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22:00
I had a lot of fun when a student asked me how bad the quotient of a manifold under a free but not proper action was.
I never remember these words ...
Proper = not silly.
Properly discontinuous? No.
G x M -> M x M, (g,m) -> (m,gm) is proper.
22:02
There's a free action of R on the torus: move a point along the irrational line it lives on. That's hardly proper.
So everything's proper in the compact setting.
is it more common practice to write a factorised expression like (u+2)(u+2) or merge them to (u+2)^2
Either way is fine, @WDUK, but most would write the latter.
The quotient is an uncountable indiscrere space.
Always? Interesting.
The examples I know are.
22:03
Right. Dense orbits lead to bad topologies. That's how I got in trouble on my topology final exam where I screwed up.
okay thanks
makes sense @Ted, I know have a question in a similar vein but I'll think about it a bit longer by myself this time
In that case, leaves were all asymptotic to a single leaf, but not dense.
Oh, you know I love a good foliation,.
Any one in here have over 200K?
Jeff Atwood on September 1, 2010
Have you ever wondered what happens when you reach 200,000 reputation?
22:06
I guess it's not obvious to me if nonproper free actions always have dense orbits. But I shouldn't think topology now.
Hell no.
Yeah, @MikeM, but you fell into my trap and let me screw up :(
@BalarkaSen Standard bad action of $\Bbb Z$ on $\Bbb R^2 \setminus \{0\}$ has closed orbits but is not proper.
over 200k you get a free gift card which you can use no where
LOL @WDUK
22:07
Damn, Balarka, it's past your bedtime again.
"have you ever wondered when you're over 9000K?"
@TedShifrin What are your favorite papers?
Hmm, I mentioned that Atiyah paper on non-multiplicative signature. Still love that. Chern-Lashof's papers were great. I liked a lot of Lawson's work, and he writes well.
A number of Griffiths's papers. Particularly Variations on a Theorem of Abel.
Some of Bryant's papers I worked through, too.
I don't know Chern-Lashof or Griffiths
Chern-Lashof were generalizing Fary-Milnor to higher dimensions :)
Really beautiful stuff.
22:13
I think you mentioned that to me once.
Topology and Morse theory and geometry all mixed together.
I think I did. But they talk about Lipschitz-Killing curvatures, so important to learn what you're learning now first :P
that sounds... bloody
Maybe I should give it a read.
Don't sidetrack yourself, @MikeM.
Good point. I don't have time for anything ever again.
22:15
Right.
Oh yeah, we also went through some of Gage-Hamilton's curve shortening flow papers in seminar at UGA ages ago. Those were fun.
J talked about that at one of the first grad student talks I saw here.
Ok, I'm now wondering whether every $\sigma$-algebra has a minimal generating family of sets, it feels like this should be true, but I know my intuition is always wrong so...
@Alessandro: Based on what we were just discussing, I vote no.
LOL @ "I know my intuition is always wrong." Easy enough to apply an algorithm. :D
@TedShifrin The surface in ex. 10 sounds like it's some sort of a cone, but with a lot of self intersections.
I was very proud of that problem, @Balarka. I stumbled into it making up a final exam about 4 years ago. Why self-intersections?
Have you ever gone to the seashore? (That's a hint.)
22:19
$uv$ can take the same values again and again.
goes to look again at the problem
Balarka is too sick to have ever gone much anywhere.
@TedShifrin I guess it's sort of a sea-shell.
Right. That's what it is. Cone on what? You're wrong about self-intersections.
@MikeMiller: I did go to a seashore a couple times, but mostly spent time on the hotel whenever my parents go on a trip. I made a decision a year ago that I won't go to family-trips anymore.
22:22
They're all Trump voters, huh?
looks like I was wrong again @Ted mathoverflow.net/questions/168535/…
I don't get the joke.
Should I?
@Alessandro: I'm puzzled that they say it works for countably generated. What if we use the usual $\sigma$-algebra generated by open intervals in $\Bbb R$?
I was thinking about that too, that $\sigma$-algebra should countably generated since every open interval can be written as countable union of open intervals with rational endpoints so it should coincide with the $\sigma$-algebra generated by those
22:26
@TedShifrin Um, a spiral?
Yeah, but I don't see any minimal generators ...
@Balarka: Um, a helix :) Yes.
neither do I
Sorry, right.
@TedShifrin It seems impossible to have a minimal generating set. I don't have a proof though.
It must be the case that it is impossible or else that question would have been closed right?
You talking about Alessandro's query?
22:28
yeah
Maybe it's some gross choice construction?
I'm no good at this kind of math.
Me neither.
Alessandro will explain it to us later :)
If anybody wants to take a look: I came up with conditions and a proof, but have no idea if it's right or even makes any sense:

http://math.stackexchange.com/questions/1968870/necessary-and-sufficient-condition-when-two-cycles-in-s-n-conjugate
I found the paper referenced in the MO question and it seems readable, but half past midnight seems to be more appropriate of a time to go to sleep than starting reading it, I'll see tomorrow if I can understand what's going on
good night for now!
22:33
Night, @Alessandro :)
Salut, @JeSuis.
@TedShifrin J'ai pensé à vous aujourd'hui, on a commencé les sous-variétés (sub-manifold)
Pourquoi à moi? J'suis pas coupable :P
22:36
car je te vois toujours parler de manifold etc :)
Ah oui ... c'est de ma faute, enfin :)
hi @Semiclassic
@Ted: I'm trying to do 12.b.
@TedShifrin but it's interesting
figured out two ways to do what I wanted, not sure if there's a simpler example
22:37
I haven't thought about it at all, @Semiclassic.
I wasn't even sure what you were after.
Sure @Balarka (even though you should be sleeping).
well, one is to do $f(x,y)=(x^2+y^2)^2+b(x^3-3 xy^2)$ for some $b$
I still have no idea what you're really looking for.
the level set tells the story:
22:38
That's a monkey saddle right near the origin.
Then the quartic predominates.
What do you think, does the definition of twisted homology $H_n(X;V) = H_n(C_*(\widetilde{X}) \otimes_{\mathbb{Z}[\pi_1(X)]} V)$ for a left $\mathbb{Z}[\pi_1(X)]$-module $V$ admit some kind of long exact sequence for a pair $(X,A)$?
so there'll be three global minima
@TedShifrin connais-tu des problèmes (théorèmes etc) qui portent sur le calcul différentiel et les équations différentielles ?
I would be worried about that, @abenthy, because $\pi_1(X,A)$ isn't always a group.
22:39
and the second term is symmetric under a 120 degree rotation
This reminds me of one of my all-time favorite questions (which @MikeM will be glad to know was first posed to me by Jerry Kazdan, of Kazdan-Warner). Give a function with a unique critical point, which is a local minimum but not a global minimum.
You told me this one.
so that means that the level curves will overall be symmetric under 120 degree rotations
that's what I was aiming for.
22:40
@JeSuis: Il y en a beaucoup.
I have heard it from you too.
@MikeMiller We can view a $\mathbb{Z}[\pi_1(X,x_0)]$ module $V$ as a $\mathbb{Z}[\pi_1(A,x_0)]$-module via the map $\pi_1(A,x_0) \to \pi_1(X,x_0)$.
not convex @TedShifrin
Sure, that's fine. But what about $H_*(X,A)$? Is that an any-of-those module?
@JeSuis: duh.
22:41
The standard example is IIRC a polynomial.
You worked one out, @Balarka. My example had an exponential in it
Ah, right
you can also do it by imitating the definition of the ovals of cassini: $$f(x,y)=((x-1)^2+y^2)^2((x+\frac12)^2+(y-\frac{\sqrt{3}}2)^2((x+\frac12)^2+(y+\‌​frac{\sqrt{3}}{2})^2$$
@MikeMiller I was thinking one could define $H_n(X,A;V) = H_n(C_*(X,p^{-1}(A)) \otimes_{\mathbb{Z}[\pi_1(X,x_0)]} V)$ where $p \colon \widetilde{X} \to X$ is the universal covering.
22:43
ugh. can't figure out why that last term isn't formatting right
But we would need a right $\mathbb{Z}[\pi_1(X,x_0)]$-module structure on $C_*(p^{-1}(A))$ for this I guess, which I don't see.
You need braces around \sqrt3
That's what I would think. I'm worried about that.
@abenthy Right, deck transformations need not preserve that.
22:44
Have you checked the standard algebraic topology books?
oh well, you see the idea
@TedShifrin beaucoup ? Je ne trouve pas d'exercices mélangeant les deux
i didn't like that example, though, because it's degree 6
and i wanted something of lower degree if possible
Yes, May, Davis-Kirk, Whitehead and Hatcher all use bundles as coefficients system when it comes to saying that twisted homology defines a homology theory with a long exact sequence
Wow, you did your homework.
22:45
@JeSuis: En anglais ça s'appelle dynamical systems. Par exemple, tu pourrais regarder Hirsch-Smale.
would it work when $p^{-1}(A)$ is path-connected?
I still don't see a natural action of $\pi_1$ on that even if you assume path connected.
f(x,y)=x^2(1+y)^3+y^4. for your example ?
@abenthy I think the main obvious assumption is that $\pi_1(A)$ is a normal subgroup of $\pi_1(X)$.
That forces $\pi_1(X,A)$ to be a group.
One can show that the image of $\pi_1(p^{-1}(A)) \to \pi_1(A)$ is the kernel of $\pi_1(A) \to \pi_1(X)$.
22:47
(Of course one needs $\pi_1(A)$ to be path connected for that to be a group, too.)
@TedShifrin Ok thanks, I need to wait to be available on the librairie
So the covering $p^{-1}(A) \to A$ is isomorphic to $\widetilde{A}/(this group) \to A$...
Ça va, JeSuis :)
library**
Why should that extend to an action of $\pi_1(X)$ in a natural way?
22:48
@MikeMiller So you don't think that with that definition of the relative homology group, one gets a long exact sequence?
@BalarkaSen :/ It doesn't, was just something I found out which I thought might be helpful to show the long exact sequence.
I think Mike's idea is more promising here.
The thing is, I found a paper where, for a special kind of subspace A, a long exact sequence is constructed and wanted to generalize it / understand it.
good night all
good night
@abenthy Actually, I think you can probably define it in a dumb way, just as $C_*(X,A;V) = C_*(X;V)/C_*(A;V)$.
Now, it's not clear that this has a module structure anymore.
But it should exist and have a differential and whatnot.
22:54
the problem is that $C_*(A;V)$ isn't a right $\mathbb{Z}[\pi_1(X,x_0)]$-module, right?
Yeah, so we can't get that sort of module structure on the quotient.
what if we require that $p^{-1}(A)$ is invariant under the action of $\pi_1(X,x_0)$?
I dunno. At this point it starts getting complicated enough that I abandon ship.
I sure understand that, thanks for thinking about it :) I now have come to the point that I think this is what is needed.
(just for the record, above I made a mistake, $C_*(A)$ needs to be a $\mathbb{Z}[\pi_1(X,x_0)]$-module, not $C_*(A;V) = C_*(A) \otimes_{\mathbb{Z}[\pi_1(X,x_0)]} V$)
23:07
You mean $C_*(p^{-1}(A))$, presumably. But yeah, that would be enough structure to get the module structure on $C_*(X,A)$.
Hello
Oh yes, thank you!
Do you guys force yourself to read a book that's boring?
you mean a novel or a scientific book?
A math text because I have confronted a book that is hard for me to push myself through
23:09
I would say I did, but every math text got interesting when I kept reading it somehow..
Are there books that you find boring?
now that you mention it, I had to work through numerical analysis things for an exam which I found pretty boring, but did it anyway.
of course, you always find something interesting, even if its just unrelated stuff like thinking about if there is anything categorial or something in it :)
which book are you trying to read?
I have to read this
I'm not experienced in algorithms.
,and the text is a bit wordy.
The topics look interesting to my taste, maybe you can read another book instead?
for example Computational Complexity: A Modern Approach is pretty good for complexity theory.
abenthy , yes the topics are indeed interesting , it's just that the textbook to me is a bit wordy...
Have you read that text I shared you?
23:17
you mean the book "Algorithm Design"? No, but I learnt about almost all of these topics in lectures.
Randomized Algorithms by Motwani is also good
Of course, if you have to prepare for a lectures, you something just have to read the reference they give you or you are doomed
@Ted: Got it. My curves lies on a sphere, so $\gamma \cdot \gamma' = 0$. Differentiate: $\|\gamma'\|^2 + \gamma \cdot \gamma'' = 0$. Assume $\gamma$ is arclength parametrized to get $\gamma'' \cdot \gamma = -1$.
That is, $k \mathbf{N} \cdot \gamma/\|\gamma\| = -1/a$, i.e. $k \mathbf{N} \cdot \mathbf{n} = -1/a$. Left side is exactly $-\Bbb{II}(\gamma', \gamma')$, so by Meusnier $\kappa \cos(\theta) = 1/a$. Since $\theta$ (the lattitude) goes from $0$ to $\pi$, it's positive and less than $1$, hence $\kappa \geq 1/a$.
I'm going to sleep. Bye!
@TedShifrin Of course, the minimum is attained for the lattitudes. Your next question is cool though.
I don't really know. I wonder if $z = x^4 + y^4$, near the origin, gives a counterexample.
Ruled surfaces have negative (or zero, in which case blah) curvature, so I can't cook up one like that either.
Also, the previous surface shouldn't give any counterexamples at all. Scrap that.
23:45
@Balarka: You can get it directly from Meusnier (without differentiating your $\gamma$) if you remember that you know what all normal curvatures are on the sphere.
So you know how a the graph of a cube is just the graph of two squares connected at common vertices? Is there a term for doing that in general to a graph?
Hmm, I have no idea @SAW.
Like given $G$, construct $G'$ such that $|V(G')| = 2|V(G)|$ and $E(G') = E(G) \cup \{u'v' | uv \in E(G) \} \cup \{ uu' \}$?
Yeah, I understand your idea. You double the graph and then put edges between corresponding vertices. But I've never studied graph theory.
Ah, okay.
Because about 3 hours or so ago I became very interested in studying this construction in order to help something else I'm doing along. xD
23:49
Have you tried searching for something like doubling a graph?
ted that is a wonderful idea
2:0) a.m. is not my strongest thinking hour
nor mine :)
wouldn't you know
first result is wolfram mathwolrd
world
Have you ever taken a class where you didn't know the material in advance?
That is, you didn't prepare for the class in advance.
I never did :)
It depends on the individual's learning style, and to some extent on the teaching style of the professors.
@SAW: any luck?
23:54
it goes into literally no detail sans a definition :0
but hey, it exists, and i know the terminology
to the arxiv
@TedShifrin Unfortunately, my professor is a bit clumsy and doesn't set very clear learning goals...
The material is interesting, but she says we have to know all the chapters in this textbook:amazon.com/Algorithm-Design-Jon-Kleinberg/dp/0321295358
@JingWeng: Is there a syllabus for the course that indicates what you're going to do each day or each week?

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