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12:00
mostly things related to noncommutative geometry.
Ohh.. then i wont understand even a shit
$\Bbb F_1$-stuff, etc.
Why so many starred messages above?
@Balarka Correct me if I am wrong...
The closed interval is open if I give $\Bbb{R}$ the discrete topology right?
It has to be Soham
In the discreet topology, EVERY subset is open
It's just the power set.
12:02
Yes just making everything is all right :p@Alec
@Rememberme by "closed interval", if you mean $\{a \leq x \leq b : x \in \Bbb R \}$, then yes.
what Alec said.
Remember a topology is just a thing $(X,\mathcal{J})$ such that the union of any things from $\mathcal{J}$ is in $\mathcal{J}$ and so forth. We just call the things in $\mathcal{J}$ "open sets"
Yes I meant that only
You could call them potatoes if that helps.
huh?
potatoes ?
12:03
They're these sort of brown things that grow underground.
I know that ... but why potatoes
@Balarka How's your sinusitis .. recovering ?
r9m
r9m
@Rememberme I must admit your city has a fantastic weather! :)
@r9m it is really dangerous weather i tell you
I encourage people to give concepts stupid names they know not to be true when they meet a concept with a familiar name, (a,b) is KNOWN to be an open interval, rather than thinking of open-ness (because of the name) it can be useful to think of it as potato in the topology induced by (metric)
r9m
r9m
@Rememberme :O why?
12:05
eesh. i missed the star war.
@Rememberme dunno. only started taking the meds a few days ago.
but probably, yeah
Be careful @r9m you can easily catch up dengue, cold , malaria , typhoid .. I have been suffering with all that
I remember when I was last on meds. I only remember taking them and the date being 9 days after that.
@BalarkaSen Good to hear that
r9m
r9m
@Rememberme I see .. !! I'll be careful then ..
12:06
@r9m this weather .. you will feel it so nice up on the outside but slowly it will start degrading you from inside
Are you feeling thirsty@r9m(asking for a specific reason)
r9m
r9m
@Rememberme my God! that sounds aweful!!
@SohamChowdhury you wasn't the one starring things, then?
r9m
r9m
@Rememberme not really .. but I did devour a lot of sweets and felt thirsty after that :P
@BalarkaSen nope.
Ya @Soham I guess you are the only one who stars our comments :p
12:08
I was on AoPS, acculi.
@Soham so, how's things?
oh, cool.
@r9m Keep drinking water .. loads and loads of it or else the consequences can be dangerous
Have any of you guys seen the notation of $S_p(n)$ = sum of digits when $n$ is written in base $p$ before?
Even if you dont feel thirsty that is
r9m
r9m
12:09
@Rememberme OKAY!!! I'll keep that in mind!!
Again ... the star war has begun
2 mins ago, by Remember me
Are you feeling thirsty@r9m(asking for a specific reason)
The beginning of a horror story worthy of Lovecraft.
@Soham So what have you been thinking about these days?
I feel irritated whenever someone says "maths" for some reason.
algebra. polynomial equations, too, after I did that torus parametrization.
12:10
Oh...
Have any of you guys seen the notation of Sp(n) = sum of digits when n is written in base p before?
@BalarkaSen "There is British English, and then there are mistakes." -- attributed to Queen Elizabeth
2
@AlecTeal subscript pls
@DanielFischer The star-panel is a mess. Can you clean it up?
12:12
I C&Ped
r9m
r9m
@AlecTeal ya .. in an olympiad combinatorics book as far as I can remember (I am not sure though) $S_p(n)$
@Balarka Mods can do stuff like these?
DanielF to the rescue!
12:14
yeah
No comedy today please @Rememberme
I never thought that topology is used in chem
What comedy @Alec
Topology is used in chemistry, economics and a lot of other things
graph theory is the study of 1-dimensional CW-complexes :P
now B will jump on me
<3
define a CW-complex.
12:15
There we go
I knew it.
I am just joking.
well, some formalization of a triangulation.
that's all I know now.
not really
12:17
well, then I don't.
@Soham In the first algebraic topology class in the uni, SB was giving a few examples of topological spaces. He asked "does everyone know what a graph is?" and when everyone said yes he asked us to define what it is. When nobody replied, I tentatively said "um, a 1-dimensional simplicial complex?" :P
hahahaha
and then everyone stares at you
I used to do things like that a lot.
wtf is this 9th grader doing here
"I am a software developer by trade, and while I can't say I've taken great interest in category theory, category theory seems to have taken great interest in my field recently." -- from the reviews of that book :P
Now i think everyone comes and takes classes from Balarka :p
12:20
gets smacked
@Soham The grad students in the alg top class were silly. Didn't recall the defn of disjoint union of top. space. duh.
you've told me this twice
@Rememberme nah. I really know very little.
really?
Thats a very good attitude to have :)
@BalarkaSen The grad students in the class were silly.
12:21
I guess I am just frustrated at the undergrad level in here.
Which institution @Balarka ISI?
Don't worry I wont come there and spy on you :p
not about attitude. look at a random "proper" math book. you'll feel humbled too, @Rem.
I agree.
Everyone knows very little in a wish to learn more
@robjohn I also developed that integral I showed you privately in 4 dimensions, and it looks great.
12:23
I guess the grad students I started to learn math with knows more math than I do now. The course also included singular cohomology, but I dropped out because of schoolwork, so I didn't get to sit on that.
Actually I have under research more such forms, all are pretty amazing.
some guy on AoPS: "before an olympiad, I google some $A^1$ htpy theory pdf. that way, the next day, I think that math is actually possible to do." :P
When did you start learning maths (as in calculus and all) (just out of curiosity) @BalarkaSen
*$\Bbb A^1$ homotopy theory
ah. didn't know.
12:25
don't recall, @Remember.
~6th grade?
no, surely not.
8th grade .. i guess
or 7th
12:25
I started learning "serious" math in 7th or 8th, I think/
I learned calculus just after class 8 started.
But I learned a lot of CS after that, instead of math.
Calculus was a good experience.
Ohh....
Calculus was truly.. and when I started thinking just on calculus i understood how continuity might work in topology...
The great epsilon delta
If someone had given me some kind of "simplified Princeton Companion" or told me a little about what "serious math" was like, I would've started far earlier.
Pain at first but fun at last
12:28
@BalarkaSen my first math-nirvana moment was probably seeing Euler's calculation of $\zeta(2)$. yours?
@SohamChowdhury Math is a pretty cool substitute for video games, right?
@BalarkaSen yes. I left gaming exactly at that time.
end of class 7.
@Soham If my teacher had given me proofs of what she stated I would have fallen in love with maths way earlier :)
@BalarkaSen Yes it is .. though I still play Halo sometimes
Speaking of proofs, where does this formula come from:
http://mathworld.wolfram.com/Riemann-vonMangoldtFormula.html
12:30
B knows analytical NT.
@Balarka You know ANT?
I saw him jabbering about Mertens' thm somewhere years ago.
@SohamChowdhury hmm. I guess it was Bring radicals.
then that led to Gal theory?
@SohamChowdhury nah, only the basics. proof of PNT, that is.
sure, quintic was a good motivation for me to learn Galois theory.
12:31
ah. I used to want to learn that once. still do.
My first such moment was Euler's fornula
and then I came upon King's wonderful book.
what book?
@SohamChowdhury are you able to reproduce any of the proofs to the calculation of $\zeta(2)$? Or maybe you have a proof of yours?
R. B. King "Beyond the quartic equation"
12:31
@Chris'ssistheartist just the simple, nonrigorous one by Euler.
the one about the zeros of $\frac{\sin x}x$.
@Balarka I would like to read that book....
The Basel problem is a problem in mathematical analysis with relevance to number theory, first posed by Pietro Mengoli in 1644 and solved by Leonhard Euler in 1734 and read on 5 December 1735 in The Saint Petersburg Academy of Sciences (Russian: Петербургская Академия наук). Since the problem had withstood the attacks of the leading mathematicians of the day, Euler's solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper On the Number of Primes Less Than a Given...
he factorizes the function in terms of the roots etc.
Even I am fascinated towards quintic but never had the prerequisites for it @Balarka
you need to know some algebra.
12:33
@Chris'ssistheartist don't you like that proof?
Yes I know that is why I am doing algebra ...
@SohamChowdhury Yes, I like. I also have 2 personal, original proofs to the Basel problem. One will appear in my book (at least).
@Balarka We require parts of Altop for ANT right?
not at all
12:35
wtf why would you
unless he means algebraic, in which case perhaps you need Nielsen-Schreier?
Don't know but Wolfram mentions Taniyama shimura under Altop for some reason
Serre: algebraic NT, geometry, topology. Badass guy.
@SohamChowdhury Nielson-Schreier is a petty group theory fact.
why would you need that to study algebraic NT?
12:37
in any case, there are purely algebraic proofs of N-S
I know that.
@Rememberme T-S is arithmetic geometry.
Okay ...
arithmetic geometry is algebraic NT, analytic NT, topology, Galois theory all punched into one.
and arith. top.?
12:38
Wow .. thats one heck of a topic
pretty cool idea, this knots=primes analogy.
arithmetic topology is a vague topic.
Hatcher has a book on it I guess
@Soham I don't think you understand the relation between knots and primes yet.
I wouldn't call something cool if I don't understand it.
12:39
Knots and primes ???
(I don't understand it either, to let you know)
well, there are prime knots. that's all I know.
How on earth do they happen to be related
analogy $\neq$ relation
oh. well, prime knots is a petty analogy.
12:40
anyhoo.
the real analogy lies in realizing $\mathsf{Spec} \, \Bbb Z$ as a 3-manifold and embeddings $\mathsf{Spec} \, \Bbb F_p \to \mathsf{Spec} \, \Bbb Z$ (which looks like embedding of circles in $S^3$)
what did you do today?
as far as I understand, at most.
Prime knots.. what are they in simple language?
yeah, I heard that primes are knots in that manifold or something?
@Rememberme knots that you can $\#$ into other knots afaik. like primes.
12:41
Schubert's theorem ...
google is nice, I know
:P
Yes I never had any idea about all these
I like all of this... but my teacher keeps on saying ...
Do JEE!Do JEE
oh, my math teacher allowed me to not work out all those lol derivatives in calss. :D
12:43
@Soham You recall the defn of $\mathsf{Spec} \, \Bbb Z$? If you don't, you should go back to the transcript where I told you about it.
It's a pretty fun object.
(I don't understand any of it)
prime ideals = open sets?
let me look
no, no. prime ideals are the points.
Now open sets come into picture!!?
@BalarkaSen ok, found it
$\mathsf{Spec} \, \Bbb Z$ what are these
12:45
Jul 12 at 18:18, by Balarka Sen
it's actually more than a topological space : at every point of $\mathsf{Spec} \, R$ (i.e., a prime ideal $\wp$ of $R$), there is an associated field $R/\wp$. so every point has a field attached to it. you can do the same with every open set (localization).
$\mathsf{Spec} \, R$ is defined to be collection of prime ideals of $R$ and the topology on it is defined by setting the closed sets to be collection of prime ideals which contain some certain ideal $I$ of $R$.
You know, Balarka, when we went to camp in Germany last year, I saw a poster written in Fraktur. :P
outside a church.
yep, and that's an interpretation of $\mathsf{Spec} \, R$ as an affine scheme.
affine scheme ... many undefined terms for me :(
@SohamChowdhury oh?
@Rememberme I just defined it.
12:47
yeah, they chose two kids from all the schools in Kol where they teach German.
No i mean stuff like prime ideals and all
oh. you'll know those when you learn ring theory.
Yes...
after doing the torus thing, I again got interested in polynomial equations. I used to be mad about all those. analytical geometry is still fun for me.
like plotting higher analogues of ellipses and stuff.
anyway, in this interpretation $\mathsf{Spec} \, \Bbb Z$ can be depicted as a "line" with each point being an ideal $(p)$ for a prime $p$ in $\Bbb Z$
12:48
(const. sum of distances from > 2 points, etc)
@Balarka I had something to ask you...
@BalarkaSen yeah. I've seen some diagrams. There's some kind of fuzz on 0.
At the very end of the line, there is $(0)$ sitting in it. Note that every ideal of $\Bbb Z$ contains $(0)$, so every neighborhood of $(0)$ intersects $(p)$ for any $p$.
So this topology in non-Hausdorff.
@SohamChowdhury Yeah, the fuzz comes from what I said above.
whoa.
"assume all useful top. spaces are Haus"
:P
the spectrum of a discrete valuation ring.. <<--- what is this
12:49
spec of a kind of ring
you have to understand what Spec is first.
and then what a DVR is.
It is a connected space .. thats the only thing I know about it
and then join the two.
And that to by WIKI
Though I dont know how it is a connected space
Okay bye ... I have to do some topology
12:51
@Soham Here's a depiction of $\mathsf{Spec} \Bbb Z[x]$
I saw the second one by Mumford on that blog.
Note that some points have big doodles around it.
$(x^2 + 1)$ has a huge doodle on it because the integral domain $\Bbb Z[x]/(x^2+1) \cong \Bbb Z[i]$ is attached to it, which is very big.
$(x)$'s doodle is smaller, because the ring $\Bbb Z[x]/(x) \cong \Bbb Z$ is smaller than $\Bbb Z[i]$
So $\mathsf{Spec} \, R$ is not just a topological space, but a topological space with points having various "fatness"
precisely what an affine scheme is.
12:55
cool link, Balarka.
I expect I shall enjoy it more when I know some geometry.
It would be cooler if you guys understood it
yeah. I think Bruyn talks about it somewhere in his blog too.
"Mumford's treasure map"
indeed!
12:57
what did you do today?
schoolwork. :(
you like going to school? @BalarkaSen
did you go to school?
@Rem lol
yes.
I did go to school.
oh, btw, you should also note that the line sticking out of $(x^2 + 1)$ forms something like a 2-sheeted covering space over the line sticking out of $(x)$.
that's something formalized by Grothendieck, I think.
13:02
Sanity check: Finitely generated free modules always have finite rank right?
ok, have fun, I have to go get back to schoolwork.
@Vrouvrou almost everywhere, but not necessarily everywhere.
@SohamChowdhury Continued Fractions can be useful for certain problems, especially for approximations by rationals.
13:17
@robjohn can you help me?
@robjohn plzzzzz*
@SaalHardali with what?
Finitely generated free modules always have finite rank right?
@SaalHardali I have no idea
oh thanks anyways ^^
13:44
it is a vectorial product a consequence of the vectorial sum? or it is an independent characteristic of a vector space?
in others words, every vector space have cross product?
Huy
Huy
14:05
@BalarkaSen: I have two definitions: Firstly, $M,N$ are locally isometric, if for all $p \in M$ exists an open nbhd $U$ of $p$ and a $q \in V$ open in $N$ such that $U$ is isometric to $V$. Secondly, a local isometry between $M$ and $N$ is a map $f:M \to N$ such that for all $p \in M$ and all $X,Y \in T_p M$ we have $$\langle df_p X, df_p X \rangle = \langle X, Y \rangle.$$ I find those definitions rather weird. Are they standard?
@BalarkaSen: Also, just looking at the notion, I'd guess that being locally isometric should be equivalent to the existance of a local isometry. But looking at the definitions, this doesn't seem to be the case. Is this just bad notion or am I misunderstanding something?
14:42
@Soham Engel is a beauty !! Thanks for recommending it..
@Agawa001 I have added some Notes to that answer as well.
 
1 hour later…
16:04
thanks robert
16:18
@Huy: It is not true that these two notions are equivalent. If $M$ and $N$ are compact, a local isometry $M \to N$ has to be a covering map. In particular this means that the relation "there is a local isometry $M \to N$" is not symmetric: take, say, $M$ and $N$ to be genus 2 and genus 3 surfaces, respectively; there is a covering map $\Sigma_3 \to \Sigma_2$ but not the other way around. These are locally isometric.
What's worse I can find you two hyperbolic 3-manifolds $M,N$ that don't have a local isometry $f: M \to N$ or $f: N \to M$; but they're both hyperbolic, hence locally isometric.
Huy
Huy
Thanks, @MikeMiller. So it's really just a bit confusing notation, at least in my opinion.
Well, what else would one call the two?
Huy
Huy
I don't know to be honest.
I think it's fairly clear notation. It is a bit annoying that nobody points out that "locally isometric" does not imply "exists a local isometry", though.
Huy
Huy
@MikeMiller: I usually just assume that implication to be true when reading definitions, like being homeomorphic means there exists a homeomorphism, etc. I'm not sure whether there are further counterexamples though?
16:34
Hi Huy; goodnight, MikeM
Fair enough. The definition of homeomorphic is 'exists a homeomorphism'. If a definition doesn't say ___ means exists a ____, just assume they're not equivalent. (Or try to prove/disprove.)
Huy
Huy
I'll do that in the future.
Hi there, @Ted.
Morning @Ted.
Huy
Huy
17:12
@TedShifrin: As an exercise, I'm supposed to show that any smooth curvature scalar $k(s) > 0$ and torsion scalar $\ell(s)$ determine a curve in $\mathbb{R}^3$ uniquely up to rigid motion. As a hint, I was told to compute $$\frac{d}{ds} \begin{pmatrix} \tau\\N\\B \end{pmatrix}.$$ I found $$\frac{d}{ds} \begin{pmatrix} \tau\\N\\B \end{pmatrix} = \begin{pmatrix} 0&k&0\\-k&0&\ell\\0&-\ell&0 \end{pmatrix} \begin{pmatrix} \tau\\N\\B \end{pmatrix}.$$ I feel like I should be almost done, what's the
last step?
17:23
Can there exist a homeomorphism between a group to some space ... If yes how does one visualize it ?
the group would have to itself be a space in order to even call the map a homeomorphism, in which case the homeomorphism doesn't care about the group structure it only cares about the topological structure and so your question reduces to asking how to visualize a homeomorphism between two spaces.
For eg this one I just saw on MSE:
3
Q: $M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$

ArteomLet's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$? Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{...

then you're talking about a homeomorphism between two spaces
Now how would I think about the group of all matrices @anon
"how do I think about X" is not a precise question. you learn how to think about mathematical objects by knowing features about it that allow you to mentally visualize it (so to speak)
17:28
According to me first the group has to be a topological space...
With a certain topology ...
yes, and your group there is a topological space
it's a subspace of ${\rm M}_{2\times 2}(\Bbb R)$, which is just $\Bbb R^4$ (topologically)
Ahh..
Are there some other examples of groups which don't include matrices?@anon
And we can construct a homeomorphism
To some topological space
(Z,+) doesn't include any matrices. lots of groups don't include any matrices.
if you want to construct a homeomorphism, you need your group to also be a space. so you need to call it a topological group.
That's something new to me..
A topological group...
note that for G to be a topological group, left and right multiplication must be continuous maps
anyway, sure, there are topological groups that aren't matrix groups
17:34
i mean, you could just put a topology on a group that has nothing to do with the group structure
: - )
Then we can think of properties like compactness and path connectedness for these topological groups
@Mike, offhand do you know of any coverings of a (sphere minus 2k open discs) for k>1?
you should be able to get them by considering higher-genus surfaces minus discs
if $\Sigma_{g,n}$ is a genus $g$ surface minus $n$ discs, figure out what its fundamental group is, and therefore what $\Sigma_{0,2k}$ it could possibly cover
then try those
(actually this tells you precisely what the covers of $\Sigma_{0,2k}$ are, since all covers of it are surfaces, and because it has free fundamental group, has covers of all degree...)
if we take A to be the category of finite graphs with valence-preserving surjections, and B the category of closed, compact orientable surfaces with homotopy-classes of covering maps, then I think there is a functor A->B which makes a sphere for every vertex on a graph, then connects the spheres via tubes (after deleting discs and identifying boundaries). which is why I ask.
bah. I just want to visualize one covering.
you have a degree 2 cover $\Sigma_{0,4} \to \Sigma_{0,3}$. maybe try that?
or if you'r eset on even number of boundary curves, try $\Sigma_{1,4} \to \Sigma_{0,4}$
i think that exists. it might be $\Sigma_{0,6} \to \Sigma_{0,4}$. who knows
17:50
heh
i probably couldn't draw any of these without some work
@anon: this is all probably best visualized in terms of branched covers over the sphere
18:21
@Huy it is the same for "locally homeomorphic" and "local homeomorphism"
@anon A sphere minus k open disks is homotopy equivalent to a wedge of k-1 circles. In particular there is a 1:1 correspondence of coverings of a wedge of k-1 circles and a coverings of a sphere minus k open disks. These also correspond to subgroups of the free group in which there is a lot of work trying to understand these (see the end of Hatcher chapter 1 or Stillwell's "...Combinatorial group theory").
Hi @PVAL.
I think the interest is in understanding the covering on the level of surfaces without passing to the wedge of circles.
For example, if you remove 2 disks from $S^2$, the space is homotopy equivalent to $S^1$ and the coverings (up to covering isomorphism) are from the map $z\to z^n$ or $z\ to -z^n$ thinking of the unit disk as the deleted disk.
Huy
Huy
@BenDover: What's the definition of a local homeomorphism?
$f: M \to N$ is a local homeomorphism if for every $p \in M$, there is an open $p \in U \subset M$ and open $f(p) \in V \subset N$ such that $f(U) \subset V$ and such that $f\big|_U: U \to V$ is a homeomorphism.
18:31
@Huy $f : X \rightarrow Y$ is a local homeomorphism if for every $x \in X$, there is an open neighborhood $U$ of $x$ such that $f(U)$ is open in $Y$ and $f|_{U} : U \rightarrow f(U)$ is a homeomorphism.
okay, I believe I can see how S^2 minus k disks is homotopy equiv to a wedge of k-1 circles
Replace 'homeomorphism' with 'isometry' and you get something equiv to your defn of local isometry
@Mike Miller Hi.
more generally $\Sigma_{g,n}$ is homotopy equivalent to the wedge of $2g+n-1$ circles.
What's your story? Grad student?
I'm a grad student at UTexas at Austin. I think if one is careful about the correspondences one can probably find out the genus of any finite index covering from any finite index subgroup of the free group (when one looks at the graph). I don't know how to do this, but I wouldnt be surprised if it was in Stillwell's book.
Actually, I think the cover can be chosen to be holomorphic and the Riemann-Hurwitz formula explicitly calculates the genus.
Amend that to holomorphic or anti-holomorphic
18:46
No, I think it can be holomorphic. If it's antiholomorphic just conjugate the complex structure on the cover, no?
I don't really see how the Hurwitz formula helps. Surely you need to know the ramification data at the puncture points. How do we get that?
Do you know Tye Lidman?
I managed to capture that for you, @PVAL. ^_^
It's a pretty interesting stat!
There exist people who, at some point in time, have been seen 0s ago and have talked never before! OMG
"seen" means at the keyboard
Moving the mouse, typing, etc.
:-)
Sorry, I don't mean to intrude.
waits for it
19:00
The ramification index should nicely correspond to the cover induced on $S^1$ from the cover of graphs. Thinking about each $S^1$ in the wedge as a boundary curve in the surface, and yes I do know Tye.
OK, I see what you mean.
19:36
@skillpatrol Ah, I thought it was some webcam hack where chat.se uses your webcam, if you have it, to see if you are present.
(What were you waiting for, @Karl?)
@KarlKronenfeld nah, nothing so diabolical :-)
19:57
That thought did cross my mind also.

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