okay i think i have fixed my backwards exposition of spin i hope XD: "It turns out that every finite dimensional representation of $SU(2)$ is completely reducible. And, for every positive integer $n = 1, 2, 3, ...$ there is a unique (up to isomorphism) irreducible representation of $SU(2)$ of dimension $n$, denoted $n$-irrep. In particular, generators of the $n$-irrep can be used to model the spin observable operators for a spin-$\frac{n-1}{2}$ system."

"The words category and functor were borrowed by mathematicians from the philosophers Aristotle and Rudolf Carnap, respectively."

At first I thought "When did Aristotle use the word functor"

is the differentiable manifold part of a Lie group mainly important for allowing the use of calculus with the group structure?

Yes

does calculus fall under geometry :P

i suppose analysis is its own thing

but people don't call lie groups algebraic and analytic

they call them algebraic and topological or algebraic and geometrical

Found out how to use emojis in latex

@SillyGoose You can define calculus in terms of geometry

As you may remember Euclidian geometry can be mapped to $\mathbb{R}^n$

is GL(L^2(R^3)) correct notation?

I guess?

GL works for any vector space technically

@SillyGoose $\mathrm{GL(V)}$ for a given vector space $V$ is the group of automorphisms $\mathrm{Aut}(V)$

When $V=\mathbb{R}^n$ we tipically write $\mathrm{GL}(n, \mathbb{R})$ instead of $\mathrm{GL}(\mathbb{R}^n)$

Linear automorphisms*

apparently I will have to include Lenin in the bibliography

Since a Lenin book was partly the motivation for some Soviet cosmology

marxists.org/archive/lenin/works/1908/mec

$\mathcal L^2(\Bbb R^N)$ is standard notation for the set of square-integrable functions on the real line @SillyGoose, equipped with vector space structure you can then consider the linear automorphisms as mentioned above, the notation $GL(V)$ is standard, just substitute $V$ with $\mathcal L^2(\Bbb R^N)$ and you're there.

@Slereah it is understood that for a vector space I mean that (?)

Yeah an automorphism is an isomorphism by definition so does technically already include information about the vector space structure

In category theory you don't bother to say that isomorphisms of vector spaces are linear

@SillyGoose The most valuable thing you gain from differentiable manifold structure in most contexts relevant to physics is that you can compose functions $f:M\rightarrow N$ on manifolds $M$ and $N$ with the chart maps and turn them into functions $(\psi \circ f\circ\phi^{-1}):\Bbb R^m\rightarrow \Bbb R^n$. It's the same with the group operation, it's a map from the Lie group $G$ to itself, compose the group operation with the chart maps and you have the entire toolbelt of real analysis for free.

Sometimes people find it confusing that words like "smooth" are suddenly being used in physics to describe things like vector fields, but in reality you're not defining anything new, you're just taking advantage of the chart maps and defining a vector field to be smooth if its component functions are smooth in all smooth charts.

Smooth can also be used to describe how they act on scalar fields

A smooth vector field is such that for any smooth function, the derivative V[f] is smooth

True there are quite a few equivalent statements

Some people do free mountain climbing. Does one really need to do that to feel great?

I think it's too much

I guess different people feel great by taking different types of risks.

@RyderRude Yes

hm do ya'll think it's okay to introduce definitions of point-set topology in the context of metric spaces even though i want to introduce these definitions to talk about matrix lie theory

it seems okay but im not sure. at least for this first definition for a compact matrix lie group, the notion of compact is that of compactness in euclidean space

@Charlie i guess it's a valid life choice. Nothing like being near death

What do you mean "is it ok"? There is no harm in knowing some basic topology

@RyderRude It's not for me but some people seem to enjoy it :P

well more like should the definitions be presented in the context of general topological spaces

I assume $R^n$ here means euclidean space because heine-borel is being used

You can also interpret them in the context of a general topological space, compactness meaning every open cover contains a finite subcover.

Yes that's correct

hm okay i provided that definition

but i need to also provide the atomic definitions of open, closed, etc. which are different if i am talking about general topological spaces or metric spaces

i gues im wondering if it matters if the ultimate goal is to talk about lie theory :P

@Charlie i think free hill climbing is understandably awesome, becuz at worst u immediately die doing what u love. But substance abuse and crimes can screw you for years. I don't find that great :P

But i support free mountain climbing even tho it scares me lol

They even say that because of Heine-Borel the definition is equivalent, a matrix Lie Group $(G,\circ)$ is compact $\leftrightarrow$ it is closed and bounded as a subset of $\Bbb R^N$

idk i guess it's kind of strange because definition 1.8 assumes that we endow the topological superset with a (topological, not general relativity) metric--here the superset is treated as $R^N$

where it seems like there is actually no natural metric for the manifold part of a general lie group

@SillyGoose Whether or not "it matters" really just depends on how mathematical you want your understanding to be

Sorry my connection is playing up so my messages are going through slowly

hm well i would like the statement to be correct--def 1.8 seems dubious. but i suppose the point is not a mathematical treatment of lie theory :P

also i know not general topology only point-set from a first course in real analysis

no worries at all

It's not dubious, they're telling you that if you consider the inclusion of $G$ into $R^{2n^2}$, you can more easily prove that it is compact by showing that it satisfies the conditions of Heine-Borel

You could also show that every open cover of $G$ contains a finite subcover but presumably the point they're making is that that's much more of a pain

Free climbers can live happily for their entire life. I cant say the same for hard drug addicts

So instead of going to the trouble of showing that (which again, if $G$ is compact you absolutely in principle could do) they just consider it as a subset of $\Bbb R^N$ and use the Heine-Borel theorem

but $M_n(\mathbb{C})$ does not inherently have a metric, no?

But free climbers also die at their first mistake, while addicts can recover and be happy again

Every matrix space is isomorphic to a vector space, the vector space of $2 \times 2$ real matrices is isomorphic to $R^4$ on which you can define a metric without issue by inducing it from the norm

but a vector space does not come equipped with a norm either

Sure, this is physics text so whenever they make reference to an additional structure on a space you are just to assume that they have equipped that space with such a structure

$M_n(\Bbb C)$, being the space of square complex matrices can be equipped with a metric, so we use it here because it's necessary to use Heine-Borel

yes but the compactness of a matrix Lie group should exist independent of assigning an arbitrary metric

It does, every open cover contains a finite subcover doesn't require a metric

All you've got there is a different way of proving it, which requires defining such a metric

hm it seems then that the purely topological way is more accurate and the metric space way is pedagogical

There is no sense in which the topological way is "more accurate"

well the topological definition requires no arbitrary choice

They're just different, the fact that using Heine-Borel requires definition of a metric doesn't make it any less credible or accurate

One can also just skydive instead of free climb. But i guess the safety factor would turn off adrenaline junkies

There are many "safe" extreme sports

It's a bit like proving a statement about vector spaces with and without reference to a basis, neither is "more correct" or accurate the other, they're just different. If you prefer to prove the statement without choosing (arbitrarily as you say) a basis then you're welcome to prefer that, but one isn't more accurate than the other as a result

hm well a basis does not really change the structure you are working with

a metric fundamentally changes the structure you are working with i feel like

I dont mind safety. I think it's a plus

In fact, it's a requirement for me

like some results to do with inner product spaces do not make sense as results for general vector spaces

e.g. just from the definition of an inner product, the cauchy-schwarz inequality doesn't have a meaning for a vector space not equipped with an inner product

I'm not doing anything which has a 1 in 500 death rate

@RyderRude thought is a shield from elemental experience, and thinking is a human speciality. But because it creates isolation from nature some of us chase such elemental experiences. Perhaps the ones with thicker shields need stronger stimuli

But u can be close to nature while being safe

But again, I think the free climb stuff is also a valid life choice

I guess being close to nature stops being satisfactory to them, so they have to get nature to almost kill them for it to feel good

You're never outside of nature. Physical location is not really at the bottom of this, pastoral settings etc. are no different than taking drugs lol (but less bad side effects)

Yes the confrontation with death pushes the regular thought process aside

You can get the same effect by jogging for one hour. Ok if you're in shape, 3 hours lol

This is not a recommendation

I have jogged for long but didnt feel much

What's long?

40 mins i guess

But i havent done it for an hour as far as i remember

But still, near death stuff must be something else entirely

But it felt good?

Yeah. It was elating.

Exercise, in general, brings that feeling

It's the same mechanism... adrenaline they call it?? :)

Yeah

Also, u dont really have to risk life to feel like u r risking ur life

I'm sure skydiving for the first time would feel as if u r risking it even tho u r pretty much not

Some people will view jogging as a risk

Especially if they're running alone in nature. What if a bear pops out?

And i think after a few years, even free climbers must stop feeling that they r risking it

That can happen in some places

Also, u can die from jogging too much i guess :P

I think the risk component is more psychological. Wanting to feel a greater achievement. But biologically probably you can get the same effect by risking less... your ego won't get as swollen but I think biologically the same effects will occur

Yes. I'm sure skydiving feels as thrilling as free climbing

Even tho it is safe

Yes, but then some people are inclined to think "Millions of people have skydived... I wanna do something GREATER"

There is something about clinging to a vertical rock at 3000 feet tho. It must be very special.

It reminds me of how Feynman admitted he sometimes did stuff just so he can tell the story, lol

Usual trekking wont give u that feeling

Yes... and if you're scared of heights you'll feel even better after doing it. That's the funny bit

I mean, if you survive

Imagine climbing Everest

lol

I still think its not worth it lol

Buzz Aldrin did say that the moon's surface was boring

It ain't. 'Cause again if you eliminate all the ego stuff which in my mind is kind of childish, there is no difference from taking Heroine, lol

I'm sure Everest's top feels the same, especially with the lack of oxygen

@Amit tbf climbers live happy lives

They have extremely special lives

I don't know if you can generalize, I'm not saying it's always ego driven, I think someone can fall in love with this stuff too

Yes, i'm talking about the ones who love it

That's fine. But maybe it is a bit silly to do it to prove something

They dont do it to prove anything

Human motives are complex, hard to make sweeping statements

Proving something to "yourself" is the same. And the psychological reasons you need to do that vary greatly

I think they just love the feeling. It's not about proving to yourself that you can do anything

They just love being at the height and free

@RyderRude Yes but really I'm responding to what you said here... it's worth it only if you like it

The way you said it implied you're looking for some other value

I dont find it worth the feeling. Tbh i would even hate the feeling itself

Because it would feel scary af

So there is nothing in it 4 me lol

Check out this climber called Alex Hannold

Cancel that free climbing course!!

lol

There was also a guy called Marc Andre. He climbed stuff no one else did

And he pretty much lived homeless

Alain Robert is cool

@Amit there goes the theory that there r no old free climbers

Dude's 60

Yeah, it's really his life

When he talks about it you can see too

is there a physical importance to the fact that the connected component containnig the identity of a lie group is a normal subgroup?

If anyone's interested, there have been a couple intriguing papers on arXiv over the last 2-3 days on SN2023ixf (1, 2, and some others). It looks like the progenitor may have had an initial mass of ~11 solar masses, a little bit lower than initial estimates. But it's holding steady in brightness in optical/UV, so if you have access to even a small telescope, you might be able to see it.

@ACuriousMind @Slereah @bolbteppa. I remember from a while back (18 months, 2 years, something like this) a quote or a statement in chat which I think was by one of you. It was something about how a clear formalism was like a clean window letting light passing through. I'm looking for the exact statement and possibly the source.

I must remember incorrectly as searching for some obvious keywords does not return what I'm looking for, but it's was really some observation about the need for good formalism.

I'm refereeing a paper where the formalism is actually used to hide the paucity of new results and stifles the clarity of the contents so I'd like to have the original quote.

^^ I must remember ** slightly ** incorrectly...

"Rigor cleans the window through which intuition shines." ~ Ellis D. Cooper , @ZeroTheHero

@Amit yeah that sounds like it... thanks.

It's one of one of the quotes on ACM's profile

Is the exponential map of a connection just the combination of two exponential maps

The exponential map $\exp : TTM \to TM$ which makes a diffeomorphism on $TM$ and then the exponential map that's just the flow from $TM$ to $M$ at the given point

Although... Not quite no

it needs to involve the full tangent bundle for it to work and only then fix a point I think

@Charlie Good find. It's actually from a fairly recent book called "Mathematical Mechanics".

@Charlie That's where I copied it from 🤣

Ah there it is : the flow of the connection gives us in $T_p M$ exactly one curve per vector, and those curves project down to geodesics on the manifold

@Amit i can ride my bike w no handlebars

Suppose i have a conductor which is surface charge. The Macroscopic electric fields can never have discontinuities right? I suspect any discontinuity that we apply as boundary conditions are simply the effect of us using a Gaussian surface that has a thickness of 5-6 molecules. Am I right in saying thaat even the macroscopic fields start changing rapidly as we move thru the conductor and go to the other side, which appears as a discontinuity?

@Amit And the book copied it from ACM's profile

hi -- i have a question. im trying to look at the behavior of a particle under some potentials. when i use these equations of motion for a particle under the influence of another particle of opposite charge, i get a figure that shows the radius increasing overtime which doesnt make any sense. shouldnt these equations give an r[t] that is in circular motion?

the potential im using is $-\frac{1}{r}$

lols

Watch out Einstein

@Slereah You know what's a really good way of thinking about/introducing a connection? Suppose you have a manifold $M$ and two vector fields $X$ and $Y$ and you really want to differentiate $Y$ in the direction of $X$. Treat them both as sections $X, Y : M \to TM$ of the tangent bundle $\pi : TM \to M$. Then, actually differentiate $Y$: take its derivative $DY : TM \to TTM$. Then write $D_X Y = DY \circ X : M \to TM \to TTM$.

But no one wants acceleration to be $TTM$-valued aka 2-flares, they want accelerations to be vectors as well. So now you observe that every $T_{(p, v)}TM$ has a natural copy of $T_pM$ sitting inside it, given by $T_{(p, v)}T_pM \cong T_pM$. Alternatively said, this is the vertical subbundle $V = \ker(d\pi)$ where $d\pi : TTM \to TM$ is the derivative of the bundle projection. Choose a fiberwise projection $\Pi : TTM \to V$. Then define $\nabla_X Y := \Pi(D_Y X)$

This is equivalent to the Ehresmann connection business, by defining $H := \ker \Pi$ to be the horizontal subbundle. $D_Y X$ landing here would correspond to zero acceleration, because $\Pi$ of that would be zero, so $\nabla_X Y$ would be zero. $TTM = H \oplus V$.

The whole apparatus of connections is defined because from the POV of global analysis, acceleration is naturally a 2-flare. But who the hell understands what a 2-flare is?

Do you mean a 2-jet

What's a flare

An element of $TTTTT\cdots TM$ (n-times) is an $n$-flare

n-jets are a bit different than this

I have read them called non-holonomic jets usually

They're closely related. But for example 1-jets naturally transform as affine covectors

Whereas 1-flare is just a vector

They're different bundles (eg count dimensions)

Yeah but non-holonomic jets are the same dimension as TTM

Holonomic 2-jet is $J^2 M$, non-holonomic 2-jet is $J^1 J^1 M \cong TTM$

I have never heard of this terminology. But your second expression doesn't work out, $J^1 M = \Bbb R \times T^*M$

Because 1st order Taylor expansion of a function is (constant) + (covector)

$J^1_0(\mathbb{R}, M)$ if you prefer

Ok, then I agree it is $T^*M$

To me a (local) $k$-jet is an element of $J^k M$, a (global) $k$-jet is a section $M \to J^k M$ which is holonomic if it is a prolongation of something ie a genuine Taylor expansion upto order $k$ of a function $f : M \to \Bbb R$ (denoted $j^k f$) and non-holonomic otherwise.

I need to write out like a dozen definition of the connection in that damn book I think :p

There's plenty of them

Should be useful

I find the one I described the cleanest

One thing I wonder is if $TTM$ relates to $J^1(\mathbb{R}^2, M)$

I have seen some people mention that $TTM$ has some relations with the derivatives of families of geodesics

Since $\partial_t \partial_s \Gamma(s,t) \in TTM$

Yes, a Jacobi field is like a vector field along a path in $TM$

So a 2-parameter gadget in $TTM$

a doodad

a thingamabob

How much of $TTM$ do families of geodesic span tho?

could you define it entirely from it?

geodesic spelunkies

also families of curves, not geodesics

Yeah you should be able to get everything by families of curves

I'm sure I could cook up like a dozen definitions for connections

well, affine connections, anyway

Oh, you're thinking there's some restriction because $[\partial_s, \partial_t] = 0$

So maybe I am wrong

Yeah that's the thing where it's the restriction to uuuuh

Frobenius

the semi-holonomic jets?

Integrability

Or that

I forget what's the condition

No that's just the second jet

$[X_1, [X_2, [X_3, [\cdots]]]] = 0$?

A wise man once wrote this : " Non-holonomic you just treat every jet coordinate independently, semi-holonomic you identify the first derivative part of the first jet with the first derivative part of the second jet, and holonomic is considering the commutation of derivatives "

Semi-Riemannian geometry

Iterated Lie brackets upto some degree is zero

So the families of curves just span $J^2 M \subset TTM$

Oh, sub-Riemannian, not semi-Riemannian

although that's enough for a torsion-free connection

@Slereah OK, I understand now. Yes, that makes sense.

Basically you can define a $k$-jet as a section of $TM$, together with a section of $TTM$, together with ..., with a section of $T^k M$ such that the obvious diagram commutes

$T^k M \to T^{k-1}M \to \cdots TTM \to TM$ being the $d^k \pi$'s

yeah pretty much

thumbs

IIRC it's related to the notion of sprays

I asked Ryan about this def of jets a long time ago and he was surprised he's never seen this before

I am likewise surprised you know this

But you're the niche man

Relativity is full of jets if u know where to look

All of this is buried in the kms book

Also Kobayashi's stuff

he wrote a lot of jet stuff

@Slereah Enlighten me

@BalarkaSen Most of the important bundles in relativity are secretly jets or built from them

samuel-lereah.com/tmp/…

Plenty of jets there

samuel-lereah.com/tmp/…

already jetpilled in 1981 damn

Coleman did a lot of interesting stuff on the topic

also that guy in modern times :

samuel-lereah.com/tmp/…

another interesting part being that connections are reductions from second order jets to first order jets

Or subsets of those, anyway

Yep

It is a nice proof of why there's only one LC connection actually

and several conformal ones

I have never seen this proof. Your point is that since it goes from 2nd order jets to 1st order jets its necessarily very overdetermined

So there cant be a lot of these

Well it's more that we're using a subbundle of those

i see

The reduction is between some subbundle of $J^2(\mathbb{R}^n, M)$ to $J^1(\mathbb{R}^n, M)$, but in fact they are both principal bundles of the same group in the orthogonal case

Related to the fact that the group prolongation of $\mathfrak{o}(p,q)$ is $\mathfrak{o}(p,q) \oplus \{ 0 \} $

So the connection gets entirely determined by the components of the metric tensor

interesting

You can find all that stuff in Kobayashi's book iirc

It's a nice book

This is Kobayashi-Nomizu or something different

another Kobayashi

well, same Kobayashi, different book

Transformation groups in differential geometry

?

yeah

Gotcha

One thing I wonder about is if there's some nice way to think about how the tangent bundle and its dual are related through a double fibration

Like you have the double fibration $\mathbb{R} \leftarrow \mathbb{R} \times M \to M$

Left fibration is curves, with the jet being the tangent bundle, right fibration is scalar functions, with the jet being the cotangent bundle

Feels like there is some Diagram Reason why they should be dual

That maybe relates to duality in twistors which also have double fibration stuff

Except they're more boring dualities because they map point to point in this case