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7:00 PM
Can you send me adress on email?:D
 
They disappear just when you need them! I'm wrapping my head around the $I$-adic topology
 
@AlessandroCodenotti Present
 
No, @Kasmir. I appreciate the thought, but ... no. :) If I end up near you in Europe next time we'll get together.
 
I dont know what you would like as present but I can be creative :D
 
LOL @Tobias
 
7:00 PM
YES! FOR SURE :D
@TobiasKildetoft it is not called present?
cadeau ?
 
Hi @Tobias!
 
hmm i was very sure it was called that let me google ._.
 
@Kasmir: Yes, that's correct.
un cadeau = a present
 
Wait are you in europe @Kasmir?
 
7:02 PM
Yes Sweden =p
Born here ._.
Kasmir is an alias btw
 
Oh, I don't why I thought you also were in the US
 
Even I knew better, @Alessandro :P
 
Only Ted Leaky and Mathein know the real name :D
 
Yeah, but Ted is old and doesn't remember names he doesn't use.
I have enough trouble every time I have to email @Fargle :P
 
that is good in this situation because I dotn want to use real identity =p
 
7:03 PM
i feel like there arent even that many of us in the US
 
Anyhow, @Kasmir, welcome back.
 
I think Ted it is a good idea for this 2 weeks am free is to watch your lectures as crash course to get in shape !
 
We've lost some of the youngsters, @EricSilva.
 
@TedShifrin Thanks alot!:D
 
@Kasmir: That way you'll catch up on your sleep :P
 
7:04 PM
haha
 
i'm still a US person, but I don't show up nearly as much now
I'm more in the h-bar lately
 
So @Tobias the idea behind the $I$-adic topology is that $I$ should be small, so $I^2$ should be smaller and so on
 
@AlessandroCodenotti right, just like p-adic
 
@Semiclassical: The exercise set looks like a mixed bag. More like an applied math/engineering course. The Fresnel integral is in there.
 
7:06 PM
@LeakyNun I'm learning about those as a special case so it's not really good to provide intuition
 
one point to note: If you argue that the result of the integral should only depend on $r=\sqrt{x^2+y^2}$, then without loss of generality one can assume $(x,y)=(r,0)$ with $r>0$
 
@Semiclassic: Your true colors are showing :P Any progress on jobs, etc.?
 
@AlessandroCodenotti I see
 
except that's not great notation given the context of the problem, blurgh
 
I'm not sure how to argue that, though, without knowing what I know.
 
7:07 PM
So an open set has a "copy" of $I^n$ around each of its points
 
@TedShifrin not really. I've been a student so long that I can't break myself out of that mindset
 
As one of my best friends loves to say, "Wait 'til you're my age!" :D
 
in particular, whenever I try to look at my resume draft my anxiety flares up bad
(i think the material in tehre is fine, but ugh formatting)
 
Oy ... take a mild anti-anxiety pill. ;)
 
yeah, uh
 
7:08 PM
For a small fee, I'll do it for you. You doing it in TeX?
 
I have. for months
@TedShifrin planning to do it in word, since I"m not aiming for just academic jobs
 
So let's think about $\Bbb Z$ and $I=(p)$ as you were suggesting, this topology is weird, no bounded set is open for example
 
the draft is in word
 
Except I can't stand Word. Controlling its tabs and margins requires godlike powers.
 
but i think it might as well be saved in notepad for how little I"ve formatted it
 
7:09 PM
@Semiclassical what identity have you named again? I can't seem to find it
 
You could do TeX and send them a .pdf anyhow.
 
@AndriiKozytskyi the one I stated actually had a typo in it
 
i assume some places probably have a policy of what document types theyd accept but that's weird
using word actually makes me wanna cry
 
Suppose $\omega_1,...,\omega_d$ are the basis for the left invariant 1-forms on Lie group G. Let $\alpha$ be a path in G, what does the notion $\int_{\alpha(0)}^{\alpha(t)}\textrm{d}\omega$ mean?
 
@AlessandroCodenotti personally i find learning special cases actually good for intuition
 
7:10 PM
oof, two typos
 
@loch That's what I started doing a couple of messages later :P
 
ah
 
i has question
 
@Semiclassic @Andrii: Perhaps the crazy Russian professor intends you to do $(x,y) = (0,0)$ in very easy fashion and then differentiate with respect to $x$ and $y$ under the integral sign? Have you learned that? I haven't tried.
 
Is it a vector valued path in the space of 1-forms?
 
7:12 PM
@quallenjäger: It makes no sense with the $d\omega$.
 
wow, I utterly butchered the identity I quoted
 
@TedShifrin well I can show you how far I've gotten with the old approach if you have time. He didn't seem to have much problem with that way
 
@AlessandroCodenotti I think you might also consider this more-specific example: if p = 2, then 1+2+4+... = -1 in the 2-adic topology/metric
 
arxiv.org/pdf/math/0507536.pdf, On page 3 with Chen-Theorem 1 and Chen-Theorem 2.
 
Well, you're going to have to write things in correct notation, @Andrii. What you were writing before made absolutely no sense.
 
7:13 PM
I don't understand these notions
 
@AlessandroCodenotti basically large powers of 2 are "small"
 
$$\sum_{n=1}^\infty T_n(x)\frac{t^n}{n} = \ln\frac{1}{\sqrt{1-2tx+t^2}}$$
where $T_n(x)$ is the nth Chebyshev polynomial of the first kind.
 
@quallenjäger: It's a complete typo.
 
Note that the above expansion is presumably only valid for $|t|<1$.
 
How should it be?
 
7:14 PM
They were writing $d\text{stuff}$ for Euclidean space and were stooopid.
 
As weird as this topology is, it is still Hausdorff for $\Bbb Z$ and $(p)$. Is the $I$-adic topology on a ring Hausdorff in general? I think there are issues if $I$ is very big
 
They mean the "vector" of $1$-forms $\omega = (\omega_1,\dots,\omega_d)$.
 
you would want the intersection of I^n to be zero @AlessandroCodenotti
 
So the integration gives me a path in the space of 1-forms?
 
What they're doing is looking at the vector of line integrals $\int_{\alpha(0)}^{\alpha(t)} \omega_i$ as functions of $t$.
 
7:15 PM
Yeah, if it isn't I can't separate two points in the intersection
 
No, you get a vector of scalar functions of $t$ once you do the path integrals.
 
@AndriiKozytskyi ahah, the DLMF has it listed here in slightly different form: dlmf.nist.gov/18.12#E9
 
@Semiclassical Thanks a lot!
 
Hmm ._. what should I have as preparation for fourir ?
 
@Andrii: Is this a course for physicists or engineers you're taking?
 
7:16 PM
the course expect us to know line integral green and stoke etc ._.
is that really all ?
 
HOW MANY FINITE NON ABELIAN GROUPS ARE THERE
6
 
To help get the problem into that form, note that the result should presumably be invariant under rotations of (x,y) around the origin
 
@Kasmir: Basic differential equation stuff.
 
@TedShifrin yep, I'm phys first year
 
@geocalc33 at least 5
 
7:16 PM
OK, @Andrii, so you actually will want to learn the stuff I was telling you. It's super important in physics.
 
only five?
wow
 
No, infinitely many
 
at least
 
@TedShifrin thanks! :D ill check your lectures first
 
as such, one can assume without loss of generality that $(x,y)=(\rho,0)$ with $\rho>0$
 
7:17 PM
@TedShifrin you can count on me
 
$\infty \geq 5$, it checks out
 
@Kasmir: I actually have one (at the end of the first semester) explaining Fourier a bit and doing a neat application or two.
LOL, ok, @Andrii ...
 
@TedShifrin PERFECT! thanks alot Ted !:)
 
no, it isn't infinitely many @AlessandroCodenotti
 
wait but how do you know there's ifnitely many
 
7:18 PM
You have to review projections onto subspaces, @Kasmir. If you didn't learn those, watch those lectures first.
 
then $\sqrt{(x-\xi)^2+(y-\eta)^2} = \sqrt{\xi^2-2\xi x+x^2}$
at which point the relation to the stated identity starts to become obvious
however, I think this approach is more of a curiousity than anything
 
@LeakyNun what
 
the morally-correct approach is, as @Ted has emphasized, to use properties of harmonic functions
 
dude $S_{n}$
 
@TedShifrin this 2 weeks will be review weeks, so ill start from the begining and skip those i feel confident with , ill focus on what you said now :D projections onto subspaces and diff equations
 
7:19 PM
Yeah, but I cannot fathom what's expected in this exercise set, @Semiclassic. I don't see anything about integrals of normal derivatives, etc.
 
@AlessandroCodenotti you should know better, as a student of set theory :P
 
@geocalc33 take Z_n for any natural n
 
@famesyasd that's abelian
 
Very abelian.
 
Dihedral groups
 
7:20 PM
sorry I misread
 
@EricSilva yeah, the symmetric group on n elements is the better example
 
How about $S_n$ for all $n\ge 3$?
 
the symmetric groups!
 
how many finite abelian groups oare there
 
i said it up there :(
 
7:20 PM
At least 7 :P
 
Eric, when do you do your geometry lectures?
 
@AlessandroCodenotti they're too big to form a set
 
@LeakyNun A proper class then. Unless you mean up to isomorphism
 
I was planning to do it today but im sick :(
 
Oh, Eric, BTW, the guy accepted my answer on currents, so I guess he was OK with it.
 
7:21 PM
@famesyasd Now you can say Z_n :)
 
there's only 7 finite abeliean groups?
 
i have the immune system of a tiny child
 
Ugh, Eric.
And you were looking forward to this stuff.
 
chronic health problems are actually the worst thing
 
There's at least 7 finite abelian groups. There could be more.
 
7:21 PM
@Semiclassical Well any way that works is useful. Thanks a whole bunch, again
 
Since my cancer, my immune system isn't as good as it once was :(
 
(there totally are more. that's the joke)
 
staying up all night reading up on some minimal surface stuff didnt help :/
but yeah im pretty bummed, i was really looking forward to talking about it
 
No, staying up all night is not good.
So what did they do today without you?
 
i guess nothing
usually tuesday is an office hours day
 
7:22 PM
Sigh.
 
but it's the last week
 
So can you still do some of it tomorrow?
 
we need to find those groups!!!
 
hopefully im better
 
where are they hiding
 
7:23 PM
if im better before i have to travel i will definitely squeeze in the talk
 
@TedShifrin I see, Is is the same the integration $\int f(x_t)dx_t$, where $f: \Bbb R^m \rightarrow L(\Bbb R^m,\Bbb R^n)$ and $x$ a path in $\Bbb R^m$
 
Think about some (family of) groups you know @geocalc33
 
So $f$ is more a less a path in the space of the linear mapping between $\Bbb R^m, \Bbb R^n$.
 
@quallenjäger: Each one is just a path integral as a function of the endpoint, @quallenjäger. At some point we can talk about left-invariant $1$-forms on a Lie group if you want. It's generalizations of $d\theta$ on the circle, for example.
 
something something Haar measure?
 
7:25 PM
That probably doesn't help, Semiclassic.
 
i guess not
 
The number of "isomorphism classes" (this doesn't really work very in ZFC but whatever) of finite (nonabelian) group is much more manageable! @Leaky
 
@LeakyNun We should be asking how much really rather than how many
 
@andrii: Here are some exercises (from my book) on what I was talking about. Probably the notations are all different.
 
actually, come to think of it
 
7:27 PM
@AlessandroCodenotti so do we have more discussions on i-adic lol
 
I'm trying to think of how this integral could show up in a physics problem
hmmmmm
 
@quallenjäger: In local coordinates, each one-form looks like $\sum f_i(x)\,dx_i$. You then put in the parametrization of the path and do a usual line integral.
 
@LeakyNun eh, they're all fun and nice, until people insist on completing them
 
heh
 
@Semiclassical simple harmonic motion
 
7:29 PM
@Semiclassic: You mean his original question? It's the 2-D electric potential due to a ring of charge.
 
What's this analysis in my algebra?
 
@TedShifrin Thanks, I'll get to them soon enough
 
LOL
 
@TedShifrin yeah, that sounds right
 
Or close.
 
7:29 PM
@AlessandroCodenotti think of it as an inverse limit
 
2D in the sense of the system being confined to a plane
 
It goes from $I^n$ is small to inverse limit of $R/I^n$ in two lines and I'm like woah calm down there
 
@TedShifrin I see, so the line integral is determined by the endpoint is because of the Gauss theorem hold for line integral right?
 
so in 3D it'd be the electric potential due to a cylindrical shell of charge
 
@AlessandroCodenotti I really think you should study p-adic before I-adic
a letter makes all the difference
 
7:31 PM
in which case the potential should be constant inside and behave like log(x^2+y^2) outside
 
I think of 10-adic numbers (I know, 10 isn't prime) as normal integers but infinite to the left
just like real numbers are infinite to the right
 
No, @quallenjäger, it depends on the particular path. These $1$-forms are not exact.
 
Like in the p-adic topology on Z what's a nonconverging Cauchy sequence?
 
oh it's quite interesting that question
 
@Semiclassic: Well, I get confused switching dimensions. $1/r$ is harmonic in $\Bbb R^3-\{0\}$ and $\ln(r)$ is harmonic in $\Bbb R^2-\{0\}$.
 
7:33 PM
because any periodic expansion is rational
 
right
 
@AlessandroCodenotti so I'd say 0! + 1! + 2! + 3! + ...
this turns out to be an important example in local class field theory
ok not really local class field theory
 
Ah that's a clever one
Obviously Cauchy (for all p even!)
 
i'm so used to thinking in terms of 3D that interpreting 2D problems as cross-sections of 3D problems is reflexive
 
right
@AlessandroCodenotti actually that's a bad example
 
7:35 PM
And I can intuitively see why it doesn't converge, nice example
 
because I don't think it is known whether it is rational
 
@TedShifrin Ohh! right, my mistake. That's why we need exterior derivatives.
 
@AlessandroCodenotti I have a better example now
 
No, no, the exterior derivative was a typo, remember?
 
@LeakyNun I don't see what shouls be rational or not?
 
7:36 PM
@TedShifrin But stoke's theorem doesn't require the 1-form to be exact
 
I think you can have a sequence that converges to 1/(1-p)
basically 1+p+p^2+...
 
Only if you do a closed path that bounds a surface can you take derivatives and apply Stokes's Theorem.
 
@LeakyNun Sure, I'd like to hear all of the examples
 
Yes, I see.
 
@AlessandroCodenotti we don't know if 0!+1!+2!+... is rational
 
7:36 PM
@quallenjäger: You can only apply Stokes if the two paths form a path that bounds. The topology of the space now becomes relevant unless you're just staying local.
 
How can I see then for a general line integral, the value depends on the end points only?
 
It doesn't. It depends on the path $\alpha$. They said so.
 
Ahh, ok, I misunderstood it.
 
They mean to integrate along the path.
 
Thanks , now it makes sense.
 
7:37 PM
Okey dokey :)
You're finally asking something I know about :P
 
@LeakyNun Yes, but I don't see the issue, it is Cauchy, so it must have a limit in the completion, am I missing something?
 
you aren't missing anything
 
@TedShifrin I missed that.:D
 
the completion is complete
I thought you were asking about a sequence that doesn't converge in Z
 
I mean if I want the line integral depends only on end points, I need stronger assumption for stokes or gauss to apply right?
 
7:38 PM
I see what you mean now
But I'd be really surprised of it had a limit in Z
 
I believe in the afterlife because I want keep doing math after I die
 
You need exact forms for that to hold, @quallenjäger. If they're invariant forms they can't be exact (because they'd have to be derivatives of invariant functions, hence derivative of a constant).
 
@TedShifrin I see, thanks.
 
@AlessandroCodenotti I don't know everything about that sequence lol
I don't know anything*
 
@TedShifrin I need to go through all the hyperbolic geometry again.
 
7:40 PM
Anyway I have to go now, thanks for your exampes! I'll read more about the p-adic numbers and I'll probably have more questions later (or in the following days)
 
Bye, Alessandro!!
 
bye
 
@quallenjäger: Where was hyperbolic geometry?
 
I think you'll see me asking a lot of things that you knows about in near future.
In the next few chapter.
 
Bye everyone
 
7:41 PM
Well, I'm no expert on hyperbolic geometry/topology, but we'll see.
 
They try to develop a path on a hyperboloid using $SO(1,3)$ groups on the origin.
 
They're thinking of the hyperboloid in Lorentz space as the model. That's cool.
 
Basically something like the (Klein??)-geometry.
 
mkay guys :D very happy to talk with you again :D
 
The restriction of the Lorentz metric to that hyperboloid gives a Riemannian metric of constant negative curvature, yes.
 
7:42 PM
kasmir will go sleep for now._.
 
Let me know how you sleep with the videos, Kasmir :)
 
See you tomorrow yall :D
 
So hyperboloid is just described by a symmetry group and the origin
 
haha i just prepared what to watch Ted =p
I need full focus for those lectures =p
mkai Gnight yall :D
 
Right, I have to proof that shortly before:D To find a riemannian structure on that model.
 
7:43 PM
@TedShifrin convert to algebra land :P
 
@quallenjäger: It's invariant under the group. So, yes, it's a homogeneous space (actually symmetric space) with that group.
 
you mean Minkowski Metric doesnt you? I am not familiar with Lorentz metric.
 
Same thing. Signature (1,n) or (n,1).
 
Yes right.
Thanks Ted.
 
My favorite way of doing this stuff uses invariant differential forms, not surprisingly :P
 
7:45 PM
Now I can keep reading:D Need to write a report for the end of august
Like 40 pages
Disguisting
 
OK, have fun :P
 
I will:D definitely
 
8:08 PM
fwiw one can indeed compute the integral from earlier using that expansion
it's not even bad
that doesn't make it the right way to do it, of course :S
 
8:20 PM
Anyone happens to be able to explain the significance of field extensions like from $\Bbb Q_p$ to $\Bbb Q_p(p^{1/p^{n}})$? I have an idea about field extensions e.g. $\Bbb Q(\sqrt{2})$ and so on and I have an idea about p-adic numbers $\Bbb Q_p$.
 
@Rudi_Birnbaum Are you reading about some of Scholze's work?
 
@TobiasKildetoft: :-) Yeah, just trying to get some flavours.
What do I gain when I extend by those specific roots?
 
That I have no idea of
 
maybe its connected with the Frobenius lift ...
map
 
@Rudi_Birnbaum Ahh, right, it is adding inverses of iterates of Frobenius
 
8:31 PM
still not closed under $p$th roots
 
@mercio Right, but the next step is to take a limit
 
and then you have $\lim_{n\to\infty} \bigcup_n \Bbb Q_p(p^{1/p^n})$
:-), exactly
 
yes it's an injective limit in the category of fields
 
I like it
:D
 
I think I have the wrong word somehow though
inductive*
?
I'm having a brainfart lol
 
8:34 PM
\colim
:D
Now what happened "meanwhile" with $\Bbb F_p((t))$, that had the Frobenius
?
OK, it got perfect.
So by those colimits $\Bbb Q_p$ gets a Frobenius, $\Bbb F_p((t))$ gets perfect and we get a tilt from one to the other.
just guessing wildly
 
8:58 PM
@rudi is this in reference to some perfectoid magic?
 
@ÍgjøgnumMeg yeah sure.
hi btw! howdy?
 
@Rudi Seeeeervus, I just finished reattaching the rear derailleur to my bike and now I'm extremely happy :)
hahaha
all good with you?
 
yeah thanks great! Nice to hear, your bike is well again!!
 
Yessss :D Very happy indeed, haven't cycled for about a week now and it's beginning to hurt my soul
lol
 
Have been cycling with an old friend of mine yesterday whom I didn't meet for 10 years. but we were training together for more than 12 years like 5 times a week ...
 
9:01 PM
hahaha nice
how was that?
 
was great, though he stayed lean and top fit ... me rather not so lean, but still can take a tough ride (apprently :-) )
 
How many km? ;)
 
just like 60
1:55 with 700 hm
I was always behind him...
almost
only downhill i was in front
 
:D Nice, 60km is still a reasonable ride for me tbh, the most I did was like 107km and that was the hardest I ever did
lol
 
lol
one better slowly approaches longer ones
 
9:04 PM
although me + bike = ~115kg
 
most important keep riding through the year. roughly about the same for me ..
thats more important than single very long or very hard units
 
the good thing is, shifting that much weight around makes your legs powerful as f*ck
hahaha
yeah :)
 
:-) yeah, with 20+ y of training I have no problems with that rather with finding trousers ...
 
hahaha
Nice
 
oh its actually 25+ (getting really old now)
 
9:07 PM
you've been cycling longer than I've been alive
 
:-) lol
 
what about your inscription plans?
 
i need to adjust the b-screw on the derailleur but I don't have the right screwdriver -_-
For master?
 
which one is it?
yes
 
9:09 PM
It's quite a small one lol
and
 
Imbus?
 
I'm gonna decline the offer to study in the wintersemester and reapply for the sommersemester instead so that I can work for a few months to save some money, and meanwhile I'm gonna apply for as many scholarships as I can so that I can finance the final three semesters of the master
lol
no it's a phillips head
 
Oh yeah sure b screw ...
I'd say a good plan!
 
:D
Right, but it means I have circa 5 months without mathematics
so I'm gonna ask an algebraic geometer at my uni if he can give me smth cool to work on for those 5 months
 
Yeah good idea, or you have some reading schedule for yourself.
 
9:12 PM
yeah I've just been looking at some general things but I haven't really gotten sucked into anything yet
 
Still the plan with Heidelberg?
 
yep :)
 
cool! Any take on the Brexit proceedings?
 
I have (I think) a really good application for some scholarships from the DAAD so I'm going to apply for those and I'm going to do the Goethe-Institut Großes Deutsches Sprachdiplom
´´and
if brexit happens my master will cost like 6000€
which is still far less than a master in England
so I can just take a loan or smth
 
@ÍgjøgnumMeg crazy ...
 
9:14 PM
mhmmmm
in England it's like 9000€ for 1 year tho
 
DAAD is quite nice, I had variuos grants there for travel and conferences ..
@ÍgjøgnumMeg Its not good I think.
 
yeee there is a scholarship called DAAD Stipendium für Graduierte aller wissenschaftlichen Fächer
 
Refereeing can be a bit rigorous at times but once you have it its nice, and they are getting less and less beaurocratic
 
and one called Stipendium für Graduierte aller akademischen Disziplinen
 
9:16 PM
the references I have are extremely strong I think
(without trying to sound arrogant or smth, I don't want that)
 
Thats good afaik at your "career stage" your personal prospects are quite important to them. Less your achievements, obviously ...
That can be good or bad. Sometimes young candidates have already done quite a bit of good work.
 
Yeahhh, I applied to study at the BMS in Berlin in like.. January last year, but the only grades I had to show them were those from my second year of studies, which were poor (due to some significant mental health issues that I had back then)
 
While appearing less like a high potential on paper ...
In such cases of a bit of mismatch its important that you explain your situation in short and concise. That can change you chances dramatically then
 
Can I send you smth and see what you think of it? :)
 
@ÍgjøgnumMeg I won't be able to say anything on the subject, but I could maybe/hopefully give you some advice on writing style, clarity etc. yeah np
 
9:21 PM
@Rudi hang on, I will send in a different room
 
9:33 PM
Does it really a big deal?
it looks like it will be as computationally expensive as trying to divide the number with primes that are smaller than its square root
 
well he can't explain the algorithm in details in a 3 minutes vid
it's a bit more involved and yes it's faster than dividing the number by everything smaller than its square root
for very very large numbers
 
Isa
Can arguments be measured starting in the imaginary axis? I think it's necessary to start an argument in the negative part of the y-axis to find the Riemann-surface of $f(z)=\sqrt{z^2+1}$
 

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