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12:00 AM
It's technically a bit harder with the approach you're doing.
If $u,w\in\Bbb R^n$, then $d\psi(u,w) = (\psi(u),d\psi_u(w))$ parametrizes easily.
Doing it with the chart (as you were starting to write) you have to worry about the domain a lot more.
 
Hmm, it seems the worries about the domains only disappear because you switch domain and codomain (essentially)
The differentiability of the inverse is still just as worrisome
 
Yup. Nah. You have an extension of $\psi^{-1}$ and then you get an extension of $d\psi^{-1}$ to everything.
 
you're right
I can not deny this
and it makes sense that this works
 
Yup, parametrizations are much more natural for submanifolds, anyhow, and then you're all set for pullbacks for integration.
 
12:16 AM
oh, there's an issue
nevermind, there's not
 
Oh, you had me in suspenders.
 
I was gonna say that, while giving us a chart, this map doesn't give us local triviality of the bundle yet
but we can just undo the identification in the first component afterwards
 
Sure, you're using the fixed $\Bbb R^n$ for the fibers.
 
I'm left with no objections
 
That, in and of itself, is objectionable.
 
12:23 AM
Are you done with homework Thorgott
 
I think A is eager to resume his lecture :)
 
I am currently calculating the tangent planes of an ellipsoid
 
Nah I am done I just have a good proof of Morse-Bott lemma that I want someone to read lol
@Thorgott OK
 
ok, done
now I shall prove the product of two manifolds is a manifold
 
Boring.
 
12:30 AM
done
 
quickboi
 
now I shall calculate the tangent space of the product manifold
I've seen this stuff before, so it's not exactly difficult
 
It's boring even if you haven't seen it, but one needs practice with the game.
 
More interesting: A neighborhood of the diagonal $\Delta \subset M \times M$ is diffeomorphic to $TM$.
 
it's in the category "good to know"
now this last exercise may prove to be a slightly ugly computation
 
12:36 AM
That's harder, A.
 
Yeah. That's the first time I realized I don't really understand products.
 
I don't know how to do that without tubular nbhd thm.
 
hmm
what does the diagonal of the torus look like in R^3
 
It's a circle
Look up (p, q)-torus knots. It's a (1, 1)-"knot"
 
yeah, ok, then I do believe a neighborhood of that is diffeomorphic to $TS^1$
by virtue of example, the theorem ought to be true
 
12:41 AM
Yes, universally.
This is relevant when you want to understand the analogue of the tangent bundle for Poincaré duality in algebraic topology.
 
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)?
 
It's because (+1)*(+1) = (-1)*(-1) = +1
More or less :)
 
Can you interpret it in terms of an almost complex structure on the tangent bundle?
You probably did a jacobian argument.
 
I don't know what that is, so I'm gonna say no
yeah, the standard atlas is oriented
because of what Balarka said
 
What do you think, A?
 
12:46 AM
Hmm.
 
Suffices to play my game on the zero section of $T(TM)$.
 
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 $M$
 
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 $(-X,0)$?
 
$TTM$ is scary
 
Well, it only works "naturally" along the zero section. I'm not being too dramatic here.
 
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 $TTM$
 
Oh, OK.
Somehow I am suspicious that we've never heard of this.
 
Me too!
 
But it should just be the fact that $V\oplus V$ is always canonically orientable as a vector space.
 
Ah OK that's what you meant. Sure.
 
And that's because $V\oplus V \cong V\otimes \Bbb C$, and so we have a complex structure on the vector space.
 
12:53 AM
That's a good way to think about it. I had never thought of that
 
Well, you can tell me tomorrow that I screwed up.
 
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 $TM$: $TTM \cong \pi^*TM \oplus \pi^*TM$
 
Sure.
It was the almost complex structure viewpoint that I don't recall hearing (but I may be wrong on that).
 
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.
 
It feels like Hamilton's equations for $T^*M$ or something.
 
1:02 AM
It seems to depend on the connection
Bizarre
 
Well, I was going to do it just along the zero section and then propagate on the fibers.
 
That makes sense
I want to write down the formula for this $J$ in terms of the chosen connection $\nabla$ now
 
Why isn't Mike here to tell us we're being silly?
 
I asked him elsewhere
 
Oh. Did he tell you we're being silly?
 
1:07 AM
Hasn't yet
 
It's funny that I never thought to make this comment years ago.
 
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
 
Yeah, should be connected somehow to the contact/symplectic structure on $T^*M$.
 
Must be
 
Agh.
 
1:11 AM
Mike says it's correct
 
So weird that none of us has ever thought of it or seen it.
 
Yeah!
 
Here's something from Google. $TM$ for any non-flat Riemannian manifold $M$ always admits an almost-Kähler structure that is not Kähler.
 
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.
 
Aha. "This is done by making use of the almost-complex structure of $TM$ owing to Sasaki."
 
1:13 AM
Sasaki, the nutcase
 
Oh?
 
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.
 
I found something as well. It claims this is integrable if the connection is torsion and curvature free
 
I'm feeling now like I should have known this years ago.
Hi, @MikeM :P
 
So for flat manifolds, $TM$ is naturally a complex manifold.
 
1:14 AM
@BalarkaSen Now that seems like a nice result.
 
@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
 
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
 
Well, all this came from my offhand comment to Thorgott. How fun.
Strange that the curvature of metric should determine whether somebody is complex.
 
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
 
His clever idea was to use exponentiation to go from (p,v) to exp_p(v), above which you know how to split TM.
 
1:17 AM
This must be buried in Kobayashi Nomizu somewhere.
 
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).
 
Ah I see
@TedShifrin Yeah. Brilliant, Ted!
 
My questions have this weird tendency to make you guys go on long tangents
 
Ok, staying up was worthwhile. I will need to think about this example
Thanks and good night, all!
 
Good night, A.
LOL, so you should shut up, @Thorgott :D
 
1:39 AM
:(
 
2:11 AM
Anyone here?
 
 
4 hours later…
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 $(S\mathbb{R}^2)([2])$?
 
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 $s_0$?
 
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.
 
6:22 AM
@Pig Can you tell me a joke :-) ?
 
 
1 hour later…
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 $$ \int_{a}^{b} f + \int_{a}^{b} g = \int_{a}^{b} (f+g)$$
 
7:59 AM
@Knight You are not using that $L$ and $U$ can be made close for finer $P$
 
8:15 AM
@robjohn Okay, they can be made as close to each other as we want. But how it would help us?
 
8:38 AM
@robjohn Thank you. So, for summarize, we can have cross product in n-space.
 
 
1 hour later…
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
 
10:47 AM
looks like I can't get commutative diagrams to work here
 
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.
 
 
2 hours later…
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"
 
hi guys i was reading about the probability and I read Bayes theorem in it,so is there any proof of the Bayes theorem?
 
@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
 
ok that's fair i read one but i have one doubt in it!
 
@Thorgott :D Die Frage war eher, ob man bei dem Ausdruck Dativ oder Akkusativ benützt hehe
Also nice, danke
 
actually i derived that using Venn diagram ,but i not known for good statistics so i was looking for analytical proof
@Thorgott
 
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
 
 
3 hours later…
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
 
 
3 hours later…
7:18 PM
Guys... you are discussing about some theorems but before you do that you should first do se baisiiicccs
 
🤯
 
Howdy, @MikeM.
 
Hi chat
 
7:35 PM
Salut, @Astyx
 
Quoi de neuf ?
 
@Logik You tell 'em, son.
 
Tout le monde se brûle, @Astyx.
 
Ah oui. Près de chez toi aussi ?
 
Pas loin. :(
 
7:39 PM
Arf. Prends soin de toi
 
Oui, je fais de mon mieux.
Qqch de maths à discuter?
 
J'ai réfléchis un peu à l'histoire de trajectoire de velo
 
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.
 
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
 
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.
 
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).
 
@Astyx, my little Mathematica notebook works for you? :P
 
@KonformistLiberal will this help? math.jhu.edu/~eriehl/ssets.pdf
 
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
 
@EnjoysMath I don't see how it helps.
 
@KonformistLiberal k, just checking :)
 
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.
 
@EnjoysMath Thank you though :)
 
Yes I should when I find the time. Right now I'm busy with my internship
 
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
 
 
2 hours later…
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?
 
it is
 
thes the example is wrong
 
the no odd-parity entries observation says they aren't equal
but they are isomorphic
 
??
odd parity?
sorry i did not understant
understand*
 
that's what you said, no
 
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?
 
it is
 
thus M and N are isomorphic modules
3.5 example
So the author is wrong?
 
yes, that is wrong
 
ok i was confused. thanks
 
That's nonsense
 
10:28 PM
which?
 
The conclusion in the pdf
 
in general, free modules of the same rank are isomorphic
 
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
 
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
 
10:30 PM
I'd try something infinitely generated
 
so started looking at infinite groups
and found that pdf and got confused
ok really thanks
 
It can't work with finitely generated groups, because of the structure theorem
 
hmm, whether that's possible in the finitely generated case seems like an interesting question, actually
right, but what about finitely generated modules generally
 
im trying like 3 hours with finitely generated abelian groups
 
Over a PID the structure theorem saves us again right? So we need to look at ugly stuff
 
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
 
so we have to do worse than Dedekind domains
 
Ah, I'm not surprised to see Jónsson and Tarski being mentioned there
 
10:47 PM
hmm, I'm not seeing an easy f.g. example
 
i cant find either
the exercise looked innocent
 
does the exercise ask for finitely generated
if not, it is innocent
 
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
 
I don't think that's true
 
(a,c/d) --> (a,dc/d)
 
10:52 PM
one of these is finitely generated, the other is not
 
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
 
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)
 
one contains an element of order 4, the other does not
 
which one?
(a,1) goes on
ohh (0,1)
same as the finite case
 
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
 
i dont know any other infinite groups
the rest are nZ
all free groups
 
free groups suffice for a counter-example
free abelian groups*
 
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
 
A+B and A+C having the same rank doesnt imply B and C have the same rank
 
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
 
11:59 PM
no, those aren't direct sums
 
only (0,0,0)
intersection
 

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