Oscar Cunningham

Blog: https://oscarcunningham.com
Pronouns: he/him
0∊ℕ: True
0⁰: 1

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npub10mu87ru8hnpc7rfaequx5r7vtmrr94ujhqv9xsrjd4nxmmg5vufsqp2h9c

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oscarcunningham_at_mathstodon.xyz@momostr.pink

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nprofile1qqs8a7rlp7rmesu0p57uswr2plx9a33j67ftsxzngpex6enda52xwycpz3mhxue69uhhyetvv9ujuerpd46hxtnfduqs6amnwvaz7tmwdaejumr0dszan4k0

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nostr-pub.wellorder.net

relay.damus.io

nos.lol

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npub10mu87ru8hnpc7rfaequx5r7vtmrr94ujhqv9xsrjd4nxmmg5vufsqp2h9c

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Last Notes

2026-10-09 20:07:07 UTC

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If a word problem is decidable, can we always prove it's decidable? Or could there be a monoid where there's an algorithm which seems to work, but we can't prove that it works?

2026-10-04 09:47:56 UTC

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In that case, is there a way to think about Napoleon's theorem using perspective? Like the arbitrary triangle in the middle is actually an equilateral triangle in 3d space that we're seeing from an angle. Hmmm, but then the equilateral triangles around the outside would actually not be equilateral.

2026-09-20 21:26:59 UTC

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Thanks both, very cool! This is exactly the kind of answer I'd hoped for, but I'm afraid I have nothing to contribute.
I did try searching the OEIS for "irreducible representation 3003" and I found that G_2, E_6 and Fi_22 all have irreps of dimension 3003. The last two also have an irrep of dimension 78, but not 14 or 15. It feels like I'm just on the boundary of mathematics and numerology.

2026-09-19 12:37:46 UTC

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You talk about the trick of identifying vectors and pseudovectors. I wonder if there's a similar trick that would identify the spaces corresponding to other repeated numbers in Pascal's triangle. For example the spaces of bivectors in 16 dimensions and trivectors in 10 dimensions both have 120 dimensions.

2026-07-13 14:55:32 UTC

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It's a blog if it has an RSS feed.

2026-05-29 18:36:40 UTC

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I don't understand how PTJ can be their only option. Surely most pure maths post-docs these days know basic category theory? There must be several at Cambridge.

2026-05-16 13:44:32 UTC

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Ah thanks, "branched cover" was the concept I was missing. I see now how each vertex looks like a cone over a pentegram. So it branches in the same way as z ↦ z².

2026-05-16 13:05:59 UTC

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🤔 The pentagram is a double-cover of the circle. But the great icosahedron can't be a multiple-cover of the sphere because the sphere is simply connected.

2026-05-07 12:04:24 UTC

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In the pilot wave interpretation of quantum mechanics the evolution of the configuration depends only on the rate of change of the phase of the wavefunction. So because the ground state of the wavefunction always has constant phase, the configuration will be "frozen". So in particular the QFT vacuum isn't a boiling sea, it's more like a frozen landscape!

2026-04-04 15:32:11 UTC

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I think people in first-world countries should be giving much more money to those who most need it.
But these articles (the ones Oxfam churns out about wealth inequality) are very misleading. Wealth is a slippery concept.
One thing is the way that debt affects the measurement. The very poorest people in the world often have a high quality of life (fucking Alex Jones). You personally have more money than the poorest 30% or the world. If we took your money we could double their wealth (Alex Jones certainly deserves it).
But I think the more fundamental problem is the way that future income is valued. Wealth in the form of stocks is counted based on the market value, which comes from the expected payout those stocks will give in the future. But future income from labour isn't included as 'wealth'. This makes the poor seem comparitively poorer than they really are.
A friend of mine switched from being an employee to a self-employed contractor doing the same job. He pays himself minimum wage from his own company and takes the rest as dividends. Did he really become wealthy overnight through a mere administrative change?
A fairer way to judge wealth is to convert everything to income. Imagine we took all of the wealth of the world's billionaires and combined it into one fund, with the interest split evenly between everyone on Earth. Each person would get about $150 per year. That would be a big help for the world's poorest (poorest in terms of consumption). But it's a much more moderate change than you might imagine from Oxfam's headlines.

2026-03-12 08:43:29 UTC

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if you wanted more accuracy could you use a multiple exposure to get a linear combination of filters?

2026-03-11 08:38:55 UTC

It looks like the ⍼ saga has finally ended!
The unicode character '⍼' was a mystery first noticed in 2022: https://ionathan.ch/2022/04/09/angzarr.html. It's in the Unicode standard as a mathematical symbol. But no one could find any examples of it having been used, and no one knew what it represented. It was like the only record of an extinct species, fossilized in the Unicode standard.
But now someone has found the document it originally came from! It represents the angle 'Azimuth', and the symbol probably comes from the path light takes through a sextant. https://ionathan.ch/2026/02/16/angzarr.html
(@npub1nf4…nqe4 might appreciate this symbol!)

2026-02-28 07:52:25 UTC

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I've heard
a/b ≤ c/d → a/b ≤ (a+c)/(b+d) ≤ c/d
referred to as the 'Batting Average Lemma', which I'm sure makes sense if you're American. Like if you succeed in a out of b attempts and then your performance improves to c out of d, then your overall performance has to be between the two.

2025-12-07 10:02:34 UTC

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Sometimes I like to wonder about how much of the standard model we'd have been able to figure out if humanity were only allowed to do chemistry experiments. This is an analogy to trying to go beyond the standard model without doing further high-energy experiments.

2025-11-27 10:03:49 UTC

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the thing is that they pack them in cubic boxes, so you're mostly paying for the cardboard

2025-11-13 10:36:22 UTC

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For the usual sphere S² is the 'reflection about p' actually a rotation of π about the axis through p? If it was an actual reflection then it would fix some vectors in the tangent space rather than negating them.

2025-11-12 06:16:46 UTC

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There are some good notes about random permutations by @npub1nf4…nqe4 here: https://math.ucr.edu/home/baez/permutations/. Looks like he answers your first question, and probably provides some information that will be useful for your second question.

2025-11-09 07:14:06 UTC

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Sometimes I notice my own posts often disappear from my timeline after a short amount of time. When I first post, the post appears in my own timeline. But if I go back to check later, it's not there. I think this is what you're observing and the boost is just an added layer of complication.

2025-10-20 20:15:19 UTC

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I think the point is that it's a particular copy of U(1) within the gauge group of the standard model. If I remember correctly the weak force is also associated to a U(1) subgroup, but they're different copies.

2025-10-11 13:47:29 UTC

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Something like this happened in real life to Igor Tamm: https://damtson.wordpress.com/2017/08/12/igor-tamm-and-the-taylor-expansion/
> […] The two were soon detained by a Red Army detachment. Neither of them had any documents with them. To the people who arrested them, it appeared obvious that they were White scouts, deserved to be shot.
> Luckily for Tamm, the commander of the detachment was a drop-out student. He grinned grimly while listening to Tamm’s explanation that he had graduated from the Physics and Mathematics Department of the Moscow University.
> “So, you’re a mathematician? You’re lying, aren’t you? No problem, we will give you a test now. Here! Derive for me the formula for the expansion of a function in Taylor series. Including the remainder term! If you can do it, you will be freed. If not – you and your friend will face the firing squad.” […]

2025-10-09 13:04:08 UTC

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I would have expected the smallest eigenvectors to be a combination of positive shells and negative shells with 0 overall mass. Then the field outside them will be very nearly 0 regardless of how the sample points are positioned.

2025-09-26 14:11:02 UTC

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Idea Adventures? Idea Analysis? I just think UOEIA is a funny acronym.

2025-09-02 12:15:41 UTC

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See also: https://oeis.org/wiki/The_empty_sequence

2025-08-28 08:55:00 UTC

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I guess to prove that something isn't Rupert you would have to write some code that takes a proposed rotation and derives a bound that shows that no rotation within some distance of that rotation can work. Then you apply that to enough rotations to produce a cover of the space of rotations.

2025-08-19 14:31:57 UTC

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Actually most of the data is in the black grid! This image shows how it works.
https://media.mathstodon.xyz/media_attachments/files/115/055/911/327/410/749/original/916dec3fc9f936a3.png

2025-08-18 11:18:23 UTC

A working QR code in the style of Piet Mondrian. Inspired @npub1mgl…620u and @npub1p33…l877.
#Art #PietMondrian #QRCode
https://media.mathstodon.xyz/media_attachments/files/115/049/472/461/955/344/original/a8149e2f0aad79f4.png

2025-07-30 10:16:02 UTC

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The paper 'Infinitesimals' (Lightstone, 1972) shows that each hyperreal can be thought of as having a decimal expansion whose digits are indexed by the nonstandard integers. But in that system not every decimal expansion corresponds to a hyperreal number. In particular there's no hyperreal whose digits are 9 for the positive standard integers and 0 elsewhere. The decimal representation of the hyperreal (0, 0.9, 0.99, …) has 9s all the way up to the ∞th decimal place, where ∞ is the nonstandard natural number given by (0, 1, 2, …).
Of course the decimal expansion whose digits are 0 for every nonnegative nonstandard integer and 9 for every positive nonstandard integer does correspond to a hyperreal, namely 1.

2025-07-30 07:05:01 UTC

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Given that hyperreals can be defined as equivalence classes of sequences of reals, you could say that 0.999… was notation for (0, 0.9, 0.99, …), which is indeed a hyperreal number less than 1. But I don't think this is a good idea, because there's not a good way to give every hyperreal a decimal expansion. Like, what would the square root of this number be?

2025-07-01 13:18:56 UTC

Knot theory news:
Mark Brittenham & Susan Hermiller: 'Unknotting number is not additive under connected sum': https://arxiv.org/abs/2506.24088
If you have a knot tied in a loop of string then the 'unknotting number' is the number of times you would need to pass the string through itself in order to untie it. It was an old conjecture that the unknotting number of two knots tied in the same piece of string was just the sum of the unknotting numbers of the two knots individually. But this result shows that it can be easier to undo two knots when they're both tied in the same loop than it would be if they were each tied in separate loops.
#Math #Maths #Mathematics #KnotTheory

2025-06-14 21:44:49 UTC

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(We need 0⁰ = 1, not 0)

2025-06-14 16:21:39 UTC

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Whereas in Analysis they'll tell you that 0⁰ is undefined, and then write nonsense like exp(x) = ∑xⁿ/n!.

2025-06-14 14:21:23 UTC

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Ugh, yeah. My least favourite version is when they give the definition for 0 and say that it's 'by convention' even though they've literally just given a uniform definition that works fine for 0.

2024-12-27 08:07:48 UTC

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I have no relevant expertise, but I think it's what you said in your first post: 'we're assuming the plasma conducts electricity so well that the electric field is zero.'. The location of the electrons is fully determined by this assumption. So we can just look at the motion of the ions, and let the electrons sort themselves out by moving to cancel any electric field.

2024-11-29 19:58:55 UTC

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Right, there's no frame-independent concept of distance, but we could still talk about the angle they subtend in Bob's vision.

2024-11-28 18:03:42 UTC

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We see the image flattened on the horizon from a distance, but we're falling towards it. So eventually it's going to stop being distant. What happens to the image as we fall towards it? We can't see it cross the horizon, because any light starting on the horizon would never reach us. But we also don't crash into the image as we cross the horizon, because what we see has to make sense locally. So where does it go?

2024-11-25 20:18:22 UTC

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So when the captain is 1m in front of the horizon do they see the 49 ships all compressed into that 1m? The ships look like they've been flattened lengthwise?

2024-11-14 18:46:16 UTC

'time is a flat circle'? what are you talking about, all circles are flat
#DifferentialGeometry

2024-11-13 08:30:44 UTC

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Whatever else you can say about Musk, he's really good at rapidly implementing his decisions. I think he's going to hate being in charge of a department that has no powers and is merely advisory. Especially when Trump simply ignores his recommendations. It's also funny that the Department of Government Efficiency will have two heads. Doesn't seem very efficient. And it will likely lead to more headaches for Musk as he tries to get anything done. I suspect he'll quit or be fired within a year.

2024-11-09 18:40:41 UTC

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Oh no, the Ceres black hole will only last 10^39 years!

2024-11-09 18:29:03 UTC

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Ah, perfect. That makes a lot of sense.

2024-11-09 18:10:31 UTC

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I tried the black hole temperature formula on the mass of Ceres, and got a temperature of about 100K. That's much hotter than the CMB, so how can asteroid-sized black holes exist without evaporating?

2024-09-26 05:47:29 UTC

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Thanks, that's a good collection of resources. I was thinking particularly of Savage's 'The Foundations of Statistics'.

2024-09-25 19:36:33 UTC

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I think ignorance is enough, without needing to bring in physics. There are various arguments showing that a rational agent must behave as though they assigned probabilities to the various possible states of the world, according to their own state of ignorance. See https://en.wikipedia.org/wiki/Dutch_book_theorems.

2024-09-23 02:35:09 UTC

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Yeah, that would work. The only improvement I can think of without making it too complex would be to assign the variables one at a time and check each condition as soon as possible.
Like instead of
for a in (0,1):
for b in (0,1):
for c in (0,1):
if (a or b) and (b or not c):
return (a,b,c)
you could do
for a in (0,1):
for b in (0,1):
if (a or b):
for c in (0,1):
if (b or not c):
return (a,b,c)

2024-09-23 02:25:44 UTC

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If there's only 10 literals it might be quicker to solve them yourself using brute force.

2024-09-16 17:27:32 UTC

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I think that's the polyhedron and its 38 closest friends.

2024-09-15 08:53:07 UTC

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Ah good. I suspected 109 was going to show up somewhere.
I have heard that the third-parties Amazon uses for on-demand printing can be poor quality. At least the print quality looks good on yours.

2024-09-15 08:16:43 UTC

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This is a really weird error. If a page was missing you would expect two numbers to be skipped since it has a front and a back. Did they print 110 twice? Or does the book just start with even numbers on the left but end with even numbers on the right?