@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"
"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
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
@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?
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
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
@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
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
@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
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
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?)
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
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
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!
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?
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
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
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
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
"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)."
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?
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