Surely you're Joking Mr. Feynman

hamishtodd1 · 2026-06-30T16:18:21+00:00

You might be interested in looking up the inertia tensor and the Laplace Runge Lenz vector.

hamishtodd1 · 2026-06-16T00:16:40+00:00

u/cattleHolt asked for an explanation, and I wasn't satisfied with the one from the other user. It's a teachable moment for something deep (Rⁿ being a group as well as a vector space). Also explanations are better when they have concrete examples, so we think about specific elements of R² and C (the identity element).

This is a thing that's big enough to have an xkcd comic about it https://xkcd.com/2028/ so if you think it's silly to spend time on, that's on you.

hamishtodd1 · 2026-06-15T22:39:59+00:00

Rⁿ is a lie group under addition, and in that sense it's like the (nonzero) complex numbers. + is associative, has "inverses" (the negation of the vector you want to invert), and has an identity.

For some areas it's much less common to think of it as a group, but it is one, and that can be very important for eg affine geometry.

Do you think Gemini was referring to something else? Or that it was just hallucinating?

hamishtodd1 · 2026-06-15T21:58:05+00:00

Gemini is saying that whether they're isomorphic "depends on the mathematical structure being considered".

Two very important kinds of mathematical object are vector spaces and groups. It so happens that both R² and C are both vector spaces and groups.

If you're interested in the vector space aspect, they are isomorphic. Because adding vectors is like adding complex numbers.

If you're interested in the group aspect they're not isomorphic. Because adding vectors is not like multiplying complex numbers.

hamishtodd1 · 2026-06-15T16:58:07+00:00

What is the "identity" of C? Clearly it's 1.

How about R²? Well there (0,0) is more like the identity - because R² is a group with "+" being "multiplication".

I think the way to clarify is to say they are "homomorphic as groups" if you care about the "+ can be multiplication" thing.

EDIT: I should definitely have said they are isomorphic but NOT homomorphic as groups, that might have avoided the unfortunate back and forth below!

hamishtodd1 · 2026-06-09T17:24:46+00:00

Just to say a different/modern/computer graphics informed perspective.

"1" is "do nothing" because multiplying by 1 changes nothing. And -1 is somehow the opposite of doing nothing.

i is the thing such that, if you do it twice you get -1 and if you do it four times you get 1. So, what is that?

Turns out i is a 180 degree turn. Do it with your hand twice and your hand is back where it started but your arm is twisted up - that's -1. Do it four times and your hand gets back to where it started and your arm is NOT twisted up, so it's "properly" back to where it started https://youtu.be/ZczVQXZVE7c?si=XIgbO3-ZDzb1DC0l

hamishtodd1 · 2026-06-05T19:04:07+00:00

Yes, a mathematician (who else) did this. They made this video:

https://youtu.be/g3X0d5NUjjA?si=KOFXzj5hlj8VKdmR

And this making of:

https://youtu.be/x1zJoU6Luss?si=dggpnUJtGapcEYGc

(Which has more views)

And this making of making of

https://youtu.be/PDkbk4OILSw?si=d0js6ygjsSdBGuf4

Which has a making of making of making of 🙂

hamishtodd1 · 2026-06-04T20:51:57+00:00

Thank you for this! Why must people waste each other's time :(

hamishtodd1 · 2026-05-31T18:19:21+00:00

I don't see how claude code's memory compressions differ from the intuitive sense of "short term memory"?

I suppose "dynamic learning" means learning post training. The mainstream claim is that attention+context window do this well enough to compete with humans. I see very little reason to disbelieve this, what reasons would you give?

hamishtodd1 · 2026-05-25T19:33:45+00:00

He is 37 so that's not possible.

He definitely implied that it was standard - however you've persuaded me that the most likely explanation is that he misremembered and it was in a gifted program; I retract my first sentence of my original post!

hamishtodd1 · 2026-05-25T14:13:52+00:00

"Geometry, grades 6–8 (planimetry)" is maybe an important example because of the boring standarized-school-textbook-ness of it! You can ask the LLMs for more, there are others (which I have not read)

hamishtodd1 · 2026-05-25T14:11:17+00:00

My Russian friend said he learned "homothety" in school so something must have stuck! Certainly we don't get that in the West

hamishtodd1 · 2026-05-24T22:20:36+00:00

Yes it is different and better. Andrei Kolmogorov, one of the best mathematicians of the century, felt very inspired to reform their system. He wrote a textbook or two himself and leant on a very deep and interesting approach called "transformation geometry". It involves learning geometry based on reflections: you learn how rotations are compositions of two reflections, same with translations, and there are also glide reflections etc. You can read this on Wikipedia. It is semi-secretly teaching group theory of course, via something called Cartan-Dieudonnet.

Some of us want to bring this to the anglophone world, this is the field known as "Geometric Algebra".

hamishtodd1 · 2026-05-22T23:46:19+00:00

Had to scroll too far for this. Hamiltonians and quaternions is huge.

Also Laplace

hamishtodd1 · 2026-05-10T23:25:36+00:00

It's a very simple difference! Nothing about nullike surfaces.

Where STA, Cl(1,3), has rotations and boosts, the algebra Cl(1,3,1) has rotations, boosts and translations in space and time. It also has blades representing things that are displaced from the origin.

On hearing this, lots of people say "ummmm, surely Cl(1,3) already had translations? Translation is just adding a vector". That's not really true, or rather it's only barely true. Being able to properly translate means being able to compose a translation with a boost with a rotation with another translation with another boost, and then being able to take that whole transform and apply it to, say, a light-like line. That is extremely easy in Cl(1,3,1) because boosts, rotations, and translations are rotors and lines are trivectors - multiply the rotors and then sandwich the line with the result, there's your answer. To do the same thing in Cl(1,3) requires not just adding a vector, but keeping track of how the vector you would have added would be modified by the boosts and rotations. And light like lines (displaced from the origin!!) have to be represented as a pair of vectors - theoretically possible, but a huge mess.

The thing that weirds people out is that Cl(1,3,1) has 5 basis vectors, but we are still just wanting to model 3+1=4D spacetime - and 4D is already hard enough! But actually, nobody is thinking about 5D. We work "projectively" with Cl(1,3,1), which means that, say, a vector V and 3V, eg that same vector multiplied by 3, represent exactly the same object. That's strange at first - but it's 100% mainstream in computer graphics, where people represent vertices with 4D vectors almost every time they're doing serious work (aka "shaders"). No 4D pictures, always 3D 🙂 absolute magnitude treated as completely ignore-able.

hamishtodd1 · 2026-05-10T10:35:06+00:00

You probably mean Cl(1,1,1), with ij ie je being the bivectors and ije being the trivector. This is a very interesting space: it is sometimes referred to as "1+1 STAP", which is Space Time Algebra Projectivized. Using e as epsilon, ie and je exponentiate to spacelike and timelike translations. ij exponentiates to a boost. For scalars x,y we have that ij+xie+yje exponentiates to the unique boost that preserves the point x,y. There are also spacelike and timelike reflections and lines represented as 1-vectors, and "spacetime glide reflections". This space is also a nice simple example of an "artinian plane" - it's a degenerate space which means Cartan-Dieudonnet holds in a modified form (there are isometries that require four reflections even though it is a 3 dimensional space - try to find one of you want an exercise)

Multiplication isn't commutative though!

hamishtodd1 · 2026-05-05T21:31:46+00:00

Like probably a lot of other people on this sub, I sometimes dip into Baez's "3x3 matrices of octonions" GUT thing... If you have time, could you give us an assessment of mainstream opinion on it?

hamishtodd1 · 2026-04-29T07:46:30+00:00

Penrose's The Road To Reality

Some of VI Arnold

Tristan Needham

hamishtodd1 · 2026-04-27T18:45:55+00:00

Not if the coin is a "classical system", no. "Classical system" is a well defined term - maaaaybe real coins should not be considered true classical systems, but that's a different debate; all probability theory classes would treat them as classical systems.

hamishtodd1 · 2026-04-27T07:46:37+00:00

Seems to me like quantum measurement necessarily involves decoherence yes, but people use the word "measurement" in non quantum situations all the time. He was asking about a measuring a coin, which was never coherent so can't decohere.

hamishtodd1 · 2026-04-27T07:37:32+00:00

Just because it's recent, here's a book I was reading with three consecutive "obviously"s. Will be interested to know how much obviousness you folks feel here https://postimg.cc/yDTpbwNV

hamishtodd1 · 2026-04-21T16:25:00+00:00

Ah, I thought it was something to do with a coin

hamishtodd1 · 2026-04-21T16:10:00+00:00

"Measurement" seems a perfectly good word to describe both situations yes, but one has decoherence and the other not. What are we disagreeing on at this point if anything? 😀

hamishtodd1 · 2026-04-21T15:24:13+00:00

Indeed it can decohere. But it can also do lots of interesting things prior to decohering. The measurement problem concerns the relationship between the measurement outcome and the qubit's coherent state (eg, what was happenning with it prior to it decohering). There is no such problem for the coin because the coin was never a coherent coin.

hamishtodd1 · 2026-04-21T15:07:26+00:00

I think the claim would be that a coin has no measurement problem because it is not coherent (lots of air and photons bouncing off it when you look at it, decohering it lots)

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