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16:02
In computers class; again.
@MikeMiller huh. so topologically n-disk bundles are isomorphic to linear ones?
(I only vaguely remember concordance spaces)
I got the book you suggested from the uni's library @Balarka, looks like there's a lot of interesting stuff covered in it
G-P? Yes!
Yes, that one
let me know if you want any help with anything. you'll learn a lot if you go through it
16:09
@Balarka don't quote me on it but I think so
I'll surely have stuff to ask you :P
alright, fair enough
I think today I'll rub off the rust from my algebraic topology
learn spectral sequences
or i dunno equivariant cohomology
both good choices. i'll go for spectral sequences since my favorite writer has a set of notes on it
will be a good refresher even if i suck at learning it
Equivariant cohomology is somethingI tried and failed to understand a while back
Mostly because I wanted to understand the Duistermaat-Heckman formula
16:19
What resource would you suggest for spectral sequences Mike/Balarka?
(Logic: A semiclassical approximation is, in the context of the path integral formulation, basically just a saddle-point approximation. So a case where the saddle-point approximation is in fact exact should amount to the semiclassical method giving the exact answer.)
Alas, I failed badly :/
Oh, I also did something silly: I answered a real analysis question :)
I probably didn't give a valid proof, though. Sigh.
@abenthy I have Hatcher's notes in mind
(I can even see an objection, but I'll wait to see if anyone comes up with it first)
@Semiclassical This is one of the few places I can recommend the big book on mirror symmetry
Yikes.
I believe you, though.
16:24
They do a very good job explaining localization
It's all for U(1) but I think that's all you should need
Might be right.
I haven't been looking at that for a while now, though.
About two years.
If I have some modules $B_1\subset...\subset\bigcup_n B_n\subset\bigcap_n Z_n \subset...\subset Z_1$ and primed versions of this together with a map $f$ between primed and unprimed so that $f\lvert: Z_n/B_n\to Z_n'/B_n'$ is an isomorphism for all $n$, is this a kind of setting I would have to look at direct and indirect systems to see that
$$f\lvert:\frac{\bigcup_n Z_n}{\bigcap_n B_n} \to \frac{\bigcup_n Z_n'}{\bigcap_n B_n'}$$
is an isomorphism?
ew
Maybe start with the simpler case of $Z_1\cup Z_2$ etc?
That might be enough for a proof by induction, since $\bigcup_{n=1}^{k+1} Z_n=Z_{k+1}\cup (\bigcup_{n=1}^{k} Z_k)$
D: its a step I think I have to do in spectral sequences, where the book just says that if we have an isomorphism between the pages of spectral sequences $E_r$ and $E_r'$ for all $r≥n$ for some $n$ we also have an isomorphism for $E_\infty$ and $E_\infty'$.
oh damn
I messed up the cups and the caps
16:40
$\cup \leftrightarrow \cap$?
$\displaystyle f\lvert:\frac{\bigcap_n Z_n}{\bigcup_n B_n} \to \frac{\bigcap_n Z_n'}{\bigcup_n B_n'}$
is correct
kk. My logic still applies. (which should tell you how shallow my insight actually is, lol)
no self-deprecation
bad
well in thise case you have $Z_k = \bigcap_n^k Z_n$^^
lol
huh. neat.
It's a filtration?
(checking my knowledge of the lingo here)
16:42
yes, both for $Z_n$ and $B_n$ but in other directions
the entire thing starting with $B_1$ ending with $Z_1$ is called a tower I think
but maybe thats just a colloqiualism
But anyways. If you can prove it for the simplest cases, you can probably work out a proof (by induction?) for the general case.
yes, I'm sure, but I didn't want to do it via induction because then I have made a big thing out of it and it is only a side-comment in a seminar talk I am preparing
16:49
is there a good way to evaluate the integral of sin(x)/x without contour integration
(also with contour integration you cannot close the loop in the top or bottom half plane because of $e^{ix}$ and $e^{-ix}$ though rihgt?)
typically one does it as the imaginary part of the integral of $e^{i x}/x$.
oh.. duh
And then that goes to zero in the upper half-plane.
But I know this question has come up on the main site.
or maybe not so duh, $e^{ix}/x$ is singular at $x=0$
16:53
Yeah, you need to do something at 0.
it looks like a rainbow
small half-circle around the origin, big half-circle closer to infinity
right.
I think the simplest proof outside of contour integration I know makes use of sin(x)/x being the Fourier transform of the box function.
Or maybe it's how you prove that sin(x)/x is the Fourier transform of the box function
you can do that with $\partial_k \theta(k)=\delta(k)$ i think
by using the fact that the box function is its own square, in conjunction with the convolution theorem.
i suspet you ultimately need the value of some integral of sin(x)/x to get the height of the box
16:57
Yeah.
I think I must be remembering how to derive $\int_{\mathbb{R}}\frac{\sin^2 x}{x^2}\,dx$ from that integral.
cute trick though, should pretty reasonably give you the values of int sin^m/ x^n
Probably. I haven't done a lot of that.
(Sounds like the kind of stuff our previous integral-enthusiast would've known)
[A weird thought born when thinking about art x mathematics] In a rough sense, when an artist is making an artwork, the details of the artwork seemed to reflect the historical background, the artist's emotion and ideas onto it. Thus it seems it can be thought as the following:

Let the set S to contain all historical context, emotions and ideas. Then the making of an artwork seemed to act like a map that maps S to the artwork such that it has some form that is a subset of S. Therefore under this framework, it seems an artwork is some kind of injective map. It is then tempting to say this is
@Secret You might appreciate this link, then: link.springer.com/chapter/10.1007%2F978-3-319-20603-5_14
(for me it falls in the realm of comedy)
17:21
@Secret you know, you've spent too much time on mappings
Quite the opposite, I don't spend enough time on mappings, otherwise I should be able to ponder about this, get the solution, and the move on to the next thing
> This book constitutes the thoroughly refereed proceedings of the 5th International Conference on Mathematics and Computation in Music, MCM 2015, held in London, UK, in June 2015. The 24 full papers and 14 short papers presented were carefully reviewed and selected from 64 submissions.
(emphasis added)
result:
Is it well-defined to talk about a "linear map" from a real vector space to a complex vector space. For example the map $\mu\mapsto \int f d\mu$, with suitable real vector space of measures and a complex function $f$. I would like to conclude that the above mapping is injective on some subspace, and to prove that I would like to use that linear mappings are injective iff kernel only contains zero.
18:16
Definitely @Martin, but your scalars should be complex in that case
hi @GFauxPas :)
Hi Mr Khan
hello sir !
are you french?
No, faux pas has become an English word
oh okay :D
i got a question for you btw since you know alot of stuff =p
18:17
I'm having coding problems and stackoverflow is so much less friendly than math.se :(
I'm no expert but I can try to help
what is the difference between pointwise convergence
and other kind of convergence
With pointwise convergence, you give me an epsilon, I'll give you a delta.
you can type the same way as math.se but then view it on "ask question section " if you type it there you can see normal
Yeah but the people there are more elitist
okay :D
18:19
So in the language of x and y coordinates
and the uniform one ? we are doing series of functions and analytic functions (intro course nothing to deep )
Draw a horizontal line at the level of the limit
Going through the graph of the function
In general
You tell me a delta, that means, you draw a box around the graph with height 2epsilon
ok, the notion of "physical speed" is that because it takes time to lift the fingers and press it down again, but in musical notation there is no rest to indicate that, thus literal interpretation result in it takes zero time to lift and place the fingers on the piano key, thus infinite physical speed
I'm sorry, I meant, height 2delta
but, isn't the time to lift fingers an assumed convention in the musical score...?
18:22
Wait no I was right the first time!
Height 2 epsilon
I draw a box around the limit point with width 2 delta
No matter how thin you make your box, I can make my box that much skinnier, and the box will contain both the graph and the limit point
Follow?
That's pointwise convergence
Oh wait, we're talking about sequences
yes but that is the normal limit idea
i know how to work with epsilon and delta
I'm sorry I was thinking of continuous limits at a point
Okay it's like this
but cant figure out is the difference between uniform convergence and pointwise convergence
:D
You draw horizontal lines around the limit point
Okay so uniform convergence is like this
Draw the curve you want to converge to
And instead of drawing two vertical lines above and below the limit point
why is the klein bottle has genus two. How is it possible to make two cuts on it and it not falling apart?
18:26
You draw two curves above and below that curve, parallel to it.
It's called an "envelope"
Then my job is now to guarantee that the sequence will converge to within a small epsilon everywhere in the envelope
Do you have a visual intuition now?
Once you have that I'll give you an algebraic defn
hmm so uniform convergence is like
the two curve coinside :D
be the same
They can be arbitrarily close everywhere
So algebraic ally
In general, with the epsilon delta game
delta will generally be a function of epsilon and L
oh okay :D thanks alot !
But with uniform convergence, delta can depend on epsilon, but not on L
So if we do the game and we end up with a box
You have to be able to slide the box anywhere along the curve without changing its size
Without the curve popping out
Hope that helps :)
it does ! its a bit too abstract and having apicture allways simplfies things =P
18:32
That's why I tried to explain it with a siding box
Maybe wikipedia or proofwiki.org has a picture
do you know what courses they talk about this in usa?
Analysis
am trying to look at some lectures in english
18:33
I can't wait to take courses in this, I've been learning from places online
this is from harvard
maybe it could be of help to you :D
You see how as the sequence converges, it never leaves the envelope
Here the sequence is converging, but some of the $f_n$'s pop out of the envelope in places
Pictures courtesy of Google Image search :)
Does that help ?
@KasmirKhaan
I realize in my explanation I was mixing up uniform continuity and uniform convergence :( but I was correct with the envelope and sliding rectangle part
:D
yes it does make sense now
 
1 hour later…
20:02
@s.harp Yeeep
To the extent that it's a playing with ideas, such things can be interesting to read. To the extent that it presents itself as something to be taken seriously, it is comic.
20:31
Is there a generalization of the residue theorem I can use on an integral like $(z^2 +1)^{-1/2}$?
Hey !
All power series converges ?
@GFauxPas I think this is the derivative of ArcSinh
@Maks no.
Yes it is @Astyx but pretend I don't know that
@ZachHauk Apparently they do
20:36
@Maks What do you mean by that ?
They all do at the point $x=c$ but that's so uninteresting we ignore that case if that's the only place it converges
For a power series around $c $
it isnt guaranteed to converge for all $\Bbb C$
hi @Semi
@GFauxPas It's got a branch cut from $i$ to $-i$, so on the face of it the answer is no.
However, if you do the substitution $w=1/z$, then the resulting contour will wind around a pole at zero and so can be computed that way.
20:41
Cool , let me try that
I should amend the above a bit: What's your initial contour?
If it winds around the branch points, then you can use that trick. If it winds around zero but not the branch points, then it's automatically zero because $1/\sqrt{1+z^2}$ is holomorphic at the origin.
Let's say I'm trying to integrate it as a real integral from 0 to 1
That's not going to help, then.
I'm pretending I don't know it's arcsinh(1)
Hmm, okay
The residue theorem is useful for when you have a closed contour.
20:43
So what about from -1/2 to 1/2?
That's not a closed contour, so it won't be directly useful
Now, you can use the residue theorem to argue that the result won't change if you allow your contour to incorporate a small imaginary part
Right but sometimes on an integral from real a to real b, a < b, I can curve around from b to a and connect to a
Sure, but that's not true for $1/\sqrt{1+z^2}$.
Yeah I just graphed it and I see what you mean. You memorized where the branch cut is there or its intuituve?
The branch points will be where the square root vanishes.
And a simple branch cut for that function will just go straight from one branch point to the other.
(your calculations shouldn't depend on how you draw your branch cut, so a lot of the time we just pair of branch points and draw line segments between them to define the cuts)
20:49
I see
I should stress that, by and large, one deals with integrals over closed contours when using the residue theorem
Well that's one of the hypotheses of it
The main exception is that sometimes we start with a contour which isn't closed, and then add on pieces which won't change the result in order to make it closed.
Like the bromwich integral
Right.
Or the keyhole contour.
20:53
I don't understand contours that start from some infinite point, come and does stuff, then goes back to the infinite point
I don't know how to do those
contours don't go to infinity, they go to a finite point, and you take a limit
@GFauxPas you take a limit
Like the contour integral of $1/\Gamma (z) $
Yeah so what does the pre-limit contour look like
Like $-1000 + \epsilon i $ to $-1000 - \epsilon i $?
A "Hankel contour"
Oh
And then the outer circle goes to zero in the limit?
Or rather, we just collect all the residues while avoiding the unpleasant places?
21:01
I tend to think of my integrals on the Riemann sphere (or other appropriate covering space) when possible for this reason.
Are all important properties preserved when you "invert" them onto the Riemann sphere?
I guess $1/z$ is a homotopy
Well
If you define $1/0$ somehow
I don't know how that works
I'm reading wikipedia on it
So If I map a contour onto the R sphere and it's partly on the northern hemisphere and partly on the southern hemisphere, it needs to change orientation at $z=0$?
Not going to be able to answer that, for the simple reason that I don't know a good way to draw pictures here :)
I have a mental image, but trying to describe it probably won't help.
21:21
Well can you just answer this question then:
Call a horizontal piece dx and a vertical piece dy
Can I apply any complex plane theorems I want on the rectangle [-dx,dx]×[-dy,dy]?
On the sphere ?
On the almost - rectangle
so long as the function is analytic on that piece, I should say so
So the mapping from the plane to the sphere is a manifold
that sentence does not make sense
I meant
The sphere can be locally treated as a plane?
I don't know what the formal word for that is
I guess that's what I just asked
21:26
the sphere minus a point is biholomorphic or conformally equivalent to the plane, which roughly means there's no difference between the two objects in the world of complex analysis
Can you define a metric that gives the same distance between points before and after the biholomorphism?
there is no metric on the sphere that restricts to the metric on the plane, if that's what you're asking
look up stereographic projection if you haven't
I don't know the words from what I'm asking,I basically am asking if there's a metric on the sphere that will give you the same distance between points as the Euclid's metric on the plane
on the sphere minus a point, sure. on the whole sphere, no
What you're asking for sounds like the Fubini-Study metric, though that might be the reverse
21:30
That's cool, I've never worked with metrics on curved surfaces before, except for arc length, but I never though to check if arc length is a metric
Actually it's more nasic than that
I don't know how to find the shortest distance between two points if I have to stat on a curved surface
As in, it gives distance in the plane based on distances on the sphere
Right
21:52
is rudin's analysis good?
Good day/evening fellas!
hi @Null
@ZachHauk I see you changed your nickname. Why?
i didn't like meow-mix
@TedShifrin o/
22:04
Hippa! M le méchant! Comment ça va?
hi @TedShifrin
Ca va bien et toi ?
I see @MikeM seems to have made it home in more or less one piece.
Hi @Zach
Oui, Hippa, plus ou moins.
i was thinking about the converse of that circle thing and i realized it was obvious
so if we have a circle in $\Bbb R^2$
Je n'ai pas vu ton frère depuis bien longtemps, M Hippa.
22:05
More or less.
@TedShifrin Ah, c'est bien possible. il a été sans ordi pendant quelque temps
Aha @Hippa. J'espère qu'il n'y a rien de louche :)
first apply a scalar transformation (or whatever you would call it; of the form $\lambda \mathrm{\; Id}$) such that the circle becomes the size of the unit circle
@TedShifrin Non il a juste cassé l'ancient :(
But, @Zach, for the converse we don't know that the locus is a circle yet.
@Hippa: Comme toi et la chaleur ...
22:06
@TedShifrin wait, what converse am i trying to prove again? just for clarification
@Zach: You're trying to prove that if you have a right angle between the two lines that the intersection point traces out a circle.
@TedShifrin Oui mais lui c'est pas la première fois, moi mon ordi avait 6 ans
6 ans ... ce n'est rien :P
@TedShifrin yeah i showed that yesterday using analytic geometry
But we're trying to see it by geometry, @Zach, so that you can generalize to any angle ...
22:08
also, what a coincidence; today in geometry class we proved that the midsegment of a triangle is parallel to it's corresponding side :P
Aha @Zach ... very good :P
I hope you're not trying to impress your poor teacher with projective geometry stuff :D
not really...
she really
to put it bluntly, doesn't care at all
That's not uncommon with high school teachers, Zach. And with some college teachers, too :(
can a function $f(x)=\sqrt[3]{x}$ be even properly defined? The question arises for me, because for positive numbers (assuming R as the domain) we get 3 possible roots, so positives can't be the domain. Negatives neither by similar reason. So the only domain possible is {0}?
In fairness to teachers, it's not like all students are like Balarka, you, and Akiva ... in terms of enthusiasm and talent.
22:11
Can't accuse me of caring.
Not 3 possible roots, @Null, unless you're doing complex numbers.
You're old, @MikeM. Shaddup.
@Null but we're considering the real roots
Ok, but then somewhere we have to say that or not?
@Null: You should be able to prove that $g(x)=x^3$ is an increasing function when you work with $\Bbb R$.
you can just say
22:11
guys i had a question about constrained optimization. why does it always work out when doing simplex method that the optimal solution is always a corner point solution?
$f: \Bbb R \to \Bbb R$
@ping: Because if you're in the middle of an edge, unless the function is constant along that edge, it will either increase or decrease as you move along the edge. Remember we're looking at linear functions.
@TedShifrin why is $x^3$ interesting here?
@Null: Because you're talking about its inverse function?
but why cant the optimal exist in the space below tho?
22:13
@Null because its the inverse
ok
What do you mean by space below, @ping?
well, i have to eat dinner, then perform for my school's band. bye!
Have a good evening, Zach.
and to you as well :)
22:14
like ok so if you do it graphically, then you have a bunch of lines representing the constraints that you draw and the feasible region is the region bounded between the axes and all of the of lines
OK, same rationale for the middle of any face, @ping. Remember that by calculus, an interior max/min would have to be a critical point, and linear functions have none.
i dont follow
Have you had multivariable calc?
Good evening everyone
Hi @Alessandro :)
Ah, Balarka lives to tell another story.
22:17
tbh no
which is something i am very ashamed of as an engineer
Yeah, engineers need solid calculus.
Hi @Ted
@TedShifrin the derivative is $3x^2$, which is never below 0, so (nonstrict) increasing of $x^3$ is shown?
But you can convince yourself, @ping, that at a point a linear function will be constant in certain directions and increase/decrease perpendicular to those directions. So you can always get bigger by moving until you hit an edge. Then along the edge, the same principle continues to apply.
the function is strictly increasing and you need it to be if you want an inverse
so you'll need to improve your result a little
22:19
OK, @Null, so that proves that you have at most one solution to $x^3=y$ for any $y$.
@TedShifrin he hasn't proved that yet!
Now you can argue that every $y$ value is taken on (something to do with behavior as $|x|\to\infty$).
@MikeMiller does this build on something like injectivity?
We're trying to establish injectivity, @Null.
I should butt out.
You convinced now, @ping?
22:20
nah, you should butt in
i'm too lazy
Let a<b. We want to show that f(a)<f(b).
$a^3<b^3$ follows directly from $a<b$, since we use an odd exponent(therefore signs are preserved).
I don't know if that is sufficient.
It sounds like assuming what we're trying to prove, unless you really can do high school algebra to prove it.
It sounds like you have a memorized rule. But why does it work?
Hi
I guessed that it will work that way.
@Astyx hi :)
Salut, @Astyx.
22:24
How are you two doing ?
@Null: So you can prove it just by rules of algebra or you can prove it by calculus. But you probably should be able to do it somehow!
@Astyx I picked up some steam :)
I haven't :p
@TedShifrin So I really only procrastinated the problem? Or is it the right direction?
I already told you.
So, @Balarka, are you still alive?
22:32
I am
I'm glad to hear it.
@BalarkaSen bold statement
@ohmy_ohmy This online editor is very good if you need it TexPaste. I hope help to you with it.
@Null I should have added "last I checked", yes.
Hi all! Take a look at this ring $A'=A[X_a]_{a\in A}/(a^2X_a-a,aX_a^2-X_a)_{a\in A}$, where $A$ is some ring (commutative with unity). I'm trying to prove that for a maximal ideal $\mathfrak{m}$ of $A'$ it holds that $A'_\mathfrak{m}$ (the localization at $\mathfrak{m}$) is a field.

It's enough to prove that the maximal ideal of $A'_\mathfrak{m}$ is $\{0\}$. So therefore I took an element $x\in\mathfrak{m}A'_\mathfrak{m}$ (the maximal ideal), and assumed that $x\neq 0$. If $x=\frac{a}{s}$ this means that there exists no z. div. of $a$ in $A'\setminus\mathfrak{m}$.
However, there is an elt. $b_0$ with a von Neumann weak inverse, such that $ab_0$ also has a von Neumann weak inverse. If $b_0\in A'\setminus\mathfrak{m}$, then $b_0(1-ab_0(ab_0)^{-w})$ is a z. div. in $A'\setminus\mathfrak{m}$, which is a contradiction.
Where $c^{-w}$ means the von Neumann weak inverse of an elt. I. e. an elt such that $c^2c^{-w}=c$ and $c(c^{-w})^2=c^{-w}$.
I want to show that $b_0\in\mathfrak{m}$ leads to some kind of contradiction, but I don't know how.
I've posted a question as well, but I've had no luck so far.
Anyone with commutative algebra skills that can put me on the right track?
22:49
@SemiC For posterity: i.sstatic.net/DGcNc.png My guess of a dire node blowing up with probability 1/4 in a specific depth is apparently very efficient in making the tree longer.

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