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11:20 AM
Hello everyone, I'm reading QFT and SM by Schwartz, and discussing Yang-Mills theories, he say "a free theory of N complex fields is automatically invariant under U(1) X SU(N)". Anyone can explain me why?
 
"The construction of the tangent bundle immediately generalizes to a construction of the frame bundle $Fr(M)$ instead. Because it is so funny, we immediately define the higher frame bundles $Fr_k(M)$."
Hilarious
@john I think he's talking about the sigma model
You can check that if you "rotates" free fields, the action stays invariant
ie if you have two fields $\phi_1$ and $\phi_2$, you can define new fields $\phi' = (\cos(\theta) \phi_1 + \sin(\theta) \phi_2)$ and so forth
and this will be about equivalent to the old theory
Basically just a change of variavles
The $U(1)$ part is because it's quadratic so just about anything will be $U(1)$ invariant since it's just $\phi^* \phi \to (U \phi)^* (U\phi) = \phi^* U^\dagger U \phi = \phi^* \phi$
 
Yeah the $U(1)$ part is clear, I don't get why also $SU(N)$
 
The sum of $N$ free fields essentially defined a $U(N)$ inner product?
Since it's $a^* a + b^* b + c^* c \ldots$
and the same with the kinetic terms since it commutes with derivatives
 
@Slereah I don't understand this
 
You could equivalently define your fields as a $\mathbb{C}^N$ vector, $(\phi_1, \phi_2, \phi_3, \ldots)$
And then your Lagrangian will be something like $L = (d\vec{\phi}) \cdot (d\vec{\phi}) + m^2 \vec{\phi} \cdot \vec{\phi}$
 
11:42 AM
Oh ok now I see. Thanks
 
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1 hour later…
12:47 PM
@ACuriousMind The tangent bundle can be defined as the space of germs of curves $\gamma : \mathbb{R} \to M$, and the frame bundle as the space of germs of diffeomorphisms $f : \mathbb{R}^n \to M$
Is there a connection between the two and the mapping of the tangent bundle to the frame bundle
What mapping between $\mathbb{R}$ and $\mathbb{R}^n$ does one need in between
 
what "mapping of the tangent bundle to the frame bundle" do you mean?
 
A frame field I guess
I guess maybe it's just the preimage of the curve in $\mathbb{R}^n$ that makes the link?
Also what are the higher order tangent bundles for the $k$-jets of curves
"However, even for the case of $n = 2$, constructing a vector bundle (for abbreviation v.b.) structure on $T^2M$ over $M$ is not as evident as in the case of $TM$. More precisely sometimes it is impossible to define a v.b. structure on $T^2M$."
Goddamnit
Why can't anything ever be trivial
 
glS
1:25 PM
@Slereah what prevents you from making a vector bundle structure there? Does it mean that there might not be a local trivialisation for it? Also, that is referring to a vector bundle $T^2 M\to M$, rather than one on $T^2M\to TM$, right?
 
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@Slereah I'm afraid I really don't know what you're thinking about there
 
@ACuriousMind Ah well, I'll figure it out
@glS If you want to learn more check out arxiv.org/pdf/1403.3111.pdf
Once in a while I'd wish a paper would just tell me "Turns out that's way easy to do"
 
glS
@Slereah " More precisely based on this lifted nonlinear connection we prove that $T^k M$ admits a vector bundle structure over $M$ if and only if $M$ is endowed with a linear connection" ah, interesting
 
Fortunately I'm not gonna do without a linear connection
Imagine those poor suckers working on topological manifolds
 
glS
1:42 PM
so.. if we write Newton's eqn in "geometric" terms with the differential, we should have $d^2 q=F\circ dq$, assuming a force field that depends on up to first derivatives, and $q:\mathbb R\supset I\to Q$. Then, changing coordinates, thus assuming this to be satisfied for $f\circ q$ for some diffeo $f:Q\to Q$, we'd get $d^2 f\circ d^2 q=F\circ df\circ dq$, and thus in the new coordinates Newton's eqn would read $d^2 q=\underbrace{(d^2(f^{-1})\circ F\circ df)}_{\equiv F'}\circ dq$
this kinda looks too nice to be true, as a compact way that contains terms that would typically be written with Christoffel symbols etc. Am I getting something horribly wrong?
 
I don't know too much but I know that the Newton-Cartan formalism certainly has Christoffel symbols for different coordinates
You have specific connections for "famous" coordinates, like the Coriolis connection
 
glS
@Slereah am I assuming some connection to make the argument? I honestly don't know
I'd be tempted to say that the choice of "force field" $F$ here kinda amounts to a choice of connection, in that it gives a way to relate different tangent spaces, but I'm not even sure whether this sentence makes much sense
 
IIRC you get essentially the same thing as GR
 
glS
meaning?
 
The connection is your gravitational field, forces are just additional terms for your EoM
but that's for Newton Cartan specifically, of course, not sure what exact formalism you're doing here
but even then, that would be a flat connection I guess?
Which you still need to change if you change coordinates
 
glS
1:56 PM
@Slereah I mean, the idea was simply to rewrite Newton's equation for a generic force field using the differential. I suppose to be more precise the $F$ in $d^2 q=F\circ dq$ should be understood as a vector field $TQ\to T^2Q$ with $F(q,v)\equiv (v,F(q,v))$, but I'm not (willingly) making any other weird choice or assumption
 
What's your manifold here?
 
glS
a generic smooth manifold I guess?
 
I mean is it time, space or spacetime?
 
glS
does that change the argument?
I mean, do I have to impose some physical interpretation to the manifold at this level?
 
Guess not
Although I guess that affects the type of coordinate change you can do
Trying to figure out how centrifugal forces would work here, but otoh those are coordinate changes of both time and space
Are there any fictitious forces you can get for coordinate changes that don't depend on time?
 
glS
2:06 PM
I'm just wondering about the general structure of te coordinate transformation, I don't really care about the physical interpretation of the manifold and the diffeomorphism
 
@glS what is $\mathrm{d}^2q$
 
glS
@ACuriousMind double differential. So if $f:M\to N$ then $df:TM\to TN$, and $d^2f\equiv d(df):T(TM)\to T(TN)$
 
I wouldn't write that like that, because you can confuse it with applying the exterior derivative $\mathrm{d}$ twice (which gives zero always)
 
glS
is there a standard alternative way to write it?
 
what's wrong with the traditional $\ddot{q}$ :P
 
2:09 PM
@ACuriousMind not diff geom enough
 
glS
@ACuriousMind well, nothing I suppose, but it looks a bit weird if the function is not time-dependent?
 
@glS what do you mean? Newton's equations are the equations for a curve $q(t)$
 
glS
I guess mostly it's that $\ddot q$ makes me think of a standard derivative, rather than the differential, which is not quite the same thing. But obviously anything goes, as long as we understand each others
 
we usually call the parameter of that curve "time", and write $\ddot{q}$
 
glS
@ACuriousMind yes, but when I introduce a coordinate transformation I'm considering a curve of the form $f\circ q:I\to Q$ where $f:Q\to Q$ is the coordinate transformation, which is not "time-dependent". Then I want the (double) differential of $f\circ q$, and I want to write it in a way that allows me to isolate $\ddot q$ (or however you want to write the double differential of $q$)
 
2:13 PM
I don't think this works like you think it works
 
glS
quite possibly, but why? =)
I mean, do we agree that I can write "Newton's equation" for a curve $q:I\to Q$ as $d(dq)= \tilde F(dq)$, where $\tilde F:TQ\to T(TQ)$ is defined as $F(q,v)\equiv ((q,v), (v, F(q,v)) )$ (working in local coordinates, or with $Q=\mathbb R^n$)
 
I guess it works since $\mathbb{R} \cong T\mathbb{R}$
 
glS
after all, locally, $dq(t)=(q(t), \dot q(t))$, and thus $d(dq)(t)=(\,\,(q(t),\dot q(t)), (\dot q(t), \ddot q(t)) \,\,)$. On the other side $\tilde F\circ dq(t)=(dq(t), (\dot q(t), F(dq(t))) )$
 
@glS no, $\mathrm{d}q(t)$ is not that - the argument to $\mathrm{d}q$ lives in $TI$, but the argument to $q$ lives in $I$ - you can't write $t$ for both of that
or, at least, you first need to explain how you're identifying the interval $I$ with its tangent space $TI$ there
 
glS
@ACuriousMind this I agree. What I should have more formally wrote is $dq(t, e_1)$, but because the tangent space to $I$ is one-dimensional, the $e_1$ bit does'nt really give much information
 
2:19 PM
@Slereah uh, what? $T\mathbb{R}\cong\mathbb{R}^2$
 
Oh yeah
Was thinking of the fiber I guess
Yeah idk it seems like a convoluted way of writing it
Differential geometry just uses dots for curve tangents
like Newton did
 
glS
@ACuriousMind yes, but given $(t,\lambda e_1)\in T_t I$, we have $dq(t,\lambda e_1)=\lambda dq(t,e_1)$, and so I can just stick to look at $dq(t,e_1)$, no?
 
@glS what's "$e_1$"?
 
glS
@ACuriousMind some choice of "basis" for the tangent space
 
there's no natural length on your tangent space
if you're just picking a random vector/number that's very odd
 
glS
2:23 PM
@ACuriousMind mh. I mean, if we are working locally, we just have $TI\simeq I\times\mathbb R$, so why can't I just literally take the tangent vector with unit length in the standard metric for $\mathbb R$?
 
why would you use that metric? then the choice of vector depends on your choice of coordinates "locally"
our time $I$ is just a manifold, I don't think we have a metric there
I think you really just need to drop trying to define this via the differential and just say "extend the curve by its 1-jet/tangent vector to $I\to TQ, t\mapsto(q(t),\dot{q}(t))$"
 
glS
why not? Does the choice matter at all? If we wanted to be more formal here we should write, I think, $d(dq):T(TI)\to T(TQ)$, where $dq(t,\lambda e_1)=(q(t),\lambda \dot q(t))$, and thus $$(\,\,d(dq)\,\,)((t,\lambda e_1), \,\, (\mu e_1, \eta e_1)\,\, ) = (dq(t,\lambda e_1), \,\, (\mu \dot q(t), \,\,\lambda\mu \ddot q(t)+\eta \dot q(t)) \,\,).$$ Ok this won't make any sense to anyone, let me try to explain wth I mean with it
 
here's how you do it properly: physics.stackexchange.com/a/54948/50583
 
glS
for a generic $f:M\to N$, we have (locally) $df(x,v)=(f(x), \partial_v f(x))$, where $\partial_v f(x)\equiv v^\mu \partial_\mu f(x)$. So then given this particular diffeomorphism $df:TM\to TN$, we should have $d(df)(\,\,(x,v),\,\, (\delta x,\delta v) \,\,) = ( df(x,v), \,\, (\partial_{\delta x}f(x), \partial_{\delta x}\partial_v f(x)+ \partial_{\delta v}f(x))\,\,)$
 
We all want to work on the one dimensional WZW model
 
glS
2:39 PM
@ACuriousMind I mean, that's very close to what I'm trying to say I think. For example, when they write "the acceleration is $(x,v;v,a)\in TTM$" that's precisely what I get when writing $d^2 q\equiv d(dq)(t)$
ignoring the one-dimensional tangent component of $I$
 
I'm afraid you've completely lost me with your notation
 
glS
@ACuriousMind yea, I guessed as much. I'm simply applying the definition of differential of a function, to the diffeomorphism $df:TM\to TN$. Assuming everything's local/flat and nice, so $f:\mathbb R^n\to\mathbb R^m$ and $df:\mathbb R^{2n}\to\mathbb R^{2m}$.
Then the "base point" for $TM$ I'm writing as $(x,v)\in\mathbb R^{2n}$, while the corresponding tangent directions as $(\delta x,\delta v)\in\mathbb R^{2n}$
 
I got as much, but now you have an even worse problem where you need to plug three random vectors into your $\mathrm{d}^2q$ (because $TTI$ is 4-dimensional now!)
what you really want is just a function $I\to TT^\ast M$ like the answer I linked gets it from the $(m\circ\dot{q})'$, not the abomination you're trying to construct here :P
 
glS
@ACuriousMind yes, but why is that a problem I don't understand. If I just pick the same element for all tangent vectors, you get the correct equation fine. In other words, I'm just defining $dq(t)\equiv dq(t,1)$ and $d(dq)(t)\equiv d(dq)((t,1),(1,1))$
 
@glS that's simply an unnatural choice - in order to pick a "1" you need to fix a coordinate system
 
glS
2:48 PM
@ACuriousMind which I am. Or even just assume $Q=\mathbb R^n$ as far as I'm concerned here
 
I thought our goal was to formulate an abstract coordinate-free version of Newton's laws
 
glS
ok, fair enough. But it is still coordinate-free, because whatever coordinate I choose, and whatever the corresponding choice of "1", everything will still work.
 
I don't see the use of both abstract notation and having to fix a coordinate system
@glS This can't possibly be true - you want the very specific choice of "1" where $\mathrm{d}q(t,1) = (q(t),\dot{q}(t))$, don't you?
and you need to choose that 1 at every $t$, so it's really a "time-dependent 1"
I really don't think this is a good way to proceed :P
it is much easier to just define the curve $t\mapsto (q(t),\dot{q}(t))$ and not bother doing stuff with the differential at all
 
glS
@ACuriousMind mh. So your point is that this formulation might not hold up when changing local chart, right?
@ACuriousMind well, ok, but if the argument was correct and you could write transformation laws as $d^2 q=(d^2(f^{-1})\circ F\circ df)\circ dq$, that would be kinda interesting imo
 
@glS I haven't even started asking you what the heck $F$ is :P
 
glS
2:58 PM
@ACuriousMind reading this again: you are referring to a coordinate system for $TI$ though, right? But isn't $TI$ parallelisable? You can just fix a single "global" coordinate system for it, so why should we worry about changing local charts for it.
@ACuriousMind in my defence, that I've seen somewhere
 
you said it has to be a function $F: TQ\to TTQ$, but you need to explain how this connects to the usual vector field $F(x,v)$ we write in lesser formulations
 
glS
that you can formulate Newton's equation as the integral curves of a vector field $TQ\to TTQ$, which is what I mean with $F$ here
 
the answer I linked does this very well for its $F: TQ\to TT^\ast Q$
 
glS
@ACuriousMind it's just the second component of the vector field: $\tilde F:TQ\to TTQ$ with $\tilde F(x,v)=((x,v),(v, F(x,v)))$, with $F$ the "standard force field"
 
@glS now that is a much better explanation of what's going on here than anything you're trying to do with the $\mathrm{d}q$ notation!
 
glS
3:00 PM
@ACuriousMind ok but is that needed? This is something I'm still to properly understand in general. Tangent and cotangent bundle should be fiberwise isomorphic, so one should be able to work with either in any situation, no?
 
note that an integral curve is just that - a curve $I\to TTQ$
not some function $TTI \to TTQ$
and that curve relates to the usual $q(t)$ just by $t\mapsto (q(t),\dot{q}(t),\dot{q}(t),\ddot{q}(t)$
 
glS
@ACuriousMind I fail to appreciate the difference between these two. $TI$ is parallelisable, and thus so is $TTI$, so we don't need to worry about changing local charts for it or anything like that. I can just make a "canonical" choice of all tangent vectors in it, and then essentially identify $I\simeq TI\simeq TTI$
 
@glS by that logic I can just identify all $\mathbb{R}^{2^n}$ with $\mathbb{R}$!
in other words, I neither get what you mean by "identify" here nor why parallelisability is relevant
 
glS
@ACuriousMind let me take a step back here. If I say that $\gamma:I\to Q$ is an integral curve for some vector field $X:Q\to TQ$, that means $(\gamma(t),\dot\gamma(t))=X(\gamma(t))$, right?
 
@glS your r.h.s. has to be $(\gamma(t),X(\gamma(t))$ if you want this as an equation in $TQ$
or, well, this is notationally confusing
 
glS
3:10 PM
well, I'm defining $X$ as $X:Q\to TQ$, so $X(\gamma(t))\in TQ$ by definition. I thought that's what a vector field being a "section" meant
 
normally we write $X(q)$ for the tangent part of the value of $X$
but sure, we don't disagree what that equation means, at least - that's correct, that's what being an integral curve is
 
glS
@ACuriousMind ok. But then you'd disagree on writing that equation as $d\gamma(t)\equiv d\gamma(t,1)=X(\gamma(t))$, I take it?
 
yes
because $\mathrm{d}\gamma(t)$ doesn't mean anything and "1" is just undefined
you could define a "$1(t)$" by $(t,1(t)) = \mathrm{d}\gamma^{-1}(\gamma(t),\dot{\gamma}(t))$
but that makes your equation completely impractical because if I already know the tuple $(\gamma,\dot{\gamma})$, why would I use that to compute this $1(t)$ instead of just using the integral curve equation for the tuple directly?
 
glS
@ACuriousMind but are you saying that it's not a good formulation, or that it's outright wrong? Because we can define it fine: $TI\simeq \mathbb R^2$, so literally just take $1$ to be the unit vector orthogonal to the first component of htis $\mathbb R^2$. That is, $e_2$.
 
@glS you can't just write $\cong$ and claim you can pick a unit vector
you need to say which isomorphism you're using for your $\cong$
and you must care which you pick, since you want that $\mathrm{d}\gamma(t,1) = (\gamma(t),\dot{\gamma}(t))$ and not e.g. $\frac{1}{2}\dot{\gamma}(t)$ in the second component, so you're not allowed to pick random "1"s otherwise your equation is just wrong
 
glS
3:20 PM
@ACuriousMind fine. Modeling the tangent space via equivalence classes of curves, we have $TI$ as the set of pairs $(t,[\gamma_t])$ with $t\in I$ and $[\gamma_t]$ equivalence classes of curves $\gamma_t:I\to I$ with $\gamma_t(0)=t$ identified by their first derivative in the usual way. I then define $(t,1)$ as the element of $TI$ of the form $(t, [s\mapsto t+s])$.
 
Thank goodness that @ACuriousMind is around
 
@glS $s\mapsto t +s $ is not a curve $I\to I$ unless $t=0$
 
Dunno who would answer all those math questions otherwise
 
glS
@ACuriousMind right. Assuming $I=\mathbb R$ maybe? =)
 
well, I (or rather its interior) is certainly isomorphic to $\mathbb{R}$
but you're essentially fixing the parametrization of the curve here
the equation $(\gamma,\dot{\gamma}) = X(\gamma)$ holds regardless of whether I choose $t$ or $2t$ or some crazier functions as my time parametrization compared to some reference choice
but you're assuming here again just that $I\cong \mathbb{R}$ without fixing the isomorphism
just because it's a one-dimensional manifold doesn't mean it's not still a manifold!
 
glS
3:27 PM
@ACuriousMind yes. But isn't it correct to say that, at least fixing this parametrisation, the equation $d\gamma(t,1)\equiv d\gamma(t,e_1)\equiv d\gamma(t, [s\mapsto t+s])=X(\gamma(t))$ makes sense? though I can see the issue, or at least the "unniceness" of having to choose a parametrisation.
@ACuriousMind I could just be assuming $I=\mathbb R$ (equal, not isomorphic). Though if one does things more carefully, isn't it enough to have the function $s\mapsto t+s$ defined on some neighborhood of $0$? In which case the definition would still make sense for any $t\in I$ for $I$ open?
 
@glS I am neither convinced that $[s\mapsto t+s]$ is well-defined nor do I see why the value of $\mathrm{d}\gamma$ at this particular tangent vector should be $\dot{\gamma}(t)$ and not some other multiple of $\dot{\gamma}(t)$
let's grant you that in your world curves are just functions $\mathbb{R}\to q$ and you don't need to worry about reparametrizing them
 
glS
@ACuriousMind mmh following the standard action of the differential, we should have $d\gamma(t,[s\mapsto t+s])=(\gamma(t), [s\mapsto \gamma(t+s)])$, and then isn't $[s\mapsto \gamma(t+s)]$ what we should mean with $\dot\gamma(t)$ really?
 
(so my first point is moot)
 
glS
maybe this can be reformulated as saying that there is some $v$ such that $(t,v)\in T_{\gamma(t)}Q$ is such that $d\gamma(t,v)=X(\gamma(t))$?
 
@glS that's correct
 
glS
3:34 PM
furthermore, said $v$ is (should be?) unique, and thus "canonical"?
 
@glS now you're just doing this:
21 mins ago, by ACuriousMind
you could define a "$1(t)$" by $(t,1(t)) = \mathrm{d}\gamma^{-1}(\gamma(t),\dot{\gamma}(t))$
 
glS
@ACuriousMind wait though, this $v$ doesn't depend on $t$?
ah no wait, it does
 
@glS of course it does
but note that this is circular - the equation for the integral curve is an equation that you're supposed to solve for $\gamma(t)$ but here you need to already know $\gamma(t)$ in order to determine your $v$ by inverting $\mathrm{d}\gamma$
 
Do we need to define curves from $\mathbb{R}$ anyway
Can't we define them via line elements
That way there is no issue of reparametrization
 
I don't know what you mean by "via line elements"
have you read too much Weyl and are now talking old-timey math speak :P
a line element to me is something I can integrate over a curve
 
3:45 PM
A one dimensional distribution, if you prefer :p
 
@glS Okay, so I think this actually works in the sense that it's not circular, just an extremely weird way to write $\dot{\gamma}(t)$
 
aka "the tangent line", which I'm sure Newton would approve of
 
glS
3:59 PM
@ACuriousMind but it kinda does in a trivial way? Define the isomorphism $TI\simeq I\times\mathbb R$ as $\phi:(t,[s\mapsto t+\lambda s])\mapsto (t,\lambda)$ (I'm assuming all elements of $T_t I$ can be represented with such a curve for some $\lambda\in\mathbb R$, hopefully that holds). Then $(t,1)\equiv \phi(t,[s\mapsto t+s])$, and the integral flow equation reads $d\gamma(\phi^{-1}(t,1))=X(\gamma(t))$
idk, maybe I'm stretching it. It it takes too much effort to make it work it might not be a great approach. I'll have a look at the post you pointed out. I wonder if using the cotangent bundle is "just" a way to fix this issue
 
4:16 PM
@glS nah, that post is using the cotangent because it recognizes that $m\dot{q}$ is a momentum and hence should live in the cotangent space
and so since $\dot{q}$ is tangent, $m$ must become a map from the tangent to the cotangent space
 
looks like something is happening here
 
things are always happening
 
whats the tl;dr
 
Newton but with forms
 
glS
@Slereah no forms, only tangent vectors!
I don't like forms
 
4:25 PM
the iterated tangent bundle is a notoriously difficult object to work with explicitly
i like them but it requires care
best to use jets whenever possible
 
@glS Vectors are just forms on the space of forms
 
glS
@Slereah lol is that saying $(V^*)^*\simeq V$?
@BalarkaSen this was the starting point
3 hours ago, by glS
so.. if we write Newton's eqn in "geometric" terms with the differential, we should have $d^2 q=F\circ dq$, assuming a force field that depends on up to first derivatives, and $q:\mathbb R\supset I\to Q$. Then, changing coordinates, thus assuming this to be satisfied for $f\circ q$ for some diffeo $f:Q\to Q$, we'd get $d^2 f\circ d^2 q=F\circ df\circ dq$, and thus in the new coordinates Newton's eqn would read $d^2 q=\underbrace{(d^2(f^{-1})\circ F\circ df)}_{\equiv F'}\circ dq$
 
Very clunky. I advocate writing the subspace defined by $F(p, q) = q''$ of $J^2 M$, the space of 2-jets on $M$.
 
glS
yea, I don't really understand jets
 
@gls something i feel like i should know: suppose i have a particular 2-qubit pure state. is there a standard algorithm for how to generate that state from $|00\rangle$ using unitaries?
 
glS
4:34 PM
@Semiclassical like, taking a unitary whose first column is the state? Or you mean algorithm as in "gate decomposition of the unitary"?
 
gate decomposition
my vague recollection is that the answer is yes, if you're happy with approximations
 
glS
@Semiclassical quantumcomputing.stackexchange.com/a/12548/55 seems like a pretty straightforward approach. Generate a state with same Schmidt decomposition with local unitary plus cnot, and then apply local unitaries to get the target state
 
glS
5:06 PM
anyway, sweeping under the rug the issues with the definition of the inputs to $dq$ and $d(dq)$, the overall equation works fine, if I guess possibly only at a formal level. By which I mean that if $f\circ q$ is the curve solving Newton's equation for $F$, or equivalently $f\circ q$ is the integral curve for $\tilde F$, then $d^2(f^{-1})\circ \tilde F\circ df$ gives the correct expression for the "modified force field" $\tilde F'$, of which $q$ is an integral curve
that is, you get the same final answer, which should be $\ddot q^\eta = \dot q^\mu \dot q^\nu \partial_\mu f^\alpha \partial_\nu f^\beta \partial_\alpha\partial_\beta (f^{-1})^\eta + F^\mu \partial_\mu (f^{-1})^\eta$ (filling out the appropriate arguments for the functions)
which I suppose is also kinda obvious, a posteriori, noticing that $d^2(f^{-1})\circ\tilde F\circ df$ is the pullback of $\tilde F$ through $df$, and so we are saying that if $dq$ is an integral curve for $\tilde F$ then $d(f\circ q)$ is an integral curve for $(df)^* \tilde F$
 
glS
5:34 PM
@ACuriousMind if I'm reading the page correctly, at least some Wikipedia editor agrees with me on this point lol en.wikipedia.org/wiki/…. They explicitly talk about choosing a "canonical" cross-section $i$ of the trivial bundle $TI\simeq I\times\mathbb R$ such that $i(t)=1$ for all $t\in I$.
 
@glS That article makes the same misstep as you did - $TI\cong I\times \mathbb{R}$ is true but there is no unique such isomorphism, so saying "1" does not actually pick out a "canonical" element of $TI$ - just of $I\times\mathbb{R}$.
but I thought about this a bit and I think there's a much simpler characterization of the "1" we're hunting here
it's the tangent vector of the identity map $I\to I$
 
glS
@ACuriousMind mh, wait, how are you modeling tangent vectors now? If you define tangent vectors at $t\in I$ as (equivalence classes of) curves with $\gamma(0)=t$, then the identity map $I\to I$ doesn't send $0\mapsto t$, no?
 
@glS I'm just saying that the identity $I\to I$ is itself a curve in $I$, and that this curve has a tangent vector at every point
i.e. $\mathrm{id} : I\to I, t\mapsto t$ has a $\dot{\mathrm{id}}(t) \in T_t I$, and that's the "1(t)" at every point
 
glS
5:50 PM
@ACuriousMind right, but you should provide one such curve for each $t\in I$, no? doesn't that essentially just give tangent vectors as $[s\mapsto t+s]$, as I was doing before?
 
@glS I'm not talking about the equivalence class of this curve
I'm not using any particular representation of tangent vectors
I'm just saying the identity is a curve in $I$, its image is all of $I$, and a curve has a tangent vector at every point of its image
what we write as $\dot{\gamma}$ usually - for other manifolds, this doesn't give you a vector everywhere because curves don't fill the entire manifold, but for $I$ this curve does
 
glS
well, so the suggestion is then to write the integral curve as $d(q\circ \operatorname{id})=X\circ q\circ \operatorname{id}$?
 
no, you can still write your input as 1 or whatever
this is just a nicer definition of that "1" that doesn't require any choice of coordinates or mucking about with explicit representations of tangent vectors
 
glS
so $dq(t,1(t))=X\circ q$ with $1(t)\equiv \partial_t \mathrm{id}(t)$
I'd be tempted to say that that's the same as I was doing, because if you write down explicitly what $\partial_t \mathrm{id}(t)$ is modeling tangent vectors with curves, you get $[s\mapsto s+t]$.
but I agree this is a cleaner way to see it
 
I agree - this is the same vector as $[s\mapsto s + t]$
 
glS
5:58 PM
so jumping back to the starting point, one can write "neatly" how "force fields" transform through a generic diffeomorphism $f:Q\to Q$ as $\tilde F\to d(df^{-1})\circ\tilde F\circ df$, which I think is nice =P
 
6:45 PM
is there any velocity limit of an inertial system in relation to another one, for which, above this limit value we use the loretz transformations ,whenever we want to jump from one to another, and the galilean ?
 
No you can always use the Lorentz transformation
Whether or not the Galilean one will do only depends on what you're willing to tolerate as error
 
I mean for very small values v/c is nearly zero no?
 
Sure
 
then it means that using the galilean trasformations, doesn't represent high error ?
 
Depends what you consider high
 
6:48 PM
I see
 
A plane's speed is enough to measure a difference on an atomic clock
 
what?
ahaa
atomic clocks, I am not familiar, but I guess what you are trying to say is that
the velocity of plane
does have considerable affect
in time
 
7:12 PM
Well my point is that "considerable" depends on what scale you're considering
Is there a generic name for the structure constructed from the jets of maps of $\mathbb{R}^k$ to $M$
I guess for $k = 0$ it's just the points, then $k = 1$ you get the $k$-th order tangent bundles, and for $k = n$ the $k$-th order frame bundles
What structure is it for $k = 2$
Sounds like it would be some string theory business
Also I used $k$ for both the dimension and the jet order
Sloppy work on my part
 
7:53 PM
@Slereah Taylor series
 
there's no Taylor bundle
 

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