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I'm typing this after reading their section on page 6 about the "pentagon axiom". They write that RxR^2, R^2xR, and R^3 are isomorphic, but not literally equal
by ajkjk 2y ago
I'm typing this after reading their section on page 6 about the "pentagon axiom". They write that RxR^2, R^2xR, and R^3 are isomorphic, but not literally equal under different constructions, because (a,(b,c)) is different from ((a,b), c), and that this is a problem. I feel like math kinda gets this wrong by treating sets as meaningful objects on their own. Physics gets it closer to correct because it tries to be "covariant", where the meaningful answers to questions have to either have units on them or be totally unitless e.g. irrespective of coordinate system.
The two spaces Rx(RxR) and (RxR)xR have many isomorphisms between them (every possible coordinate change, for instance). But what matters, what makes them "canonically isomorphic", is not isomorphisms in how they are _constructed_ but in how they are _used_. When you write an element as (a,b,c), what you mean is that when you are asked for the values of three possible projections you will answer with the values a, b, and c. Regardless of how you define your product spaces, the product of (a), (b), and (c) are going to produce three projections that give the same answers when they are used (if not, you built them wrong). Hence they are indistinguishable, hence canonically isomorphic.
This is exactly the way that physics always treats coordinates: sure, you can write down a function like V(x), but it's really a function from "points" in x to "points" in V, which happens to be temporarily written in terms of a coordinate system on x and a coordinate system on V. We just write it as V(x) because we're usually going to use it that way later. Any unitless predictions you get to any actual question are necessarily unchanged by those choices of coordinate systems (whereas if they have units then they are measured in one of the coordinate systems).
So I would say that (a,(b,c)) and ((a,b), c) are just two different coordinate systems for R^3. But necessarily any math you do with R^3 can't depend on the choice of coordinates. There is probably a way to write that by putting something like "units" on every term and then expecting the results of any calculation to be unitless.
- generationP 2y agoThis is exactly the "specification by universal property" that the author is talking about. In your case, it's saying "a 3-dimensional vector space is a vector space with three chosen vectors e, f, g and three linear maps x, y, z such that each vector v equals x(v) e + y(v) f + z(v) g". But as the author points out, not everything follows from universal properties, and when it does, there is often several universal properties defining the same object, and that sameness itself needs to be shown.
- ajkjk 2y agoYes, I know it's the meaning of it, but I'm saying that the presence of "units" allows you to sort of... operationalize it? In a way that removes the ambiguity about what's going on. Or like, in theory. I dunno it's a comment on the internet lol.
- eru 2y agoUnits help with some common cases, but units still don't allow you to distinguish between, say, energy, work and torque.
- ajkjk 2y agoI think that is due to insufficient imaginativeness. For example, energy and work are the same units but energy is an absolute quantity while work is a delta. So arguably work should be a sort of tangent element of the energy, rather than the same thing. There's no distinction in flat space but if you e.g. changed coordinates such that energy was on the surface of a sphere, then work is a spherical displacement instead, which is a totally different class of objects. Likewise, torque is only in the same units because we don't regard radians as a unit, but we should. They are distinctly different.
- eru 2y agoRadians can be seen as a ratio of two lengths, the length of your arc and the length of the circumference of the unit circle. All units cancel in such ratios. > There's no distinction in flat space but if you e.g. changed units such that energy was on the surface of a sphere, then work is a spherical displacement instead, which is a totally different class of objects. Well, maybe. But in other circumstances you want to treat eg heat and work interchangeably. Just look at https://en.wikipedia.org/wiki/Work_(thermodynamics) https://en.wikipedia.org/wiki/Work_(thermodynamics) and https://en.wikipedia.org/wiki/Work_(physics) https://en.wikipedia.org/wiki/Work_(physics) and https://en.wikipedia.org/wiki/Work_(electric_field) https://en.wikipedia.org/wiki/Work_(electric_field) Basically, how much you _want_ to encode in your type system depends on the needs of your application. (Approximately all type systems can be made to work for all applications. But they differ in the degree of convenience and error proneness.)
- yau8edq12i 2y agoI think it's the first time that I witness this comic https://xkcd.com/793/ https://xkcd.com/793/ happening but with math as a target.
- mathgradthrow 2y agoYou use R^3 in your example of why coordinates don't matter. R^3 can be covered by one chart. Maybe your argument would be more convincing if you pick a different manifold. I have no idea what your complaint is otherwise.
- ajkjk 2y agoI'm not talking about charts of R^3; I'm talking about the different isomorphic constructions of products like ((a, b), c) and (a, (b, c)) as being a sort of 'choice of coordinate system' on the isomorphic class of objects.
- mathgradthrow 2y agoyes, and these choices dont matter individually, but how these choices glue together does, in fact, depend on all of them collectively.