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16:00
Probably need to argue that you can try a mobius transform
And figure out under what the conditions that'll map the unit circle to itself
And how would you do that?
shrug
remember, I got that answer by cheating :P
I guess you could try cross-ratio
possibly
i think you can assume that $z=g(w)$ has $g(w)\sim (w-1/2)$
16:02
i.e. turn the question around and look at the inverse mapping with an explicit zero.
somehow that seems more productive?
but I dunno. Blashcke products are something I know of but don't know enough about
You'd probably need to somehow argue that $\frac{1-w/2}{w-1/2}g(w)$ must be a constant map
along with figure out why that's the thing to look at :/
Suppose $\mathbb{F}^{n}_{q}$ be the n-dimensional metric space over the Galois field GF(q), where $q = p^m$ p is a prime. then Find the sizes of $\mathbb{F}^n$ and $\mathbb{F}^{n}_{q}$? Is the answer for $\mathbb{F}^{n}_{q}$ to be $q^n$
@YOUSEFY vector space, not metric space
and yes
@arctictern yes it is vector space
@Semiclassical that'd be about equivalent to saying (a0+b)/(c0+d)=1/2 implies it's (az+1)/(cz+2) wlog I imagine
probably
I mean, there's two problems. One is to find the answer, the other is to show it's the only answer
16:08
@arctictern if it were metric space, then what would happen?
if it were a metric space, you'd have to have a metric :/
@YOUSEFY the phrase "metric space over a field" doesn't mean anything
@Lozansky Hmm, ths Schwarz-Pick theorem here looks relevant: en.wikipedia.org/wiki/Schwarz_lemma
@Semiclassical I found sth in Khan Academy !
16:11
@arctictern But, another question saying "Find the number of vectors in a sphere of radius r in $\mathbb{F}^n$ and $\mathbb{F}^{n}_{q}$" so I'm assuming there are the same thing
um
What the heck would a sphere of radius $r$ in a field be?
@YOUSEFY they are defining the sphere by an equation (sum of squares of coordinates equals r^2), not a metric.
there is no meaningful metric on a finite vector space
@arctictern Are you saying that definition of sphere is in vector space?
Probably it means it in the sense "Count the number of vectors satisfying $x^2+y^2+z^2<r^2$."
No idea how you'd do that, though.
@Semiclassical Ok....
16:26
I know field theory in physics, not field theory in math :P
@Semiclassical I've studied Abstract Algebra, but still is a very big topic you cannot cover every thing on it. It is important in Quantum Mechanics, Combinatorics, Algebraic Computational Complexity, Information Theory, Coding Theory and other topics but these just my major of study.
yeah. math be huuuge.
@Semiclassical that's a ball, not a sphere, and can't be defined without ordering on the scalar field
@YOUSEFY Huh?
oh, I see your point
in as "contained within" versus "distance less than"
i.e. how many vectors satisfy $x^2+y^2+z^2=r^2$
@Semiclassical Howdy!
16:38
oh hey
long time no hear
yeah indeed, been away for a while
how's things?
fine, fine
TAing an intro physics course this semester. not terrible yet
not too much grading?
yeah, though I've got lab reports to look at
16:40
how about you?
I don't have to teach this semester so I'm free for a while
I'm in Belgium now
hmm, exciting
finished my PhD ayyyy
16:41
ahah
postdoc or something now?
postdoc yeah
I'm at KU Leuven
neat
what're you working on lately?
still trying to learn RH stuff
gotcha
i've moved away from that for the time being
we're doing a working group / reading course through Deift's new book
16:43
oh, neat
have you seen Tom Trogdon's book on numerical RH stuff?
yeah I saw it on arXiv I think
really interesting, don't have the time to read it atm though
hmm, chapters 5-9 of that link look appealing
ch 9 is our goal
16:45
right
not sure if 1 chapter a week is feasible yet but we'll see
I might want to look at chapter 11
just to know wth a determinental point process is
there's a lot of probability stuff going on around me that is still a mystery to me
I never properly learned probability
not surprised
I never really figured out wth free probability theory is
I tried to read that section of Tao's RMT book actually
didn't get it
16:47
I tried to read the part of it in his blog
and yeah, didn't get it :S
lol
my strategy now is just to leave the probability aspect to my peers and focus on the mechanics
the mechanics of the analysis I mean
in terms of actual projects I might be looking at some asymptotics of some eigenfunctions related to electrostatics thingies (I never learned physics either) or some exceptional orthogonal polynomials stuff
16:49
right now I'm trying to do Picard-Fuchs stuff, so not really any asymptotics at the moment
I remember reading through that electrostatics stuff as well, though I forget what I learned :/
@Semiclassical With a view toward number theory? On the wiki page it says it's related to elliptic curves.
with a view towards integrals on Riemann surfaces
we've done stuff with that before
16:51
but what I'm trying to do now seems a lot harder than what I did back then
mainly because the curves are non-hyperelliptic :/
or at least, are not presented in hyperelliptic form
which makes things a good deal hairier
in the hyperelliptic case, you've got stuff like $y^2=P(x)$ with $P(x)$ a polynomial (usually with distinct roots)
uh huh, so your branches are easy and such
in the present case, you get stuff like $(x^2+y^2-1)^2=mx+b$ (to give a case I remember off the top of my head)
in that particular case, the Riemann surface is genus-1 with two pairs of degenerate points
but if I replace $mx+b$ with something cubic in $x$ and $y$, that's generically genus 3
...so a nonhyperelliptic genus-3 curve.
and I want to consider how 1-forms behave on that surface as I vary the parameters (e.g. $m$ and $b$)
it's...pretty bad.
hah
sounds interesting though
alright well I'm gonna go make dinner
cya around :)
16:57
later
17:11
haha, i like what Dylan is doing to the nobel prize committee.
17:23
hey @BalarkaSen
I just want to verify something
this should be $(x - y) \in Y$ right ?
not $z \in Y$
Huh? No. Should be $y \in Y$.
It's infimum over the distance between x and y, y varying through Y.
yes but why does it follow from definition ? $y \in \bar{x}$ means that $x - y \in Y$
?
Sure. You're just reformluating. Write $z = x - y$.
That thing has a typo, is all.
17:32
yeah I understand
17:59
Doing exercises for some courses really does me good.
What are you working on ?
Symplectic geometry.
Well, well, look who the complex geometry cat dragged in!
Hi @TedShifrin.
Hi @Balarka
18:01
Been thinking about that problem of yours for a while.
@Danu And not just for courses! :D
And you'd rather be doing high school homework, @Balarka? :D
Nah, homework's over for now.
Just need to keep up with stuff, which is not a hard thing to do.
Unless you procrastinate like some people ...
@TedShifrin lol
@TedShifrin :'(
The nice thing about courses is that the exercises are semi-guaranteed to correspond to what I know
I did some small exercises today, working out what the HELL Huybrechts' notation for curvature tensors means
it's terrible
Well, that's not an acceptable excuse.
I don't know his notation. What is it?
18:05
It's probably not unclear to you.
But I think this is a travesty:
Procrastination is essential. I can't live without it.
Everything is $\cdot$, then suddenly $\wedge$.
OK, so, I would write the structure equation $\Omega = d\omega - \omega\wedge\omega$, where $\omega$ is the (matrix-valued) connection $1$-form.
$\Omega$ is the (matrix-valued) curvature $2$-form.
I got lucky that I had this lecture on "Riemannian geometry" that did this stuff too
He has clearer notation
For instance, that $A(A(s))$ is wedge in the form part and concatenation in the bundle part is something he should've said (IMO)
The typical Riemannian geometry book/course doesn't use differential forms. They write $R(X,Y)Z = \nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X,Y]}Z$. I much prefer forms. :)
18:08
The lecture I took did both, which was nice :)
Of course, so do I. But that is awful notation.
Hey, Mike agrees with me :D
I don't like what Huybrechts did. First of all, $d^2(s)$ doesn't even make sense.
Right!
And most of his $\cdot$s should be $\otimes$s ...
Not to mention some should be $\wedge s$.
18:10
Yeah!
Terrible.
Hey guys, listen, I discovered a new function, I call it the "Zipper-function". It is defined as $f(x,y)=ln(7x^2+3x-sin(x)-sin(y))$
So I had a bad time with that for 3 hours
It Looks like a zipper
He's being lazy. That needs to be done in a trivialization and then one needs to verify that curvature actually is well-defined, giving a $2$-form that transforms by the adjoint representation (conjugation) of the gauge group.
18:11
Oh, but that is in a trivialization, this much I must give him
@Ted: Mhm, gauge group, not group of changes of frame. :D
for once not algebra in this chat
Oh, you talk about gauge groups, that has something to do with physics.
@0celo7 You must not come here often
@PhysicsGuy, it's actually almost half of a pair of pants. Cute.
18:12
@Danu you're right, I do other things
I'm here essentially every day talking about complex geometry :P
@Danu that's what I meant by algebra.
But we're talking about that right now, smartass.
@MikeM: I am speaking Danu's language. And I don't refer to "the group of changes of frame," just an element as a change of frame. Now go to your room.
Looks like you're defining curvature on vector (?) nuclea
18:13
Anyways, right now I wanna wanna wanna find a symplectomorphism that's not an embedding.
Nuclea = bundles.
But I only know very few examples of immersions that are not embeddings.
And I think they're all *immersions of one-dimensional manifolds into 2-dimensional ones, so no symplectomorphisms.
I had enough diff geometry for today
There are lots of such things, @Danu
Well, can you bootstrap up some dimensions, @Danu?
18:15
Immersions that are not embeddings I mean
@TedShifrin The only ones I know are the ones from G&P, I think.
There are some cool parametrized surfaces in differential geometry that cross themselves.
Even minimal such. But you need to map $2$-dimensional into $4$-dimensional for your question to make sense, I assume.
Also this seems interesting
In mathematics, more precisely in differential geometry, a soldering (or sometimes solder form) of a fiber bundle to a smooth manifold is a manner of attaching the fibres to the manifold in such a way that they can be regarded as tangent. Intuitively, soldering expresses in abstract terms the idea that a manifold may have a point of contact with a certain model Klein geometry at each point. In extrinsic differential geometry, the soldering is simply expressed by the tangency of the model space to the manifold. In intrinsic geometry, other techniques are needed to express it. Soldering was introduced...
@TedShifrin Yeah, I guess. Or equal dimensional ones, I don't know.
You can create infinitely many such examples from the ideas in G&P.
The solder form is coming from the mysterious part of an affine connection.
We were discussing this in here a bit about a month ago.
@Balarka: So what's your current reckoning on tubes?
18:18
@TedShifrin My problem is partly that I don't know so many symplectic manifolds.
@TedShifrin Hmm? Solder form is connection-independent.
I guess I know plenty in dimension two (orientable surfaces).
I don't believe that's correct, @0celo. It's related to the torsion of the connection.
But I'm not going to think about this now
It depends only on the bundle projection.
@TedShifrin: I haven't made much progress, but only just started thinking math a few hours ago. I have to show that the surface normal and the curve normal makes constant angle along a line of curvature and I am convinced I need Codazzi for this. For one, $k_1$ being constant along a line of curvature means $E_v = 0$ by Codazzi. Let's see what that means.
18:20
Oh... What abouuuuuut an immersion $S^2\to \Bbb C\rm P^2$...
Meh, I guess I only know the standard embedding
@TedShifrin The analysis exam was easy, nothing crazy on it.
Would someone like to try a guess at the following question? (this is not homework, I made up the question myself): i.imgur.com/coX6ta1.png
How do I make this an image here?
There is an "upload" button.
I'm trying to find out if this homomorphism is injective, so I think I need its image first.
@Danu You can make an immersion by taking the degree 2 map $\Bbb{CP}^1 \to \Bbb{CP}^1$ and then slightly perturbing it inside $\Bbb{CP}^2$ (the self-intersection set will be a point I believe) :)
Not sure if that helps you though!
18:24
@BalarkaSen No idea how to check if that remains a symplectomorphism
I duck out because I don't know what those are. Though, I believe one can make this immersion a map of algebraic varieties: leaving as a comment in case that helps.
@Balarka: In particular it has to preserve area.
Was that to my $E_v = 0$ thing?
No, it was what symplectomorphism means.
Ah, alright.
18:27
Proceed on your track re $E_v=0$.
Okie-dokie. Thanks.
Symplectic geometry seems like perhaps the most geometric flavor of geometry I've heard about.
I guess I don't know Riemannian well enough.
@TedShifrin what is sympletic geometry ?
@Danu Hmm?
@Danu What's more geometric than lengths and angles?
Manifold together with a non-degenerate $2$-form, Karim.
Yeah, @Danu, it's less geometric to me than everything else.
18:29
I see
@TedShifrin Lol.
Projective geometry, hermitian geometry, Riemannian geometry have more "geometry."
Complex doesn't :D
complex geometry is also cool
Complex geometry is algebra, from what I've seen.
18:29
Absolutely it does. You're just not seeing it.
^ is that my fault, or because of what I'm reading?
That's like saying Riemannian geometry is all algebra because it's built out of tensors. Such comments are stooopid.
You misspelled stupid.
I spelled it the way @Pedro taught me in chat 3 years ago.
It is amazing how much "geometry" comes from the algebraic properties of the Riemann tensor.
Schur's lemma, for instance.
18:32
Hermitian metrics and curvature are highly geometric, @Danu. And now that you're getting to curvature, finally, I should remind you about my observation (exercise you'll ignore) that in the complex world curvature decreases for submanifolds (or subbundles) and increases for quotient bundles. So, in particular, any complex submanifold of $\Bbb C^n$ is non-positively curved.
@TedShifrin Oh, actually this is in Huybrechts.
I'm getting to positive curvature
@Ted I'm just kidding with you.
Well, that's definitely geometry.
Yeah, that's right. But it took 190 pages.
Huh? @MikeM
18:33
Re gauge group.
I don't like Huybrechts's style, @Danu.
Ohhh, re gauge group :P
@TedShifrin That's a pretty cool fact.
That was so long ago and I was in so many other fights, I'd forgotten.
So far, the nicest geometric idea I saw was this general concept of replacing holomorphic functions with holomorphic sections, which then correspond to hypersurfaces.
That was neat.
@Balarka: It gives you upper bounds on curvature of any complex submanifold of $\Bbb CP^n$, of course.
That's just the point that bundles allow you to patch local data to global, @Danu, but, sure, very important.
18:35
@TedShifrin Well, this concept of there being too few holomorphic functions is pretty essentially complex
It's the small things :'(
Of course. Similarly the relationship between having lots of holomorphic sections and positive curvature of the bundle (which you'll get to eventually).
true dat, after all holomorphic functions are defined for complex functions
@TedShifrin Yeah, that's presented as a corollary of the negative curvature thing.
Anyhow, there is plenty of neat topology in symplectic geometry; I'm just not sure how "geometric" it is. But I try to distinguish perhaps too much what "geometry" means.
Perhaps I'm just betraying that I like topology
18:37
Is there someone who wants to take a look at my image from above and tell me if he thinks intuitively wether the homomorphism is injective or not? :)
I certainly think it comes to me easier than geometry
she's are also welcome :)
But I also spent more time on it
@abenthy: I don't look at things that algebraic.
In other news, I'm still at uni and stores have closed for today
No dinner confirmed :\
18:38
Ted Shifrin: Its the same idea from ordinary singular homology, identifying $C_*(A)$ as a subcomplex of $C_*(X)$. Don't you need this to define relative homology?
@Danu: Can't you get a sandwich at a bar?
I like pictures. But I stumble a lot when the pictures are hidden behind some (essential) algebra or analysis.
@TedShifrin I guess I could, but I'd planned to avoid junk food.
Oh, well, go to a better restaurant that's not too expensive, then :)
I still have some rice at home, I can just cook that and... then improvise :P
18:39
Hello @TedShifrin
Let $u(x)$ be harmonic in $\Omega$ and over an open smooth part of $\partial{\Omega}$ it holds that $u=\frac{\partial{u}}{\partial{\mathcal{v}}}=0$, $\mathcal{v}$ is unit external normal in $\partial{\Omega}$.

How can we show that $u(x) \equiv 0$ in $\Omega$ ? Could you give me an idea how we can show this?
@Evinda Use some averaging property?
We've talked about this lots of times, @Evinda. It's the maximum principle (which follows from the average value property you've used lots).
@Ted There are topological and geometric parts of symplectic geometry. Lagrangian submanifolds are still geometry.
I guess I'm not sure why they're geometry, @MikeM, even though they've figured in papers I've written.
Actually, holomorphic curves is probably the better example of geometry.
Lagrangian submanifolds don't have an h-principle but they do have lots of perturbations.
18:41
To me geometry should involve some sort of curvature or connection, but I guess there's a derivative in there somewhere :P
That's what Chern wrote in the preface to Sharpe's book
I guess the merging of symplectic topology and algebraic geometry gets to be geometry. Fair enough.
Sharpe's book has to be the most unreadable book I ever owned. (Next to Hermann Weyl, maybe.)
Yeah?
I did that kind of stuff, and I still found Sharpe almost impossible to read.
I lost interest when he wanted me to do something with Bianchi submodules
18:43
Well, there's nothing wrong with Cartan connections, but I just found his exposition unenlightening. I wonder what Robert Bryant thinks of the book.
Anyhow, long gone.
@Ted To me the difference is between flexibility and rigidity. Anything with a finite-dimensional moduli space is geometry.
Or small automorphism groups.
Interesting interpretation
@MikeM: I can't apply that criterion, as most of my interest in geometry is local, even though things like Chern classes then give you something globally meaningful in, say, the compact case.
But in this case , we have conditions for an open smooth part of $\partial{\Omega}$ and not for $\partial{\Omega}$... :/ @TedShifrin
Yeah, it's hardly perfect. But I think it's ok.
18:49
Oh, I missed that, @Evinda. It can't be an "open" part of the boundary. Continuity says it has to be on a closed subset.
It puts the study of psc metrics in topology but Einstein metrics in geometry. I'm ok with that.
I don't see what it has to be true, honestly, @Evinda. Why can't I insulate part of the boundary and make it 0 temperature, and then let the temperature be higher on the rest of the boundary? That would lead to a nonzero temperature distribution inside.
Oh, no, I'm thinking heat equation, not harmonic. Hmmm ....
I guess steady-state that couldn't happen, as there would be heat flow along the boundary to even it out in the long run. ... I don't know how to do this, offhand.
Presumably you've been taught stuff that I don't know.
And I no longer have my hundreds of books.
good evening
19:02
Hi @Alessandro
Hi, can anyone who has taken the GRE mathematics subject test tell me which books are most helpful for practicing calculus?
@Danu The easiest example I can think of for your question is to pick a symplectic embedding of a higher-genus curve (take hypersurfaces in $\Bbb P^2$, say) and then precompose with a covering map.
@MikeMiller I'll try to think about what that means. Thanks!
19:24
Hey, if anybody here is conversant with bounding procedures for integrals and would look at my question I would be happy, the current answer has a mistake which I think cannot be lifted:
1
Q: Speed of convergence of $\frac1n\int_0^1 f(x/n)dx\to0$?

s.harpLet $f\in L^2(0,1)$, this implies that $$\frac1n\left|\int_0^1 f(x/n) dx\,\right| =\left|\int_0^{1/n} f(x) dx\,\right|=\left|\langle \chi_{[0,1/n]}, f\rangle_{L^2}\right|≤\|\chi_{[0,1/n]}\|_{L^2}\cdot \|f\|_{L_2}=\frac{\|f\|_{L^2}}{\sqrt{n}}$$ My question is whether there is a better bound on th...

 
1 hour later…
20:31
@TedShifrin I made more progress. Think I can solve it soon.
Fiddling with the Christoffel symbols were actually helpful :P
Hi @PVAL
hi
The easiest non-projective compact symplectic manifolds are surface bundles over surfaces (as long as the fiber has genus $\ne =1$ such a thing is always symplectic.)
Hi @PVAL-inactive! Are you talking to me? :P
someone was asking about examples of symplectic manifolds
I think it was you.
I was
Actually, I was asking for symplectomorphisms that fail to be embeddings---but thanks anyways!
You don't have to give me an example
it's an exercise for one of my classes (though it doesn't count for any grades or such)
@PVAL-inactive Out of curiosity, is that true for noncompact cases too?
20:42
Do you mean local symplectomorphism?
@PVAL-inactive No, just plain simplectomorphism. The exercise was to first show they're immersions (which I did), and then to give an example that shows they may fail to be embeddings.
@Danu A symplectomorphism is by definition a diffeomorphism.
Derp. Sorry.
I want something that pulls back the symplectic form to a symplectic form
But just smooth, sorry
Take a covering map
I guess that's right.
20:47
Coming up with a compact one that isn't a covering map is probably NOT possible.
@TedShifrin OK, pretty sure I am done. You were right that some nontrivial work was needed.

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