I may take a look, but the fact is that it is not so important

it is essentially just to make a remark on a seminar I will do in a couple of weeks

@0celo7 Kinda. First sheaf cohomology, then I attended a talk on positive TFTs.

so it is not worth spending too much time on it

if the people at math.SE come up with an answer, I may include the remark, if else I won't :D

@yuggib Um...the "configuration space" is commonly the manifold itself - the cotangent bundle is the phase space, the tangent bundle is where Lagrangian mechanics lives.

@ACuriousMind Well ok, I intended the space of position and velocities for the tangent, and the phase space in the other case

but probably you're right, with config space just manifold is intended

@yuggib I think that the tangent bundle is a vector space if the manifold was a vector space is kind of a red herring - that's formally true, but really you shouldn't be adding points on the manifold at all - the vector space structure is completely disjoint from the geometrical structure that one wants to capture by the notion of a manifold.

Am I overly nerdy, or is this really going to appear as an offensive choice of a username:

?

@ACuriousMind Well, I am interested on the properties of the tangent space as a top vec space

@TheDarkSide Not sure what "nerdy" about it, but yeah, it is kind of offensive.

not on the geometry :P

you know that I do not like geometry so much :D

@yuggib If you're not interested in the geometry, why are you looking at manifolds? :P

@ACuriousMind just to make a smart remark in a talk I have to give (if possible)

Hehe, alright, let's see:

That it is a topological vector space at least demands that it be (co)homologically and homotopically trivial

@ACuriousMind Yes, maybe that's not the right word. Maybe it is something like moral police or thereabouts.

Came across it while physics.stackexchange.com/review/suggested-edits/106184

First question is: What manifolds have a chance to not be $\mathbb{R}^n$ and still be trivial.

The only thing I can think of are balls.

^ Taken out of context, that made me giggle.

Which are homeomorphic to $\mathbb{R}^n$, which makes them unsuited.

:P

@ACuriousMind So it is a necessary condition for the vector bundle to be a top vec space that it is a trivial bundle?

Hm, I think being homologically and homotopically trivial might already force the manifold to be homeomorphic to $\mathbb{R}^n$.

Surely there are pathological situations where "by accident" the tangent bundle can be endowed with a vec space structure even if it is not the trivial bundle

at least in infinite dimensions

What's your notion of "infinite-dimensional manifold"?

frechet manifold

Is the tangent bundle ever the trivial bundle

Isn't it supposed to have the rotation group or somesuch

Ah, it seems there are contractible manifolds which are not $\mathbb{R}^n$.

At least in $d>4$.

in the case of $M$ vec space it is (I have been told so)

Although I guess the tangent bundle to $\mathbb{R}$ would be

@Slereah Of course - the tangent bundle to every Lie group is trivial

By "trivial" are we saying "the associated group is the identity"

@Slereah Trivial means it is globally isomorphic to $M\times\mathbb{R}^n$.

What associated group are you talking about?

The group of the fiber bundle?

anyways it is nice how it works on math.SE...they pointed out silly things (that was in my opinion clear from the context) with unuseful comments, and then nobody came up with an answer

@Slereah What does that mean? You always have $\mathrm{GL}(n)$ acting on the fiber.

The structure group acting on the transition function

anyways, my question was actually mostly on the homeomorphism between $T^*M$ and $(TM)^*$ when $TM$ has a top vec space structure

@TheDarkSide Only to people who get offended easily.

@Slereah Again, I think the structure group of the tangent bundle is always $\mathrm{GL}(n)$. (Which gets reduced to $\mathrm{GL}^+(n)$, i.e. matrices with positive determinant for oriented manifolds, the existence of the reduction is the obstruction to being orientable)

@ACuriousMind : Oh okay

Because when I read "trivial bundle", I think more that the structure group is $I$

No, that's not what it means. A bundle $B$ is trivial if the local isomorphism $B\to M\times F$ for $F$ the fiber extends to a global isomorphism $B\to M\times F$.

Hm, isn't that the same tho?

@Slereah Not at all, as the example of trivial tangent bundles shows.

I guess I need to reread some Steenrod

send me Steenrod

I'll actually appreciate it

Triviality may mean that you can perform a reduction to the trivial group $1$, I don't know enough about reductions to answer that.

Topology for Christmas

@Slereah yes pls

Sorry tho, I need me Steenrod :V

For bundles on the go

isn't "Christmas" offensive

It is to heathens

@ACuriousMind I can choose coordinates so that the metric determinant at a point is unity, right?

That's just Riemann coordinates

@Danu @Danu I have no idea of all the tags.

I keep getting Cepheids.

wtf are you talking about

@0celo7 Planet Hunters

planethunters.org

@ACuriousMind I'm taking a coordinate free proof from a book and translating it into index notation for an answer I'm writing

This hurts me

But it's easier to prove things like this

however, I preview that no one will answer my question...so no smart remark in the talk T__T

@yuggib: I think it's going to be difficult to show what a manifold must be such that the tangent bundle is a vector space: Manifolds can have isomorphic tangent bundles without being themselves isomorphic, and contractibility of the tangent bundle doesn't imply contractibility of the base AFAICS. So there's no obvious condition on the manifold to have a vector space as tangent bundle, and I suspect this is not investigated, since there's no geometric reason to think about adding points.

ahhhhhhh I have to get out HE

that damn book

I think HE is book of the dammed for GR

@ACuriousMind yes, but at least is it true (even for Fréchet manifolds) that if $TM$ is a top vec space then $T^*M\simeq (TM)^*$?

at the very least for trivial $TM$ and $M$ itself a vector space

@yuggib I don't think so, see nLab,MO,math.SE: A Frechet vector bundle is itself locally a Frechet space, but the dual of a Frechet space is not in general a Frechet space, so the dual of the tangent bundle is not a Frechet vector bundle, hence it cannot be the cotangent bundle.

This could only be true if $TM$ is itself a Banach space.

What is a Banana space anyway?

is it true for Banach manifolds?

Possible, but I have no idea

@0celo7 a space that is contractible to a banana

@yuggib are all bananae contractible to points?

that is a question for ACM, but probably yes

@ACuriousMind ?

but maybe they have holes on the external part

@0celo7 Yes, unless a worm or some other thing has eaten a hole through them

@ACuriousMind Yes...

Thus, healthy banana spaces are contractible :P

@ACuriousMind but why the cotangent bundle should be a frechet bundle? even locally, it is defined as the dual of the tangent bundle on each fiber

and on the fiber, the dual of a frechet space could be non-frechet

but anyways, it is definable

@ACuriousMind I like a bit of worm juice with my fruit

@yuggib Hm...you're right.

It's ugly

I love ugly things

@yuggib do you love your mother?

If the cotangent bundle isn't a bundle, what's its point

@ACuriousMind it is a bundle, but not with a frechet structure

I think...

it is the dual bundle

this seems to suggest to me that "Frechet manifolds" are not the actually correct setting to consider "infinite-dimensional manifolds" in.

But you probably don't care for such considerations

the ones I am thinking of are Banach, or even Hilbert manifolds

and actually trivial

hey ppls

but why not frechet

?

would one of you mind reading this answer

and telling me if it is written acceptably

@yuggib Well, given a X-manifold, it seems very natural to me that the tangent and cotangent bundle should be X-manifolds as well

I have no clue how my formal proof skills are

@ACuriousMind :-D

If taking the bundles is not an operation within your category, it's a crap category, and the category of manifolds is already very ugly to begin with

but the infinite dimensional world is so much complicated

and reflexive/self-dual objects are few

@yuggib Well, the suggestion is not to restrict the notion necessarily - this might suggest you should consider a generalization of Frechet manifolds as the proper notion.

I.e. find the "smallest" generalization of Frechet manifolds such that both bundles are again generalized Frechet manifolds

ahahah ok

but the result on which I would like to do my smart remark works only on locally convex spaces

I think this might be one of the motivations behind "diffeological spaces" and "smooth spaces".

so at most I would be ok with "locally convex manifolds", provided they exist and bundles can be defined on it

@yuggib Hm, then you might have to settle for the ugly things, as long as you're happy with them ;)

anyways, I would already be happy to say that for a Hilbert space $H$, seen as a manifold, $T^*H\simeq(TH)^*$

what are the charts for a Hilbert space

what even is a Hilbert space

and also, to find what Hilbert space has $TH=L^2(\mathbb{R}^d)$ (probably itself)

@yuggib Isn't that just $(TH)^* = (H\times H)^* = H^* \times H^* = H\times H = T^* H$ because bundles on vector spaces are trivial?

Or, even without knowing the fiber is $H$: $(TH)^* = (H\times T_x H)^* = H^*\times (T_x H)^* = H \times T_x^* H = T^* H$.

@ACuriousMind ok, so for Hilbert spaces sounds trivial

at least

Without $H=H^*$, this seems to fail, though.

and I assume also for reflexive banach space

@0celo7 concerning your answer, a small remark: it is better to put the hypotheses and setting of a theorem before, and not after the statement

;-)

ok, so it works for reflexive Banach spaces...better than nothing

@yuggib ok, that it?

@0celo7 what "that it?" means?

@yuggib anything else?

well, for the proof I leave the opinion for someone more versed in the subject ;-)

so @ACuriousMind needs to weigh in

hmmmm...now which (reflexive) banach space $B$ has $B\times T_x B=L^2(\mathbb{R}^d)$ o.O

@0celo7 Potentially confusing notation/terminology: $b^1,b^2,b^3$ are not the components of $b^a$, right? What is $T'_x\mathcal{M}$? What's a "triple scalar product" (if you mean $\vec a \cdot (\vec b\times \vec c)$, it is much better to write that than just use the name, imo)? Unclear formulation: How does uniqueness follow from calculating the things?

@ACuriousMind $b^i$ are the components wrt. the basis $X,Y,Z$

$T_p\mathcal{M}=\operatorname{span}(u)\oplus T_p'\mathcal{M}$

Okay, that's just a case of me hating abstract index notation, then :P

@0celo7 Well, that belongs into the answer!

it's the orthogonal complement to the span of $u$

> in the subspace of Tp orthogonal to ua (henceforth known as T′p)

what is unclear about that

Ah, overlooked that, sorry

"henceforth"

@ACuriousMind what about it

@ACuriousMind haha

abstract indices sucks sometimes

for instance a vector in abstract indices is really $v^a=\sum_i v^i(e_i)^a$

I chose $X^a=(e_1)^a$, etc.

what is wrong with "henceforth" :(

@0celo7 It's not wrong.

It sounds somewhere between funny, overly formal and pretentious to me. (And yes, I'm a hypocrite, three of my answers also contain it :P)

@ACuriousMind for the last comment, that's a page of boring

it boils down to just plugging and chugging

the energetic reader may verify :)

@0celo7 I don't want you to do the calculation - but I think the comment is slightly unclear: You say that $q^a$ is unique because one can calculate $q_a v^a$. It's not clear what the result of that calculation has to do with uniqueness at all, at least to me.

@ACuriousMind ah

we have the decomposition in the OP

well

hmm

@ACuriousMind If there were another $q$ that accomplished that decomposition, it would also have to satisfy $A(v,u)=q_av^a$

but because the scalar product on the RIGHT is nondegernate, this $q$ must equal the first one

That does make sense

@ACuriousMind I can't left/right lol

I didn't even notice :D

I added in an explanation

Springer is broken again

@Slereah Such manifolds are called parallelizable.

@BalarkaSen Indeed.

As @ACuriousMind noted, Lie groups are paralleizable : take a basis for the tangent space at a point, and push that off to the whole group by multiplication.

@ACuriousMind Henceforth is a great word

What's the alternative?

"From now on"?

"from now on known as $T'_p\mathcal{M}$" sounds stupid

@0celo7 strike the "known as"

@ACuriousMind that doesn't even make sense

hm...it would in German :P

Or at least, everyone would understand that.

I guess I would prefer to just write $T'_p\mathcal{M}$ after "subspace" there and omit the bracket completely.

Pls translate

I can't think of what it would be

@0celo7 translate what?

Von jetzt an

@ACuriousMind Depends on what you mean by homotopically/homologically trivial. You mean all the homotopy and homology groups vanish?

@BalarkaSen Yes, but I already found out that's not true.

It'd be contractible (by Whitehead), but certainly not homeo to R^n, yeah.

@0celo7 I don't think "henceforth" is pretentious enough

Homotopy and homology only detects homotopical infos about your space. And homotopy theory is very very different from topology (all the homotopical info about your space can be expressed combinatorially, e.g.). So it's extremely unlikely they'll say anything about homeomorphism type of your space.

@0celo7 I think you should replace it with "(from this moment forth it shall be denoted by ____ by authorial decree)"

Henceforth is great. "From here on" is also OK. As a native speaker I don't know any other ways to say the same thing without being clumsy.

@NeuroFuzzy royal authorial decree

I'm the king of cats

@Slereah proof?

@0celo7 gonna go out on a limb and say $dt=dz=dr=0$, so it's just $ds^2=L d\varphi^2$...

@NeuroFuzzy that's not rigorous

@0celo7 proof?

@ACuriousMind the $\mathrm{d}$s are not numbers that can be set to 0

@0celo7 It's what effectively happens for a curve of constant $r,z,t$, though.

And I'm pretty sure that he means such a curve with "integral curve of $\phi$".

@0celo7 Lol okay, $\dot{x}^i g_{i j} \dot{x}^j$ w/ $\dot{x}^r=\dot{x}^z=\dot{x}^t=0$

well, I've backed myself into a hole

I was never serious about this

::merely pretending::

@NeuroFuzzy >using middle Latin indices in GR

Ah, the old "No, that's not an error, I was just looking if you were paying attention"

> doing wat i waant

With that, you've lost any right to be indignant when I call you a troll or be sad when I ask you if that's a serious question :P

@ACuriousMind Stop calling me a troll D:

And you can't delete it from my memory :)

In fact, the removed there is even more telling :P

Maybe you'll get a concussion again and forget it

I think I'm gonna take a screenshot

@ACuriousMind of?

I'm being discriminated against

you hate German-Americans

Saved as "OcelotTroll.png"

wtf

I've done nothing wrong!

there is no evidence

That's for the jury to decide.

you're going to put me on trial

who is the jury

I demand peers, i.e. IQ=80 German-Americans

@0celo7 So you can try to intimidate them? Nice try.

me and @Slereah will be on jury duty

(I accept bribes)

::silenced gunshot; @AngusTheMan is dead on the floor::

:o

:o

oh no

looks like a jury member decided to shoot himself in the forehead

oh well

@0celo7 three times?

@Loong don't be silly

that would be a waste of good ammunition I don't have that kind of money

did I manage to encase myself in concrete and push myself out to sea as well?

@AngusTheMan you did fall out of a helicopter

into the Sahara

funnny how that happened

I'm talented aren't I

you were

fun fact for you all, if you google holly holm and click on images, "amy childs vajazzle" is a suggested term... wth internet ... wth

@ACuriousMind why does the metric of the sphere contain a term $\sin^2\theta$

don't tell me to calculate it

Stop trolling :P

what's the intuitive reason

I'm asking a genuine question here

@AngusTheMan Not for me, must have something to do with your search habits ;)

but this is my "clean" computer

great now @ACuriousMind thinks I'm a troll

it's a genuine question ._.

he's never gonna answer

sigh

I think ACM is trolling

@ACuriousMind Seriously

Great.

@ACuriousMind Also why is there a $\sin^2\theta$ piece in the polar coordinate metric

@0celo7: Hint: $r\sin(\theta)$ is the radius of the circle of constant $\phi$ at $\theta$.

What

@0celo7 You bring that on yourself. Too much, too dry humor and people stop being able to tell the difference between the serious you and the trolling you.

@ACuriousMind I know that, but what does it tell me

@0celo7 figured out an alternative to "henceforth"

@AngusTheMan I object to the lack of capitalization in my moniker there, BTW ;P

$C$, (the number formerly known as ___)

@NeuroFuzzy Taking bets whether 0celo7 gets the reference or not now.

No

And whenever people say it I don't get it.

@Slereah I have a feeling Vadim's plan is going to get everyone killed.

@ACuriousMind am I supposed to get the reference

I was hoping to get some help developing a model for the emf produced by a magnet/solenoid generator. I have a preliminary one but can't figure out why it isn't very accurate when I run tests. Would that be an appropriate question for here?

@jkeuhlen I'm not sure. Do you have a precise question, or rather "This model doesn't work very well, can you figure out why"?

And now no one will explain the reference

Great.

A little bit of both. I could ask it as: I have a spherical magnet with known magnetic moment m and a solenoid with N loops and area A. How do I piece everything together to get the emf from a free fall through the solenoid. Or I can walk through what I already did (assuming kinematics, no rotation, no friction) and ask if anyone can find the problem.

Sounds off-topic as homework-like to me in both variants, I'm afraid.

Yeah, it's for an engineering class I'm in so it is rather homeworky. It isn't an actual assignment, just a part of a project.

@ChrisWhite Cool, thanks.

@ACuriousMind I still don't get the metric thing

or that weird reference

Whawwwww. I thought I'd done a really good job on Why do we obtain classical physics by taking the limit of Planck's constant to zero? and the OP likes anna v's answer better. ::sniffle::

@ACuriousMind lol

There’s an awful cost to getting a PhD that no one talks about / quartz

@ChrisWhite Are you also interested in Planet Hunter?