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1:47 AM
Hi guys, I have a question regarding the Levi-Civita tensor, $\epsilon_{abcd}=\sqrt{-g}\hat{\epsilon}_{abcd}$, where $\hat{\epsilon}_{abcd}$ is the Levi-Civita symbol.
I want to show that $\nabla_f \epsilon_{abcd}=0$, and I have been given a hint to go to an inertial coordinate system where the christoffel symbols vanish. I know that this can always be done at a point, and I am hence able to show that $\nabla_f \epsilon_{abcd}=0$ at this point, but I don't know how to conclude that it holds globally.
Clearly $\nabla_f \epsilon_{abcd}=0$ is itself a tensor, and transforms like one, but I don't think the vanishing of a tensor at a single point of a manifold allows you to conclude that it vanishes everywhere (although the "covariant taylor series" on p.g. 39 of this paper suggests it does sciencedirect.com/science/article/pii/0370157385901486)
How can I finish the argument ?
3 hours later…
5:11 AM
@EmilioPisanty Haha, just because I like Git doesn't mean I'm happy that other tools are depricated.
@AaronStevens If I ever saw that I'd just edit it.
1 hour later…
6:28 AM
@SigmaAlpha Just use the product properties of the Levi Civitta symbols
$$\varepsilon_{abcd} \varepsilon^{abcd} = -24$$
Or use the defininition
The Levi Civita is defined by a sum of products of kronecker deltas
Kronecker deltas can be written as metric tensor contractions
6:41 AM
@Slereah, that would work for the symbol $\hat{\epsilon}_{abcd}$, but I want to show it for the tensor, which is the symbol multiplied by $\sqrt{-g}$.
I have proven it using a different method. But I have been told specifically to use the hint of going into a locally inertial frame, and so I would like to know how to do it this way
Well, if you feel like it
You can also use the tensor definition of $g$
The important part is that every definition of $\varepsilon$ and $g$ is entirely defined by the metric tensor
Therefore their covariant derivative can very easily shown to be 0
7:15 AM
OK, Thank you @Slereah
2 hours later…
8:53 AM
You know I think the best page to exemplify what nlab is like is to look up how numbers are defined
Well technically these are natural number objects
Just objects that act like numbers
But still fun
What's a good definition of Noether's theorem
Somewhere between Wikipedia's handwave and nlab's "consider the variational bicomplex of the infinite jetbundle of the action functional..."
1 hour later…
10:23 AM
@DanielSank yeah, I know ;-)
still, I'm pretty annoyed that the standard is cementing this quickly around git, given that there still aren't any good GUI solutions
(by which I mean: free to use (ideally FOSS), stable, usable, intuitive, cross-platform GUIs.)
10:49 AM
@DanielSank I always do
11:22 AM
My lecturer defined the Heisenberg algebra as the three-dimensional complex Lie algebra, generated by x,p and h-bar (I'll just denote by h), such that [x,p]=ih, and [x,h]=[p,h]=0.

He then described a particles spin in the x,y,z directions as corresponding to what amounts to the direct sum of three copies of this Heisenberg algebra, where we identify the three h-bars that we obtain.

Then, he defines some quantised angular momentum operators from these J_x,J_y,J_z, and takes certain linear combinations of these to construct apparently the generators for sl(2). However, he defines e,f,h for
I can go into more details (define all the operators) if necessary, but maybe this is a general thing that one does
12:15 PM
Slereah, vzn and Danielsank may be interested
Long story short: A Bell theorem which demonstrate how under a sufficiently large set of probability theories including quantum mechanics, can cause a product state to become entangled when gravitationally time dilated near a superposition of mass
Side note: One cannot do time travel with this because the measurement that establishes the correlation happens after the clocks became correlated. In particular, the events A and B are timeline separated regardless of ordering
12:43 PM
@NovaliumCompany why do you need 1500v? I thought e papers where meant to be super low voltage?
@MathCubes but we already know the universe is bigger than the observable universe...
1:24 PM
@JackDon If you can type it up in a more focused way, I would suggest asking your question on the main site
1 hour later…
2:36 PM
Q: Is there a specific structure to the automorphism set of a field theory with respect to internal v. spacetime symmetries?

SlereahI'm trying to work out what it means exactly for a field to be transformed, without referring to gauges for now. As far as I can tell, from a rigorous perspective, a transformation of the field is an automorphism of the relevant vector bundle, ie, for a vector bundle $$\pi : E \to M$$ A transf...

plz help I am very dumb
1 hour later…
3:40 PM
@JackDon the angular momentum operator $\mathbf{J} = \mathbf{r} \times \mathbf{p} = - i \hbar \mathbf{r} \times \nabla$ has the $\hbar$ constant in it yet the dimensional quantity $\hbar$ should not be in there if you're talking about $sl(2,\mathbb{R})$ or $su(2)$
i.e. on dividing out $\hbar$ the $\mathbf{J}$ operator combinations give the $sl(2,\mathbb{R})$ generators
4:25 PM
thinking about it
I'm pretty sure that the two symmetries don't span the whole space
ie for a point particle in 1D, the vector bundle is $\mathbb{R}^2$, and I don't think two copies of $\text{Diff}(\mathbb{R})$ make up $\text{Diff}(\mathbb{R}^2)$
Q: How to place a bounty on a slight modification of a question that is not mine?

frogeyedpeasI have been thinking a lot about the question: If a mass moves close to the speed of light, does it turn into a black hole? But I didn't find the answers satisfactory. I have two options: Ask the question "WHAT happens if a mass moves close to the speed of light", cite this question, and then ...

4:45 PM
@EmilioPisanty It's all because of GitHub, I think.
Thinking about it
The internal symmetries can just arbitrarily change every value of the bundle
Although... no, that wouldn't work for every function
@ACuriousMind you're a smarty pants
What do you think
string theorist to the rescue!
@Slereah Indeed, $\mathrm{Diff}(M\times N)$ is not isomorphic to $\mathrm{Diff}(M)\times\mathrm{Diff}(N)$.
Yeah I'm guessing so
Do you think those two subspaces of the automorphism group have any interesting properties?
This already fails for linear maps on vector spaces - rotations on $\mathbb{R}^2$ are not obtained from individual linear maps on the 2 $\mathbb{R}$s!
4:56 PM
@ACuriousMind true!
I mean I guess at least they contain all the interesting symmetries, since I think the Coleman-Mandula ones live in there
Also somewhat related
I'd say that the more "general" automorphisms are unnatural in physics, since either your bundle is a tensor bundle and has its transformations naturally induced by diffeomorphisms, or it is a gauge bundle and the transformations on the fibers shouldn't mix with spacetime transformations at all.
If within the set of vertical automorphisms, there is a subgroup
Can we associate a principal bundle to it?
@ACuriousMind I'm kinda trying to approach it the other way here
First start out with generic transformation
And then see which ones leave the action invariant
Well I suppose it would need to be a Lie subgroup, 'course
I really need to get that book on the standard model from bundle stuff
it seems to contain that kind of business
@Slereah That's probably going to be difficult since there's no generic version of Coleman-Mandula for arbitrary classical theories afaik
Well I'm not planning on doing it in general
Although I may try it for the 1D point particle
That seems doable
$\mathbb{R}$ bundle on $\mathbb{R}$
Find out which $\text{Diff}(\mathbb{R}^2)$ leave $\dot{x}$ invariant
Hopefully it's $\mathbb{R}^2 \times \mathbb{Z}_2$
string theorist saves the day again :D
5:05 PM
that's point theory!
I wonder who came up with the idea that point particles can be field theories in 1D
God bless his heart
you need to understand points before you can graduate to strings
The best one of course is $0$-dimensional quantum field theory
@ACuriousMind would you like to be called string-man or string-theory-man? Or maybe ABAPman or COBOLman... hmmm
@Slereah If your bundle is a vector bundle of rank $n$, you always have a corresponding $\mathrm{GL}(n)$ principal bundle (the "frame bundle")
Whether you can reduce that to a subgroup of GL(n) depends on the existence of a global section and is typically obstructed by topological properties of the base, e.g. orientability for SO, or spin structure for Spin
What about, let's say, a complex bundle $\mathbb{C}$
Is there a possibility of a $U(1)$ bundle
That sort of business
Oh I do remember that the diffeomorphism group is roughly equivalent to a choice of a section of the $GL$ bundle, right?
or is that physicist fancy talk
5:15 PM
@Slereah U(n) bundles for complex vector bundles always exist, cf. math.stackexchange.com/questions/2410268/…
@Slereah Well, a section of the GL-bundle basically gives you a Jacobian at every point, but it is not the same as a diffeomorphism
Is there some isomorphism between the two, tho?
Or are diffeomorphisms larger
or something
The physicist's failure to properly distinguish the two notions is the root cause of the popular misconception that GR is special because it is "diffeomorphism invariant"
Well I don't buy that anyway because special relativity is also diffeomorphism invariant
given the proper context
5:18 PM
@Slereah The diffeomorphisms are larger - consider that translations on $\mathbb{R}^n$ all have the same Jacobian.
It ain't
I think I'm experiencing semantic satiation for the word morphism
They would transform the Minkowski metric
@ACuriousMind is it some quotient?
5:19 PM
@bolbteppa into another flat metric
Still SR
@Slereah Are the first derivatives of functions on $\mathbb{R}$ some "quotient" of functions on $\mathbb{R}$? :P
I shouldn't have said that the diffeomorphisms are larger
SR only works in Cartesian coordinates. #cartesianmastercoordinates
@ACuriousMind Well it's not wrong!
So I guess it's more that a diffeomorphism induces some infinitesimal transformation which is ~ the GL section?
Or something
@Slereah It is wrong: there are GL(n)-sections that are not Jacobians - on compact manifolds, only those with integral 1 are: mathoverflow.net/a/82590
So no great relations between the two
I wonder why so many physics books go on about it
5:28 PM
Because for soldered bundles it is natural to look at the vertical transformations as induced by the diffeomorphisms of the base.
I know all those words, but
It ain't easy
yo bro, you need 3 m's at the end of hmmm
I'm not a pervert
@enumaris what about 'mhmm'?
5:30 PM
@ACuriousMind you gotta do the head motion
::does the head motion::
I suppose that the diffeomorphism acts on the coordinate base by rotating it so
is that right
I'm not sure what you mean by that
Trying to decrypt what you said
5:33 PM
I'm afraid from where I'm standing that's more like encryption :D
homomorphic encryption?
Given some section of the bundle of linear frame, the diffeomorphism induces a rotation of those, itself a section of the bundle of linear frames?
@Slereah Yes, that's it (the matrix of the "rotation" is the Jacobian)
ah yes, there we go
I wonder if finding the 1D free particle symmetries from that process is doable, rly
Fairly small automorphism group and symmetry group, but still
Maybe I can show that it must be an affine transformation and that would simplify things
although if that is so, I can do it for arbitrarily many dimensions
What would be the automorphism for the 1D free particle, anyway
$\text{Diff}(\mathbb{R}) \times \text{Diff}(\mathbb{R})$ would be $x(t) \to f(x(g(t))$, but what would be $\text{Diff}(\mathbb{R}^2)$
Oh is it like... $x(t) \to f(x(g(x,t), t)$
Or something
I guess writing them as function doesn't help here
I wonder if there's a good argument for using only $\text{Diff}(V) \times \text{Diff}(M)$ instead of $\text{Diff}(V \times M)$ because that would certainly help
I'm not sure what kind of wonky ass action would have such an invariant
6:07 PM
hey guys, whats up?
the world
@Slereah You're using more than just the product - the vertical transformations consist of one automorphism of the fiber for every point, which is a function $g : M -> \mathrm{Aut}(V)$
Also, note that you're also not using the full $\mathrm{Diff}(V)$ - e.g. for a vector bundle you only use $\mathrm{Aut}(V) = \mathrm{GL}(V)$
Oh the automorphisms have to respect the vector space?
Well that does help certainly
Well, they have to respect whatever structure you have put on your theory
so what does the automorphism look like?
in coordinate form
6:13 PM
If this $V$ is a representation space for some gauge group $G$, then you have just a function $M\to G$
@Slereah I'm not sure I understand the question. E.g. for the GL(n) transformation induced on the tangent bundle by a diffeomorphism, it's still the Jacobian, but for a gauge theory you would have a collection of local functions into the gauge group (with "gluing constraints" on the overlaps of the local patches)
I mean if I have said section $x(t)$, what is the action of an automorphism on it?
although someone answered on the math channel
hopefully correct
in Mathematics, 50 secs ago, by Mike Miller
$\text{Diff}(\Bbb R) \times C^\infty(\Bbb R, \Bbb R^\times)$, with $(\varphi, f)$ acting on $\Bbb R^2$ by $(\varphi,f)(t,x) = (\varphi(t), f(t) \cdot x)$. The fact that you can define an automorphism as a map to $\Bbb R^\times$, instead of as a section of some bundle, is because the automorphism bundle of a line bundle is canonically trivial --- the only automorphisms of a line are scaling.
@Slereah Uhhh...just by concatenation? :P I mean an automorphism is by definition a map from the bundle to itself, and the section is a bundle-valued function, so what exactly are you looking for here?
You cannot, in general, talk about these bundle things in a single set of coordinates
Either your bundle is always trivial as the direct product of base and fiber, in which case it is completely unnecessary to speak of bundles at all, or it can be non-trivial and you need to use collections of local patches
E.g. you cannot write down a vertical transformation of a generic bundle in single coordinates, but instead you get a collection of functions $g_i : U_i \to G$ for subsets $U_i$ of the base where the $g_i$ are related on all $U_i\cap U_j$ by a condition similar to "they have to be gauge transformations of each other".
How accurate would a DIY spectrometer like this be? After I calibrate it with known lasers.
@ACuriousMind Well I'd prefer to start with something simple to start with
Also, I was thinking to modify it as to analyze the spectrum of a distant (say 5 meter away) surface. Would there be any obvious obstacles in doing so?
6:25 PM
@Slereah I'm just cautioning against getting too used to thinking of bundles as merely a fancy way to speak about the direct products.
@ACuriousMind well it is locally fine :p
I know it's not the same but still
Thinking about it
You could make a non-trivial example
By taking the moebius strip spacetime
But that sounds like a bad idea
@Slereah Yes, and if you keep saying that to yourself you end up as the physicists who'd rather talk about "gauge transformations at infinity" instead of talking about bundles. :P
@ACuriousMind noooooo
Haunt me no more spirit
6:39 PM
gosh dang it
why is it so hard to find a leaky integrate and fire model of the neuron with the same conventions as the hodgkin-huxley model...
@ACuriousMind does that math answer seem correct to you, btw
@Slereah Yes (but when in doubt, you should trust Mike more than me anyway :P)
in Mathematics, 30 mins ago, by Mike Miller
Your section is irrelevant, but if you want to know how this acts on a section $(t, x(t))$, you just plug into the formula for the action on the bundle: the new section is $(\varphi, f)(t, x(t)) = (\varphi(t), f(x(t)) \cdot x(t))$. This is not written in the correct form (the right should be written as a function of $s = \varphi(t)$, probably), but you may rewrite it as $$(s, f(x(\varphi^{-1}(s)) \cdot x(\varphi^{-1}(s))$$ if you want
Is mike a wizard
He's a (mathematical) gauge theorist, this is exactly the stuff he does
$f$ is an element of the general linear group for that vector space right?
@ACuriousMind Got lucky there
If he was a gardener it might have been harder
6:52 PM
@Slereah Well, the value of $f$ is, yes. His $C^\infty(\mathbb{R},\mathbb{R}^\times)$ is the space of maps $M\to \mathrm{Aut}(V)$ for this particular example
Ah yes
I really need to focus on something, really
my website has like 55 draft articles and none finished
Some of them finish mid-sentence
2 hours later…
9:08 PM
Q: Why did all the comments on my answer vanish so quickly?

RuslanI had a couple of comments under my answer (this one), and was about to reply to a comment of the question author when I noticed that the author had accepted the answer and deleted his account. Then, after I hit the Add Comment button to submit the reply, I noticed that all the comments vanished....

10:52 PM
@Slereah let me know when you find the magical solution
11:10 PM
Cubics aren't meant to be solved unless they're trivially factorable

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