What is a physical transformation that cannot be modeled as an element of a group

@SillyGoose why should there be such a thing?

anything that deserves the name "transformation" should be a) composable with other transformations and b) reversible. That's essentially the axioms for a group

I mean you can drop reversibility I guess

A retract is still a transformation kinda

the only case where that happens off the top of my head is the renormalization semi-group

Or a projection

while these are "mathematical transformations", I would not consider them what we usually call "transformation" in physics

@SillyGoose All elements of a group need to have an inverse. Thus, all irreversible physical transformations qualify, like anything increasing entropy in a closed system.

Althought you can say from all physical transformations, it is element of the group { transformation } and the group operation is the identity mapping

Plenty of transformations that aren't reversible then

On a vector space it's the linear maps as opposed to invertible linear maps for instance

You're not requiring the determinant to be 0

@ACuriousMind First :) :) :) I can not really remember a not bad or evil decision from their side since 2012

@peterh If you'd read the whole conversation I was replying to, you'd understand that I meant this is the first of several bad decisions I see coming. It is not a statement about any past events.

is it the robot apocalypse

(no, I will not elaborate what the others are because there's a chance they don't come to pass)

@ACuriousMind O.k.!

@ACuriousMind : Rugby!

Honk

I guess i was thinking about why groups are so effective as structures modeling things

Like for rotations sure they happen to follow what we call the group axioms

I mean groups were made to model such things

But even closed under group operation is kind of a strong requirement but maybe not

IIRC Sophus Lie came up with Lie groups as relating to symmetry groups of differential equations

Did he call them Lie groups back then

@SillyGoose The group axioms are really just: There are some operations that a) can be composed (the group multiplication) b) can be reversed (existence of inverse) c) when you compose more than two of them, it doesn't matter which composition you consider first (associativity)

Would have been a bit pretentious

I still don’t quite understand the associativity clearly. In the context of rotations acting on a regular n-gon it certainly makes sense. And so one can abstract how the rotations compose to retain the same associative structure without considering the object. But is there a different way of seeing it?

Specifically for rotations?

For an abstract arbitrary group

Ah, then the reasoning will also be abstract :) I mean, I don't think there's any intuition behind associativity when you don't have a concrete group at hand

I mean if you're interested in algebraic structures in general

Remember the diagram

Also is it accurate to think that: let there be a group. Endow group with a “scalar product” taking values from some ring. An algebra is born.

If you want to consider more general structures just look them up

Endow an abelian group with a scalar product and a module is born

Well not abt general structures but just about what the associative axiom has to do with an arbitrary group

Algebra or just a vector space?

Algebra since the group can be nonabelian in general

(Or so i think)

Oh, right

So really modules are particular cases of algebras?

Yes, groups can definitely be nonabelian, like the rotations

@Amit hm i see

I never realized until a discussion a few days ago that commutators in any context probably originate from some group structure :P

One fine day, I'll understand Lie algebras, lol

Is there a lucid explanation of the Heisenberg group’s importance? I briefly looked through Woit’s section in his book but I didn’t really see much physical exposition but I might have missed it

@Amit me too :)

@SillyGoose The Heisenberg group is the group corresponding to the 1d CCR (in the sense of the Lie group-Lie algebra correspondence)

The words SvN theorem and irreps are states and CCR are thrown around but i haven’t founda description clearly connecting it all

The Heisenberg group is related to the Poisson algebra IIRC

Hm so i probably missed this by either not having learned mechanics yet or elsewise but what is the importance of the CCR

@SillyGoose You know quantum mechanics, don't you? :P

The CCR are just $[x,p] = \mathrm{i}$

I suppose xD but seldom have i used the CCR

At least i think 0.0

I mean you use them every time you consider some operation in QM that's a function of them and compute a commutator of that operator e.g. with the Hamiltonian

they're essentially the only thing you need for spinless QM

I suppose i have been stuck in the land of qubit systems :P

well, sure, for qubits they're entirely irrelevant

But okay i see

Hm so is the heisenberg group supposed to contain like every non-spin related unitary transformation there can be in quantum?

no

just like the algebra of $\{x,p,1\}$ doesn't contain every observable

but the representation of that algebra (and hence of the group) induces a representation of the algebra generated by it - i.e. if you know how $x$ and $p$ act, you know how arbitrary polynomials in them and even arbitrary functions $f(x),g(p)$ act

and that's what we want - on the classical phase space, observables are just functions of $x$ and $p$, so in spinless QM we also don't expect any relevant observables that are not functions of $x$ and $p$

Okay so it is onlh every unitary transformation generated by x and p

I was trying to fit it into the generators of transformations framework and this makes sense then :D

Okay so then the irreps as in other cases do probably correspond to the states wrt to some eigenvalue of some observable?

what irreps?

of the Heisenberg group?

Yes

Although it seems this situation is more complicated than SU(2) 0:

that's the neat thing: there's only one unitary irrep: The one physicists know and love, where the Hilbert space is $L^2(\mathbb{R})$, $x$ acts as multiplication and $p$ as differentiation

Although i guess the Heisenberg group has a not complicated representation as matrices

Oh

Wait the whole Heisenberg group has a single irrep?

yes

it's neat

Hm so if you would like to represent the heisenberg group on L^2(R), you would need to direct sum uncountably infinite many of the irrep?

No, $L^2(\mathbb{R})$ is the irrep

the group is non-compact (like the Lorentz group), so its unitary irreps are necessarily infinite-dimensional

Oh

So what leads to having to do functional analysis over infinite dimensional linear algebra

functional analysis is infinite-dimensional linear algebra

Oh i forgot that L^2(R) is still a vector space .-.

Or is that true

sure, it's a vector space

you can add functions

you can multiply them by numbers

so they're vectors

Hm

What to call the transport operator

is it $P$ or is it $\mathrm{tra}$

$\Gamma$ is also a common notation

Also $\tau$ apparently

Will probably go with tra since it is less likely to be ambiguous

Also IIRC Visser used $g$, which was a weird choice

youtube.com/shorts/qrWcjTSV6HA

VSauce just made a claim that the visible part of a flame is solid???

touch it to check

@Slereah what if it's just really not dense

also i think someone did put something to the flame and soot condensed on the surface

@PM2Ring : You're right. I just saw it on Phys.Meta.

reddit's killing third party apps, SE's making more awful decisions, youtube's banning ad blockers

feels like all the websites I use are going to shit

Ch ch ch ch ch changes (I hope your ad blockers work, lol)

@LeakyNun what does VSauce actually say? Why are hydrogen flames always nearly invisible but flames of carbon-rich materials can have bright yellow areas? Incandescence of the solid particles due to slow combustion. That's why a rocket engine burning kerosene and oxygen (from two tanks) is super bright on the ground and almost invisible in space - no residual oxygen for the soot particles to continue to burn and incandesce.

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