thing is, a "scalar function" is a section of the trivial line bundle $M\times\mathbb{R}$ and the Leibniz rule for your connection forces $\nabla (fs) = (\mathrm{d}f)s + f\nabla s$ for all scalar functions $f$ and all sections $s$
but since your sections are interchangably scalar functions, this means $d=\nabla$
first, there's no "angle" because there's no metric, and second the characteristic of the connection form is just that its kernel is the horizontal subspace
note that parallel transport is defined by the lift of curves being horizontal - it doesn't matter what exactly the vertical space is
For a connection $\nabla$ on a line bundle, in a local trivialisation, the connection looks like a one form $\nabla s=ds + sa$ but this is not a proper one form cause it depnds on the choice of local trivialization.
Why does the curvature $da$ not depend on the choice of local trivialization??
...
@Slereah no, that's just the 1d case of the proof that curvature $F$ is well-defined as a global form on the whole manifold even though the gauge fields only exist in local trivializations
that the bundle is a line bundle is wholly irrelevant there
"for a metric connection on a bundle isometric to the trivial line bundle, this is the only choice of 𝐷, because the metric uniquely determines the flat sections as those of constant norm."
Hm
I guess maybe you can get non-trivial connections if the connection isn't LC
I do vaguely recall that the exterior derivative changes if you have torsion
no, I think what you need to do is "forget" that the sections are scalar functions
and then you can have non-trivial connections, but you will no longer be able to switch between the idea of a "scalar function" and a section of the line bundle freely - the thing that swaps between the two essentially "fixes" a trivialization/horizontal choice in the line bundle
i.e. the answer to your question is: No, scalar functions can't have non-trivial connections, but line bundles can.
mathematicians aren't from another planet, they even call vector fields "vector fields"! :P
@Slereah sure, because in proper terminology all the other "real functions" are sections of line bundles and not just real-valued functions and hence there's no need for the additional word scalar as opposed to physics where you can have "real functions" that are pseudoscalars or whatever
but they'll understand what you mean by a scalar just fine :P
the "what the hell is going on here" is the reaction of someone who had a first course in diff. geo. and then tried to read a GR textbook while someone who has done diff. geo. for a decade will likely have encountered a lot of other sloppy notation and fuzzy arguments before
(over-generalization, of course, but "mathematician" is an even broader generalization so I don't feel very guilty about it :P)
I've dealt with billion dollar companies whose entire project management method was a gigantic excel script written by an army of interns and people who do not know how to program
I've never seen any code take so long to do so little
On the other hand when National Nuclear Labs wanted a button putting in a silly place to export a silly CSV file that will never be used I said "Of course, that's half a day's work" :-)
Imagine the poor guys who had to port a bunch of Minitel services to the internet
"To kick-start the process, the PTT ordered millions of Minitel terminals (built by French manufacturers such as Telic-Alcatel and Matra) and made them available at no cost to everyone in the country who had a telephone line."
Those were the days
Free government internet computers
It was a weirdly simple system for the era
compared to 80's internet
Although that meant that everyone had the same terminal
My tutor (older student who holds a smaller class in addition to the lecture) said that if we have a (hypothetical) function $E=\frac12mv^2(t)+E_0t$, then $$\frac{\mathrm d}{\mathrm dt}E=ma(t)v(t)+E_0$$ but $$\frac{\partial}{\partial t}E=E_0$$
I didn't fully understand the reason but apparently because the first term in the function does "only depend indirectly" on $t$, it's treated as a constant for a partial derivative but not for the total/"normal" derivative
Can someone give me an intuitive reasoning for why the field tensor for a generic gauge theory has the extra $[A_\mu,A_\nu]$ term? I have checked mathematically that adding a commutator term makes the field tensor transform covariantly under unitary transformations, but why on an intuitive level should it be the commutator term...
I mean if I don't know that it would be the commutator term(and hence don't know whether adding it will make the definition transformation nice), how can I arrive(heuristically or otherwise) to the fact that it would be the commutator of the two gauge fields?
If you write $E$ as $E = E(t,v(t),a(t))$ we have in general that $$\frac{d}{dt} E = \frac{\partial E}{\partial t} + \frac{\partial E}{\partial v} \frac{dv}{dt} + \frac{\partial E}{\partial a} \frac{da}{dt} \neq \frac{\partial E}{\partial t}$$ In your example where $E(t,v(t),a(t)) = m a(t) v(t) + E_0 t$ we have $$ \frac{\partial E}{\partial t} = E_0 , \frac{\partial E}{\partial v} = m a(t) , \frac{\partial E}{\partial a} = m v(t)$$
@ManasDogra If you mean commuting two $D_{\mu} = \partial_{\mu} + i A_{\mu}$ type terms then if $A_{\mu} = A_{\mu}^a T^a$ where the $T^a$ satisfy $[T^a,T^b] = i f^{abc} T^c$, that term wont vanish when you take the commutators of two $D_{\mu}$'s unless the $f^{abc}$ are zero
@Slereah Has anyone in history ever found e.g. the polar coordinate Christoffel symbols starting from this definition
@ManasDogra how are you defining the field tensor in the first place?
if you start from the idea that the curvature/field tensor is the commutator of two covariant derivatives, then there's no need to "add" anything, that definition just gives you the correct formula for both Abelian and non-Abelian field strengths
see, that's what I mean - if you don't take that (with or without the last term) as a definition, but $F_{\mu\nu} = [D_\mu, D_\nu]$, then you don't run into this problem
and it is perfectly reasonable to define $F$ like that in the Abelian case - $F$ measures how strongly non-commutative covariant derivatives are, that's an interesting quantity either way
it's also manifestly covariant that way, much nicer than writing down a "random" combination of derivatives of $A$ and trying to find one that's covariant :P
Let we have a vector bundle over a manifold $p : E \rightarrow M$ and there is a section $s$ on it and also $\nabla$ is a connection on this vector bundle. Is there any relation between $ds : TM \rightarrow TE$ (as an exterior derivative of a map between two manifolds) and $\nabla s$ i.e if $X$ i...
Is $TE = H \oplus V$ just saying, for a vector field $\mathbf{A}$, that $\partial_i \mathbf{A} = (\partial_i A^j) \mathbf{e}_j + A^j \partial_i \mathbf{e}_j$ where $\partial_i \mathbf{e}_j = \Gamma_{ij}^k \mathbf{e}_k$ is the horizontal part involving the connection
@Slereah A projection always depends on the splitting into two parts - you can't just "project" onto a subspace $V_1 \subset V$, you need $V = V_1\oplus V_2$ in order to say how the projection works
It looks like the vertical map thing just something like $(\partial_i A^j) \mathbf{e}_j = \pi(d A^j) \mathbf{e}_j$ where $\pi$ just gets rid of the $dx^i$ in $d A^j = dx^i \partial_i A^j$
the thing is that the $\mathrm{d}s(X)$ in the definition is in general not in the fiber, i.e. not Lie-algebra/representation-space valued. You need to "project it back" in order to get something that is
in terms of the connection form $A$, the horizontal space is the kernel of $A$, so applying $A$ to something is the projection: Only the vertical parts survive this
that's why the covariant derivative is $\mathrm{d} + A$
If an electron, travelling in a straight line, emits a photon perpendicular to its direction of motion (as seen from the inertial lab frame), will the frequency of the photon be the same in the inertial lab frame as that in the rest frame of the electron?
For a connection $\nabla$ on a line bundle, in a local trivialisation, the connection looks like a one form $\nabla s=ds + sa$ but this is not a proper one form cause it depnds on the choice of local trivialization.
Why does the curvature $da$ not depend on the choice of local trivialization??
...
