Mathematics - 2020-05-31

this reminds me, I proved earlier that $TM$ is always orientable. is there a good reason for why this should be true (other than the calculation being really straightforward)?

Balarka Sen

It's because (+1)*(+1) = (-1)*(-1) = +1

More or less :)

Ted Shifrin

Can you interpret it in terms of an almost complex structure on the tangent bundle?

You probably did a jacobian argument.

Thorgott

I don't know what that is, so I'm gonna say no

yeah, the standard atlas is oriented

because of what Balarka said

Ted Shifrin

What do you think, A?

Balarka Sen

12:46 AM

Hmm.

Ted Shifrin

Suffices to play my game on the zero section of $T(TM)$.

Balarka Sen

You're saying $TM$ has a natural almost complex structure?

I guess $TTM \cong TM \oplus TM \cong TM \otimes \Bbb C$ as bundles over $M$

Ted Shifrin

I'm using $T_{(x,0)}(TM) \cong T_xM\oplus T_xM$. So can I sent $(X,0)$ to $(0,X)$ and $(0,X)$ to $(-X,0)$?

Thorgott

$TTM$ is scary

Ted Shifrin

Well, it only works "naturally" along the zero section. I'm not being too dramatic here.

Balarka Sen

12:51 AM

I was just thinking of choosing a connection to set up the first isomorphism. Then pullback the fiberwise almost complex structure on $TM \otimes \Bbb C$ to $TTM$

Ted Shifrin

Oh, OK.

Somehow I am suspicious that we've never heard of this.

Balarka Sen

Me too!

Ted Shifrin

But it should just be the fact that $V\oplus V$ is always canonically orientable as a vector space.

Balarka Sen

Ah OK that's what you meant. Sure.

Ted Shifrin

And that's because $V\oplus V \cong V\otimes \Bbb C$, and so we have a complex structure on the vector space.

Balarka Sen

12:53 AM

That's a good way to think about it. I had never thought of that

Ted Shifrin

Well, you can tell me tomorrow that I screwed up.

Balarka Sen

I think it's actually OK. $\pi : E \to M$ be any bundle projection, then $TE \cong \pi^*TM \oplus \pi^*E$ as bundles over $E$ (noncanonically, we only have a short exact sequence $0 \to \pi^* E \to TE \to \pi^* TM \to 0$ and that splits by a choice). Plugging $E = TM$, we get an isomorphism of bundles over $TM$: $TTM \cong \pi^*TM \oplus \pi^*TM$

Ted Shifrin

Sure.

It was the almost complex structure viewpoint that I don't recall hearing (but I may be wrong on that).

Balarka Sen

So $TTM \cong \pi^*(TM \oplus TM)$ as bundles over $TM$, and $TM \oplus TM \cong TM \otimes \Bbb C$ so we get an almost complex structure on $TTM$ by pulling back the one on $TM \otimes \Bbb C$. So $TM$ is an almost complex manifold -- that is so strange

I really want to believe something is wrong.

Ted Shifrin

It feels like Hamilton's equations for $T^*M$ or something.

Balarka Sen

1:02 AM

It seems to depend on the connection

Bizarre

Ted Shifrin

Well, I was going to do it just along the zero section and then propagate on the fibers.

Balarka Sen

That makes sense

I want to write down the formula for this $J$ in terms of the chosen connection $\nabla$ now

Ted Shifrin

Why isn't Mike here to tell us we're being silly?

Balarka Sen

I asked him elsewhere

Ted Shifrin

Oh. Did he tell you we're being silly?

Balarka Sen

1:07 AM

Hasn't yet

Ted Shifrin

It's funny that I never thought to make this comment years ago.

Balarka Sen

I am pretty confident this is not wrong, but this is bizarre. Yeah, seems like a natural example which is not in textbooks if it is correct, and that surprises me

Ted Shifrin

Yeah, should be connected somehow to the contact/symplectic structure on $T^*M$.

Balarka Sen

Must be

Ted Shifrin

Agh.

Balarka Sen

1:11 AM

Mike says it's correct

Ted Shifrin

So weird that none of us has ever thought of it or seen it.

Balarka Sen

Yeah!

Ted Shifrin

Here's something from Google. $TM$ for any non-flat Riemannian manifold $M$ always admits an almost-Kähler structure that is not Kähler.

Mike Miller

It's just the statement that you get a splitting $TTM \cong E \oplus E = E \otimes \Bbb R^2$. The bundle $E$ being two apparitions of $\pi^*TM$.

I don't think it's any good for anything.

Ted Shifrin

Aha. "This is done by making use of the almost-complex structure of $TM$ owing to Sasaki."

Balarka Sen

1:13 AM

Sasaki, the nutcase

Ted Shifrin

Oh?

Mike Miller

Sasaki would make sense as he's also the guy who came up with the natural splitting of T(TM) to give TM a canonical metric.

Balarka Sen

I found something as well. It claims this is integrable if the connection is torsion and curvature free

Ted Shifrin

I'm feeling now like I should have known this years ago.

Hi, @MikeM :P

Balarka Sen

So for flat manifolds, $TM$ is naturally a complex manifold.

Mike Miller

1:14 AM

@BalarkaSen Now that seems like a nice result.

Balarka Sen

@MikeMiller Yeah, that's what my nutcase comment was referring to

Yes, I would like to compute this out

Hot damn this is a nice example

Mike Miller

I think people naively believe the metric on TM is obvious but even if you have a connection it's not clear what the horizontal subspace of T(TM) is away from the zero section

Ted Shifrin

Well, all this came from my offhand comment to Thorgott. How fun.

Strange that the curvature of metric should determine whether somebody is complex.

Balarka Sen

I bet for flat $M$ that makes $T^*M$ a Kahler manifold, equipped with the symplectic form, the Sasaki metric, and this weird ass complex structure

Mike Miller

His clever idea was to use exponentiation to go from (p,v) to exp_p(v), above which you know how to split TM.

Ted Shifrin

1:17 AM

This must be buried in Kobayashi Nomizu somewhere.

Mike Miller

His clever idea was to use exponentiation to go from (p,v) to exp_p(v), above which you know how to split T(TM).

Balarka Sen

Ah I see

@TedShifrin Yeah. Brilliant, Ted!

Thorgott

My questions have this weird tendency to make you guys go on long tangents

Balarka Sen

Ok, staying up was worthwhile. I will need to think about this example

Thanks and good night, all!

Ted Shifrin

Good night, A.

LOL, so you should shut up, @Thorgott :D

Thorgott

1:39 AM

:(

Knight

2:11 AM

Anyone here?

Konformist Liberal

5:52 AM

Hi everyone, I read that n-th component of a singular simplicial set should decompose into ker $d_0 \oplus$ im $s_0$, but for example, for the case of $(S\mathbb{R}^2)([2])$, I can't see how a non-degenerate 2-simplex is in the kernel of $d_0$. What am I missing here?

Pig

what is $(S\mathbb{R}^2)([2])$?

Konformist Liberal

I wanted to mean the second component of the singular simplicial set that "belongs" to $\mathbb{R}^2$

Pig

what is your $d_0$ and $s_0$?

Konformist Liberal

6:08 AM

I thought they have [a standard]/[only one] definition? And I think I understood what I am missing. A non-degenerate simplex is indeed not in the kernel but, it can be written in that form by a simple trick of subtracting a degenerate simplex from it (if chosen carefully, it makes the whole thing be in the kernel of $d_0$) and then adding that degenerate simplex again.

Knight

6:22 AM

@Pig Can you tell me a joke :-) ?

Knight

7:30 AM

@robjohn Sir you there?

I want to prove that the integral of a sum is equal to the sum of integrals from the very first definitions of integrals.

Let $f$ be an integrable function on $[a,b]$ and $g$ is also an integrable function on $[a,b]$. We have $$ L(f,P) \leq \int_{a}^{b} f \leq U(f,P) $$
$$L(g,P) \leq \int_{a}^{b} g \leq U(g,P)$$

And we know $$ L(f+g, P) \geq L(f,P) + L(g,P) \\
U(f+g, P) \leq U(f,P) + U(g,P)$$

And since, $f+g$ is also integrable, we have
$$ L(f+g, P) \leq \int_{a}^{b} f+g \leq U(f+g, P)$$

If we add the first two inequalities, we would have
$$ L(f,P)+L(g,P) \leq \int_{a}^{b}f + \int_{a}^{b} g \leq U(f,P) + U(g,P) $$

And from this information

11 mins ago, by Knight

And we know $$ L(f+g, P) \geq L(f,P) + L(g,P) \\
U(f+g, P) \leq U(f,P) + U(g,P)$$

we can have
$$ L(f,P) + L(g,P) \leq \int_{a}^{b}(f+g) \leq U(f,P) + U(g,P) $$

But even coming to this far I'm unable to conclude that $$ \int_{a}^{b} f + \int_{a}^{b} g = \int_{a}^{b} (f+g)$$

robjohn

7:59 AM

@Knight You are not using that $L$ and $U$ can be made close for finer $P$

Knight

8:15 AM

@robjohn Okay, they can be made as close to each other as we want. But how it would help us?

user777

8:38 AM

@robjohn Thank you. So, for summarize, we can have cross product in n-space.

Mike Miller

10:01 AM

so long as you very carefully explain what you mean by "cross product" then yes, since there's a number of properties you might demand of a higher dimensional cross product that are not satisfied

Thorgott

10:47 AM

looks like I can't get commutative diagrams to work here

Thorgott

10:59 AM

I was gonna point out that the differential functor and the product functors gives rise to a square diagram of categories with corners $\mathbf{Diff}^2,\mathbf{DiffVecBun}^2,\mathbf{DiffVecBun},\mathbf{Diff}$, which does not commute, but commutes up to a natural isomorphism, given by the identification $T(M\times N)\cong TM\times TN$ canonically.

Edward Evans

12:46 PM

@Thorgott Seeervus, ich hätte nochmal eine Sprachfrage hahaha, sagt man eigentlich "etwas auf einer anderen Weise beweisen" oder "etwas auf eine andere Weise beweisen"

Yuvraj

hi guys i was reading about the probability and I read Bayes theorem in it,so is there any proof of the Bayes theorem?

Thorgott

@Edward letzteres, es ist "die Weise", feminin

@Yuvraj yes, you should be able to find one by just googling

it follows directly from the law of total probability

Yuvraj

ok that's fair i read one but i have one doubt in it!

Edward Evans

@Thorgott :D Die Frage war eher, ob man bei dem Ausdruck Dativ oder Akkusativ benützt hehe

Also nice, danke

Yuvraj

actually i derived that using Venn diagram ,but i not known for good statistics so i was looking for analytical proof

@Thorgott

Konformist Liberal

1:44 PM

Hi everyone, I asked a short question just above, chat.stackexchange.com/transcript/message/54527132#54527132, and then found an answer, but a verification or any kind of comment would be great

geocalc33

4:15 PM

Say you have to add two vector fields $X_1=\big(x\log(x),-y\log(y)\big)$ and $X_2$, where the two vector fields represent boosts (a certain type of Poincare symmetry). A condition is that $X_1 + X_2$ yields a vector field with $0$ x-component

apparently you can't add two boosts (it's done via composition)? I'm confused then, because I thought that adding the vector fields would mean that the boosts are being added

I think it might have to do with lie algebras maybe, but I'm not very familiar with those

Logik

7:18 PM

Guys... you are discussing about some theorems but before you do that you should first do se baisiiicccs

🤯

Howdy, @MikeM.

hi

Hi chat

7:35 PM

Salut, @Astyx

Astyx

Quoi de neuf ?

NoName

@Logik You tell 'em, son.

Ted Shifrin

Tout le monde se brûle, @Astyx.

Astyx

Ah oui. Près de chez toi aussi ?

Ted Shifrin

Pas loin. :(

Astyx

7:39 PM

Arf. Prends soin de toi

Ted Shifrin

Oui, je fais de mon mieux.

Qqch de maths à discuter?

Astyx

J'ai réfléchis un peu à l'histoire de trajectoire de velo

Konformist Liberal

Quick question: Is Kan condition not only sufficient but also necessary to make an equivalence relation out of the simplicial maps being homotopic? It seems like so, but I can't find a reference.

Astyx

En particulier ce que je disais était faux : si on regarde des trajectoires paraboliques $(t,\alpha t^2)$, au passage à l'origine la roue arrière du velo doit faire marche arrière pour $\alpha$ "suffisament grand"

Et on sort de l'enveloppe convexe de la trajectoire

Ted Shifrin

Ah, il me fallait me rappeler notre histoire de vélo :P

@KonformistLiberal: That sounds like a heavy-duty alg top question and I don't think I ever knew the Kan condition.

Konformist Liberal

7:49 PM

@TedShifrin I just thought maybe it's trivial or someone can send me a quick reference (I couldn't find any so far).

Ted Shifrin

@Astyx, my little Mathematica notebook works for you? :P

EnjoysMath

@KonformistLiberal will this help? math.jhu.edu/~eriehl/ssets.pdf

Astyx

Yep, I used it to do some testing, thank you. I didn't find the time to think about it too much though, but I think it has to be a relation between the curvature of the trajectory and the size of the bike. Will have to check when I find the time

Konformist Liberal

@EnjoysMath I don't see how it helps.

EnjoysMath

@KonformistLiberal k, just checking :)

Ted Shifrin

7:54 PM

Oh, you should do the exercise of finding the front path in terms of the back (the initial condition is what angle the handlebar is turned). Curvature plays a prominent role. I assumed the distance was $1$ between the (point) wheels.

Konformist Liberal

@EnjoysMath Thank you though :)

Astyx

Yes I should when I find the time. Right now I'm busy with my internship

Alessandro Codenotti

8:15 PM

Turns out that regular and second countable implies normal, this seems very useful, I'm surprised I've never seen it before

Manolis Lyviakis

10:19 PM

any1 here?

Let M and N be free modules over Z . $M={(a,b)|a,b \in Z}, $ $N={ (2a,2b)|a,b \in Z } $. If M and N were
vector spaces, they would be isomorphic. However, because our scalar multiples are from
a ring that does not have multiplicative inverses, N has no vectors with odd-parity entries.
Thus, M and N are not isomorphic.

isnt f((2a,2b))=(a,b) a linear isomorphism?

Thorgott

it is

Manolis Lyviakis

thes the example is wrong

Thorgott

the no odd-parity entries observation says they aren't equal

but they are isomorphic

Manolis Lyviakis

??

odd parity?

sorry i did not understant

understand*

Thorgott

that's what you said, no

Manolis Lyviakis

10:24 PM

its an example from a pdf saying these 2 modules are not isomorphic

and my question is isnt the f i prupose an isomoprhism?

Thorgott

it is

Manolis Lyviakis

thus M and N are isomorphic modules

buzzard.ups.edu/courses/2017spring/projects/…

3.5 example

So the author is wrong?

Thorgott

yes, that is wrong

Manolis Lyviakis

ok i was confused. thanks

Alessandro Codenotti

That's nonsense

Manolis Lyviakis

10:28 PM

which?

Alessandro Codenotti

The conclusion in the pdf

Thorgott

in general, free modules of the same rank are isomorphic

Manolis Lyviakis

ye

Alessandro Codenotti

It's like saying that $2\Bbb Z$ and $3\Bbb Z$ are not isomorphic as abelian groups because one doesn't have elements divisible by 3 and the other one does

Manolis Lyviakis

yes

thats how i came up with the isomorphism

2Z and Z

IM trying to come up with A, B C modules such B and C not isomorphic but A product B isomorphic to A product C

im looking at finitely genererated abelian groups

but cant find

Thorgott

10:30 PM

I'd try something infinitely generated

Manolis Lyviakis

so started looking at infinite groups

and found that pdf and got confused

ok really thanks

Alessandro Codenotti

It can't work with finitely generated groups, because of the structure theorem

Thorgott

hmm, whether that's possible in the finitely generated case seems like an interesting question, actually

right, but what about finitely generated modules generally

Manolis Lyviakis

im trying like 3 hours with finitely generated abelian groups

Alessandro Codenotti

Over a PID the structure theorem saves us again right? So we need to look at ugly stuff

Manolis Lyviakis

10:33 PM

but even if u get the number of elements right

on group will have an element with order that the other group does not have

Infinite Z-modules looks like a good place to start

Thorgott

link.springer.com/chapter/10.1007/BFb0059566

so we have to do worse than Dedekind domains

Alessandro Codenotti

Ah, I'm not surprised to see Jónsson and Tarski being mentioned there

Thorgott

10:47 PM

hmm, I'm not seeing an easy f.g. example

Manolis Lyviakis

i cant find either

the exercise looked innocent

Thorgott

does the exercise ask for finitely generated

if not, it is innocent

Manolis Lyviakis

i want A+B= A+ C whith B not isomorphic to C

it doesnt ask

for finitely generated

just general modules

ohh

Z+Z= Z+Q?

as Z modules

Thorgott

I don't think that's true

Manolis Lyviakis

(a,c/d) --> (a,dc/d)

Thorgott

10:52 PM

one of these is finitely generated, the other is not

Manolis Lyviakis

oh yeah

i need (1.0) (0,1)

Reals+Z =Reals +Q

(x,n) -> (x,n/1)

(x, a/b) -> ( bx, ba/b)

doesnt look 1-1

Manolis Lyviakis

11:10 PM

almost 4 hours cannot find anything i feel stupid i give up... at least for today

or maybe Q+Z_4= Q+V(klein group)

Thorgott

one contains an element of order 4, the other does not

Manolis Lyviakis

which one?

(a,1) goes on

ohh (0,1)

same as the finite case

Thorgott

In general, I wouldn't advise trying to find an example with infinitely generated abelian groups, as those can be quite complicated

or rather, I wouldn't advise to try do it with groups like Q and R

Manolis Lyviakis

i dont know any other infinite groups

the rest are nZ

all free groups

Thorgott

free groups suffice for a counter-example

free abelian groups*

Manolis Lyviakis

11:40 PM

free abelian groups will just be a product o n=the sum of the ranks coppies of Z say Z+Z^4=Z+(Z^3xZ) = ZxZxZxZxZxZ so if i have A+B=A+C where all free abelian free groups C=B since they must have the same rank and will be the same coppies of Z

Thorgott

A+B and A+C having the same rank doesnt imply B and C have the same rank

Manolis Lyviakis

ok didnt thought of that

so it must have the rank of A

and B and C contibute nothing

contribute nothing

but if use a subspace of A then its just the inner product which is not isomorphic to the exterior product since their intersection is not just zero

$A \oplus B \cong A \oplus C $ is not an inner product its a general product

ok consider the set $A={ (a,0,b)| a,b \in Z}$ , $B=(a',c,0)| a',c \in Z$ , $ C={ (0,l,0)| l\in Z}$ $A \oplus B \cong A \oplus C \cong Z^{3} $ B has rank 2 C has rank 1

thanks

Thorgott

11:59 PM

no, those aren't direct sums

Manolis Lyviakis

only (0,0,0)

intersection

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$Mathematics$ Mathematics