My childhood self would have been so happy to see a picture of Pluto at night. It still delights me!
This was snapped by the New Horizons spacecraft just 19 minutes after its closest approach. Pluto is a mere 21,000 kilometers away, while the Sun is far, far away: 4.9 billion kilometers!
You can clearly see Pluto's hazy atmosphere. It's mostly nitrogen, with a bit of methane — but under the slow influence of cosmic rays, these gases react to form more complex hydrocarbons and organic cyanide compounds, technically called 'nitriles', which condense and fall to the ground... like a faint mist of brown snow.
The twilit landscape you see here includes the nitrogen ice plains of Sputnik Planitia and rugged mountains of water ice in the Norgay Montes!
The god of the underworld, now exiled to the cold darkness of space, slumbering....
October 4, 2025
For example, yesterday I learned about 'crown ethers': beautiful crown-shaped molecules of carbon, hydrogen and oxygen. Above is a ball-and-stick model of a crown ether called 15-crown-5.
Why is it called that? I guess because it has 5 oxygens (in red) and 10 carbons (in black) for a total of 15 heavier atoms. Then there are a bunch of hydrogens poking out. It has perfect 5-fold symmetry.
Yasui Yoshio won the Nobel Prize in 1987 for discovering crown ethers around 1967. He's usually known as Charles J. Pedersen. He was born in Korea but later moved to Japan and then the US.
One use of crown ethers is to hold metal ions. A sodium atom with one electron missing can snugly fit inside 15-crown-5. So can a transition metal with two electrons missing, like cobalt (Co²⁺), nickel (Ni²⁺), copper (Cu²⁺), or zinc (Zn²⁺).
Other metal ions fit better in larger crown ethers. For example, here is 18-crown-6 holding a potassium ion. Beautiful 6-fold symmetry!
The 6 bonds here come from 6 oxygens connecting to the central potassium atom in purple. I don't know, but I bet each oxygen donates 1/6 of an electron to the potassium atom, which is missing one.
1/6 of an electron?!?
You can't actually chop an electron into parts, but you can have a single electron whose wavefunction is smeared out around all 6 oxygen atoms and the central potassium.
So, this ball-and-stick model is pretty misleading! The molecule doesn't really have sticks in it. Let me show you another picture of it:
This more chubby depiction is called a space-filling model. This makes it clear how the potassium fits snugly inside the crown.
Another thing you can put in 18-crown-6 is hydronium, H3O+:
Hydronium is the ion formed by a water molecule and an extra proton. We see here that its 3-fold symmetry fits nicely into the 6-fold symmetry of 18-crown-6.
But when 18-crown-6 does not have an ion inside it, it folds into a shape that lacks 6-fold symmetry:
Here are some more crown ethers:
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They are:
1) 12-crown-4,
2) 15-crown-5,
3) 18-crown-6,
4) dibenzo-18-crown-6,
6) an aza-crown ether, meaning one that contains nitrogen.
For a final burst of fun, here's an aza-crown ether holding a nickel atom in the center, and two chlorines (in green):
Cl2.png)
It's called transNi(cyclam)Cl2.
The diversity of highly symmetrical structures in chemistry is truly wondrous! And needless to say, I didn't draw any of these pictures. I got them from Wikicommons, and you can find out where by clicking on them. Most are in the public domain, thanks to the generosity of their creators.
October 6, 2025
Dubois-Violette and Todorov noticed that the Standard Model gauge group is the intersection of two maximal subgroups of \(\mathrm{F}_4\). I'm trying to figure out if this is more than a coincidence.
Very roughly speaking, \(\mathrm{F}_4\) is the symmetry group of an octonionic qutrit. Of the two subgroups I'm talking about, one preserves a chosen octonionic qubit, while the other preserves a chosen complex qutrit.
A precise statement is here:
Over on Mathstodon I'm working with Paul Schwahn to improve this statement. He made a lot of progress on characterizing the first subgroup. \(\mathrm{F}_4\) is really the group of automorphisms of the Jordan algebra of \(3 \times 3\) self-adjoint octonion matrices, \(\mathfrak{h}_3(\mathbb{O})\). He showed the first subgroup, the one I said "preserves a chosen octonionic qubit", is really the subgroup that preserves any chosen Jordan subalgebra isomorphic to \(\mathfrak{h}_2(\mathbb{O})\).
Now we want to show the second subgroup, the one I said "preserves a chosen complex qutrit", is really the subgroup that preserves any chosen Jordan subalgebra isomorphic to \(\mathfrak{h}_3(\mathbb{C})\).
I want to sketch out a proof strategy. So, I'll often say "I hope" for a step that needs to be filled in.
Choose an inclusion of algebras \(\mathbb{C} \to \mathbb{O}\) All such choices are related by an automorphism of the octonions, so it won't matter which one we choose.
There is then an obvious copy of \(\mathfrak{h}_3(\mathbb{C})\) sitting inside \(\mathfrak{h}_3(\mathbb{O})\). I'll call this the standard copy. To prove the desired result, it's enough to show:
Part 1) should be the easier one, but I don't even know if this one is true! \((\text{SU}(3) × \text{SU}(3))/\mathbb{Z}_3\) is a maximal subgroup of \(\mathrm{F}_4\), and Yokota shows it preserves the standard copy of \(\mathfrak{h}_3(\mathbb{C})\) in \(\mathfrak{h}_3(\mathbb{O})\). But he shows it also preserves more, seemingly: it preserves a complex structure on the orthogonal complement of that standard copy. Is this really 'more' or does it hold automatically for any element of \(\mathrm{F}_4\) that preserves the standard copy of \(\mathfrak{h}_3(\mathbb{C})\)? I don't know.
But I want to focus on part 2). Here's what we're trying to show: any Jordan subalgebra of \(\mathfrak{h}_3(\mathbb{O})\) isomorphic to \(\mathfrak{h}_3(\mathbb{C})\) can be obtained from the standard copy of \(\mathfrak{h}_3(\mathbb{C})\) by applying some element of \(\mathrm{F}_4\).
So, pick a Jordan subalgebra A of \(\mathfrak{h}_3(\mathbb{O})\) isomorphic to \(\mathfrak{h}_3(\mathbb{C})\). Pick an isomorphism A ≅ \(\mathfrak{h}_3(\mathbb{C})\).
Consider the idempotents $$ \begin{array}{ccl} e_1 &=& \text{diag}(1,0,0) \\ e_2 &=& \text{diag}(0,1,0) \\ e_3 &=& \text{diag}(0,0,1) \end{array} $$
in \(\mathfrak{h}_3(\mathbb{C})\). Using our isomorphism \(A \cong \mathfrak{h}_3(\mathbb{C})\) they give idempotents in \(A\), which I'll call \(f_1, f_2, f_3\). Since \(A \subset \mathfrak{h}_3(\mathbb{O})\) these are also idempotents in \(\mathfrak{h}_3(\mathbb{O})\).
Hope 1: I hope there is an element \(g\) of \(\mathrm{F}_4\) mapping \(f_1, f_2, f_3 \mathfrak{h}_3(\mathbb{O})\) to \(e_1, e_2, e_3 \in \mathfrak{h}_3(\mathbb{C})\) ⊂ \(\mathfrak{h}_3(\mathbb{O})\).
Hope 2 Then I hope there is an element \(h\) of \(\mathrm{F}_4\) that fixes \(e_1, e_2, e_3\) and maps the subalgebra \(g A\) to the standard copy of \(\mathfrak{h}_3(\mathbb{C})\) in \(\mathfrak{h}_3(\mathbb{O})\).
If so, we're done: \(h g\) maps \(A\) to the standard copy of \(\mathfrak{h}_3(\mathbb{C})\).
Hope 1 seems to be known. The idempotents \(e_1, e_2, e_3\) form a so-called 'Jordan frame' for $\mathfrak{h}_3(\mathbb{O})$, and so do \(f_1, f_2, f_3\). Faraut and Korányi say that "in the irreducible case, the group $K$ acts transitively on the set of all Jordan frames", and I think that implies Hope 1.
As for Hope 2, I know the subgroup of \(\mathrm{F}_4\) that fixes \(e_1, e_2, e_3\) contains \(\text{Spin}(8)\). I bet it's exactly \(\text{Spin}(8)\). But to prove Hope 2 it may be enough to use \(\text{Spin}(8)\)/
Let me say a bit more about how we might realize Hope 2. It suffices to consider a Jordan subalgebra \(B\) of \(\mathfrak{h}_3(\mathbb{O})\) that is isomorphic to \(\mathfrak{h}_3(\mathbb{C})\) and contains $$ \begin{array}{ccl} e_1 &=& \text{diag}(1,0,0) \\ e_2 &=& \text{diag}(0,1,0) \\ e_3 &=& \text{diag}(0,0,1) \end{array} $$ and prove that there is an element \(h\) of \(\mathrm{F}_4\) that fixes \(e_1, e_2, e_3\) and maps the subalgebra \(B\) to the standard copy of \(\mathfrak{h}_3(\mathbb{C})\) in \(\mathfrak{h}_3(\mathbb{O})\). (In case you're wondering, this \(B\) is what I was calling \(g A\).) Hope 3: I hope that we can show \(B\) consists of matrices $$ \left( \begin{array}{ccc} \alpha_1 & z^\ast & y^\ast \\ z & \alpha_2 & x \\ y & x^\ast & \alpha_3 \end{array} \right) $$ where \(\alpha_1, \alpha_2, \alpha_3\) are arbitrary real numbers and \(x, y, z\) range over 2-dimensional subspaces \(V_1, V_2, V_3\) of \(\mathbb{O}\). This would already make it look fairly similar to the standard copy of \(\mathfrak{h}_3(\mathbb{C})\), where the subspaces \(V_1, V_2, V_3\) are all our chosen copy of \(\mathbb{C}\) in \(\mathbb{O}\).
If Hope 3 is true, the subspaces \(V_1, V_2, V_3\) don't need to be the same, but I believe they do need to obey \(V_1 V_2 \subseteq V_3\) and cyclic permutations thereof, simply because \(B\) is closed under the Jordan product.
So, we naturally want to know if such a triple of 2d subspaces of \(\mathbb{O}\) must be related to the 'standard' one (where they are all \(\mathbb{C}\)) by an element of \(\text{Spin}(8)\), where \(\text{Spin}(8)\) acts on the three copies of \(\mathbb{O}\) by the vector, left-handed spinor, and right-handed spinor representations, respectively — since this is how \(\text{Spin}(8)\) naturally acts on \(\mathfrak{h}_3(\mathbb{O})\) while fixing all the diaonal matrices.
This is a nice algebra question for those who have thought about triality, and more general 'trialities'.
So, that's where I am now: a bunch of hopes which might add up to a clarification of what I mean by "the subgroup of symmetries of an octonionic qutrit that preserve a complex qutrit".