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12:01 AM
not seeing why the situation in this question is evidently more subtle than I thought
 
@JuanSebastianLozano So, we didn't finish our conversation.
If you still care, that is.
 
@Semiclassical I do not see the subtlety.
 
I should probably go to sleep though.
 
well, the accepted answer makes use of the heat equation. but to me it seems more trivial than that, i.e. (1) implies (2) directly
 
I agree.
 
12:13 AM
ah.
 
@Semiclassical I do want to procrastinate, so I'll tell you what I was thinking about earlier
 
hah, sure
 
(if you want to listen, of course)
 
it's fine. not sure how awake my brain is, mind
 
Consider the north pole and the south pole on the sphere, and the height function on it. One can consider the set of all gradient flowlines from the north pole to the south pole
there's an action of $\Bbb R$ on this space, sending $\gamma(t)$ to $\gamma(t+s)$
(reparameterizing the flowline); mod out by this and you have a nice space of unparameterized trajectories
for this example, the space is $S^1$ - gradient flowlines are in bijection with points on the equator (sending a gradient flowline to the one it goes through)
 
12:15 AM
aka each line of longitude on a globe is labelled uniquely by an angle?
 
right.
a more complicated setting is the height function on the following "tilted torus"
between, say, the bottom-most and second-to-bottom (or third-to-bottom) point, there are precisely two gradient flowlines
now consider the gradient flowlines from the very top point to the very bottom point. (i actually don't know what this space is, though i have good guesses.)
unlike the previous example, this space will not be compact - you could have a sequence $\gamma_n$ of gradient flow lines that go closer and closer to one of the critical points in between the top and bottom
 
there's sort of a cell-structure there
 
those blue lines are gradient trajectories
 
right.
 
so imagine a sequence of trajectories that, in the limit, starts at the top, goes down the "top slightly to the right of the hole" trajectory, then down the "bottom front" trajectory
does that parse?
 
12:19 AM
would that amount to a bifurcation of the flow lines?
wait, no
 
not sure I know what that is, being an ignoramus
such a thing is called a "broken trajectory", since instead of being a flowline between two critical points, it stops somewhere in between
it's not an actual flowline, just two flowlines one after another
 
yeah, i see what you mean. to put it another way, the middle two critical points are 'unstable'
 
so I said earlier that the space of trajectories, due to this, is not compact. but we can add these in (so that we allow 'broken trajectories' to be points in the space). then the result is compact.
right right
given as fact: the original space of trajectories (the noncompact one) is a manifold.
 
what i was getting at re: bifurcation is that i can pick two trajectories at the top which are nearly the same initially but which split by the end
 
natural question: what kind of object is the compacfication?
 
12:22 AM
lemme think for a bit
 
oh, this wasn't a question i was going to propose one solves, but sure
lol i just got an email from the conference i'm going to on... thursday... with a formal offer, and deadline of acceptance = tomorrow
 
i kind've feel like one should be able to get that in a simple way by using the broken trajectories to define a cell structure, and then take the dual structure of that (so that all flowlines which are locally 'the same' are identified)
that's not quite right, i know
 
the trouble is that these trajectory spaces do not obviously reflect the topology of the original manifold
you can, in the end, assemble them that way, but it's nontrivial to see how it works
 
ah. so there's some legwork involved
 
yeah. i'd just like to understand what this space looks like, before i try to understand how they all fit togethert
 
12:26 AM
right.
 
let's label our points a,b,c,d. a is the top, d the bottom, b,c in between
so the answer is that it's a manifold with boundary, and the boundary is in bijection with $\mathcal M(a,b) \times \mathcal M(b,d) \sqcup \mathcal M(a,c) \times \mathcal M(c,d)$
 
what i find tricky is 'seeing' the broken trajectories
 
yeah, i agree with that.
the idea is to pick an initial "tangent vector" at the north pole to say what direction you're going to go in
that's very very close to the flowline from the top to b
 
aside from the parts going from a->b and c->d, i mean
 
this should ultimately be something that is very close to a->b b-> d
oh, they drew the lines going from b to d in the back of the picture
 
12:28 AM
that helps
what i'm having a hard time telling from that picture is which trajectories are 'generic' and which ones correspond to broken trajectories
 
i suggest trying to look up front instead of behind, because behind is hard for me to see precisely
the broken ones are just the ones that "stop" and have a "kink' in them
 
okay, so the ones which start/stop at b and c?
 
do you see precisely where b and c are in the picture?
 
think so
 
they're solid dots
one of the solid dots is "in the back" so you have to look a bit
 
12:31 AM
they correspond to saddle points, yes?
of the height function, i mean
 
yeah
more or less vertically below a, and in the backside, is b
similarly d = bottom point looks to be vertically below c, and is also in the "backside"
 
main thing i wanted to confirm is that the two blue lines at the farthest right and left aren't broken trajectories, but all of the rest are
 
that's correct
gotta run, back later
 
1:19 AM
The Legesbue measure obeys $\lambda^*\left(\left\{ \mathrm{T}\vec{u} \, \vert \vec{x} \in C \right\}\right) = \left\lvert \det \left(\mathrm{T}\right)\right\rvert \lambda^*\left(\left\{ \vec{u} \, \vert \vec{x} \in C \right\}\right)$. What does \lambda^*\left(\left\{ \vec{\mathrm{a}} \cdot \vec{u} \, \vert \vec{x} \in C \right\}\right)$ equal?
Sorry, $\lambda^*\left(\left\{ \vec{\mathrm{a}} \cdot \vec{u} \, \vert \vec{x} \in C \right\}\right)
Sorry $\lambda^*\left(\left\{ \vec{\mathrm{a}} \cdot \vec{u} \, \vert \vec{x} \in C \right\}\right)$
 
How long can I sit on someone's error without writing them an email?
 
user147690
You can edit your messages by hovering over them and on the left clicking the down arrow and edit. (Don't know anything about your question though sorry)
 
user147690
@PVAL Depends how much you like them :P
 
I don't know them
 
user147690
Are you a reviewer?
 
1:27 AM
no
They posted a preprint with what at this point I'm relatively sure is a fatal error.
I looked at it, and found it, but I got other things to do..
 
user147690
No idea what to do there sorry. If you don't care about attaching your name to the email, you can just email from a dummy account, and then not worry about it again
 
Likely I will send an email asking "I don't understand how this works, can you explain it to me" being relatively certain it doesn't work
 
1:54 AM
@PVAL Just do it, as an author I would prefer it the sooner someone did, if they intend to do it.
@SemiC Anyway "gluing" is the process by which you take a broken trajectory and "glue them together" to show that there's an honest trajectory nearby that goes from a to c, and that the space of trajectories is a manifold with boundarh with appropriate boundary set. In higher dimensions you get the structure of a manifold with corners instead.
In the gauge theory context you would more or less like to get similar results but the answers tend to be much more gross. Eg there's no reason you couldn't get a Hawaiian earring.
There's enough structure that you can work with it but it's still bad.
 
unpleasant
looking at that picture from before, the part that makes my head heart trying to visualize it are the trajectories on the 'inner' surface
 
2:10 AM
how do I know what journal to publish in?
 
Ugh can't sleep
 
can I do AMS or famous?
and how much money do I get?
:)
 
talk with your advisor
 
@BalarkaSen sorry I have same problem
because I get so excited about math I cant sleep
 
any good ideas about what do to when I'm bunking sleep?
 
2:12 AM
@ForeverMozart If you are unaffiliated probably about -$100. Usually your university will pay the fees though.
 
math is one option.
lol -$100
 
@PVAL so I will not be made rich? WHAT
 
you'll be made rich. backwards.
 
ok well what are the most famous journals
 
annals, Inventiones, Acta, JDG ...
 
2:16 AM
I never heard of those
 
let's not forget vixra
 
oh wait annals is AMS right?
 
no
Annals is published by Princeton University.
 
Fundamenta Mathematicae is good right
Topology and its Applications
 
I've never heard of the first
 
2:22 AM
61
Q: Which are the best mathematics journals, and what are the differences between them?

Scott MorrisonSuppose you have a draft paper that you think is pretty good, and people tell you that you should submit it to a top journal. How do you work out where to send it to? Coming up with a shortlist isn't very hard. If you look for generalist journals, it probably begins: Journal of the American Ma...

I guess you were right
ok another question
if papers about a similar topic have been published in the same journal, is that the best place to put it?
 
I don't know exactly where Topology and its Applications ranks. It's certainly a reputable journal but not on the level of Geometry and Topology or JDG. There's maybe 2 or 3 people on the board I recognize compared to literally every single one on G&T.
@ForeverMozart The best place to put it is on the arxiv
 
why arxiv?
 
That is if you after have had an experienced researcher look at it first.
And you have LaTeX'd it up very carefully.
 
what is the point of arxiv? To let everyone know not to copy you?
 
The point is for people to read it.
That is the end goal.
 
2:29 AM
but what if they copy it
 
Well you have a submission to arXiv to defend against that.
 
is it prestigious?
 
It's a preprint server.
 
so anybody can post there?
 
Don't you have an advisor to talk about this sort of thing? That's part of what they're for.
 
2:31 AM
yes I meet tomorrow, I just wanted to know more beforehand
 
I'd prefer if we discussed more about math, though. But that's just my taste.
 
Oh my colleagues just put up a paper I've been hearing about for a few months.
I should probably read that.
 
ok but last question: can anybody put a paper on arxiv or do they review it?
 
@Semiclassical You're here?
 
You need someone to endorse you, but that's quite easy to get. There are no reviewers.
Endorsement just means "Yeah this guy can post stuff"
Maybe I'll read this when I'm on the same flight as him and can just ask him questions.
 
2:35 AM
it should bring me fortune and fame
 
If you want those you probably shouldn't be doing mathematics.
Maybe drive in NASCAR or play basketball.
 
Your colleague has written quite a lot
 
"Why wants money or fame? Gimme a new theorem."
 
If you asked him he would tell you they're all terrible papers.
 
A result of his was mentioned in a fellow student's candidacy here last week.
 
2:37 AM
I think the Floer ones are quite nice. I have no opinions on the ones he did during an REU.
 
I don't believe anyone can write more than this guy.
 
I guess probably his connected sum calculation?
I guess his calculation of Ciprian's Pin(2)-invariants for Seifert-fibered spaces is probably more obviously relevant.
 
Essentially that the Pin-(2) invariants don't obstruct embeddings into S^4 in any way that wasn't already known for Seifert fibred spaces over S^2. I.e. if an older gauge-theoretic invariant vanishes these do as well.
Ya thats the one.
 
His paper from yesterday shows that the same is true of any invariant you get from subgroups of Pin(2).
Although I'm skeptical because I don't believe the $h$-invariant is known for enough SF spaces for that to be true.
 
is he famous?
 
2:41 AM
I assume all REU papers are trash.
 
Does Bob have people thinking about homology cobordism?
 
@balarka back now
 
My colleague is a grad student. There is probably one 'famous' grad student.
 
No I am Bob's only student.
 
Who's thinking about this?
 
2:41 AM
Ahmad Issa
Cameron's student
 
@Semiclassical You think you want to discuss about maximizing things along integral curve problem?
 
not at the moment
 
Aright, sure.
 
I've heard of his name before but I don't remember why.
 
I probably told you.
He was at GTGC
 
2:43 AM
Probably.
Oh yeah I probably skipped one of his talks.
 
I don't think he talked.
 
Then I definitely did not see him talk.
 
The colleague you mentioned is terrified of what's going to happen to him after graduation. I think he'll be fine.
 
There was an undergrad from UT who talked there about a paper of Bob and Scharlleman and said Gompf et. al in the abstract.
 
2:44 AM
shelah is famous
 
The first author got an NSF postdoc with Danny R with only his first paper.
That's a bit much.
 
@anon Holy crap.
 
I don't think I could do anything to incite him more than that.
 
can I do that?
will Obama give me NSF
 
Bob gets really mad about proper attribution I take it?
 
2:47 AM
Literally the only thing I've ever seem him offended about was crediting him for work that was also due to other people
I don't know if mads the word.
I don't think he gets mad.
 
My roommate rode back from a conference with him once. Apparently he was very quiet.
 
Mainly him being attributed for other peoples work.
Yes he is.
 
who are we talking about shelah?
 
I am pretty sure your colleague is in the situation where literally every school will be making him offers before he applies to places.
 
One of Ciprian's other students, same year, will be in a similar situation.
The one graduating this year is going to MIT. He probably did not need to ask, but I don't know.
 
2:52 AM
If I ever want to feel really bad, I just remember John Pardon is the same age as me.
 
At my school people think about Francesco Lin.
 
he is slightly younger than me so thanks not I feel bad too
 
John Pardon's a friend of prof.
 
I don't remember where he's going. Probably just across the street.
 
He's at Stanford still I think
He was offered a tenured position there in his 3rd year of grad school.
 
2:54 AM
I will not be that famous I do not think, but maybe famous to topologists
 
I mean Francesco
 
someday I will be called the King of Counterexample
 
Oh I actually need to read one of Lin's papers.
 
Which one?
Oh, nevermind, I know.
 
wow how many papers do you read?
I usually only read a paper if it is related to something I'm trying to solve
 
2:56 AM
I was told that was a result of Murphy and I couldn't find the paper.
 
positive music for thinking youtube.com/watch?v=TW8Zo0NDkA8
 
inb4 rick roll response
 
how is Shelah so productive? he has magical wizard powers
 
 
1 hour later…
4:09 AM
@Ras_al_Ghul you killed Han solo you bastard
oh wait no you were in batman
 
lol
 
got my movies mixed up there
@BalarkaSen you should watch this youtube.com/watch?v=wN4yLPPvRBg
 
Ah, yeah, I have seen it before.
 
you probably seen every math documentary like me
 
Nah.
 
4:18 AM
have you heard of the Erdos number?
it is number of mathematicians you are away from someone who worked with Erdos
 
Yup.
 
my advisor met him once
 
4:45 AM
I just made that
autocad is so useful
 
Is that the Lelek thing again.
 
yes kindof
I need a high resolution image for my paper so I am going to make them in autocad
 
BTW, did you check if your example worked?
 
I think it does, tomorrow I will talk to someone who doesn't take any crap
so hopefully it is still good afterwards
 
Let's hope it is.
 
4:48 AM
i mean he is very precise
 
5:03 AM
listen to something good instead of that stuff, @Forever.
 
I know, I just had that stuck in my head for some reason
like what?
 
@BalarkaSen You're very patient with your covering space friend.
 
He's kind of annoying when he does that but he's a friend nonetheless.
 
I agree with you that there is no subtlety.
 
hey now
 
5:05 AM
Yay.
@ForeverMozart What kind of things do you like to hear?
 
Well there is in the sense that "Count the number of degree $n$ covering maps onto $S^1$" ends up giving the number of partitions of $n$, but this is not a super interesting result. It's only interesting that the number it gives you is maybe momentarily surprising.
 
Bpb Dylan and Mozart
 
@MikeMiller I agree with it being momentarily surprising, but not being super interesting.
 
and some electronic music
 
@ForeverMozart Dylan is great.
 
5:09 AM
steely dan and neil young
 
"stuck inside the mobile with the Memphis blues again", if you haven't heard it already.
 
blonde on blonde
I listened to that many times
 
5:23 AM
this guy is a true intellectual mathoverflow.net/questions/237311/…
Let me add, as a corollary to the point I make in the post, is that we shouldn't be impressed with a difficult mathematical argument simply because it is difficult. Rather, let us criticise difficult arguments for being obscure and try to improve them.
wow so smart
I remember being envious of carpenters and artists, who after an achievement can point at what they've done, its value obvious; while in mathematics our very achievements can undermine our perception of their value.
this is very true in my situation
Metaphysics has become mathematics, ready to form the material for a treatise whose icy beauty no longer has the power to move us."- Andre Weil
WHAAAT that is insane
@BalarkaSen can you believe this?
 
Believe what?
 
that he said that
 
I dunno what metaphysics is.
 
Do you know what meta.*.SE is?
 
Oh, I do.
I certainly do not believe meta.physics.se has merged with math.se
 
5:29 AM
Surprised those words came from Dr. Weil
still not sure what that quote means exactly but
w/e
 
to me it means have we become too advanced in our understanding of the universe to appreciate the mysteries
 
@BalarkaSen Understand their relationships to *.SE. Now remove the .SE.
 
his proof of a very simple statement is only understood by very few people, so it kindof makes sense
 
Your construction gives surprising result for meta.SE, in which case removing the SE gives "meta".
 
I might argue that the statement is contestable
 
5:33 AM
In many mathematical realms, the actual achievement in research is that certain issues and ideas become easy to understand.
 
metaphysics is too philosophical, too holistic IMO to be associated purely w/ math
perhaps he meant theoretical physics, of the cosmological flavor
 
yes probably
 
just a subtlety
 
there is a proof of some major classification theorem in group theory that takes up like 200 pages
 
"Or maybe he just wanted to write something fun which is incomprehensible" - me thinking.
 
5:34 AM
the statement itself is very simple
but almost nobody understand the proof
 
I think it takes more than 200 pages.
 
is this relatively recent?
 
I think this is the problem he is talking about
 
If you're talking about classification of simple group, I think it takes 200 pages each major group theorist working on that at the time I can think of.
 
so the beauty is undermined because we really gained no understanding from the proof
 
5:37 AM
The original CSG proof is ~10k pages. I think that considerable progress has been made in reducing that number, but I don't think it's in triple-digits yet.
Mozart, do you think the same for the four-color theorem, then?
 
This discussion is too metamathematical for my tastes.
 
like we all do, of course: in terms of the instanton web homology of embedded trivalent graphs in $\Bbb R^3$
 
four color theorem proof, smartass }:P
 
did they prove it like that?
@EricStucky People are trying to prove four colors using instanton stuff.
 
@BalarkaSen sometimes I like to step back and look at the forest
 
5:39 AM
I don't know if they made it though. Apparently it's a hot topic, or so I think.
 
Hot topic = two people are doing it.
 
I stand corrected.
 
But the two are the chairs of the MIT and Harvard math departments.
 
probably need some younger ideas in there
 
5:40 AM
I revert back to my original standing position.
 
My understanding is that there has been no further progress since the original papers, because otherwise I would have heard about it.
 
Hmm, I see.
 
@MikeMiller i wanted to follow up on that gluing map stuff from earlier
is there a good answer in general to what kind of object the 'compactified space of flowlines' will correspond to for a given manifold?
or is it always fairly involved?
 
@balarka , i may decide to weave thru Ted's book this summer
 
Nice. I'd be happy to help you if you want any.
 
5:44 AM
@Semiclassical as long as you've chosen the function you're thinking about carefully enough (the "Morse-Smale conditions"; pretend this just means that the space of flowlines is a manifold), then you always get a "manifold with corners" as the compactification
and the strata of this thing (for instance, the "boundary" - the codimension 1 part of the corners - corresponds to trajectories that break once; the codim 2 part corresponds to twice-broken trajectories; etc) are what you would expect from the above parenthetical
 
thanks. absolutely @balarka
but i'll have to restrain myself
 
i'd have to think about that quite a bit more than i'm prepared to right now, but it's an interesting story
 
perhaps wait on a tough question for a few days, research or think it out
 
"Morse theory", in its most refined form, should say something like "you can reconstruct the manifold from these trajectory spaces"
 
before i bother you or Ted about it, or else I'll be poking constantly
 
5:46 AM
:P
 
i'd be curious what happens if one moves into the domain of Riemann surfaces
pretty sure it'll be related to the quadratic/strebel stuff i've mentioned
 
this theory doesn't naturally live in a holomorphic world, i don't think
 
@Brody I'd like to be poked constantly, but it's good to think through thinks yourself and/or come up with good and thoughtful questions instead of bad ones.
You'll find a lot of food for thought in Ted's book (and his lectures!)
 
it'd probably fall into the realm of Picard-Lefshetz theory (not that i know what any of that means)
 
@BalarkaSen looking forward to it
 
5:49 AM
yeah, i believe that. this gluing theory shows up in a lot of modern contexts. whenever you have a moduli space of some flavor (for symplectic manifolds, moduli spaces of holomorphic curves) you'll have some "breaking" into simpler kinds of curves; that's more or less the same phenomenon
and one needs a gluing theory there too. there are other, more complicated phenomena in this setting, though
 
the reason i say it's 'probably' related is because quadratic differentials generate geodesic flows on riemann surfaces
and that connects to the triangulation stuff which i only sort've know about
 
i once hoped to take Ted's class if I attended UGA (except I didn't know the instructor, just the course outline)
 
@Brody What're you studying now?
 
never thought I'd run into him here lol
 
mind that gradient flows are usually not geodesics
 
5:50 AM
true
they were in the spherical case, but of course that's special
 
@Brody Life takes strange turns! Did you talk to him? He comes in the chat often.
 
@BalarkaSen elementary ODEs and multivariate calc. stuff is very rote and/or ad hoc
 
yeah, that was strange luck. i guess someone has probably thought before about when that's true in general
seems like a natural question to ask
 
haven't had a sustained convo with Ted, but maybe a comment here and there
 
i remember reading on that a while back
 
5:51 AM
He's been away from the chat for a while. I e-mailed him a couple days ago - he said he'll come back soon. You should grab a conversation with him, perhaps.
 
phrases like "eikonal equation" showed up which i should really know as someone who does semiclassical stuff
 
23
Q: Functions whose gradient-descent paths are geodesics

Joseph O'RourkeLet $f(x,y)$ define a surface $S$ in $\mathbb{R}^3$ with a unique local minimum at $b \in S$. Suppose gradient descent from any start point $a \in S$ follows a geodesic on $S$ from $a$ to $b$. (Q1.) What is the class of functions/surfaces whose gradient-descent paths are geodesics? Certainly if ...

@BalarkaSen did he say why he left? did i annoy him too much?
 
yeah, i saw that question
 
I remember Ted telling me that his diffgeo notes have that problem.
"When are gradient flow lines geodesic?"
 
@BalarkaSen sure. might help to overcome my reservedness too
 
5:53 AM
found the answers a bit hard to penetrate easily
 
@MikeMiller I asked him if he left annoyed by me, he said it wasn't I who annoyed him. So it seems he is annoyed, but didn't directly mention you.
 
why are mathematicians offices so messy
 
I am trying to keep him posted about what I do to persuade him into coming back :P Maybe you may want to shoot him an e-mail too, I dunno.
 
fair enough, though i think that clarifies it. i don't particularly intend to apologize so there's no value in sending an email for either of us.
 
my undergrad advisor's office was legendary. Stacks of books and papers up to 6 feet tall, and only a narrow path from the door to his desk. My current advisor's office is not much different
 
5:54 AM
hopefully he comes back and just ignores me or whatever.
 
@ForeverMozart was expecting a pun joke to the question
 
well, you'd expect us to be well organized
 
@Semiclassical the comments are pretty insightful. they point out that, in particular, $f$ must be rotationally symmetric. i think this should mean that the only shapes you have are spheres and $\Bbb R^2$-type things.
 
well-ordered?
 
5:56 AM
functor-ganized (unorganized?)
 
that's fair. he'll come back (he said so...), and i don't think he'll ignore you.
 
doesn't seem workable...
 
well, assuming it's a Morse function, at least that should be your conclusion, i think.
 
that'd pretty much mean that geodesic flow can only be generated by gradient flow in cases where there's literally no other possibility
 
functorginzed is a nice word. I plan to use it.
 
5:57 AM
"even I don't understand the dead cat, the math is really how it works"
 
@Semiclassical no, you could make very strangely shaped spheres
like the boundary of a neighborhood of a "U"-shape in $\Bbb R^3$
 
not following, but i'm not exactly alert right now
 
@balarka a new entry in the category theorist's dictionary
 
i just mean in this case gradient flow will definitely not be geodesic flow
even though it's topologically a sphere. i think it would be very rare for gradient flow to be geodesic flow
oh, but I think I misinterpreted your statement.
 
ah. i had in mind the fact that if $f$ is rotationally symmetric, then the flows would just have to be rays emanating from the origin
 
5:59 AM
totally
 
i'd have to think about it
and right now i'm replaying a chapter of Hotel Dusk i lost because i dropped my ds and the game froze.
 
yikes
 

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