The h Bar

Silly Goose

00:00

what is this cover

moss on a tree?????????

ACuriousMind

I would assume it's something like an electron microscope picture, but I can't find any information about that specific image, either

qwerty

00:17

it's weird I ran a reverse image search

this looks like the flipped image to me

carinemasutti.bandcamp.com/album/learn

but no further info either

the artist gives herself as design credits idk if that refers to the image

Silly Goose

00:35

interesting...

also would it be accurate to call a free fermion hamiltonian "gapless"?

I have some $H = \int dq \epsilon(q) c^\dagger(q) c(q)$ (in a CMT, not QFT, context)

and I am looking at $\epsilon(q) \sim q$ or $\epsilon(q) \sim q^2 + a$

er actually so I think with quadratic dispersion, there should be a gap in the ground and first excited state

imbAF

01:09

If H is a field operator then is this true : $\langle H*\rangle=\langle H \rangle*$ ?

SnoopyKid

01:39

Any thoughts on this? Hello, which one can obliterate bones to very fine ashes or powder (not vaporized into gaseous state as in nuclear explosions) that are totally unrecognizable as human bones in less than 10 seconds? The temperature of cremators or conventional explosives such as TNT or dynamites?

PM 2Ring

04:07

@SnoopyKid Didn't you ask this yesterday? But anyway, I suspect that you need more than 10 seconds, &/or much higher temperature. I expect it's pretty easy to identify human crematorium ashes, and cremation certainly takes a lot longer than 10 seconds.

As for TNT or dynamite, I doubt you'd get a better result. Especially if you just drop the explosive onto a pile of bones. But maybe if you do it inside some strong container, with a lot of explosive...

OTOH, these explosives don't generate a huge temperature. They produce a lot of expanding gas in a very fast reaction. If you want high temperature, try thermite reactions.

Allie

goldstein is dense

the book, not the guy

Tobias Fünke

07:32

morning :)

Tobias Fünke

07:49

@imbAF what does your star denote?

@SillyGoose some turbulent flow or so?

SnoopyKid

08:17

@PM2Ring yes I did ask yesterday. Thanks for the response. I thought the energy of, perhaps, around 20k tonnes of TNT alone is enough to turn bones into unrecognizable powdery remains because first it will obliterate bones into very small pieces then burn them to very fine ashes. Is this make sense

Loong

Just use hydrofluoric acid.

SnoopyKid

@PM2Ring yes I just find out crematoriums can't actually burn whole bones into fine ashes. It just burns the organic parts of bones so whats left is just the hydroxyapatite, and the remains of cremator are apparently still very recognizable as human bones. They turn the calcified bones into ashes by crushing them in a machine.

@Loong I heard that some people accidentally fall into acid lakes and everything including bones literally become one with the acid

Loong

08:39

Fortunately, there are no hydrofluoric acid lakes.

But there are pickling baths.

John Rennie

You don't HF to dissolve hydroxyapatite. HCl will do.

SnoopyKid

@JohnRennie hydroclorine? Is that the full term

So, can a conventional explosive like 20,000 tonnes of TNT turn bones into unrecognizable very fine ashes using its tremendous energy alone

John Rennie

@SnoopyKid HCl is hydrochloric acid

Signor Feynman

09:25

And remember that $\mathrm{HF}$ can melt ceramic, if this is some kind of real life Breaking Bad

@SillyGoose you usually call the spectrum "gapped/gapless"

Loong

@SignorFeynman I once accidentally used a glass measuring pipette for hydrofluoric acid because I was focused on something else in the workflow. Of course, I immediately noticed that something was wrong and pulled the pipette out again; but then it was already turning white.

Signor Feynman

A typical example of gapped spectrum is the Bogoliubov spectrum $E_k=\sqrt{k^2+|\Delta|^2$

@Loong damn, is it a fast process? Does is take much to pierce through the glass?

Loong

It is fast. But I pulled the pipette out again right away, so the acid ran out and the reaction didn't go any further; so it didn't go through the glass. However, it was already white then.

I can also still see a small scar where a drop has passed through two gloves.

qwerty

@Loong is the correct tool to use... plastic? or something else?

Loong

yes, plastic

Tobias Fünke

09:40

@SillyGoose gapless usually means that in the thermodynamic limit there exists no finite difference from the ground to the first excited state

but it can be more complicated, actually, when the ground state is degenerate. so it depends on your precise definition.

see e.g. this: nature.com/articles/nature16059

72

Q: What does it mean for a Hamiltonian or system to be gapped or gapless?

I've read some papers recently that talk about gapped Hamiltonians or gapless systems, but what does it mean? Edit: Is an XX spin chain in a magnetic field gapped? Why or why not?

Signor Feynman

@TobiasFünke I'm not sure that the thermodynamic limit is relevant here. Even the relativistic particle spectrum is gapped. So long as arbitrarily small energies are not enough to excite the system, you get a gap

Probably I sound more assertive than I should. I'm speculating about my understanding

Tobias Fünke

09:58

@SignorFeynman it is relevant

see e.g. the link to the PSE thread

where an expert explicitly states that, too.

I mean OK, you can define a term like you want, but to my knowledge this really is the usual understanding (in condensed matter)

Signor Feynman

Oh, okay. I'll check that. Sorry for being rash

Tobias Fünke

ah no problem.

again, it might depend on the context and or definitions. but when I encountered the term, it always referred to the thermodynamic limit ^^

Signor Feynman

I tend to associate the thermodynamic limit to $N\to\infty$ along with volume. In the case of that answer, he's only considering an infinite volume (as I understand) to remove the discreetness of the spectrum, isn't it?

The accepted answer, on the other hand, is more like the usual physics argument I've often heard. I wonder if there is any loophole

Tobias Fünke

well, I guess they are talking about lattice/spin models, no?

where the number of particles is just the number of lattice sites (?)

give me a second

imbAF

How does spontaneous symmetry breaking occurs ?

Signor Feynman

10:19

@imbAF for a finite system it doesn't because the real ground state is a superposition of all the candidate ground state configurations corresponding to the minima; when the system is infinite dimensional, this is no longer possible.

It turns out that the tunneling between such minima is zero and so each of them is a legit ground state. When the system is in one of such minima, since it's infinite, it takes infinite energy to e.g. flip all the spins, that's why there is no tunneling. Effectively the system "choose" one minimum

imbAF

what would the system be? What would the ground state be? I though when SSB is not taking place, the ground state, which I have read, that it's the vacuum state is not degenerate.

But hold on

if SSB occurs in the cases of infinite dimensional systems, then how is that possible in reality? In reality every system, is finite, and in reality SSB apparently takes place

so that would imply that in reality infinite dimensional system has to be a thing, since SSB is a real thing

But infinite dimensional system can't be a real thing, while SSB is one

Tobias Fünke

same for phase transitions

a good argument for: model is not reality, but still can work

naturallyInconsistent

@Allie that's part of why it is recommended. It has all the main stuff crammed into a short text; and classical mechanics is not what people really want to do. Most physicists want a short text because they want to quickly get students to get up to speed on the quantum stuff that they want them to work on.

Signor Feynman

10:46

@imbAF well, the reason why those tunneling amplitudes - which are the reason of SSB - go to zero is that the tunneling $\propto\exp{-\mathrm{cost}\times V}$, so even for a real system that is not infinite, a macroscopic volume is enough to make this amplitude neglibibly small

PM 2Ring

11:49

@Loong Hydrofluoric acid is scary stuff. From geo.utexas.edu/geosafety/hf-accident.html

A 37-year-old male laboratory technician was performing acid digestion of oil well core and ditch samples with 70% w/w concentrated hydrofluoric acid in a fume cupboard. He was believed to be seated when he knocked over a small quantity (100-230 ml) of hydrofluoric acid onto his lap, splashing both thighs. The only personal protective equipment (PPE) worn was two pairs of wrist length rubber gloves and a pair of polyvinyl chloride (PVC) sleeve protectors. As a result of the fact that the technician was working alone, it is unclear whether the spill was from the digestion cup or the 2-l bulk…

This accident occured in Western Australia in 1995. I knew a chemist who was a friend of the family. It was a devastating tragedy.

Loong

I know a laboratory technician who opened a microwave digestion too early, before it had cooled down properly. She was just able to turn away; but with the marks on her back, she didn't wear bikinis after that.

PM 2Ring

For those who arent chemists: Hydrofluoric acid is actually a milder acid than hydrochloric, and in small quantities a HF burn doesn't seem that bad. But it has a strong affinity for calcium, which it pulls out of your bones and nerves.

Hydrofluoric acid is used to make uranium hexafluoride, which is used in the standard techniques for enriching uranium. This is an acidic gas which decomposes, releasing HF and fluorine, when it reacts with moisture. So you need to be very careful when enriching uranium... Amazingly there were no accidents with it during the Manhattan project, AFAIK.

Loong

12:07

@PM2Ring There was one during the Manhattan Project with maybe three dead or so.

PM 2Ring

Ah, ok.

Loong

S-50 (Manhattan Project)

The S-50 Project was the Manhattan Project's effort to produce enriched uranium by liquid thermal diffusion during World War II. It was one of three technologies for uranium enrichment pursued by the Manhattan Project. The liquid thermal diffusion process was not one of the enrichment technologies initially selected for use in the Manhattan Project, and was developed independently by Philip H. Abelson and other scientists at the United States Naval Research Laboratory. This was primarily due to doubts about the process's technical feasibility, but inter-service rivalry between the United States...

PM 2Ring

Ftom nationalww2museum.org/war/articles/…

The first major accident came in September of 1944 at the Philadelphia Navy Yard, where three men were attempting to fix a clogged tube containing liquid uranium hexafluoride. Unexpectedly, the tube exploded, causing a huge blast in the liquid thermal diffusion transfer room, killing two of the three men and sending uranium hexafluoride into the air.

Loong

Yes, that's the one.

PM 2Ring

From the Wiki link:

> An investigation found that the accident was caused by the use of steel cylinders with nickel linings instead of seamless nickel cylinders because the army had pre-empted nickel production.

Military intelligence...

I suspect there may have been some incidents with UF6 in Iran...

SnoopyKid

12:57

@PM2Ring so a 20k tonnes of TNT can turn bones into unrecognizable fine powders using its energy to obliterate each bones into tiny bits then burn them into fine powders?

Loong

Why are you so fixed on 20 000 tons of TNT? There has never been a coordinated explosion of so much conventional explosives.

Signor Feynman

13:17

So, either this is some vswiki thing (ewww) or something more grim and unsettling is looming

Or good old trolling?

Silly Goose

14:18

@SignorFeynman @TobiasFünke do you guys know any resources that touch on the gap in 2D ising model? (The “hamiltonian” is gapped outside the critical T and gapless at the critical T)

Signor Feynman

@SillyGoose I think Sachdev somewhere discusses (it's a QPT)

Probably also Altland and Simons that you checked (?)

You should check statistical field theory books in general

Tobias Fünke

14:37

nope sorry

imbAF

15:00

what is the physical interpretation of the VEV of a field operator?

Silly Goose

I am trying to understand how (1) gap $\implies$ exponentially decaying ground state correlation functions and (2) 2D ising below the critical temperature has a gap and does not have exponentially decaying ground state correlation functions.

okay i will check out sachdev

Signor Feynman

15:30

@SillyGoose I have a handwavy idea for (1)

Silly Goose

oh wait sorry i mean to take (1) as a fact

and so that (2) is seemingly contradictory given (1)

or actually

Signor Feynman

What are you calling (1)?

I'm talking about the exponential decay

User198

1. What is the Lie group of rotations in 3D space called? SO(3) ?
2. What is the manifold that is invariant under rotations?
3. What does this mean:

"Recall the matrix differential equation: $\frac {dA} {dt} = \Omega \cdot A $. This equation can be integrated to give: $A(t) = e^{Wt}A(0) $ which shows a connection with the Lie group of rotations. Where $\Omega$ is the angular velocity tensor.

Silly Goose

(1) is that "a gap in the spectrum implies exponential decay of ground state correlation functions"

Signor Feynman

Okay, so you want to take that as a given, I see

Silly Goose

15:33

yes

ACuriousMind

@User198 1. Yes. 2. What? 3. What's unclear?

User198

2.) Since SO(3) is a Lie group. By definition, Lie group is a group that is also a differentiable manifold.
What is that differentiable manifold in the case of the SO(3) Lie group?

Silly Goose

@User198 the group

User198

Hm

SO(3) is a manifold

How do I visualise it?

Silly Goose

it is the same as all maths. there is an underlying set $X$ with additional structure $A,B,C,...$

ACuriousMind

15:39

@User198 That's an entirely different question from what you wrote. The Lie group manifold is not "invariant" under the group, it is the group.

Silly Goose

the underlying set $X$ here can be viewed as the set of all $3 \times 3$ special orthogonal matrices

ACuriousMind

If you're looking for a more "geometric description" of SO(3), it's double-covered by the three-sphere $S^3$ (this is intuitively easiest to see if you phrase rotations in terms of Euler angles).

Signor Feynman

I'm sorry, Silly G, it's been a few months since I was mildly interested in all of this. What you talk about is remiscent of the KT phase transition, which happens in the XY model (not Ising, though) where there is an algebraically decaying correlation function

Silly Goose

this set has a natural group structure, differentiable manifold structure, etc.

User198

@ACuriousMind What is "double-covered"?

ACuriousMind

15:42

@User198 There is a map $S^3\to\mathrm{SO}(3)$ under which the preimage of any point is two points. If you're not familiar with the theory of covering spaces, that's part of what you will have to learn to fully appreciate Lie theory.

User198

@ACuriousMind Isn't in the case, of say, unit circle. The unit circle is a Lie group, and also it is invariant under rotations?

ACuriousMind

@User198 I don't know what you mean by the circle being "the" Lie group

Silly Goose

any group is isomorphic to itself under conjugation, if that is what you are referring to

SnoopyKid

@Loong I am curious if a conventional explosive can turn bones into "unrecognizable as human remains" remains but with different way from nuclear explosions that used tremendous heat to turn bones into unrecognizable remains

User198

@ACuriousMind I edited the message sry.

ACuriousMind

15:44

If you're talking about the Lie group of 2d rotations being isomorphic to the circle and the circle - as the 2d unit "sphere" - being invariant under those rotations, that's an accident.

Silly Goose

i.e. given a group $G$, for any $g \in G$ the map $\text{conj}: G \to G$ defined by $x \mapsto gxg^{-1}$ is an isomorphism

ACuriousMind

SO(3) is not the 2-sphere, which is the set of unit vectors in 3d that are invariant under 3d rotations

and neither are any of the higher SO(n)s the respective $S^{n-1}$. That $\mathrm{SO}(2)\cong S^1$ is an exceptional isomorphism, not something revealing a general rule

User198

Hm

Is the $\mathbb{R}$ a Lie group?

ACuriousMind

what do you think?

User198

I think yes

Because it is a smooth manifold

And invariant under something, and that something is linear translation.

ACuriousMind

15:47

I don't know where you get that "invariant" thing from

the definition of a Lie group does not contain any claims about "invariants"

A Lie group is a group that is also a smooth manifold in such a way that the group operation is smooth.

User198

But isn't a group a set with some axioms, and that set contains some "operations" that keep something the same. It is a set with simetries.

ACuriousMind

No?

User198

Hm ok

ACuriousMind

Where did you get that strange definition from?

If you don't even understand the technical definition of a group that retroactively explains a lot of confusion :P

User198

I tought that is what groups are for. For systemizing simetries.

ACuriousMind

15:50

@User198 And you never thought to look up the actual mathematical definition?

User198

@ACuriousMind I guess so. :/

@ACuriousMind I did. I guess I missinterpreted it.

"The symmetries of an object form a group, called the symmetry group of the object, and the transformations of a given type form a general group."

ACuriousMind

I have a hard time understanding how you could misinterpret the group axioms. There is nothing in there about invariances or symmetries or whatever, just a bunch of algebraic properties.

User198

Yes, yes. I overlooked the general definition and went straight to where it could be applied to. So I somehow connected in my mind that groups are about symetries.

But In general they don't have to be.

As we see from the group axioms.

User198

16:17

I got it now. Thanks for clearing it up for me.

For eg

The group $(\mathbb R, +)$ is a group because it satisfies all the axioms.
The group $(\mathbb R, \cdot)$ is not a group because $0$ doesn't have an inverse, i.e. $0 \cdot \frac{1}{0}$ is not defined.

imbAF

18:03

Is there a physical interpretation as to what the scalar value of the VEV of a field operator is?

For example in QM the expectation value of the H- operator is the expected energy value of the system

ACuriousMind

18:21

QFT is QM, the meaning of expectation values is not different - for any given operator, it's just the expected value of that operator. I don't understand the question.

imbAF

18:44

I gave the example

there is an interpretation of the result in QM, the scalar that you get can be interpreted as the expectation value of the energy, when the operator is the Hamiltonian

but in QFT you have "two different types" of operators, you have something like the Hamilton operator, which is expressed via ladder operators, and I believe it's VEV would give the energy. While the VEV of the field operator $\phi$ $\psi$ also gives a value, and what would the intepretation be?

ACuriousMind

@imbAF but you didn't ask about any specific field in QFT

your question is equivalent to asking what the meaning of the expectation value of "an operator" in QM is

imbAF

No, I didn't but I could also do the same in QM

ACuriousMind

to which I really don't know what kind of answer you expect

Signor Feynman

Well, remember also that the fields aren't typically even Hermitian

imbAF

not give a specific operator, but just be general and say that the expectation value of an arbitrary operator would be the expectation value of the physical quantity that this operator represents.

Instead of saying all this convoluted sentence

I just cut it short with an example

Anyway what I want is the physical meaning of the result of VEV of $\phi$ where $\phi$ is a field operator

Signor Feynman

18:47

Before asking what's the physical meaning of the VEV of the field, have you wondered if the field itself has a "meaning"?

ACuriousMind

@imbAF QFT is still QM, so this is of course true in QFT still. Self-adjoint operators are (in principle) measurable and the expectation value represents the expected value of measuring them.

imbAF

you mean $\phi$ ?

Signor Feynman

Yes

(I, not ACM)

imbAF

Ofc I have and is one of the things that I don't have fully grasp. All I can say is that a field operator assigns an operator at any spacetime point

Idk how helpful that should be

But again, all I want is the physical meaning of the result of VEV of $\phi$ where $\phi$ is a field operator

ACuriousMind

Again, without telling us more about the field operator $\phi$, this is a meaningless question.

imbAF

18:50

@ACuriousMind Ok so then if you have $\langle 0|\phi|0\rangle=c$ where c is some constant. What does this constant represents? What is the physical intepretation of this value?

Signor Feynman

Think of a harmonic oscillator. The position operator has a meaning, the momentum operator has a meaning (they're hermitian and thus associated to observables) and so do their expectation values; what meaning do $a$ and $a^\dagger$ have?

Beware, knowing how they act doesn't give you a "meaning" :P

imbAF

@ACuriousMind This is like saying that if I do not specify the operator in QM (X,P,H etc) you can't say what the meaning of $\langle \psi| \hat O |\psi\rangle$ is

But you certainly can

just the expectation value of the physical quantity represented by the operator $\hat O$

Signor Feynman

@imbAF So all operators represent physical quantities?

imbAF

if they are hermitian yes

that is implied

I didn't think I had to also emphasize on that too

ACuriousMind

@imbAF The identity is a Hermitian operator. What is the "physical meaning" of $\langle \psi \vert 1 \vert \psi\rangle = 1$?

Signor Feynman

18:53

Except that as I said, quantum fields are not hermitian

Except the case of the real scalar field

imbAF

@ACuriousMind it has none

ACuriousMind

@imbAF So why do you think without narrowing down what $\phi$ is we can answer your question, if you have to admit that even in QM you cannot give a "physical meaning" for every self-adjoint operator $O$, as the example of the identity shows?

imbAF

is 1 an hermitian operator or a scalar?

ACuriousMind

As should be obvious from my sentence "The identity is a Hermitian operator" just before that, by the first $1$ I mean the identity operator, not the number 1.

imbAF

Since I can't find any physical quantity that 1 represents, I would say that that expectation value has no meaning

ACuriousMind

18:58

So clearly not all QM expectation values have "physical meaning" without specifying more about the operator in question. So why should you expect there to be a generic meaning for "a field $\phi$" without specifying anything more about it?

imbAF

Fine

What is the physical meaning of the VEV of the real scalar field operator?

ACuriousMind

I don't know in what sense you think this question is more specific. What is "the real scalar field"?

There's no real scalar field in the Standard Model, for instance (except the two "inside" the complex Higgs)

imbAF

All I am asking is if the "c" has a meaning

the same way $\langle \phi| H | \phi\rangle=c$ and c is the (expectation) value of the energy of the system

ACuriousMind

And all I'm telling you repeatedly is that it depends on what the $\phi$ is in your concrete theory.

imbAF

Does the text specify that?

All I can tell you is that below, he considers a complex scalar field

ACuriousMind

19:04

@imbAF No.

imbAF

I don't have any further info

@ACuriousMind Well , I can't provide any further details.

I am just parroting what the book says

ACuriousMind

I'm not asking you to provide further details, just to understand that it's unreasonable to expect an answer to your question with the details provided, just as it would be unreasonable to expect a "physical meaning" for an operator $O$ in QM without providing any further details about what $O$ is.

imbAF

Can you give me a specific example, that allows you to give an intepretation?

ACuriousMind

The text will probably end up showing you the Higgs mechanism, and in that context the $c$ will end up in the formulae for the masses of the resulting massive particles.

imbAF

Ok, but what is the specific thing you considered to come to this conclusion?

What I mean

ACuriousMind

19:10

I don't know what kind of answer you expect other than "I understand the Higgs mechanism and how it's usually presented in textbooks" :P

imbAF

In qm if I provide the detail that $O=H$ you can interpret the result of the expectation value as the energy. What is the equivalent of $O=H$ here?

ACuriousMind

@imbAF I mean "O=H" is just a bunch of symbols. What you're saying is a tautology: Of course if $O$ is the energy operator, then its expectation value is the expected value of energy.

Likewise, if it's the angular momentum operator ("$O=L$"), then its expectation value is the expected value of angular momentum, etc.

imbAF

You said I have to be specific for you to be able to give a meaning to the result of the expectation value. Now that I am, it's a tautology

ACuriousMind

I'm trying to get you to understand that there isn't really a lot of distinction between saying what an operator is and what its expectation value means

if you understand "what the operator is", such as it being the Hamiltonian or angular momentum or whatever, then you understand what its expectation value is, and vice versa

imbAF

Then what you seek with this: ", just as it would be unreasonable to expect a "physical meaning" for an operator $O$ in QM without providing any further details about what $O$ is."?

ACuriousMind

19:16

if you understand this, you should understand why it's so weird to ask for the meaning of the "expectation value of $\phi$"

If you don't understand what I'm trying to say, I'm sorry, I can't really think of a different way to say it

imbAF

I don't. This is going to linguistics now, so I will not bother with what I wanted to find

I don't understand

User198

20:00

1.) The unitary group U(1) is isomorphic to the set $\{ z \in \mathbb{C} \mid |z| = 1 \}$, under multiplication. Yes?
1.a) I presume this Unitary group is used in QM and connects to unitarity matrix?

2.) The orthogonal group O(1) is isomorphic to the abelian cyclic group $\mathbb Z_2$. Yes?

3.) When I specify a group, I have to specify a set and an operation that works on the elements of the set such that they all remain in the set. But sometimes that operation is not explicitly denoted.
Is that operation in most cases a multiplication? Or is it usually always denoted explicitly what operation is it in the case?

4.) Connected to 1.) I also see the group SO(3), the set of all real orthogonal matrices with determinant 1. It is also isomorphic to the unit circle?

Simmilar like U(1)? Is that a coincidence that they are both isomorphic to the same object?

ACuriousMind

@User198 What does it mean for an operation to be "a multiplication"? Also, we've already been over the fact that SO(3) is not isomorphic to any $S^n$ (but is double-covered by $S^3$), why are you now asking whether it's isomorphic to $S^1$ again?

User198

@ACuriousMind My bad. I meant to write SO(2)

ACuriousMind

Well, $\mathbb{R}^2\cong\mathbb{C}$ (as real vector spaces), so you should be able to figure out why 2d rotations and 1d unitary operations are the same

User198

Ok yes. Makes sense

And abelian means that that a group commutes?

Do you have a favourite group?

ACuriousMind

why would I have a favourite group :P

User198

20:15

Idk. It sounds like there are so many of them and with various properties. Kind of like animals in the animal kingdom.

But I have a really blurry image and I don't see the trees for the forest

But maybe some people have favourite theorems, I tought it was the same thing with groups. xD

ACuriousMind

perhaps

User198

Eg. the Monster group sound cool

ACuriousMind

but mostly I remember specific groups either because they occur often or because they're specific counterexamples

User198

It would be cool to have a visual map (if some groups can be visualised or be represented somehow with a drawing) of them:

ACuriousMind

The usual visualizations for Lie groups/algebras are Dynkin diagrams. They don't really depict "the group", though.

User198

20:46

And are there any "general" rules that give the isomorphisms between arbitrary groups that are well known?

Like is there a rule that: "A group G is isomorphic to another group D."

Or: "A group G is isomorphic to B x C."

Or is that a too broad statement (question to ask) because there are many specific group and you have to denote the isomorphism specifically for groups when you encounter it?

Signor Feynman

@ACuriousMind Come on, how is not $\mathrm{U}(1)$?!

ACuriousMind

@User198 I'm not really sure what the exact question is. Can you give examples for such rules in other (non-group) cases?

@SignorFeynman it's just a circle :P

Signor Feynman

I like abelian gauge theory

User198

@SignorFeynman Why would U(1) be your favourite? I mean I know nothing about group theory, but it sounds trivial even to me. Cool for maybe a starting example.

Unless I am mistaken and it has some deeper meaning and beauty...

@ACuriousMind Hm. Let me see.

Like maybe Generalized Stokes theorem. Some theorem that lets you switch your point of view from one side to another. Not necesarilly an isomorphism.

Meh idk. Scratch that.

I am looking for something like a "Fundamental theorem of group theory"? Is there a theorem(s) like that? That is essentially my question.

ACuriousMind

21:09

Really group theory is not that uniform - the kinds of things you do with finite groups are very different from the kinds of things you do with Lie groups and so on

User198

I found this: "The fundamental theorem of finite abelian groups states that every finite abelian group
G
{\displaystyle G} can be expressed as the direct sum of cyclic subgroups of prime-power order"

Is that important? It has Fundamental in its name xD

@ACuriousMind Ah I see.

Practically 2 different disciplines. Descrete and continous groups that is.

User198

21:38

@ACuriousMind Just one more question. xD sorry. Is the lie algebra of SO(3), namely so(3) the vector space of angular velocity and the appropriate Lie bracket is the cross product?

ACuriousMind

in some sense yes, in some sense no :P

if you just look at SO(3) as matrices acting on $\mathbb{R}^3$, you get the algebra $\mathfrak{so}(3)$ also as matrices on $\mathbb{R}^3$, and there certainly is no cross product of matrices

Signor Feynman

@User198 I'm a fan of expressing trivial things in a fancy way

No content, 100% appearance :P

User198

@ACuriousMind Ok thank you.

@SignorFeynman :)

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