Mathematics - 2016-09-03

gcy-rolle

04:28

hello,everybody.

MathWanderer

04:42

I am going to do a reading course in algebra with Lang's Algebra!

So excited!

Martin Sleziak

07:13

Several recently starred posts mention how to ignore another user in chat (1, 2, 3). Maybe eventually everybody will talk only to himself :-)

Martin Sleziak

07:35

in Functional Analysis Study Group, 42 secs ago, by Martin Sleziak

For the users who remember the user t.b. I will add link to lecture notes in functional analysis which I found here.

Balarka Sen

08:36

Darn it, I missed the showdown. Popcorns wasted. :(

@MartinSleziak This is what I am afraid of.

Popcorn jokes apart, many of the people (temporarily/permanently) left this chat recently, mostly good people who used to contribute constructively and actually talk about math. I feel like this is a fault of the community: with those people gone 80% of the conversations have either become off-topic or streetfights. I confess that I feel responsible and ashamed for this. I think we all should, and try to make this chat a better place.

I wouldn't want this community to fall apart - I used to learn a lot from being here.

Martin Sleziak

08:52

@BalarkaSen Isn't it a bit too soon to say that many users have left? I guess that in general there might be less activity in this room during summer.

Balarka Sen

I am quite certain the people I am thinking of actually left, and not just because they are busy with their own life.

For example, they seem very active on math.SE.

Martin Sleziak

I think there was a bit of similar drama in the past. Quote from this meta post: "This hypothetical situation is actually very concrete: there exists a chat user who has so far caused 2 chat regulars and good mathematical contributors (on main) to leave chat for good."

Balarka Sen

Yes, I have heard of that.

Martin Sleziak

Maybe some of the users will start to discuss more in Algebraic Theory chatroom (which was rather inactive lately or abstrat algebra chatroom (which is not much better).

Or maybe they will simply come back to chat later.

Secret

09:09

From what I can tell, periodic table, here and h bar are all quite inactive recently, possibly it has something to do with the start of the school year in the nothern hemisphere

Mary Star

10:27

Hello!!
I want to show that $\lim \inf a_n + \lim \sup b_n \leq \lim \sup (a_n + b_n)$.
Do we use for that the property $\lim \inf a_n\leq \lim \sup a_n$ ?
But then we have $\lim \inf a_n + \lim \sup b_n \leq \lim \sup a_n + \lim \sup b_n$. Does this help?

Martin Sleziak

@MaryStar There are several posts about this inequality on the main. Like this question and probably several other questions linked there.

I think that you should try to use the definition: $\limsup\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} \sup\limits_{k\ge n} x_k$

Or some of the other equivalent definitions of limit superior.

Now when I think about how to prove it, probably some of the other definition might indeed be more suitable.

But perhaps it works even with this definition. We should be able to prove $\inf\limits_{k\ge n} a_k+\sup\limits_{k\ge n} b_k \le \sup\limits_{k\ge n} (a_k+b_k)$, right?

@MaryStar If you prove the above inequality, then you are basically done. (It only remains to take the limit.)

Now I see that what I wrote here is basically the same thing as in the second part of my answer in the linked post.

I do math art

11:39

@BalarkaSen I don't feel neither responsible not ashamed for the decisions others take. People come and leave, that's a real fact, and then there are many reasons that maybe many of us aren't aware for which people are not here. As an example, on my hottest work period on my book I think I missed here some months. On the other hand, if you suggest that some users make other users to leave this chat, OK, maybe, but at the same time, they may come back anytime,

open their room and chat there all day long, and I bet no one would disturb them, or coming here and ignoring the users they don't like. They had enough options not to be disturbed if they wanna stay here.

Give me a break with these airs of too important to stay here.

Anyway.

BBL

One thing, did any get some cool way here?

$$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$

BBL (I'm around working on some challenging problem)

deostroll

Hello folks, why is the taylor series expansion always defined around a point $a$?

Taylor series

In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. The concept of a Taylor series was formulated by the Scottish mathematician James Gregory and formally introduced by the English mathematician Brook Taylor in 1715. If the Taylor series is centered at zero, then that series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin, who made extensive use of this special case of Taylor series in the 18th century. A function can be approximated...

Balarka Sen

12:16

@Idomathart I did not ping you while I wrote that particular message.

Hey @Soham.

Soham Chowdhury

hi

just stopping by

Balarka Sen

What's up?

I do math art

@BalarkaSen True. But since you referred to all users when you said what we should do I thought I should tell you my opinion not to include me in all the next time.

Sorry, I have a lot of stuff to do now.

BBL

Soham Chowdhury

@BalarkaSen just writing some stuff up, and waiting for Canadian prof to reply to an email

Balarka Sen

@Idomathart I said what we should do from common sense of community ethics. If you don't care for it, sure, feel free not to.

@Soham Whatcha writing

I do math art

12:27

@BalarkaSen The community already offers all you need if you really want to stay away from a user, like ignore this user. Some simply don't like some, and I don't think the community shoud do efforts such that all fall in love with each other. Imho, one has at hand all tools needed to avoid other users.

Again, imho.

BBL

Balarka Sen

@Idomathart That sounds irrelevant to whatever I said. Oh, and if you don't want to be included in "all", you should leave the community. As long as you're in the community, I will include you too - you're no more special than an arbitrary user to be not included.

Martin Sleziak

Since ignoring users is now trending topic here, I will ask whether there is also a possibility to hide a particular user from the transcript. (If you simply ignore user, the transcript still shows their messages - I have just test this.)

Balarka Sen

You should do your stuff than writing a reply to this. Writing a long reply and "BBL" everytime you're only pretending to be working.

I do math art

@BalarkaSen It's relevant for the simple fact we do different mathematics, we have different visions on almost anything about math and hence all this possible disputes. Community cannot improve that, if you want to understand me, because I'm stubborn on my visions, and also the others are stubborn on their visions, and so on. It's not that you included me in all, not that I'm special, not the case in these circumstances,

but you said what we all should do as if you were a decisional factor of the site, but that is just your opinion, and you can express your opinion without including me, especially in this case where I see no logic to do that.

Balarka Sen

No I said what I think we should do. I explicitly wrote "I think".

Yes, it is my opinion. You're by no means forced to follow that, that's perfectly fine.

But I am weary of these conversations, so I abandon it.

I do math art

12:41

@BalarkaSen Maybe the others also felt offended saying that 80% of the math conversations are gone because of one, two users leaving.

Users come and leave, for who may know what reasons.

Balarka Sen

I do not think so, because it is true that 80% of the conversations here are nonmathematical.

I do math art

@BalarkaSen You bring your contribution to it right now. Do you want to increase that percentage you gave?

Balarka Sen

I did not say "because two users left".

However I did mean to imply because of these constant dramas.

I do math art

@BalarkaSen I and robjohn did a lot of math in the past until some things happened around, like this math is not that good as other areas some practie.

Soham Chowdhury

13:52

@BalarkaSen bit of AG, mostly to have the ideas sit correctly in my head

Balarka Sen

@Soham What sort of AG?

Agawa001

14:11

i really dont get the point why u people are raising the tension in internet, i suggest you to save some keypresses for whatever useful rather than electricty/time/lifespan tension can retract human average life expectability as you know especially when it is virtual and the source of offense is ambiguous and far to reach

Balarka Sen

lol

Soham Chowdhury

@BalarkaSen just defining schemes for now

Balarka Sen

@Soham What's a scheme.

Soham Chowdhury

Wait, you actually don't know?

I find that hard to believe

Balarka Sen

Nope. Only vaguely.

Soham Chowdhury

14:16

Well, you should read the post when it comes up :)

Balarka Sen

Post?

Soham Chowdhury

I'm reviving my blog

Balarka Sen

Aha. Nice, I will.

Soham Chowdhury

:)

Balarka Sen

I know a scheme is something which is "locally an affine scheme" but I am unsure how to make that rigorous.

Noah P

17:34

Hi all

Any ideas for a maths programming project?

I've previously made programs for the fibonacci sequence, solving quadratics, and seeing if a number is prime or not

Secret

18:10

It depends on what topic you are most proficient and your passion

Agawa001

have you heard about that AI based system that take judgments in beauty contests ?

well if you ask my opinion, it works freaking awfully great!

beauty.ai

Martin Sleziak

18:30

In Calculus and Analysis chatroom there was a request whether somebody would check self-answer to In proving primitive by substitution, we demand $x=\phi(t)$ to be a bijection. In Integration by substitution we do not. Why?.

Since this room is much more frequented than the one where the request was posted, I though that it might be a reasonable idea to repost it here. Just in case there is some volunteer.

Tobias Kildetoft

@BalarkaSen Well, do you know what an affine scheme is to get started?

Balarka Sen

18:42

@TobiasKildetoft Yes, yes. That much I am aware of :)

Tobias Kildetoft

@BalarkaSen Ok, so you know what a locally ringed space is?

Balarka Sen

Vaguely. Space with a sheaf of rings?

Tobias Kildetoft

right, and where the stalks are local rings

(affine schemes are in particular locally ringed spaces)

Balarka Sen

Of course.

Tobias Kildetoft

or at least that is one way to view affine schemes (not the way I view them)

Balarka Sen

18:45

(I know it should not be too hard for me to learn the definition of a scheme itself, I was just looking forward for Soham to teach me about it. But thanks for telling me about them)

Tobias Kildetoft

So being locally affine just means that the space has an open cover such that each open set in that cover gives you something which is isomorphic to an affine scheme when you take that sub sheaf

Of course, if you instead define an affine scheme as a representable functor from the category of $k$-algebras to sets, then defining what it means to be so locally gets messy

(but defining what a group scheme is becomes much simpler)

Balarka Sen

@TobiasKildetoft So, e.g., $\Bbb P^n$ is locally affine wrt that definition? Take the cover by the complements of the coordinate hypersurfaces.

I guess there's an issue with the generic points. Bleh.

Also, I suppose by isomorphism you mean sheaf-isomorphism there.

Tobias Kildetoft

isomorphism of locally ringed spaces (which is just sheaf isomorphism since the local part is automatic for isomorphisms)

Balarka Sen

Mhm, yup.

My biggest issue with affine schemes is that they have that weird point which is dense in the whole space. Modulo that, $\Bbb A^n$ is an affine scheme so $\Bbb P^n$ would be a scheme (hence, in particular, any projective variety).

Tobias Kildetoft

Right, varieties "are" schemes

Balarka Sen

18:54

What I wonder is what can one say about the manifold analogues of such schemes. Eg., locally isomorphic to $\Bbb A^n$ (or $\text{spec} k[x_1, \cdots, x_n]$ if you prefer).

Certainly that's a much smaller class.

Tobias Kildetoft

No idea. I never really felt like I understood non-affine schemes very well (nor for that matter the affine ones that were not group schemes)

I would even need to look it up to give the definition of a scheme in the functor sense

Balarka Sen

Me neither.

Functor-of-points is scary stuff.

Tobias Kildetoft

as I said, it makes group schemes much easier to understand

Balarka Sen

I see

Tobias Kildetoft

you just replace the functor with one to groups such that composing with the forgetful functor to sets gives an affine scheme

Balarka Sen

19:00

Hi @Alyosha

Alyosha

Hello Balarka.

Balarka Sen

What're you learning?

Alyosha

Does anyone understand the Abel-Jacobi map?

(That, at the moment, though only since today)

I'm only interested in the Riemann Surface case.

Balarka Sen

I have vaguely heard of the Albanese.

Alyosha

That is, does anyone know about the correspondence between divisors and line bundles?

Balarka Sen

19:02

But I don't know crap about it.

@Alyosha I do know about that a bit.

What do you want to know?

Alyosha

Supposedly under this map there is a correspondence between effective divisors and line bundles with a choice of preferred section up to multiplication by $\mathbb{C}^*$.

But I can't find a friendly enough definition of the Abel Jacobi map to test this.

Balarka Sen

I do not know this description of the correspondence, but there's a more topological way to see it.

Alyosha

The way I've seen it done is to do it for divisors $D=p$ that correspond to points, then take tensor products to do it in general.

Balarka Sen

That's for Riemann surfaces though. There's an explicit description of a correspondence between homology classes of hypersufaces in a manifold and line bundles on it.

Alyosha

I'm more interested in just Riemann surfaces, unless the other description is more intuitive.

Once one understands one case it's probably easy to get the rest.

Balarka Sen

19:10

I don't know about intuitive. The idea is to note that homology class of a hypersurface (in your case, a point), lives in $H_{n-1}(M; \Bbb Z/2)$. Poincare dual of that lives in $H^1(M; \Bbb Z/2)$ - that gives a map $\pi_1(M) \to \Bbb Z/2$. That's a permutation representation of a double cover over $M$. Any double cover automatically gives you a line bundle (the transition maps $U_{ij} \times S^0 \to U_{ij} \times S^0$ extend canonically to maps $U_{ij} \times \Bbb R \to U_{ij} \times \Bbb R$)

Alyosha

You mean the correspondence between $H^1(X,\mathcal{O}^*)$ and the line bundles?

Balarka Sen

Probably, yes. But I am working with singular cohomology, not sheaf cohomology, and my description is topological, not algebraic.

Not sufficient to give you a holomorphic section though, in general, I believe.

For one, working $\Bbb Z/2$ completely forgets about orientation, whereas every Riemann surface is oriented canonically. That's the reason you have a topological section for the tautological line bundle on CP^1, but no holomorphic ones.

Speaking of, I still need to work out an explicit topological section for the tautological line bundle w/ intersection number -1 (I know one exists by a intersection-theory argument).

Alyosha

Is this obviously the same as the construction I mentioned earlier?

I forget what the isomorphism in Poincare duality actually is.

Or at least it seems hard to work out explicitly.

Balarka Sen

Cap with the fundamental class. Equivalently integrate a cohomology class over a chain.

Alyosha

OK, I think I can dimly see the equivalence.

Balarka Sen

19:17

@Alyosha Well, in my case, you're working with real line bundles.

Alyosha

Yes, OK.

And then you take the tensor product after that, I assume?

Balarka Sen

Um, why would I do that?

It directly constructs a bundle with divisor any given hypersurface (or more precisely, homology class of a hypersurface).

Alyosha

You've constructed a line bundle corresponding to a hypersurface

Balarka Sen

Yes.

Alyosha

Yes, but divisors are sums of hypersurfaces.

I realise that this is a silly point.

Balarka Sen

19:22

Hypersurface means codimension 1 manifold: my construction works fine as the disjoint ones also represent homology classes in $H_{n-1}$.

So the tensor product bit it unnecessary work.

Alyosha

Oh right.

Balarka Sen

:)

Yes, in essence, if you were working with only points. Tensor product "concatenates twists", so if you have constructed a line bundle corresponding to a point, tensor product makes that same construction again and again and each time you get an additional zero.

So if you have a line bundle $\ell$ corresponding to homology class of hypersurface $M$, and a line bundle $\ell'$ corresponding to homology class of hypersurface $N$, $\ell \otimes \ell'$ corresponds to homology class of $M \sqcup N$.

Or I think it should. I haven't worked out a rigorous proof.

arctic tern

Question on my mind before I do yardwork. If $R^+$ is a (commutativ, unital) semiring and $R$ the ring made from it (Grothendieck construction, so formal differences of elements from $R^+$, like $\Bbb N\subset\Bbb Z$), then are ideals $I\triangleleft R$ of the form $J-J$ for ideals $J\triangleleft R^+$? (My particular context is Burnside rings associated to symmetric groups, since they describe combinatorial species.)

Alyosha

I'm trying to work out what gives you a preferred section if $D$ is effective.

Balarka Sen

I don't really know what effective means.

Alyosha

19:30

All coefficients are positive.

Balarka Sen

Oh, I see. The claim is that only effective divisors correspond to line bundles, yes?

If so, that makes perfect sense. Negative coefficients mean different orientations - not a thing on complex manifolds.

Alyosha

Have you not constructed a line bundle in either case?

Balarka Sen

I am not sure what you mean. I have not done this construction on the holomorphic category - only on the topological category.

Alyosha

Are line bundles and sections not both topological objects?

I don't know what you mean here.

Balarka Sen

Eh, complex line bundles are not on principle the same as real 2-plane bundles. And sections of line bundles over complex manifolds are required to be holomorphic, not just continuous.

What did you mean by "have you not constructed a line bundle in either case"? I am confuzzled.

Alyosha

19:38

I meant a real line bundle.

I'm still trying to read through the construction and wanted to check whether the analogous statement about sections held here.

Balarka Sen

What do you mean by "the analogous statement about sections"?

I am not sure if I saw any statement about sections before.

Alyosha

Above, you have stated a correspondence between divisors and real line bundles. Under this correspondence, are effective divisors sent to a line bundle with a natural choice of section?

Balarka Sen

I do not see why the adjective "effective" is relevant for real line bundles. Also, no, I don't think one gets a natural choice of section under that construction - we are passing off to homology, so that would only give a generic section even if it does.

Oh, hi, @MikeMiller.

Alyosha

I'm inclined to agree at this point, the two situations seem somewhat different.

Balarka Sen

They somewhat are, agreed.

Mike Miller

19:47

@anon: Does the KAN decomposition of a connected semisimple Lie group always provide a diffeomorphism $K \times A \times N \to G$? (In particular, are $K, A, N$ always closed subgroups?) Do you know someplace with a good exposition of KAN?

Maybe also @TobiasKildetoft

Hi @BalarkaSen.

Balarka Sen

No, I mean, Poincare duality is.

Alyosha

Wait, that makes no sense.

How do you get that map, then?

I'm happy with Poincare.

Balarka Sen

The isomorphism $H^1 \cong \hom(\pi_1, \Bbb Z/2)$ just comes from the universal coefficient theorem. I do not know if this has a differential forms interpretation.

Mike Miller

@Alyosha I feel like you're talking about complex geometry and Balarka is talking about topology.

@BalarkaSen How could $H^1(M;\Bbb Z/2)$ have a differential forms interpretation?

Alyosha

OK. I've never looked at the UCT properly.

Balarka Sen

19:50

@MikeMiller Eh, you got me. I had in mind $H^1(M; \Bbb R) \cong \hom(H_1(M), \Bbb R)$.

Alyosha

@MikeMiller Do you know anything about the effective divisors giving sections part?

Mike Miller

I didn't look much at the conversation. What is your question?

Alyosha

There's some construction giving a line bundle from any divisor.

Supposedly if the divisor is effective then this gives you a preferred section.

I've not looked enough at the construction to know why it's true, so I'm curious about
1. why there is a preferred section
2. how to think about the construction.

Balarka Sen

The counterexample to this when your divisor is not effective, is the tautological bundle. But that's as constructive a comment I can give you on this.

I don't really know how to explicitly construct a section. Mike probably does.

Alyosha

Thanks for that, though.

Mike Miller

19:55

It seems silly to ask these questions without reading the construction. Go grab a copy of Griffiths and Harris.

Alyosha

I suppose I was being lazy asking here.

Anyway, thank you!

@BalarkaSen you reminded me that I've not thought about fundamental classes properly. :P

Balarka Sen

That's how I think about homology classes :)

Tobias Kildetoft

@MikeMiller I never really studied Lie groups, so no idea

I am fairly certain the corresponding statement for schemes is true

Mike Miller

ok, that's good to hear.

Tobias Kildetoft

Assuming the subgroups are the ones I would expect

Mike Miller

20:05

@BalarkaSen Your question about 'manifold schemes' is not the correct notion. There are only two Riemann surfaces locally isomorphic to $\Bbb C$. You need to work with open subsets of $\Bbb C$, which doesn't seem to me to admit a good generalization to varieties (or else compact Riemann surfaces would all be algebraic for more obvious reasons).

Tobias Kildetoft

(they are not called by those letters in algebraic groups)

Mike Miller

$A$ is for abelian, $N$ for nilpotent, $K$ for... Kompact

I feel reasonably well that if they're closed subgroups and the obvious map is a bijection, you can show it's a diffeomorphism

Tobias Kildetoft

Hmm, not sure what the corresponding ones would be then. Probably maximal torus and the unipotent radicals of the positive and negative Borels

but that is more symmetric in two of them than yours

Mike Miller

Doesn't sound right, $K$ will properly contain the maximal torus.

well, most of the time

Tobias Kildetoft

Yeah, really not sure what the corresponding ones should be (there are some for which a statement like that holds)

Balarka Sen

20:10

@MikeMiller That's a nice point.

Yes, I agree, there are not too many varieties even in dimension 1 which are locally isomorphic to A^1.

Agawa001

i hardly believe this is english i m reading

Mike Miller

@TobiasKildetoft Are there maximal proper subgroups or something?

Tobias Kildetoft

@MikeMiller For the algebraic groups, one would take the ones I mentioned above (unipotent radicals of the positive and negative Borels and the maximal torus)

arctic tern

20:29

@MikeMiller dunno what the conditions are. I'm only familiar with the decomposition for some classical matrix groups. (KCd talks about SL(2,R) for instance.)

I'm sure some or other standard lie groups textbook will state the relevant result in some section on the Iwasawa decomposition.

Tobias Kildetoft

@MikeMiller Just found the statement in the notes by Libine. It is Theorem 106 of arxiv.org/abs/1212.2578

Alyosha

@MikeMiller Thanks, Griffiths and Harris explained it well.

Mike Miller

20:51

@arctictern Thanks, I'll try. I'd like to have an authoritative source for the fact that any Lie group is diffeomorphic to $K \times \Bbb R^\ell$, $K$ the maximal compact subgroup. It is fairly easy to reduce this to the case of connected semisimple, which should follow from KAN but I didn't know a source for that particular corollary.

Tobias Kildetoft

@MikeMiller Did you see the notes I linked?

Mike Miller

Oh, I hadn't checked. I thought you were still talking about the algebraic group stuff.

Tobias Kildetoft

Ahh, no these are notes on real Lie groups and has the precise statement you needed

Mike Miller

Awesome, thanks.

Unfortunately not the best typeset thing I've ever seen but it does the job.

Balarka Sen

@MikeMiller Referring to the new idea I mentioned, it is to do the proof for the trivial bundle $p : X\times \Bbb R^n \to X$: that, I think, can be done by taking the section $s$ with zero oriented intersection number, homotoping it away from the zero section, and then making it a section again (doing the last step should be analogous to genericity of Lefschetz maps).

Then doing this locally for points of opposite intersection number for arbitrary vector bundles, by taking a nbhd containing them over which the bundle trivializes. Do you think this is workable?

(I feel like this idea is pretty ad-hoc; there should be something simpler or more apparent)

Mike Miller

20:58

Something like that.

I'm out. See ya.

Balarka Sen

Byes. I'll try to make this precise.

Mike Miller

@TobiasKildetoft Looks like proofs are deferred to Knapp, Lie groups beyond an introduction. I have a copy of that lying around somewhere. Thanks.

Tobias Kildetoft

@MikeMiller Ahh, yeah, I think I skimmed that at some point

Agawa001

21:55

gd night

Semiclassical

22:22

Pretty sure this is the largest comment I've ever seen: math.stackexchange.com/q/1913600/137524

0celo7

22:36

@Semiclassical I'm going insane

Given a first order ODE initial value problem, can two solution curves ever cross?

@Semiclassical how does that not exceed the max comment length?

the existence and uniqueness theorem says that there always exist solutions and that each initial value has ONE solution

it doesn't say anything about curves crossing

Semiclassical

i presume it's because the max comment length is based on characters, and Did somehow managed to squeeze it through that

re: curves crossing, i'm not sure. i would suspect the point is that, since it's a first order ODE, if they have the same values at one point they should have the same first derivative there

0celo7

Seems reasonable, so?

Semiclassical

or even more basically, come to think of it, the only info one should need for a first order ODE is its value at one point in time. that'd seem to fix the solution before and after. so there'd be no way to have two distinct solutions, since that's tantamount to having the intersection point determine two solutions.

0celo7

@Semiclassical I was thinking that the solution curves crossing is inconsistent with uniqueness applied to the "time reverse" ODE.

@Semiclassical AH!

Yes, that's right

Semiclassical

right

0celo7

22:45

Picard-Lindelöf gives a solution on $(t_0-\epsilon,t_0+\epsilon)$

so you can look at uniqueness before and after $t_0$

@Semiclassical Thanks.

Semiclassical

np

0celo7

23:36

@Semiclassical Consider a homogeneous ODE. If we pick the initial value 0, is the whole solution zero?

Semiclassical

do you have a first-order ODE in mind?

0celo7

Parallel transport equation.

But I want baby steps

I realized I was too quick to accept "existence and uniqueness of ODEs" as a proof in Riemannian geometry.

Semiclassical

well, if it's not first-order then having the initial value be zero isn't enough

0celo7

@Semiclassical Yeah, first order and even linear.

Semiclassical

hmm. so first-order, linear, and homogenous

0celo7

23:39

Yes, so of the form $\dot{\mathbf{v}}(t)+S(t)\mathbf v(t)=0$ where $S(t)$ is a matrix

Semiclassical

in that case I think the answer is yes, since the ODE would have to be of the form $y'(x)=f(x)y(x)$.

oh, a first-order system

0celo7

yeah

I want to know if $\mathbf v(0)=0$ implies $\mathbf{v}(t)=0$ for all $t$ where the solution is defined.

Semiclassical

I would think the answer is yes, since you'd necessarily have $\dot{\mathbf{v}}(0)=0$

and then it being locally constant is enough to keep it at zero everywhere

something like that

but I'm always paranoid about counterexamples

0celo7

what're some examples we could check out

how many first order linear homogeneous ODEs are there

wait, aren't those all solvable?

Semiclassical

well, there are a lot of matrices $S(t)$ :)

0celo7

23:45

there's always an integrating factor

@Semiclassical I want to check 1D first.

Semiclassical

they should have the formal solution $\mathbf{v}(t)=e^{-\int S(t)\,dt}\mathbf{v}(0)$

0celo7

well, that settles it, right?

Semiclassical

eh, I think so. the only thing i'm paranoid about is whether $S(t)$ could fail to commute with $v(0)$ and create problems that way. probably not, though.

bolbteppa

What do you mean, how could solution curves cross, then they would not be unique?

If you can prove Picard-Lindelof, no reason why you can't prove the inverse function theorem (which is used in Sard's theorem which you can prove apparently), same technique and idea

0celo7

No, I'll just block you.

@Semiclassical fail to commute?

How does a vector commute with a matrix

Semiclassical

23:50

point.

yeah, definitely no issue in that case

0celo7

@Semiclassical This exercise in ODEs might be interesting, but I figured out my problem.

And it might explain the ODE thing

My problem was showing that if the parallel translate of a vector vanishes, then that vector must necessarily be zero.

Semiclassical

ah

0celo7

This is equivalent to showing that if the initial data is zero, then the solution is zero

Because you can just move to the point where it's zero, and "re-solve" the equation

You should get the same solution

But for a homogeneous ODE, clearly the zero solution is valid

Semiclassical

right.

0celo7

So invoke uniqueness: the original solution must be zero as well

So, now we know that the kernel of the parallel translate operator is just 0, so it's injective. Apply rank-nullity to get surjectivity. Thus parallel translation is an isomorphism

Praise Jesus.

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