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12:00 AM
@TedShifrin Hmm...
FWIW, I asked something similar-ish to my supervisor and he also didn't talk about pullback/inverse image sheaves
 
Actually, she clucked "méchants garçons" :P
 
he tried to go through the vector bundle map
 
Let me check Huybrecht's definition when I get a chance. I'm still fiddling with getting my new computer working.
 
The possibly relevant sheaf-theoretic definitions are literally the very last bit of the last appendix.
 
I'll do that for tomorrow.
 
12:04 AM
No haste :)
What's best is that I'm finally done with chapter 2 of Huybrechts
Major achievement!
 
Moving on ... :)
OK, have a good weekend :)
 
@TedShifrin I think I'll try to have a look at one of the tpossible topics...
 
OK :)
 
Sent Huybrechts my typo compilation for chapters 1-2
About 30 mistakes
Let's see if he appreciates it or not... :P
 
12:21 AM
Maybe not.
 
@MikeMiller Yeah, I can imagine him just being like "yeah, just mind your own business bro"
@MikeMiller Hmm this way of proving things by resolutions is really nice once you know the method
So powerful
What is the "Frobenius integrability principle" supposed to be?
I know this theorem giving a criterion for integrability of a distribution---does it go further than that?
 
futilely trying to come up with a method to solve $a_{n} = f(a_{n-1})$ atm ...
 
12:37 AM
can be hard if $f$ is complicado
 
am assuming it is differentiable, though not much help
i guess it's really just $f$ iterated $n$ times onto $a_{0}$ now that I think about it
$f(f(...f(a_{0})...))$ ... xD
I suppose a more interesting application would be $a_{n} = f(a_{n-1}) +g(a_{n-2})$ and so on in general. xD
 
1:01 AM
Hello
 
@Danu It gives an incredibly simple criterion for that.
 
1:26 AM
my question from earlier re: $(y^2-1)^2=x^4-1$ was actually not quite what I wanted, it turns out
what I really wanted to figure out is how one actually thinks about the branch-cut structure in that case
and it's kind've confusing :/
 
1:55 AM
Gromov boundary of free group?
Does anyone have idea about this ?
 
@Semiclassical I have a question about WKB if you're around
 
@Semiclassical Sakurai p. 112
Sakurai's way of doing WKB is different than usual
usually I would expand the exponent as a power series in $\hbar$
then compare terms in the SE
I'm worried about a sign in 2.5.45
namely the sign in front of the $\frac{i}{2}\hbar\cdots$ goes away in the last line
 
i'm loading a pdf copy of Sakurai, but it's being slow so I can't immediately comment
 
Well you could save it so it wouldn't take so long :)
And I'm busy with math, so I can wait
 
2:02 AM
fair
Deriving WKB stuff tends to be a pain, at least in the standard matching of asymptotics
 
@Semiclassical Sakurai's not doing that
He's doing...something.
 
2:55 AM
sakurai finally loaded, but i'm perplexed: on page 112 etc. he's doing path integrals, not WKB.
 
3:07 AM
@Semiclassical
I have the second ed
 
3:26 AM
@Danu Thanks. It's in Hatcher too, apparently.
@Mambo I know what the Gromov boundary is, and what it is for the free group. What do you want to know?
 
@0celo7 isn't that just $(\pm)(\pm)=1?$
 
The two signs are not the same
 
you take square roots twice
once in the 0th order term, and once in the 1st order
 
right.
 
3:39 AM
I think they're independent
 
I don't think that can be quite right, since if so the sign would not be well-defined.
 
hmm?
 
well, i mean, if there's just one sign choice involved, you can argue that it's just a matter of whether $x$ is increasing or decreasing
but if there's two choices, that'd be four.
and that doesn't seem right.
my guess is that the issue is settled by carefully considering what is intended by the square root.
 
what do you mean?
 
that's a bit vague, isn't it
hmm
i guess I mean that, while one is accustomed to saying "there's two sign choices in taking a square root", that doesn't necessarily mean they're both valid.
 
3:45 AM
what would make one invalid?
 
therein lies the issue i'm carefully avoiding giving an actual answer to :P
namely because i don't see it off the top of my head
one other bit about that presentation, though it's a fairly typical abuse
$\ln[k(x)]$ isn't dimensionally valid.
 
Ok, what should one write?
Introduce some constant?
 
yeah. i mean, that just amounts to some overall constant in the wavefunction.
so it's not actually a big deal.
 
ah, ok
 
one observation: it makes a lot more sense to me for the wavefunction to have $\frac{1}{\sqrt{k(x)}}$ as prefactor than $\sqrt{k(x)}$
simply because $k(x)\to 0$ at a turning point.
so the former implies that this solution must break down in the vicinity of a turning point, whereas the latter wouldn't
and that's indeed the case.
but, to the matter at hand...hmm
one can write equation 2.5.44 as $$W_1'(x)^2 = W_0'(x)^2+i \hbar W_0''(x)=W_0'(x)^2\left(1+i\hbar \frac{W_0''(x)}{W_0'(x)^2}\right)\approx W_0'(x)^2\left(1+\frac{1}{2}i\hbar \frac{W_0''(x)}{W_0'(x)^2}\right)^2$$
hence $W_1'(x)^2\approx (W_0'(x)+\frac{1}{2}i\hbar W_0''(x))^2$.
 
4:00 AM
yeah
 
i feel like that's a better form to work with, but it doesn't quite resolve the issue.
meh. i'm not seeing it right now.
let me glance at my copy of shankar's qm to see if it's handled in there
 
differently (better)
 
hopefully.
 
not really
 
hmm. i've found their equivalent to 2.5.46, but not the derivation leading up to it. nm, found it
 
4:04 AM
I understand Shankar
I don't understand Sakurai
it's a completely different approach
 
one thing to note is that, in general, getting prefactor right is a pain
hrm. this is weird. both shankar and sakurai cover much the same ground, but snankar's treatment explicitly goes into the perturbative expansion
but the ODEs they get for what amounts to $W_1(x)$ are not the same
 
@BalarkaSen You there?
@Semiclassical Are they at least similar?
 
sakurai's is $W_1'(x)^2 = W_0'^2(x)+i\hbar W_0''(x)$, whereas shankar's would be $i\hbar W_0''(x)=2i W_1'(x) W_0'(x)$
 
Oh, that's right
They define $W_1$ differently.
$W_1$ in Sakurai is the 0th and 1st order terms together
Shankar keeps them separate
 
ew.
that's kind've terrible.
 
4:14 AM
why?
 
lumping both together. i much prefer shankar since then one can do things by matching coefficients
 
@0celo7 Yes, but I have to go in a minute. Ask what you have to ask quickly.
 
in fairness, though, shankar's got a typo in the ODE i just cited
 
@BalarkaSen Locally Euclidean, Hausdorff, $\sigma$-compact implies second countable?
All I've been able to show is that the space is a countable union of sets homeomorphic to balls.
 
what's sigma-compactness
 
4:15 AM
should be $i\hbar W_0''(x)=2 W_1'(x) W_0'(x)-W_0'(x)^2$, where i've accounted for the difference in definition as well
 
space is a countable union of compact sets.
 
meh
 
comparing that to sakurai seems to then amount to whether $2W_1'(x)W_0'(x)$ is comparable to $W_1'(x)^2$.
which it would be if it weren't for that factor of two. hrm.
 
comparable?
 
dunno off the top of my head.
 
4:18 AM
I found an MSE post on the topic, which I'm pretty sure is wrong without more work.
 
i'm just pretending that i can equate the two formulations and see what that'd imply
 
Countable union of countable sets can be uncountable.
 
but i did a typo anyways. should've been $i\hbar W_0''(x)=2 W_1'(x) W_0'(x)-2W_0'(x)^2$ for shankar a la sakurai
 
@0celo7 eh?
 
don't tell me that's wrong
 
4:19 AM
It is
 
well, crap
Ok, then I'm done
the proof isn't hard
I just have to review some stuff about countable sets...
 
comparing that with sakurai's $i\hbar W_0''(x) = W_1'(x)^2-W_0'(x)^2$ actually seems fairly plausible then
 
Ok, google has showed me the proof
 
since the two being identical would amount to $W_1'(x)^2-2 W_1'(x)W_0'(x)+W_0'(x)^2=(W_1'(x)-W_0'(x))^2=0$
which amounts to claiming that the correction to the first derivative is small.
so while they're not the same, they're not as far apart as they might seem.
that said, shankar's doesn't require square roots whereas sakurai's does. hence, shankar is better :P
 
@Semiclassical Well since their answer for the wave function in the end is the same, I'm sure their equations are close
 
4:23 AM
yeah.
 
but I don't think you've explained the signs
 
yeah. my point is more that sakurai's presentation is really quite similar to shankar's, but shankar's approach avoids the sign issue entirely whereas sakurai creates one and doesn't explain it.
so if you trust Shankar, then you don't have to worry about it in the first place.
I don't find that terribly satisfying, but i'm also a bit tired of thinking about it :/
 
My prof emailed me and said I should come to his office if I'm not convinced by Tuesday
 
good call.
 
He doesn't really understand it either
(so why did he use this approach in class???)
 
4:28 AM
because that's the book he chose to use, and he preferred not to deviate from the text?
 
I guess, but how can anyone like that approach to WKB
 
it does seem a bit sloppy.
i'll confess, though, that I find WKB derivation to be tiresome in general. i really like the results they let you obtain---tunneling, quantization, etc.
but the derivations just kind've suck.
 
5:10 AM
hello everyone
 
5:41 AM
so I'm bout to start the Byzantine task of writing out my CV that too in Latex
 
 
2 hours later…
7:41 AM
hows it going
 
8:38 AM
just completed i
*it
and huge thanks to latex stackexchange
 
@BalarkaSen What is?
 
9:24 AM
@Danu Cell structure of Grassmannians.
 
what do you think the best way to convince someone the earth i is a sphere? assuming they have knowledge of math but no physics other than what they can directly observe
ground up cant trust newton, cant trust anything type style-
lets say they believe the earth is flat and is the stereo-graphic projection of the sphere such that the arctic surrounds the outside
 
9:44 AM
Ah, okay
@MikeMiller Huybrechts replied early this morning---he was very happy :D
> This is an impressive list of typos (also a little depressing).
hehe
 
good morning
 
@Danu Cool
Alright, gonna study some more vector bundles.
 
10:06 AM
Solve 1
Method 1: The standard method
Method 2: The way too overthinking method

Both methods give the same answer, however please stick to method 1
 
10:29 AM
@BalarkaSen I'm having a brainfart: Why is the zero set of a transversal section of a codimension one subbundle of $TM$ a one-dimensional manifold?
The dimension of the image of the section is... $\operatorname{dim}M$?
 
@Danu A global section is literally an embedding of $M$ in $TM$, so it has dimension $\dim M$, yes.
 
Yeah, right? Ok. So why is the set of zeros of dimension one, and not zero?
Wouldn't that be intersecting two transversal sections, both with codim =dim M?
 
You're doing the intersecting inside the subbundle.
 
@BalarkaSen So codim = dim M-1
Ah, okay
 
10:42 AM
So we have a map $M\to E$ where dim E=2dim M-1, and we're looking at something transveral to a codim= dim M-1 submanifold
The preimage has the same codimension, hence dimension 1
love it
adding numbers
I should be a number theorist
I like that I can actually understand the first paragraphs of this paper :D
 
doing arithmetic is the hard stuff
which paper, @Danu?
 
The one for the first topic
 
I would've expected it to get crazy out of hand immediately
I understand the introduction pretty ok.
I think all the hard work comes from this paper:
 
good to hear
 
10:46 AM
index theorems how do I
 
don't know crap about them
 
me neither
It is looking like my seminar on Char. Classes will be essential
in understanding the topics
(if I go for option 1 or 2---3 is a wholly different beast)
 
I think I'm going to study it from Milnor-Stasheff after I finish this chapter in Hatcher.
 
which chapter are you doing? 4? 5?
 
I'm working on Hatcher's VBKT, not his AT book.
 
10:49 AM
ah ok
K theory
another one of these epic-sounding things
 
just vector bundles for now :) I attended a seminar where a grad student gave an exposition on the Bott periodicity from VBKT. I didn't find it exciting.
it seemed like a technical theorem. I think it admits a homotopy-theoretic translation which does sound interesting however
but Hatcher's exposition doesn't go through that line
 
what properties of a sphere are different from a plane in which the top and bottom "borders" are connected and the right and left "borders" are connected
 
do you mean the torus?
 
i suppose i do, thank you
 
that's what you get when you paste the upper and lower edges, and then the left and right edges of a square
 
10:58 AM
but perhaps not, for example the plain has no inherit curvature at any point
 
Then I don't understand what you said.
 
Need some help in understand the proof here:
http://dbfin.com/topology/munkres/chapter-1/section-3-relations/problem-13-solution/
> Besides, for any other lower bound x′ of S , x′∈L , therefore, x′≤x .
@danu But aren't we already showed that $x\leq x'$ since x is the lowest upper bound of $L$ (which is also a lower bound of $S$), thus how can $x' \leq x$ ?
 
@shaihorowitz Yeah, and the torus doesn't either!
 
@Danu careful with that
flat torus doesn't.
 
Intrinsically
 
11:00 AM
flat torus
 
@Danu fair enough
 
thanks guys i now have something i can google to further clarify what i'm thinking about
 
assuming that's what you mean, the sphere and the torus have a whole world of differences. for a start - the former is simply connected, the latter isn't.
 
you couldn't tell that locally though i don't think, which i guess is also a property i'm looking for
 
locally any two equidimensional manifolds are topologically the same
so that's not very interesting
 
11:04 AM
true the question in which this arises isnt to interesting for essentially that reason
 
But then you can think about curvature and then the sphere is different from the torus
I'm out for now---cya guys!
 
I am out of ink and I can't find a new pen. Darn it.
bye
 
the original question was how to prove with math and no physics that the earth is not flat, the easiest of course is to measure a bunch of shadows at the same time... but any finite number seems to be also represenntable by a displaced sun on a flat earth.
 
that's not really a math question, so hard to answer without any physics :)
 
truth
also physics says the earth is flat
as long as you approach near c :P
 
11:17 AM
@BalarkaSen , are you available to help me understand a proof about the least upper bound property?
 
I'm busy right now, sorry
 
@secret i'll do my best
 
ok
http://dbfin.com/topology/munkres/chapter-1/section-3-relations/problem-13-solution/
> Besides, for any other lower bound x′ of S , x′∈L , therefore, x′≤x .
@shaihorowitz But aren't we already showed that $x\leq x'$ since x is the lowest upper bound of $L$ (which is also a lower bound of $S$), thus how can $x' \leq x$ ?
 
if $x' \leq x$ but x' is a lower bound then x' is in L (by definition) and lower than the lowest upper bound hence contradiction
 
 
1 hour later…
12:47 PM
@Secret O_O Very interesting pictures.
 
That's basically the 4D to 3D to 2D projection of that equivalence relation treated as a hypersolid in 4 variables
 
It just looks cool IMO
 
They are indeed quite pretty
However be aware that projections can be misleading sometimes especially if the object is highly asymmetric
 
Mhm, I see. But that's ok, since I can't understand what your talking about anyways.
Say, there's some interesting algorithms for $\pi$ and $e$:
38
Q: Mirror algorithm for computing $\pi$ and $e$ - does it hint on some connection between them?

Yuriy SBenoit Cloitre offered two 'mirror sequences', which allow to compute $\pi$ and $e$ in similar ways: $$u_{n+2}=u_{n+1}+\frac{u_n}{n}$$ $$v_{n+2}=\frac{v_{n+1}}{n}+v_{n}$$ $$u_1=v_1=0$$ $$u_2=v_2=1$$ $$\lim_{n \to \infty} \frac{n}{u_n}=e$$ $$\lim_{n \to \infty} \frac{2n}{v_n^2}=\pi$$ ...

 
I do need to beef up my knowledge on sequences later, so that I will be less confused on the symmetries and other things in the partial sums that are responsible for the outcome of the sequences
 
1:13 PM
Hi @Krijn
 
Heyooo @Bala
 
Sup
 
Grading homework, meh
 
yeah, bleh
 
The only fun thing about grading is the mistakes some students make
 
1:24 PM
XD That's not a good thing @Krijn
 
@Krijn eg?
 
$f$ is injective so $f(x_1) = f(x_2)$
Uhhhh, you mean that $f$ is constant?
 
What kind of homework you grading?
 
@Krijn yikes.
 
@SimpleArt Introduction to Mathematics
So all sorts of stuff on functions, relations, sets etc.
 
1:34 PM
@BalarkaSen ^QFT 101
 
lol
 
written by a mathematician (Fields medalist!), too ^
 
Borcherds is a cool guy
 
1:57 PM
What is the following integral: $\int_0^\pi u^ne^{iu}du$?
Need a tad bit of help
 
Hi nerds
 
@Ramanujan Hi, worshipper of dead mathematician :)
@SimpleArt Integrate by parts
 
XD:)
 
@Danu Aren't we all?
 
@Krijn I certainly am.
 
2:07 PM
@Danu Who's your favorite?
 
@Krijn Imposssilbe to say.
I like Thurston and Grothendieck a lot.
But there are so many...
 
Grothendieck?
I thought he wasn't really your flavor
 
yes.
@Krijn I don't know anything about his work on a technical level.
 
I like Bourbaki
 
@BalarkaSen That's cheating.
 
2:10 PM
I've read part of his memoirs (Récoltes et Semailles)
I was very impressed by his way of speaking about mathematics.
 
@Krijn OK, I like Lang
 
(of course you have to read past his paranoid accusations etc to appreciate it)
 
I still have A Mathematician's Apology on my reading list
 
It's not by Grothendieck though :P
It's pretty unimpressive to someone who lives in the current age, IMO.
I read it and wrote an essay on the issue it tries to discuss---beauty in mathematics.
 
Is it outdated then?
 
2:14 PM
I live in the current age and I liked the bit I actually read
 
@Krijn Quite badly so, IMO. But as you see, opinions differ.
 
But then I didn't read much, so.
 
Rota has a better piece on beauty in mathematics
 
Hmm, I'll check it out one day
For now, I have to catch a train
 
I think that his conclusion that it's mostly about enlightenment is sound.
 
2:19 PM
While I like the poetic nature of Grothendieck's description of his views on mathematics a lot, I get the impression that he had an overly grand view about mathematics. I can't say I share that feeling (I should mention I haven't read much of Grothendieck, his mathematical or mathematico-literary works alike).
 
I don't understand what you mean by "overly grand view". Any examples/concrete things?
Hi @Mike
 
I had to cite Bourbaki in my paper.
 
Not off the top of my head right now, I'm afraid.
@MikeMiller You're writing a paper?
 
@MikeMiller You didn't want to?
 
@Balarka Apparently I have been for 3 months.
@Danu Have you ever actually tried to read it?
 
2:24 PM
@MikeMiller Bourbaki?
No.
I feel like I'm not proficient enough in anything to make it worthwhile.
 
I didn't know that; nice. May I ask what it's on?
 
@Danu i need general solution approximated
4
Q: Construct more simple functions by restricting the domain

JohannesI need a function that has the same values as $f(x)=\sin\left(\frac{x}{2}\right)\cdot\sin\left(x^2\right)$ between $x=0$ and the following root of $f(x)$ at $x > 0$. At any other point, the function may differ from f (the reason I need this function is that I need to get the area under $f$ betw...

$$\frac1{2\beta}\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^{\pi}\left(\frac u\beta\right)^n\sin(u)du$$
$u=\beta x^2$
 
 
1 hour later…
3:35 PM
Is it true that among all rectangles inscribed in a right angle triangle,square has the largest area?Does it always have to be a square?Can't it be any other rectangle?
 
Huy
3:56 PM
prove it
 
hi @Huy
 
Huy
hi Balarka
 
what're you upto
 
Huy
mostly teaching, just one more week till holidays
 
same here. strange; what kind of holiday is it there?
 
3:58 PM
He is posting everywhere only "The question was answered in the comments above." as answers. And someone voted up too. Please, have a look this guy

http://math.stackexchange.com/users/215955/moritz
 
Huy
autumn holiday
 
interesting
 
Huy
2 weeks off
 
3 for me
 
Huy
spoiled kid
 
3:58 PM
=P
i need this holiday, i'm all drained up because of schoolwork
 
Huy
same
but mostly because of teaching :P
 
right
 
Got hyperdrive again at $0x=0$
2
Q: It is possible to show/prove that the cancellation property is necessary to prove $0x=0$ for $x\not\in \mathbb{Z}^+ \cup \{0\}$?

SecretFor any nontrivial algebraic structures with additive identity 0 and multiplicative identity 1 (and binary operation defined by "juxaposition of its arguments"), and at least one sided distributive law holds, one can easily prove that 0 is an absorber, i.e. $0x=0$ for $x=\{0,1\}$ as follows: $$0...

 

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