Mathematics - 2016-10-13 (page 3 of 5)

1:00 PM

@0celo7 I guess it should have been obvious to me that those are actually the curvature of the lines. Just writing down the definition makes it fall out. Thanks.

shai horowitz

@abenthy give the question a shot dont ask to ask just ask

Balarka Sen

@0celo7 Curvature of a curve is the reciprocal of the radius of the circle which fits it the best?

At that point.

0celo7

No, of a surface

DHMO

@BalarkaSen what's the curvature at the middle point of a saddle? infinity?

0celo7

you can compute the curvature by looking at the circumference of little circles on the surface

Balarka Sen

1:02 PM

@DHMO -1

abenthy

Can latex be used here? $1+1$ test

DHMO

@BalarkaSen wait, negative curvature?

Balarka Sen

@0celo7 No, not yet.

@DHMO Yep.

DHMO

@BalarkaSen how do they work?

Balarka Sen

I am not sure what that question means.

DHMO

1:03 PM

what does negative curvature even mean?

how can a circle have a negative radius?

0celo7

@BalarkaSen I must admit, it's not obvious to me off the top of my head why the eigenvalues give the curvatures of those lines.

But I know that's the correct interpretation

shai horowitz

@abenthy tinyurl.com/cfqcvpc for latex

0celo7

But maybe Ted has some more equations that I don't remember.

Balarka Sen

Just write down a proof, @0celo7. You can do it.

@DHMO A circle doesn't have negative curvature... in any case we're talking about curvature of surfaces here.

abenthy

Okay, so for a space X with fundamental group pi, universal covering X~ and a Z[pi]-module V one defines the homology with local coefficients in V of X as H_n(X;V) = H_n(C_*(X~) tensored with V over Z[pi]). Now my question is, is there a long exact sequence ... -> H_n(A;V) -> H_n(X;V) -> H_n(X,A;V) -> ... for pairs (X,A) ?

0celo7

1:05 PM

@BalarkaSen Probably not. I know the shape operator off the top of my head in terms of covariant derivatives, it would get very messy :P

Balarka Sen

Negative curvature at a point intuitively means the surface curves away from itself at that point.

abenthy

Because I have a construction of this in mind, but at one point, it doesnt work...

DHMO

@BalarkaSen curves away... from itself??

Balarka Sen

@DHMO Or rather, curves away from itself at two different directions.

shai horowitz

it doesnt look like i can help @abenthy but for refrence latex does work you just need to render it with javascript the links can be found in the top right

DHMO

1:08 PM

@BalarkaSen I see

Balarka Sen

Just take the saddle point as a local model for negatively curved surfaces.

Otherwise understand the definitions.

abenthy

@shaihorowitz okay, thanks for the note

Balarka Sen

If you take that as a local model, that should tell you - intuitively - the hyperboloid is negatively curved at certain points, the torus is negatively curved at two points, the pseudosphere is negatively curved, etc.

@0celo7 I wonder, to what extend is that true? Suppose I have two surface and they have the same constant curvature near a point of each. Does that mean they are locally isometric near those points?

0celo7

Relevant

Balarka Sen

@0celo7 I don't see how. I said local isometry, that is, there exists an isometry between two parameterized subsurfaces of the surface containing the points of interest.

Danu

1:21 PM

@BalarkaSen heyo

0celo7

@BalarkaSen I don't think there are local isometries between the middle parts of the cigar there

Balarka Sen

@Danu Hi.

Danu

So since SW numbers are intersection numbers, I'd like to understand a geometric idea of what SW classes are

So that I can see why it should say something about cobordism

0celo7

@BalarkaSen I have to run to class, bye

No, you're right.

I'll explain later.

Balarka Sen

@Danu I don't think their Poincare dual is easier as a geometric model here. The idea is the following: take a cobordism $W$ from $M$ and $N$ - then $TM$ is stably $TW|_M$ (by direct summing with the trivial vector bundle that gives the normal field) and $TN$ is stably $TW|_N$.

So morally $TM$ and $TN$ should have the same "Stiefel-Whitney things", as they are stably restrictions of the same bundle. The formalization of that is SW numbers.

Danu

1:24 PM

I know the proof :P

Balarka Sen

@Danu The proof in itself gives intuition for why SW numbers give bordism invariants to me.

Danu

Yeah, I guess your way of phrasing it is suggestive

The emphasis on SW classes as being stable

Thanks

Balarka Sen

:) Glad that helped.

I think it's an interesting question what the PD's of SW classes are. I don't know the answer. Ted asked me that for $w_1$, which is easy. Also $w_n$ is PD to Euler class.

Mike asked for $w_2$, but I haven't thought about it.

Danu

Don't know Euler class :(

0celo7

@BalarkaSen Does the hyperbolic space have a transitive group of isometries?

Balarka Sen

1:28 PM

@Danu I'll phrase it differently for a special case: if $M$ is $n$-dimensional, $w_n(M) = \chi(M) \pmod{2}$.

This generalizes to any rank $n$ bundle by Euler class.

@0celo7 Interesting question.

0celo7

@BalarkaSen If you can prove th statement for hyperbolic space, it will be true in general

Balarka Sen

Since I can lift to universal cover?

Danu

Whoa, Bob Dylan winning the Nobel for literature

2

awwwyis

0celo7

Exactly

Balarka Sen

@Danu Oh wow

Danu

1:31 PM

Bold move by the committee

Balarka Sen

great decision

0celo7

It takes some proving, but any constant curvature surface has a covering isometric to one of the model surfaces

Balarka Sen

@0celo7 Isom(H^2) does acts on H^2 transitively.

0celo7

It's clearly true for the sphere and plane

@BalarkaSen but does it have the property we want?

Balarka Sen

Which property?

0celo7

1:33 PM

Namely given $p,q\in H^2$, there are open sets $U,V\subset H^2$ containing $p$ and $q$, respectively, and an isometry $U\to V$.

that might be automatic

Balarka Sen

I just said Isom(H^2) acts on H^2 transtively. That means, by definition of transitivity, there is an isometry taking any point to any other.

0celo7

Yes, yes

Ok, so what you said is true

Mike Miller

@Balarka I never asked what $w_2$ is pd to. Just what it does.

Balarka Sen

Ah, ok, misremembered.

0celo7

@BalarkaSen Have you heard of the "geodesic deviation" equation?

Balarka Sen

1:38 PM

nope

0celo7

It's a tool physicists use to describe the spread of geodesic packets

The spread depends on the sign of the curvature

that's another way to understand it

s.harp

do you mean the jacobi equation/jacobi fields?

Danu

@BalarkaSen Also, is this doable if I just know the axioms, not how they're actually constructed?

Mike Miller

You will never ever get anything out of the construction.

Balarka Sen

I read the construction/proof of existence from Hatcher, it wasn't exciting.

Danu

1:42 PM

That's weird

But I'd have no idea how to think about its Poincare dual and stuff with just the axioms

I mean... The axioms only tell you something about the classes of $\Bbb RP^n$

Balarka Sen

It shouldn't be hard to see what $w_1$ is PD to from the definition.

By not hard, I mean it should be doable from the axioms.

Mike Miller

I mean, you probably want to know an iota more about w_1. But sure.

Danu

I guess I'm either being dumb or you know something I don't know

Balarka Sen

Why do you think you're dumb if you can't solve a problem a minute after it's stated?

Danu

You repeatedly said it's easy plus I just need the axioms

Mike Miller

1:48 PM

It took him half a day or something.

Balarka Sen

^

I explained what I meant by easy.

Mike Miller

And I also think he had more info than you did.

Balarka Sen

Hmm.... actually maybe you need stuff from the next section.

classifying vector bundles by maps to Grassmannians, that is.

Danu

lol

okay

0celo7

@s.harp yes

Balarka Sen

1:53 PM

Maybe I should ponder on what $w_2$ means.

Mike Miller

Compare to $w_1$, too.

shai horowitz

i just need to say something to get rid of the astrix

Secret

2:31 PM

Work in progress algebraic structure. Found no contradictions so far, unless there is still something overlooked:

Interesting properties include:
1. The + cayley table forms a subsemigroup which is isomorphic to the left zero semigroup
2. There exists an element $a\in\textrm{WIP(3)}$ such that $(aa)a\neq a(aa)$
3. Distributive and + associative laws are controlled by the subsemigroup
4. The only 6 cases of nonassociativity can be combined into one rule

shai horowitz

what is the absorb-er for addition on the extended reals?

Secret

Extended real number line

In mathematics, the affinely extended real number system is obtained from the real number system ℝ by adding two elements: + ∞ and – ∞ (read as positive infinity and negative infinity respectively). These new elements are not real numbers. It is useful in describing various limiting behaviors in calculus and mathematical analysis, especially in the theory of measure and integration. The affinely extended real number system is denoted R ¯ {\displaystyle {\overline {\mathbb...

$\pm\infty$ almost does that except it does not absorb itself

shai horowitz

it kind of does but it cant absorb it negative

@secret i think there may be a contradiction

you have (00) = q

q0=1

(00)0=1

(00)(00)=0=qq=1*0

(00)(00)(00)=qqq=?

Secret

2:52 PM

Hmm, I have no idea, the most nonassociative systems I have worked with in the past are lie algebras which they still have the Jacobi identity to breaks things apart...

shai horowitz

in short the 0*q vs q*0 symmetry breaking is weird

Secret

But clearly, there seemed to be no rules on how brackets should be handled for the expression (00)(00)(00)

shai horowitz

yeah i simply constructed it to show the table only tells us until we multiply 3 things together

Secret

It does seemed the sturcture can get not closed very quickly unless there's some extra axiom to control the growth of the nonassociative terms...

(well, at least we don't get cases like 0=1, 0=q and 1=q)

Secret

Btw what is a magma with identity and not all elements with inverses called?

Danu

3:00 PM

@BalarkaSen what story did you end up translating?

Balarka Sen

@Danu "House of Asterion" by Borges. I asked; translating already translated stories doesn't seem to be an issue.

shai horowitz

loop i think

Danu

@BalarkaSen But it ain't right :P

Secret

ok

Balarka Sen

Yeah, it's a bit odd. But w/e, as long as I have finished my work with the least effort.

Anubhav

3:04 PM

@BalarkaSen

hii

Balarka Sen

hi

Danu

@BalarkaSen High school spirit is strong in this one :D

Anubhav

I went to my home, after a week, I again came back here :p

6

Q: 3-manifold examples of homomorphisms between fundametnal groups that are not induced by continuous maps

I am looking for some examples of closed, orientable 3-manifolds $M$ and $N$ and a homomorphism $\phi : \pi_1(M) \to \pi_1(N)$ such that $\phi$ is not induced by any continuous map $f : M \to N$. Are there examples of this occurring for lens spaces? I do not believe that this phenomenon can occ...

Balarka Sen

@Danu except that I want my effort to be spent on mathematics instead, not dating.

Anubhav

this question looks interesting

Danu

3:06 PM

@BalarkaSen that's just you being insane

:)

HAVE YOU NO HORMONES? ;)

shai horowitz

judge not anothers utility function lest yee be judged

Danu

Judge me baby

Balarka Sen

@Anubhav fun.

Lozansky

Reminds me of optimizing weight vectors for utility functions in matlab

Balarka Sen

@Danu meh

shai horowitz

3:09 PM

you complain to much about your position of authority that many would kill for just because you need to condense information @danu

feel teh burn

Danu

hahaha

> position of authority

I wish :'(

shai horowitz

I feel you there haha, its more homework not more power

power = the ability for one to accomplish ones utility function

thats why the utility companies make bank

Secret

Non-associative algebra

A non-associative algebra (or distributive algebra) is an algebra over a field where the binary multiplication operation is not assumed to be associative. That is, an algebraic structure A is a non-associative algebra over a field K if it is a vector space over K and is equipped with a K-bilinear binary multiplication operation A × A → A which may or may not be associative. Examples include Lie algebras, Jordan algebras, the octonions, and three-dimensional Euclidian space equipped with the cross product operation. Since it is not assumed that the multiplication is associative, using parentheses...

Hmm, see if any 4 element rules can help me out here... NB just proved jacobi identity is not obeyed by WIP(3) due to the + left zero subsemigroup basically causing only the leftmost term to survive

Danu

hahaha

You sound like a funny guy (gal?)

Evinda

Hello!!!

Let $L=\{ (x,y): x^2+y^2<5\}$.
Suppose that we have a function $v(x,y)$ that is harmonic in $L$ and $\Omega \subset L$ an arbitrary space such that $(0,0)$ belongs to that space.
Suppose that $v$ is equal to $1$ at the boundary of $\Omega$.
How can we compute the integral $\int_{x^2+y^2 \leq 4} v(x,y) dxdy$ ?

First I thought that we could use the second version of the mean value theorem. But then we couldn't use the fact that $v$ is equal to $1$ at the boundary of $\Omega$. Could we? If not, could you give me a hint how else we could compute the integral?

Mary Star

3:15 PM

Hello!!

Does the equation $x^2+2y^2=1$ have non-zero solutions?

shai horowitz

guy, good not assuming @ danu

DHMO

@MaryStar both terms of the left hand side are non-negative

(assuming that both variables are real)

Danu

@shaihorowitz So your profile says you work on math: What kind of math are you interested in?

Secret

The solutions are those on the ellipse formed at z=1

Mary Star

What do we get from that? @DHMO

Evinda

3:19 PM

Hello @DanielFischer
Did you see my question? Do you have an idea?

shai horowitz

i only published once and it was in non linear dynamics, fell out of acadamia due to depression just getting back in. research solo time i was alone mostly on discrete maths

Danu

Okay

DHMO

@MaryStar are the variables limited to integers?

Mary Star

We have that $x,y\in \mathbb{Q}$. @DHMO

shai horowitz

@Evinda your approach sounds correct,

DHMO

3:24 PM

@MaryStar let x=a/b and y=c/d, then we have $(ad)^2 + 2(bc)^2 = (bd)^2$

shai horowitz

fun fact euclids elements 4 axioms were not nearly enough as shown by godel an example of a missing axiom is that a line contains at least two points. a second is the axiom that space is continuous and any circles that look like they over lap do in fact overlap at a point, any else think of any missing axioms

Danu

@shaihorowitz I'm personally not so interested in the foundational matters

DHMO

@MaryStar $a^2 + 2(bc/d)^2 = b^2$

I don't think my approach is helpful

shai horowitz

@danu what's your thing?

DHMO

please find others sorry

Danu

3:26 PM

Geometry/topology of manifolds

shai horowitz

@MaryStar what was your question

what overall topology would you give the universe?

Balarka Sen

K, got physics done.

Mary Star

My question is if the equation $x^2+2y^2=1$ has non-zero solutions. @shaihorowitz

Balarka Sen

Back to math.

shai horowitz

@marystar are we working in the real numbers?

Mary Star

3:28 PM

We have that $x,y\in \mathbb{Q}$. @shaihorowitz

Danu

@shaihorowitz Topology? No idea... Black hole singularities n shit make it hard

Balarka Sen

@Danu A nice fact is that the origin of hyperbolic geometry was in the proof that Euclid's fifth axiom cannot be proved from the four others. Just sayin' - it's a nice example of concrete things coming from foundational logic.

Danu

I knew that

shai horowitz

well consider the following cases x,y negative. x negative y positive, xy both negative, x zero y positive, x zero y negative xy both zero. @MaryStar

Balarka Sen

cool

Danu

3:30 PM

I don't know anything about Riemannian geometry, sadly

Balarka Sen

me neither

Danu

I saw the problem sets for last semester's course on it and it looked super cool

Mike Miller

@Anubhav Nice question. There's at least one example. I suspect when the codomain is a connected sum examples are extremely easy to come by.

Danu

when I did it we were just doing handicapped characteristic classes

shai horowitz

do you think the universe is discrete or continuous?

Danu

3:33 PM

I don't think that's a meaningful question yet

Balarka Sen

lol... disjoint union of uncountably many holy grails?

shai horowitz

fair enough but it will be,

uncountably many white holes

Danu

I'm not sure it will be, any time soon.

Balarka Sen

@Danu Actually, now that I think about it, there is an easier way to understand $w_1$ than classifying vector bundles.

shai horowitz

do you realize how small a transistor is

Krijn

3:35 PM

no

How small is a transistor

shai horowitz

7 atoms thick

Evinda

@shaihorowitz The theorem is the following:

If $u$ is harmonic in $\Omega$ and $\overline{x} \in \overline{\Omega} \setminus \partial{\Omega}$ then $u(\overline{x})=\frac{n}{ w_n r^n} \int_{|\overline{x}-\overline{\xi}| \leq r} u d{\overline{\xi}}, r \in (0,d), d=dist(x, \partial{\Omega})$.

So we have $u(\overline{x})=\frac{3}{ 4 \pi \cdot 8} \int_{|\overline{x}-\overline{\xi}| \leq 4} u d{\overline{\xi}}$.

For $\overline{x}=(0,0)$ we have $u(0,0)=\frac{3}{32 \pi} \int_{|\overline{\xi}| \leq 4} u(\xi_1, \xi_2) d{\overline{\xi}}$.

Danu

@shaihorowitz Do you understand the issues we're facing with quantum gravity right now (I sorta do, because I'm a high energy physicist by training!)? It's tricky!

shai horowitz

yeah but the trickyness makes it fun, besides i believe in quantum immortality so i got some time to figure it out

Danu

@BalarkaSen Tell me about it

Mary Star

3:38 PM

- $x=0, y>0$
We have $2y^2=1 \Rightarrow y=\frac{1}{\sqrt{2}}\notin \mathbb{Q}$. So, this case is rejected.

- $x=0, y<0$
We have $2y^2=1 \Rightarrow y=-\frac{1}{\sqrt{2}}\notin \mathbb{Q}$. So, this case is rejected.

- $x=y=0$
This case is not possible, since that would mean $0=1$.

Are these cases correct?

What can we say about the cases:
- $x,y<0$
- $x<0, y>0$
- $x>0, y>0$
? @shaihorowitz

shai horowitz

try examples @MaryStar the point is the squaring makes everything positive, and yeah the cases look good so far

@Evinda i'm out of my depth so i'm prone to error by the way but it looks right

Evinda

@shaihorowitz Ok. And how do we use the fact that the given harmonic function equals to 1 at the boundary of $\Omega$ ?

shai horowitz

@MaryStar forgot case y=0 x positive... trivial solutions....

Balarka Sen

@Danu I'll tell you the statement. $w_1(E) = 0$ implies $E/M$ is orientable, and vice versa. I don't think my easier argument is complete yet, but it's fiddling with the determinant bundle. Maybe you'd want to think about that.

Lozansky

Is $Log(z^2) = 2Log(z)$ ?

I'm guessing no

shai horowitz

3:47 PM

@Lozansky yeaeh

Adeek

@BalarkaSen

hi

Lozansky

It's not, right?

shai horowitz

@Lozansky thats how logs work

it is right

Lozansky

Sorry forgot to add

$z \in \mathbb{C}$

And Log is the principal branch of log

The problem is that $2Arg(z)$ might be outside of $(-\pi, \pi]$

shai horowitz

@Lozansky it seems right at least up to a complex conjugate

Krijn

3:51 PM

Probably mod $2 \pi$ or something

shai horowitz

yeah that would be my guess, or a way of defining a principle value

Lozansky

Hmm

Wait

Krijn

Reading the wiki, it seems to be harder than that

0celo7

@BalarkaSen I was going to read Hatcher but now I want to read Kreyszig

Lozansky

I have a counterexample

shai horowitz

3:54 PM

?

Balarka Sen

Hi @Adeek

@0celo7 Do whatever that makes you happy

0celo7

@BalarkaSen Well I want to reread it because you're reading about surfaces

Balarka Sen

Kreyszig's which book?

0celo7

I remember he defines Gaussian curvature directly in terms of the curvature of the curves through the point

very concretely

then shows it's the determinant of the "shape operator"

Lozansky

$$z^2 = (-z)^2 \rightarrow 2Log(z) = 2Log(-z) \rightarrow Log(z) = Log(-z) \leftrightarrow z=e^{Log(z)} = e^{Log(-z)} = -z+2 \pi i k$$

0celo7

3:56 PM

@BalarkaSen Diff geo

Mike Miller

do whatever makes you happy

Adeek

hey @BalarkaSen so I just have a quick question suppose I proved that

0celo7

@MikeMiller Ok then I will play video games

Mike Miller

gooed

Adeek

$Aut(D_{2n}) \semieq Z_n \rtimes Aut(Z_n)$

here the above is isomorphism

I proved that

Balarka Sen

3:57 PM

@Adeek Don't feel like thinking about group theory today. Ask arctic maybe.

Adeek

@arctictern here ?

Mary Star

We have the following:
$x^2=1-2y^2\Rightarrow x=\pm \sqrt{1-2y^2}$

$1-2y^2\geq 0\Rightarrow 2y^2\leq 1 \Rightarrow y^2\leq \frac{1}{2}\Rightarrow -\frac{1}{\sqrt{2}}\leq y \leq \frac{1}{\sqrt{2}}$

For $x,y<0$, what rational numbers are there for example in the interval $(-\frac{1}{\sqrt{2}},0)$ ?

@shaihorowitz

Adeek

oke

Lozansky

@Krijn @shaihorowitz

See above

Adeek

@arctictern I would like to determine the map $\psi$ that gives us the semi-direct product above

shai horowitz

3:58 PM

@Lozansky yeah as i said up to the complex conjugate and principle value k=1 usually