my impulse would be to go to index notation because I'm not sure how to mix cross-products with matrix multiplication

So the point is that the vertical stuff is precisely the subspace you cannot get as such a graph.

Hm ok I see that I can obtain $H_b$ as such a graph, but I'm not seeing how to get the local section I want from it

Just take a constant linear map in that local trivialization, @Alessandro.

Well, that doesn't quite make sense.

Just choose a chart on $F$ near $b$ and then it makes sense.

the linear map whose graph is the desired subspace

Right.

Hi, @Rithaniel.

Am I not getting a local section on $B$ rather than $M$ this way?

@TedShifrin Heya Ted

Oh, I've got your notation backwards. Most people use $B$ for the base. Duh.

You're using $B$ for the bundle. So $B=M\times F$ locally.

Power went out today. Getting all my internet tabs re-opened

Right I have $B=M\times F$ locally

Ok but now I see what you mean

So using charts on $M$ and $F$, do what I said :P

yeah I see now

Is the reduced relative homology defined as $\ker(H_n(X,A) \rightarrow H_n(\{\mathrm{pt}\},\{\mathrm{pt}\}))$? $(\{\mathrm{pt}\},\{\mathrm{pt}\})$ is the terminal object in topological pairs right

of course that means that its just the same as the usual relative homology, but that's what I want to justify

@Semiclassic: Somewhere on here I wrote an answer showing the right way to handle that orthogonal/cross-product thing. It is not unrelated to the spirit of your pet peeve.

hmm

Let me see if I can find it easily.

thx

what happens there

that's not the pic I posted or know

@TedShifrin I'm still slightly confused by this

although the music selection seems weirdly interesting

internet is breaking

Here, @Semiclassic.

last try

Sorry, Alessandro, I mean constant linear map once you've trivialized, don't I?

ok wtf

stackexchange broke

@TedShifrin what do you mean constant

hmmmmm

You take a linear map $T\colon\Bbb R^n\to\Bbb R^m$ whose graph is the desired linear subspace at the origin. Now just take $f(x)=T(x)$ as your section.

lmao now somebody else accidentally posts a picture of algebraist kid referencing nlab in the exercise session

I think that's what reduced relative homology should mean, ye

seems like an implication of that answer would be that, if $M$ is a proper orthogonal matrix, then $Mv_1\times Mv_2=M(e_1\times e_2)$, which goes against the grain of what you said above

the point being that the differential of a linear map is pretty much just the linear map itself then?

Right, @Alessandro.

(because det M = 1 and $(M^{-1})^\top=M$)

@Thorgott thx

Ok makes sense, thanks a lot!

Yes, I said that, @Semiclassic.

Do you happen to see why any two such sections must be the same class in $J^1B$ as well?

"Nope"?

Well, you need orientation-preserving.

ah

so $A(u\times v)=\pm Au\times Av$

Right.

fair enough

If they have the same tangent space as the image, that means the 1-jets are the same, Alessandro.

But maybe I'm not sure what you're asking.

sensible too, come to think of it: if you do a genuine rotation, then the cross product of the rotated vectors should be the rotation of the original cross product

That's all related to why you physicists call the cross-product a pseudovector.

right

The exterior product really is the right math notion.

yeah, i don't disagree.

I just thought of that post, though, when you pet peeved.

Ah, I get to be impolite prof now that polite proofs is here.

I have the biggest noob question, but why does $(\sin(t),\cos(t))$ make a particle go clockwise, intuitively?

think about small t

Think about where it starts, too.

Hm I'm a little confused now. Geometrically I agree that it should be obvious that same tangent means a first order contact means same 1-jet. But my definition of 1-jet says that in local coordinates all of their first derivatives agree, and that seems stronger to me than asking for just the images of the tangent map to agree

also, if you understand why $(\cos t,\sin t)$ is counter-clockwise, then the effect of reversing the functions should be obvious

Oh, you're right, @Alessandro. What was the precise statement again?

Interchanging the functions reflects across what line, @polite?

I want to see why if I take two local sections $\sigma_b,\rho_b$ such that $\sigma_b(x)=b=\rho_b(x)$ and $Im(T_x\sigma_b)=H_b=Im(T_x\rho_b)$, then $j^1_x\sigma_b=j^1_x\rho_b$

@TedShifrin the whole vector/pseudovector thing is weird in physics because it's not a matter of the notation we use but of how they transform under coordinate transformations

whereas with the exterior product it's manifest i guess

Wait

It is $y=x$

I doubted myself for a while there for some reason.

Yes, it is.

(My deleted message did say that)

@Alessandro: You're right, of course. Just think about the trivial bundle case. Two linear maps can have the same image without being the same. Your complaint stands.

That's why I asked if there were further conditions on the definition. In the principal bundle case, we have further conditions, of course.

math.stackexchange.com/questions/3991778/… please someone clarify this?

OK, lunchtime for me. Bye.

wiki does claim that the correspondence holds in general, with neither proof nor reference of course

I can give you an unspecific reference

99% sure you can find this in Kolar-Michor-Slovak, but I don't know where

That's the reference wiki gives for the definition of Ehresmann connection actually

of course it would be

@user2103480 judging from this question, this may not just be your problem: math.stackexchange.com/questions/3991927/…

let me test it myself

looooooool

welp

@Thorgott I'm trying to see if the correspondence is also in it

imgur seems broken

so meta stack knows about it

ah, I think it's the start of ch-17

ok, the claim is there, but not the argument

yeah but it just says "it's canonically identified with a section"

Someone's having fun

ok, I think the issue is the definition of jet

KMS define jets by post-composing with curves and demanding derivatives agree up to r-th order and with that definition, it's clear that first order contact means the same thing as equal tangent spaces, no?

ok no, it's not

it's clear that their definition implies equal tangent spaces, but the converse doesn't seem true to me

yeah right, I'm also a bit puzzled, though I'm sure that's on me

lmao the algebraist kid argued that if $G \oplus \Bbb Z \simeq \Bbb Z^n$ then $G \simeq \Bbb Z^{n-1}$ by the classification theorem of finitely generated $\Bbb Z$-modules and the topologist lecturer was visibly perplexed

thorgott brain

I mean, he's right?

that's silly, but only because this is a lemma you effectively prove before classifying f.g. abelian groups

but I don't think there's anything wrong with it further

the one thing you shouldn't prove using a fundamental lemma is a result you need to prove said lemma

i like the plural of lemma, sounds like some yellow sea creature: lemmata

@LeakyNun still seems like an overkill. the rest he did pretty clearly and imo better than the official solution

do you have an argument better than "subgroups of free abelian groups are free abelian"

nah

$G$ is isomorphic to the subgroup of $\Bbb Z^n$ given by the obvious isomorphism into $\Bbb Z^n$ and then it must be finitely generated free abelian so some $\Bbb Z^k$ and I guess $\Bbb Z^{k+1} \simeq \Bbb Z^n$ as groups implies $k+1 = n$

replace "I guess" by tensoring with $\mathbb{Q}$

That's a nice fact

ah nice

that removes

Here's another nice one

all the exponents

or do the fancy GGT approach and look at their growth rates

If Q is a bilinear form on Z^n with n vectors v_i so that Q(v_i) = 1, then Q is diagonalizable positive definite aka similar to the all-1 matrix

prove that tensoring with $\Bbb Q$ is functorial first man

@user2103480 obvious

@Thorgott yes I will do that definitely

groups taught me that I don't care at all about symmetry

you should do that, it's fun

damn these jazz albums that stackexchange accidently posted instead of my image are straight fire

although one is just the best of thelonious monk which is too easy to give credit for

Hey guys. Can someone help me visualize something? Need topology/differential geometry.

The manifold R x S2 is homogeneous but nowhere isotropic. How do we see that?

@AlessandroCodenotti I mean there are mixing maps which aren't minimal, so you can put some positive invariant measure along the invariant subset and get non-ergodic invariant measures

Denjoy is one such example if you haven't seen it

@Ted @Thorgott math.stackexchange.com/questions/3992095/… I ended up asking on main

@BalarkaSen I haven't, where can I read it?

nice

counterexamples in topology be like

it's all the hawaiian earring?

always has been

@user2103480 the only place where i've managed to find weird counterexamples interesting is in dynamical systems

besides that, pathological cases are just kinda....bleh

the Hawaiian earring is an immensely interesting space

pshh don't summon the alessandro

both topologically and algebraically

@AlessandroCodenotti Suppose graph of two linear maps are the same sets. What can you say about the linear maps? This will answer your question.

he will get mad at your ignorance for pathologies

@AlessandroCodenotti No clue. Look up "Denjoy counterexample"

@BalarkaSen watch TheSerfs now

I can't

no, tomorrow

@Astyx LMAO

happy inauguration

@Semiclassical I generally don't remember many counterexamples. I personally like the "nowhere monotone, differentiable function" and "meagre set of full measure" a lot

i know that sin(1/x) generates a lot

@BalarkaSen hmm

but ugh

hmmmmm

not sure

The first one only came to my attention after wondering why the proof that brownian motion is nowhere monotone doesn't imply it's nowhere differentiable

But I'm not fully thinking about it right now

@Alessandro You stated it wrong. Section of $J^1B \to B$, not over $M$.

@AlessandroCodenotti they're the same linear maps

brownian motion is something i should understand better than I actually do

And this proves the book is correct

like, random walks, yay

but when I start seeing ito calculus I'm like "welp, i can't say anything"

@Semiclassical should you? I'm not sure how well one actually needs to understand it in most cases

tbh it's more just the frustration of seeing people talk in a calculus language that i don't understand

Howdy, a @Balarka.

@Semiclassical that's fair, I'm also still adapting to the calculus

Hi, Ted!

Hi Ted

Hi Astyx

stuff like $dX_t$ where $X_t$ is brownian motion

i know that that language makes sense but

for me it mostly signifies a boundary on my knowledge

@AlessandroCodenotti You can pass to coordinates and reduce the problem to the jet bundle of R^n x R^m -> R^n. Sections of this are graphs of functions R^n -> R^m; asking their tangent maps at 0 are the same is asking their derivatives have the same graphs hence derivatives are the same hence they define the same 1-jet

These are all integral equations and hard to work with if one doesn't exactly know what $dX_t$ manipulations do to the integrals

right

i also know that i'm more versed on differential equations than on integral equations

To solve exercises, I transform this back to the integral equation anyways, since the differential manipulations often hide some stochastic fubini 'n shit

gotcha

But the profs just work with the differential representation, and very fluently so

1-jet equivalence and order 1 osculation are the same thing. I think this breaks down in higher dimensions but @Ted would know this way better than I do.

@BalarkaSen Uhm I disagree? (x,y)->(x,y,0) and (x,y)->(y,x,0) as maps R^2->R^3

How do they have the same graphs

Alessandro: Wiki says you got it wrong, and fixing the base space fixes it, I think.

@BalarkaSen they have the same image

So what?

@TedShifrin wait what do you mean

Section over B, not M.

I'm not following. I always had sections over $B$?

Where exactly do you mean?

Oh, we were doing sections over M. Look carefully at our discussion .

no but that was to see that $\sigma_b$ exists

Balarka out here using "graph" to mean tuple $(x,f(x))$ or what

But once I have $\sigma_b$ I use it to build a section $B\to J^1B$ given by $b\mapsto j^1_x\sigma_b$

What? You don't know what graph of a function means?

They're sections of the trivial bundle projection

You have two sections of R^n x R^m -> R^n, with tangent map at 0 same image. That forces their derivatives equal.

In virtue of being a graph

Two linear maps with graphs equal as sets are same as linear maps

any two functions with graphs equal as sets are equal

Yes but this proves what you want

i dont see how

but im also not thinking

This is calculus. I don't know how to explain it better

Tangent map of a section (x, f(x)) : R^n -> R^n x R^m has image the graph (v, df_0(v)) : T_0 R^n -> T_0 R^n x T_f(0) R^m

Which implies the image determines df_0. Shrug.

Did you see my comment here @Balarka? chat.stackexchange.com/transcript/message/56797363#56797363