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12:00 AM
@bolbteppa Yeah I'm just "expanding" the terms or whatever to show that both sides are the same
not really a proof
The proof is half a line, $\nabla \times (\nabla \times \mathbf{A}) = \epsilon_{ijk} \epsilon_{k pq} \partial_j \partial_p A_q = 2! \delta^{pq}_{ij} \partial_j \partial_p A_q = \partial_i (\partial_j A_j) - \partial^2 A_i = \nabla (\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A}$
The question is whether you are happy dealing with these $\epsilon_{ijk} \epsilon_{kpq}$ identities
@SillyGoose The c/a operators are just specific combination of the field operators, cf. e.g. this answer of mine
I'm not sure what you mean by the "if I hand you a Hermitian operator" part - the c/a operators' usefulness is specific to free fields
also see this post physics.stackexchange.com/a/437583/50583 @SillyGoose
(A quick way to prove that identity is here math.stackexchange.com/a/3659638/82615 a longer more general way is here math.stackexchange.com/a/919177/82615 )
12:21 AM
found my mistakes, also this document is quickly getting out of hand
how important is number theory for fundamental physics right now?
i havent seen it anywhere. do physicists learn it
very minimally I'd guess
it looks like an extremely rich subject. hopefully the next fundamental theory can get it into the center
Witten says our final theory may be like number theory, where we know what it is but we can never grasp it all
12:51 AM
"final" theory hmm
what a silly thing to do
camelot is a silly place
is a dimensionful constant just a way to generalize units for some dimension
not sure if I should assume it's like $\frac{\text{charge}}{\text{area}\cdot\text{ time}^2}$
to cancel out one time unit
wait nvm $\Phi_B$ has time in its dimensions not electric field flux
nvm again I guess it does have time. volt-meters have 1/s^2
1 hour later…
2:17 AM
@SillyGoose a dual mice is an object that eats a mice a spits out a scalar
no wait ur right
2:48 AM
am i reading this wrong or does $\widetilde{\gamma}_*X$ not make any sense?
@lucabtz Lol
$X \in T_{\gamma(0)}M$, which can be thought of as a function $X: C^\infty(M) \to \mathbb{R}$. Then, $\tilde{\gamma}_*: T^*P \to T^*[0,1]$?
but these maps do not seem compatible
oh wait i mixed up pushforward and pullback
this is a pushforward so $\tilde{\gamma}_*: T[0,1] \to TP$, which then actually makes sense i think
well really it seems we want the ? mapping. so we should really be considering $\tilde{\gamma} \circ \gamma^{-1}$? and the pushforward of this map is the appropriate map?
3:09 AM
You mean pushout?
3:28 AM
Either way, you might've confused pushouts to cofibrations, or pullbacks to fibrations.
Hm Iโ€™m not familiar with that language
i guess i am just wondering what map $\tilde{\gamma}_*$ is
It doesnโ€™t seem to be actually mean what those symbols mean as defined in this text (nakahara)
Nakahara defines $f_*: TM \to TN$ to be the pushforward of $f: M \to N$
3 hours later…
6:53 AM
@ACuriousMind I will say though it's a little ironic it starts off with a definition of gauge theory as where dynamical variables defined with respect to a "reference frame" without defining what they mean exactly by a reference frame. in my head, a reference frame is a tetrad.
@lucabtz the denominator is fine of course. The "mystery" is the $(-s)^J$ factor in the numerator for the $t$-channel of a spin $J$ particle "exchange"
They say it comes from contracting the momenta at each vertex (derivative interaction) , yet what you get is much more complicated (see e.g. scalar QED, the t-channel also has a $u$ in the numerator)
So $\frac{(-s)^J}{t-m^2}$ is an asymptotic form, I don't know how to derive it though :P
(I used scalar QED as an example because we're considering spin $J$ exchange, $J=1$ in this example, for a scalar field)
7:13 AM
okay...is this an accurate understanding of what physicists call a gauge transformation...
wait i think it should a specification of a set $\{t_{ij}\}$
1 hour later…
8:16 AM
@Mr.Feynman oh i see i didnt notice it was s in numerator and t in denominator
@SillyGoose $t_{ij}$ are transition functions not gauge transformations
physicists often call everything a gauge transformation though
@lucabtz oh no what are gauge transformations ;_; i thought what I thought from reading ACM's answer here: physics.stackexchange.com/questions/795405/…
@SillyGoose are these your own notes?
@qwerty yes
@SillyGoose currently wondering exactly this haha
me too I mean
@qwerty i hope we both find our peace xD
@qwerty they are essentially just nakahara's sections on gauge theory + some of tu's text on differential geometry, ignoring many technical details and trying to fix imprecise statements where they appear and focussing on what the objects are and do and why they naturally can be constructed
8:26 AM
I'll take a look at Nakahara but I think it's a bit too hardcore for me
not familiar with tu
I was just reading Terry tao's take on it on his blog earlier
8:38 AM
@lucabtz apparently you can easily derive it (GSW say ๐Ÿ’€๐Ÿ’€๐Ÿ’€)
"it does not make sense because it does not make sense" XD
1 hour later…
9:53 AM
is the gauge field a map from the bundle space into the cotangent space on the base space?
for some reason i thought it was a map from the base space into the cotangent space on the base space, but i am looking back at my notes and it seems like not
$\mathcal{A}_i \equiv (\text{id}_\mathfrak{g} \otimes \sigma_{i*})\omega: P \to T^*U_i$ if I am not mistaken
$\omega: P \to \mathfrak{g} \otimes T^*P$ and $\sigma_i: U_i \to P$
oh wait sorry the codomain of $\mathcal{A}_i$ should be $\mathfrak{g} \otimes T^*U_i$
@Obliv $$\nabla^2\vec A\neq(\nabla^2A_x,\nabla^2A_y,\nabla^2A_z)$$ in general; what you wrote down is only specific to Cartesian coördinates, and in that case, you would not have had to do all these maths.
@SillyGoose Physicists don't call a "choice of transition functions" a choice of gauge, where did you get that from? I thought we had already discussed gauge conditions
10:11 AM
@Obliv I know Im way too late to this party, but you should somewhat see it as a definition of the vector Laplacian: $$\vec\nabla\,^2\vec A=\vec\nabla(\vec\nabla\cdot\vec A)-\vec\nabla\times(\vec\nabla\times\vec A)$$ such that both terms on the RHS are vector fields, so the LHS is also a vector field outcome. That is, a scalar Laplacian is div grad, but a vector Laplacian is grad div - curl curl. Totally different things, giving you differerent things.
It just so happens that under Cartesian coördinates, the scalar Laplacian and vector Laplacian looks similar.
@SillyGoose You have the global connection forms on the bundle and the local connection forms on the base. They are in each case ordinary $\mathfrak{g}$-valued 2-forms on their respective domains. You just need to plug in the proper definition of the pullback, which also precomposes with the original section to turn a form on $P$ into a form on $U$.
ooh okay wait so was my mistake was doing some strangeness by trying to pullback a global object using the local definition of pullback
but really i should convert the objects into their local forms, apply the definition of pullback (which at least in nakahara is defined locally) and then reconvert into a global object?
I don't understand the question - the local $A_i$ on the base do not glue together to some "global object", that's the whole reason we start talking about them so much
10:23 AM
oh wait i see
actually nvm i dont think that makes sense
so if my principal bundle is trivial, then i can define a global gauge potential, right?
still by $\sigma^*\omega$, but this time we get a map $A: M \to \mathfrak{g}\otimes T^*M$
okay cool beans
maynnn i should just work through a diffe g book at this point...
10:44 AM
I hope I'm not hijacking the discussion but what's an example of a principal bundle? maybe in a gr or em context
the physics begins >:D
"stationizing" is an idiosyncratic choice of verb :p
i guess optimizing would be a bit better xD
I admit "make stationary" is a mouthful though
im not sure why optimize is not used in the actual statements of principle of "least" action
10:48 AM
@Mr.Feynman ive seen this formula before and i always just trusted it mostly, i always sorta understood it in the way its done in the answer by "counting the indices" (or more fancily one would use representation theory), i never wondered too much for the reason there is s in the numerator. I just though the other stuff would be subdominant in the high energy asymptotics but i never wondered much about the calculations
hmm, is optimisation actually synonymous with ti make stationary?
hm im not sure. i am just thinking back to when i learned calculus we called finding the extreme points of a function optimization
what is $\mathrm{Tr}(X,Y)$
@qwerty in GR the frame bundle - the fibers are ordered bases of the tangent space at each point, and ordered bases are equivalently $\mathrm{GL}(n)$-matrices, so the fiber is a group.
matrix trace here; at least that is what i have gathered from other resources
10:50 AM
@SillyGoose why the comma?
oh oops yeah that's not necessary XD
@SillyGoose oh then most likely it is just the Killing form
@SillyGoose I think it is... given an entire field is called optimisation you raise an excellent alternative...
the general definition if you want to avoid matrices should be $\langle X, Y \rangle = \mathrm{Tr}(\mathrm{ad}_X \circ \mathrm{ad}_Y)$
why is there this overload of weird notations like $\overset{\wedge}{,}$ here? everyone else writes the Chern-Simons Lagrangian just as $\mathrm{Tr}(A\wedge\mathrm{d}A + \frac{2}{3}A\wedge A \wedge A)$
10:53 AM
and $\mathrm{ad}_X: \mathfrak{g} \longrightarrow \mathfrak{g}$ is the adjoint representation of the algebra $\mathrm{ad}_X(Y) = [X, Y]$
@ACuriousMind yeah i also am very weirded out
is one supposed to be able to visualise algebra
there is this theorem : "the quotient of a group by the commutator group is abelian"
@RyderRude what does visualize mean?
the proof is easy [ab] and [ba] are the same equivalence class
@RyderRude whats a commutator group
shud i leave it at this or should i pursue visualisation
10:56 AM
@lucabtz it's the group generated by elements of the form $ab b^{-1}a^{-1}$
@ACuriousMind I've never worked with "ordered bases", wiki says they're a "frame"? like similar to a tetrad?
@qwerty yes, exactly - the tetrad fields are sections of the frame bundle
is it ok to move forward with just an algebraic manipulation understanding
@RyderRude i mean the statement seems somewhat clear
10:57 AM
a quotient sets everything in the quotiented-out group to zero, so of course if you quotient out all commutators you get an Abelian group - all the commutators are now zero
of course if you mod out the commutators the group you get is abelian
you just need to have developed the intuition for the quotient
yes. it sets all the commutators to identity
why are they "ordered"...? is it so you can construct a tetrad field or something like that
@RyderRude so what more visualization do you need
@ACuriousMind yes. becuz $[ab][ab]^{-1}$ is now identity
@lucabtz like something in terms of group action on an object
11:00 AM
@qwerty well, if you think about how we use bases to express vectors in components, you should notice that we actually always also have an order attached to the basis - otherwise how do you know which component is the first one?
@RyderRude i dont know the statement seems pretty obvious i wouldnt even spend time proving it
and if you switch the order you can turn right-handed systems into left-handed systems
@RyderRude my point is that if you know what a quotient does then there's not really anything to prove
@ACuriousMind ahh that makes sense. Ive never seen this pointed out though!
i guess i dont have a feel for quotient
@RyderRude but you even said it so what's not clear
an abelian group is one where the commutator between any two elements is the identity
11:02 AM
and as you set the quotient identifies any commutator with the identity
so its abelian by definition
@ACuriousMind im not sure how to interpret the action written this way
@RyderRude you said it here, it seems already clear to you
@SillyGoose why
i dont know how to decide when something is clear
@SillyGoose where's the problem?
11:04 AM
@lucabtz well my guess would be to literally take the wedge products then trace out the lie algebra elements (if matrices)
but im not sure if thats the same
i know that $[a][b][a^{-1}][b^{-1}]$ is identity.. so all the elements commute
but i dont have a feel for what equivalence classes mean i guess
i dont have a feel for quotient
it's really weird
@lucabtz I wish I could but that's the second damn equation. I'll set the question aside for the time being but it bothers meow
@Mr.Feynman usually thats just a historical motivation for string theory i dont think it is so important for the rest of the book
it's really mechanical to me.. i dont have a feel
@SillyGoose no, you need to do the trace as endomorphisms, like lucabtz said: The thing inside the trace is Lie-algebra valued, and so you can take the trace of its adjoint action
11:08 AM
@ACuriousMind I'm still not sure when the quantisation of gauge systems book starts off talking about dynamical variables defined with respect to a (I think much more generic) "reference frame" how I could say it in more exact or mathematical language. actually someone linked me to the idea of torsors the other day and it reminded me a lot of that.
welp that's why i write it this way XD
@qwerty I'm not sure what you mean - where does QoGS talk about "reference frames"?
well how could one know that that is how you interpret that expression a priori :P
@SillyGoose removed my previous yes as answer because ACMs answer is clearer and SilliGoose values precision a lot so i guess the other answer is better
SilliGoose my evil twin
11:12 AM
@SillyGoose well, you can't, but I don't see the point of inventing a lot of completely new notation instead of just writing this as a caveat under the definition :P
but yeah that the correct way to take the trace of a lie algebra element
@ACuriousMind first line in the book? 1.1
@SillyGoose in reality the only ambiguity may be the trace but there isnt much more than you can do so
oh, in the intro
"a gauge theory may be thought of as one in which the dynamical variables are specified with respect to a "reference frame" whose choice is arbitrary at every instant of time."
"a transformation of the variables induced by a change in the the arbitrary reference frame is a called a gauge transformation"
11:14 AM
well, they just mean that - as is explicitly shown later - that at each instant of (space)time, you have a choice of arbitrary functions in the solutions to the equations of motion that don't influence the physics.
you might call these functions a "reference frame", since you need to know someone else's choice to compare their values for the variables to your values
there is no direct relation to (reference) frames in geometry
@SillyGoose lol
hmmm it is a little reminiscent of co-ordinate systems, which I assume is why they chose that language
@qwerty if by torsor you mean a space with a free and transitive group action, basically the principal bundles we have been speaking about consist in attaching a torsor to each point of space
so its very related
@lucabtz yes it was on John baez's blog i can find a link
@qwerty well,. it will turn out that freedom of (space)time coordinate choice is a special case of this - see the chapter on generally covariant systems
11:18 AM
@qwerty its okay its the standard definition of torsor
@ACuriousMind sure. I follow that. I m just wondering if they're correct maths terminology for it since I seem to be sort of following the gist of the physics but the maths is going over my head
*there is
@lucabtz thanks this is helpful
why do i have to take the trace of this adjoint rep of the lie algebra?
@SillyGoose the trace is something you apply to a linear map
the algebra elements are not linear maps
at least not in general
@qwerty the correct terminology is just talking about these "arbitrary functions of (space)time", which mathematically are sections of the principal bundle, which in other language still is essentially a G-torsor at each point - but note that QoGS remain firmly "local" in their discussion and will never even mention bundles
@SillyGoose well, how else are you going to get a number?
actions are $\mathbb{R}$-valued
but then why choose the adjoint representation
11:24 AM
however if you have a representation $\rho: \mathfrak{g} \longrightarrow \mathrm{End}(V)$ you can say take $\mathrm{Tr}(\rho(X))$
thanks . that's very useful
also this chern-simons stuff seems like a real rabbit hole :P
@SillyGoose the only data you have is the Lie algebra and ad is the standard representation on the Lie algebra
okay well i can accept that it is a natural rep i suppose
also the CS form should transform in that representation
11:25 AM
but isn't every lie algebra in some sense a matrix lie algebra, so can't we just somehow wiggle around and use the concrete definition of trace
maybe up to total derivatives or something
@SillyGoose you can choose any other representation, too, if you want, but the adjoint one is "intrinsic" since its on the Lie algebra itself
also for simple Lie algebras it doesn't matter since every symmetric bilinear form is a multiple of the Killing form
im not sure about my last statement about the CS form so take it with a bunch of salt
@ACuriousMind what's the rationale for remaining "local" in the exposition?
@SillyGoose you can definitely relate the trace of the adjoint with the trace of the matrix themselves if in a matrix lie algebra
11:28 AM
@SillyGoose you're not taking the trace in the standard sense of matrix algebras! Note that many algebras like $\mathfrak{su}$ are traceless in their matrix form, but their Killing forms are not zero!
how to get a feel for equivalence classes
you need to take the trace of the matrices that represent their adjoint action on the algebra, not the trace of the algebra elements themselves
well okay maybe i should ask: our action should be a number--that's okay. but what is compelling us to use this "$\text{Tr}$" operation a.k.a. trace of adjoint representation of argument
@qwerty because it's complicated enough already I guess? :P
@SillyGoose nothing, except that you don't have a lot of other options to get a number from this thing
hm impelling could have worked there too actually
11:30 AM
ahahaha so I can continue to be the noob in the back when @SillyGoose talks about principal bundles
@ACuriousMind for the killing form though should be something like $<X,Y> = \frac{1}{2}\mathrm{Tr}(X Y)$ in a matrix Lie algebra or I remember wrongly
@ACuriousMind but then it is possible to cook up some exotic version of the already exotic chern-simons theory? :P
@lucabtz well, that's the "everything is a multiple of the Killing form" again
I will spend the next 100hours reading this book and still have no idea what a gauge theory is
@qwerty im just trying to understand an action :P not gauge theory
11:31 AM
@SillyGoose I mean, again, for simple Lie algebras all the symmetric bilinear forms you might try to use here are multiples of each other
@ACuriousMind right right just making sure
so I don't really see in what sense something else would be "exotic Chern-Simons theory" and not just an entirely different theory :P
are there lie algebras that are not simple and not semi-simple?
@SillyGoose not in physics :P
no i mean of course there are
well i guess i didn't notice that we should restrict to simple lie algebras; though all mentioned cases i have seen (u(1), su(2), etc.) seem to be simple
11:33 AM
of course there are - the Poincaré group is not semi-simple
well now that's just hard :P
but i see
@SillyGoose you don't have to! Just for non-simple algebras you might indeed be able to find other bilinears to use here that are not multiples of the Killing form; I'm just arguing that because there is no choice at all in the simple case, there's little point in immediately worrying about special cases :P
lurks in the background of this lively convo
well i might as well specialize to a simple group :P already specializing to compact, connected, and simply-connected...
if i have a trivializable principal bundle $(P, M, \pi, G)$ does this mean $P \cong M \times G$ (where $\cong$ is whatever bundle isomorphic would mean)? or what is the precise statement. i tried looking up "trivializable" a bit ago but it didn't seem like standard terminology or smth because not much came up
if you already found out that "trivializable" is not standard terminology, then how do you expect anyone to tell you what it means? :P
11:43 AM
@SillyGoose thats a trivial bundle
not trivializable
a bundle is trivial iff its isomorphic to $M \times G$ with the action $(x, g_1).g_2 = (x, g_1 g_2)$
i think i'll leave you all to your bundles. really nice to chat physics. ciao!
hm i guess i should see what freed calls what
it just means trivial
12:01 PM
@lucabtz tell that to my intrusive thoughts :P
i dont feel good about my math and physics understanding
12:30 PM
@SillyGoose Asking a question like this is like someone trying to study say string theory then asking why the indices $j$ need to be the same when you multiply two simple matrices $A_{ij} A_{jk}$, I just don't understand why people are trying to use the biggest most abstract language possible and not just studying this stuff normally first, and as far as I can tell you just scared someone away from physics thinking this bluff language is to be taken seriously
12:40 PM
@RyderRude The commutator subgroup is the set of all elements which are obstructions to commutativity, i.e. all non-trivial elements $(a,b)$ for which $ab = (a,b)ba$ holds, so if you all yourself to interpret any $(a,b)$ as equivalent to the identity, you have $ab = ba$, i.e. you're just setting up an equivalence relation on a group and the quotient group is the set of equivalence classes associated to your equivalence relation i.e. the chunks of things equivalent to one another
@bolbteppa Who do you think you'll convince if you are just so dismissive of what other people are doing? How would you feel if I called the way you talk about physics a "bluff language"?
(also I don't see anyone getting "scared away" anywhere here)
@bolbteppa yes . i get that this holds.. maybe im overthinking it
Do the manifolds on which compactification is done in such a way that there is no low energy SUSY although there is SUSY in higher dimensions, have a name?
Where can I find a discussion of such string compactifications?
@Sanjana uh, just "manifolds"? :P
I mean...yeah a generic manifold should break all the SUSYs. But did people study such compactifications anticipating that there might be no low energy SUSY at all?
12:53 PM
@Sanjana string theorists generally believe SUSY is broken below the compactification scale, so you will not find a lot on this
by which I mean - string theory on a compactification manifold that does not preserve SUSY is actually believed to be inconsistent
If someone is trying to learn say qft without knowing basic calculus, it's just reckless to start enabling them explaining their questions drip by drip instead of just telling them to learn everything properly, similarly if someone doesn't understand why you need this simple trace term in the action and are asking about such simple things in the middle of quoting papers on kappa symmetry in Chern-Simons theory they clearly have never studied basics things like Yang-Mills theory,
Encouraging them to ignore the basics and enabling them is just absolutely reckless and it's been going on for a while now, it's clear that this poster barely understands why the 4-potential of electromagnetism is a connection yet we're constantly making things as abstract and general as possible peppered with the most basic questions in between, this negative type of pedagogy would not be possible without the internet, it's just crazy
since compactification isn't actually a "process", if the compactification manifold doesn't preserve SUSY, you just don't have SUSY, and hence you break string theory since the SUSY is important for UV finiteness etc.
@ACuriousMind But why is that? Does it actually violate some physical principle or it is just a well motivated belief amongst theorists?
see my next message :P
@ACuriousMind You just time travelled to answer my question in the past :)
12:59 PM
@bolbteppa i think SillyGoose is ok with their style of study. unless they find any difficulties, there's no reason to make these comments
@ACuriousMind Hmm...But can't there be SUSY in 10 dimensions for UV finiteness, absence of tachyon etc and when we descend down to 4D instead of the phenomenologically interesting $\mathcal{N}=1$, we get $\mathcal{N}=0$?
everyone approaches study differently @bolbteppa
I'm trying to help people by telling them to learn things properly and even trying to explain things, if that is ignored and being interpreted as me being negative that fine I'm not going to lose out by simply ignoring it
@Sanjana but there is no "SUSY in 10d" if your compactification manifold doesn't carry parallel spinors!
I thought its the other way round. If there is SUSY in 10 D and the compactification manifold has parallel spinors then there is some SUSY in 4 D.
@bolbteppa I personally think I learn a lot by reading your replies to other posters. I am sure there are a lot of ppl out there like me. So please keep replying when you are free ^_^
1:04 PM
@bolbteppa i didnt think u were being negative. by "there's no reason to make these comments", i only meant ur last comment where u r dismissing SillyGoose's preferred style of study
you are helpful generally
@Sanjana obviously preserving/having parallel spinors is the special situation that encourages a definition, the default is that this is not the case, why put a name on that
@bolbteppa Frankly: Your technical replies might be helpful but your abrasive attitude is not. You're pretty much constantly toeing the line of slipping into outright insults ("Bluff language", "crazy", "should not be taken seriously").
Also in this particular instance I think you profoundly misunderstood the nature of the question (because you refuse to entertain abstract lines of thought as equally valuable) and interpreted it in the most uncharitable way possible.
@Sanjana well, the problem is that you'd have to solve the supergravity equations, which is a horrible task when you don't have the assistance of SUSY. I think in the end we just don't know any non-supersymmetric string vacua
@Sanjana there is a tiny bit of discussion of non-supersymmetric stuff in 10D vs 4D here which may or may not be useful
i think it's all in good fun to have multiple viewpoints about the usefulness of abstraction. i dont find it remotely insulting.
1:19 PM
Abstraction is fine when you understand what you're abstracting
@bolbteppa abstraction is also the point when you're learning math
im not sure if you have noticed, but the past few days i have been reading a text titled Geometry, Topology, and Physics. In particular, chapters titled "Fibre Bundles", "Connections on Fibre Bundles", and today "Holonomy", which if not is the literal definition of building up the pre-requisites to understanding the object of learning I had in mind from the beginning, then is pretty close
some people like more abstraction, some like less. it's good to understand all viewpoints and not be completely dismissive
@bolbteppa I disagree with your first paragraph. one of my profs being willing to entertain elementary discussions about representation theory got me interested in the subject and how it plays a role in qm in the first place. also, i am not sure what i have quoted from the paper you mention. i linked it after googling because it was an example of chern-simons theory being extended from a simple structure group to a not simple structure group, which that concept itself is not hard to understand
I agree with ACM's observation that you also seemed to misinterpret my question. Would you like to write out a proof that the $\text{Tr}$ (not matrix trace, but the operation appearing in the chern-simons action) is precisely the only operation I can perform to obtain a number out from the action for a general chern-simons type theory?
If you would instead like to say it's the only reasonable option or is an option, I already got that from Witten in 1989 and it moreover explains nothing.
I think both the ACM and the bolbteppa line of thought (dubbed like this for the sake of clarity) have a point: on the one hand the way (rigorous or not whatsoever) one learns about math/physics is personal and should not be questioned, on the other hand one should also understand the basics, which is what I've suggested e.g. concerning DG in the past few days.
My only advice to @bolbteppa is to take it easy. This is not meant to attack you but your message sounded spicy. Other than that, I agree with what bolteppa is saying, in general (not talking specifically of this discussion)
1:34 PM
i think if you feel particularly feel positive or negative about a certain type of pedagogy i can understand that, but i do not understand what exactly your problem with my approach is. it's been working for me personally (at least in my opinion), and i do not push it on to other people (at least i do not try to)
i also do agree that doing computations is meaningful and is ultimately what most people have to do and be familiar with.
2:14 PM
@SillyGoose i just asked a local mathematician who called calling finding stat points "optimization" an abuse of terminology... so uh, continue with the "stationizing" ;)
@qwerty extreme points are "optimising" in some sense. Stationary-only does not imply optimisation. i.e. both extremising and optimising are strict subsets of stationarising
2:43 PM
yes i was clarifying that it seems optimizing and extremising are synonymous, not optimising and stationising
@SillyGoose I'll risk coming off negative one more time. You are asking why, in an incredibly advanced paper by Witten, he is using (a multiple of) the Killing form in an action constructed from simple Yang-Mills fields aka connections.
The Killing metric is used because it's an invertible non-degenerate metric on the finite-dimensional simple Lie algebras which is invariant (as explained in the wiki) and (as mentioned in the wiki) it's the only such metric, you study this as part of studying 'Lie Algebras'.
In infinite dimensions or for non-simple Lie algebras things get far more complicated and, as usual, despite all the supposed generality they are completely ignoring these deep generalizations by putting special conditions on things while conveying the appearance of total generality...
In that case we can also nitpick: extremising means max and min are both accepted, whereas optimising only one of the two is wanted.
What he's then doing is studying (in the non-abelian case), instead of the usual $F^2 = F_{\mu \nu} F^{\mu \nu}$ Maxwell action, he's studying $\tilde{F}^{\mu \nu} F_{\mu \nu} = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma}F_{\rho \sigma} F_{\mu \nu} $, which in the Abelian case this becomes a total derivative aka a 'topological term' but in the non-abelian case it becomes non-trivial.
One usually encounters this stuff when first defining the Maxwell action and one has to consider all possible (quadratic) actions, leading you to $F^2$, so he's just re-doing this analysis in the non-abelian case, and he's actually excluding the $F^2$ term because it depends on the metric $g^{\mu \nu}$, he's allowing himself the use of $\epsilon^{\mu \nu \rho}$.
However one has to study the non-abelian generalization of $F^2$ properly first, i.e. one has to study basic Yang-Mills theory or one will get lost by simple things, which is usually done in qft. He just assumes the reader knows all this
@bolbteppa this is my favorite line of this week
@bolbteppa by the way, IIRC there is this argument on L&L 2 and they do some magic to single out the usual Maxwell term
That's exactly what I was referencing
2:53 PM
I wanted to show off my knowledge of L&L
That second volume is still the best of the series for me
When you do particle physics and study yang-mills fields you basically have to re-do this analysis and you find it gets more complicated
LL2 does not cover qft stuff?
it's all classical?
@qwerty no
It covers classical electrodynamics and some GR
Classical field theory, basically
yes i'm asking why this is being covered in ll2?
My only 'negative' point is it is obviously a mistake to spend weeks/months agonizing over the simplest 'prerequisite knowledge' part of some incredibly advanced paper, getting distracted by supposedly big language in the set-up and misled into thinking this is really important (despite being told its not), instead of spending that time studying the prerequisite material properly which the author assumes the reader knows.
2:55 PM
@qwerty Because we're talking about classical stuff at this point
Instead of the bundle language, if I simply say $A^{\mu}$ is a Yang-Mills field, simply by knowing basic electromagnetism I know this transforms as a connection, I know the gauge transformations, I know its defined on some manifold with some metric $g^{ik}$ if I'm doing a curved space version, if I take the bundle language seriously I have to worry about a ton of irrelevant useless things
I mean, L&L considers the Maxwell action, which is the same term appearing in the QED lagrangian and generalization of it appears in the Yang Mills lagrangian
It's like thinking you can't do any quantum mechanics until you understand differential geometry so you can then think of a wave function as a section of a complex line bundle, so before you can study e.g. the hydrogen atom now have to specify transition functions and local trivializations, has any person in the world ever even discussed any of this for the continuous spectrum of the hydrogen atom
oh i thought you were talking about yang mills
The lagrangian is still a classical object, though :P
Although you call it "QED lagrangian", it's still a classical theory of electrodynamics. It becomes quantum when you quantize it
@bolbteppa I think the comparison would work better with functional analysis stuff in the case of QM, but you have a point :P
At least DG is cool to learn, FA is so technical it gets boring. If I could erase rigged Hilbert spaces from this world I'd sure do it
I don't want to bother learning that stuff and yet I guess I have to know some of it
3:01 PM
Yeah that's another great way to get distracted by irrelevant stuff, why don't we just write everything in the language of axiomatic set theory symbols in a language a computer will understand if we're going to strive for rigor, or are we going to admit we're happy assuming obvious things and focusing on what really matters
@Mr.Feynman do you really have to know some of it?
@naturallyInconsistent I mean, there are many things you don't have to do :P
first, i'd like to clarify that @bolbteppa i personally do not find your comments in general negative, and it is in fact nice to see different approaches to the same material in this chat. separately, i think maybe my intentions/interests are not clear since i have never stated them. (1) i am doing this chern-simons stuff for fun. (2) practically, i have wanted to learn differential geometry since i think it is the last "foundational" mathematics (other than complex analysis maybe) that i have
no familiarity with
so it is not that i am trying to even learn physics or learn physics efficiently. but the chern-simons business is interesting and i can survey some practical tools in the process of learning about it
@Mr.Feynman but that is not why I am nitpicking you on that. ACM said that the whole rigged Hilbert spaces shebang fails to actually be of any use in understanding the continuous spectrum of QM and QFT, that it is actually a dead end, so there is really no physical reason to be interested in that deep lore.
@bolbteppa in that case I guess it's about how one approaches it. If you see somebody using plane waves and say "you know, this is delicate and it works because [...]" then it's fine. On the other hand, if you say that plane waves are nonsense, you're taking it too far
3:05 PM
it seems even quite a gap to relate this chern-simons stuff to physical stuff. and it is not like it plentifully models stuff in the real world either. i have only found that it has been found relevant in fractional quantum Hall effect models and topological quantum computer models. other applications in mathematics and etc. of course
@naturallyInconsistent Oh, sure there isn't any physical content. Yet, I'm curious as to why things are different
@Mr.Feynman He does not think plane waves are nonsense. But he thinks that by focusing on Hilbert spaces we are defining things so that the continuous spectrum "eigenfunctions" are actually illegitimate because they do not live in Hilbert spaces, i.e. that momentum eigenstates are not suitable quantum states, and he has a problem with that.
@naturallyInconsistent I'm sorry, who's "he"?
@Mr.Feynman ah, the curiosity is gonna crack the pot this time. Good that it is not killing kitties
@Mr.Feynman bolbteppa, obviously
we write a group using its list of generators and the relations. then taking the quotient is the same as adding some relations where the subgroup elements are identity
is this correct
3:09 PM
@naturallyInconsistent Oh, my example was not about anyone here. I was making a point about how people sometimes exaggerate stuff
I mean, I was replying to your reply to him, so if I do not mean you, then I mean him.
@Mr.Feynman ... maybe this is genius...
@RyderRude you are adding/imposing an equivalence relation on the group, where $g$ is equivalent to $h$ i.e. $g \sim h$ if some condition holds, and you want this to be compatible with the group multiplication, so that $g h^{-1} \sim e$ or $e \sim g^{-1} h$ thus you get a subgroup $H$ of elements equivalent to the identity and you can see that $gH = H g$ will have to hold
yes.but it's hard to visualise an operation on equivalence classes.
i feel the generator formulation makes it intuitive
and thus the chat descends into abstract concreteness
or concrete abstractness
whatever causes more damage
An equivalence class is represented by a single element from that class, so operations on the equivalence class are just operations on an element from the class, it's no different from chopping the real line into 24 hour periods and working with hours like 2pm in a 24 hour period instead of 'the equivalence class of all numbers equivalent to 2pm'
3:13 PM
yeah... one can also just visualise representatives
but representatives are not closed under the group operation. one has to go back to the equivalence class
e.g. if i pick 2 as the representative of even and 3 as the representative of odd, then 2+3=5 which is not a representative
so i have to instead do [2]+[3]=[5]=[3]
The plane is described by a line through the origin (subgroup equivalent to the identity), along with all parallel lines in the plane, where your cosets are the parallel lines, each line can be represented by a vector starting from the origin, where you really only care about the component of that vector which is perpendicular to the line through the origin
is this about the grassman manifold
A: Why the term and the concept of quotient group?

Jack Schmidt Quotient groups $A/B$ count cosets of $B$ inside $A$. The counting even works well with addition. The Cartesian plane forms a group A, and a line through the origin is a subgroup B. The cosets of B inside A are all the parallel lines. How many are there? Suppose B is the line: $$ B = \{ ...

oh this is something else
@bolbteppa yes. if the perpendicular component is the same, then the points r equivalent
a term like $\text{Tr}[X \wedge Y] = 0$, right? since in index notation this term is the product of the usual trace of a commutator (always $0$) times some form related stuff?
3:28 PM
You should write it out explicitly to check, section 12.8 of this has a nice little discussion of these CS terms making the indices and traces explicit, Nastase's books on Classical and Quantum Field Theory also explain these things if you want to get into it
3:51 PM
the generator formulation of a group is non-unique, right? e.g. the cycle group can be formulated as , A|A^3=I and $A,B|A^2=B, B^2=A, AB=BA=I$
i think i did the variation correctly...
4:16 PM
bgm: Goldbach Variations, by Glenn Gloud
00:00 - 17:0017:00 - 00:00

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