The h Bar

lucabtz

12:32 AM

@Mr.Feynman should put a measure upside down first

@SillyGoose (i) and (ii) are equivalent

Also the connection is just the choice of the horizontal, the vertical you always have already

Silly Goose

1:00 AM

@lucabtz oh

do you happen to know of a resource that shows that (10.4) and the $H_pP$ in definition $(10.1)$ are equivalent?

lucabtz

@SillyGoose yeah I studied this stuff on Loring Tu book. He explains how to obtain omega if you start from the horizontal distribution and vice versa

Silly Goose

is it a springer text on differential geometry?

lucabtz

Loring W. Tu. Differential Geometry: Connections, Curvature and Characteristic
classes. Springer, 2018. isbn: 9783319855622

Silly Goose

it is a little unfortunate that it seems nakahara's definitions do not necessarily jive well with a mathematics textsbook

okay thank you

lucabtz

Look around theorem 28.5

Silly Goose

1:12 AM

oh perfecct thank you

does the condition $C = A \oplus B$ imply that $A \cap B = \{0\}$? where $A, B, C$ are vector spaces

some sources state that two vector spaces are complementary in $C$ when $A \oplus B = C$. others include the condition that $A \cap B =\{0\}$

okay i think i see. What Nakahara calls a connection Tu calls a horizontal distribution. Then, Tu defines a natural $\mathfrak{g}$-valued 1-form from the data of a horizontal distribution, called the Ehresmann connection. Then, Tu proves that given an Ehresmann connection, you can recover a horizontal distribution, showing the objects are in a sense equivalent

lucabtz

1:58 AM

@SillyGoose yeah you can either start from the connection form $\omega$ and find the horizontal right-invariant distribution as the kernel of the form, or start from the horizontal right-invariant distribution and get the connection form from it

Silly Goose

okay i see

lucabtz

they are both the same data basically, so you can call the connection whichever

note that distribution just means a subbundle of $TP$

Silly Goose

also so when we naturally construct the vertical subspaces, we define the natural projection $\nu: TP \to VP$ at this point right? Essentially collecting all vertical subspaces $V_pP$ and defining a subbundle $VP$ of $TP$

well i guess maybe not? $\nu$ does not seem like a projection from a bundle space to a base manifold...so it is a different sort of projection than a bundle theoretic projection?

lucabtz

the vertical distribution $VP$ is the subbundle of $TP$ defined by $VP = \mathrm{ker} \ d\pi$

notice that since $\pi: P \longrightarrow M$ you have $d\pi: TP \longrightarrow TM$ and thus $\mathrm{ker} \ d\pi \subset TP$

however there is no natural vertical component map $v: TP \longrightarrow VP$ until you define the horizontal distribution (hence until you pick a connection)

to be stupid consider $\mathbb{R}^2$ and say the vertical subspace is the one generated by $(0, 1)$

you have to pick a horizontal subspace (another basis element) to tell me which is the coordinate of a vector along $(0, 1)$

say i consider the vector $(1, 1)$

if i choose as horizontal complement the space generated by $(1, 1)$ then $(1, 1) = 0 (0, 1) + (1, 1)$ and the vertical component of $(1, 1)$ is zero

instead say i choose as horizontal complement $(1, 0)$, then $(1, 1) = (1, 0) + (0, 1)$ and its vertical component is $1$

Silly Goose

hmm i see

lucabtz

2:12 AM

in particular you might be tempted to consider the orthogonal complement when thinking about vector spaces like $\mathbb{R}^2$ because you naturally picture Euclidean spaces, but consider there is no metric on $TP$ so no notion of orthogonal

Silly Goose

does being in $\ker d\pi$ means that locally your base manifold point does not change as you take a directional derivative?

lucabtz

@SillyGoose i picture it as all vectors tangent to the fibre and thus whose projection on the base manifold vanishes

Silly Goose

what is meant by "whose projection on the base manifold vanishes"?

lucabtz

@SillyGoose that it is in $\mathrm{ker}\ d \pi$

$\pi$ is the projection map

Silly Goose

i guess can i think of it as a curve $\gamma(t)$ in $P$ and in $G_x$ locally looks like $\gamma(t) = (x, g(t))$. this projects to $x$ for all $t$, so is a constant curve in $M$. the tangent vector associated to a constant curve is $0$?

lucabtz

2:23 AM

yes

Silly Goose

okay awesome possum

lucabtz

im a bit confused the fibre should be called $P_x$ in my opinion but the concept is correct

its the vectors tangent to the fibre (tangent to curves that only move along the fibre as you said)

Silly Goose

oh. my notation might be strange. i am assuming a principal bundle $(P, M, \pi, F \approx G, G)$ where i am accepting nakahara's notion that we are just going to write the fiber as the structure group itself for a principal bundle

lol i am finally getting to some physics

i feel like diffe g is confusing because every single object is really like 7 different objects at the same time lol

but i guess that's kind of cool lots of structure

is there a usual notation for the tangent bundle to a neighborhood $U \subset M$?

i would like to write the domain for the locally defined gauge potential $A_i$ over $U_i \subset M$

$A_i$ itself can be thought of as an element of $\mathfrak{g} \otimes \Omega^1(U_i)$, the set of $\mathfrak{g}$-valued differential $1$-forms over $U_i$. but i am not quite sure how to write the domain of such differential forms is when restricting to a neighborhood

lucabtz

2:51 AM

@SillyGoose the forms in $\Omega^1(U_i)$ are sections of the bundle $TU_i \longrightarrow U_i$ thus they are functions $U_i \longrightarrow T U_i$

Silly Goose

wait should it be $T^*U_i$ in both cases?

lucabtz

yes

Silly Goose

this is separate but related. so a connection $1$-form is formally an element $\omega \in \mathfrak{g} \otimes \Omega^1(P)$? So it is $\mathfrak{g}$-valued and it is a $1$-form over $P$ (a section of the bundle $TP$)

lucabtz

3:37 AM

a section of $T^\ast P$

anyhow what is the question there

Silly Goose

oh sorry yes a section of $T^*P$. i was wondering if i correctly described the mathematical objec that $\omega$ is

lucabtz

oh yes seems correct

Silly Goose

hm something seems incorrect about how i wrote this. $\omega$ is a map from $TP \to \mathfrak{g}$. but $\sigma_i^*$ is a map $TP \to TU_i$?

I am wondering what i am missing...

lucabtz

4:07 AM

@SillyGoose $\sigma_i^\ast$ is the pullback, it is a contravariant functor (cofunctor) in the language of category theory (since you're doing algebraic topology)

so the arrow gets reversed

$\sigma_i: U_i \longrightarrow P$ then $\sigma_i^\ast: T^\ast P \longrightarrow T^\ast U_i$

the contravariant functor is the $T^\ast$, similarly as $T$ is a covariant functor

123

5:19 AM

Hello Everyone...

Silly Goose

6:12 AM

@lucabtz oh oops i forgot taking the dual. i was more wondering about how $\omega$ has codomain $\mathfrak{g}$ but $\sigma^*_i$ has domain $TP$, which doesn't match

Silly Goose

6:36 AM

i have: a connection $1$-form is a $\mathfrak{g}$-valued section of the bundle $T^*P$. In other words, $\omega: P \to \mathfrak{g} \otimes T^*P$.

and: a local section of the base manifold $M$ is a map $\sigma_i: U_i \to P$. its pullback is then a map $\sigma_i^*: T^*P \to T^*U_i$

but, the codomain of $\omega$ clearly does not match the domain of $\sigma_i^*$, so how can $\sigma_i^*\omega$ be defined?

there is probably a natural way to include $T^*P$ into $\mathfrak{g} \otimes T^*P$--is this what is being done?

Silly Goose

6:50 AM

am i understanding correctly: (1) a gauge field is a set $\{A_i\}$ of local $\mathfrak{g}$-valued $1$-forms, i.e. local sections of the bundle $\mathfrak{g} \otimes T^*P$ defined in terms of a connection $1$-form via $\sigma^*_i\omega$ where $\sigma^*_i$ are local sections of base manifold $M$. (2) a choice of gauge is a choice of local sections $\{\sigma_i\}$. (3) gauge transformations are transformations on the space of local sections of the bundle $TP$?

@lucabtz oh that is a nice way to think of it

ACuriousMind

8:14 AM

@SillyGoose If you're going to be pedantic about the notation you first have to set up the notation correctly ;) $\mathfrak{g}\otimes T^\ast P$ doesn't really mean anything, you can't take the tensor product of a vector space with a bundle, you can only take tensor products of two vector bundles. So there's the trivial bundle $P_\mathfrak{g} = \mathfrak{g}\times P$ and the form is a section $P\to P_\mathfrak{g} \otimes T^\ast P$.

Now the pullback you're looking at is the function $\sigma^\ast_i \otimes \mathrm{id}_\mathfrak{g} : T^\ast P \otimes P_\mathfrak{g} \to T^\ast U_i \otimes P_\mathfrak{g}$

Silly Goose

ah wonderful thank you

i need like a differential geometry thesaurus. a tangent vector is a X is a Y is a Z is a...

ACuriousMind

@SillyGoose I would be careful with the words "choice of gauge", since this clashes with the usual terminology in physics where by "a choice of gauge" we usually mean a set of conditions that pick out a specific set of $\sigma_i, A_i$, not the local sections themselves

and also "gauge transformations" depend on the context - since the typical physics text does not use the bundle language, it will call different things that all act "the same" on the local gauge fields "gauge transformations" even if they are different things

the mathematical approach usually uses "gauge transformation" for fiber-preserving automorphisms of the bundle, but the "physics gauge transformations" are also often local transition functions, not such global automorphisms; see also this answer of mine arguing that the "transforms as" definitions of physicists are really specifications of transition functions

Silly Goose

@ACuriousMind oh wait so is this like, setting the lorenz gauge $\partial^\mu A = 0$ is setting a constraint on the $\{A_i\}$, which induces a set of local section $\{\sigma_i\}$?

er i guess they're both specified by the choice of "gauge" at once?

where're the global quantum field theorists in this world!

ACuriousMind

@SillyGoose I expressed myself badly: Actually, no, the condition doesn't pick out a specific choice of local sections, you really have the set of all global connection forms $A$ modulo gauge transformation and the gauge condition picks (ideally) exactly one representant from each of the equivalence classes

Mr. Feynman

@lucabtz you're taking me too literally :P

Silly Goose

8:27 AM

okay so a physics choice of gauge selects a unique connection $1$-form, $\omega$ in my notation above, the globally defined data associated with a gauge field

ACuriousMind

@SillyGoose No, it selects a unique connection 1-form from each gauge orbit

Remember that there's an $A$ obeying both your gauge condition (e.g. Lorenz) and the Maxwell equations for any possible source current $j$ for the Maxwell equations

the solution to the Maxwell equations alone is not unique - the set of solutions $A$ to any given Maxwell equation is a gauge orbit (if everything behaves nicely), and the gauge condition picks ideally exactly one $A$ from that set, so that the combination Maxwell equations + gauge condition has unique solutions

the point of a gauge condition is to "destroy" the gauge symmetry (in the Hamiltonian formalism, by turning a first-class constraint into a second-class constraint, see also this answer of mine)

Silly Goose

oh okay i think i see. so would my previous comment be correct if i fixed the current for my maxwell eqn so that i was now talking about a single maxwell eqn defined on my bundle rather than a set of maxwell eqns (with arbitrary current)

also just to clarify it is a connection $1$-form that appears in an action of the physical theory?

ACuriousMind

The $A$ is the (local) connection form, yes

Note that while the $A_i$ are only defined locally in the bundle trivializations (and you only need them locally to have a variational principle), the Lagrangian $\mathrm{Tr}(F\wedge{\star}F)$ and hence the action is globally well-defined

Silly Goose

does local connection form mean the connection 1-form as written in a neighborhood or doess it mean the pullback by a local section of the connection 1-form

or is it unambiguous because your gauge field should be a function of the base manifold, which requires the pullback

ACuriousMind

yes

Silly Goose

8:40 AM

ah i see

@ACuriousMind why is that so?

also so do we really need the lagrangian to be globally well-defined per your comment? is it just a coincidence that it is globally well-defined

ACuriousMind

@SillyGoose Because $\mathrm{Tr}(F\wedge{\star} F)$ is gauge-invariant, so it doesn't change under the "gauge transformation" by a transition function on an overlap, so you can stitch its local values together to a global function on the base manifold

@SillyGoose I'm not sure what characterizes a "coincidence" in this case; it's just a fact :P

Silly Goose

i see

ACuriousMind

and it's a reassuring fact, since for U(1) you already have that $F$ itself is gauge invariant so that means the electric and magnetic field of classical EM are indeed globally well-defined even though $A$ may not be

Silly Goose

what would a not globally defined field even mean physically?

ACuriousMind

exactly

Silly Goose

8:46 AM

for instance, the chern-simons action is not invariant under what witten calls a gauge transformation, but which is (it seems) a naive (not in an offensive way, but as in a first thought sort of idea) application of the gauge transformation from electromagnetism to the chern-simons action

well if we have a not globally defined field could we see this as a fuzziness of information analogous to what happens in usual quantum mechanics

ACuriousMind

not really

it's not quantum

also, the Chern-Simons action is not invariant under gauge transformations, but the change is a total derivative, i.e. gauge transformations are still symmetries of the theory

Silly Goose

oh ._. i see

it is a total derivative bc it just changes by a constant and a constant is exact (to use the differential form language)?

ACuriousMind

oh, are we talking about large gauge transformations that actually change the value of that action? that's a whole other can of worms

Silly Goose

hm i dont know what a large gauge transformation is

ACuriousMind

it's what I call a "false gauge transformation" in this answer of mine and this is exactly the ambiguity of physicists using "gauge transformation" both for automorphisms of the bundle and for transition functions

Silly Goose

9:00 AM

@ACuriousMind also wait is it the action that can change up to a total derivative? i thought it was the lagrangian (or lagrangian density)

blebs

ACuriousMind

it's the Lagrangian, yes

Silly Goose

okay i see then yes the situation i am talking about is probably the second thing yo usaid

and per our discussion a few weeks ago indeed that's why i described what witten called a gauge transformation as a naive gauge transformation since it seems like it is just modeled after what happens in EM and not a genuine gauge transformation

hm so then what is the witten paper considering? just some interesting transformations on the chern-simons action?

ACuriousMind

9:38 AM

I mean it's not not a gauge transformation in the physical sense :P

as I said that topic is itself a bit of a can of worms if you want to make it mathematically precise, but the path integral in QFT essentially has to sum over all the possible bundles ("instanton sectors") - contrary to what the mathematical formalism would like, we don't actually have a single fixed bundle

and these "false" gauge transformations switch between bundles

Ryder Rude

10:04 AM

hi

bolbteppa

12:02 PM

Amazing how this had a very deep answer

Explains so much

@SillyGoose you wont really understand what a connection is until you learn the language of gerbes

I'm going to have a heart attack reading all this language aimed at obscuring the idea of taking a simple derivative :p

Connection form

In mathematics, and specifically differential geometry, a connection form is a manner of organizing the data of a connection using the language of moving frames and differential forms. Historically, connection forms were introduced by Élie Cartan in the first half of the 20th century as part of, and one of the principal motivations for, his method of moving frames. The connection form generally depends on a choice of a coordinate frame, and so is not a tensorial object. Various generalizations and reinterpretations of the connection form were formulated subsequent to Cartan's initial work. ...

This entire article is very triggering, can't just take a simple derivative, no we have to re-define it as a map using bundles and call the derivative a connection, then we give the game away and admit connections arise from taking derivatives of the basis in a particular frame, then we repent and throw holy water over our back burning off the sin of going to a particular frame to repent

Mr. Feynman

@bolbteppa that subtle sarcasm kills me :P

@bolbteppa I don't understand why you despise this so much: I mean, it's not like the Differential Geometry Police is raiding your house to have you use this formalism

I mean, plenty of people don't give a damn about bundles in physics

naturallyInconsistent

12:20 PM

@Mr.Feynman oh, you underestimate how many people have already fallen to the Differential Geometry Police. However, not as many as the extremely active King Genuflection Bureau is at making sure people respect strings.

Mr. Feynman

I want to learn strings to be allowed to disrepect ST

But I think I'm falling in love in the process

naturallyInconsistent

ah, the pot is cracking

Mr. Feynman

Periodic motion, they say

bolbteppa

@Mr.Feynman I will never stop mocking the differential geometry morality police

Ryder Rude

12:38 PM

@bolbteppa hi. did u try to learn this alien language stuff and found that it didnt have a pay off?

im interested in this language, which is y im asking

bolbteppa

Yes

Ryder Rude

so it only overcomplicates basic ideas

i was recently reading the category theory definition of a cartesian product, and it was over complicating a basic idea

bolbteppa

Absolutely

Ryder Rude

if theyre going to talk about pairs of functions anyway, y not just talk about pairs of elements

it's overcomplicating the cartesian product

bolbteppa

Category theory is supposed to be a good idea, supposedly showing how different theorems in different subjects are all coming from the same underlying ideas, however that promise completely fizzles when you actually study it

Ryder Rude

12:42 PM

oh

Mr. Feynman

@bolbteppa I disagree and I'm glad to read that because these are the best dissings of the century

If you dislike rigorous differential geometry, which is arguably one of the most intuitive fields of mathematics, what do you think about measure theory? :P

Or Real Analysis

Please, don't tell me you prefer Newton/Leibniz style calculus

Slereah

Measure theory is just saying if something is big or not

bolbteppa

With measure theory you know what you're getting yourself into, you've signed up for the rigor, it's unavoidable and necessary

Slereah

How unintuitive could it be

ACuriousMind

12:57 PM

@Slereah how big is the Cantor set

bolbteppa

You are absolutely forced to deal with the contradictions arising from naive probabilistic thinking, it has earned the rigor, with real analysis you are talking about simply filling in a few tiny gaps in basic calculus, you can frame it in a grand confusing way and there are around 5 equivalent ways to do this because completeness of the real numbers can be frames in these different ways making it very confusing to study if you don't know that

Slereah

@ACuriousMind somewhat

bolbteppa

This is what I mean about the 5 equivalent ways

This watches slightly differently in light of the Greiner stuff

My dream died, and now I'm here

►

Ryder Rude

im trying to figure where vectors and dual vectors really come from

bolbteppa

1:14 PM

You know what a vector and a dual vector are

Ryder Rude

without using co ordinates, we take a function and a curve at a point. these two can be mapped to a number. so we hav a function $D(\gamma, f)$

@bolbteppa kind of

and when we use co ordinates, we can compute $D(\gamma, f)$ using tuples identified with $\gamma$ and $f$. these tuples we call a vector and a dual vector

Mr. Feynman

@Slereah that's why it gets so technical

No new ideas, just that very same concept discussed with all pathological stuff

Ryder Rude

@bolbteppa she is very unbiased in this video

Dune Part 2 is now on streaming

Ryder Rude

1:40 PM

is the smoothness of a manifold a property of its atlas, and not a property of the manifold?

Ryder Rude

1:50 PM

ooh it must b a property of the atlas becuz one can define both a smooth and a non-smooth atlas for the same manifold

so the "smooth structure" is assigning an atlas having this property

Ryder Rude

2:11 PM

i hav a 1D manifold which is the real line, $R$. i define two atlases : Atlas 1 : $\psi (x)=x$. Atlas 2: $\psi (x)=2x, x\geq 0 ; \psi(x) = -3x, x<0$

why arent these two atlases two different smooth structures?

the first one is a smooth structure becuz it's a homeomorphism onto $R^1$ and there's only one chart so the compatibility of charts holds. the second one is also smooth for the same reason

Ryder Rude

2:27 PM

sorry i meant $\psi (x)=3x ; x<0$ for the second atlas

im asking this becuz it says there's only one smooth structure on 1D, 2D and 3D manifolds

ACuriousMind

what you have is not an atlas

Ryder Rude

but theyre continuous and invertible

and there's only one chart in each atlas, so compatibility holds @ACuriousMind

ACuriousMind

if you want to do math you need to be precise: An atlas consists of charts, and charts consist of functions between open subsets of a manifold $M$ and open subsets of $\mathbb{R}^n$.

but anyway your central error lies somewhere else entirely: Yes, there is more than one smooth structure on $\mathbb{R}$. This does not contradict anything, you simply need to read the theorem about uniqueness of smooth structures more carefully.

bolbteppa

$[0,\infty)$ is not an open set either

Ryder Rude

2:46 PM

@bolbteppa i had corrected it to $\psi (x)=3x, x<0$

@ACuriousMind it says theyre unique upto diffeomorphism. but my atlases arent diffeomorphic, which is whats bothering me

ACuriousMind

@RyderRude how do you know they are not diffeomorphic? Prove it!

also, it's not the atlases that are supposed to be diffeomorphic (that doesn't mean anything), it's the two manifolds with their different smooth structure induced by the atlases

Ryder Rude

i thought "up to diffeomorphism" meant functions that are smooth in one atlas must be smooth in the other atlas

bolbteppa

With the fix it looks like you have basically put three charts on the manifold $\mathbb{R}$

Ryder Rude

no, it's two atlases each with one chart

bolbteppa

Atlas 2 with the fix does not cover the real line

ACuriousMind

2:55 PM

@RyderRude no it doesn't

bolbteppa

Also you are ignoring the overlap, everything has to be nice on the overlap, it doesn't look like your 'charts' are nice on the overlap is it not discontinuous

Ryder Rude

@bolbteppa it is continuous. it says 2x for the positive half and 3x for the negative half

and it's only one chart

@ACuriousMind oh

lucabtz

@Mr.Feynman neat notation to make both the parties happy $\int xp f(x) dx$ the p looked like upside down d

Spanish notation

ACuriousMind

What you are trying to say is this: We have two manifolds, $\mathbb{R}_a$ with the standard smooth structure and $\mathbb{R}_b$ with the smooth structure induced by the global chart $\phi : \mathbb{R}\to \mathbb{R}_b, x\mapsto \begin{cases}2x & x\geq 0 \\ 3x & x<0 \end{cases}$.

$\mathbb{R}_a$ and $\mathbb{R}_b$ are diffeomorphic via the diffeomorphism $f : \mathbb{R}_b \to \mathbb{R}_a,x \mapsto \begin{cases} \frac{1}{2}x & x\geq 0 \\ \frac{1}{3}x & x < 0\end{cases}$

lucabtz

@RyderRude the categorical product captures way more than just the Cartesian product

Ryder Rude

3:20 PM

@ACuriousMind oh

and this is a diffeomorphism because it's the identity transformation between the charts, which is a smooth matp?

@lucabtz yeah people did say that. do u find it useful

Ryder Rude

3:35 PM

is it relevant for GR that $R^4$ has inequivalent smooth structures

Mr. Feynman

3:48 PM

@lucabtz what is this monstrosity :P

Not even Sakurai could come up with something worse

Ryder Rude

20

Q: Is there a relation between 4-dimensional general relativity and exotic smooth structures on $\mathbb{R}^4$?

Let's say General Relativity is the study of the Einstein equation on smooth Lorentzian manifolds, i.e. pseudo-Riemannian manifolds of signature $(n-1,1)$. I've heard more than once people say that the reason why there are many non-diffeomorphic smooth structures on $\mathbb{R}^n$ only for $n=4$ ...

Obliv

4:24 PM

Is stellar aberration a specific case of something more fundamental i.e. aberration causes by relative movement b/t lens and object

If u look out of the window of a vehicle going moderately fast, unless u focus directly on the object with ur eyes, your view will appear streamy/blurry so i am guessing this is the same thing more or less

actually no because with telescopes we still get the aberration despite focusing on moving stars. hmm idk @SirCumference maybe u know

maybe the blurriness is related to our brains/eyes ability to process images like the baseline refresh rate or whatever

I wonder if faster animals have better refresh rates

Silly Goose

@RyderRude i feel like one can say this about anything that is not too exotic in math. All concepts exist because they’re trying to solve some problem and those problems are quite simply stated. But trying to prove statements about such concepts without formalizing a definition is quite hard

And the already proved (even unproved) properties of such objects allows physicists to mindlessly use them, right :D

@bolbteppa at the very least it’s nice to be able to just write $(X, \tau) \in \text{Top}$ instead of is a topological space hehehe

Marco

4:42 PM

Hello everyone, everything good? Does anyone here have any familiarity with machine learning?

Sanjana

6:03 PM

UC San Diego physicist McGreevy has released a paper preprint with the title Nishimori's self-tuning as evidence for the existence of God.

Silly Goose

Lol the god i want to believe in is the one that constantly performs von Neumann measurements on the world

bolbteppa

Lunacy

Silly Goose

I think the title is probably just a form of clickbait the authors probably do not actually believe it is evidence for god…

unless using god in the loose sense which they define

hence clickbait

But yay for any time decoherence appears hehehe

bolbteppa

Coincidentally?

John McGreevy

John T. McGreevy (born 1963) is an American historian who has been serving as Charles and Jill Fischer Provost of the University of Notre Dame since July 1, 2022. He was formerly the dean of the College of Arts & Letters at the University of Notre Dame from 2008 until 2018. McGreevy earned his Bachelor of Arts degree in history from the University of Notre Dame and his Doctor of Philosophy degree in history from Stanford University. He has been on the Notre Dame faculty since 1997. He is the author of Catholicism and American Freedom. == Books == Parish Boundaries: The Catholic Encounter with...

Silly Goose

Maybe they’re trying to get funding from the John templeton foundation;)

So who uses genuine bundle theory anyways

ACuriousMind

6:20 PM

uh, mathematicians? :P

Silly Goose

oh i mean in terms of predicate

ACuriousMind

I don't really know what you expect as an answer to that question

Silly Goose

lol physicists*

ACuriousMind

Bundles are simply part of the language of differential geometry; you can do your best to avoid it or you can embrace it

it's a bit like asking who uses analysis or the theory of probability spaces anyways :P

at a physical level of rigor we often do stuff like interchanging integrals and derivatives where in rigorous analysis we'd have to check more carefully when the theorems that allow this apply. So is that "using analysis"?

Silly Goose

hm i guess i just haven't seen (probably because i haven't even learned qft) results of bundle theory being used for physics. in contrast to representation theory which seems pretty fundamental to non-rel quantum theory and which (i at least think) is necessary to begin to understand quantum theory

@ACuriousMind i would say that is using analysis

well i haven't really learned classical field theory either which is where it should first appear

ACuriousMind

6:30 PM

@SillyGoose I really think this isn't a useful way to think about what's happening. Plenty of physicists "do representation theory" without ever formalizing it in the way representation theory actually does, especially in the infinite-dimensional case. Likewise, you can "discover" by mere computation without any theory of bundles that $\int F\wedge F$ (even writing it in that form is anathema to those who cling to the index formalism!) is an integer and an invariant -

in "bundle theory" you would learn it is the second Chern class of the bundle, and that's the "mathematical" way to look at it, really. So is someone who doesn't call it a Chern class "using bundle theory"?

Silly Goose

hm then i guess i would associate "using representation theory" to making the conjecture that part of the physical theory (e.g. quantum mechanics) actually fundamentally is representation theory. so if we are define a physical theory by writing up a dictionary of phenomena to mathematics, that representation theory would appear in the language on the right hand side.

ACuriousMind

ah, but "fundamentally" is such a wonderfully objective-sounding word that's really so subjective :P

Silly Goose

so okay the physicist can crank through a million computations to find their particular results, but the mathematical physicist will make the observation that all these computations come from a single result and i would call that a usage of the mathematics referenced in the observation

@ACuriousMind hm well maybe this is an example of what i am looking for

ACuriousMind

I mean then I don't really understand the question - as you've seen, the language of bundles is how you describe mathematically what's happening with a gauge field

maybe your issue is simply that you haven't seen the physics that's being formalized here; have not been mystified by the nature of instantons or large gauge transformations; have not wondered why the topological term is an integer or why the Chern-Simons currents have the specific form they do

Silly Goose

yes i think that might be the issue

bolbteppa

6:37 PM

You mean, the 'mathematical physicist' will write an essay before they even write down a vector, then they will be too afraid to compute a derivative they will first have to define a map to compute that derivative, then they will be too afraid to admit the vector has coordinates so they'll invent more maps with more essays to fix the fact the problem a derivative has in curved spaces, and on and on

ACuriousMind

as much as I love the rigorous approach to physics I do think it is rarely the best way to first encounter a physical subject

Silly Goose

well the closest i have come to something that would actually be formalized through differential geometry is reading about the aharnov-bohm effect (and electromagnetism)

but the physical write up (e.g. in sakurai) of the aharnov bohm effect just doesn't make sense :P

which is why i cling to the structure of the underlying mathematics...

bolbteppa

Exactly, when people get confused by a simple physics idea they run to math thinking it will fix the problem and it doesn't

ACuriousMind

because you won't have any sudden moments of clarity of the mathematical structure bringing order to the things you've already seen and also because many of the mathematical texts aren't really written with the physics in mind because, well, they're written by mathematicians :P

Silly Goose

@bolbteppa well i more meant: sakurai simply does a computation and finds an interesting result. this doesn't really amount to an understanding of what's going on to me

@bolbteppa Lol well this sort of text I too skip over

in my current survey through the differential geometric stuff, i have utterly skipped any technical details, which you may take solace in

@ACuriousMind usually i have an itch to know how one can manipulate the objects being talked about in a physical theory which can be answered by learning some of the underlying maths side by side with the physics :P

ACuriousMind

6:51 PM

@SillyGoose well, in the language of bundles the A-B effect is: You have a flat connection ($F=0$) that is not isomorphic via a gauge transformation to the trivial bundle with zero connection ($A=0$); flat connections are classified by their holonomies around non-trivial loops $\mathrm{e}^{\mathrm{i}\int A} = \mathrm{e}^{\mathrm{i}\Phi_B}$, which is the flux through the solenoid, and the holonomy is the phase a particle incurs when it moves a full loop around the solenoid.

Silly Goose

also for A-B effect, this answer suggests that the bundle theory picture actually does make things clearer: physics.stackexchange.com/questions/305497/…. and that the "popular A-B effect" as presented in sakurai wrongly emphasizes it as depending on using the adiabatic approximation

ACuriousMind

The larger theory here explains some things not obvious from the direct computation, e.g. that this effect can happen if and only if you have non-simply-connected regions of zero field in which the particle moves, which follows from the classification of holonomies by non-trivial loops

Silly Goose

i am a bundle theory embracer

bolbteppa

7:07 PM

In the actual answer bundles are irrelevant, it just says bundles magically explain things, what you're getting confused about is that whenever you write $\int A_{\mu} dx^{\mu}$ we see this quantity transforms up to a constant shift when $A_{\mu} \to A_{\mu} + \partial_{\mu} \lambda$, which we call a gauge transformation, this is the same transformation law that a connection $\Gamma^{\mu}_{\nu \rho} = (A_{\nu})^{\mu}_{\rho}$ transforms under if we treat the $\mu,\rho$ indices as matrix indices

(where these indices have one trivial index in the abelian case)

The constant shift is obviously a phase factor in the wave function case, in classical electromagnetism the constant shift is ignored in the action, so in both cases the vector field we had transforms more generally as a connection because of how the model is set up

No amount of formalism is going to give something deeper than the surprising fact that in electromagnetism when you assume particles interact with EM fields via the $-\frac{e}{c}\int A_{\mu} dx^{\mu}$ type interaction (which leads to Maxwell's equations), it admits gauge transformations, which the potential form of Maxwell's equations also admit, we can rationalize this as a transformation on the trivial additional matrix indices we attach to our vector giving a connection interpretation

This stuff starts failing when you introduce mass terms which we do in QFT and we're left with vectors which was what we cared about in the first place, by biasing yourself to connections in the hope of false generality you start throwing away serious stuff like massive vector fields in the standard model coming from the exact same starting point

Silly Goose

7:52 PM

@ACuriousMind what is the logic behind these statements other than the classification statement? (1) it seems that the base manifold $M$ being simply-connected implies some statements about bundles over it being trivial. (2) it seems that a non-trivial bundle requires one to define a notion of parallel transport in order to take into account the non-triviality of the bundle structure, and having zero curvature allows one to compute the holonomy in a nice way?

ACuriousMind

@SillyGoose careful, I didn't say the bundle was non-trivial!

I said we have a bundle with flat connection that is not isomorphic to the trivial bundle with zero connection

the point is that a flat connection is fully classified by its holonomies around non-trivial loops, i.e. the assignment of holonomies $h : \pi_1(M) \to G$ is equivalent to the connection data. So in a simply connected space, you have that $F=0$ implies automatically that you are gauge-equivalent to $A=0$ because $\pi_1(M) = 1$, but in a non-simply connected space you can have flat connections that are not gauge equivalent to $A=0$

the flux through the solenoid is the holonomy map: For flux $\Phi_B$, you have a phase of $\mathrm{e}^{\mathrm{i}\Phi_B}$ around a loop, i.e. $\Phi_B = \oint A$, the holonomy map is $h : \pi_1(M)\cong \mathbb{Z} \to \mathrm{U}(1), 1 \mapsto \mathrm{e}^{\mathrm{i}\Phi_B}$

This is the mathematical answer to the question "How can we have an effect from the vector potential when the field strength is zero?" - "zero field strength" means "flat", but in non-simply-connected cases, "flat" does not mean "gauge-equivalent to the zero gauge field"

Charlie

8:10 PM

I know that the curvature two-form and the Riemannian curvature tensor(s) are related, but does having vanishing curvature two-form mean flat geometry in the GR kind of curvature sense?

i.e. $F=0\rightarrow \text{Riem}=0$?

ACuriousMind

the Riemann tensor is the curvature tensor for GR interpreted as a GL(n) gauge theory, yes

but here "curvature" is more generally $F = \mathrm{d}A+A\wedge A$ without it necessarily having anything to do with Riemannian geometry

Silly Goose

@ACuriousMind hm I am not understanding how $\pi_1(M) = 1$ implies that $A$ is gauge-equivalent to $A = 0$. So in the simply-connected case, we have $h$ defined by $h(1) = 1 = e^{i2\pi n}$ for $n \in \mathbb{N}$. Which says that $\oint A = 2\pi n$ for $n \in \mathbb{N}$?

oh wait

i think i am thinking of the $h$ map incorrectly

ACuriousMind

@SillyGoose I'm trying to give a high-level overview here, not every detail :P Maybe phrased another way: The gauge-equivalence classes of flat connections are in bijection to maps $h : \pi_1(M) \to G$. When $\pi_1(M) = 1$ is the trivial group, there is only one such map, so only one equivalence class.

Silly Goose

it should be the map $h: \{1\} \to U(1)$, which maps $1 \to e^{2\pi}$, without the $n$

i think i can maybe see it. the loop integral $\oint A$ is gauge invariant right?

so when we have fundamental group $\pi_1(M) = \mathbb{Z}$ we get a bijection between $\mathbb{Z}$ and the gauge equivalent classes via $\oint A = 2\pi n$ for $n \in \mathbb{Z}$ where $\oint A$ corresponds to a gauge equivalent class?

ACuriousMind

no

there's no quantization here (yet)

the map is $h : \mathbb{Z}\to \mathrm{U}(1), 1 \mapsto \mathrm{e}^{\mathrm{i}\Phi_B}$, where $\Phi_B\in\mathbb{R}$ is the flux through the solenoid

i.,e. the bijection is between gauge equivalence classes and $\mathbb{R}/2\pi\mathbb{Z}\cong\mathrm{U}(1)$

you map an $A$ to its $\mathrm{e}^{\mathrm{i}\oint A}$, and $\oint A$ is just any real number

Silly Goose

8:23 PM

isn't it a particular real number though if we are talking about integrating around an integer number of loops around a solenoid

ACuriousMind

we're not integrating around an "integer number of loops"

you're integrating around exactly one loop

Silly Goose

oh i am conflating loops

ACuriousMind

and the value you get from that determine the equivalence class of the connection

Silly Goose

okay i see. so we have a set of gauge-equivalent classes $\{A\}$; this class is in bijection with a flux $\Phi_B(A)$, which is conceivable from ordinary electromagnetism.. this flux is some real number. but the number of classes is separately in bijection with $\mathbb{Z}$ by a non-trivial classification theorem that is applicable to this situation

ACuriousMind

no, the classes are not in bijection with $\mathbb{Z}$!

they're in bijection with maps $\mathbb{Z}\to\mathrm{U}(1)$

Silly Goose

8:27 PM

oh no

ACuriousMind

and $\mathrm{Hom}(\mathbb{Z},\mathrm{U}(1))\cong\mathrm{U}(1)$

all I'm saying is that the math tells you this: flat connections can have non-trivial holonomy classified by some real number modulo $2\pi$, and if they do, they are not gauge-equivalent to "no connection". The physical input here is that that classifying number is the flux

Silly Goose

i seee

okay so what is the reason for needing to define parallel transport? in particular, (in the context of bundles) is the notion of parallel transport only necessary to define when we are talking about a non-trivial bundle?

ACuriousMind

no, the connection defines the parallel transport

you can think about the connection as modifying the derivative as $\partial_\mu + A_\mu$ or as defining a parallel transport operator along paths $\gamma$ via $\mathrm{e}^{\int_\gamma A}$

the data encoded in these is equivalent

Silly Goose

so is parallel transport defined to construct paths in the bundle space that move across fibers?

ACuriousMind

8:42 PM

when you lift a path in the base to the bundle, you could choose many different paths - parallel transport is picking one of these paths and saying "that's the one"

Silly Goose

Okay i see

Charlie

Is there a reason GR is formulated in terms of the Riemann tensor instead of the curvature form?

ACuriousMind

The Riemann tensor is the curvature form

you'll have to be a bit more specific about why you think it isn't for me to answer that more usefully

Charlie

:0

wait actually?

oh that's so cool, I hadn't realised

ACuriousMind

the thing that might be confusing you is its 4 indices, but two of those are "secretly" the indices of $\mathfrak{gl}(n)$ so it maps directly onto the curvature form notion of $F_{\mu\nu}^a$, it's just that in this special case we choose to express the algebra index $a$ by the two indices for the components of an n-by-n-matrix

Charlie

9:05 PM

Ohhhh

user 85795

@Loong thanks for the clarification

Charlie

That's a mind blow for me

ACuriousMind

at this point I feel like I'm linking this once per week but I explain this in a bit more detail (in particular footnote 2 and the expansion of $F_\omega$) in this answer of mine

Charlie

Oh right, I've actually seen that answer of yours before but because I didn't know anything about connections etc. I didn't really get far into it

Ty

ACuriousMind

that answer is the result of me fighting with the sentence "GR is a gauge theory" for several years :P

Charlie

9:18 PM

well, keep up the good fight :P

several years 🤯

Hi everyone.

hi

@SillyGoose I am again going to point out that all this magical formalism still has left you confused about parallel transport which is a very simple idea, I'll waste my time explaining it in detail since this has been asked about in here for I think years now

Dannyu NDos

Forgive me for not being a physics student, but I got an idea. If rotation involves acceleration, and if general relativity concerns the acceleration, would that mean we can detect graviton from an object (say, an atom) that rotates really fast?

Ryder Rude

9:32 PM

@DannyuNDos hi

bolbteppa

Or maybe I will wait and let other people give their attempt at a simple explanation first it's just a lot to explain

ACuriousMind

@DannyuNDos just because GR can have something to do with acceleration, that does not mean acceleration produces gravitons; electromagnetism "concerns" electric and magnetic fields but you won't detect photons from just any electric or magnetic field

Dannyu NDos

Oh.

Nevermind. This kind of curiosity is why I really want to re-enter the univ, this time to the department of applied physics.

I have a master's degree in applied math, so

ACuriousMind

also, you are forgiven for not being a physics student :P

Ryder Rude

@SillyGoose does this help one understand why the Aharonov Bohm effect is non local

Silly Goose

9:58 PM

@bolbteppa is it not as (how i interpreted) ACM described: a reasonable prescription to assign a path in the bundle space $P$—that is “tangent” to fibers—to a path in the base space?

@RyderRude What do you mean by non local?

Ryder Rude

@SillyGoose i mean the particle sensing the magnetic field in a region that is out of its reach

Nakahara is saying : "If the union of two atlases is again an atlas, then it means the atlases are compatible. Compatibility is an equivalence relation on atlases and this equivalence relation defines whether or not two atlases define the same smooth structure

this is, strictly speaking, incorrect, right?

becuz the union of the two atlases i gave is not an atlas. it wud imply that my two atlases define different smooth structures according to Nakahara's definition

ACuriousMind

@RyderRude and again, that example does define two different smooth structures

it just so happens that the two structures are diffeomorphic

Ryder Rude

oh

Silly Goose

@RyderRude im not too sure. mathematically, it seems the A-B effect is precisely a consequence of the presence of the flux tube/solenoid

Ryder Rude

yeah

so apart from Nakahara's equivalence relation, we also use diffeomorphism between smooth structures to determine whether or not two smooth structures are the "same" in another sense

bolbteppa

10:13 PM

@SillyGoose I have no idea what that means

ACuriousMind

@RyderRude it's just two entirely different things: There's two atlases being equivalent and there's two smooth manifolds being diffeomorphic

bolbteppa

$\mathbf{A}(\mathbf{x})$ is a vector field, if you want it to follow some path then parametrize the path by say $t$ and you have $\mathbf{A}(\mathbf{x}(t))$ as a vector field on the curve, we can find its tangent vector easily

Ryder Rude

right..the latter is instead an equivalence relation on manifolds rather than atlases

the latter means all calculus results on one manifold carry over to the other manifold

bolbteppa

Here is one way parallel transport arises. If you have a vector $A^n$ in some $N$ space with coordinates $y^n$, and a surface embedded in the space with $x^{\mu}$ coordinates, $\mathbf{A} = A^n \mathbf{e}_n = A^n \frac{\partial \mathbf{r}}{\partial y^n}$ is a general vector, and $A^n \frac{\partial x^{\mu}}{\partial y^n} \frac{\partial \mathbf{r}}{\partial x^{\mu}} = A^{\mu}(x) \mathbf{e}_{\mu}$ is the restriction of that vector to the surface. Thus
$$A^{\mu}(x) = A^n e_n^{\mu}(x) = A^n \frac{\partial x^{\mu}}{\partial y^n}$$

The quantity $U(x,y)^{\mu}_{\nu} = e^{\mu}_n(x+dx) e^n_{\nu}(x)$ is sometimes called the comparator. Look how concrete this is compared to the waffle most books bring up to discuss parallel transport. The point of all this is really for covariant derivatives. Taylor expanding in $dx$ we get
$$U(x+dx,x)^{\mu}_{\nu} = \delta^{\mu}_{\nu} + [dx^{\rho} \partial_{\rho} e^{\mu}_n]e^n_{\nu} = \delta^{\mu}_{\nu} - dx^{\rho} e^{\mu}_n(\partial_{\rho}e^n_{\nu}) = \delta^{\mu}_{\nu} - dx^{\rho} \Gamma^{\mu}_{\rho \nu}.$$

Ryder Rude

ok so R admits many different smooth structures in one sense, but once we modulo out by diffeomorphisms, we get only one smooth structure. and this is what the theorem says by "up to diffeomorphism"

bolbteppa

10:17 PM

Our vector $\mathbf{A}$ doesn't have to live in the same space as the coordinates, we can do this dance with some $V^{M'} = e^{M'}_{M}(x) V^{M}(x)$ and get a
$$U(x+dx,x)^M_N = \delta^M_N - dx^{\mu} (A_{\mu})^M_N,$$
where $A_{\mu}$ is a connection with matrix indices $M,N$. One should study the transformation law of a connection in both cases, and note that the $M,N=1$ case should reproduce electromagnetism gauge transformations, and in general Yang-Mills, the only issue is that the $\mu$ vs $M,N$ now transform differently. In the $M,N=1$ case we now obviously have $U(x,y) = e^{-\int A_{\mu}…

All of this is really just another way to talk about taking the partial derivative of a vector field, $\partial_{\mu} \mathbf{A} = \partial_{\mu} (A^{\nu} \mathbf{e}_{\nu})$, we are forced to express $\partial_{\mu} \mathbf{e}_{\nu}$ as a linear combination of the original basis defining a connection, $\partial_{\mu} \mathbf{e}_{\nu} = \Gamma_{\mu \nu}^{\rho} \mathbf{e}_{\rho}$, and we find $\partial_{\mu} \mathbf{A} = (\partial_{\mu} A^{\nu} + \Gamma^{\nu}_{\mu \rho} A^{\rho}) \mathbf{e}_{\nu}$. The above way just does it all in terms of coordinates, but we really just snuck a basis in via…

Ryder Rude

it is wild to know that there's uncountably many smooth structures on $R^4$ even after moduloing out diffeomorphisms

this wud imply that the physical universe comes with a choice of smooth structure

and the question "why this smooth structure" is unexplained by physics

bolbteppa

Plenty of 'rigorous' books simply lie and say you can't compare vectors in different tangent spaces, and you need to shift things to the same tangent space to be able to compare vectors, well I just compared vectors in different tangent spaces when I took a derivative of the basis vectors and defined the connection. Even the wiki I linked above on connection forms literally uses this same idea as the starting point then dresses it up pretending its all abstract and magical

Ryder Rude

i find this question a twin to "why this topology"

yet another "free parameter" for the physics theory

but the thread i linked did say that there r trails to obtain a unique smooth structure for physics

something about "physical requirements"

bolbteppa

If you're going to work exclusively in terms of coordinates, you can't just blindly consider $A^{\mu}(x+dx) - A^{\mu}(x)$, you need to factor in the change in the basis as well which is what the tetrads do, the parallel transport is basically moving the vector from $x$ to $x+dx$ and accounting for how the basis changes as you move it from $x$ to $dx$, leaving you to be able to consider how the coefficients $A^{\mu}$ themselves change, i.e. to work out $A^{\mu}(x+dx) - A^{\mu}(x)$

Silly Goose

@bolbteppa how is the second equality in the second equation obtained?

bolbteppa

10:30 PM

From $A^{\mu}(x+dx) = e_n^{\mu}(x+dx) A^n$, we now re-express $A^n$ in terms of a vector at $x+dx$ via $A^n = e^n_{\nu}(x+dx) A^{\nu}(x+dx)$, so that $A^{\mu}(x+dx) = e_n^{\mu}(x+dx) A^n = e_n^{\mu}(x+dx) e^n_{\nu}(x+dx) A^{\nu}(x+dx) = \delta^{\mu}_{\nu} A^{\nu}(x+dx)$

Silly Goose

oh wait then is the $e^\mu_n(x)$ a typo

bolbteppa

I don't see a typo

Silly Goose

im comparing

to what you wrote just now above

bolbteppa

As a check, if you re-do everything for a covariant vector $A_{\mu}$ you should get the correct sign in the covariant derivative

@Mr.Feynman show me the bundle version of the above that is more illuminating

Silly Goose

i guess i don't understand where $e^\mu_n(x)(x+dx)$ comes from in your original expression (in the screenshot)

bolbteppa

10:37 PM

Oh sorry, there is a typo, $e^{\mu}_n(x)(x+dx)$ should be $e^{\mu}_n(x+dx)$

Silly Goose

oh okay makes sense then

Ryder Rude

do u people find it useful to learn about different smooth structures on $R^4$ that are not diffeomorphic

and the theorem that proves this

Silly Goose

@bolbteppa to check my understanding, the last expression here is expressing $\textbf{A}$ in the basis of tangent vectors over $x + dx$?

bolbteppa

Yes, well a modified version of $\mathbf{A}$ accounting for the change of the basis as you move from $x$ to $x+dx$

Ryder Rude

there's so much to learn...i feel i cant get into the details of everything

Nakahara skips over any examples of these smooth structures

Silly Goose

10:48 PM

ah yes so we have $\textbf{A}$ defined over a some surface coordinitized by $x^\mu$, so we have $\textbf{A}(x^\mu)$. so the last expression is writing $\textbf{A}(x^\mu + dx^\mu)$ in the basis of the tangent space over $x^\mu + dx^\mu$?

@bolbteppa wait is $\textbf{A}$ a vector or a vector field

bolbteppa

$\mathbf{A}(x)$ is a vector field, the definition of a vector field is that the vector $\mathbf{A}$ depends on the coordinates $x$ of your space to give a field of vectors, the last expression is really the combination $[A^{\mu}(x) - dx^{\nu} \Gamma^{\mu}_{\rho \nu} A^{\nu} ]\mathbf{e}_{\mu}(x+dx)$ which is not just $\mathbf{A}$ at $x+dx$ it's some complicated expression telling you how $\mathbf{A}$ changes as you move it along the surface from $x$ to $x+dx$

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