Greg Egan

I am a science fiction writer and computer programmer.
Latest novel: MORPHOTROPHIC.
Latest collection: SLEEP AND THE SOUL.
Web site: gregegan.net

Web site: https://www.gregegan.net
Location: Perth, Western Australia

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npub12cuzzqzv8e8y7na77a9zv47mx2v6pec60qs5h7takzmv5p0mw0fsmgd38p

NIP-05 Address

gregegansf_at_mathstodon.xyz@momostr.pink

Profile Code

nprofile1qqs9vwppqpxrunj0f7l0wj3x2ldn9xdquud8sg2tl97mpdk2qhah85clrftml

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npub12cuzzqzv8e8y7na77a9zv47mx2v6pec60qs5h7takzmv5p0mw0fsmgd38p

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Last Notes

2025-05-04 14:14:55

In base 10, there’s an integer whose reciprocal has a repeating block with all 10 digits exactly once:
1/72,728 = 0.000(0137498625) ...
I found examples for all even bases from 2 to 34 except 8, 16 and 32. And for a few days I thought “If these are the only examples, surely it can’t be hard to prove there are no others.”
But then I stepped back and looked at the simpler things in number theory people have been trying to prove for centuries.
For example: are 3, 5, 17, 257 and 65537 the only primes of the form 2^(2^n)+1?
https://en.wikipedia.org/wiki/Fermat_number
If Euler couldn’t prove this, and nobody since has managed to do it either, it seems very unlikely that I’ll ever prove a conjecture about the bases in which the reciprocal of an integer has a non-redundant pandigital reptend.
But check out base 34!
https://gregegan.net/SCIENCE/Reptends/Reptends.html
https://media.mathstodon.xyz/media_attachments/files/114/449/976/483/147/941/original/b6d0d08b7ea93622.png

2025-03-26 22:18:00

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Yes, I phrased that badly; I knew the other two conditions followed from the first, so I should have said that explicitly, rather than making it sound like they were independent requirements.

2025-03-18 00:04:36

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I wasn’t arguing against anything you’re saying, I was just trying to spell out all the requirements on D(a) for the sake of performing concrete calculations.
@npub1nf4…nqe4 @npub1ums…7ra6

2025-03-17 11:26:26

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I should just add that, while the equation:
D(a c) = a D(c) + D(a)
makes the map:
x ↦ a x + b
↓
x ↦ a x + (b q + D(a))
into an automorphism (where q is any invertible element), when you *also* need it to be class-preserving and outer, if you compare this with what happens to
x ↦ a x + b
conjugated by:
x ↦ c x + d
Namely:
x ↦ a x + (b c + d (1-a))
you have the additional constraint that:
D(a) = δ(a) (1-a)
where δ(a) is not a constant function.

2025-03-16 22:50:02

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The crucial properties that D(a) needs to satisfy in order to get an automorphism are:
D(a c) = a D(c) + D(a) = c D(a) + D(c)
D(1) = 0
while D(a) = 0 for all a is ruled out, because then the automorphism is inner.

2025-03-16 05:20:06

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I think I’ve found a class-preserving outer automorphism of Aff(ℤ/8)!
You map:
x ↦ a x + b
to:
x ↦ a x + (b + D(a))
where
D(1) = D(3) = 0
D(5) = D(7) = 4
I don’t think this is unique. But I’m still searching through the (4!)^8 choices of permutations of the odd and even parts of ℤ/8, for each value of a.
Conjugation preserves the parity of b, so rather than there being 8! choices of permutations of ℤ/8 for each value of a, there are (4!)^2.

2024-11-10 07:33:56

I’m a sucker for doppelgänger stories with ironic twists — like Nabokov’s “Despair” [and the Fassbinder movie] — and “A Different Man” taps that vein.
I mostly enjoyed it, but it felt a bit thin for a ~2h feature film, and towards the end the tone slid from committed-to-its-premise to jokily perfunctory.

2024-10-20 07:19:06

Here’s a fun puzzle for fans of the classical differential geometry of curves and surfaces.
Suppose a curve γ(s):R→R^3, parameterised by arc length s, is a geodesic on a surface of constant Gaussian curvature of –1.
Suppose γ''(s_0) = 0, but γ'''(s_0) ≠ 0.
What does that tells us about the torsion of γ(s) at s=s_0?

2024-10-16 11:09:45

“We have a science adviser from MIT, but we broke his spirit and now he just nods everything through.”
— The Franchise

2024-10-15 13:06:02

“But, but … humans hallucinate all the time! How dare you set the bar for AI so much higher!”
Yes, we all remember the many stories written by human journalists that accused Woodward and Bernstein of breaking into the Watergate Hotel.
https://www.theregister.com/2024/08/26/microsoft_bing_copilot_ai_halluciation/

2024-10-13 11:14:06

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FWIW, my favourite Coppola movie is still “Rumblefish”: a visceral, dreamlike evocation of youth, with spinning clocks, racing clouds, and the witness to it all frantically scribbling in a notebook, trying to capture every evanescent moment.

2024-10-13 10:36:25

“Megalopolis” wasn’t terrible, but I wish it had had much less repackaged Roman history and lazy (often literal) quotes from Shakespeare, and more batshit magic materials and Art Deco futurism saving the world. And hoorah for the New Pledge Of Allegiance and The The’s “Lonely Planet” over the credits, but nothing in the film itself engaged much with any real struggle for hope or justice.

2024-10-12 07:50:39

Woke this morning expecting a totally crap day after insomnia at 2 am. A moving truck rolled up next door promising hours of noise.
Went and cut the back lawn, cut my hair, then cleaned the shower.
With the truck gone, in the silence, fixed some code and it ran 1200 times faster.

2024-10-11 05:23:03

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Of course this is all just a hypothetical exercise in which white-hat hackers give a proof of principle, but nobody malicious would actually ... oh, wait:
https://www.abc.net.au/news/2024-10-11/robot-vacuum-yells-racial-slurs-at-family-after-being-hacked/104445408

2024-10-08 22:49:22

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Mine knows, but labels it “Australian English informal”.

2024-10-06 04:01:21

One reviewer, in more innocent times, protesting about the deranged narcissistic billionaire protagonist of my story “Crystal Nights”:
“… unstable people like that don’t get to run companies!”
https://www.gregegan.net/MISC/CRYSTAL/Crystal.html

2024-10-05 22:34:34

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I recently discovered that people who study satellite “constellations” and rendezvous in orbit call the equations for free-fall in this frame the Clohessy–Wiltshire equations:
https://en.wikipedia.org/wiki/Clohessy–Wiltshire_equations

2024-10-05 22:23:26

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Yes, and specifically I say in the post that we’re measuring things in the reference frame of a tidally locked body that always points towards the central mass, so we know exactly how fast it’s rotating around its own axis.
Of course you could work this out in any frame you like, but I find it intuitively easier to linearise the central force after choosing a specific frame, and it’s one whose test-particle trajectories I studied very closely when I was looking at the Newtonian, Kerr and Schwarzschild metric versions for “Incandescence.”
And in this case we also have the advantage that the triangle the spacecraft form lies on a circle that is fixed in the “tidally locked hollow asteroid” frame.

2024-10-05 11:47:03

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The measurement techniques used for a system like this, when the distances along the arms of the interferometer are unequal, and constantly changing, are quite different than those used by ground-based detectors.
https://arxiv.org/abs/gr-qc/0409034

2024-10-05 11:24:04

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How does this triangle formation look for the actual orbits? (Size of the triangle here is much larger than that for LISA!)
https://media.mathstodon.xyz/media_attachments/files/113/254/555/370/289/961/original/1e8349ee59ec9d38.mp4

2024-10-05 08:55:04

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“The gravitational waves that LISA is designed to observe have typical timescales of hours. So long as the distance between the satellites is smoothly changing over these time scales, the gravitational waves can be observed as an additional modulation on top of this smooth change. Each satellite is in an independent Keplerian orbit around the Sun with the plane of the triangle inclined at 60 degrees to the plane of the ecliptic. Over the course of the mission, the nominal 2.5 million kilometer distance between each satellite will vary by hundreds of thousands of kilometers. LISA will be able to measure the absolute distance between the satellites to a few centimeters and will measure hour-scale fluctuations at the level of several picometers (1pm = 1 trillionth of a meter), the level required to detect gravitational waves.”
https://lisa.nasa.gov/faq.html

2024-10-05 08:51:17

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The change in the distance between two points as the same gravitational wave passes is directly proportional to the distance itself. But yes, LISA is aimed at measuring lower frequencies / longer wavelengths than LIGO, and I think that’s the main reason for putting it in space.

2024-10-05 08:37:46

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The resolution of the absolute distance will be much coarser than terrestrial detectors, but because the arms are about a million times longer, they’re also measuring a much larger change.

2024-10-05 07:18:29

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For more on the precise parameters of the LISA orbits, see: “Trajectory Design for the ESA LISA Mission”, Waldemar Martens and Eric Joffre
https://arxiv.org/abs/2101.03040

2024-10-05 06:56:41

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But we also need to include the Coriolis force due to the rotation of the asteroid. That contributes an acceleration of 2 ω y'(t) in the x direction and –2 ω x'(t) in the y-direction.
The final equations this gives us for the xy plane are:
x''(t) = 3ω^2 x + 2ω y'(t)
y''(t) = –2ω x'(t)
There are runaway solutions for these equations, but there are also bounded ones, where a free particle traces out an ellipse that is exactly twice as long in the y-direction as in the x-direction. (In my novel “Incandescence”, aliens in orbit around a black hole are able to tell that Newtonian gravity isn't a good fit for their situation, because they can measure this ellipse as departing from the 2-to-1 ratio.)
That still doesn't give us an equilateral triangle. But if we combine this 2-to-1 elliptical motion with just the right amount of bouncing up and down along the z-axis, we can make a free particle trace out a perfect circle, inclined at 60° to the xy-plane.
And then we can place three particles on that circle, spaced out at 120° intervals, so that they always form an equilateral triangle.
Of course, all of this is just a linear approximation to the exact orbits, only valid for very small distances. In reality, the LISA spacecraft will fly in a triangle that “breathes” and “flexes”, changing its shape and size. So unlike its ground-based counterparts, it will need to perform its extraordinary feats of measurement while the arms of its laser interferometer are not even fixed in relation to each other!

2024-10-05 06:56:15

The LISA gravity wave detector will comprise three spacecraft, all in free-all in heliocentric orbits, that always lie on the vertices of an equilateral triangle. But how can they do that?
The answer is: they can't! Not precisely. Of course, 3 spacecraft at 120° intervals around a perfect circular orbit could do it, but the LISA craft need to be much closer together.
So, the question is: how do they fly in a formation that *approximates* this goal?
To see how, the best way is to imagine what forces you would feel if you were inside a hollowed-out asteroid in a circular orbit, tidally locked so that one axis always pointed towards the sun.
Call the sunwards axis the x-axis, the direction pointing around the orbit the y-xis, and the direction out of the plane of the orbit the z-axis.
Along the z-axis, you would feel a “tidal squeeze” of –GM/r^3 z pulling you back towards the plane of the orbit. A free particle displaced along the z-axis would bounce up and down as if it was on a spring, with a periodic motion (not coincidentally) the same as the period of the orbit.
In the xy plane, things would be a little more complicated. You would feel a “tidal squeeze” of –GM/r^3 y along the y-axis, but that would be exactly cancelled out by a centrifugal force of GM/r^3 y due to the tidally locked rotation of the asteroid, which has an angular rate of ω = √(GM/r^3).
Along the x-axis, you would feel a “tidal stretch” of 2GM/r^3 trying to tear you away from the centre of the asteroid, *plus* a centrifugal force of ω^2 x = GM/r^3 x, for a total of 3GM/r^3 x.
https://media.mathstodon.xyz/media_attachments/files/113/253/504/608/004/523/original/9ed4207940b0e5f4.mp4

2024-10-04 13:52:02

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In fact in the counterexample, only 3 out of 7222 vertices are joined!

2024-10-04 13:41:20

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You're right! Timothy Gowers actually said that you join the corresponding vertices only for a subset, but then the blog post seemed to say you joined them all. But looking more carefully at the paper, it does specify that you take a fixed subset of the vertices.

2024-10-04 10:16:47

Suppose you take a graph G (a collection of nodes joined by edges), make an exact copy of it, G', and join G and G' with edges along some of their corresponding vertices.
If you randomly delete edges from G and G' with the same probability (but don't delete any of the edges you added between G and G'), can it ever be *less likely* that you can still get from a vertex x in graph G to another vertex y also in G than it is to get from x to the copy of y in G'?
The conjecture that no choice of G can make it less likely is known as the Bunkbed Conjecture. And three mathematicians have just found a counterexample!
H/T Timothy Gowers, Javier Campos
https://igorpak.wordpress.com/2024/10/01/the-bunkbed-
conjecture-is-false/
https://arxiv.org/abs/2410.02545

2024-10-04 05:05:23

A great analysis of the physical constraints on interstellar quantum communication. It’s always struck me as highly likely that if advanced civilisations are sending transmissions through space it will be more quantum internet than “I Love Lucy” spillage.
https://arxiv.org/abs/2408.02445

2024-10-04 03:58:20

“[Security researcher Dennis Gliese] had recently found a way to take control of a long list of Ecovacs robots, including lawnmowers and Deebot vacuum cleaners, armed with only a smartphone … entirely over Bluetooth, from up to 140 metres away.”
https://www.abc.net.au/news/2024-10-04/robot-vacuum-hacked-photos-camera-audio/104414020
“After he first revealed the vulnerability in public, the company’s security committee downplayed the issue, saying it requires “specialised hacking tools and physical access to the device”.
It’s hard to square their statement with the reality. All it had taken was my $300 smartphone, and I hadn’t even laid eyes on Sean’s robot until after hacking into it.”
“Ecovacs did in fact have the X2 tested – and certified as secure – by a German company called TÜV Rheinland.
TUV Rheinland’s Alexander Schneider directed me to a digital certificate, which contained an almost complete absence of detail about how it was actually tested.”

2024-10-03 06:15:21

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Where the fruit flies sing:
“[W]e demonstrate that hearing regulates aggression in Drosophila males. Further, we show that courtship and aggression songs differentially affect aggression, indicating that hearing contributes to the context-dependent regulation of aggression.”
https://www.pnas.org/doi/full/10.1073/pnas.1605946114

2024-10-03 04:11:16

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Apparently some people spent their COVID lockdown fixing errors in the first, imperfect draft of the connectome … and also helping to classify 4,581 new types of neuron that were found in the micrographs.
“But these tools aren’t perfect, and the wiring diagram needed to be checked for errors. The scientists spent a great deal of time manually proofreading the data — so much time that they invited volunteers to help.”

2024-10-03 03:43:07

“Murthy hopes to eventually have a male fly connectome, too, which would allow researchers to study male-specific behaviours such as singing.”
Uploaded female fruitfly seeking choral companion to share the long, icy eternity of virtual immortality.
https://www.nature.com/articles/d41586-024-03190-y

2024-09-29 12:58:34

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Thanks!
Nothing says user-friendly like forcing the user to go into the command line for a basic settings change.

2024-09-29 07:27:21

Does anyone else with a Mac with multiple keyboard layouts defined keep seeing a little blue menu offering them the chance to switch keyboards popping up — apparently at random, and not in response to any mouse action or key presses as far as I can tell — in all kinds of different apps, since upgrading to Sonoma?
I have no idea what triggers this, or how to switch it off, and it’s driving me crazy. Maybe I’m accidentally pressing some magic key combination in bouts of clumsy typing, but I can’t see anything in the Settings that would explain this. In Keyboard > Input Sources I’ve switched off using the Caps Lock key to “switch to and from last used Latin input source”, and there’s nothing else that seems relevant.

2024-09-29 00:12:45

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My novella was “The Four Thousand, The Eight Hundred”. It’s in this collection, “Phoresis and Other Journeys”:
https://www.gregegan.net/BIBLIOGRAPHY/Ebooks.html#Phoresis
It’s a while since I watched “The Expanse” (I didn’t finish it before it moved from Netflix to Prime, and I keep meaning to go back to it), but on Wikipedia the plot summary for the first episode reads:
“The series opens with Julie Mao in a spacesuit, alone aboard a spaceship. On the dwarf planet Ceres, Detective Miller is assigned to find Mao and return her to her rich parents on Luna … ”
In the second episode:
“On Ceres, Miller investigates the theft of water, now severely rationed due to the Canterbury's failure to return … ”

2024-09-28 23:51:28

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I was lucky to write a novella about Ceres in 2015, late enough for it to already be clear that they had a lot of water … in the story, they have so much that they send ice to Vesta, which is water poor.
If I recall correctly, in the TV series “The Expanse” (based on books written somewhat earlier) there are water shortages on Ceres.

2024-09-27 09:16:43

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I’m being handed a set of graph nodes by the rustworkx package, so I have no choice but to sort them.

2024-09-27 08:20:11

I’m a novice in Python compared to the languages I’ve used for decades, so it took me a while to jam the message into my brain: if you want your code to give identical results each time it’s run, never iterate over the members of a set without turning it into a sorted list first.

2024-09-27 04:34:09

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Another take:
https://www.abc.net.au/news/2024-09-27/megalopolis-film-reviews-francis-ford-coppola/104401160
FWIW, I plan on seeing it for myself. I’ve sat through enough movies I loathed that the critics all loved, and vice versa, that this seems well worth the gamble.

2024-09-26 23:06:34

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Don’t look in the trunk!

2024-09-26 21:33:23

Breaking: leading AI moguls have predicted that by 2029, their bots will not only be indistinguishable from humans, they will be able to prove the Riemann Hypothesis with all three-and-a-half legs tied behind their backs.

2024-09-25 02:53:11

I checked how long my pet goldfish stared at this research paper, and it was about as long as it stared at other papers that have won the Ig Nobel prize.
https://www.science.org/content/article/could-monkeys-predict-u-s-election

2024-09-24 03:08:57

This looks hugely valuable. It’s the kind of resource the government should have provided; it’s incredibly frustrating that they’ve lost interest in reducing Covid, but kudos to the people who have stepped in to fill the gap here.
https://www.abc.net.au/news/2024-09-24/covid-safety-schools-course-sick-days-teachers-long-covid/104319032

2024-09-22 13:08:36

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I haven’t watched the video, but I wonder if they’re confusing equipotential surfaces and surfaces of equal |g|?

2024-09-22 03:16:32

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It’s fun to spell out why this works, i.e. why do the thousand numbers of the form:
999 + 999 k, for k=0,…,999
all have the same digit sums?
We can rewrite the terms in the arithmetic progression as:
999 + 999 k
= 1000 k + (999–k)
Multiplying by 1000 separates out the digits of k from those of 999–k, so the total digit sum is just the sum of the digit sums of k and of 999–k.
D(999 + 999 k) = D(k) + D(999–k)
In general, D(a)+D(b) is *not* D(a+b), but in this case we have a guarantee that there can be no “borrowing” from higher columns when we subtract k from 999, because no digit we subtract can be bigger than the one we are subtracting from, and equally there can be no need to carry when we add k and 999–k.
So we *do* have:
D(k) + D(999–k) = D(k + 999 – k)
= D(999)
for all k=0,…,999

2024-09-21 02:43:18

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I don’t think this reduces the patient dose required, which would be more related to the sensitivity of the detectors, and the fundamental physics of the interaction of the X-rays with the tissue being imaged.

2024-09-21 01:44:11

I really hope this pans out: X-rays from carbon nanotubes shrink the radiation source from 20kg to a few hundred grams, and the aim is make a CT scanner a third the cost of the standard kind, and that can fit in an ambulance.
https://www.abc.net.au/listen/programs/healthreport/xray-microx-nanotechnology/104374808

2024-09-20 11:45:09

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The first 1000 multiples of 999 all have digit sums of 27.

2024-09-16 21:55:01

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Surely it was his commitment to build a tau lepton collider in his first week in office.

2024-09-15 21:28:20

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The angular radius of a water rainbow is 42°.
The angular radius of an ethane radius is 56°.
The angle 2φ, in the Wikipedia article, is the angular radius: the angle between the direction of the sun and the refracted light forming the rainbow.
Don't be fooled by the 2 in 2φ. That is just the way they have defined some things in their construction. They happen to solve for φ, which then needs to be doubled to obtain the angular radius.

2024-09-15 21:04:47

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No, you didn't screw up. For n=1.25, the angular radius of the rainbow is 56°.
@npub1fzg…30u3 seems to be confused; they claim the opposite relationship between refractive index and rainbow size that what I showed in those tweets, maybe because I show smaller rainbows for some exotic cases with *higher* refractive indices. But ethane is exotic in the opposite manner to the examples I gave.

2024-09-14 04:24:33

MacOS spellchecking, 2024. (Yes I know, there is an alternative spelling “neurones” in UK and Australian English, but actual dictionaries allow both, and “neurons” has been far more common in Australia for decades.)
https://media.mathstodon.xyz/media_attachments/files/113/133/996/818/223/419/original/b5edd39a98c7289e.png

2024-09-13 14:44:01

Today I was prompted (ha ha) by a question Anders Sandberg asked the latest GPT: at what angle should you throw a projectile — subject to uniform gravity, and ignoring air resistance — in order to maximise the length of its trajectory?
[The Chatbot got quite close, but fell over at the end, being too lazy to compute an easy derivative, and then screwing up its numerical solution to a needlessly difficult formulation of the problem.]
It’s well known that to maximise the horizontal distance travelled, you should launch the projectile at 45°. But to maximise the distance along its parabolic path before it hits the ground, the angle turns out to satisfy the beautiful transcendental equation:
s arctanh(s) = 1
where s = sin(θ)
This is solved by:
s ≈ 56.47°

2024-09-12 12:26:22

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For anyone who is accustomed to subtitles, they are the very opposite of distracting. On rare occasions they're poorly done (bad translations, or insufficient contrast between the text and the background), but in most cases they’re like a magical heads-up display that grants you the power to understand, effortlessly, what the actors are saying without losing any of the expressiveness of the original performance.

2024-09-12 11:52:29

Does anyone really watch foreign-language programs dubbed into English, when subtitles are available? I hadn't been hit with the old lips vs dialogue mismatch for about 50 years, but suddenly I’m seeing Apple TV promos as hilarious as those dubbed spaghetti Westerns from the 70s.

2024-09-12 10:40:33

The first chapter of “AI Snake Oil” by Arvind Narayanan and Sayash Kapoor" is free to read onine:
https://press.princeton.edu/books/hardcover/9780691249131/ai-snake-oil#preview

2024-09-07 13:07:33

Most nuclei require X-rays or gamma rays to raise their energy levels, but it can happen to thorium-229 with a far-ultraviolet photon of 8 eV (wavelength 150 nm).
This was predicted in the 1970s, but the precise frequency was only measured recently, and it might eventually be the basis of a vastly more accurate form of clock:
https://www.nature.com/articles/d41586-024-02865-w

2024-09-05 01:33:45

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“As expected, small asteroid 2024 RW1, discovered on September 4th, 2024 morning, entered the atmosphere at 20.8 km/s above Northern Philippines islands the same day, at 16h 39min UT.
Some witnesses reported their observations, sometimes illustrated with pictures and videos ... ”
https://www.imo.net/2024-rw1-atmospheric-entry-in-pictures-videos/

2024-09-04 12:58:30

“A small asteroid – approximately 1 meter wide – will strike Earth’s atmosphere today (September 4, 2024) over the Philippines around 16:46 UTC according to the European Space Agency.”
https://earthsky.org/space/small-asteroid-hit-earth-philippines-sept-4-5-2024/

2024-09-04 12:45:09

“Sunny” was frustrating at times — and the way the robots work doesn’t stand up to much scrutiny — but if you’re willing to suspend disbelief and go along for the ride, it’s fun.

2024-09-01 04:28:52

“Why A.I. Isn’t Going to Make Art”
by Ted Chiang
“Many novelists have had the experience of being approached by someone convinced that they have a great idea for a novel, which they are willing to share in exchange for a fifty-fifty split of the proceeds. Such a person inadvertently reveals that they think formulating sentences is a nuisance rather than a fundamental part of storytelling in prose.”
...
“The task that generative A.I. has been most successful at is lowering our expectations, both of the things we read and of ourselves when we write anything for others to read. It is a fundamentally dehumanizing technology because it treats us as less than what we are: creators and apprehenders of meaning. It reduces the amount of intention in the world.”
https://www.newyorker.com/culture/the-weekend-essay/why-ai-isnt-going-to-make-art

2024-08-20 11:08:35

“Scavengers Reign” is something else — it’s as if Miyazaki decided to remake Tarkovsky’s “Solaris”. I have absolutely no idea what’s going on, but as a gloriously hallucinatory xenobiological dream, I’m happy to just sit and stare at all the strange creatures chittering around.

2024-06-04 08:45:56

Wow, lucky people in Sydney are seeing a fairly rare kind of soliton cloud known as a “roll cloud”: tubular clouds that appear to be rolling across the sky.
https://www.abc.net.au/news/2024-06-04/nsw-sydney-roll-clouds-explainer/103934424
https://en.wikipedia.org/wiki/Arcus_cloud#Roll_cloud
https://media.mathstodon.xyz/media_attachments/files/112/557/471/530/557/763/original/c9f73f577defe390.png

2024-06-02 01:45:52

China’s Chang’e-6 probe successfully lands on far side of the moon
https://edition.cnn.com/2024/06/01/china/china-change6-moon-landing-intl-hnk-scn/index.html

2024-04-26 11:07:01

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Here’s a construction of a portion of the honeycomb that I think can be extended indefinitely.
The black point at the centre corresponds to the identity matrix I.
The 6 cyan and 6 yellow points are the 12 nearest neighbours of I in 𝔥₂(𝔼):
\[\left(
\begin{array}{cc}
2 & (-\omega )^k \\
(-\omega )^{6-k} & 1 \\
\end{array}
\right)\]
and
\[\left(
\begin{array}{cc}
1 & (-\omega )^k \\
(-\omega )^{6-k} & 2 \\
\end{array}
\right)\]
where k=0,...,5.
Grey lines join all nearest neighbours among these 13 points.
The 6 blue vertices that form a hexagon centred on the black point are all equidistant from the black point, one magenta point and one yellow point. They are:
\[\frac{1}{\sqrt{6}}\left(
\begin{array}{cc}
3 & (1-\omega ) (-\omega )^k \\
(1-\omega ) (-\omega )^{5-k} & 3 \\
\end{array}
\right)\]
We then map that central hexagon to 12 others, using Lorentz transformations that take I to each of its 12 nearest neighbours in 𝔥₂(𝔼), and that take one hexagon edge to the opposite edge.
Using the action:
A ↦ gAg*
the 12 matrices we use for g are:
\[\left(
\begin{array}{cc}
1 & (-\omega )^k \\
0 & 1 \\
\end{array}
\right)\]
and
\[\left(
\begin{array}{cc}
1 & 0 \\
(-\omega )^{6-k} & 1 \\
\end{array}
\right)\]
https://media.mathstodon.xyz/media_attachments/files/112/337/099/950/335/253/original/fa2acb4b4eebbfad.mp4

2024-04-24 10:30:25

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It’s not hard to check that the determinant of my $C(j,n)$ matrices is:
\[1+(j^2+4 j n+4 n^2)(1+\omega+\omega^2) = 1\]
Also, the trace can be written as:
\[\left(j+\frac{3 n}{2}\right)^2+\frac{3 n^2}{4}+2 \gt 0\]

2024-04-24 03:57:28

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Here are the coordinates for all the vertices on one horosphere.
As conventional 4-vectors in the hyperboloid model, (t,x,y,z), define:
\[\begin{array}{l}V(j,k)=\$\frac{j^2+j k+k^2+5}{2 \sqrt{6}},\frac{1}{2} \sqrt{\frac{3}{2}} j,\frac{j+2 k}{2 \sqrt{2}},\frac{j^2+j k+k^2-1}{2 \sqrt{6}})\end{array}\]
For any real values of j and k this 4-vector satisfies:
\[\begin{array}{rcl} g(V(j,k),V(j,k))&=&-1\\
g(V(j,k),(1,0,0,1))&=&-\sqrt{\frac{3}{2}}\\
g(V(j,k),V(j+1,k))&=&-\frac{5}{4}\\ g(V(j,k),V(j,k+1))&=&-\frac{5}{4}\\ g(V(j,k+1),V(j+1,k))&=&-\frac{5}{4}\end{array}\]
So this lets us embed a Euclidean \(A_2$ lattice in the horosphere, and getting the vertices of a hexagonal honeycomb is just a matter of omitting some points.
The equivalent 2×2 Hermitian matrices are:
\[\frac{1}{\sqrt{6}}\left(
\begin{array}{cc}
j^2+j k+k^2+2 & \omega (j+2 k)+2 j+k \\
-\omega (j+2 k)+j-k & 3 \\
\end{array}
\right)\]
where $\omega$ is the cube root of 1 with positive imaginary part.
We omit points where $j-k=3n.$ We get the hexagon centres by taking those points, summing their 6 nearest neighbours and normalising.
\[\begin{array}{l}C(j,n)=\\
(\frac{1}{2} \left(j^2+3 j n+3 n^2+2\right),\frac{j}{2},\frac{1}{2} \sqrt{3}
(j+2 n),\\\frac{1}{2} \left(j^2+3 j n+3 n^2\right))
\end{array}\]
or in matrix form:
\[\left(
\begin{array}{cc}
j^2+3 j n+3 n^2+1 & \omega (j+2 n)+j+n \\
-\omega (j+2 n)-n & 1 \\
\end{array}\right)\]
These hexagon centres lie on a parallel horosphere, with:
\[\begin{array}{rcl}
g(C(j,n),(1,0,0,1))&=&-1\\
g(C(j,n),C(j+1,n))&=&-\frac{3}{2}\\ g(C(j,n),C(j-1,n+1))&=&-\frac{3}{2}\\ g(C(j-1,n+1),C(j+1,n))&=&-\frac{3}{2}\end{array}\]

2024-04-23 12:30:55

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I just want to add something about the horosphere and ideal points. I was a bit confused about that, but now I think I understand what’s going on.
A given horosphere will have a single limit point on the boundary of the Poincaré ball model. In fact, *all* the horospheres that are defined via all the affine planes that share the same normal null ray will have the same limit point in the ball model.
For example, all the horospheres defined by planes normal to (1,0,0,1) will have a unique limit point of (0,0,1) in the ball model.
But this does *not* mean that the defining equation of the horosphere:
g((1,0,0,1),(t,x,y,z)) = α
α < 0
has a unique null vector solution.
That equation is solved by a whole 2-parameter family of null vectors:
(t, x, ±√(–x^2–α(2t+α)), t+α)
But what matters, in terms of the ideal point, is the limit as t→∞ of the projection into the Poincaré ball:
(x, ±√(–x^2–α(2t+α)), t+α)) / (1+t)
which is (0,0,1).

2024-04-23 10:40:04

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As far as I’ve always understood the terminology, when someone talks about the “normal” to a hyperplane, N, that means they’re describing the set of points x such that:
x·N = c
for some constant c.
If the hyperplane passes through the origin, then c=0, but for a general affine hyperplane, c need not be zero.

2024-04-23 10:29:44

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“So what the heck do they mean?”
Strictly speaking those statements in the Wikipedia articles are correct, but they’d save a lot of confusion if they added the word “affine”!

2024-04-23 09:43:59

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I went crazy last night trying to make sense of the claim that a horosphere comes from a vanishing dot product with a null vector … then finally realised that it’s nonsensical.
Suppose the null vector is (1,0,0,1). Its dot product with a generic vector (t,x,y,z) is:
d = –t+z
so the only vectors for which d=0 will take the form:
w = (t,x,y,t)
But the dot product of w with itself is:
x^2+y^2
So w is either null or spacelike! There is no intersection with the hyperboloid, which consists of timelike unit vectors!
[Edited to add: I don’t actually see the claim that the dot product *vanishes* in Wikipedia.
From the article on the horosphere,
https://en.wikipedia.org/wiki/Horosphere#Models
“In the hyperboloid model, a horosphere is represented by a plane whose normal lies in the asymptotic cone.”]

2024-04-23 06:50:42

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These are the 6 vertices whose hexagon centre is the identity matrix, multiplied by √6 to give Eisenstein-integer matrices.
The dot product between a hexagon centre and any of its vertices is –√(3/2).
The dot product between adjacent vertices is –5/4.
The dot product between nearest-neighbour hexagon centres is –3/2.
These particular 6 vertices lie on the two horospheres defined by dot products of –√(3/2) with the two null vectors (1,0,0,±1).
Empirically, all the vertices in the honeycomb seem to be matrices with Eisenstein integer entries divided by √6. The diagonal entries are always multiples of 3/√6.
https://media.mathstodon.xyz/media_attachments/files/112/319/177/099/999/204/original/87115d5d0ee20872.png

2024-04-22 23:17:52

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–3/2 is the dot product between two nearest neighbours among the set of points on the hyperboloid whose 2×2 matrices have Eisenstein integer entries.
So this is consistent with your guess that those points are hexagon centres in the honeycomb.

2024-04-22 23:13:15

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I figured out the dot product between hexagon centres, when all the *vertices* of the hexagons lie on the same horosphere and the dot product between vertices is –5/4 ... and it is indeed –3/2!

2024-04-22 14:44:47

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Right, sorry, I was confused about that. I’ll see if I can figure out the correct distance between the hexagon centres.

2024-04-22 14:00:05

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Because the hexagons all lie in a Euclidean plane, I think the distance between hexagon centres should be √3 times the edge length of the hexagons.
But that makes the *distance* between the hexagon centres:
√3 arccosh(5/4)
and the dot product
–cosh(√3 arccosh(5/4))
which is nothing you can get from Eisenstein integers.
The vertex-vertex dot product of –5/4 looks much more promising! But it’s still not obvious to me how to get the vertex coordinates.

2024-04-22 13:41:30

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In Coxeter’s “Regular Honeycombs in Hyperbolic Space”
on the Wayback Machine at:
https://web.archive.org/web/20160610043106/http://www.mathunion.org/ICM/ICM1954.3/Main/icm1954.3.0155.0169.ocr.pdf
Coxeter gives the edge length of this honeycomb to be 2φ where cosh^2 φ = 9/8, which I believe means the dot product between the Lorentzian unit timelike vectors corresponding to adjacent vertices should be:
–cosh(2 arccosh(3/(2√2)))
= –(2 cosh(arccosh(3/(2√2)))^2–1)
= –(9/4–1)
= –5/4
I’m posting this now in case I don’t get any further, but I ought to be able to figure out some more things that follow from this.

2024-04-22 11:43:43

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I don’t know if this helps, but the identity matrix / aka the vector (1,0,0,0), has the following 12 nearest neighbours among 2×2 hermitian matrices with determinant 1, positive trace, and all entries Eisenstein integers.
https://media.mathstodon.xyz/media_attachments/files/112/314/692/770/242/063/original/36c4ad7c867c7cfa.png

2024-04-15 09:57:23

This article in [@QuantaMagazine](https://mstdn.social/@QuantaMagazine) is astonishing. In some viral infections, many or even most viruses have depleted genomes; they rely on their full-genome cousins to co-infect the same cell to provide the missing proteins they need in order to replicate … then benefit from replicating faster because they’re smaller!
https://www.quantamagazine.org/viruses-finally-reveal-their-complex-social-life-20240411/