{ \binom{n}{0} = 1, \qquad n \ge 0 }

\displaystyle{ \binom{\binom{n}{k}}{1} = \binom{n}{k}, \qquad \qquad 0 \le k \le n}

and the only nontrivial one: the Lind–Singmaster family involving the Fibonacci numbers F_i where F_0 = 0,\ F_1 = 1:

\displaystyle{ \binom{F_{2i+2}F_{2i+3}}{\,F_{2i}F_{2i+3}\,} \;=\; \binom{F_{2i+2}F_{2i+3}-1}{\,F_{2i}F_{2i+3}+1\,}, \qquad i = 1,2,3,\dots }

The first three equations in the Lind–Singmaster family are these:

\begin{array}{ccc} \displaystyle{ \binom{15}{5} } &= &\displaystyle{\binom{14}{6}} \\ \\ \displaystyle{\binom{104}{39} } &= &\displaystyle{\binom{103}{40}} \\ \\ \displaystyle{ \binom{714}{272} } &=& \displaystyle{\binom{713}{273}} \end{array}

I’ll explain the Lind–Singmaster family later. But here’s the question I’m most interested in:

Is there any good explanation for the seven binomial coefficient coincidences?

Today my collaborator Paul Schwahn found a beautiful explanation of the first one, namely

\displaystyle{ \binom{10}{3} = \binom{16}{2} }

His explanation uses representation theory. The Lie algebra \mathfrak{so}(10) has a 10-dimensional representation, the ‘vector’ representation V_{10}, and also two 16-dimensional representations, the ‘left and right-handed spinor’ representations S^\pm_{10}. There’s an isomorphism of representations

\displaystyle{\Lambda^2 S^+_{10} \cong \Lambda^3 V_{10} }

and similarly for S^-_{10}, but we might as well work with S^+_{10}. Here \Lambda^k means the kth exterior power. For any vector space X we have

\displaystyle{ \dim(\Lambda^k X) = \binom{\dim X}{k} }

Thus, taking dimensions, the isomorphism of representations

\displaystyle{\Lambda^2 S^+_{10} \cong \Lambda^3 V_{10} }

instantly gives

\displaystyle{ \binom{16}{2} = \binom{10}{3} }

It is not super-easy to prove this isomorphism of representations, but it’s still nice to find a deeper layer of meaning underlying what might otherwise seem like a meaningless coincidence!

Can we find representation-theoretic explanations—or other explanations—for the other six coincidences?

I have not succeeded, but let me tell you about two failed tries.

We can look for isomorphisms like

\displaystyle{\Lambda^2 S^+_{10} \cong \Lambda^3 V_{10} }

involving representations of \mathfrak{so}(n) for larger n. In fact this isomorphism is part of a pattern! The next one involves the vector and left-handed spinor representations of \mathfrak{so}(12). But it’s this:

\displaystyle{ \Lambda^2 S^+_{12} \cong \Lambda^4 V_{12} \oplus \Lambda^0 V_{12} }

so it gives

\displaystyle{ \binom{32}{2} = \binom{12}{4} + 1 }

or

\displaystyle{ 496 = 495 + 1 }

So we fail to get an equation between binomial coefficients: we’re off by one.

The second paper I cited, Binomial collisions and near collisions, presents a list of cases where two binomial coefficients differ by one. This is on the list. So we failed to explain an equation between binomial coefficients, but explained a near-miss.

Here’s another failed attempt at explaining an equation between binomial coefficients. The equation

\displaystyle{ \binom{78}{2} = \binom{14}{6} = 3003 }

is fascinating to anyone who knows their exceptional Lie groups. 78 is the dimension of \mathrm{E}_6, while 14 is the dimension of \mathrm{G}_2. \mathrm{G}_2 is a subgroup of \mathrm{E}_6 because \mathrm{G}_2 is the automorphism group of the octonions and \mathrm{E}_6 is the isometry group of the bioctonionic plane. We’d get the above equation if the 2nd exterior power of the adjoint representation of \mathrm{E}_6, upon being restricted to \mathrm{G}_2, were isomorphic to the 6th exterior power of the adjoint representation of \mathrm{G}_2.

Amazingly, it seems these two representations of \mathrm{G}_2 are not isomorphic even though their dimensions are the same: both 3003.

Even more amazingly, \mathrm{E}_6 and \mathrm{G}_2 both have irreducible representations of dimension 3003, but they are not the representations I just mentioned.

I would be happy for someone to check these two claims.

If anyone knows good explanations of the remaining six binomial coefficient coincidences, please let me know!

The Lind–Singmaster family

Lind and Singmaster were trying to find all n,k with

\displaystyle{ \binom{n}{k} = \binom{n-1}{k+1} }

I’ll rapidly sketch the key steps of their argument. Simplifying the equation above we get

n(k+1) = (n-k)(n-k-1)

or

n^2 - (3k+2)n + (k^2+k) = 0

Solve for n using the quadratic formula. This formula turns out to have

\sqrt{5k^2 +8k+4}

in it. So we need 5k^2 +8k+4 to be a perfect square!

Now we’re trying to find integer solutions of

5k^2+8k+4 = m^2

A quadratic diophantine equation! Multiply by 5 and complete the square:

5m^2 = (5k+4)^2 + 4

y = 5k+4 is an integer when k is, so we need to find integer solutions of

y^2 - 5m^2 = -4

This is a ‘Pell equation’, and people know how to solve these. In this particular case we get all the solutions from this fact:

L_n^2 - 5F_n^2 = 4(-1)^n

where F_n are the Fibonacci numbers 0, 1, 1, 2, 3, … and L_n are the Lucas numbers 2, 1, 3, 4, 7, …. These are two sequences satisfying the same famous recurrence relation, just with different initial conditions.

We want n odd, to get

L_n^2 - 5F_n^2 = -4

It turns out y = L_n, m = F_n with n odd give all solutions of the Pell equation

y^2 - 5m^2 = -4

However, remember I said y = 5k+4 is an integer when k is. But the converse isn’t always true, and we need k to be an integer! This clearly happens iff y \equiv 4 \bmod 5.

So we need to know when L_n \equiv 4 \bmod 5 Apparently this happens iff n \equiv 3 \bmod 4. I won’t think about this now… but this is the last hard step.

In summary, we’ve seen

\displaystyle{ \binom{n}{k} = \binom{n-1}{k+1} }

if and only if y = 5k+4 is a Lucas number L_n with n \equiv 3 \bmod 4. We could quit here, but people like to use the identity

L_{4i+3} - 4 = 5F_{2i} F_{2i+3}

to get a formula for k in terms of Fibonacci numbers. This is gilding the lily, I’d say, but that eventually leads to the formula I showed you:

\displaystyle{ \binom{F_{2i+2}F_{2i+3}}{\,F_{2i}F_{2i+3}\,} \;=\; \binom{F_{2i+2}F_{2i+3}-1}{\,F_{2i}F_{2i+3}+1\,}, \qquad i = 1,2,3,\dots }

The takeaway message is: our problem can easily be reduced to a quadratic diophantine equation, then put in Pell form… and it’s known that the sequence of integer solutions of a Pell equation obeys a linear recurrence relation! We luck out in this case and get solutions connected to Lucas and Fibonacci numbers.

There may be a simpler argument, but this is what I’ve seen.

19 Comments | mathematics | Permalink
Posted by John Baez

The E6 Root Polytope

6 September, 2026

I’ve been thinking about the exceptional Lie algebra E6, as a spinoff of my project on E7, so I want to get a good mental picture of the E6 root polytope. This is 6-dimensional polytope with remarkable symmetry.

Let’s climb up to it, starting with some of its 4-dimensional faces, which are called 4-demicubes because you get them by taking a 4-dimensional cube, or tesseract, and removing every other corner. The 3-demicube is just a tetrahedron, since you can fit two tetrahedra in a 3-dimensional cube like this:

The 4-demicube builds on this fact in a surprising way.

I’m going to use the technology of Dynkin diagrams, or technically Coxeter diagrams: they’re closely related, and the difference is invisible here. I won’t explain them, just use them. I explained them here:

• Symmetry and the fourth dimension: part 3, part 4, part 5, part 6.

Let’s dive in!

The 4-demicube lives in 4 dimensions. It has 8 vertices.

You get it from a 4-dimensional cube, which has 24 = 16 vertices, by keeping every other vertex, throwing away half. That leaves 8.

What are its top-dimensional faces, aka ‘facets’? Surprise: there’s only one kind! All of them are regular tetrahedra.

In higher dimensions the demicube has two kinds of facet. You get a simplex-shaped facet from every other vertex, formed when you remove it. And you get a demicube-shaped facet from each of the cube’s facets. But in 4 dimensions the two kinds happen to be the same shape!

Eight of them are tetrahedra. These appear at the 8 corners you sliced off: one per removed corner.

Eight more come from the 8 faces of the 4-dimensional cube. These are 3-demicubes. But as we’ve seen, the 3-demicube is also a tetrahedron!

So the 4-demicube is especially symmetric: it has 16 tetrahedral facets. You can find coordinates where its vertices are

(±1, 0, 0, 0),   (0, ±1, 0, 0),   (0, 0, ±1, 0),   (0, 0, 0, ±1)

It’s actually one of the 4-dimensional regular polytopes, sometimes called the 4-orthoplex. It’s also called the 16-cell because it has 16 facets. It’s the 4-dimensional cousin of the octahedron, which has 8 triangular facets.

You can read some of these facts off the D4 Dynkin diagram, if you know what you’re doing. As you can see above, this diagram has a central node with three arms, each just 1 edge long: a perfectly symmetric three-pronged star. To get the 4-demicube, you ring the tip of any one arm.

To get the facets of the 4-demicube, delete an unringed node so the piece still holding the ring stays connected, and see what diagram survives. There are two choices: you can delete the tip of either other arm. But either way, what’s left is a straight chain of 3 nodes—the so-called A3 diagram—with a ring at one node at the end. This gives the tetrahedron.

Both choices give the same shape of facet, a tetrahedron, because all three arms of the D4 Dynkin diagram are interchangeable. That ceases to be true in higher dimensions!

 

Next, the 5-demicube. This lives in 5 dimensions and has 16 vertices.

You get it from a 5-dimensional cube—which has 25 = 32 vertices—by keeping every other vertex, throwing away half. That leaves 16.

What are its top-dimensional faces, or ‘facets’? There are two kinds!

Sixteen of them are 4-dimensional analogues of the regular tetrahedron, called 4-simplexes. These appear at the corners you sliced off: one per removed corner.

The other ten come from the ten faces of the 5-dimensional cube. After you take every other vertex, they become 4-demicubes. These are precisely the 4-demicubes we saw in the last section!

You can also read these two kinds of facets from the D5 Dynkin diagram. As you can see above, this diagram has three arms of lengths 2, 1, 1 (edges from the central branch node). To get the 5-demicube, you ring the tip of either length-1 arm. That ringed diagram encodes the whole polytope.

To get the facets, delete an unringed node so the piece still holding the ring stays connected, and see what diagram survives.

There are two choices.

If you delete the tip of the other length-1 arm, what’s left is a straight chain of 4 nodes—the diagram whose polytope is the 4-simplex. That gives the 4-simplex faces.

Or you can delete the tip of the length-2 arm. Then what’s left is a shorter branching diagram, the one I showed you in my last post! That gives the 4-demicube faces.

So the 5-demicube has both 4-simplex and 4-demicube faces.

Next let’s go up to the 6th dimension, which was my goal all along.

 

The E6 root polytope lives in 6 dimensions. It has 72 vertices.

What are its facets? You can read them straight off the E6 Dynkin diagram, using the same procedure we’ve been using so far.

As you can see, the E6 Dynkin diagram has three arms of lengths 2, 2, 1 (edges from the central branch node). To get the root polytope, you ring the node that’s the tip of a length-1 arm. That fact is not obvious, but let’s go ahead and do that.

Then, to get the facets, delete any unringed node such that the piece still holding the ring stays connected, and see what diagram survives.

There are two choices: the two other nodes at tips of the Dynkin diagram.

However, deleting either of these nodes leave a D5 diagram with a ring on one node, and this gives the 5-demicube we saw last time: a 5-cube with alternate vertices removed.

So the facets of the E6 root polytope are all the same shape: 5-demicubes!

With more work, we can count the facets of the polytopes we’ve been studying:

• The E6 root polytope has 54 facets, all 5-demicubes. They come in two kinds, because we had two choices of which node to delete, so there are really 27 ‘positive’ 5-demicube facets and 27 ‘negative’ 5-demicube facets.

• The 5-demicube has 16 4-simplex facets, one for each vertex that we removed from the 5-cube to create this demicube, and 10 4-demicube facets, one for each facet of that 5-cube.

• The 4-demicube has 8 3-simplex facets, one for each vertex that we removed from the 4-cube to create this demicube, and 8 3-demicube facets, one for each facet of that 4-cube. But both the 3-simplex and the 3-demicube are the familiar tetrahedron. So in fact the 4-demicube has 16 tetrahedral facets. Indeed, the 4-demicube is the 4-dimensional analogue of an octahedron: the so-called 4-orthoplex, or 16-cell.

Using some fancier math I explained here, we can count all the faces of the E6 root polytope. This polytope, is also called 122 due to the shape of its Dynkin diagram: the ring is on a branch of length 1, not counting the central node, while the other two branches have lengths 2. You can look up all this information on the Wikipedia page 122 polytope:

Faces of the E6 root polytope, or 122
dim faces count
5 5-demicubes 54 = 27 + 27
4 4-demicubes = 4-orthoplexes 270
4 4-simplexes 432 = 216 + 216
3 3-simplexes = 3-demicubes = tetrahedra 2160 = 1080 + 1080
2 2-simplexes = triangles 2160
1 1-simplexes = edges 720
0 0-simplexes = vertices 72

The 5-dimensional facets are all 5-demicubes, but as we’ve seen, they come in two kinds: that is, they lie in two orbits of the symmetry group. We can call 27 of them ‘positive’ 5-demicubes and 27 of them ‘negative’ demicubes. Of the 4-dimensional faces, 270 are 4-demicubes and 432 are 4-simplexes. Moreover the 4-simplexes come in two kinds: 216 are faces of positive 5-demicubes while 216 are faces of negative 5-demicubes. Let’s call the first kind of 4-simplex ‘positive’ and the second kind ‘negative’. The 3-dimensional faces are all tetrahedra, but they come in two ‘kinds’: 1080 of them are faces of positive 4-simplexes, and 1080 are faces of negative 4-simplexes. None is the face of both a positive and negative 4-simplex.

If you’re curious about how to count these things, see how some of us counted all the faces of the E8 root polytope here:

• John Baez, Integral octonions (part 5), The n-Category Café, September 3, 2013.

Here is a table of faces for the E7 root polytope, which is also called 231:

Faces of the E7 root polytope, or 231
dim faces count
6 221 polytopes 56
6 6-simplexes 576
5 5-orthoplexes 756
5 5-simplexes 4032
4 4-simplexes 16128 = 4032 + 12096
3 3-simplexes = tetrahedra 20160
2 2-simplexes = triangles 10080
1 1-simplexes = edges 2016
0 0-simplexes = vertices 126

Its 4-dimensional faces are all 4-simplexes, but they come in two ‘kinds’: that is, they lie in two orbits of the symmetry group of this polytope. Of the 4-simplexes, 4032 are the face of three 5-orthoplexes, while 12096 are the face of one 5-orthoplex and two 5-simplexes.

Here’s the E8 root polytope, also called 421:

Faces of the E8 root polytope, or 421
dim faces count
7 7-orthoplexes 2160
7 7-simplexes 17280
6 6-simplexes 207360 = 138240 + 69120
5 5-simplexes 483840
4 4-simplexes 483840
3 3-simplexes = tetrahedra 241920
2 2-simplexes = triangles 60480
1 1-simplexes = edges 6720
0 0-simplexes = vertices 240

There are two kinds of 6-simplex faces: 138240 of them each lie in one 7-simplex and one 7-orthoplex, while 69120 of them each lie in two 7-orthoplexes (and no 7-simplex).

2 Comments | mathematics | Permalink
Posted by John Baez

The Mantle

22 August, 2026

As we descend from the base of Earth’s crust through the mantle, the rock does not remain unchanged. Pressure and temperature rise inexorably, and the minerals that thrive at the surface are forced, step by step, into new and denser crystallographic arrangements. This is the story of those transformations.

In this tale, I’ll act like I know a bit about minerals. I actually don’t: there are a bewildering variety, and I can never remember them. So don’t worry: when you come across a jargon-filled patch of prose, just power through it. You might learn a little… or you can just ignore it. The overall point here is that the Earth is made of beautiful crystalline structures that change character in complex ways as we descend.

The Mohorovičić discontinuity

Our story begins at the boundary where Earth’s crust, rich in feldspar and quartz, gives way to the denser mantle beneath. We see this boundary through its effect on seismic waves, and it’s called the Mohorovičić discontinuity or “Moho”. The Moho does not lie at one fixed depth: it’s 5–10 kilometers below the seafloor, but 30–50 kilometers below most continents, and as much as 70–80 below young mountain belts like the Himalayas.

The mantle just below the Moho mainly consists of a rock called peridotite, which is made mostly of olivine and pyroxene, with smaller amounts of garnet (or, at shallower depths, spinel). Peridotite has a delicious coarse green appearance:

More precisely, this is what peridotite looks like up here. But when geochemists talk about the bulk composition of the upper mantle, they often use an idealized model called pyrolite—not a rock you can pick up, but a hypothetical recipe Ted Ringwood proposed in the 1960s for the primitive upper mantle.

Why? Since the Earth has had a convecting mantle, solid mantle rock wells up in places. As it does, the pressure drops, and a bit of it melts: the minerals with lower melting points. This melt flows upward. It’s called basalt. It builds the Earth’s crust. But it leaves a residue behind, made of minerals with higher melting points.

In Ringwood’s theory, which for expository purposes I’ll assume is true, pyrolite is what mantle rock is like before any partial melting depletes it of basaltic ingredients. The name is a portmanteau of pyroxene and olivine, the two dominant minerals. Pyrolite is about 60% olivine; the remaining 40% is mostly pyroxenes plus garnet.

• A pyroxene is a mineral built from single, unbranched chains of corner-sharing SiO₄ tetrahedra, with metal cations—chiefly Mg, Fe, and Ca—linking the chains together. The general formula is XY(Si,Al)₂O₆, where X and Y are those cations.

• Olivine is a green silicate, (Mg,Fe)₂SiO₄:

Its crystal structure in the upper mantle is an orthorhombic arrangement of isolated SiO₄ tetrahedra knit together by magnesium and iron in octahedral sites. It’s called the α-phase because we’ll see some more compressed phases as we descend.

• A garnet is built from separate SiO₄ tetrahedra held together by cations, but assembled into a dense, hard, characteristically cubic-symmetry crystal. There are different kinds of garnet, but the general formula is X₃Y₂(SiO₄)₃: three divalent X cations, two trivalent Y cations, and three isolated silica tetrahedra. The mantle’s garnet is largely pyrope, Mg₃Al₂(SiO₄)₃.

As we descend, the pyroxenes and garnet gradually dissolve into each other, producing a new high-pressure mineral called majorite. Here’s a rare sample from a meteorite fall in Canada:

So even before the dramatic change 410 kilometers down, the rock is no longer the simple olivine-pyroxene-garnet assemblage we had further up.

The 410-kilometer discontinuity

Roughly 410 kilometers down, the pressure reaches about 13,000 atmospheres and the temperature hovers around 1,400°C. Olivine can no longer hold its familiar shape. It transforms to its β form: wadsleyite, a mineral with the same chemical formula but a fundamentally different atomic arrangement. Instead of isolated SiO₄ tetrahedra, wadsleyite contains paired Si₂O₇ groups, and the oxygens pack more densely. The density jump is sharp enough to be detected globally by seismologists as a reflector of earthquake waves.

Wadsleyite has a remarkable property: it can hold several weight percent of water locked within its crystal structure. The transition zone may thus contain more water than all the oceans combined! However, very little wadsleyite has been seen on the Earth’s surface. Here’s a bit from that same meteor fall in Canada:

The 520-kilometer discontinuity

Descend further, to around 520 kilometers, and the temperature goes up only a little, to roughly 1500–1600°C, since convection here is strong. The pressure goes up to about 175,000 atmospheres. At this point wadsleyite transforms into the γ form of olivine: ringwoodite. This is denser, still chemically Mg₂SiO₄, but now with cations packed into tetrahedral and octahedral holes in a close-packed oxygen framework—the most efficient packing geometry that nature offers for this composition:

Ringwoodite is named for the great Australian geochemist Ted Ringwood, who studied these transitions. Here’s an artificially manufactured sample:

For a long time the mineral’s existence in the mantle was purely hypothetical. But in 2014, a tiny grain was discovered as an inclusion inside a diamond brought up from the deep mantle by an eruption, providing the first direct proof of its existence in Earth’s interior.

The 660-kilometer discontinuity

At a depth of 660 kilometers and a pressure of roughly 230,000 atmospheres, the most dramatic phase transition of all occurs. Ringwoodite does not merely rearrange into a still more dense form! Instead, it decomposes into two entirely new minerals: bridgmanite (MgSiO₃) and ferropericlase (MgO). The majorite garnet also decomposes, yielding davemaoite (CaSiO₃), which is stable through the rest of the lower mantle:

The 660-kilometer discontinuity is sharp, globally consistent, and marks the conventional boundary between the upper and lower mantle. One reason it’s important is that enormous slabs of colder, denser rock sink through the upper mantle until they hit this discontinuity, where the phase change between ringwoodite and bridgmanite creates a kind of barrier.

These slabs are 30–100 kilometers thick and hundreds to a thousand kilometers across! Some punch straight through into the lower mantle and keep sinking. But many flatten out when they hit the barrier, sometimes lying there and piling up for tens of millions of years. You can see this in seismic images beneath Japan and the Marianas. Numerical models suggest that they pile up until they overwhelm the barrier and flush down in a comparatively sudden avalanche—lasting mere millions of years.

The lower mantle

This is the realm of bridgmanite, probably the most abundant mineral in the Earth. Bridgmanite is a beautifully symmetric cage of corner-sharing SiO₆ octahedra, with Mg tucked into the large cavities between them. It accommodates enormous pressure because there is very little void space left to compress.

It is a striking fact that while bridgmanite is the most abundant mineral on the planet, it went unnamed until 2014, simply because no natural hand-sized specimen had ever been recovered. Everything we know about it comes either from high-pressure laboratory synthesis, from microscopic grains in shocked meteorites, or from the indirect testimony of earthquake waves that have traveled through 2,000 kilometers of it.

For over 2,000 kilometers of descent, from 660 to roughly 2,700 kilometers down, bridgmanite and its companion ferropericlase reign without significant further phase change. Seismic velocities increase steadily, but there are no dramatic discontinuities.

The D″ discontinuity

As we approach the core-mantle boundary—at depths around 2,700 kilometers, pressures of approximately 120,000–125,000 atmospheres, and temperatures of 2,200–3,7000°C—even bridgmanite yields. It transforms into the post-perovskite phase. Post-perovskite is a layered, sheet-like structure of SiO₆ octahedra, quite different from bridgmanite’s three-dimensional cage, making it potentially much weaker and more prone to flow.

This transition is believed to be responsible for the seismic D″ discontinuity observed at 2,900 kilometers depth. The D″ layer is a highly dynamic region, likely the site of storage of subducted materials and the source of deep mantle plumes.

A summary of the descent

The table below summarizes the major transitions:

Depth (km)        Minerals
0–410 olivine (α) + pyroxenes + garnet
410 → wadsleyite (β)
520 → ringwoodite (γ)
660 → bridgmanite + ferropericlase + davemaoite
660–2700 bridgmanite dominates
~2700 → post-perovskite
2900 → liquid iron core

The interesting thing about this story is that it was told first by seismology—the sharp jumps in wave speeds at 410 and 660 kilometers were detected long before geologists could reproduce those pressures in the lab—and only later checked by diamond-anvil cell experiments squeezing tiny mineral samples to millions of atmospheres. The rocks never rise to the surface to tell their story directly, so much of the tale above is just theory.

Which minerals are there the most of?

We can estimate how much of the Earth is made of wadsleyite, ringwoodite, and bridgmanite using known shell volumes, estimated densities, and mineral proportions from the pyrolite model.

Step 1: Earth’s mass budget by layer

The Earth’s total mass is M⊕ ≈ 5.972 × 1024 kg. The mass budget by layer is approximately:

•    Crust: ~0.4% of Earth’s mass
•    Upper mantle + transition zone (35–660 km): ~18% of Earth’s mass
•    Lower mantle (660–2,891 km): ~49% of Earth’s mass
•    Core (outer + inner): ~32.5% of Earth’s mass

Step 2: The transition zone (410–660 km)

Using PREM densities averaging ~3,760 kg/m3 across the transition zone, and the volume of each spherical shell:

Wadsleyite zone (410–520 km):
Shell volume ≈ 4.8 × 1019 m3
Shell mass ≈ 1.76 × 1023 kg
Fraction of Earth’s mass ≈ 2.9%

Ringwoodite zone (520–660 km):
Shell volume ≈ 5.9 × 1019 m3
Shell mass ≈ 2.24 × 1023 kg
Fraction of Earth’s mass ≈ 3.8%

In the pyrolite model of mantle composition, forms of olivine (wadsleyite and ringwoodite) make up roughly 60% of the transition zone by mass, with the remaining ~40% being majoritic garnet. Applying this correction:

Wadsleyite: 0.60 × 2.9% ≈ 1.8% of Earth’s mass
Ringwoodite: 0.60 × 3.8% ≈ 2.3% of Earth’s mass

These estimates carry roughly 20–30% uncertainty, mainly from the assumed 60% olivine proportion in the transition zone, which varies with local temperature and bulk composition.

Step 3: Bridgmanite (660–2,700 km)

The lower mantle holds about 49% of Earth’s mass—it is an enormous shell! Bridgmanite constitutes approximately 80% of the lower mantle mineral assemblage (by mass) in the pyrolite model:

0.80 × 49% ≈ 39% of Earth’s mass

This is consistent with the well-cited literature figure that bridgmanite comprises approximately 38% of the planet’s mass—making it the single most abundant mineral in the Earth by a vast margin.

Mineral Depth (km) Fraction of Earth’s Mass
Wadsleyite 410–520 ~1.8%
Ringwoodite 520–660 ~2.3%
Bridgmanite 660–2,700 ~38–39%
All three combined 410–2,700 ~42%

Thus, these three minerals—all members of the same Mg₂SiO₄/MgSiO₃ chemical lineage—together constitute roughly 42% of Earth’s entire mass. All other named minerals on Earth, including quartz, feldspar, calcite, diamond, and the roughly 3,800 others known to mineralogists, divide up the remaining scraps.

13 Comments | chemistry, geology | Permalink
Posted by John Baez

Three Generations in E7

12 August, 2026

It’s long been a mystery why there are 3 generations of quarks and leptons: three sets of particles, apparently identical except for how they interact with the Higgs boson. It would be nice if there were some good physical explanation. Nobody knows one. Barring that, it would be nice if some beautiful mathematical structure made this pattern seem natural. That’s what my new paper is about.

It’s my third paper about exceptional algebraic structures and the Standard Model. When you classify famous gadgets in algebra, beautiful gadgets with fancy names like ‘simple Lie algebras’ and ‘Euclidean Jordan algebras’ and ‘positive hermitian Jordan pairs’, you tend to get infinite series of them—together with a few exceptions that can be built using the octonions. This is a bit spooky, so I’ve been interested in this for a long time.

A few physicists have hoped that these exceptions are good for something. For example, maybe the quirky features of our best theory of particle physics, the Standard Model, aren’t accidental. Perhaps they fall out naturally from some exceptional algebraic structure.

It’s a long shot, but we’ve been stuck on figuring out new fundamental laws of particle physics for so long—roughly since the early 1980s—that it’s worth a try.

In 2018, Michel Dubois-Violette and Ivan Todorov noticed that the gauge group of the Standard Model falls out as symmetries of the so-called ‘exceptional Jordan algebra’ together with some ordinary Jordan algebras sitting inside it. I tried to clarify that here, with a huge amount of help from an excellent young mathematician:

• John Baez and Paul Schwahn, The Standard Model gauge group from the exceptional Jordan algebra. (Blog article here.)

It’s very nice, because the Jordan algebras in question arise naturally when you try to axiomatize the foundations of quantum physics. It would be so cool if something about quantum physics made the Standard Model seem mathematically natural!

But really this result only concerns the gauge bosons in the Standard Model: the photon, gluons, and the W and Z bosons. It says nothing about the fermions—that is, the quarks and leptons. And it seems quite hard to get those into the picture.

In 2020, Latham Boyle tried to solve this problem by tensoring the exceptional Jordan algebra with the complex numbers. This made one generation of fermions appear quite naturally! But the connection to the foundations of quantum physics seemed lost: tensoring the exceptional Jordan algebra with the complex numbers seems at first like it might be just a formal trick.

This spring, Latham and his student Endre Bokor and I showed the connection to quantum physics is not lost:

• John Baez, Endre Bokor and Latham Boyle, Jordan pair quantum theory and the Standard Model. (Blog article here.)

The idea is to work, not with Jordan algebras, but with more general things called Jordan pairs, which have been studied by mathematicians since at least 1975. We showed that you can still do quantum physics with Jordan pairs. And we showed that there’s an ‘exceptional’ Jordan pair that naturally contains the Standard Model gauge group and one generation of fermions!

This Jordan pair is built from the bioctonions: the octonions tensored with the complex numbers. And it’s closely related to an exceptional Lie algebra called \mathfrak{e}_6.

This is nice because the work of Dubois-Violette and Todorov used a smaller exceptional Lie algebra called \mathfrak{f}_4. Going up to \mathfrak{e}_6 gives the room to include one generation of fermions.

There’s an even larger exceptional Lie algebra you can use to build a Jordan pair: it’s called \mathfrak{e}_7. Bokor, Boyle and I tried using this to get three generations of fermions. There are things that make this tempting: not just the fact that \mathfrak{e}_7 is bigger, but the fact that the Jordan pair you get from it has a kind of three-fold symmetry. But we couldn’t get it to work.

Around this time I got very interested in some work that someone had sent me in October 2025. My inbox is packed with new theories of physics. Since the rise of large language models the inflow has increased: I get about two emails a day from someone telling me they’ve made a revolutionary discovery in physics. Practically none of these theories appeal to me. But this paper, and this thesis, were different:

• Benjamin Nasmith, An exceptional combinatorial sequence and Standard Model particles, 2020.

• Benjamin Nasmith, Tight Projective 5-Designs and Exceptional Structures, Ph.D. thesis, Royal Military College of Canada, 2023.

He claimed to fit three generations of fermions into the exceptional Lie algebra \mathfrak{e}_7.

When I started seriously trying to understand this paper, I wound up translating it into a language I’m more comfortable with, and expanding on the ideas a bit. So I wrote this:

• John Baez, Three generations in \mathfrak{e}_7.

Here’s the basic idea.

The idea

There is a standard way to fit the Lie algebra of the Standard Model gauge group, which I call \mathfrak{g}_{\text{SM}}, into the Lie algebra \mathfrak{e}_7. You can construct a Lie algebra L that fits between them:

\mathfrak{g}_{\text{SM}} \subset L \subset \mathfrak{e}_7

As a vector space we have

\mathfrak{e}_7 \; \cong \; L \oplus V

for some vector space V of dimension 3 \times 32.

Moreover, the Lie algebra \mathfrak{g}_{\text{SM}} acts on V, via the \mathfrak{e}_7 Lie bracket, precisely as it does on three generations of Standard Model fermions and their antiparticles, including right-handed neutrino and its antiparticle—but ignoring spin!

There is, in fact, a very interesting three-fold symmetry built into \mathfrak{e}_7, which is revealed when we put the Standard Model Lie algebra \mathfrak{g}_{\text{SM}} into it. It permutes the three generations.

Like Nasmith, I am not proposing a theory of physics. I’m only observing a fascinating mathematical pattern that might (or might not) be of some use in physics.

There are lots of things this pattern does not include: basically, everything I didn’t already mention. It does not include the spin of the fermions and gauge bosons. It does not include the Higgs boson, though in some sense it comes close (see the paper). It does not include a Lagrangian, so it doesn’t say anything at all about particle masses or interactions.

I could say a lot more about this… most importantly, where this Lie algebra L comes from. The details are very interesting. There’s also the curious role of the right-handed neutrinos. But I’ve already spent weeks explaining all these things in my paper, so I won’t do it here. Instead let me say a bit about how I wrote the paper.

Writing the paper

I’ve been wanting to keep up with how AI is transforming math. About a year ago a friend gave me a subscription to Claude Pro. I wanted to test it out, despite my many misgivings, including how large language models are contributing to global warming and income inequality. Given the amazing things that people have recently done in math using large language models, I didn’t think that never trying them out would put me in the best position to make good decisions about the future.

So, I wrote this paper with help from Claude Opus 4.8.

I started by giving it Nasmith’s paper and asking a long series of questions about that paper over several days. The results were very interesting and helpful. Eventually I asked it to summarize and expand on our conversation. It quickly spat out a 10-page paper.

This paper was written in a breezy, pleasant style—but also quite hard to understand in detail, since it mixed Nasmith’s terminology with the Lie algebra terminology I prefer, and the proofs skipped over some steps.

It took me about three weeks of hard work to fully understand and re-express all the ideas a way that I like. For a while I felt dumb and frustrated, because when I asked Claude to fill in the gaps in proofs, it used math I was not very competent in, like the theory of regular subalgebras, and the theory of minuscule representations. But I learned this math, and everything turned out to be basically correct—in part, I’m sure, because Nasmith’s original work was correct.

For several weeks I checked, reorganized, expanded and completely rewrote this material. By the end everything was written in a style I like, emphasizing the ideas I consider important, proving things fairly carefully, and adding a lot of expository material—for example, explaining the theory of regular subalgebras.

Almost no traces of Claude’s original writeup remain, even though I was deeply influenced by them. My proofs make few references to deep theorems, though they assume solid familiarity with simple Lie algebras and their root systems. The proofs also require no brutally hard computations—though Claude was eager to do such computations to check things.

Any mistakes in this paper are my own.

I’m not sure what conclusions I draw from writing this paper. I’m writing another math paper now, with a human coauthor, and I have no desire to get help from a large language model. For work on my own it could be very helpful. Jacob Tsimerman says it roughly doubles his productivity. Would using it be so bad for the environment, or so bad for society, that I should avoid it? Maybe. I deliberately stuck with Claude Opus 4.8 instead of something more powerful, to see what I could do with what you get from a $20/month subscription. But maybe that’s still bad.

I avoid flying to conferences, which in some ways cripples my ability to keep up with new trends and influence people—but I don’t mind that. It gives me more time to think.

I will think carefully about my next move.

38 Comments | mathematics, physics | Permalink
Posted by John Baez

Jordan Triples and the Standard Model

22 July, 2026

I don’t usually talk about particle physics here. I have a whole series of articles about octonions and the Standard Model on my other blog. But I’m kind of excited about this new paper, so I’ll talk about it here too:

• John Baez, Endre Bokor and Latham Boyle, Jordan pair quantum theory and the Standard Model.

Jordan algebras were introduced by Jordan, von Neumann and Wigner in 1934 in an attempt to formalize algebras of observables in quantum theory. They come in 4 infinite series—but there’s one more, the ‘exceptional Jordan algebra’, consisting of 3 × 3 self-adjoint matrices of octonions. For years physicists sought to find some use for it.

In 2018, Todorov and Dubois–Violette noticed that the symmetries of the exceptional Jordan include the Standard Model gauge group in a nice way. But it was unclear how to bring in the fermions—the quarks and leptons. That’s what our new paper does.

To do this, we need to go beyond Jordan algebras. Jordan pairs and Jordan triples are two closely linked formalisms that generalize Jordan algebras. Our paper explains them in detail—and how they’re connected to geometry and quantum mechanics. But here I will mostly skip that wonderful story, so I can quickly explain the connection to the Standard Model.

Here’s how the Standard Model gauge group, together with its representation on one generation of fermions, drops out of a Jordan triple.

The bi-Cayley triple

Let

\mathbb{O}_\mathbb{C} = \mathbb{C} \textstyle{\otimes}_\mathbb{R} \mathbb{O}

be the bioctonions: octonions with complex coefficients. Write \mathbb{O}_\mathbb{C}^2 for the space of column vectors with two bioctonion entries.

\mathbb{O}_\mathbb{C}^2 has a certain triple product

[x,y,z]=\frac{1}{2}(x(y^{\dagger}z)+z(y^{\dagger}x))

which obey the axioms of a gadget called a ‘positive hermitian Jordan triple’. It’s called the bi-Cayley triple.

Now, every positive hermitian Jordan triple gives rise to a \mathbb{Z}_2-graded real Lie algebra

\mathbf{k} = \mathbf{k}_0 \textstyle{\oplus} \mathbf{k}_1

Not a Lie superalgebra: a plain old-fashioned Lie algebra with a \mathbb{Z}_2-grading!

How does this work? We take the hermitian Jordan triple itself to be \mathbf{k}_1. The Lie algebra \mathbf{k}_0 consists of all linear maps from \mathbf{k}_1 to itself that are of this form:

x \mapsto [a,b,x] - [b,a,x]

for some a,b \in \mathbf{k}_1. These maps are called real inner derivations. They form a Lie algebra since the commutator of two such maps is another such map. With a bit more work we can define other operations making all of \mathbf{k} into a \mathbb{Z}_2-graded Lie algebra.

So, we get a big Lie algebra \mathbf{k}, and a Lie subalgebra \mathbf{k}_0 sitting inside it. From this we get two Lie groups: a big one K whose Lie algebra is \mathbf{k}, and a subgroup K_0 whose Lie algebra is \mathbf{k}_0.

The quotient is K/K_0 is a nice kind of manifold called a hermitian symmetric space. Conversely, any compact hermitian symmetric space give rise to a positive hermitian Jordan triple!

This geometric picture is revealing. The group K acts transitively as symmetries of our hermitian symmetric space, while the stabilizer of any point is isomorphic to K_0. Our original Jordan triple, \mathbf{k}_1, is then the tangent space of that point! So, K_0 acts on this Jordan triple. This action preserves the triple product, and we call K_0 the real inner automorphism group of our Jordan triple.

Here’s another great thing about the geometric picture: hermitian symmetric spaces were classified by Eli Cartan (who seems to have spent his life classifying things). As a result we also know the classification of positive hermitian Jordan triples. They come in four infinite series together with two exceptions. One is the bi-Cayley triple, and other is the Albert triple, which is the complexification of the exceptional Jordan algebra. The bi-Cayley triple is a subtriple of the Albert triple. It’s these two exceptions that are connected to the Standard Model. But we’ll start with the bi-Cayley triple.

The 3-graded Lie algebra coming from the bi-Cayley triple is the compact real form of \mathfrak{e}_6:

\mathfrak{e}_6 = \big[\mathfrak{so}(10) \textstyle{\oplus} \mathfrak{u}(1)\big] \textstyle{\oplus} \mathbb{O}_\mathbb{C}^2

The even part of this Lie algebra is in brackets. The corresponding hermitian symmetric space is called the bioctonionic plane (\mathbb{C}\otimes\mathbb{O})P^2. The even part of our 3-graded Lie algebra, \mathfrak{so}(10)\oplus \mathfrak{u}(1), generates the stabilizer of a point in the bioctonionic plane. The odd part, our friend \mathbb{O}_\mathbb{C}^2, is the tangent space of that point.

Here’s the first big surprise. The even part transforms as the adjoint representation of \mathrm{Spin}(10), while the odd part itself transforms as the 16-dimensional complex spinor representation of \mathrm{Spin}(10). Ignoring the extra \mathrm{U}(1) for a moment, this is exactly what we see in a \mathrm{SO}(10) grand unified theory: gauge bosons in the adjoint representation, and one generation of fermions in the 16-dimensional spinor representation.

So before we do anything, the bi-Cayley triple already smells like it contains the ingredients of an \mathrm{SO}(10) grand unified theory.

Tripotents

In a Jordan algebra the important elements are the idempotents, e^2 = e. In a Jordan triple W their role is played by tripotents: elements e with

[e,e,e] = e

A tripotent always lets us split W into three parts via something called its Peirce decomposition. The operator w \mapsto [e,e,w] has eigenvalues 0, 1/2, and 1, so W splits into the corresponding eigenspaces

W = W_0(e) \textstyle{\oplus} W_{1/2}(e) \textstyle{\oplus} W_1(e)

which are called the Peirce 0-space, Peirce 1/2-space and Peirce 1-space of e. A tripotent is called minimal when its Peirce 1-space is one-dimensional. Two tripotents e_1, e_2 are called colinear when each lies in the other’s Peirce 1/2-space.

I can’t resist explaining some of the quantum physics here. In a hermitian Jordan triple, the triple product [-,-,-] is linear in the first and last slot, but conjugate-linear in the middle slot. So, if you multiply a tripotent by a phase \alpha, you get a new tripotent:

[\alpha e, \alpha e, \alpha e] = \alpha \overline{\alpha} \alpha e = \alpha e

This should remind you of how when you multiply a unit vector in a Hilbert space by a phase, you get a new unit vector. In Jordan triple quantum mechanics, minimal tripotents take the place of these unit vectors. The hermitian symmetric space K/K_0 that I was talking about earlier is the same as the space of minimal tripotents mod phase! So, it generalizes the familiar space of ‘pure states’ in quantum mechanics: unit vectors mod phase.

But let’s get back to the Standard Model.

A chain of Jordan triples

From here on, the single fact driving everything is this: in any hermitian Jordan triple, any minimal tripotent’s Peirce 1/2-space is itself a hermitian Jordan triple!

If we run this starting from the bi-Cayley triple, we get this chain of hermitian Jordan triples, where each row’s 1/2-space is the next row’s triple:

Jordan triple Lie algebra \mathbf{k}_0 \oplus \mathbf{k}_1 (even part in brackets)
W = \mathbb{O}_\mathbb{C}^2 \mathfrak{e}_6 = [\mathfrak{so}(10) \oplus \mathfrak{u}(1)] \oplus \mathbb{O}_\mathbb{C}^2
W' = \mathfrak{a}_5(\mathbb{C}) \mathfrak{so}(10) = [\mathfrak{su}(5) \oplus \mathfrak{u}(1)] \oplus \mathfrak{a}_5(\mathbb{C})
W'' = \mathrm{M}_{3,2}(\mathbb{C}) \mathfrak{su}(5) = [\mathfrak{g}_{\mathrm{SM}}] \oplus \mathrm{M}_{3,2}(\mathbb{C})

Here \mathfrak{a}_5(\mathbb{C}) is the Jordan triple of antisymmetric 5\times 5 complex matrices, \mathrm{M}_{3,2}(\mathbb{C}) is the Jordan triple of 3\times 2 complex matrices, \mathfrak{g}_{\mathrm{SM}} = \mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus \mathfrak{u}(1), and

G_{\mathrm{SM}} = \mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3)) \cong (\mathrm{SU}(3)\times\mathrm{SU}(2)\times\mathrm{U}(1))/\mathbb{Z}_6

is the true Standard Model gauge group.

The gauge group from two tripotents

Start with the bi-Cayley triple. Choose two colinear minimal tripotents e_1, e_2. Descend the table twice:

• Start with W = \mathbb{O}_\mathbb{C}^2, which has real inner automorphism group (\mathrm{Spin}(10)\times\mathrm{U}(1))/\mathbb{Z}_4.

• Fix e_1. Its Peirce 1/2-space is W' = \mathfrak{a}_5(\mathbb{C}), with real inner automorphism group \mathrm{SU}(5)\times\mathrm{U}(1).

• Fix e_2 (colinear with e_1, so living in W'). Its Peirce 1/2-space in W' is W'' = \mathrm{M}_{3,2}(\mathbb{C}), with real inner automorphism group exactly G_{\mathrm{SM}}.

In other words, the subspace of the bi-Cayley triple colinear with both e_1 and e_2 is a Jordan triple whose real inner automorphism group is the Standard Model gauge group.

The choice of e_1 and e_2 also pins down how G_{\mathrm{SM}} sits inside the original group \mathrm{E}_6. At each we step take the subgroup that acts with determinant 1 and preserves the chosen tripotent up to a phase; this gives a chain of subgroups whose members are \mathrm{Spin}(10), \mathrm{U}(5), and G_{\mathrm{SM}}, so we get the embeddings

G_{\mathrm{SM}} \subset \mathrm{SU}(5) \subset \mathrm{Spin}(10)

In particle physics, this is the classic chain taking us from the so-called \mathrm{SO}(10) grand unified theory down to the \mathrm{SU}(5) grand unified theory down to the Standard Model. And it’s well known that restricting the 16-dimensional complex spinor representation of \mathrm{Spin}(10) along this chain gives precisely the Standard Model representation \rho_{\mathrm{SM}} on one generation of fermions! So we get one generation of Standard Model fermions this way.

The six particles types as Peirce spaces

We have gotten the representation of the Standard Model gauge group on one generation of fermions without any fuss. But it’s also fun to peer into the details, and see how the different kinds of fermions emerge.

For any tripotent e, we have projections P_0(e), P_{1/2}(e) and P_1(e) onto its three eigenspaces: its so-called Peirce projectors. Since we get the Standard Model structure using two minimal tripotents e_1 and e_2 in the bi-Cayley triple \mathbb{O}_{\mathbb{C}}^2, there are nine composites of two Peirce projectors we can apply to this triple. This is how we pick out the different kinds of fermions!

As a representation of the Standard Model Lie algebra

\mathfrak{g}_{\mathrm{SM}} = \mathfrak{su}(3) \textstyle{\oplus} \mathfrak{su}(2) \textstyle{\oplus} \mathfrak{u}(1)

any generation of Standard Model fermions transforms as the direct sum of six irreducible representations:

\rho_{\mathrm{SM}} = (3,2,\tfrac{1}{6}) \textstyle{\oplus} (\bar 3,1,\tfrac{1}{3}) \textstyle{\oplus} (\bar 3,1,-\tfrac{2}{3}) \textstyle{\oplus} (1,2,-\tfrac{1}{2}) \textstyle{\oplus} (1,1,1) \textstyle{\oplus} (1,1,0)

These correspond to the six types of left-handed fermion: q_L, \overline{d_R}, \overline{u_R}, \ell_L, \overline{e_R}, \overline{\nu_R}. Six irreducible pieces, six particle types.

It turns out these are exactly the six nonzero components of the Peirce decomposition of \mathbb{O}_\mathbb{C}^2 with respect to both e_1 and e_2. Those six match up one-to-one with the particle types:

Peirce projector representation of G_{\text{SM}} particle type
P_{1/2}(e_2) P_{1/2}(e_1) (3, 2, +1/6) q_L
P_{1/2}(e_2) P_0(e_1) (\overline{3}, 1, +1/3) \overline{d_R}
P_0(e_2) P_{1/2}(e_1) (\overline{3}, 1, −2/3) \overline{u_R}
P_0(e_2) P_0(e_1) (1, 2, −1/2) \ell_L
P_1(e_2) P_{1/2}(e_1) (1, 1, +1) \overline{e_R}
P_{1/2}(e_2) P_1(e_1) (1, 1, 0) \overline{\nu_R}

The remaining three combinations—P_1(e_2)P_1(e_1), P_1(e_2)P_0(e_1), and P_0(e_2)P_1(e_1)—all vanish, which is why we land on six pieces and not nine.

So the whole package—the gauge group G_{\mathrm{SM}}, the embedding G_{\mathrm{SM}} \subset \mathrm{Spin}(10), the representation \rho_{\mathrm{SM}}, and even the split of one generation into its six particle multiplets as distinct Peirce components—all comes out of the single object \mathbb{O}_\mathbb{C}^2 once you choose two colinear minimal tripotents.

And if you prefer to start one level up, with the Albert triple \mathfrak{h}_3(\mathbb{O}) \otimes \mathbb{C}, you get the same result by choosing three mutually colinear tripotents instead of two—but for that, read our paper!

2 Comments | mathematics, physics | Permalink
Posted by John Baez

Galilean Limits of Electromagnetism

18 July, 2026

Maxwell’s equations are invariant under Lorentz transformations. The usual equations of fluid flow are not! Like the rest of Newtonian mechanics, they’re invariant under Galilean transformations like

t' = t, \quad x' = x - vt

So, if we simply slap these two theories together, we get a mess! How can we study electrically conductive fluids—like plasma—without bringing special relativity into the game?

We can use a limiting case of Maxwell’s equations where we ignore terms that become tiny when all the particles are moving much slower than light.

There seem to be at least two ways to do this: there’s an ‘electric limit’ of Maxwell’s equations and a ‘magnetic limit’. Both are invariant under Galilean transformations. The original derivation of these limits by Le Bellac and Lévy-Leblond in 1973 used the version of Maxwell’s equations including the electric permittivity \varepsilon_0 and magnetic permeability \mu_0 of the vacuum, whose product is 1/c^2. This is convenient but not necessary, as explained here:

• Jose A. Heras, The Galilean limits of Maxwell’s equations.

In the magnetic limit of Maxwell’s equations, we throw out effects due to time-varying electric fields:

People often use the magnetic limit when studying nonrelativistic electrically conductive fluids. In this situation they often consider a version of the magnetic limit where the charge density \rho is zero, since this is typically close to true in a plasma. However Heras does not do this, nor does the original paper:

• Le Bellac and Levy-Leblond, Galilean electromagnetism.

In the electric limit of Maxwell’s equations, we throw out effects due to time-varying magnetic fields:

It’s fun to compare the magnetic and electric limits.

The magnetic limit has been called ‘pre-Maxwellian’, because it’s like electromagnetism before Maxwell added the extra term that makes a changing electric field create a curl in the magnetic field. Without this term there is no light!

In the electric limit you also can’t have light, because it’s missing the term that makes a changing magnetic field create a curl in the electric field.

In the magnetic limit you can’t have capacitors, because those store energy in the electric field, and in the magnetic limit the energy density is just \mathbf{B} \cdot \mathbf{B}/2.

Similarly, in the electric limit you can’t have inductors, because inductors store energy in the magnetic field, and in this limit the energy density is just \mathbf{E} \cdot \mathbf{E}/2.

It’s all nicely symmetrical! But still somewhat mysterious to me. All the derivations of these limits that I’ve seen involve too many parameters for my taste, and too much talk. But that’s how I often feel when I’m just starting to study a piece of physics.

Besides the two papers mentioned in my last post, I’ve been looking at this:

• Giovanni Manfredi, Non-relativistic limits of Maxwell’s equations.

There’s a lot I haven’t explained here. I haven’t even said how the electric or magnetic fields transform under Galilean boosts in these limiting theories! I find this subject fairly confusing, and I’d probably have to redo all the calculations to really understand them. As Feynman said, “what I cannot create I do not understand”.

Someday I should dig deeper into this subject and explain how the two limits work in a way I find satisfying. I should also draw the connections to this earlier article of mine:

• Magnetohydrodynamics.

12 Comments | physics | Permalink
Posted by John Baez

Octonions and the Standard Model

16 June, 2026

Paul Schwahn and I have come out with a new paper about octonions and the Standard Model:

• The Standard Model gauge group from the exceptional Jordan algebra

It builds on things I’ve discussed here, but it goes further. Let me explain a bit.

A bit is just a binary alternative: 1 or 0, true or false. That’s how it works in classical logic. We could also have a ‘trit’, meaning 3 alternatives.

In quantum physics we instead have qubits and qutrits.

Qubits and qutrits are usually described using complex numbers. The algebra of observables of a qubit is the Jordan algebra \mathfrak{h}_2(\mathbb{C}), consisting of 2 \times 2 self-adjoint complex matrices. Similarly, the algebra of observables of an qutrit is the Jordan algebra \mathfrak{h}_3(\mathbb{C}), consisting of 3 \times 3 self-adjoint complex matrices.

We can also study systems with more than 3 alternative ways to be. They work the same way, using the Jordan algebras \mathfrak{h}_n(\mathbb{C}) with n > 3.

But we can also do quantum mechanics using other number systems! The options have been mapped out, and the largest allowed number system for this purpose is the algebra of octonions.

A weird thing is that Jordan algebras built using octonions can describe qutrits, but not quantum systems with more than 3 alternative ways to be. The algebra of observables of an octonionic qutrit is the so-called ‘exceptional’ Jordan algebra \mathfrak{h}_3(\mathbb{O}), consisting of 3 \times 3 self-adjoint octonion matrices. What makes it exceptional is that \mathfrak{h}_n(\mathbb{O}) is not a Jordan algebra when n is bigger than 3.

So, there’s something special about octonionic qutrits—and it turns out that every symmetry in the gauge group of the Standard Model is a symmetry of an octonionic qutrit!

Not every symmetry of an octonionic qutrit is a symmetry of the Standard Model. But those that do have a simple description. They are those that restrict to give symmetries of an ordinary qutrit sitting inside the octonionic qutrit… and an ordinary qubit sitting inside that!

That sounds exciting, but also vague, so let me make it precise.

While lots of people say the gauge group of the Standard Model of particle physics is \text{U}(1) \times \text{SU}(2) \times \text{SU}(3), in fact a certain subgroup of this acts trivially on all known particles. If we mod out by that, we’re left with a group called \text{S}(\text{U}(2) \times \text{U}(3)), which is

\Big\{ x \in \text{SU}(5) : x = \left( \begin{array}{c c c c c} \ast & \ast & 0 & 0 & 0 \\ \ast & \ast & 0 & 0 & 0 \\ 0 & 0 & \ast & \ast & \ast \\ 0 & 0 & \ast & \ast & \ast \\ 0 & 0 & \ast & \ast & \ast \end{array} \right) \; \Big\}.

and this is the group I’m talking about.

We proved two theorems describing this group in terms of the symmetries of an octonionic qutrit. The group of automorphisms of the exceptional Jordan algebra \mathfrak{h}_3(\mathbb{O}) is a 52-dimensional Lie group known affectionately as \text{F}_4—so that’s what I mean by the symmetries of an octonionic qutrit.

Here’s our main result:

Theorem 1. Suppose X,B are Jordan subalgebras of \mathfrak{h}_3(\mathbb{O}) such that

X \cong \mathfrak{h}_2(\mathbb{C}), \;\; B \cong \mathfrak{h}_3(\mathbb{C}), \;\; X \subset B.

Then

\text{Stab}(X) \cap \text{Stab}(B)_0 \cong \text{S}(\text{U}(2) \times \text{U}(3)).

Here \text{Stab}(X) is the stabilizer of X—that is, the subgroup of \text{F}_4 consisting of elements that map X to itself—while \text{Stab}(B)_0 is the identity component of the stabilizer of B.

This ‘identity component’ business is rather sneaky, but it turns out that guys in \text{Stab}(B)_0 are symmetries of an ordinary qutrit that can be described as unitary operators on \mathbb{C}, while \text{Stab}(B) also contains those symmetries that are described by antiunitary operators. The CPT symmetry of the Standard Model is antiunitary, for example.

Theorem 1 emerged from a related result, which grew out of the work of Todorov and Dubois-Violette:

Theorem 2. Suppose A,B are Jordan subalgebras of \mathfrak{h}_3(\mathbb{O}) such that

A \cong \mathfrak{h}_2(\mathbb{O}), \;\; B \cong \mathfrak{h}_3(\mathbb{C}), \;\; A \cap B \cong \mathfrak{h}_2(\mathbb{C}).

Then

\text{Stab}(A) \cap \text{Stab}(B)_0 \cong \text{S}(\text{U}(2) \times \text{U}(3)).

Todorov and Dubois–Violette proved this for a certain standard choice of subalgebras A and B. Thus, the challenge in proving Theorem 2 was to show that every other choice can be mapped to this standard choice using the action of \text{F}_4. This shows that the theorem is not an artifact of a specific choice, but rather a general fact.

How do we prove these results?

We start by constructing the octonion product from \text{SU}(3)-invariant operations on \mathbb{C} and \mathbb{C}^3. We then use this description to reprove Todorov and Dubois–Violette’s special case of Theorem 2. Then we show that \text{F}_4 acts transitively on the set of subalgebras of \mathfrak{h}_3(\mathbb{O}) that are isomorphic to \mathfrak{h}_3(\mathbb{C}). We also show every Jordan subalgebra of \mathfrak{h}_3(\mathbb{O}) isomorphic to \mathfrak{h}_2(\mathbb{C}) is contained in a unique Jordan subalgebra isomorphic to \mathfrak{h}_2(\mathbb{O}). This lets us prove that \text{F}_4 acts transitively on the set of pairs of Jordan subalgebra A, B \subset \mathfrak{h}_3(\mathbb{O}) with A \cong \mathfrak{h}_2(\mathbb{O}), B \cong \mathfrak{h}_3(\mathbb{C}) and A \cap B \cong \mathfrak{h}_3(\mathbb{C}). Theorem 2 then follows from Todorov and Dubois-Violette’s special case. We conclude by using these results to prove Theorem 1.

However, if you want to get into the details of the physics, the interesting part is how the strong force gauge group \text{SU}(3) and the electroweak \text{S}(\text{U}(1) \times \text{U}(2)) show up from the relation between octonionic qutrits, complex qutrits and complex qubits. You’ll see that in the proof of Lemma 4.

And if you want to get into the details of the math, the main interesting thing here is the use of Jordan algebra technology like ‘Peirce decompositions’ and ‘Jordan frames’ to figure out what it must be like when you have a Jordan algebra \mathfrak{h}_2(\mathbb{L}) or \mathfrak{h}_3(\mathbb{L}) sitting inside \mathfrak{h}_3(\mathbb{K}), where \mathbb{L} is some normed division algebra contained in a bigger normed division algebra \mathbb{K}.

What it all ‘really means’, if anything, is a question for later. It could be just a coincidence. Of course I hope not.

14 Comments | mathematics, physics | Permalink
Posted by John Baez

Interview with Micah Zarin

2 June, 2026

I’m not completely happy with this interview with Micah Zarin. It was nothing he did, it was me. I forgot to say that current-day AI wastes a lot of energy, and companies hope to use it to lay off people, and oligarchs are using it to extract lots of money from everyone. While obvious, these things are tremendously important and I should have emphasized them.

I was distracted by Micah’s fear that AI would make a career in math pointless, which really surprised me. So instead of giving my general thoughts on AI, I focused on putting myself in his place and imagining what to do in that situation. I suggested doing math with the help of AI as a way to overcome his fear and go ahead doing math while keeping abreast of new developments. If AI overtakes humans in math in his lifetime, which is far from certain, this could be a way to keep productively participating in math throughout this process. But I warned him to be very critical of what LLMs say, to lessen the danger of getting caught up in the ‘AI vortex’ that is turning many people into crackpots.

Mathematics, in case you haven’t been paying attention, is different from some other subjects because it’s an area where LLMs have shown some truly impressive problem-solving ability: read the various mathematicians’ comments in Remarks on the disproof of the unit distance conjecture where they grapple with this. But nobody really knows where this is going. So far LLMs have not shown much ability to invent new theories of mathematics, so it would be jumping to conclusions to assume AI will soon overtake humans in that realm. It would also be jumping to conclusions to assume it won’t.

Whatever happens, the real danger is not that AI will become too good, but that it will become too evil—most likely because of the oligarchs, corporations and governments behind it. I wish I had emphasized that point, which is always on my mind.

I think I succeeded in making another point, which is that life will not become pointless simply because some other entity gets better than humans at something and knocks us off our throne. To think that the meaning of life resides in our superiority is a childish attitude.

13 Comments | risks | Permalink
Posted by John Baez

Summing the Reciprocals of Primes

1 June, 2026

The sum of the reciprocals of the primes diverges, but very slowly. The sum of the reciprocals of the first 100 primes is

2.106…

The sum of the reciprocals of the first 1,000 primes is

2.457…

For the first 10,000 it’s

2.709…

And it keeps creeping up, ever more slowly. To get the sum to reach 6, you need to add up the reciprocals of the first 3 × 10¹³² primes—far more than the number of atoms in the observable universe! Luckily there is no shortage of primes.

Here’s how you can see that the sum diverges, and that it diverges very slowly. First, remember Euler’s product formula for the Riemann zeta function:

\displaystyle{ \sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_{p \text{\, prime}}\left(1-\frac{1}{p^{s}}\right)^{-1} }

This diverges logarithmically when s = 1. Taking logs we see

\displaystyle{ \sum_{p \text{\, prime}}\ln\!\left(1-\frac{1}{p}\right)^{-1} \;\approx\; \sum_{p \text{\, prime}}\frac{1}{p} }

must diverge too—but only log-logarithmically!

A deeper result, called Merten’s Second Theorem and proved here, says that

\displaystyle{ \sum_{p \le n} \frac{1}{p} = \ln\ln n + M + o(1) }

for some constant M. This constant is called the Meissel–Mertens constant. You can think of it as a fancier relative of Euler’s constant \gamma, which is defined by

\displaystyle{ \sum_{k = 1}^n \frac{1}{k} = \ln n + \gamma + o(1) }

But it’s much less widespread in mathematics than Euler’s constant, much as primes are less widespread than natural numbers.

Here’s how it works:

What’s a bit surprising, given the slowness of convergence and the somewhat erratic behavior of the primes, is that people can compute the Meissel–Mertens constant very precisely:

M \approx 0.26149721284764278375542683860869585905\ldots

The trick, of course, is to use another formula for this constant, which lets you compute it much more efficiently than the definition. Here it is:

\displaystyle{ M = \gamma + \sum_{n = 2}^\infty \frac{\mu(n)}{n} \ln \zeta(n)}

where \mu is the Möbius function and \zeta is the Riemann zeta function. By the way, this formula shows that calling M a fancier relative of \gamma is not just talk.

When you know how to compute the Meissel–Mertens constant, you can compare the sum of reciprocals of primes to \ln\ln n + M, and the agreement is very good:

In 1983, Guy Robin proved the curve goes above and below the actual sum infinitely many times, i.e.

\displaystyle{\left(\sum _{p\leq n}{\frac {1}{p}}\right) -\ln \ln n-M}

changes sign infinitely many times. You can see a bit of that happening here:

Puzzle. Can you find a naturally occurring sum that diverges even more slowly, for example like \ln \ln \ln n ?

Acknowledgements

The pictures were created by Dcoetzee, Marek Wolf and Saroad, respectively, and placed into the public domain on Wikicommons. Click on the pictures for more details.

16 Comments | mathematics | Permalink
Posted by John Baez

From Pentagons to Pentagrams

29 May, 2026

I recently showed you that if you take the regular icosahedron:

considered in a coordinate system based on the golden ratio, and then replace √5 by -√5 in all your formulas, you get the great icosahedron:

But this fact isn’t an isolated one-off! If we do the same for the regular dodecahedron:

we get the great stellated dodecahedron:

There’s also a star polyhedron called the great dodecahedron:

and if we play the same game, replacing \sqrt{5} by -\sqrt{5} in all the formulas, we get the small stellated dodecahedron:

These six polyhedra form a family; the four nonconvex ones are called the Kepler–Poinsot polyhedra. I never understood what was so great about them, though of course they look ravishingly attractive. So it was nice to learn that if we include the convex ones, they come in three pairs related by the operation of replacing \sqrt{5} by -\sqrt{5}, which is called Galois conjugation. This is mentioned near the end of this book:

• John Horton Conway, Heidi Burgiel and Chaim Goodman-Strauss, The Symmetries of Things, A K Peters, Natick, Massachusetts, 2008.

These authors spend more energy describing three other relations among this family of polyhedra:

But I’m more interested in Galois conjugation, which carries each polyhedron in this picture to the one at the opposite corner of the hexagon. I got interested in Galois conjugation because it interchanges two kinds of quasiparticles that propagate in icosahedral quasicrystals, called phonons and phasons:

• Phasons in Quasicrystals.

But there’s some simple geometry behind it, which I’d like to discuss here.

You’ll notice that in all the examples I gave, Galois conjugation takes regular pentagons to regular pentagrams. And that turns out to be a general fact!

Let \mathbb{Q}(\sqrt{5}) be the golden field: that is, the set of all numbers

a + b \sqrt{5}

with a,b rational, equipped with the usual addition, multiplication, subtraction and division. Define Galois conjugation

f \colon \mathbb{Q}(\sqrt{5}) \to \mathbb{Q}(\sqrt{5})

by

f(a + b \sqrt{5}) = a - b \sqrt{5}

This map preserves all the field operations, and if you apply it twice you get back where you started. Thus, it’s like complex conjugation in many respects.

The golden field gets its name because it contains the golden ratio

\displaystyle{ \Phi = \frac{1 + \sqrt{5}}{2} = 1.6180339\dots }

If we apply Galois conjugation to the golden ratio, we get its negative reciprocal:

\displaystyle{ -1/\Phi = \frac{1 - \sqrt{5}}{2} \approx -0.6180339\dots }

This suggests that Galois conjugation should somehow map regular pentagons to regular pentagrams! Why? Well, in a regular pentagon, each exterior turning angle is 2\pi/ 5 = 72^\circ:

while in a regular pentagram, each exterior turning angle is 4 \pi /5 = 144^\circ:

The cosine of the exterior turning angle for the pentagon is

\cos(2\pi/5) = 1/2 \Phi

and we apply Galois conjugation to this, we get the cosine of the exterior turning angle for the pentagram!

\cos(4 \pi/5) = -\Phi/2

This is not quite a proof that Galois conjugation turns regular pentagons into regular pentagrams—indeed, we have to clarify what we even mean by that claim. But it’s a key ingredient of the proof.

To be more precise, let’s consider a regular pentagon in \mathbb{R}^n whose vertices lie in \mathbb{Q}(\sqrt{5})^n. Beware: such a pentagon is impossible in the plane!

Puzzle. Show this.

But it’s possible in 3 or more dimensions. For example:

\begin{array}{ccl} v_1 &=& (1,1,1) \\ v_2 &=& (0, 1/\Phi,\Phi) \\ v_3 &=& (-1,1,1) \\ v_4 &=& (-1/\Phi,\Phi,0) \\ v_5 &=& (1/\Phi,\Phi,0) \end{array}

taken in cyclic order, are the vertices of a regular pentagon in 3 dimensions. And once you can get one, you can get plenty, by translations and rotations. It may take a bit of thought to dream up rotation matrices with entries in the golden field, but there are lots: even rotation matrices with rational entries are dense among all rotations.

Now, Galois conjugation acts on \mathbb{Q}(\sqrt{5})^n coordinatewise; let’s abuse language and call this map

f \colon \mathbb{Q}(\sqrt{5})^n \to \mathbb{Q}(\sqrt{5})^n

If we take a regular pentagon and apply this map to its vertices and edges, what do we get? A regular pentagram, I claim!

We can simplify the proof by noticing that only the cyclic ordering on the vertices is needed to distinguish a regular pentagon and a regular pentagram.

Theorem. Let v_1, \dots, v_5 \in \mathbb{Q}(\sqrt{5})^n be the vertices of a regular pentagon, listed in cyclic order. Then f(v_1), \dots, f(v_5), listed in the same cyclic order, are the vertices of a regular pentagram.

Proof. Define the edge vectors

e_i \;=\; v_{i+1} - v_i \;\in\; \mathbb{Q}(\sqrt{5})^n, \qquad i = 1, \dots, 5 \pmod 5

All these have the same squared length

e_i \cdot e_i = L^2 \in \mathbb{Q}(\sqrt{5})

The exterior turning angle at each vertex of the pentagon is 2\pi/5, so this is the angle between the consecutive edge vectors e_i and e_{i+1}. Since

\cos(2\pi/5) = 1/2\Phi

we have

e_i \cdot e_{i+1} \;=\; \cos(2\pi/5) L^2 \;=\; (1/2\Phi) \, L^2

Now let v_i' = f(v_i) and e_i' = f(e_i) = v_{i+1}' - v_i'. It is easy to check that the usual dot product of v, w \in \mathbb{Q}(\sqrt{5})^n obeys

f(v) \cdot f(w) = f(v \cdot w)

so

e_i' \cdot e_i' \;=\; f(L^2)

and

e_i' \cdot e_{i+1}' \;=\; f(1/2\Phi) \, f(L^2) \;=\; -(\Phi/2) \, f(L^2)

Therefore the cosine of the angle between consecutive edge vectors e_i' is

\displaystyle{ \frac{e_i' \cdot e_{i+1}'}{\sqrt{(e_i' \cdot e_i')(e_{i+1}' \cdot e_{i+1}')}} \;=\; \frac{-(\Phi/2) f(L^2)}{\sqrt{f(L^2)^2}} \; = \; -\Phi/2 }

Since

\cos(4\pi/5) \; = \; -\Phi /2

it follows that the angle between consecutive edge vectors is 4\pi/5.

But wait! The above calculation secretly assumed f(L^2) is positive, because we claimed that the usual positive square root of f(L^2)^2 equals f(L^2), and we also felt free to divide by f(L^2). Why is f(L^2) positive? Writing

e_i = (c_1, \dots, c_n)

with c_j \in \mathbb{Q}(\sqrt{5}), we have

L^2 = \sum_j c_j^2

and thus

f(L^2) = \sum_j f(c_j)^2

This is a sum of squares of real numbers, hence nonnegative. It is strictly positive because f is injective, so the f(c_j) are not all zero.

The five points v_i' are coplanar, since coplanarity amounts to the vanishing of certain 3 × 3 minors in the matrix of coordinate differences, which is a polynomial condition over \mathbb{Q}(\Phi), hence preserved by f. These points are also distinct, since f is injective. The edges e'_i thus form a planar closed 5-gon with equal edge lengths \sqrt{f(L^2)} and constant exterior turning angle 4\pi/5 at each vertex. This is a regular pentagram.   █

The same style of argument shows that applying Galois conjugation to a regular pentagram with vertices in \mathbb{Q}(\sqrt{5})^n, we get back a regular pentagon. And if you’re worried about what happened to the plane, fear not! We can draw regular pentagons in the plane whose vertices have coordinates in a certain quadratic extension of \mathbb{Q}(\sqrt{5}). This larger field again has an automorphism that carries regular pentagons to regular pentagrams.

We can also play similar games with heptagons and the like, using different fields.

Acknowledgments and addenda

The red pictures of polyhedra were made using were created using Robert Webb’s Stella software and placed on Wikicommons. The diagram of Kepler–Poinsot polyhedra was created by Tilman Piesk and placed on Wikicommons.

For a description of the small stellated dodecahedron and great dodecahedron as Riemann surfaces—branched coverings of the sphere—try this:

• John Baez, Small stellated dodecahedron, Visual Insight, 15 June 2016.

I do not know how these branched coverings are related to the Galois theory perspective given here!

On Mastodon, J. M. animated an icosahedron morphing into a great icosahedron:

and a dodecahedron morphing to a great stellated dodecahedron:

Here’s another fun example. If you take a rhombicosidodecahedron with vertex coordinates all in the golden field:

and apply the Galois transformation \sqrt{5} \mapsto -\sqrt{5}, the pentagons turn into pentagrams, while the squares stay squares and the equilateral triangles stay equilateral triangles. It gets messy if we draw everything, but if we draw just the pentagrams it’s beautiful:

Interestingly the squares and triangles, not drawn, stay the same size—because the squared lengths of their edges are rational! But the squared lengths of the pentagon edges involve \sqrt{5}, so they change as they become pentagram edges.

5 Comments | mathematics | Permalink
Posted by John Baez

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