Mathematics - 2016-06-18 (page 1 of 4)

user174558

00:12

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I am in the 200 club.

skill patrol

Nice to see @JasperLoy

:)

user174558

There is so much math we will never finish studying it in one life.

skill patrol

Indeed.

user174558

Yet people can say 'Oh your degree in math takes 3 years. Is there so much to study in math?' ROFLMAO.

user174558

I wonder why people don't say the same for medicine, for example.

user174558

00:18

@skillpatrol You should call yourself Ski/ull Pa/etrol.

skill patrol

I don't think / is allowed.

user174558

I am a bit pissed with math books which omit too many proofs.

user174558

It is OK to state a few theorems without proofs, at the advanced levels.

user174558

But it is not OK if there are too many.

user174558

Then there really is no point reading the book.

skill patrol

00:20

Left as an exercise for the reader?

user174558

Not referring to that.

user174558

More of 'beyond the scope'.

skill patrol

I see.

user174558

Maybe around 5 is OK. But 10 is too much.

skill patrol

Look for a better book.

user174558

00:23

My favourite calculus book is now Kaplan/Lewis.

user174558

That is because it has a beautiful chapter on the elementary transcendental functions.

user174558

Something you don't find treated rigorously in most calculus/analysis books.

user174558

If there is a book on calculus of variations, and you can't even prove that the shortest distance between 2 points is the straight line, then that is sad.

user174558

Perhaps that is what math books today have become.

user174558

Developing lots of machinery but in the end leading to nothing of real significance.

skill patrol

00:27

I gotta run @JasperLoy nice seeing you :)

user174558

@skillpatrol See you pal.

Chinatsu-creepy-chan

01:46

hello

can somebody please help me with this system?
$x^2 = yz$
$y + z = 29$
$x,y,z \in \mathbb{N}$

The main goal is to find $x$

Mike Miller

02:22

Hi @Ted.

Ted Shifrin

@Chinatsu-creepy-chan So for what positive integers $y$ up to 28 is $y(29-y)$ a perfect square?

Hi @MikeM

user19405892

hi

Can someone explain how the solution gets $26460$?

Chinatsu-creepy-chan

@TedShifrin ugh, idk :c. i tried using the fermat's little (lesser?) theorem and i was reading about quadratic residue

Ted Shifrin

Way too fancy. Just write down numbers or think about how a product of two integers can be a perfect square.

Chinatsu-creepy-chan

Aw c'mon. I didn't think of that in the test. 4 and 25.

but alright, what if i need a solution without trial and error?

Ted Shifrin

02:33

Don't be mad at me! :)

Think about prime factorization? Note 29 is prime.

Chinatsu-creepy-chan

@TedShifrin sry, i'm not mad at you. i'm mad at me cuz i didn't finish this one in the math olympiad.

@TedShifrin yup. what about it? i thinked on it... nothing came out of it D:

Ted Shifrin

You can conclude that both $y$ and $29-y$ must be perfect squares.

Mike Miller

@TedShifrin Someone asked a question about Hodge theory on semi-Riemannian manifolds. I told them that at least the standard theory of the Laplacian etc fails miserably (it's not elliptic!) But do you know a way in which people have salvaged it? Perhaps in the Lorentzian setting on a manifold with cylindrical timelike ends, or something?

Chinatsu-creepy-chan

yes, yes. I thinked on it but... I want a deterministic way. :p

Ted Shifrin

no clue, Mike ...

Chinatsu-creepy-chan

02:36

btw @ted thank you so much!

it was just a dumb thing i didn't even realize... if i had taken the "boring" way i would have found it so easily...

Ted Shifrin

If not, to have a perfect square, a prime factor of $y$ would also have to be a factor of $29-y$. Can that happen?

Chinatsu-creepy-chan

no.

i thought about that - different parities

Ted Shifrin

that's not good enough

Mike Miller

@TedShifrin Found an authoritative answer by Willie Wong (who else?)

Ted Shifrin

Cool ... I would have asked Rafe Mazzeo. What's Willie's answer?

Chinatsu-creepy-chan

02:43

oops, oops. i didn't get the perfect square part. i knew that the parities were different but my model was (y,z) = (k²n, n) where k is even and n is odd

so yz would be a perfect square

Ted Shifrin

Important here that 29 is itself prime.

Mike Miller

@TedShifrin "Jesus, absolutely not. What were ya thinkin?"

Ted Shifrin

To you?

Chinatsu-creepy-chan

@TedShifrin i don't get it. should it say that if y + z = 29 then y and z are perfect squares? D:

Ted Shifrin

Reread what I've written.

Jake1234

02:50

Hey guys, not sure how to show that $\| \int_{a}^{b} f \| \leq \int_a^b \|f\|$ for $f:\mathbb{R} \rightarrow \mathbb{R^n}$, $\|\cdot \|$ is the standard euclidean norm. Any ideas please?

Ted Shifrin

One way is triangle inequality with Riemann sums.

Chinatsu-creepy-chan

@TedShifrin sorry, i cant get why there's no factor of y in 29 - y :/

Oh. Modular arithmetic shows why. Ok, fine.

Ted Shifrin

Can a prime $p$ Divide both $y$ and $29-y$?

Mike Miller

@TedShifrin Nah. This answer

Chinatsu-creepy-chan

Yeah, that's what I thought about. y = 29 - y = 0 (mod p) -> 29 = 0 (mod p) so p = 29.

Ted Shifrin

02:55

Ok, sure.

Chinatsu-creepy-chan

if there was, y or z would have to be 0.

This part is ok. But I don't get how you got the perfect squares.

They could possibly be primes between then. Like 3x5 and 2x7

(yeah, I know the sum isn't 29, just example)

Ted Shifrin

Huh?

Chinatsu-creepy-chan

Why do z and y have to be perfect squares?

There's no actual theorem of proof, idk u.u

Ted Shifrin

If $y$ and $z$ aren't both perfect squares, every non-square factor of $y$ must also appear in $z$.

Chinatsu-creepy-chan

Why?

Forever Mozart

03:00

when the chrome was thick and the women were straight

Ted Shifrin

How else is the product a perfect square?

Chinatsu-creepy-chan

Oh...........

Boy, that is really smart.

:P

What if I wanted to find these pairs without "brute force"?

Any theorem, rule, algorithm?

Ted Shifrin

Factorization in the Gaussian integers gives you the ways of writing 29 as a sum of squares.

Leaky Nun

@TedShifrin $(5-2i)(5+2i)$?

Ted Shifrin

Yup.

Leaky Nun

03:07

But the problem is

how do you factorize $29$?

Ted Shifrin

Because 29 is prime, there's only one solution essentially.

You just did.

Leaky Nun

@TedShifrin Yeah, but that is because I already know how to write 29 as a sum of squares

but how would a factorization algorithm look like?

Ted Shifrin

I don't know the answer.

Leaky Nun

Alright.

Ted Shifrin

You need to draw a circle of radius $\sqrt{29}$ and look for integer points on it.

Leaky Nun

03:13

That is essentially equivalent to expressing it as a sum of squares

and it only applies to factorization of the kind of $(5+2i)(5-2i)$

How would you factorize $5+2i$, for example?

Ted Shifrin

Sure. There are also formulas for Pythagorean triples.

Leaky Nun

Sure.

Ted Shifrin

$5+2i$ is prime because the square of its length is prime and square length is multiplicative.

Leaky Nun

Oh, right

So polar form is clearer in this context

$$(r_1e^{i\theta_1})(r_2e^{i\theta_2})=(r_1r_2)e^{i(\theta_1+\theta_2)}$$

Leaky Nun

03:36

1, 2, 4, 6, 8, 10, 12, 16, 18, 20, 22, 24, 28, 30, 32, 36, 40, 42, 44, 46, 48, 52, 54, 56, 58, 60, 64, 66, 70, 72, 78, 80, 82, 84, 88, 92, 96, 100, 102, 104, 106, 108, 110, 112, 116, 120, 126, 128, 130, 132, 136, 138, 140, 144, 148, 150, 156, 160, 162, 164, 166, 168, 172, 176, 178, 180, 184, 190, 192, 196, 198, 200, 204, 208, 210, 212, 216, 220, 222, 224, 226, 228, 232, 238, 240, 250, 252, 256, 260, 262, 264, 268, 270, 272, 276, 280, 282, 288, 292, 294, 296, 300

What do these numbers have in common?

Jake1234

@TedShifrin Thanks for the reply.

I was hoping for a more direct way though. I thought about minkowski's or holder's inequality, but I can't think of a way to use them.

Guys, any ideas on showing $\| \int_{a}^{b} f \| \leq \int_a^b \|f\|$ for $f:\mathbb{R} \rightarrow \mathbb{R^n}$, $\|\cdot \|$ is the standard euclidean norm. , in some direct way?

Forever Mozart

04:22

is anybody hererererereer

this is a nice paper: projecteuclid.org/download/pdf_1/euclid.bams/1183510788 :)

@LeakyNun i cannot figure out the pattern

Leaky Nun

@ForeverMozart It's the image of the totient function

Forever Mozart

ugh, I thought it was a simple pattern you could figure out

i'm baking some biscuits

and watching cape fear

@how is your Friday?

@BalarkaSen

Balarka Sen

04:38

I'm seriously ill.

Forever Mozart

mentally or physically?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

those^ are not mutually exclusive :P

how serious? @BalarkaSen

Balarka Sen

I don't know. I have a fever, which I think classifies as physical illness. So physical, mostly.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:(

Forever Mozart

drink some nyquil

advil pm is good too

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

04:42

see a doctor

Forever Mozart

if you have no other symptoms, its probably just a virus and you can treat it

Balarka Sen

@Skull: Done so.

It's bacterial. On a couple antibiotics.

Forever Mozart

then you will be better in 2 days

watch netflix or something

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

05:22

@BalarkaSen be sure to completely finish your antibiotic prescription even if you feel better. A lot of people make the mistake of not finishing the prescription and get ill again. Get plenty of rest.

2

user174558

07:47

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Thank you Dr Skull for the timeless advice.

user174558

07:58

There are several definitions of computability but I cannot find any book giving complete proofs that they are all equivalent. The book by Boolos at least shows the equivalence between 3 definitions. Is there a better source, or does one have to resort to reading the research papers themselves?

user174558

@huy You look very young.

user174558

@TedShifrin Speaking of Pythagorean triples, very few number theory books mention the fact that the radius of a circle inscribed in a Pythagorean triangle is an integer. This is something I found truly remarkable!

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

08:37

@JasperLoy :D

Huy

10:52

@JasperLoy so do you

user174558

11:11

@AlexClark I think I found a proof of Chern-Gauss-Bonnet in Spivak Vol 5. Maybe one day I will get all 5 volumes.

user174558

11:27

Does anyone know the difference between Matsumura's Commutative Algebra and Matsumura's Commutative Ring Theory? I am wondering why they are two separate books.

robjohn

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Thanks. I have had to be online less frequently recently, so the progress is slow.

user174558

12:36

@robjohn Hey there! I am still alive!

user2617526

13:13

Hello people, if anyone is very enthusiastic about probability problems, do you mind having a look at my question on Maths SE. It didn't get any answers or hints. I hope its okay to post here once, and am not breaking any rules. Thanks a lot. math.stackexchange.com/questions/1829413/…

JeSuis

14:24

Still stuck on the following problem : Let $f:\Bbb{R}^2\to \Bbb{R}^2$ be a $1-$Lipschitz function, how can I prove, the $n-$th iterated function noted as $f^n$, that $f^n/n$ converges.

Albas

@Semiclassical you there?

Semiclassical

'yeah

Leaky Nun

Roughly, what is the condition that 1/a1 + 1/a2 + ... converge?

I'm guessing f(n) > O(n^2) [informal notation]

Albas

Why does electromagnetic induction actually happen? By that I mean to say that I know magnetism takes place due to moving charges. So if i bring a magnet close to a conducting loop in the magnet's reference frame yes the loop is moving and hence it would produce a magnetic field but this is in with respect to the magnet moving but magnetism is a global phenomenon which takes place in thus experiment. I do not get how magnetic field is produced in thus case @Semiclassical

Leaky Nun

@Albas relativity.

How Special Relativity Makes Magnets Work

►

It is possible to derive the magnetic formulae with coulomb's law and lorentz transformation.

Semiclassical

14:43

@albas there's a fairly concrete version of this, namely the question of what happens when a magnet moves through a coil of wire

if you take the perspective of the magnet, then: there's a static magnetic field and the coil is moving. consequently the electrons in the coil are moving at some speed perpendicular to the magnetic field, and therefore the Lorentz force will cause the electrons to circulate creating a current

Does that perspective make sense?

Albas

Oh. That kind of makes sense... Actually that makes beautiful sense@Semiclassical

Semiclassical

mmkay. now, suppose we take the perspective of the coil. then it's at rest, and the magnet is moving. hence the magnetic field will change over time.

here we seem to run into a quandary. the charges in the coil start at rest, and a magnetic field only acts on moving charges. so how can a current be generated?

Albas

Yes. Same problem I am facing

Semiclassical

the only way it can be generated is if, somehow, an electric field is also present.

Albas

Yes

Semiclassical

14:55

and the only thing that can generate that is the changing magnetic field.

and that's precisely the idea of electromagnetic induction: a time-varying magnetic field generates an electric field.

Albas

But why doesn't a constant magnetic field also do the job?

Semiclassical

And here is where I realize I left out a part of the story I usually give when I do the first perspective.

One which is kind've important right at the moment.

Albas

Due to the constant magnetic field there will be a force acting and wouldn't that cause the charges to move

What is it?

Semiclassical

keep in mind, a magnetic field can't make things start moving.

all it can do is make stuff circulate.

it changes the direction of moving charges, but it can't get them to start moving.

but, back to the first perspective for a moment.

Albas

Okay.

Semiclassical

14:59

okay. Going back to the 'magnet at rest' scenario for a bit, suppose we go to the moment where the coil is at the center of the picture

the electrons in the coil are all moving straight down. But what direction is the magnetic field in that case?

J. M.

I am having a hard time resisting the temptation to bring up ICP and their thoughts on magnets... ;)

Semiclassical

lol. it's too bad they didn't say "F***ing magnets, how do they do work?"

because that's actually an interesting question :p

Albas

Opposite to the direction of the magnetic field produced by the bar magnet

Semiclassical

sure. in other words, the magnetic field experienced by the electrons in the coil would be straight down as well

iwriteonbananas

@TedShifrin One love! <3

Mike Miller

15:04

yeah, but they wouldn't want to talk to a scientist, lying and getting them pissed and whatnot

Leaky Nun

@Semiclassical Quick question: in my view, when the two currents are travelling in the same directions, both the electrons and the nuclei should repel each other, then why do they not?

Semiclassical

what two currents?

Leaky Nun

let's say I have two metal wires

and I generate a current inside them

Albas

Yes correct @Semiclassical

Leaky Nun

such that they travel in the same direction

Semiclassical

15:05

@leakynun work out your question, and i'll come back to it once I'm done with albas

iwriteonbananas

Howdy @MikeMiller

Leaky Nun

ok, thanks

Semiclassical

@albas but that means that, when the coil in the middle of the magnet, there won't be a Lorentz force acting on the electrons. there can't, because the electrons would be moving in the same direction as the magnet's field.

Mike Miller

hi

Semiclassical

so that means that, when the coil is at the very center of the magnet, we don't expect to see a current.

now, suppose we instead look at when the coil is at the 'top' of the magnet---say, where the magnetic field points horizontal and outward.

Mike Miller

15:08

what are you up to today

Semiclassical

in that case, we do get a Lorentz force as I said earlier and we do get a current. conversely, if you go to the bottom of the magnet, the magnetic field points horizontal and inward. so we again get a Lorentz force and a current, but now opposite to what it was at the top.

Leaky Nun

$\displaystyle\sum_{r=0}^\infty\dbinom{2r}{r}=\frac{36+2\sqrt{3}\pi}{27}$

Semiclassical

and if we go to when the coil is far away from the magnet, then the field lines all point vertically upward and so no current is generated.

iwriteonbananas

i was productive this morning, preparing seminar talks i need to hold in 3-4 weeks. now i should be working on my thesis...

what're you up to?

Semiclassical

that means that the only times when we get a current are when the coil is in the vicinity or the top or bottom of the magnet, and the two scenarios produce opposite currents. right?

Albas

15:10

Ahh... Correct

iwriteonbananas

it's alternating between beautiful and sunny weather and the-world-is-ending rain storms here

Mike Miller

spent yesterday trying to read some homotopy theory I didn't understand. it spooked me. so now I'm reading something more down to earth by my advisor before I go back into spooky territory again

J. M.

@iwriteonbananas That can't be good; you'll get moldy bananas that way. ;)

Semiclassical

If we now translate this to the 'coil at rest' scenario, this tells us that the only times when we get an electric field (and therefore an electric current) are when the magnet's field is changing

iwriteonbananas

modern homotopy theory is a spooky field

Semiclassical

15:12

in other words, it's the time-varying nature of the magnetic field that gives rise to the electric field. and that's induction

iwriteonbananas

@J.M. all my bananas are rotten already. i've had these since the 50's. i live in an underground bunker which was built for the case of nuclear war

2

Mike Miller

apparently the only way to define certain flavors of equivariant homology is to understand spectra a little, so i gotta work on that.

Albas

Yes I am starting to get it now

Mike Miller

but jumping into something May wrote is a hell of a start

here it's a heat wave

not as bad as it is where I grew up, but still nasty hot for LA

Semiclassical

It's not that hot up here (MN) but just kind've stuffy

iwriteonbananas

15:14

i find anything by May terribly difficult to read

Mike Miller

it'll be 97 on Monday

@iwriteonbananas I think I'm going to try to assemble the ideas from elsewhere and get someone who's good with spectra to brave the book with me

iwriteonbananas

did you grow up in death valley?

Mike Miller

97 is for LA. I grew up in the Coachella valley, which is another desert

121 on Monday there

Semiclassical

Just read a lot of stuff by April first. After all, April showers bring May flowers.

gets hit by a moldy banana

iwriteonbananas

jesus that's intense

Mike Miller

15:15

you do not go outside much

or you die

whichever you prefer

iwriteonbananas

i read something about spectra the other day. there's a little something in chapter 5 of hatcher

Semiclassical

Spectra of what kinds of operators?

Mike Miller

different kind of spectra

Semiclassical

mmkay.

Mike Miller

i'm find with spectra of linear operators :) a spectrum in the context of homotopy theory is a way of assembling the idea of a "stable homotopy type" into a coherent idea

iwriteonbananas

15:17

starts on page 583

right

Semiclassical

is spectrum in the homotopy context a generalization of the operator notion, or just a different thing altogether?

Mike Miller

like if you think $M$ and $N$ should be equivalent if $\Sigma^k M \simeq \Sigma^k N$ for some large $k$ ($\Sigma$ is suspension - take two cones on $M$ and glue them together along the bases), then you work "stably"

a spectrum is a generalization of that idea, and afaict has literally nothing to do with any other use of the word spectrum I know

eg spectral sequence, which has nothing to do with either of these

Semiclassical

weird.

Albas

@Semiclassical can you suggest me any mathematically rigorous book on induction. My book doesn't talk much about it

Semiclassical

not off the top of my head, I'm afraid. You should note, though, that none of the stuff I've said above was particularly 'rigorous.'

iwriteonbananas

15:19

i read something about where the terminology 'spectral sequence' comes from a couple weeks ago...i'll try to find it

Mike Miller

as a nontrivial example of stable equivalence, the "Poincare homology sphere" $SO(3)/I$, where $I$ is the isometry group of the icosahedron, has $\Sigma^2 P = S^5$

so even though this space is interestingly different from $S^3$, they're "stably" equivalent

iwriteonbananas

"A question that often comes up is where the term
“spectral” comes from. The adjective is due to
Leray, but he apparently never published an explanation
of why he chose the word. John McCleary
(personal communication) and others have speculated
that since Leray was an analyst, he may have
viewed the data in each term of a spectral sequence
as playing a role that the eigenvalues, revealed one
at a time, have for an operator. If any reader has
better information, I would be glad to hear it."

Semiclassical

@Albas Anyhow, what this establishes is that if the two reference frames for the conductor-magnet system are to be two descriptions of the same experimental outcome, then we need to have electromagnetic induction for sake of consistency.

Mike Miller

interesting

Semiclassical

Now, here's something kind've amusing. @albas

iwriteonbananas

15:21

@MikeMiller that's cool, i didn't know that twice suspending that thing gives a sphere

Mike Miller

anyway, when you work stably, you invert a lot of the usual complications when doing homotopy theory. so people like to work stably. (this is why people talk a lot about 'stable homotopy groups of spheres' - we're better at calculating those than the unstable ones!)

Balarka Sen

replace that by any homology sphere and it works

iwriteonbananas

really?

what about once suspending?

Semiclassical

@albas the first paragraph of the introduction here summarizes the above discussion

Balarka Sen

"double suspension theorem". i don't know how to prove it.

Leaky Nun

15:22

Would anyone tell me roughly how this is proved?

$\displaystyle\sum_{r=0}^\infty\dbinom{2r}{r}=\frac{36+2\sqrt{3}\pi}{27}$

Balarka Sen

@iwriteonbananas i emailed tim chow about that once

Mike Miller

@iwriteonbananas up to homotopy you only need to do it once, since $\Sigma M$ will have the homology of the sphere one dimension up, and be simply connected, so get a map $S^{k+1} \to \Sigma M$ by using Hurewicz; it's necessarily a homotopy equivalence.

Semiclassical

read that paragraph first, then scroll up to the top and look at the title/author :)

Mike Miller

But $\Sigma M$ is never a manifold when $M$ is a manifold unless $M$ is a sphere, so it's not homeomorphic.

iwriteonbananas

@BalarkaSen en.wikipedia.org/wiki/Tim_Chow ? Why did you email a soccer player about something like that?

Balarka Sen

15:24

different person

Mike Miller

The double suspension construction was the first place anybody had constructed non-combinatorial triangulations (triangulations that don't come from a PL structure).

iwriteonbananas

@MikeMiller I see...well, maybe not quite. where do we get that map from again?

Mike Miller

$\pi_{k+1}(\Sigma M) = H_{k+1}(\Sigma M) = H_k(M) = \Bbb Z$

Balarka Sen

i don't remember what he replied back with.

Semiclassical

The punchline being, that very example of two descriptions of a conductor-magnet system was used by Albert Einstein as a motivating example in the paper he wrote introducing Special Relativity to the world :)

iwriteonbananas

15:26

Ah of course

and the fact that it's a htp equivalence comes from whitehead's thm?

Mike Miller

I guess it's not obvious to me why the map is degree 1, actually

Balarka Sen

yes

Mike Miller

If we have that then we get it from Whitehead's

Albas

@Semiclassical I am so happy to see finally some consistency in these few days

iwriteonbananas

right

Semiclassical

15:27

Yeah. And what I find really funny is that that very example of 'consistency' is something that Einstein cited in his discovery of special relativity.

Balarka Sen

I don't remember what the Hurewicz isomorphism does. Sends a map S^k --> X to the homology class of the singular cycle it represents in X?

iwriteonbananas

sends $[f:S^k\to X]$ to $f_*([S^k])$

Balarka Sen

That's exactly what I asked :) OK.

iwriteonbananas

oh, right

Mike Miller

lol, which also makes it obvious that the map is degree 1. my b.

Leaky Nun

15:33

@Semiclassical Sorry for bothering, but could you answer my question?

Semiclassical

state the question, and i'll see what i can do

Balarka Sen

good to know fever didn't screw my mind up.

Mike Miller

but in any case you can see why stabilization is a homotopically valuable tool; for a homology sphere, the fundamental group is the only 'obvious' difference between it and the sphere. stabilization makes them the same.

Leaky Nun

29 mins ago, by Leaky Nun

@Semiclassical Quick question: in my view, when the two currents are travelling in the same directions, both the electrons and the nuclei should repel each other, then why do they not?

Semiclassical

You started saying the question, but you really didn't finish.

Leaky Nun

15:35

Well, let's say there are two metal wires perpendicular to each other

and I generate currents in them to the same direction

Then, the electrons are moving, so the electron density is higher than the nucleus density

so a negative charge is generated

so they will repel each other

right?

iwriteonbananas

that's true

stabilization is kewl

Semiclassical

hmm. the currents will also produce magnetic fields.

Leaky Nun

@LeakyNun *parallel

Balarka Sen

how I think of spectra, not knowing much about them, is what represents a (extraordinary) cohomology theory.

probably a wrong-headed way to think of them.

iwriteonbananas

you'll need to teach me about stabilization at some point, Mike

Mike Miller

15:38

It is rarely useful to have precisely one perspective on an object. To only think of them as stable homotopy types and not at all as representing cohomology theories would also be unhelpful.

I just did!

Semiclassical

and if memory serves, those induced magnetic fields will cause those two loops to attract one another @albas

Balarka Sen

agreed.

iwriteonbananas

well, true...but i mean you should fly to my country and organize a bootcamp on stabilization for me

Semiclassical

more generally, two parallel wires will attract one another due to the magnetic fields they produce

Mike Miller

like i know shit about it

iwriteonbananas

15:40

haha

Semiclassical

Either that or produce a Skype bootcamp :p

Mike Miller

i'm busy with this, which apparently explains flavors of equivariant homology in a way poor li'l me will understand

Semiclassical

I got interested in the word 'equivariant' at one point when the phrase 'Duistermaat-Heckman formula' started showing up in places

I entirely failed at understanding it.

user 1618033

Yesterday I posted here a picture of Ramanujan and there was no star (doing it three times).

Balarka Sen

equivariant just means upto a group action.

Mike Miller

15:42

that seems like you think you just mod out and go on your way? the word you wanted there means "respecting" a group action

r9m

@user1618033 you say 'with Ramanujan' .. yet I see a single person in this picture ... (unless you have your image hidden via steg .. :P )

iwriteonbananas

looks fun, mike

Balarka Sen

ah, you're right. "respecting" is better wording.

user 1618033

@r9m OK, of ... Does it seem to me or you are in a bad mood? :-)

Semiclassical

mostly because the DH formula seems to give examples where semiclassical approximations would be exact.

user 1618033

15:44

@r9m hehe, not really ;)

Thinking to ask some question on main ...

Mike Miller

localization theorems are nice but I don't really understand the yoga yet

Semiclassical

But it was in the context of a certain spin system, which meant that the semiclassical phase space would be the surface of a sphere (or something like that, i forget details)

r9m

@user1618033 I am in a 'troll' mood :P

Semiclassical

and so the group action would have to do with that. somehow.

I really never understood it, though.

Danu

Hi guys

user 1618033

15:45

@r9m why don't you star the picture of Ramanujan? :-) Just a little star ... :D

Mike Miller

@iwriteonbananas it's eminently readable

r9m

@user1618033 *ed

Eric Stucky

So I know that if you look at the Bruhat order as a geometric structure, you get the permutohedron, and if you look at the Tamari lattice as a geometric structure, you get the associahedron: is it possible to look at any (bounded, finite) poset as a geometric structure in an analogous way?

Balarka Sen

i googled why the double suspension theorem was true. this stuff is intense.

user 1618033

@r9m GREAT!:-)

iwriteonbananas

15:47

i don't know what localization is but reading the subsection "localization of spaces" in chapter 5 is on my bucket list. think i'll need that to say something about torsion subgroups of $\pi_k(S^n)$

@BalarkaSen that thing seems terribly formatted

Semiclassical

@MikeMiller I remember trying to read this paper and just failing :P

Mike Miller

@iwriteonbananas We're talking different types of localization. Localization in equivariant homology says something like "If you invert the action of $H^*(BG)$ on $H^*_G(X)$, then you just get the equivariant cohomology of the fixed point set $H^*_G(X^{fix})$"

250 pages nope nop enope

Semiclassical

Mostly because the idea of path integrals which could be computed exactly sounded pretty nice.

Balarka Sen

@iwriteonbananas it is

Semiclassical

lolyes

iwriteonbananas

15:49

okay, i see

Semiclassical

the length of that paper didn't help matters either, no

Mike Miller

I wanted to read this but was also spooked by length

Semiclassical

yeah, that's a bit much

though there are bits in the one i cited which you'd probably like, or at least appreciate

r9m

@user1618033 nice problem here (+1) :) Didn't see it until now ..

Semiclassical

e.g. section 3.5, where they talk about the height function on a Riemann surface as an example of equivariant localization

user 1618033

15:52

@r9m hehe, thanks! That one is an inequality I love a huge lot! Very, very nice. Do you have in mind some different approach? I'm curious about anything else possible there.

Mike Miller

I assume you mean on the Riemann sphere?

r9m

@user1618033 am thinking about it now :)

user 1618033

@r9m Without a careful approach it is painful - it seemed to me so at the beginning.

(referring to getting some very elementary way)

Semiclassical

well, they titled it with 'Riemann surface' in general.

Mike Miller

looks like they pointed out my complaint on p51: there's no circle action on a higher genus Riemann surface

Semiclassical

15:55

but they spend the first part of it focusing on the sphere

right.

Danu

@MikeMiller Vafa?! :D

(famous physicist)

Mike Miller

I know that. It's in the paper.

r9m

@user1618033 it's hard to think about something else .. :O my natural reaction would be to square both sides and then yea cauchy-schwarz and the usual telescopic upper bound .. just like Jack did :| .. what else we can do I wonder !! ^^'

Semiclassical

anyways, there might be some stuff in there you would appreciate more than you'd expect.

alongside quite a lot of chaff, to be sure

Mike Miller

probably a bit too long for me, unforch

looks like it would be a good place to learn some things

Semiclassical

15:58

yeah.

a more condensed version of that would be nice

user 1618033

@r9m Yeap, agree.

Transcript for

$Mathematics$ Mathematics