For my November 2020 diary, go here.

Diary — December 2020

John Baez

This Timurid Quran copied on Ming Dynasty gold-printed colored paper sold for £7m at Christie’s in June. Before the discovery of this one, only four other Qurans on Chinese paper were known, including one in the Detroit Institute of Art and one in the Türk ve Islam Eserleri Muzesi in Istanbul. While it's heart-stoppingly beautiful, there are questions about its provenance.

December 2, 2020

Yesterday Trump's former National Security Advisor Michael Flynn, who was kicked out after 22 days for his involvement in the Russia scandal, came out in favor of a coup. He tweeted his support for a group We The People, which has called on Trump to "immediately declare a limited form of Martial Law, and temporarily suspend the Constitution and civilian control of these federal elections, for the sole purpose of having the military oversee a re-vote."

December 3, 2020

Today the governor of California announced a Regional Stay Home Order which will go into effect within 24 hours in regions with less than 15% ICU availability. It prohibits private gatherings of any size, closes operations except for critical infrastructure and retail, and requires 100% masking and physical distancing in all others.

The order will remain in effect for at least 3 weeks and, after that period, will be lifted when a region’s projected ICU capacity meets or exceeds 15%.

California is divided into 5 regions in this plan; we're in Southern California. No region currently has less than 15% intensive care unit availability. We have 20.6% right now.

State officials predict all regions but the Bay Area will drop below 15% availability in early December, and the Bay Area will follow later in December.

December 4, 2020

On the other hand, Orange County Sheriff Don Barnes said in a statement Saturdaqy that his office would “remain consistent in [its] approach” and would not respond to reported violations of the rules on face coverings, social gatherings or stay-at-home orders. And the sheriff of my county, Riverside County Sheriff Chad Bianco, released a video statement today saying he wouldn't enforce the governor's new rules either. It's depressing — yet another sign of how Trump has politicized the virus.

Needless to say, Lisa and I are following the rules. They're only changing our behavior in one way. Every couple of weeks we'ld been having dinners outdoors where we sit 6 feet away from our friends. No more.

December 8, 2020

On top of that, the Supreme Court just threw out a lawsuit from Pennsylvania:

The Supreme Court on Tuesday refused a request from Pennsylvania Republicans to overturn the state’s election results. The justices said they would not block a ruling from Pennsylvania’s highest court that had rejected a challenge to the use of mail ballots in the state. The Supreme Court’s order was all of one sentence, and there were no noted dissents.

The Supreme Court’s order was all of one sentence, and there were no noted dissents. But it was nonetheless a major setback for Mr. Trump and his allies, who have compiled an essentially unbroken losing streak in courts around the nation. They failed to attract even a whisper of dissent in the court’s first ruling on a challenge to the outcome of the election.

His ever-dwindling chances and undiminished spewing of lies are driving some Trump supporters crazy.

As Trump Rails Against Loss, His Supporters Become More Threatening

Nick Corasaniti, Jim Rutenberg and Kathleen Gray, New York Times, December 8, 2020

With a key deadline passing Tuesday that all but ends his legal challenges to the election, President Trump’s frenzied campaign to overturn the results has reached an inflection point: Certified slates of electors to the Electoral College are now protected by law, and any chance that a state might appoint a different slate that is favorable to Mr. Trump is essentially gone.

Despite his clear loss, Mr. Trump has shown no intention of stopping his sustained assault on the American electoral process. But his baseless conspiracy theories about voting fraud have devolved into an exercise in delegitimizing the election results, and the rhetoric is accelerating among his most fervent allies. This has prompted outrage among Trump loyalists and led to behavior that Democrats and even some Republicans say has become dangerous.

Supporters of the president, some of them armed, gathered outside the home of the Michigan secretary of state Saturday night. Racist death threats filled the voice mail of Cynthia A. Johnson, a Michigan state representative. Georgia election officials, mostly Republicans, say they have received threats of violence. The Republican Party of Arizona, on Twitter, twice called for supporters to be willing to “die for something” or “give my life for this fight.”

“People on Twitter have posted photographs of my house,” said Ann Jacobs, the chair of the Wisconsin Elections Commission, who alerted her neighbors and the police about the constant threats. She said another message mentioned her children and said, “I’ve heard you’ll have quite a crowd of patriots showing up at your door.”

The algebraic K-theory spectrum of the integers has homotopy groups \(\pi_0 = \mathbb{Z}, \pi_1 = \mathbb{Z}/2, \pi_3 = \mathbb{Z}/48\), and so on. These groups are called the algebraic K-theory groups of the integers, \(K_n(\mathbb{Z})\).

The algebraic K-groups of the integers, and other rings, are a big deal. A lot of fairly recent progress on understanding them is due to Voevodsky, Rost, Rognes and Weibel, whose work culminated in a proof of the Bloch–Kato Conjecture connecting K-theory to Galois cohomology. But lots of other deserve credit too.

Three quarters of the algebraic K-groups of the integers are known. That is, they're all known except for the groups \(K_{4n}(\mathbb{Z})\), which will be known if someone can prove the Kummer–Vandiver Conjecture. This is a conjecture about prime numbers.

The Kummer–Vandiver Conjecture been verified for all primes less than 163,000,000. Nonetheless, some unimpressed Wikipedia editor says

... there is no particularly strong evidence either for or against the conjecture.

• Charles Weibel, Algebraic K-theory of rings of integers in local and global fields.

What I'd like to understand is the relation between stable homotopy groups of spheres and algebraic K-theory groups of the integers. For example, the third stable homotopy group of spheres is \(\mathbb{Z}/24\), and this is connected to \(K_3(\mathbb{Z}) = \mathbb{Z}/48\). Topologists understand this connection: there's a homomorphism from the \(n\)th stable homotopy group of spheres, called \(\pi_n^s\), to \(K_n(\mathbb{Z})\). But I'd like to understand it.

If I had to cook up such this homomorphism, here's what I'd guess. The stable homotopy groups of spheres are the homotopy groups of a space called the 'sphere spectrum'. So it's enough to get a map from the sphere spectrum to the algebraic K-theory spectrum of the integers. And the sphere spectrum is built from a symmetric monoidal category just like the algebraic K-theory spectrum is! It's built from the category of finite sets \(0,1,2,\dots\) and bijections between these, made symmetric monoidal using disjoint union. So, I'd use the obvious functor sending any finite set \(n\) to the group \(\mathbb{Z}^n\), and any bijection of finite sets to the corresponding isomorphism of groups. This gives a map of symmetric monoidal categories, and thus a map of spectra, and thus a bunch of homomorphisms \(\pi_n^s \to K_n(\mathbb{Z})\).

Is this right? Are these the right homomorphisms?

If I knew a lot more about this stuff, I'd also understand this claim: the reason \(K_3(\mathbb{Z})\) is \(\mathbb{Z}/48\) is that 24 is the largest natural number \(n\) such that every number relatively prime to \(n\) squares to 1 mod \(n\). Let's check that: $$1^2 = 1$$ $$5^2 = 25 = 24 + 1$$ $$7^2 = 49 = 24 \times 2 + 1 $$ $$11^2 = 121 = 24 \times 5 + 1 $$ $$13^2 = 169 = 24 \times 7 + 1 $$ $$ 17^2 = 289 = 24 \times 12 + 1 $$ $$19^2 = 361 = 24 \times 15 + 1$$ $$23^2 = 529 = 24 \times 22 + 1$$ Yup!

Notice anything interesting about these numbers less than and relatively prime to 24?

Yes! Every natural number \(\le 24\) relatively prime to 24 is prime — except for 1, which is an honorary prime.

A very round number is a natural number \(n\) such that every natural number \(\le n\) relatively prime to \(n\) is prime, except for $1$. These are all the very round numbers: $$ 0,1,2,3,4,6,8,12,18,24,30 $$ It is easy to see that there's no very round number bigger than \(30\). First, it would have to be divisible by \(2, 3\), and \(5\), or else some product of two of these primes would be relatively prime to this number and smaller than it, but not prime. So the first option is \(60\). \(60\) is not very round because \(7^2\) is relatively prime to it but not prime. In short, the problem is that there's a prime that does not divide \(60\) whose square is less than \(60\). So, any round number bigger than \(30\) actually needs to be divisible by \(7\), too, which means it must be bigger than \(2 \times 3 \times 5 \times 7 = 210\). But \(11^2 < 210\) and \(13^2 < 210\), so any round number bigger than \(30\) must actually be bigger than \(2 \times 3 \times 5 \times 7 \times 11 \times 13\). And so on: this is clearly a losing battle. At this stage, knowing there's always a prime between \(n\) and \(2n\) is enough to guarantee that we can keep up this argument by induction.

There's a spooky appearance of very round numbers in the theory of Coxeter groups (see the answer to Puzzle 4 near the bottom of this page.

For example: if you get as many equal-sized balls as possible to touch a single ball of the same size in 8 dimensions, namely \(240\) of them, you get a pattern called the \(\mathrm{E}_8\) root polytope. There's no wiggle room.

The so-called Coxeter number of \(\mathrm{E}_8\) is \(240/8 = 30\). Here are the numbers between \(1\) and \(30\) that are relatively prime to \(30\): $$1,7,11,13,17,19,23,29$$ They're all prime. And yes, there are 8 of them! If we add one to each, we get the magic numbers for \(\mathrm{E}_8\): $$ 2,8,12,14,18,20,24,30 $$ There's a lot you can do with these, but the simplest is to multiply them all: $$ 2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30 = 696,729,600 $$

This gives the order of the Weyl group of \(\mathrm{E}_8\): that is, the symmetry group of the pattern formed when you get as many equal-sized balls as possible to touch a single ball of the same size in 8 dimensions!

December 10, 2020

I want to spend more time doing research and writing expository papers and books, and I've saved up enough money to do this. I'll still do serious work, like trying to save the planet with applied category theory. But I'll also delve into all sorts of puzzles that I haven't had enough time for yet.

Here's one. You may have heard about the funny way the number 24 shows up in the homotopy groups of spheres: $$ \pi_{n+3} (S^n) \cong \mathbb{Z}_{24} $$ whenever \(n\) is big enough, namely \(n \ge 5\). If you try to figure out where this comes from, you're led back to a map $$ S^7 \to S^4 $$ called the quaternionic Hopf fibration. This by itself doesn't make clear where the 24 is coming from — but you can't help but notice that when you pack equal-sized balls as densely as is known to be possible in the quaternions, each one touches 24 others.

Coincidence? Maybe! But it's also true that

$$ \pi_{n+7} (S^n) \cong \mathbb{Z}_{240} $$

whenever \(n\) is big enough. And if you try to figure out where this comes from, you're led back to a map $$ S^{15} \to S^8 $$ called the octonionic Hopf fibration. And you can't help but notice that when you pack equal-sized balls as densely as possible in the octonions, each one touches 240 others!

Maybe these facts are 'coincidences'. There are lots of obstacles to understanding them. But we can at least try to get some idea of where the numbers 24 and 240 are coming from, in the homotopy groups of spheres. They're connected to Bernoulli numbers.

What follows is some very superficial stuff, just to get the pump primed. I just want to state some cool facts and illustrate them with a few examples.

The Bernoulli numbers are connected to a certain well-understood 'chunk' of the stable homotopy groups of spheres, called the image of the J-homomorphism: $$ J_k : \pi_k(\mathrm{O}(\infty)) \to \pi_k^s $$ Here $$ \pi_k^s = \lim_{n \to \infty} \pi_k(S^n) $$

is the \(k\)th stable homotopy group of spheres, and \(\mathrm{O}(\infty)\) is the infinite-dimensional orthogonal group, generated by rotations and reflections in \(\mathbb{R}^\infty\). I'll explain the J-homomorphism later, but first let's see what people do with it.

The homotopy groups of \(\mathrm{O}(\infty)\) are well-understood thanks to Bott periodicity, and you can remember them by singing this to the tune of "Twinkle Twinkle Little Star": $$ \mathbb{Z}_2 , \mathbb{Z}_2, 0, \mathbb{Z}, 0, 0, 0, \mathbb{Z} $$ This means that \(\pi_0(\mathrm{O}(\infty)) = \mathbb{Z}_2\), and so on up to \(\pi_7(\mathrm{O}(\infty)) = \mathbb{Z}\)... and then it repeats, with period 8! It's a boring song that goes on forever — the kind that annoying children sing on long road trips. But the highlights are the two \(\mathbb{Z}\)'s, which are connected to the quaternionic and octonionic Hopf fibrations: $$ \pi_3(\mathrm{O}(\infty)) \cong \mathbb{Z}, \qquad \pi_7(\mathrm{O}(\infty)) \cong \mathbb{Z} $$ The J-homomorphism maps these to — and in fact as it happens onto — the 3rd and 7th stable homotopy groups of spheres: $$ \mathbb{Z}_{24} \cong \mathrm{im} J_3 \cong \pi_3^s $$ $$ \mathbb{Z}_{240} \cong \mathrm{im} J_7 \cong \pi_7^s $$ So in these two cases, knowing \(\mathrm{im} J_k\) tells you everything. But the J-homomorphism is not onto in general, and its image is easier to understand than the rest of the stable homotopy group \(\pi_k^s\), so I'll focus on that.

Note: the cases we care about happen when \(k\) is one less than a multiple of 4. Frank Adams showed something wonderful: when \(k = 4n -1\), the image of the J-homomorphism is a cyclic group whose order is connected to Bernoulli numbers!

Namely, the order of \(\mathrm{im}J_{4n-1}\) is the denominator of $$ B_{2n}/ 4n $$ where \(B_{2n}\) is a Bernoulli number. Here are some examples:

• \(n = 1\). Here \(B_{2n} = 1/6\), so the denominator of \(B_{2n}/4n\) is \(4 \times 6 = 24\), so \(\mathbb{Z}_{24} \cong \mathrm{im}J_3 \subseteq \pi_3^s\). In this case \(\mathbb{Z}_{24}\) is all of the stable homotopy group \(\pi_3^s\).
• \(n = 2\). Here \(B_{2n} = -1/30\), so the denominator of \(B_{2n}/4n\) is \(8 \times 30 = 240\), so \(\mathbb{Z}_{240} \cong \mathrm{im}J_7 \subseteq \pi_7^s\). In this case \(\mathbb{Z}_{240}\) is all of the stable homotopy group \(\pi_7^s\).
• \(n = 3\). Here \(B_{2n} = 1/42\), so the denominator of \(B_{2n}/4n\) is \(12 \times 42 = 504\), so \(\mathbb{Z}_{504} \cong \mathrm{im}J_{11} \subseteq \pi_{11}^s\). In this case \(\mathbb{Z}_{504}\) is all of the stable homotopy group \(\pi_{11}^s\).
• \(n = 4\). Here \(B_{2n} = -1/30\), so the denominator of \(B_{2n}/4n\) is \(16 \times 30 = 480\), so \(\mathbb{Z}_{480} \cong \mathrm{im} J_{15} \subseteq \pi_{15}^s\). In this case \(\mathbb{Z}_{480}\) is not the whole stable homotopy group \(\pi_{15}^s \cong \mathbb{Z}_480 \times \mathbb{Z}_2\). Just when you were getting complacent!

This theorem says the denominator of \(B_{2n}\) is the product of all primes \(p\) such that \(p-1\) divides \(2n\). Let's try it out!

• \(n = 1\). The primes \(p\) such that \(p-1\) divides \(2n = 2\) are 2 and 3, so the denominator of \(B_{2n}\) is \(2 \times 3 = 6\), and the denominator of \(B_{2n}/4n\) is \(4 \times 6 = 24\).
• \(n = 2\). The primes \(p\) such that \(p-1\) divides \(2n = 4\) are 2, 3 and 5 so the denominator of \(B_{2n}\) is \(2 \times 3 \times 5 = 30\), and the denominator of \(B_{2n}/4n\) is \(8 \times 30 = 240\).
• \(n = 3\). The primes \(p\) such that \(p-1\) divides \(2n = 6\) are 2, 3 and 7 so the denominator of \(B_{2n}\) is \(2 \times 3 \times 7 = 42\), and the denominator of \(B_{2n}/4n\) is \(12 \times 42 = 504\).
• \(n = 4\). The primes \(p\) such that \(p-1\) divides \(2n = 8\) are 2, 3, and 5 so the denominator of \(B_{2n}\) is \(2 \times 3 \times 5 = 30\), and the denominator of \(B_{2n}/4n\) is \(16 \times 30 = 480\).

We could ask if the weird pattern I mentioned at the start keeps on going. That is: is there some nice lattice packing of balls in dimension 12 where each ball touches 504 others? Is there one in dimension 16 where each ball touches 480 others, or perhaps twice as many?

I don't know. The Coxeter–Todd lattice is a cool lattice in 12 dimensions, but each ball touches 756 others. Like 504, this number has 2, 3 and 7 as prime factors. Does that count for something? The Barnes–Wall lattice is a cool lattice in 16 dimensions, but each ball touches 4320 others. Like 480, this number has 2, 3 and 5 as prime factors. Does that count for something? Who knows!

There's a lot left to understand... the stuff about lattices may be a dead end, but the connection between \(\mathrm{im} J\) and Bernoulli numbers is worth a lot more study. I just wanted to lay some facts on the table.

Oh yeah — and what's the J-homomorphism?

You can get the orthogonal group \(\mathrm{O}(n)\) to act on an \(n\)-sphere. It's obvious how it acts on an \((n-1)\)-sphere — use the unit sphere in \(\mathbb{R}^n\). But if you take \(\mathbb{R}^n\) and add a point at infinity you get an \(n\)-sphere, and since \(\mathrm{O}(n)\) acts on \(\mathbb{R}^n\) it acts on this \(n\)-sphere. Even better, it acts in a way that preserves a point, namely the origin — or if you prefer, the point at infinity. So we get a map $$ j \colon \mathrm{O}(n) \times S^n \to S^n $$ which is basepoint-preserving.

This lets us take information about the homotopy groups of \(\mathrm{O}(n)\) and get information about the homotopy groups of spheres. That's the idea behind the J-homomorphism! It will be a homomorphism $$ J_k \colon \pi_k(\mathrm{O}(n)) \to \pi_k(S^{n+k}) $$ It goes like this. Let's work in the world of nice topological spaces and basepoint-preserving maps; this has a [smash product](https://en.wikipedia.org/wiki/Smash_product) \(X \wedge Y\) where we take the usual cartesian product \(X \times Y\) and smash down everything like \((\ast, y)\) and $(x,\ast)\). Since our map above is basepoint-preserving, it gives a map $$ j \colon \mathrm{O}(n) \wedge S^n \to S^n $$ Now, if we have an element of \(\pi_k(\mathrm{O}(n))\), we can represent it with a map \(\alpha \colon S^k \to \mathrm{O}(n)\) and then take the smash product $$ \alpha \wedge 1 : \; S^k \wedge S^n \; \to \; \mathrm{O}(n) \wedge S^n $$ But \(S^k \wedge S^n \cong S^{n+k}\), so we get a map $$ S^{n+k} \to \mathrm{O}(n) \wedge S^n $$ Now we can compose this with \(j\) and get a map $$ S^{n+k} \to S^n $$ This represents an element of \(\pi_{n+k}(S^n)\). And because that element does not depend on our original choice of a representative \(\alpha\), we've gotten a map $$ J_k \colon \pi_k(\mathrm{O}(n)) \to \pi_{n+k}(S^n) $$ It's actually a homomorphism: the J-homomorphism. Then we can take the limit as \(n \to \infty\) and get $$ J_k \colon \pi_k(\mathrm{O}(\infty)) \to \pi_k^s $$ This is great — it was discovered by George W. Whitehead, who taught me homotopy theory at MIT. But I've always found it a bit twisty, so I'm glad to see some slicker viewpoints on the nLab. For example, we can take the action $$ j \colon \mathrm{O}(n) \times S^n \to S^n $$ and express it as a homomorphism $$ \mathrm{O}(n) \to \mathrm{Aut}(S^n) \hookrightarrow \Omega^n S^n $$ where \(\Omega^n S^n\) is the topological monoid of basepoint-preserving maps from the sphere to itself. And this induces a homomorphism $$ \pi_k(\mathrm{O}(n)) \to \pi_k (\Omega^n S^n) \cong \pi_{n+k}(S^n) $$ which is the J-homomorphism.

Or, if we want to go straight to the stable case, we can work with the spectrum \(\mathrm{O}\), which is a nice way of thinking about \(\mathrm{O}(\infty)\), and the sphere spectrum \(\mathrm{S}\), which is a way of thinking about \(\Omega^n S^n\) in the limit as \(n \to \infty\). Using these we can think of the J-homomorphism as a map of spectra $$ J \colon \mathrm{O} \to \mathrm{GL}_1(\mathrm{S}) $$ Here \(\mathrm{GL}_1(R)\) makes sense for any ring spectrum \(R\): it's roughly the invertible \(1 \times 1\) matrices with entries in \(R\). The sphere spectrum is a ring spectrum — the most fundamental ring spectrum of all — and its invertible elements come from guys in \(\mathrm{Aut}(S^n)\) for any sphere \(S^n\). So this is essentially just a slick packaging of what I'd said before.

Anyway, I've drifted away from my original goal, which is to understand how the Bernoulli numbers get into the game... but with luck I'll get back to this someday, maybe after I retire.

December 11, 2020

I have three papers with students to finish and I’m trying hard to finish two by the end of this year. I’m looking forward to some freedom, where I can wake up some days and do whatever I want.

I plan to travel a bit and talk to good mathematicians when the plague ends. Brendan Fong and David Spivak are trying to set up an institute for applied category theory in the Bay area — the Topos Institute — and I hope to work there. But it may turn out that I need grad students to stay happy. It’ll be interesting to see.

On a different note, the Supreme Court shot down the case Trump called "the big one" — a crazy attempt by the attorney general of Texas to throw out the election results in four other states, for no particulary good reason except that it help let Trump win an election he actually lost. Shockingly, over 60% of Republicans in the House of Representatives have endorsed this effort.

Supreme Court Rejects Texas Suit Seeking to Subvert Election

Adam Liptak, New York Times, December 11, 2020

WASHINGTON — The Supreme Court on Friday rejected a lawsuit by Texas that had asked the court to throw out the election results in four battleground states that President Trump lost in November, ending any prospect that a brazen attempt to use the courts to reverse his defeat at the polls would succeed.

The court, in a brief unsigned order, said Texas lacked standing to pursue the case, saying it “has not demonstrated a judicially cognizable interest in the manner in which another state conducts its elections.”

The order, coupled with another one on Tuesday turning away a similar request from Pennsylvania Republicans, signaled that a conservative court with three justices appointed by Mr. Trump refused to be drawn into the extraordinary effort by the president and many prominent members of his party to deny his Democratic opponent, former Vice President Joseph R. Biden Jr., his victory.

It was the latest and most significant setback for Mr. Trump in a litigation campaign that was rejected by courts at every turn.

Texas’ lawsuit, filed directly in the Supreme Court, challenged election procedures in four states: Georgia, Michigan, Pennsylvania and Wisconsin. It asked the court to bar those states from casting their electoral votes for Mr. Biden and to shift the selection of electors to the states’ legislatures. That would have required the justices to throw out millions of votes.

Mr. Trump has said he expected to prevail in the Supreme Court, after rushing the confirmation of Justice Amy Coney Barrett in October in part in the hope that she would vote in Mr. Trump’s favor in election disputes.

“I think this will end up in the Supreme Court,” Mr. Trump said of the election a few days after Justice Ruth Bader Ginsburg’s death in September. “And I think it’s very important that we have nine justices.”

He was right that an election dispute would end up in the Supreme Court. But he was quite wrong to think the court, even after he appointed a third of its members, would do his bidding. And with the Electoral College set to meet on Monday, Mr. Trump’s efforts to change the outcome of the election will soon be at an end.

Things are a bit nasty in DC, but so far it's not spiralling out of control:

Multiple People Stabbed After Thousands Gather for pro-Trump Demonstrations in Washington

Emily Davies, Rachel Weiner, Clarence Williams, Marissa J. Lang and Jessica Contrera, New York Times, December 12, 2020

Thousands of maskless rallygoers who refuse to accept the results of the election turned downtown Washington into a falsehood-filled spectacle Saturday, two days before the electoral college will make the president’s loss official.

In smaller numbers than their gathering last month, they roamed from the Capitol to the Mall and back again, seeking inspiration from speakers who railed against the Supreme Court, Fox News and President-elect Joe Biden. The crowds cheered for recently pardoned former national security adviser Michael Flynn, marched with conspiracy theorist Alex Jones and stood in awe of a flyover from what appeared to be Marine One.

But at night, the scene became violent. At least four people were stabbed near Harry’s Bar at 11th and F streets NW, a gathering point for the Proud Boys, a male-chauvinist organization with ties to white nationalism.

The victims were hospitalized and suffered possibly life-threatening injuries, D.C. fire spokesman Doug Buchanan said. It was not immediately clear with which groups the attackers or the injured might have been affiliated.

The violence escalated after an evening of faceoffs with counterprotesters that took place near Harry’s, Black Lives Matter Plaza, Franklin Square, and other spots around downtown.

At first, officers in riot gear successfully kept the two sides apart, even as the groups splintered and roamed. In helmets and bulletproof vests, Proud Boys marched through downtown in militarylike rows, shouting "move out" and "1776!" They became increasingly angry as they wove through streets and alleys, only to find police continuously blocking their course with lines of bikes.

"Both sides of the aisle hate you now. Congratulations," a Proud Boy shouted at the officers.

But before long, the agitators determined to find trouble were successful—and posturing quickly turned into punching, kicking and wrestling.

Again and again, officers swarmed, pulling the instigators apart, firing chemical irritants and forming lines between the sides. At Harry’s Bar, an ambulance arrived, but the extent of injuries was unknown.

Each time a fight was de-escalated, another soon began in a different part of town.

In Orange County, a 50-bed field hospital will be built at Fountain Valley Regional Hospital, a 25-bed field hospital at St. Jude Medical Center in Fullerton and a 50-bed field hospital at the University of California-Irvine.

Meanwhile San Bernardino County is suing to have the governor's coronavirus restrictions lifted.

December 20, 2020

She proposed the modest title “Theoretical Physics in the 21st Century”. I like this idea because it would give me a chance to think about the ways in which theoretical physics is stuck, the ways it's not, and the ways theoretical physics can help us adapt to the Anthropocene. So, I could blend ideas from these two talks:

• Unsolved mysteries in fundamental physics, Cambridge University Physics Society, October 3, 2018.
• Energy and the environment — what physicists can do, Perimeter Institute, April 17, 2013.

Of course we can all resolve to fly less, etc. — but none of those suggestions take advantage of special skills that physicists have. Anna Knörr correctly noted that many theoretical physicists have trouble seeing what they can do to help our civilization adapt to the Anthropocene, since many of them are not good at designing better batteries, solar cells, fission or fusion reactors comes easily. To the extent that I'm a theoretical physicist I fit into this unhappy class. But I think there are more theoretical activities that can still be helpful! And I have more to say about this now than in 2013.

One lesson I may offer is this:

If something is not working, try something different.

Important discoveries await the next generation of accelerators. QCD and the electroweak theory need further confirmation. We need to know how b quarks decay. The weak interaction intermediaries must be seen to be believed. The top quark (or the perversions needed by topless theories) lurks just out of range. Higgs may wait to be found. There could well be a fourth family of quarks and leptons. There may even be unanticipated surprises. We need the new machines.

On the other hand, we have for the first time an apparently correct theory of elementary particle physics. It may be, in a sense, phenomenologically complete. It suggests the possibility that there are no more surprises at higher energies, at least at energies that are remotely accessible.

Proton decay, if it is found, will reinforce belief in the great desert extending from 100 GeV to the unification mass of 10¹⁴ GeV. Perhaps the desert is a blessing in disguise. Ever larger and more costly machines conflict with dwindling finances and energy reserves. All frontiers come to an end. You may like this scenario or not; it may be true or false. But, it is neither impossible, implausible, nor unlikely. And, do not despair nor prematurely lament the death of particle physics. We have a ways to go to reach the desert, with exotic fauna along the way, and even the desolation of a desert can be interesting.

But we're not seeing anything beyond the Standard Model: no "exotic fauna".

Glashow's "new frontier" was the "passive frontier": non-accelerator experiments like neutrino measurements, and this is indeed where the progress came since 1980: we now know neutrinos are massive and oscillate, and there is still some mystery here and room for surprises—though frankly I suspect that neutrino masses will work very much like quark masses, via coupling to the Higgs. (This is in a sense the most conservative, least truly exciting scenario.)

So, very little dramatic progress has happened in particle physics since 1980 — except for a profusion of new theories that haven't made any verified predictions. I'll argue that physicists should turn elsewhere! There are other things for them to do, that are much more exciting.

December 23, 2020

The testing site is on Canyon Crest north of campus, right next to the baseball field. They take drive-ins, not walk-ins.

We felt a bit sick right near the start of the pandemic, and it'd be nice to know if we'd gotten coronavirus. But also, our county really wants more people to start regularly testing.... if people only test when they feel sick we'll never get the true picture of what's going on.

December 26, 2020

Jeff Bezos' wealth went from $115 billion in March to $185 billion now. Mark Zuckerberg's wealth went from $54 billion to $105 billion. Elon Musk went from $25 billion in March to $143 billion now—a really dramatic increase.

For my January 2021 diary, go here.

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