8:01 AM
@ACuriousMind Another dumb question, is the "trivial" Cartan connection $\mathrm{GL}(n) \hookrightarrow \mathrm{GL}(n)$ a $G$-structure?
It is getting hard to find much informations about it
I find people saying that the affine structure is what defines a general connection, which sounds plausible but I'm not sure
Different papers seem to define an affine manifold a variety of ways
there's this notion that the set of prefered frames on a $G$-structure is like $Fr(M) / G$, does it mean that there are no preferred frames on an affine manifold
Every frame equivalent
And so to speak no structure except the affine connection, which is left purposefully large
"affine structure is a type of geometric structures which is interesting (and worth looking at) but which does not fit in the general theory of G-structures, at least not in an obvious way, and not in the way we discussed it so far. "
Why is nothing ever easy
It's a bit hard to follow because every other $G$-structure is defined delightfully by some nice little canonical Klein model and then they tell you "Oh the affine structure is defined by the connection, actually"

8:49 AM
"In more concrete terms, this amounts to having an operation that associates to any ordered pair of points a vector and another operation that allows translation of a point by a vector to give another point; these operations are required to satisfy a number of axioms (notably that two successive translations have the effect of translation by the sum vector)."
Hopefully reading about the Klein model will elucidate things a bit

9:02 AM
I have Lichnerowicz big book on connection theory but it's entirely written on a typewriter

9:20 AM
@Slereah no
a G-structure has to be a Cartan connection of the form $G\to G\rtimes\mathbb{R}^n$

Well that is what it is I think, yes
$GL(n) \to GA(n)$

In general, you need that $G/H$ for a $H\to G$ connection has the dimension of your manifold because you want to solder $\mathfrak{g}/\mathfrak{h}$ to the tangent spaces

But it seems to be usually defined slightly differently from the other structures

I don't know what $GA(n)$ is

General affine group

9:22 AM
ah

hence the name

well, you wrote $GL(n)$ both times in your "dumb question" above
so did you really mean GA(n) the second time?

@ACuriousMind That's because $G$ structures not only have the Cartan inclusion, but they also have the inclusion into the frame bundle!
As far as I can tell on a synthetic level it's a geometry with the only interesting bit being that you can define parallelism, which makes sense I guess for something connection-related
but it's hard to piece everything together
I have seen somewhere that the set of $G$-structures on a given manifold is the quotient $GL(\mathbb{R}^n) / G$, such as the set of all metrics being $GL(\mathbb{R}^n) / O(n)$, but how does that work out with affine structures
Does that mean that there is only a single affine structure?
How does that square with the wide variety of affine connections that exist, in what space do they take their values
[I am aware that I used $G$ instead of $H$ here but the stabilizer is noted $H$ for Cartan connections and $G$ for $G$-structures so bear with me]
Although I guess that since we have $O \subset GL$, but metric connections are smaller than affine connections, maybe the relationship runs the other way round
Since we have $GL/GL \subset GL / O$
IIRC the torsion part is related to the translational part of the geometry, but then I guess that means that the non-metricity tensor is related to whatever the difference between the two is?

9:40 AM
@Slereah You see...as a G-structure, the affine structure is just...nothing. You can't "reduce the structure group" to GL(n), it already is GL(n)!

@ACuriousMind well yes, that's why I am having some issues!
I guess it's maybe more "Having less structure gives more freedom for the connection"

but note that the equivalence is not between Cartan connections and G-structures, it's between Cartan connections and G-structures together with a choice of frame and a G-principal connection
so the number of G-structures is completely irrelevant - you still have that choice of connection

I see
Why is the number of connections restricted for fancier structures, though?
Are the different connections equivalent under the action of the stabilizer?
What makes a non-metric connection unfit for a metric structure, for instance

I don't really understand the question
what do you mean by the number of connections being "restricted"

From what I understand the affine structure lets you have basically any connection you want
Go wild
But this is not true for structures which are reductions of the frame bundle, like a metric structure
They have to have metric connections, $\nabla g = 0$
what does that condition correspond to in the structure part?
Wait is that even true
ncatlab says that the connection of the metric structure is the affine connection?

9:59 AM
@Slereah from what I'm seeing, the connection for an affine structure has to be flat and torsion-free, no?

@ACuriousMind I have also seen such things, but I'm not 100% sure?

careful "affine connection" does not mean "affine structure"
it's terrible terminology

Yeah idk, there's a million vaguely related things called "affine"'
I've seen that the affine connection has to be flat and torsion free to be integrable
but then the same is true for metric structure
The connection being integrable just means Euclidian space
I guess that's for the manifold being like the model space?
also to make things worse there are a few different notions of the connection being integrable, too
also there's first order integrable v. fully integrable or something
From what I can grok, first order integrable for a metric structure is "There's Riemann normal coordinates" and fully integrable means "It's Euclidian space"
I guess the metric structure can also have whatever connection it wants, but not if you want it to be integrable?
and maybe for first order integrability, there's no constraints on an affine structure?
otoh the obstruction seems to be only torsion for first order, so maybe not
argh

math.stackexchange.com/questions/3777214/… if I ask the same but for physics, would it be off-topic?

@AdilMohammed yes, in general study advice is off-topic on our main site

10:12 AM
you can try 't Hooft's list
a fine list

@AdilMohammed The JEE is actually a pretty good introduction to physics.
Especially if you give the JEE Advanced.

Oh thanks @Slereah <3
@JohnRennie indeed

If you got a good score in the JEE physics paper then I think you are well set up for first year undergrad physics.
Maths is a bit different because a maths degree is very different from the way maths is studied at school.
But I don't think this applies to physics.

The thing I kinda hate about JEE, is that they expect you to know college-level physics so you can go to a college

There are a lot of things to hate about the JEE, but on the plus side it does prepare you well for a physics degree.

10:18 AM
that is very true

10:37 AM
I wonder if you could write a program for a Turing machine to generate all possible molecules.

@JohnRennie Probably, but it would probably take a while to run

That's OK, we probably have infinite time

Are we sure

I can be reasonably sure I will never observe the universe ending.

I am angry because every paper on Cartan geometries just sort of magically take the canonical space's structure out of their hat
So I went to nlab to check out how to actually derive it
$$\array{ X &&\stackrel{h}{\to}&& \mathbf{B} O(n) \\ & {}_{{\lambda}}\searrow &\swArrow_{E^{-1}}& \swarrow_{{\mathbf{c}}} \\ && \mathbf{B}GL(n) } \,.$$
I believe in magic now

1 hour later…
12:09 PM
@JohnRennie we probably understand quantum chemistry still a bit too poorly to do that in full generality
like, we can fold proteins well, but do we understand folding for arbitrary large molecules in a way that you could write folding code not specialized to the "type" of molecule

Just solve the QED system of that system

12:40 PM
Chemistry doesn't even consider any other interaction outside of EM, surely it can't be that hard

1:37 PM
When we have Grassmann valued spinors classically for fermions, don't we already have so much of the 'quantum' structure in at that point i.e ordering matters

Well it is a good start, yes
But you still need to quantize your Poisson brackets

@DIRAC1930 not really - the quantum structure is after all turning the data from the Poisson bracket into commutation relations
if by "ordering matters" you just mean the trivial observation that anti-commuting objects don't commute, then sure, "ordering matters"
but the point of ordering procedures is that any naive quantization prescription will depend on which of the classically equivalent expressions for an observable you choose to replace the variables with observables in

Also remember that field operators neither commute nor anticommute in QFT
They deviate from that by some $\hbar$ term
you are still $\hbar$ away from quantum theory!
$\hbar$ is pretty small, but still
One calorie not quantum enough

All we have to do is enforce $\{ \psi_i , \psi_j^\dagger\} = \delta_{ij}$ and $\{\psi_i , \Pi_j\} = \delta_{ij}$ it seems

although there is a formalism in QM where we just "twist" the product between classical quantities to obtain QM
It's called deformation quantization
but the product is much more complicated
$$a \ast b = \sum_{n=0}^\infty B_n(a,b) t^n$$
I'm afraid there is no royal road to quantum mechanics

1:59 PM
What I'm saying is that this all seems like a massive trick that for fermions and that it is so close to the classical theory that it isn't very enlightening

Does it
The same is true for bosons
Bosonic states commute and so do their classical counterparts

I think it may be confusing because I don't know what is going on properly

Quantum theory isn't just "the operators have a non-commutating algebra"

We have the Hilbert space aswell but what else?

Well a big difference between classical mechanics and quantum mechanics is what's called contextuality
The order of measurements matter

2:09 PM
What exactly are we saying when we say 'we enforce canonical commutation relations'. How do we know that the space changes to a Hilbert space?

Well the space of configurations being a Hilbert space is an axiom of quantum theory
there are historical reasons why it was decided so, but there's no deriving it from classical mechanics

So, how do we know that after cannonical quantizations, the operators we have quantized even act on a Hilbert space?

Well that is the procedure that we are doing
If you're asking about obstructions, that is another matter

I don't think 'fermions' (have to?) anti-commute classically

@bolbteppa Define "have to"
I think @ACuriousMind knows some stuff about obstructions to quantization?

2:16 PM
I'm pretty sure they don't and they only do when you quantize, and that's it just a formality to treat them as anti-commuting e.g. in susy before you 'quantize' (is it already always quantized?)

Well you can have spinors that commute yes
You can even have scalar fields that anticommute
Go wild

In L&L they treat them as commuting until they quantize
Including in the Fierz identities which is fun

It's just an algebra, you can slap it on whatever vector space

they "have to" anticommute if you want a nice quantization map
if you just want to construct the quantum theory and introduce the anti-commutation relations ad hoc in the quantization step, you can do that
@DIRAC1930 Canonical quantization is not a well-defined procedure
it's an ad hoc trick to get a quantum theory
if you try to understand canonical quantization "rigorously", that's impossible because canonical quantization doesn't actually work

@ACuriousMind Bit harsh
I think you mean "it doesn't work for a general class of theories"!
plenty of cases where it does work!

2:19 PM
the Groenewold-von Howe theorem essentially says that all the nice properties we would wish to have for a quantization map, and that would make canonical quantization work, are impossible to realize
@Slereah no, I don't mean "there are cases where it fails", I mean that "canonical quantization" doesn't actually give you a map from classical observable to quantum observables

Strange

So cannonical quantization isn't just enforcing the CCR's? You have to do all other stuff like enforce the operators to act on a Hilbert space

Although apparently there's 8 conditions to fulfill for that theorem to hold

@DIRAC1930 what is "enforcing the CCRs" supposed to mean? What, exactly, is "commuting" in your CCR if not operators on a Hilbert space

Seems a bit too greedy imo

2:21 PM
@Slereah that's the point - all the rigorous quantization methods have to drop some of the constraints on the quantization map

@ACuriousMind I mean can't I have non-commuting operators on a different space?

@DIRAC1930 what sort of "different space"?

@DIRAC1930 You can define the algebra of operators abstractly if you want

It seems arbitary that we assume that it will be a Hilbert space

the $C^*$ algebra
But you can associate a Hilbert space to that

2:22 PM
Umm you can even have non commuting operators on Euclidian space

via the uuuuuh

Rotations or something

whatever process
What's it called

A solution of the Schrodinger equation $\psi = \sum_n a_n \psi_n$ is a scalar, it commutes with everything. You can promote it to a quantum field by sending $a_n \to \hat{a}_n$ where the $\hat{a}_n$ satisfy the CCR's. The Dirac equation is no different, it can be written as a Schrodinger equation, we can solve it with a $\psi = \sum_n a_n \psi_n$, all commuting scalars here, and then promote $a_n$ to an operator $\hat{a}_n$ satisfy the CCR's, or CACR's, where is the grassmann fermion here

@DIRAC1930 I feel we're putting the cart before the horse here

2:22 PM
@ACuriousMind what's that thing called

the thing is why do you want operators in the first place?

I don't know

because you know quantum theory has the Born rule
which is about a) operators and their eigenvalues and b) probabilities from an inner product

The GNS construction
that's the one

you have to know how quantum theory works before you can go look for quantization maps
@Slereah that's just getting to a Hilbert space with extra steps - you have to explain why you're looking for representations on Hilbert spaces of your abstract algebra, too!

2:25 PM
So really, we are converting them to operators that act on a Hilbert space

The paragraph I stated is solid, whether you can treat the classical fields are grassmann from the beginning is basically a formal question, I'm not sure what happens if $\hat{a}_n$ is anti-commuting and $\psi_n$ is grassmann, when you work with grassmann stuff I think it's implicitly $\psi$ so that the $a_n$ become grassmann not the $\psi_n$'s, and you just promote the grassmann $a_n$ to grassmann operators with the extra CCR aspect to it?

@ACuriousMind Representation? I am just innocently looking for a polynomial star algebra obeying those properties, good sir
No ulterior motive

But how do we even interpret what the operator even does on the Hilbert space
Seems like someone just guessed it

hm?

It's very confusing, we're really talking about multi-particle systems, so one should ask why the one-particle solution $\psi = \sum_n a_n \psi_n$ is even relevant

2:29 PM
no one's "guessing" operators here

See, things aren't so bad

a quantization map is given by a Hilbert space $H$ together with a map from (a subset of) classical phase space observables to self-adjoint operators on that space

That's all you need baby

I mean the interpretation that $\hat{\psi}(X)$ creates a particle at $X$ is very handwavy in most introductory QFT books

you keep switching between very general ideas about quantization and very specific questions about quantum field theory
this is not only confusing for us, this is also confusing for you
"Why does a field create a particle" is not a question about quantization

2:31 PM
Assume everything I am talking about is introductory field theory

quantization cannot tell you anything about particles because classical field theory doesn't have particles
for a free field, you can just write down the mode expansion into c/a operators

@ACuriousMind Do you know any books that spell things out like this?

mostly mathematical monographs, few physicists care about formalization of quantization

I've found all text books that go into this stuff either too advanced or too handwavy

but back to the particles: since the c/a operators fulfill the algebra for ladder operators, they create something from the vacuum that you can count
there are several ways to argue that this something is usefully thought of as a particle

2:34 PM
Also calling them "particles" is a bit generous

Okay lets go back to basics. What is a vacuum in classical field theory?

classical field theory doesn't have that notion, really

the vacuum is the zero field in CFT
It's the zero section
nothing fancy

maybe you want to call "vacuum" when you have no field, i.e. all fields are zero, but that's not a useful use of the word

yeah nobody really uses that word in classical theory
idk if you can have fancy vacuas in classical theory?

2:36 PM
if you're already poisoned by QFT, you might want to call classical solutions to the equations of motion "vacuum", but really, classically, there's no need for that word

What exactly is needed for a quantization map?

A good question
There's like a dozen quantization process

depends on which of the assumptions of the GvH theorem you want to drop

also not all quantization processes require a Hilbert space!
Although I guess they do secretly
IIRC even the rigorous path integral has a secret Hilbert space in there

e.g. deformation quantization drops the requirement that commutators must equal Poisson brackets, geometric quantization drops that you need to map all observables to operators, plenty of ad hoc approaches will do a mixture
canonical quantization just does "whatever works" which I why I said earlier it doesn't actually "work" because it doesn't give you recipe for what to do when you run into trouble

2:40 PM
@ACuriousMind Just do like a physicist and say it's probably not important

I would advise being less obsessed with quantization if your goal is to understand QFT, really :P

Well, if you do, you can always try the nlab page on the topic

nothing in QFT depends on how, exactly, you quantized your fields, or whether there even is such a thing as rigorous quantization of fields
we have a classical Lagrangian, we wave a magic wand that grants us quantum versions of the fields that fulfill the CCR, and then QFT goes brrr

This is interesting though
Most of QFT calculations are very boring lol

It's all simple as you can see

2:42 PM
@DIRAC1930 I don't disagree, I'm just saying this isn't very useful if what you actually want to understand is how people do QFT

People mostly do QFT by looking up the big table of Feynman diagrams and doing the sums

@Slereah Slerah keeps on dissing physicists lol

just follow the chart

@DIRAC1930 there is nothing inherently bad about just doing computations
in fact, experiments need a lot of people who don't get lost in the clouds and instead just know how to sit down and compute stuff :P
and this isn't trivial work - if it was trivial, we'd long have taught computers how to do all of it

I know but I think those people find those long QFT calculations satisfying in a way that I never found

2:58 PM
@ACuriousMind What happens if you try to quantize the free 1D Hamiltonian canonically
Is it $x^4$ that's fucking up
I remember $x^4$ being a troublemaker

one of the higher polynomials will

Goddamn $x^4$
fortunately there's no apparatus for measuring hypervolumes
Reading some site on Klein geometry but I think it was written in the 90's
The math is all written out in ASCII characters
the horror
"We know that the euclidean symmetry group of a parabola has order two, while those
of an ellipse and hyperbola have order four (and that the latter each contain a half-turn)."
Do I know that?

2 hours later…
4:56 PM
What's the Euclidean symmetry group of a parabola

@Slereah you do! I believe in you!

5:15 PM
Classically $[x,p]=0$. Are we saying that during canonical quantization, we impose $[\hat{x}, \hat{p}] = \imath \hbar$ and look for a representation of these operators on a Hilbert space?
I just want to get my language correct

that's one way to look at it
but keep in mind that the classical "$[x,p]=0$" isn't a commutation relation of linear operators on a vector space
(unless you're doing Koopman-von Neumann mechanics but let's not go into that, too :P)

Lol

1 hour later…
6:40 PM
@bolbteppa I think it means that it's symmetric by reflection along the center

7:07 PM
@Slereah my brain borks whenever I try to read typewriter books

7:27 PM
Hm
One nice definition of a $G$-structure is that given some canonical tensor on the model space $\mathbb{R}^n$, the $G$-structure is the group that leaves a canonical tensor on it invariant
or I guess a structure on it in general
That notion probably generalizes nicely enough to Cartan structures, I guess?
like leaving the lightcone of the projective Minkowski space invariant or something
I guess for the affine group that would be like the zero vector or the entire space itself, idk