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18:00
when i drew it the circle is on the top and ellipse is on the bottom
(When $y\ge 0$, yes.)
its a 3/4 of the top unit circle
from pi /4 to pi
So travel along the line $y=x$ from the circle to the ellipse. That point on the ellipse does not have $t=\pi/4$.
but they both end at ( -1, 0 ) right?
18:01
Yes.
@Maks chances are someone else can answer your question if you just write it
So if you set $\cos(t) = \sin(t)/\sqrt3$, this means that $\tan t = \sqrt 3$, so what is $t$?
why would i equate those?
I get pi /3 that way
oh
So that's the starting point on the ellipse.
because they are on y=x
18:03
But their solution is totally sloppy. You do need to adjust by the line integral along that little piece of $y=x$.
However, that line integral is $0$. Why?
you mean from (1/2 , 1/2 ) to ( 1/sqrt(2) , 1/sqrt(2) ) ?
Right.
Well it is not clear to me why without doing that line integal
Well, the vector field is orthogonal to that line segment along the line segment.
in the solution this were not even mentioned
18:05
Yeah, so they're wrong.
ahh because of -y, x
Right.
okay that is smart :D
but one thing pls !
what did you mean by not normal polar
and how do I do in those situations
When someone isn't busy, does there exist f such that $g(x)=\int_a^x f(t)\ dt$ is defined for all $x\in\mathbb R$ but $g$ is not continuous? Any type of integral is fine.
I will post a pic from my diff geo notes. Hold on
18:07
@AlessandroCodenotti do you have other exercises like the one you just gave?
Thanks alot @TedShifrin You are THE best! :)
2
Oh, I don't have it in there. Hold on.
Take your time ! :)
that wasn't really an exercise, just an assertion in a thing I'm reading that was regarded as obvious but I wasn't seeing
The ellipse is outside instead of inside, but you can see what I'm talking about.
On the right, the angle to the point on the ellipse is not $t$, but $t$ is the parameter you're using.
18:09
Okay I see why the line y=x there does not mean pi/4 now :D
it is different period kind of
You're parametrizing the ellipse by stretching (in your case vertically, not horizontally) the parametrization of the circle. So angles are distorted.
Very sloppy correction of that exam indeed
@BalarkaSen I meant what book are you learning from, if any
Okay thanks again @TedShifrin :)
Sure, @Kasmir.
Back later ...
18:11
have a nice day @Ted
@AlessandroCodenotti alright, thanks
@DHMO thanks
nice day @TedShifrin
18:28
@PaulP Candel and conlon
18:43
@MikeMiller Objections regarding what? I just had the wrong counterexample
by the way, regarding the problem you gave me, is the codimension 1 foliation supposed to be co-orientable?
It's a nice book but I wish there were more cool pictures
Hi. I'm trying to turn this cubis equation ( x^3-9x^2+(1/2)x-(9/2)=0) into a depressed equation whit this formula(http://www.texpaste.com/n/vcqsur2h) for the depressed equation x^3+px+q
But for some odd reason the two equations doesn't have the same roots?
Yah, I am wondering if there is some software, or software packages, where you could somewhat easily have some visualizations of foliations (at least for surfaces it would be nice)
@DanielGuldbergAaes what are your roots?
18:48
I think I have a decent collection of examples of foliations. What I struggle with is converting a symbolical proof into a picture-proof (the latter is the only way I can make sense of things)
I only need one root btw. and the root or the solution to the equation should be x=9.
@DanielGuldbergAaes and what did you get instead?
@PaulP There aren't many surfaces which admit foliations, right? I guess you mean singular foliations
@DHMO please how to prove that $\forall \varepsilon, ]-\varepsilon, \varepsilon[\cap \{\frac1n, n\mathbb{N}^*\}\neq \emptyset$ ?
18:50
What kind of pictures are you looking for?
@Vrouvrou You might be looking for Archimedean property
I guess I am actually more interested in seeing some concrete visualizations of the dynamics and what is happening to curves @BalarkaSen
Are quotes a good way of showing that you are defining a term? What about using "so-called"?
@PVAL-inactive An integer is said to be even if it is an integer multiple of 2.
18:52
@PVAL You can also italicize it, IMO
Here's the passage
However, if we assume a 4-manifold admits a symplectic form and the boundary is a certain contact manifold (with a natural compatibility condition), classification problems suddenly become tractable. We will show how such classification results of Eliashberg, McDuff and Lisca in the case of lens spaces can be used to put topological restrictions (completely determining the Betti numbers and signature) on these symplectic fillings
I'd like it to be clear that the meaning of "symplectic fillings" corresponds to the hinted at definition in the first sentence.
so-called sounds like I am disapproving of the terminology
@DHMO here is the soulutions and a graph of bot equations owncloud.techknight.eu/index.php/s/iZpekAu1QVqmVWq
and quotes don't really work.
how about: [...] on these, known as symplectic fillings
So if you have a foliation, invarient under a pseudo Anosov, that would correspond to a teichmuller geodesic, and I want to see what is happening to curves as you travel up the geodesic, which is streching the foliation. This could be fun to play around with, so you can see how complex these curves get
18:56
@BalarkaSen Nah. In that case it just fibers over the circle.
@PaulPlummer Well there are these things called measured laminations ya know...
@DanielGuldbergAaes sorry, no idea
@PaulPlummer I think you probably want to look at Danny Calegari's foliations book
@PVAL-inactive wouldn't have figured that to be your thing
@DMHO Honest anwser thoug ;) I just can't grasp that it isn't the right anwser.
@MikeMiller What if the same leaf hits the back the transversal once again?
18:57
I directed the reading seminar here toward it so I could take a passive role.
But they do matter to me I think.
I guess by compactness of the leaves it would fiber over a covering space of circle so that's ok
@BalarkaSen Then it's not co-orientable.
@PVAL-inactive @MikeMiller I was were talking about a software to see what is happening easily
I'm pretty sure people like Wand use hard MCG stuff, and I definitely care about the stuff they do.
and have pictures/video
18:58
Oh, nevermind, I see your point.
The transversal could cross the leaf infinitely many times if you want.
I know Kaloti uses some like Thurston stuff specifically in his thesis.
Can you do that with compact leaves? I think you required that in your question
I should know all this but I don't.
@BalarkaSen Let $\varphi$ be a diffeomorphism of the circle corresponding to (some smoothing near the end of) $x \mapsto x^2$ on $[0,1]$. Then there are no periodic points of this diffeomorphism, and only one fixed point. The induced foliation of the mapping torus has a transversal going through a non-fixed point comes back around to phi(x), then to phi(phi(x)), etc
@Paul Somehow Thurston's proof of classification of PA autos is just seeing how it acts on a generalization of foliations called measured laminations
@Mike Is the passage I wrote clear?
Yes
If you really want to make it more clear you could write something like "on these bounding manifolds, known as symplectic fillings".
But you've already put more effort into this sentence than most authors do their intros.
19:02
its an abstract
for a talk
Then you're totally fine.
@DHMO thank you, please is $\{\frac1n, n\in\mathbb{N}^*\}\cup \{0\}$ is compact in $(\mathbb{R},|.|)$ ?
@Vrouvrou que veut dire "compact"?
for the paper I feel fine talking about terms that are common in the literature. But I don't want to use foreign terminology to generic topologists/geometers when giving a talk directed at them.
@MikeMiller Ah fair. The transversal is free to be noncompact, even if the leaves are not
Thanks.
19:04
uh
The transversal should be compact
At least in the interesting cases I know.
In mathematics, and more specifically in general topology, compactness is a property that generalizes the notion of a subset of Euclidean space being closed (that is, containing all its limit points) and bounded (that is, having all its points lie within some fixed distance of each other). Examples include a closed interval, a rectangle, or a finite set of points. This notion is defined for more general topological spaces than Euclidean space in various ways. One such generalization is that a space is sequentially compact if any infinite sequence of points sampled from the space must frequently...
@PVAL-inactive I am somewhat aware, Thurston introduced train-tracks to help understand these laminations, and somehow it is involved in the classification. But I am thinking about how it could help see what is going on when a pA acts on the curve graph (vertices homotopy classes of curves, edges if there are representative which are disjoint)
(I mention train tracks because I was told I should learn about train tracks soon)
@Vrouvrou is your set closed and bounded?
@PVAL If you look at the mapping torus of a diffeomorphism S^1 --> S^1 and start at a point x which is not a periodic point then the transversal is not compact
is it bounded i think @DHMO
19:06
@Vrouvrou prove that it is
$0<\frac1n\leq 1$
You mean if you foliate the torus by quotients of lines of irrational slope?
There is certainly a compact transversal there.
@Vrouvrou good. then is it closed?
no it must be finite , i think !
No, that's not what I meant. The leaves are not compact
also there's a compact transversal
19:08
@Vrouvrou closed does not mean infinite
The example I gave the leaves are not compact either.
I literally mean foliate by the circles fibers in the mapping torus of a diffeomorphism of S^1 with at least one non-periodic point, like the example Mike gave
In fact each leaf is dense
But that was not my question. I wanted compact leaves but noncompact transversal
Well the standard Reeb foliation should have a non compact transversal
but no compact one.
19:10
Lots of noncompact leaves though
but somehow anything has a noncompact transversal because you have locally trivial charts
GSF
GSF
Could anyone help me with a question regarding algebraic groups and schemes?
I think in lower dimensions all compact leaves probably implies a fiber bundle structure thought that's kind of unclear in higher dimensions.
Actually you can see the irrational line example from a different perspective too. Foliate te torus by the tranvsersals of the irrational leaves instead (which are just the meridians) :)
It's probably just the mapping torus example with diffeomorphism being rotation by irrational angle
(which in particular has no periodic point).
19:13
Right.
OK, I'm happy. Thanks.
@PVAL-inactive Do these make it a lot easier to see what is happening to a curve?
Measured laminations?
I don't know how it works. I just know thats how the original proof of Niel-Thur goes. I am learning it passively now.
@PaulPlummer What's a pseudo Anosov map?
19:17
I think I got something out of reading the last chapter of B&F though.
APTTMCG
I think they talk about Thurston's original proof there.
Yah I just started looking at it since you mentioned it, and it does look like that is basically what train-tracks are for (and they are connected to laminations)
A map is pseudo anosov if it acts on transverse measured foliations by scaling one by $\lambda$ and the other by $\lambda^{-1}$
Wow I forgot how to spell lambda
They are the "interesting" elements of the mapping class group
scaling which?
You have two measured foliations which are transverse, on the surface, and the measures are being scaled
The mapping tori of pseudo-anosov maps end up being hyperbolic
19:25
Oh cute
That is an (appearantly, I don't know the proof) difficult theorem of Thurston (of course)
@DHMO i don't know if $\{\frac1n, n\in\mathbb{N}^*\}\cup \{0\}$ is closed , i just know that $\{0\}$ is closed
I'll probably talk about that theorem
later in this semester
so maybe I'll figure out why its true
@Vrouvrou what are the limit points of the set?
$\{0\}$
19:28
does it contain its limit points?
These days whenever I ask a question, or hear a theorem, the answer to the question "Who proved it?" turns out to be Thurston
@DHMO i don't understand ?
@Vrouvrou does the set contain its limit points?
I thought $0$ is in the set
19:29
ohh yes yes
That is what happens when you start learning about foliations @BalarkaSen and basically anything in lower than 4 dim topology and geometry
then it is closed
is there an other method , to prove that it is closed ?
what method do you want?
you can prove that its complement is open... it's a tedious task trust me
i can calculat the adherence
it is the same set then it is closed
right ?
19:32
yes
@BalarkaSen A PA map is a map which is not isotopic to a periodic map, and does not fix any disjoint union of curves. (this is somehow the point)
Hi @Semi
@PVAL-inactive I think the interesting thing is that all maps with those properties are pA in the way I mention, making the classification theorem more impressive, because theoretically there could have been other types of mapping classes
@Vrouvrou bye
19:37
hmm, I wonder if you'd know a name for the following scenario @PVAL-inactive
@PVAL Got it.
That's a good picture
classification: reducible (fixing some finite set of curves), periodic(finite order), things which aren't those are pA
@PaulPlummer Ya certainly the fact that those maps can be described in a nice way is what makes the theorem meaningful.
@BalarkaSen pA are the things which act hyperbolically on the curve graph
That does make sense.
19:41
Consider a family of complex projective curves such that the generic fiber is smooth, but there is a singular fiber where it is reducible to a set of projective lines in general position.
How did the capacitor timing lab go?
Hasn't happened yet. First one is tomorrow afternoon, and the other the following morning. @skullpetrol
@Semiclassical Lefschetz fibration?
Huh. Is that what a Lefschetz fibration is?
19:43
Is the family parametrized over CP^1?
I don't know the higher dimensional stuff but the 4-dim real stuff, that is just a Lefschetz fibration. Locally you are saying near the singular point of the map the fiber looks like an intersection of complex lines
Hmm. In principle it's parametrized over a larger projective space than that, but I can probably restrict to a CP^1 subspace.
@PVAL-inactive To be clear: When you say 'intersection' do you mean that all the lines intersect at a common point, or just a bunch of lines in general position?
I'm guessing the latter.
ah I guess in the Lefschetz fibration case (the generic case) there's only two.
but that is what you mean, you have two intersecting locally over a given fiber.
19:46
Yeah, for what I'm doing I've got at least four lines at the singular fiber.
I think.
Ya somehow you can do similar arguments to the real case to get (generically) maps which locally look like z_1z_2=\epsilon with epsilon deforming to 0
I should probably look at this: ams.org/notices/200508/what-is.pdf
yah he only cares about the smooth case though
So somehow everything hes talking about is already pretty generic
19:50
Yeah.
You are aware thats my adviser right?\
huh.
nooope, I did not know that.
Ya if you are getting multiple (more than 2) lines the critical points of the map to the parameterization space are going to be higher degree than 2
Hmm, makes sense.
That's true even if the multiple lines don't have a common intersection, I presume?
More than just the obvious double points, I mean.
It shouldn't matter what the arrangement is
19:54
@AliCaglayan Did you get into Cambridge?
Really? That's a bit surprising to me, since it would translate into the singular fiber having singularities that are more than just nodes.
If you have 3 lines in a special position or in a general position the map to the space your parameterizing over is of higher degree.
I don't know and I have to leave.
Do all Steiner Triple Systems of order $\geq 9$ satisfy the parallel line postulate?
20:19
@DHMO thank you very much @DHMO
 
1 hour later…
21:21
Hello everyone ! $:)$
I never saw this chat room is a state of inactivity, something strange is going on ..
0
Q: Do all Steiner Triple Systems of order greater than 7 satisfy the Parallel Line Postulate?

SAWbladeA Steiner Triple System is a set $\mathcal{S}$ of $v \geq 3$ elements together with a set $\mathcal{B}$ of $3$-subsets (triples) of $\mathcal{S}$ such that every $2$-subset of $\mathcal{S}$ occurs in exactly one triple of $\mathcal{B}$. As an example, the following forms a Steiner Triple System o...

21:47
@Mahmoud Everyone seems to be engaging in their recreation. Which, based on this chat, I guess is also math.
Eh, it occasionally gets into actual research stuff. (I'd count what PVAL and I were talking about in that category.)
i should be doing math but i am not
@Fargle I just came from a gaming server, to be honest :P
@Mahmoud I know that feeling. I can't stop playing Overwatch.
@BalarkaSen Right, what's wrong with doing math there days :?
21:49
My sin lately is old GB roms
Luckily I've done all my math homework (some simple induction proofs).
@Fargle My case is more complicated, it's Minecraft ...
I don't have a big grading pile until tomorrow, though.
Oh man, @Mahmoud. I didn't realize people still played that.
Any games in particular got your fancy, @Semi?
Mostly ones I had when I was a lot younger.
Right now it's Final Fantasy Adventure.
21:51
I just started playing online poker again (for play chips) and went from 2.5 buyins to 45 in the span of a weekend.
Which is old even for GB, since it dates back to 1991.
@Fargle It's very unique and precise, if you like details and creative gaming in a free platform, well you're going to love this one.
@Mahmoud Nothing? I'm just lazy
Might download Link's Awakening at some point as well, or another one of the GB Zelda games.
@Mahmoud I played for several years.
21:53
(I'd do SNES ones, but I'm trying to resist going off the deep end.)
@BalarkaSen Yeah .. that also counts.
@Fargle Minecraft ?
Happy Valentine's Day
Indeed, @Mahmoud. I (read: my parents) bought it during alpha.
Happy corporate-sanctioned relationship day :P
Also reading Dante but that's not relevant
21:54
@BalarkaSen Had you been reading Eliot a bit?
@Fargle It's way richer now, (especially during the recent updates).
What's the LaTeX…
So I've heard. I don't know if I need to go down that rabbit hole again, lol.
$\huge\color{Red}{\heartsuit}$
I can't imagine any math being more tedious than reading Dante.
21:55
@Semiclassical I did. Not much anymore.
Mmkay.
Eliot liked Dante, which is evidenced by it showing up in multiple places in his work.
@Fargle What do you mean ?
@PVAL-inactive I found the Comedy to be a pretty engaging read.
In terms of poetry I carry around Tarkovsky's book with me all the time
@Mahmoud I fear my studies will suffer quite a lot if I start playing again.
21:56
@Semiclassical Ahh
Dante did interesting things. He just did it in the most boring way imaginable.
[shrug] Cuique suum
I know exactly one line of untranslated Dante because of Eliot.
@Fargle Believe me, they'll surely do .. Any suggestions for getting outside the hole ?
which I'll now butcher because I know how it's said rather than written:
21:57
I have only read translated so thats all I know
@Mahmoud Spam right-click while looking down.
but I can't imagine calling the comedy engaging.
I'm reading the translated work of course.
Poise escus nel foco chegli afina
Now to see how it's actually written...
"Poi s'ascose nel foco che gli affina"
Not terrible for a guess based on pronunciation

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