The h Bar

9:13 AM

@ACuriousMind Is the function space of group inclusions $\phi : H \hookrightarrow G$ basically indexed by the quotient space $G / H$

and for every $x \in G/H$ there is a corresponding inclusion $\phi_x$

ACuriousMind

@Slereah unless $H$ is normal in $G$, there is no $G/H$

Slereah

Well let us say there is

For bundle reduction purpose

ACuriousMind

you mean "let us assume $H$ is normal"?

because you can't just assume the existence of a quotient that you can't actually form :P

Slereah

Yes

ACuriousMind

hm, wait

Slereah

9:22 AM

Checking if I understand the reduction $\sim$ section of the associated homogeneous space bit

ACuriousMind

this doesn't even work

Slereah

Although I think the issue is slightly more complicated with bundles because I don't think $G/H$ has to be a group quotient

ACuriousMind

when you write $G/H$, you have fixed one particular embedding of the "abstract $H$"

but normal and non-normal subgroups can be isomorphic, so writing $G/H$ doesn't make a lot of sense

@Slereah you'll have to be more specific what kind of quotient it is, then

Slereah

A very good question

ACuriousMind

I mean, if you define $G/H$ as the space of distinct inclusions $H\to G$, then there's no problem, but that's also not what we usually mean by a quotient

take $H=G$, then this just looking for group automorphisms

but we would expect $G/G$ to be trivial in any "nice" notion of quotient, so I think you're on the wrong track here

Slereah

9:37 AM

Ah it's not a quotient, but a coset space

Except they use the same notation because why not

ACuriousMind

well, if $H$ is normal then the left and right cosets are exactly the quotient :P

so it's acceptable notation

but what I said above about $G=H$ is still true, I don't think this classifies inclusions without further constraints

Slereah

Hm

How does the reduction correspond to a section on the coset space, then?

ie

If I have let's say the frame bundle $GL(n)$ and I pick a section of $\mathbb{Z}_2 = GL(n) / GL^+(n)$, does that not pick an inclusion of $GL^+(n)$ in either the right or left handed set of frames?

ACuriousMind

@Slereah the cosets are not subgroups (except for the coset associated with the identity), so while this picks an inclusion of sets it does not pick an inclusion of groups

Slereah

So does it span the inclusions as a function of sets?

ACuriousMind

no, of course not

as sets there are much more inclusions

Slereah

9:50 AM

Dagnabbit

ACuriousMind

and of course the inclusions of groups that there are are also inclusions of sets

to see that this doesn't classify inclusions of sets, just consider that a set with n elements has $n!$ possible bijections to itself, but again we would arrive at $G/G = 1$

Slereah

Does the coset space classify anything related to the inclusions in some sense?

Or am I on the wrong track

It seems somewhat important since it's the space that generates the actual structures in GR

ACuriousMind

I think you're entirely on the wrong track :P

Slereah

What does the cross section of that bundle imply?

The existence implies that the reduction exists, but what do specific cross sections imply as far as that reduction goes?

I would guess that different metrics of $GL(n)/O(n)$ select different equivalence classes of frames

ACuriousMind

10:07 AM

I think what confuses you is the following statement: The possible reductions of a $G$-principal bundle to a $H$-principle bundle are given by global sections of a $G/H$-bundle.

Slereah

It certainly does

ACuriousMind

This has nothing to do with inclusions $H\to G$ directly, but simply with how the reduction works on the level of the transition functions

Slereah

Doesn't it imply anything on a local level?

ACuriousMind

sure, locally it tells you the transition functions

Slereah

I mean point-wise, not on a neighbourhood

ACuriousMind

10:09 AM

the requirement that what you do locally to reduce the bundle must be globally consistent gives you in the end the requirement that there be a global section

@Slereah I don't really know what you mean by "pointwise"

Slereah

for $p \in M$ there is some value in the fiber $G / H$

ACuriousMind

sure, but as I said that value comes essentially from thinking about transition functions

the value of a transition function at a single point means nothing

Slereah

I see

Slereah

10:47 AM

One thing I wonder sometimes

Is picking a left or right action actually important

Or do you just pick one and keep it consistent

Slereah

11:01 AM

Sometimes a definition will just say "It has a left/right action", and I am like sure

satan 29

11:28 AM

whats the deal with so many users having names like "leftOverMonica" or "reinstateMonica"
..?

ACuriousMind

@satan29 it's a reference to the events surrounding judaism.meta.stackexchange.com/q/5193, just search Meta Stack Exchange or the wider internet for more info

glS

4:10 PM

so, ncatlab [writes](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Liouville-Poincar%C3%A9+1-form) that the canonical one-form $\theta\in\Omega(T^* X)$ is defined as the unique such form such that $\sigma^*\theta=j(\sigma)$, for any smooth section $\sigma\in\Gamma(T^* X)$, and with $j$ the isomorphism $j:\Gamma(T^* X)\to \Omega^1(X)$.

Anyone understand why this makes sense? For one, I thought differential one-forms were *defined* as smooth sections of the cotangent bundle, so wouldn't $\Gamma(T^* X)=\Omega^1(X)$? And then, $\sigma:X\to T^* X$ and $\theta:T^*X\to T^* T^* X$, so $\s…

I feel like I'm missing something obvious here

ah, wait, I got it. They mean the isomorphism connecting the two ways to understand 1-forms: as smooth sections $X\to T^* X$, or as smooth maps $TX\to\mathbb R$. So $\theta:TT^*X\to\mathbb R$, $\sigma:X\to T^* X$, $\sigma^* \theta=\theta\circ d\sigma:TQ\to \mathbb R$, and $j(\sigma):TQ\to\mathbb R$, so it checks out

ACuriousMind

yes, exactly

Slereah

It's the secret behind why Lagrangian mechanics is awful

the jet space v. the regular space

ACuriousMind

that's where the name "tautological" for this one-form comes from - if you choose to define the 1-forms as sections, then you can omit $j$ and you just get $\sigma^\ast \theta = \sigma$

glS

is it standard to understand $\Omega^1(X)$ this way though? I don't see them using that definition when introducing differential forms. I've seen this switch done "silently" often, but I always assumed it was just some abuse of notation used to simplify expressions

Slereah

wikipedia does a decent demonstration of it

Jet bundle

In differential topology, the jet bundle is a certain construction that makes a new smooth fiber bundle out of a given smooth fiber bundle. It makes it possible to write differential equations on sections of a fiber bundle in an invariant form. Jets may also be seen as the coordinate free versions of Taylor expansions. Historically, jet bundles are attributed to Charles Ehresmann, and were an advance on the method (prolongation) of Élie Cartan, of dealing geometrically with higher derivatives, by imposing differential form conditions on newly introduced formal variables. Jet bundles are sometimes...

glS

4:24 PM

@Slereah wait, is the $j$ in jets the same one used here?

ACuriousMind

@glS well, because $j$ is an isomorphism and because there is no consensus on which of the two sides really is "the" definition, whether or not you're abusing notation, "omitting $j$" or just doing the correct thing is a matter of viewpoint in any given situation

@Slereah what does any of this have to do with jets?

the tautological 1-form is just a nice thing that appears when you think too hard about Hamiltonian mechanics :P

@glS the nLab article is a bit weird in that it seems to assume it's clear that we define differential forms via their action on the tangent bundle, but you're just equally weird in assuming that it's clear we define differential forms as sections of the cotangent bundle ;)

glS

@ACuriousMind in my defense, that's how I've mostly seen them defined? Even though they then often end up using the $TX\to\mathbb R$ definition when actually calculating things

I'm actually happy to have finally seen the mapping between the two notations explicitly written down (or at least, its existence acknowledged)

@ACuriousMind the tautological one-form is something I've been trying to get a good grip on for some time. Though every time I end up getting lost in some tangentially needed definition or some corner of ncatlab. I'll get there eventually

the fact that it seems like every source writes it differently doesn't help lol

Slereah

@ACuriousMind I may be thinking of a different canonical form

Too many things called the canonical X

glS

is there an easy way to see why the definition $\theta(\alpha)=\alpha\circ d\pi\big|_{T_\alpha T^* M}$, where $\pi:T^*M\to M$ is the canonical bundle projection and $\alpha:TM\to\mathbb R$, is consistent with the local expression one often sees as $\theta=\sum_i p_i dq^i$?

ACuriousMind

4:58 PM

@glS where have you seen that definition?

glS

@ACuriousMind you mean the coordinate-independent one? It's on wikipedia for example en.wikipedia.org/wiki/Tautological_one-form

Tautological one-form

In mathematics, the tautological one-form is a special 1-form defined on the cotangent bundle T ∗ Q {\displaystyle T^{*}Q} of a manifold Q . {\displaystyle Q.} In physics, it is used to create a correspondence between the velocity of a point in a mechanical system and its momentum, thus providing a bridge between Lagrangian mechanics with Hamiltonian mechanics (on the manifold Q {\displaystyle Q} )....

ACuriousMind

I don't see what the definition there has to do with what you wrote down

what's $T_\alpha T^\ast M$?

why is $\theta$ eating a function $TM\to \mathbb{R}$ and not a vector field on $T^\ast M$?

glS

it's in the linked section of the wiki page. They write it as $\theta_m=m\circ d\pi_m$, where $m:T_q Q\to\mathbb R$

ACuriousMind

I'm saying I have no idea what your notation means, but I wholly agree with the definition on Wiki :P

bolbteppa

That's a lot of work just to take a derivative

glS

5:04 PM

@ACuriousMind I'm probably mixing the two ways to define differential forms, cue previous discussion. In my mind, a differential form on the cotangent bundle is a map $\theta: T^*M\to T^* T^* M$, and $\alpha\in T^* M$, so that $\theta(\alpha)\in T^* T^* M$, and thus $\theta(\alpha): T(T^* M)\to \mathbb R$

ACuriousMind

this is already getting extremely confusing because you've been using $M$ for the base space but Wiki uses $Q$ for the base and $M=T^\ast Q$

glS

@ACuriousMind sorry, that's my fault. Yes the wiki is using $M=T^* Q$. I'm just using $M$ as base space (I kinda find it confusing to use $M$ as they do)

ACuriousMind

here's how Wiki's definition relates to the local expression: At $m=(q,p)$, $\mathrm{d}\pi$ projects out any $\partial_p$ and keeps only $\partial_q$. And the map that "$m$ represents" is just applying the co-tangent vector $p$ to the resulting vector on $Q$. If you take a vector field $\alpha = \sum_i \alpha_q^i\partial_{q^i} + \sum_i \alpha_p^i \partial_{p_i}$ and apply the local expression for $\theta$ to it, the result is $\sum_i p_i \alpha_q^i$

which is the same you get from applying $p$ to the "$q$-part" of $\alpha$ that's kept after applying $\mathrm{d}\pi$ to it.

glS

@ACuriousMind $\pi:T^* M\to M$, so $d\pi:TT^* M\to TM$, and $d\pi\big|_{T_\alpha T^* M}$ is the restriction of the map to the fiber in $TT^*M$ over $\alpha\in T^* M$. Is this a weird notation?

ACuriousMind

oh, I see, the part of your notation that's weird is writing $\theta(\alpha)$ for "the value of $\theta$ at the point $\alpha$"

when I see $\theta(\alpha)$ for $\theta$ some 1-form, I always assume first that $\alpha$ is a vector field being fed to the form

glS

5:09 PM

@ACuriousMind but isn't that what a section of the bundle $T^* (T^*M)\to T^* M$ should do?

ACuriousMind

feeding a point to it is $\theta_\alpha$, like Wiki does with $\theta_m$

@glS yes, but that's not how people usually write it :P

we usually don't put this emphasis on them being sections

anyway, just notational differences then, and my reply above about the local expression applies, regardless of which notation you use :P

glS

@ACuriousMind so, in that notation, $d\pi_m (\alpha)=\alpha^i_q \partial_{q^i}$ and $m=p^i \partial_{q^i}$ (which is shorthand for $m_q=p^i \, \partial_{q^i}\big|_q$ I suppose), and thus $\theta_m (\alpha) = p^i \alpha^i_q(q)$, where $q=\pi(m)$

ACuriousMind

why is your $m$ expanded in $\partial_q$s?

glS

@ACuriousMind right. Should be $m_q=p^i \, \mathrm dq_i\big|_q$?

ACuriousMind

it's a point in the cotangent bundle, and so $m = (q,p)$, with $p = p_i\mathrm{d}q^i\in T^\ast_q Q$, and this $p$ is the map they mean

glS

5:21 PM

ok, yes, that's what I was trying to write lol

so, the idea is that $\theta$ takes curves in $T^* Q$ passing through $m$, and looks at the increments of the curves when projected on the base space, and multiplies that with the "height" (i.e. the output values) of $m$ iself?

so kinda, just uses $m$, which normally eats curves in $Q$ (modeling tangent space with curves), only making it work on curves on the larger space $T^* Q$, but "forgetting" about its "vertical" structure somehow?

ACuriousMind

I would just say "it takes velocities $\partial_{q^i}$ and replaces them by momenta $p_i$"

that's the idea, all the rest is technicality

glS

@ACuriousMind but isn't that just what $m$ does?

ACuriousMind

$m$ is just a point in the cotangent bundle, it does that just at a single base point $q$

$\theta$ is the thing that does this "globally"

glS

@ACuriousMind is it that simple? When you say "replace velocities with momenta", you are assuming to replace velocities with some values of the momenta. But $\theta$ takes cotangent vectors and sends them to the map replacing velocities with the corresponding momenta?

I mean, $\theta$ can be fed with any covector $m$, so it can give us any "replacing velocities with momenta" map

ACuriousMind

5:39 PM

Look again what it does to our vector field $\alpha$ up there: It forgets about the $\partial_p$ part, and then it takes the $\alpha_q^i\partial_{q^i}$ and just replaces the $\partial_{q^i}$ by $p_i$

it doesn't replace the velocities with any specific momentum - the result is a scalar function on $T^\ast M$

it's action at any specific point $m=(q,p)$ is to replace the velocities at $q$ by the momenta $p_i$

glS

6:04 PM

@ACuriousMind hang on. So now we are using a third way to understand what "differential form" means no? A "differential one-form on $Q$" could mean (1) a section $Q\to T^* Q$, (2) a map $TQ\to\mathbb R$, or (3) a map $\Gamma(TQ)\to C(Q,\mathbb R)$, which is what you now mean? Takes a vector field and gives a functional

except we are doing it on $M\equiv T^* Q$, so we have a map $\theta:\Gamma(TT^* Q)\to C(T^* Q,\mathbb R)$,

ACuriousMind

I don't really think of (2) and (3) as distinct, but you're technically right

what's happening here is very similar to currying

you can think about a 1-form as having "2 slots" - one that accepts points and one that accepts vectors

depending on which slots you're filling or not, your view on what a 1-form is changes

note that the operation that goes from (2) to (3) is just concatenation - a vector field is a section $Q\to TQ$, and so we can just concatenate it with our map $TQ\to \mathbb{R}$ to get a scalar $Q\to\mathbb{R}$

glS

@ACuriousMind that makes sense. Definition (1) fits in there too I'd say no? With (1) we isolate base points, and get functionals on tangent vectors. With (3) we (effectively) isolate tangent vectors, and get functionals on base points. With (2) we directly assign numbers to each pair of basepoint+tangent vector

ACuriousMind

yes

glS

so, going for example by the second definition, as a map $TQ\to\mathbb R$, which in our case is $TM\equiv TT^* Q\to\mathbb R$, we understand $\theta$ as a map sending $(m,\alpha)\in T_m T^* Q$ to $m_2\cdot \alpha_1\in\mathbb R$ (where with $m_2$ I mean something like "second coordinate of $m$ etc), i.e. $p_i\alpha^i_q$

nice

or going with (3) above, we get a map sending each "velocity" $\alpha$ to a map $T^*Q\to\mathbb R:m\mapsto \theta_m(\alpha)$, which we like to think of as "momentum"

glS

6:31 PM

so.... to connect these definitions above, given $\alpha:Q\to T^*Q$ 1form as a section, $j(\alpha):TQ\to\mathbb R$, and $\ell(\alpha):\Gamma(TQ)\to C^\infty(Q,\mathbb R)$, we have $$\color{red}{(}\alpha(q)\color{red}{)}(X(q))=\color{red}{(}j(\alpha)\color{red}{)}(X(q))=\underbrace{\color{red}{(}(\ell(\alpha))(X) \color{red}{)}}_{\equiv \alpha(X)}(q)\in\mathbb R,$$ where $q\in Q$, $X\in\Gamma(TQ)$, $X(q)\in T_q Q$. This is evil lol

Slereah

6:44 PM

You know what I hate

The electroweak interaction

We have unified the $SU(2)$ gauge and the $U(1)$ gauge into an $SU(2) \times U(1)$ gauge!

ACuriousMind

the deceptive thing about that is that the unbroken EM $\mathrm{U}(1)$ isn't actually the right-hand factor in $\mathrm{SU}(2)\times\mathrm{U}(1)$

it sits a bit strangely in there, and that's why you can't usefully think about a separate SU(2) and U(1) above the unification scale

Slereah

I know :p

Slereah

7:10 PM

"To see explicit examples of the abstract formalism that follows, one may want to glance from time to time at the examples of Sect. 3."

Can't they put the examples during the explanation

More Anonymous

7:55 PM

Ummm ... So basically we need to displace the Hamiltonian density by say $c$ and not the Hamiltonian by $c$ to make the connection to the cosmological constant?

How does one go to the cosmological constant from this? One cannot. $c$ will only displace Hamiltonian by a constant factor. But we can't say that $c$ will also displace Hamiltonian density by a constant factor. Cosmological constant should be viewed as a space time density. — KP99 1 hour ago

bolbteppa

8:24 PM

Cosmological constant is non-trivial in curved space where the factor is $\sqrt{-g} \Lambda$ not just $\Lambda$

More Anonymous

Doesnt the $\sqrt{-g} $ come from making it a volume integral?

bolbteppa

If the metric is constant, $\sqrt{-g} \Lambda$ is constant so the integral in the Lagrangian can be ignored, otherwise it can't be ignored it's a non-trivial contribution because of the dynamical $\sqrt{-g}$ part, which is needed for making a coordinate invariant volume integral

More Anonymous

So I cant add a term to this stress energy tensor and the spacetime around it?

$ T^{\mu\nu} = \sum_{n} \frac{p_n^\mu p_n^\nu}{p_n^0} \delta^3(\vec{x} - \vec{x}_n(t)) $

More Anonymous

9:00 PM

Hmm ... Okay ignore me I think it makes sense

rb3652

9:14 PM

Hey folks, quick question.

Can I ask a question about optics here?

ACuriousMind

sure

rb3652

If a beam of light comes back to the observer, will the observer see his/her own reflection?

ACuriousMind

I mean, how else would seeing one's reflection work?

rb3652

Something like this:

@ACuriousMind Ah, ok.

I was thinking of a funhouse-mirror kind of scenario

Or a room of mirrors

In a room of mirrors, then, why does every line of sight land on your own reflection? Shouldn't only certain light paths come back to themselves?

Not sure if my question makes sense, but I'm open to clarifying.

ACuriousMind

every different reflection you see in such a scenario corresponds to a distinct path the light can take from you to the mirrors and then back to you

if "every direction" has your reflection, then someone simply engineered the room such that there is such a path in every direction

rb3652

9:19 PM

So, to clarify, every reflection you might see in a hall of mirrors (for instance) is a beam of light "coming back to where it originated from"

ACuriousMind

yes

or, well

rb3652

Wow

That's powerful

That's really powerful

ACuriousMind

you don't really radiate light, so the light does not "originate" from you

rb3652

I see

ACuriousMind

you're just reflecting light from some other light source

but the idea is the same

rb3652

9:20 PM

@ACuriousMind Thank you! You answered my question!

imbAF

for a qm. harmonic oschillation the energy is $E= \hbar \omega (n + frac 1 2)$. Now, if $\omega$ is constant, what do we observe when $n$ gets bigger values?

ACuriousMind

uh, we observe higher energies? :P

imbAF

yes

which physically translates to what?

more oschillations over time?

with the given constant frequency ?

ACuriousMind

not really

imbAF

if $\omega$

is constant

ACuriousMind

9:33 PM

what is even "more oscillations over time with constant frequency" supposed to mean?

imbAF

how, what changes physically that implies higher energy?

ACuriousMind

it's a quantum system

you don't get some ontological description of "what" it "is"

imbAF

look

ACuriousMind

if you draw the wavefunction, you'll see it gets more nodes, i.e. looks like higher "overtones" the higher the energy gets

imbAF

but more nodes doesn't imply a different value of $k$ ?

higher value

ACuriousMind

9:35 PM

what's k

imbAF

wavenumber

ACuriousMind

wavenumber of what

imbAF

in a line with N atomes

a solid

ACuriousMind

that's not a single harmonic oscillator

imbAF

thats multiples

but the idea behind what I need is the same

For example in a solid with N particles in 3D, you have 3N modes. In a mode you have the energy $e_j=\hbar \omega_j(n_j + frac 1 2)$ where $j$ implies a mode. Now $n_j$ is the number of phonons. You can consider an oschillation through the solid, as a phonon with a specific $\vec k$ and polarisation

therefore $n_j$ is the nr. of phonons, in other words the number of oschillations

that propagate in the solid

ACuriousMind

9:45 PM

yes, but there the $n_j$ means something completely different from the $n$ for a single oscillator!

you're saying you have $n_j$ times the first excited state

imbAF

I am considering multiple non interacting oschillators

ACuriousMind

while for a single oscillator, you just have the $n$-th excited state once

imbAF

yes

but if we agree to have $omega$ constant, even for one oschillator

while $n$ changes values

first of all is that possible ?

ACuriousMind

I don't understand what you're saying, sorry

imbAF

ok

if you have two identical oscillators, in the same exact mode, but with different $n$ values, how can you tell that one has bigger $n$ value then the other

?

fqq

10:00 PM

you measure the energy?

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