Mathematics - 2016-09-21

Josué

00:20

hi

user228700

01:49

@s.harp Oh, does it say that on the side-bar? I use mobile chat so it is not visible to me. Sorry about that. I'll ask away directly from next time.

Geralt of Rivia

04:18

Is there an easy example of 3x3 symmetric positive definite matrix that the Jacobi iterative method doesn't converge ?

Chris

05:22

Anyone here at the moment? In particular, anyone skilled at analysis?

user243301

06:53

@SathasivamK Just a comment about one of your (answered) questions, because you are asking about a problem on prime numbers, from a chapter dedicated to the least common multiple and the greatest common divisor. Then you can think with help of such answers how do a new proof using $gcd(A,B)lcm(A,B)=AB$: you need think what is A and B in your problem and follow a of those answers. Is not required a response of this comment, good luck.

user36188

07:08

@anyone please unblock my acount,i am restricted to ask a question in the forum Help me Please

@I want Help,Help Help

shai horowitz

09:27

anyone online?

user228700

12:53

Wow, I see that this chat room is a little inactive, compared to some of the most active chat rooms here, such as "The h bar"...

Danu

13:23

So I was watching this video on youtube:

Problems in Topology, Post-Perelman - Stephen Smale

►

And for some reason, I drifted towards the comments. Near the top, I found this:

> Why are smart people always so fucking ugly. With all due respect but Christ, there's a -1 correlation between IQ and looks. They all seem to have speech impediments or other physical anomalies (this professor with his slissing and swallowing, kip thorn with his vocal tic, hawking who has such a curved spine any polynomial would be jealous, perelman who looks like jesus on meth etc etc). Seems like all the gene expressions went to the brain and not to the body.

Too bad nature works like this and that's why there are so few really smart people: sexual selection just leads to them not repr…

So ridiculous (and of course offensive to those mentioned), but I really had to laugh at the apparent rage of this youtube user :)

arctic tern

@MikeMiller if C is a centralizer of a closed subgroup H of a Lie group G, then it centralizes any one-parameter subgroup through H, so it fixes the associated tangent vectors. all such centralizers C should be intersections of point-stabilizers of G's adjoint action on its lie algebra. not sure if that helps.

Anderson Felipe Viveiros

Is every orientation-preserving homeomorphism from $S^2$ onto $S^2$ homotopic to identity? Any references to it? (I mean, if it is true...)

arctic tern

13:45

yes

the wikipedia article on the mapping class group (homeo(M) mod connected component) states MCG(S^2) is Z/2Z corresponding to orientation preserving and not. it gives a reference.

Anderson Felipe Viveiros

Thank you, @arctictern!

Balarka Sen

14:12

@Danu lol

@AndersonFelipeViveiros Something more general is true: any map $f: S^2 \to S^2$ is homotopically determined by an invariant known as the "degree" of the map. Homologically, that's the image of the generator $1 \in H_2(S^2) \cong \Bbb Z$ by the map $f_*: H_2(S^2) \to H_2(S^2)$. Modulo the identification with $\Bbb Z$ that's a number. Any orientation preserving homeomorphism has degree $1$, hence homotopically equivalent to the identity map.

Mike Miller

@arctictern That helps a lot with the general case (which I don't need but is interesting). For $SU(2)$ eg, the only positive-dimensional subgroups are $U(1), \text{Pin}(2), SU(2)$, which we can explicitly work out the centralizers of pretty easily.

So the main trouble is classifying centralizers of finite subgroups, which I only know how to do via a classification of the finite subgroups, which is known but unpleasant.

arctic tern

ah, forgot about 0-dim subgroups

Mike Miller

separate fact: I still find it very counterintuitive that the map $x \mapsto x^2$ on $O(2)$ (or $\text{Pin}(2)$) is a covering map on one component and constant on the other.

was convinced I'd made a calculation error for a bit

arctic tern

heh

Balarka Sen

interesting

arctic tern

14:22

if the elements of G have a nice enough geometric interpretation, then so too will conjugation, in which case we can compute centralizers of arbitrary elements. then hopefully it's restrictive enough finite intersections can be classified, even ignoring a finite set of elements being a subgroup. for instance, if R is a rotation in a bunch of planes by a bunch of angles, then SRS^-1 is the same but one applies S to said planes.

Mike Miller

that's a nice interpretation

arctic tern

(that particular one gets tricky when the angles associated to planes are the same, i.e. centralizing i acting on C^n as a real space gives centralizer U(n))

Mike Miller

course

I would expect this to be unreasonable to carry out for general SU(n) (probably?) but possible for small n

I can do it for the only two groups I can use in applications anyway

arctic tern

it may also help to find a nice set of conjugacy class representatives

everything in SU(n) is conjugate to a diagonal one

Mike Miller

mhm

arctic tern

14:29

centralizers of them should look like S(U(k_1)x...xU(k_r)) I think,

Mike Miller

actually, might be interesting to know this for SL(2,C) too

no idea whether it's useful but who cares

arctic tern

use Iwasawa decomposition to find a nice set of con. class reps for SL(2,C)

Mike Miller

thanks a bunch

arctic tern

I wonder, if an elt of G is in sufficiently "general position" (so, like, not in a discrete center or whatever), then maybe there's a unique one-parameter subgroup through it (up to scaling), and maybe its centralizer centralizes that too

(btw Iwasawa decomp for classical matrix groups is much easier than the wikipedia article seems to make it - I don't even know what a cartan involution is)

Mike Miller

nah, take the diagonal subgroup of $SO(3)$. it is its own centralizer, but definitely does not centralize the circles through each

arctic tern

14:39

true

that's a weird instantiation of klein-four

well, I was talking about individual elements.

Mike Miller

alternatively, in a positive dimensional group, it would be easy to argue that infinite order connected group elements are generic (they're comeager), and the closure of such a thing in a compact group is a positive-dinensional subgroup, which we do indeed centralize

that Klein-4 has an important role in the recent proof attempt of 4-color via differential geometry

arctic tern

wut

Mike Miller

kronheimer and mrowka construct an invariant of embedded trivalent graphs in R^3, closely related to the space of representstions of the fundamental group of its complement into SO(3) (which correspond to flat connections on a certain bundle on a certain orbifold)

the ones that reduce to flat V_4 connections in that subgroup correspond bijectively to Tate colorings of the graph

4CT is known to be equivalent to the statement "every trivalent planar graph without a "bridge" has a Tate coloring"

they can prove that for such graphs their invariant is nonzero; they conjecture that for planar graphs, it counts the number of Tate colorings. essentially it says these V_4 representstions dominate the invariant, somehow

if they can prove that, they've proved 4CT

arctic tern

cool

Mike Miller

stuck on that part tho right now

Balarka Sen

17:26

@Danu Weren't you asking for an example of a SES of holomorphic vector bundle which doesn't split a few days ago? I stumbled upon something which constitutes as an example $0 \to O(-1) \to \Bbb{C}^2 \to O(-2) \to 0$ ($\Bbb{C}^2$ is the rank 2 trivial bundle on $\Bbb{CP}^1$). The map $\Bbb{C}^2 \to O(-2)$ is given by the differential of the quotient map $\Bbb{C}^2 \to \Bbb{CP}^1$ ($T\Bbb{CP}^1 \cong O(2)$).

Sorry for the horrible notation though.

Hmm, I am not sure anymore why that doesn't split. erk. One needs to come up with a section of $O(-1)$ for a contradiction. I thought I had one; oh well. I'll look at it later.

Tobias Kildetoft

@BalarkaSen You mean a global section?

Balarka Sen

Yes.

Global holomorphic section.

Tobias Kildetoft

I thought the ones with negative entries had no global sections (or maybe these things mean something completely different than what I thought)

Balarka Sen

You're correct, they don't; that's why that'd be a contradiction.

Tobias Kildetoft

Ahh, now I see what you mean

Balarka Sen

17:44

Sorry, it's not hard to see it's not split. If $\Bbb C^2$ is isomorphic to $O(-2) \oplus O(-1)$, take a global nowhere zero section of that using triviality of $\Bbb C^2$: that's a map $\Bbb{CP}^1 \to O(-2) \oplus O(-1)$ given by $p \mapsto (f(p), g(p))$. Then $\Bbb{CP}^1 \to O(-1)$ given by $p \mapsto g(p)$ is a fine hol. section.

Just too ill to not say that immediately. Was thinking of nonzero sections.

Krijn

Hey!

Mike Miller

@Balarka Topologically every sequence of vector bundles splits, so you would have just given an isomorphism $\Bbb C^2 \cong O(-1) \oplus O(-2)$. But the first Chern class of the latter is $-3$, so I don't think I believe you.

On the other hand, I agree with what you said. So I'm confused.

Well, there's no quotient map $\Bbb C^2 \to \Bbb{CP}^1$.

Eh, but I see your point.

I'll let someone who knows this stuff figure it out.

PVAL

18:24

@Danu I think Smale looks fine for being 80+ and unshaven.

Danu

Hi!

@PVAL I think he looks pretty okay, too.

I just loved the rage in the comment---not the content so much :P

0celo7

"perelman who looks like jesus on meth"

Agreed.

Danu

More questions... Why can I view a degree $k$ homogeneous polynomial on $\Bbb C^n$ as a section of $\mathcal O(k)$? I see how it descends to $\Bbb P^n$---should I be looking at the transition functions somehow?

I need $f_j=\psi_{jk}f_k$ right, where $f_j$ is the restriction to $U_j$---yes, this solves it.

Krijn

@Danu I do think there are a lot of mathematicians that are not ugly, but just care too little about their looks

Oh, the beards I've seen...

Danu

I agree that some tend to neglect certain aspects :P

Adeek

22:06

hi

usukidoll

22:57

hi

Balarka Sen

23:15

@MikeMiller But the splitting won't be holomorphic, is the point, not? They are isomorphic as 4-plane bundles - not as holomorphic vector bundles.

Adeek

@BalarkaSen good your here

I have a quick questions

Balarka Sen

Oh, you're saying you don't believe they are isomorphic as 4-plane bundles. Hm.

I guess I don't really understand Chern classes to say anything: I thought they were for complex line bundles.

Adeek

@BalarkaSen here ?

Balarka Sen

Go ahead.

Adeek

Suppose we consider $S_p$ I want to calculate the number of p subgroups of $S_p$

I can see that only p cycles

Balarka Sen

23:18

@MikeMiller Yeah, that was the bad notation. By $\Bbb C^2$ I meant the complex vector space this time, not the 2-plane bundle.

Adeek

will be a subgroup of order p

but why

Balarka Sen

I apologized for it.

@Adeek Not really interested in doing algebra right now

Mike Miller

@BalarkaSen No, the problem is that 0 doesn't map to anything is what I was complaining about.

Balarka Sen

Oh sorry. $\Bbb C^2 - (0, 0)$, I meant.

Mike Miller

the bundle $\mathcal O(-1) \oplus \mathcal O(-2)$ is definitely not topologically trivial. I was confused about the topology.

Balarka Sen

23:26

Doesn't that contradict the fact that the SES I wrote down topologically splits?

Mike Miller

Yes. So you understand my confusion.

Balarka Sen

Weird.

shai horowitz

things get dangerous when mathematicians say "Weird."

Mike Miller

It just means I don't believe his sequence is all.

shai horowitz

by the way i'm not really contributing to this conversation have not been around for long enough to know what exactly you guys are talking about

usukidoll

23:29

how do I prove this? http://prntscr.com/cku2iq
that's like the matrices of 2 x 2 is the characteristic of R which by definition there's a least positive integer m such that $m \cdot a=0$ for all $ a \in R$ and if there's no such m then the characteristic of R is 0. ANy hints?

Mike Miller

No worries.

Balarka Sen

I wonder if it should have been $O(2)$ instead of $O(-2)$ at the end.

shai horowitz

does $e^Jx =-1$ where J is the Hamiltonian unit vector?

Adeek

hey

usukidoll

.-.

shai horowitz

23:44

hey

and i meant Quaternion unit vector

$e^([{0,1},{-1,0}]x)$ plugged into mathematica and it doesnt seem like it

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