Mathematics - 2015-07-03 (page 3 of 4)

2:01 PM

@TedShifrin I picked up your DG notes (which are wonderful, by the way, thank you) and I think there's a missing 's' on the page with Green's theorem: "We will have plenty of occasion to use [...]" It's pretty trivial, but ... I can't help myself.

Soham Chowdhury

@Balarka, sorry, I was away.

$v_p(x+y) \geq\min(v_p(x),v_p(y))$.

Balarka Sen

yes, but that's stronger than what I asked you to prove :)

Soham Chowdhury

it follows from there, since then $|x+y|_p \leq\max(p^{-v_p(x)},p^{-v_p(y)})\leq p^{-v_p(x)}+p^{-v_p(y)} = |x|_p+|y|_p$.

Balarka Sen

sure, true.

Soham Chowdhury

so that's a metric.

Balarka Sen

2:07 PM

right. so we have a new metric space $(\Bbb Z, |\bullet, \bullet|_p)$

Soham Chowdhury

@pjs36 I'm not sure "occasion" can't be used as a "mass noun" or "uncountable noun".

But this is a math chat, I'd better not say "uncountable" that way.

Balarka Sen

so one might say : why should I care about this stupid metric space?

Soham Chowdhury

why, indeed?

Balarka Sen

for that, we need to know the algebraic side of the story.

Soham Chowdhury

because it sounds wacky is good enough for me :P

@BalarkaSen okay.

Balarka Sen

2:09 PM

In algebraic number theory, one thing you always want to do is to "enlarge" your domain of integers so as to get more information about your diophaintine equations.

Soham Chowdhury

like the Christmas theorem uses Gaussian integers?

Balarka Sen

For example, consider the Fermat's equation for $n = 3$. $x^3 + y^3 = z^3$.

Soham Chowdhury

hmm

wait, are you going to tell me a proof?

makes excited noises

Balarka Sen

If you enlarge $\Bbb Z$ to $\Bbb Z[\zeta]$ where $\zeta$ is the $3$-rd root of unity (one of the solution to $\zeta^2 - \zeta + 1 = 0$), $x^3 + y^3$ factors as $(x + y)(x + \zeta y)(x + \zeta^2 y)$

Soham Chowdhury

hey, Sayan :)

@BalarkaSen yes. go on.

Remember me

2:12 PM

@BalarkaSen Is that the p-adic metric

hey @Soham

Balarka Sen

@Soham No, I am not going to tell you the proof. But the general idea is to factor it up into pieces like I wrote down above, and develop a notion of primes in ring of integers of an algebraic number fields (e.g., $\Bbb Z[\zeta]$) and exploit unique factorization.

Soham Chowdhury

oh, all those ideals and stuff.

anyway, go on about p-adics.

Balarka Sen

At least, that's what Kummer did (which worked for a restricted class of primes $p$).

pjs36

@SohamChowdhury You're right, I definitely see phrases like "on occasion", but this seemed different. Ted definitely knows his grammar, I'm sure he'll set me straight if I need set straight :)

Balarka Sen

Anyway, so algebraic number theorists always want to "expand" their domain to get more information.

Soham Chowdhury

2:15 PM

regular primes, yes. I looked up the defs, pretty weird, especially the links with $B_n$s.

@BalarkaSen yes.

@Balarka, still there?

Balarka Sen

Now here comes the (algebraic) definition of $p$-adic numbers : $\mathbf{Z}_p$ is the ring $\{(a_0, a_1, a_2, \cdots) \in \prod \Bbb Z/p^k\Bbb Z : a_{i+1} \mod p^{i} = a_i\}$

I hope I have got the indices right.

Soham Chowdhury

oookay. explain that to me.

Balarka Sen

Right, let me explain that beast.

Consider the equation $x^2 + 1 = 0$ in $\Bbb Z$. This has no solution.

Soham Chowdhury

right.

Balarka Sen

However, this has a solution in, say, $\Bbb Z/5\Bbb Z$. It's $x = 2$

Soham Chowdhury

2:20 PM

oh, sure

Balarka Sen

@Soham Yes, but that's the classical approach. We won't take the route (I'll explain why later)

Anyway, to it has a solution in $\Bbb Z/5\Bbb Z$.

Now we want to find a solution to $x^2 + 1 = 0$ in $\Bbb Z$ that is of the form $2$ modulo $5$

Soham Chowdhury

hmm.

Balarka Sen

So plug in $x = 2 + 5k$, from which you get $25k^2 + 20k + 5 = 0$.

Soham Chowdhury

yeah.

anon

@MikeMiller replied on main

Balarka Sen

2:23 PM

We look for solutions mod $5^2$.

anon

probably it will be much later if I read Farb that I will understand most of the answer

Balarka Sen

We see that $k = 1 \pmod{25}$ does the job.

Soham Chowdhury

@BalarkaSen so are we working in Z or Z/25?

Balarka Sen

Thus, $x = 2 + 5\cdot 1 = 7 \pmod{25}$.

@Soham We were working with Z, then we worked with Z/5Z, then we are in Z/5^2Z

Soham Chowdhury

@BalarkaSen yes.

oh, I think I see where this is going.

Balarka Sen

2:24 PM

So in a sense, we are "localizating" more and more, shortening our domain as we move along.

Soham Chowdhury

localizing?

anon

BTW @Balarka you just need to read the third component (the one with the picture) in my one answer to visualize braid groups, the second component was just addressing the OP's specific hangup and giving a low-tech proof valid for n=2, and the first component was apologizing for the original answer being incorrect.

Soham Chowdhury

how?

Balarka Sen

@Soham We are not adjoining anything to Z, but quotienting Z by it's prime ideals.

Soham Chowdhury

Dunno what prime ideals are :(

@BalarkaSen aren't we actually increasing the size of the domain? Z/25Z is bigger than Z/5Z, for instance.

Balarka Sen

2:26 PM

well, quotienting by normal subgroups, then.

@Soham by cardinality, sure.

Ted Shifrin

@Stan: I consider vectors to have their tails at the origin. Of course, for drawing pictures for vector addition and subtraction, we are allowed to move them ("free vectors").

Balarka Sen

but I am not talking about cardinality :)

Soham Chowdhury

so how are we "shortening our domain"?

deostroll

People, why fractals always concern topology?

Balarka Sen

@SohamChowdhury OK, forget about it, it'll take some while to explain. The point is that instead of adjoining something to Z, we look at Z/p, Z/p^2, etc.

Soham Chowdhury

2:30 PM

Got that.

Balarka Sen

That's the kind of thing you do when you want to prove that a diophantine equation has no solution say. If it has no solution mod p, say, then it has no solution in Z.

anon

(but not conversely)

Soham Chowdhury

got both of you.

Balarka Sen

But, it turns out we can "path-up" these "local" informations to get something larger than Z.

Here goes :

Soham Chowdhury

oh?

Balarka Sen

2:32 PM

$x^2 + 1 = 0$ has solution $x = 2$ mod 5, $x = 7$ mod 25, etc.

Ted Shifrin

hi @anon @Balarka @Soham

Hi @Ted

oh, heya @Fargle

hey, @Ted.

@BalarkaSen the local-global principle can be a bit fragile - e.g. it works for quadratics (i.e. minkowski's theorem) but not cubics (e.g. selmer's counterexample)

Balarka Sen

2:32 PM

So simply write the infinite tuple $(2, 7, \cdots) \in \prod_k \Bbb Z/5^k\Bbb Z$. This is a solution to $x^2 + 1 = 0$ in the "$5$-adic numbers"

anon

hi ted

Soham Chowdhury

$$\bigcup_{x,y\in \text{chat}} \rm{hello}(x,y)$$

whenever anyone enters the chat :P

@BalarkaSen okay.

skill patrol

Hi Professor @TedShifrin, did you get your results from the medical yet?

Balarka Sen

@Soham If you have followed me so far, see if the definition makes sense

17 mins ago, by Balarka Sen

Now here comes the (algebraic) definition of $p$-adic numbers : $\mathbf{Z}_p$ is the ring $\{(a_0, a_1, a_2, \cdots) \in \prod \Bbb Z/p^k\Bbb Z : a_{i+1} \mod p^{i} = a_i\}$

skill patrol

I hope all is good.

Balarka Sen

2:34 PM

(ps : the ring operation is just termwise addition and termwise multiplication)

Ted Shifrin

Thanks for asking, @skull@skill ... So far so good, but waiting for some biopsies.

Soham Chowdhury

@BalarkaSen yup

seems the op is well-defined as well. don't ask me to verify this is a ring :P

Balarka Sen

That's why we care about $p$-adics : it gives us a very simple technique of extending $\Bbb Z$ by looking at it's "localizations" at various ideals, and then taking the "projective limit"

Soham Chowdhury

ah, category theory.

Balarka Sen

So, anyway, how the heck does this relate to the topological thing we talked about before?

Soham Chowdhury

2:36 PM

what about the metric?

@BalarkaSen how?

Ted Shifrin

Well, others would say it's because you can do analysis (metric space type) and avail yourself of such techniques.

Balarka Sen

Right. So it turns out you can give $\Bbb Z_P$ a topology : $\Bbb Z_p \subset \prod \Bbb Z/p^k\Bbb Z$. Give the $\Bbb Z/p^k\Bbb Z$s the discrete topology, and give the product the product topology, and finally give $\Bbb Z_p$ the subspace topology.

This makes $\Bbb Z_p$ into a topological space. Turns out that this is precisely homeomorphic to the metric completion of $(\Bbb Z, |*, *|_p)$

Soham Chowdhury

hmm, like R is the metric completion of Q?

anon

The idea, @Soham, is that given an unknown integer x (positive for simplicity), the bigger a number n that you know x's residue mod n for, the closer you are to knowing what x is. by the chinese remainder theorem, its residue mod n is essentially the same information as its residue mod a bunch of prime powers, so let's just pick one prime. what happens if we specify an integer's residue mod every power of a prime?

well, the sequence of residues (represented by least positive integer representatives, LPR, mod p^k) will eventually be constant, once p^k is bigger than x is. and these residues will be "compatible" in the sense that the residue mod p^k is obtained from the one p^(k+1) in the obvious way. but what happens if you have an infinite collection of residues which is "compatible" in this sense but whose LPRs are not constant?

then you have a sequence of things that are "approximating" a nonexistent thingie "to arbitrary precision," exactly as a cauchy sequence of rationals approximates irrationals to arbitrary precision. just as one can construct the reals by identifying them with (classes of) cauchy sequences, (i.e. pi by 3, 3.1, 3.14, 3.141, 3.1415, ...) we can construct the p-adics by identifying them with these "compatible" sequences of residues mod powers of p.

Soham Chowdhury

interesting. very interesting.

Fargle

2:41 PM

I will say that I just woke up so I'm not taking all this in very well, but this is perhaps the best set of explanations of what the p-adics are, @Balarka and @anon

Soham Chowdhury

@anon yes, it feels like the bigger an $n$ you know the residue mod $p^n$ for, the more you know about the number.

Balarka Sen

bows, @Fargle. But I know very little about them.

@anon is out resident p-adic analyst/algebraist/number theorist

Soham Chowdhury

so if its not an int, the residues don't have to constant out. wow, this seems trippy.

Mike Miller

@anon: I responded. You should feel free to conflate $\text{Homeo}(\mathcal S,n)$ and $\text{Diff}$ for surfaces, if one or the other improves your intuition. I tend to think of the latter because I have more tools when working with it.

anon

of course, residues mod p^k have a base-p expansion with digits in {0,1,...,p-1}, and notice that the residue mod p^(k+1) will have the same base-p expansion but will have a new p^k digit. if we continue on indefinitely, this means we can identify a p-adic number (the sequence of compatible residues) with an infinite power series expansion in p!

Fargle

2:41 PM

I do too. I'd like to find a textbook for it, rather than Wikipedia and webpages.

Soham Chowdhury

so what else will you tell me about these things, @Balarka?

what's cool about the top. space of the p-adics?

Mike Miller

If you can prove an Earle-Eeles theorem for surfaces with marked points, tell me. It shouldn't be too hard, probably just a simple extension of the standard proof of Earle-Eeles.

Balarka Sen

I don't plan to tell any more right now. p-adics are places for all kind of fun.

anon

@MikeMiller there is a part of my brain that is scared about homeos instead of diffeos (things like everywhere-continuous-nowhere-differentiable functions scare me, and exotic differential structures make me think there can be disconnects), but I am good at ignoring my conscience.

Soham Chowdhury

hmm. what kinds?

Ted Shifrin

2:43 PM

Goodnight, @Mike. Imagine my surprise to see you here.

Mike Miller

@anon: I've only read the book up to about chapter 4. There's surely plenty of content later but the thing I'm saying there is just the obvious generalization of the Birman exact sequence.

Soham Chowdhury

what can you prove with p-adics?

Balarka Sen

you can take the fraction field of $\mathbf Z_p$ to get the field $\mathbf Q_p$. people do "local" number theory all over in that field.

skill patrol

Hi @KarimMansour

Soham Chowdhury

decide what font you're gonna use for Z/Q/R, man.

@BalarkaSen what apart from NT?

anon

2:44 PM

@MikeMiller What is Earle-Eeles? Did a cursory google and got a paper reference that I shelved following through on.

Mike Miller

@anon: These groups can be very different for higher-dimensional manifolds. For surfaces they're homotopy equivalent. The answer I gave should carry over almost identically, though I haven't looked to see if so.

@anon: The components of $\text{Homeo}(\Sigma_{g,k})$, $g \geq 2$ or $k \geq 1$, are contractible.

anon

@MikeMiller ah, so when you combined those two facts you were just able to use your knowledge of diff groups :-)

Mike Miller

Right

Balarka Sen

@SohamChowdhury well, they are made for number theoretic purposes, so most of the interest is number theoretic.

Mike Miller

The proof in the paper I linked (ignore the rest of it, just look at appendix B) is elementary except for the disc, where you just use Smale's theorem, which is less elementary but very cool

Balarka Sen

2:45 PM

but I guess people do dynamics with them too

Mike Miller

@TedShifrin I'm responding to a ping.

Soham Chowdhury

@BalarkaSen even then. they seem crazy interesting.

anon

@MikeMiller I assume it suffices for the connected component, i.e. ones isotopic to the identity

Ted Shifrin

Haha @MikeM

Balarka Sen

hello @Ted, @Mike, @anon

Mike Miller

2:46 PM

right

anon

the presence of $k\ge1$ should mean some rigidity (things can be "pulled back" to identity around it), but it's interesting $g$ must be bigger than $1$ as well, which leaves the 1-punctured torus as an interesting case

Balarka Sen

@SohamChowdhury p-adic numbers are interesting.

Soham Chowdhury

very :)

Mike Miller

the "point" of $g \geq 2$ is to restrict to hyperbolic things - I think the punctured torus is hyperbolic, but Ted should correct me

Balarka Sen

oh, ps : p-adics are homeomorphic to cantor set.

Mike Miller

2:47 PM

at least, in my intuition - not so much in the proof

anon

aight

Balarka Sen

not that it's interesting, just gives you a way to visualize them.

Mike Miller

@anon: In general homeo and diff are very different. There are simply connected smooth 4-manifolds for which the kernel of the map $\pi_0 \text{Diff}(M) \to \pi_0 \text{Homeo}(M)$ is infinitely generated

Semiclassical

@BalarkaSen that's actually where the connection to symbolic dynamics i was saying earlier comes in

Soham Chowdhury

@BalarkaSen as if what you'd said wasn't enough, haha

Mike Miller

2:48 PM

For $n \leq 3$-folds the inclusion Diff -> Homeo is a homotopy equivalence - Cerf proved this

Balarka Sen

@Semiclassical i see.

well, I'd have to go now. bye, everyone.

anon

@MikeMiller that's crazy

Fargle

Bye @Balarka

Mike Miller

yeah, it comes from a great paper by Danny Ruberman

it's probably a 4-dimensional phenomenon. you'll have to ask the high-dimensional people if there are finiteness results (ie that kernel is finitely generated) for higher-dimensional $M$

anon

your third comment missed a golden opportunity to say that working smoothly works more smoothly

we aren't friends anymore

Mike Miller

2:53 PM

darn

Karim Mansour

Hi @skillpatrol

how are you doing

Ted Shifrin

See, @MikeM, you and anon are on the same page now.

anon

well, I'm getting there

BTW @Mike I think any choice of nbhd around $n$ points (in $X$) that is homeomorphic to $Y$ induces a group homomorphism $B_n(Y)\to B_n(X)$

Semiclassical

a question re: the algebraic discussion of the p-adics given above

Mike Miller

it should

configuration spaces are functorial

skill patrol

2:58 PM

Fine thanks @KarimMansour how are you?

Semiclassical

namely, that it's related to looking at a sequence of quotients of the integers by normal subgroups

Mike Miller

ping me if you have other questions

Ted Shifrin

hi @Semiclassic

Semiclassical

morning @ted

Karim Mansour

good just about to go pay bills then come and study @skillpatrol

Semiclassical

2:59 PM

does that construction generalize to sets other than $\mathbb{Z}$?

Karim Mansour

hi @TedShifrin @Semiclassical

Ted Shifrin

hi @Karim

Semiclassical

hi @karim

anon

@Semiclassical yes: inverse limits

Semiclassical

ahh

anon

3:00 PM

so, it generalizes to categories that have them

in particular, AbGrp, Grp, Rng, TopGrp, TopRng, Top, R-Mod, etc.

Semiclassical

are there some useful examples of such? (other than p-adics)

anon

one can get extensions of p-adics by doing the same thing with rings of integers of number fields with respect to powers of prime ideals

plus there's absolute galois groups which are inverse limits of galois groups of finite galois extensions

Semiclassical

where i guess i mean 'useful' in the same way as p-adics are useful for diophantine equations as alluded to by balarka above

i suppose i'm looking for an example outside of number theory

anon

the formal power series ring is an inverse limit (K[x] mod powers of x)

Semiclassical

hmm, okay

anon

3:06 PM

the idea (for a linearly ordered inverse system of groups) is that an onto group homomorphism $G\to H$ means that $H$ is a "collapsed" version of $G$, one we get by squinting our eyes and blurring anything in the same coset of $H$ to be blurred into the same pixel. so if $G_0\leftarrow G_1\leftarrow G_2\leftarrow \cdots$ is a sequence of onto maps, the idea is that they are all collapsed versions of some limiting object, the inverse limit $\varprojlim G_i$.

Semiclassical

which i think can actually sort of translate into a physics language: the fock representation of bosonic states is the direct sum of 0-particle states, 1-particle states, 2-particle states, etc.

not saying it's a useful translation, mind :P

anon

I don't quite understand what you said, but be careful: infinite direct sums are direct limits of finite sums, and infinite direct products are inverse limits of finite direct products.

Semiclassical

Fock space

The Fock space is an algebraic construction used in quantum mechanics to construct the quantum states space of a variable or unknown number of identical particles from a single particle Hilbert space H. It is named after V. A. Fock who first introduced it in his paper Konfigurationsraum und zweite Quantelung. Informally, a Fock space is the sum of a set of Hilbert spaces representing zero particle states, one particle states, two particle states, and so on. If the identical particles are bosons, the n-particle states are vectors in a symmetrized tensor product of n single-particle Hilbert spaces...

Karim Mansour

yh

anon

oh yeah, hilbert spaces use different direct sums

not sure if they're inverse limits

Karim Mansour

3:09 PM

i heard about this from my physics prof

Semiclassical

the main reason i made that connection is that formal power series in one variable provide a nice analogy for states of a quantum harmonic oscillator, with a sum $x+x^3$ corresponding to a superposition of a first excited state and a third-excited state

and then adding more oscillators just amounts to adding more variables in the power series

i don't want to force the analogy, though

thinking back to what was in my head when i asked my initial question, i think what i'm really looking for is an example with differential galois theory

i.e. extensions of differential fields

anon

I am not sure how inverse limits could show up there, but I know next to nothing about the field

Soham Chowdhury

@BalarkaSen bye, thanks for all the sweet stuff about p-adics. I'll write a blog post about that.

Semiclassical

i really don't either, but the analogy sounded right

Chris's sis the artist

3:50 PM

I just finished another question in the spirit of Ramanujan (yeah, glad and proud to say that).

skill patrol

Proud?

Chris's sis the artist

@skillpatrol proud of my performance :-)

skill patrol

ok

Remember me

Hello@skillpatrol

skill patrol

Hi pal @Rememberme how are you?

Remember me

4:06 PM

Fine.. How about ya?@skillpatrol

skill patrol

Fine thanks @Rememberme

Chris's sis the artist

@Semiclassical do you see a neat way of calculating this integral without using complex numbers at all? $$\int_0^{\infty } \cos(x) e^{-a x^2} \, dx$$

Semiclassical

well, obviously if you do use complex numbers it's pretty trivial, but without...hmmmm

skill patrol

Why don't you like using complex numbers @Chris'ssistheartist?

Chris's sis the artist

@skillpatrol I love them! I'm just curious about real approaches.

skill patrol

4:14 PM

ok

Semiclassical

i think i'd go at it like so. let $f(a,b)=\int_0^\infty \cos(b x)e^{-a x^2}\,dx$

then, denoting partial derivatives by subscripts, one has $f_{bb}(a,b)=f_a(a,b)$

r9m

@Chris'ssistheartist I'm in shambles .. I can't figure out how to get the $\operatorname{Li}_3\left(\frac{1}{4}\right)$ here

Semiclassical

so $f(a,b)$ must solve a PDE. additionally, one can compute $f(a,0)$ exactly since it's just a gaussian integral.

Chris's sis the artist

@r9m :-))))))))) Think some more, don't give up. :-)

r9m

@Chris'ssistheartist I can't get rid of the $\operatorname{Li}_3\left(\frac{2}{3}\right)$ term :(

Chris's sis the artist

4:19 PM

@r9m Does it appear in Lewin's book? Or in other source (you're aware it)?

r9m

@Chris'ssistheartist not in that form .. :( I have no idea

Semiclassical

oh, and for $b>0$ one has $f(a,b)=f(a/b^2,1)$ via $x\to x/b$ substitution

Chris's sis the artist

@r9m Maybe you might like to ask sos ....

r9m

@Chris'ssistheartist don't do that to me ... I've had enough trashing from my professors this week ..

Semiclassical

probably don't need that $b>0$, actually

Chris's sis the artist

4:20 PM

@Semiclassical or we might use series representation of $\cos(x)$.

Semiclassical

eh, where's the fun in that :P

Chris's sis the artist

:-)

@r9m Really? No vacation there?

Semiclassical

i would like to work out a differential equation route, though

r9m

@Chris'ssistheartist vacation?!! they are not going to give me a degree because I performed poorly in a computer science course .. :(

Chris's sis the artist

@Semiclassical That way should work too.

Semiclassical

4:23 PM

and while it's equivalent, it's probably easier for me to think in terms of the integral $\int_0^\infty \cos(bx)e^{-x^2}\,dx$

Chris's sis the artist

@r9m Ah, sorry. How would you fix that?

skill patrol

Summer school?

r9m

@Chris'ssistheartist no fixes ,,, I have to stay back for another year and repeat the whole course .. (while I hate the guts of that subject ,,, it has nothing to do with pure maths .. )

skill patrol

:O

Must have been an important course.

Chris's sis the artist

@r9m This should have been your last year there?

r9m

4:25 PM

@Chris'ssistheartist yes ,, :((

skill patrol

Which programming language?

r9m

I must have been born with some kind of wretched luck ...

Chris's sis the artist

@r9m No worry, you're a very clever person. You'll do it! It's not the end of the world. I had to give up many universities, but in your case things are far different. :-)

r9m

@skillpatrol mathematical logic course .. :|

skill patrol

-_-

r9m

4:28 PM

@Chris'ssistheartist well I did get an offer to join another nice place meanwhile .. now it's totally ruined

@skillpatrol :P well I'm a dumb student .. can't help it :P

Chris's sis the artist

@r9m Well, believe me, in life a lot of other bad things may happen. I think you're fine, you have all you need to excel in your uni. :-)

@r9m No need to discourage yourself, keep your optimistic mood! :-)

r9m

@Chris'ssistheartist I hope I'll make it out of this place next sem .. it's torturous to me when I'm forced to study stuff that I don't like :((

skill patrol

You're just not interested in it yet pal @r9m give it time.

Chris's sis the artist

@r9m I have exactly the same problem! I simply cannot learn what I do not like.

Ben Dover

@r9m you got owned by the computer science pro

f

skill patrol

4:32 PM

Who are you @BenDover?

r9m

@skillpatrol that's what I told myself ... but I can't see how that course is going to be of any use to me with regards to what I really wanna study ,, (its like an additional burden to me)

Ben Dover

i guess i coud ask you the same question @skillpatrol

skill patrol

Askaway pal :-)

Chris's sis the artist

@r9m I don't know how your professors consider you, but I see in you a little genius, and I'd recommend you to never give up, never lose your trust in you.

You can excel if you really want to.

r9m

@Chris'ssistheartist genius is too much of an adjective for me to handle .. I can't stand in a planet with that kind of gravity :P

skill patrol

4:35 PM

@r9m you need to develop what they call a "well rounded" education by being able to learn anything thrown at you :-)

Chris's sis the artist

@r9m Well, it's up to me the way I wanna see things. ;)

r9m

@BenDover :-)

Chris's sis the artist

Back in 30 min.

r9m

@skillpatrol well I can handle any math course they dish out to me .. I was never interested in comp science (neither was I given much of a choice in that regard)

Antonio Vargas

Hey @Semiclassical

Semiclassical

4:39 PM

hey

skill patrol

But that's not what being an educated person is about. @r9m

r9m

@skillpatrol perhaps what you are saying is true .. but it's almost akin to me as being forced to sit in a history lesson

Soham Chowdhury

ah, r9m. you had Haskell too, I guess?

r9m

@SohamChowdhury ya! well I passed that :P (that was not that bad)

Soham Chowdhury

not bad?

Haskell is love, Haskell is life

r9m

4:43 PM

@SohamChowdhury I liked complexity and that part .. so I was okay with it :)

Balarka Sen

@Semiclassical yep, projective limits of things.

all sorts of categories are closed under those

Semiclassical

kk

Balarka Sen

@Soham you're welcome. hopefully I have been able to give a good enough survey so as to intrigue you about number theory.

skill patrol

@r9m Edward Witten attended the Park School of Baltimore (class of '68), and received his Bachelor of Arts with a major in history and minor in linguistics from Brandeis University in 1971.

Paul Plummer

My mouse stopped working :(

Balarka Sen

4:47 PM

you don't need mouse to chat, @Paul :P

i just got myself glasses. feels weird.

Semiclassical

what i'm curious about is what the applications of that construction would be, in analogy with your point re: diophantine equations

e.g. an application of projective limits outside of number theory

Paul Plummer

Yup, although it is a pain to reply or navigate. Need to figure out how to the the vim-like commands working without a mouse @BalarkaSen

Balarka Sen

@Semiclassical inverse limits? well, inverse limit of things in Grp are called profinite groups, and there's the whole load of theory about those

as i said, inverse limit of things in Top are useful in dynamics.

r9m

@skillpatrol I knew you'd say sth like that .. I meant some subject that probably has very little connection wrt what someone wants to do .. like making a musician attend a marine training camp .. or a jujitsu practitioner attend a kickboxing gym :P

Balarka Sen

@PaulPlummer oh yeah, fair enough.

yikes

Paul Plummer

4:51 PM

Inverse limit of finite groups are profinite

Balarka Sen

right. sorry. finite, discrete groups.

wacky topologies won't give you profinite groups either

Paul Plummer

For some reason I thought that finite groups could only have the discrete topology, maybe that was just $S_n$...

Remember me

Hello@PaulPlummer

anon

any topology on a finite group is discrete modulo a normal subgroup (the connected component of the identity). meaning, the open sets are all unions of cosets of said subgroup. see this MSE question.

skill patrol

Do you think History and Linguistics has anything to do with string theory @r9m?

Paul Plummer

4:54 PM

Hello @Rememberme

Balarka Sen

off-topic : I visualize inverse limits as seeing a chain of things (your inverse system) from far off. The analogy works wonders.

r9m

@skillpatrol perhaps .. I'm simply not knowledgeable enough to see such connections

skill patrol

Have you read his Wikipedia page @r9m?

r9m

@skillpatrol nope .. wait I will ..

Chris's sis the artist

5:09 PM

@r9m I also wonder if all this happened to you because of spending too much time in the area of integrals and series. At least you did what you liked. :-)

r9m

@Chris'ssistheartist perhaps :P but it happened by my choice ,, so I accepted every f***** consequence of that :P

Chris's sis the artist

@r9m :D

skill patrol

We all do what makes us happy.

LeGrandDODOM

5:25 PM

@skillpatrol not sure about that

Paul Plummer

@anon Ah thanks, I must have heard that finite groups did not have Hausdorff topology other than the discrete topology, but forgot the Hausdorff part

skill patrol

Why not? @LeGrandDODOM

LeGrandDODOM

@skillpatrol when you're in school you're merely compelled to study stuff you don't necessarily like

skill patrol

True, that's why most students don't try their hardest, right?

@LeGrandDODOM

Among other reasons :P

Hippalectryon

5:52 PM

Hopefully later on with some luck you get to do what you like :D

skill patrol

That's ^ the spirit

Short term pain for long term gain.

Remember me

6:17 PM

@Semiclassical I have a physics doubt .... Willing to take a look?

Semiclassical

willing to hear it out, sure

Remember me

I have a problem with the definition of mass and matter @Semiclassical

Semiclassical

ok?

Remember me

My book says:
Matter is anything that occupies space and has mass
Definition of mass: The amount of matter present inside a body....

Why are the two definitions interlinked.....?
Now if someone asks me while defining matter define mass.... The I cannot use the term matter again in the definition of mass @Semiclassical

Hippalectryon

That definition of mass seems poorly worded to me

Semiclassical

6:21 PM

that is circular, yes

Hippalectryon

Wiki's definition is much better

Semiclassical

though i wouldn't say i know a good definition of it; it's a largely philosophical point, and one i don't worry about too much

Balarka Sen

6:45 PM

high school physics in india = philosophy

cross out "physics" with "any science in general"

Transcript for

$Mathematics$ Mathematics