Mathematics - 2016-08-05 (page 1 of 3)

uoɥʇʎPʎzɐɹC

00:49

What's the name of the function that's f(x) = 1 + 2 + ... x ?

A Nerd - I

01:08

triangle numbers

Balarka Sen

01:22

@Akiva Yeah, but I am not convinced degree won't contribute to # of connected components.

Joshua Ciappara

05:01

I have a question which is so simple I don't really want to post it. Given a scheme X with global section s, suppose the stalk of s is zero in every residue field. Should s = 0? If so, why?

Joshua Ciappara

05:21

Nevermind, it's obvious when you look at it affine-locally.

Tobias Kildetoft

07:45

Cool. I got a facebook message from a friend that he had noticed my name mentioned in one of the talks at the conference he was at. So there are actually people out there who have heard of my work.

Mithlesh Upadhyay

I have not researched for any poster. This poster seems to me when he/she posted. He/she got 5 likes and more than 3 upvotes in 2-3 minutes without having any attempt?

http://math.stackexchange.com/questions/1883140/an-integration-of-product-1-xn

Balarka Sen

08:51

@TobiasKildetoft Very nice. You're quickly gathering fame and money, then? :D

Tobias Kildetoft

@BalarkaSen Possibly fame, but not much money yet (I am highly confident that this too will come soon).

Balarka Sen

Good. You owe us a treat when it comes. :P

@Semiclassical I just realized that the wave equation comes from the Hooke's law. I am sort of excited by this observation, but it's probably very trivial.

Actually I am not yet quite sure how to formulate the picture rigorously, but it should just be motion of a spring after you lift some force from it. I'll think about it a bit.

Hi @arctictern

arctic tern

09:08

hi

Balarka Sen

What's up?

arctic tern

watching the man from nowhere and mr robot

enjoying random fruity vodka cocktail

Balarka Sen

huh, never heard of those two

Tobias Kildetoft

Nice description "We can also think of e as a roadmap for a gentle stroll through the Bruhat graph (with much pausing to admire the scenery)"

My internet connection is terrible, and chat is not very good at figuring out whether it managed to send a message.

Balarka Sen

Bad internet connection is annoying, yes

Agawa001

09:35

@akiva you cant escape the fact one from either arabic or hebrew plagiarized from the other i think hebrew is the older right ?

Agawa001

10:03

hey @rob

Huy

10:20

mine's been terrible too the last few weeks, I surely will change provider when I move flats

robjohn

@Agawa001 hi

Agawa001

@robjohn how are days

robjohn

@Agawa001 still around 24 hours...

Agawa001

@robjohn i believe they turned out shorter

even with short nights not as in the two poles

Secret

11:26

Hi I am trying to do a proof by contradiction for the statement "a set of vectors of size < dim(V) where V is the vector space are always linearly dependent", but it seems I can get a contradiction to arise

Consider the following set S={v1,v2,v3,v4} and S is a subset of V and dim(V)=3. Suppose S is linearly independent, then there exists li such that

l1v1+l2v2+l3v3+l4v4=0 implies li=0 for i=1,2,3,4

Now suppose we have a basis set {a1,a2,a3} of V. Then any vi=sum_j cij aj where cij not necessary all zero is in the reals (as a zero vector cannot form a basis set). Since it is a basis, therefore the vectors in them are linearly independent i.e. p1a1+p2a2+p3a3=0 implies pj=0 for j=1,2,3

Therefore

and then using the assumption I got 0=0, which does not lead to any contradiction

Typo: I cannot get a contradiction to arise

Hippalectryon

11:39

@Secret surely you mean >dim(V) ?

Secret

yeah, I am trying to prove a set S of vectors with size > dim(V) must be linearly dependent via proof by contradiction, but I cannot seemed to get a contradiction to arise

Hippalectryon

Oh ok (you wrote <dimV earlier)

Secret

The best I get is that the ls can be not all zero, but all ls=0 is still a solution, therefore I am not sure if I have already caused a contradiction

Hippalectryon

I'm gonna be busy for 30-ish mins, I'll be back later if no one has helped you

Secret

ok

Huy

12:02

when does one use algebraic vs algebraical, topologic vs topological, geometric vs geometrical?

PVAL

12:44

I just use algebraic, topological and geometric.

Huy

why? when does one add al ?

Hippalectryon

12:58

@Secret back

@Secret I haven't looked at your proof yet, but afaik the direct proof is much simpler (as opposed to a proof by contradiction)

Why chose a proof by contradiction ?

Secret

I kinda forgot the sketch of the direct proof, i.e. do I start with a basis set {v1,v2,v3} and show that v4=linear combination of v1 v2 v3?

One reason I tried a proof by contradiction is I am trying to see "a set S of vectors with size > dim(V)" must lead to "linearly dependent" with no other possibility (e.g. is there exists a result that "linearly independent" such that "a set S of vectors with size > dim(V)" leads to "linearly independent"). Because if this is true then any assumption of other possibilities should lead to a contradiction thus disproving the existence of such possibilities

Hippalectryon

13:13

Oh for the direct proof you don't even need a basis

Balarka Sen

@Huy because "topologic" sounds weird.

Huy

who says that

Balarka Sen

ask any normal people

Huy

in German it's always algebraisch, topologisch, geometrisch

no complications

sorry I don't know any normal people

Balarka Sen

in german, though

Hippalectryon

13:21

@Secret Owait nvm :( give me a min

Indeed you need a basis but it's fairly simple

@Secret Have you seen matrices ?

Secret

yes i do

Hippalectryon

And the Rank–nullity theorem ?

0celo7

@BalarkaSen hey

Balarka Sen

Hi

0celo7

Suppose I have a smooth function $f:S^2\to M$, and I identify this with a function $f:\Bbb R^2\to M$ via stereographic projection from the north pole

If I give $\Bbb R^2$ the coordinates $(x,y)$, is there a reason why $\partial_xf=\partial_yf=0$ "at infinity"

Hippalectryon

13:31

@Secret Take a set {v1,...,v4} and the matrice A=[v1,...,v4] (the matrice whose columns are v1,...,v4) then its rank is less or equal to 3 by the rank-nullity theorem therefore the set is linearly dependant

Balarka Sen

You're restricting $f$ to $S^2 - p$ where $p$ is the north pole, and then composing with the homeomorphism $\Bbb R^2 \to S^2 - p$, that is to say?

0celo7

@Hippalectryon By matrice do you mean matrix?

Hippalectryon

@0celo7 Ah yes my bad

0celo7

@BalarkaSen Diffeomorphism, but yeah.

Balarka Sen

@0celo7 That's false in general, so I am not sure what you mean.

0celo7

13:33

@BalarkaSen typo

I basically need $\partial_x f=\partial_yf=0$ "at the north pole"

Which is "at infinity" on the plane

Balarka Sen

Once $f$ is a function from a chart missing the north pole, I don't know what derivatives being 0 at north pole is supposed to mean.

0celo7

You can probably put that point back by continuity

Secret

@Hippalectryon ok I understood the case where rank(A) < 3 (because then nullity(A) > 0 by rank nullity), but for the case where rank(A) = 3 would it cause a problem in that for the cases where rank(A) then the statement "a set of vectors of size > dim(V) where V is the vector space are always linearly dependent" will be false in general because there exists counterexamples in the form of matrices with rank 3?

Balarka Sen

@0celo7 That extends to the original map $f : S^2 \to M$. That may or may not have zero derivative on the north pole...

Secret

Soorry keep making that copy paste typo...

Balarka Sen

13:36

In that sense, it's false.

0celo7

It's apparently supposed to have zero derivative, which is confusing me

Balarka Sen

Maybe your map is something specific.

0celo7

No, it's general

Hippalectryon

@Secret I don't see the problem. When rank(A)=3 then by definition A has exactly three linearly independent columns

Balarka Sen

I don't know the context, so I can't help.

What you said so far is either vague or false.

0celo7

13:38

You're right, it is vague.

But that's all I have

I might have to track down the paper they mention

Balarka Sen

OK.

I read a bit about framed cobordism from Milnor. Cool stuff.

0celo7

Yup

Balarka Sen

I got back some motivation and started on Poincare-Hopf.

0celo7

Now they invoke Riemann-Roch. I refuse to learn algebraic geometry, book

Balarka Sen

I have yet to decide if I want to follow Milnor's proof on that one or G-P's. I don't really like the Lefschetz fixed point theory business.

0celo7

13:40

Milnor's proof is pretty sketchy at times, tbh

And it requires Morse theory

not so much Riemannian geometry

I take it back, you need geodesics

Balarka Sen

:S

0celo7

But nothing horrible like the variational theory

Secret

When rank(A)=3 then by definition A has exactly three linearly independent columns -> <insert a deduction or sentence> ->therefore since the set has 4 vectors, it must be linearly dependent

I am not sure if I overthink this, but somehow I felt I am missing a step in the logical flow

0celo7

This Ricci flow book randomly invoked Morse theory on the path space

my face kinda looked like :S

Balarka Sen

Maybe it won't hurt to learn all of that but given the amount of prereqs it seems better to learn the apparently easier proof from G-P first.

0celo7

13:42

yeah

Balarka Sen

Grr, I don't feel like reading Lefschetz fixed point theory.

Hippalectryon

@Secret By definition of the rank, if a matrix has exactly $n$ linearly independent vectors then it has a rank of n

Balarka Sen

I need more motivation for that.

0celo7

why not?

also holy crap look at the time

Hippalectryon

If {v1,v2,v3,v4} was a linearly independent set then rank(A)=4

Balarka Sen

13:43

I am not sure. Maybe because studying fixed points for maps between manifolds doesn't excite me as much, but I am to blame for that, if any.

0celo7

To fully understand this section it seems I need Shoen & Yau's book on Harmonic Maps

I am not ready for that

Balarka Sen

rip

Hi @JuanFran

Juan Fran

hi @BalarkaSen

0celo7

@BalarkaSen they use complex algebraic geometry and harmonic map theory to prove a sphere theorem

Balarka Sen

"rip" is then even more applicable

Secret

13:45

@Hippalectryon ok I thinnk I see it now, thanks

0celo7

oh and the Hurewicz theorem

whatever that is

Hippalectryon

@Secret :-)

Juan Fran

@0celo7 what is so surprising about this?

complex AG has very strong methods

0celo7

I don't know anything about it

Balarka Sen

@0celo7 if $\pi_k = 0$ for all $k < n$, then $\pi_n \cong H_n$.

0celo7

13:47

cf. Hatcher, I assume?

Balarka Sen

well, given $n > 1$.

Yep.

it's all in there

0celo7

Something I find annoying about these sphere theorem things is that they always give the pinching constants as decimals

Juan Fran

i think you Need an abelianization for n=1

0celo7

0.87 seems really random

Balarka Sen

Yes, @Juan.

0celo7

13:48

but it's probably the solution of $\tanh \pi x=\sin 2\pi x$ or something weird

Balarka Sen

$H_1$ is always isomorphic to $\pi_1^{ab}$.

Juan Fran

very true. it might be this Statement to which the book is referring to

idk

i guess there are more theorems called "Hurewicz thm"

Balarka Sen

huh, odd, because this is the easiest version of the theorem

0celo7

This proof is pretty crazy. They use existence theorems from harmonic theory, some AG theorem on complex line bundles, Poincare duality, universal coefficient theorem, Hurewicz theorem, morse theory

h cobordism

Balarka Sen

What is le theorem

Juan Fran

13:52

true it is the easiest version, but it is very basic and useful

Balarka Sen

sounds like a good one, given the amount of machinery they use

@JuanFran sure, fundamental stuff are useful

Juan Fran

so how do these things interact

i wanna see that proof

Balarka Sen

but I don't think it's worth giving it the name "Hurewicz theorem". it's just what homology is.

0celo7

Let $(M,g)$ be a compact, simply connected Riem. manifold with dim $n\ge 4$. Suppose that $\mathrm{Riem}(\zeta,\eta,\bar\zeta,\bar\eta)>0$ for all $p\in M$ and all lin. indep. vectors $\zeta,\eta\in T_p^\Bbb CM$ satisfying $g(\zeta,\zeta)=g(\zeta,\eta)=g(\eta,\eta)=0$. Then $M$ is homeomorphic to $S^n$.

Balarka Sen

By Riem(...) do you mean curvature?

Juan Fran

13:54

Riemann tensor

0celo7

Yes.

Balarka Sen

Cool. Sounds like a nice theorem.

0celo7

You use Morse theory to show that it's $(n-1)$-connected.

Juan Fran

actually i don't find it surprising that a proof uses poincare duality and UCT. arguably these are the most useful general theorems

in particular, UCT just gives base change

Balarka Sen

Do we know something analogous for diffeomorphism?

Juan Fran

13:55

so important in Algebra in general

Krijn

Hey all

0celo7

@BalarkaSen Yes.

Balarka Sen

Hi @Krijn

0celo7

That's the point of this book, actually.

Balarka Sen

Ah?

0celo7

13:56

Let $(M,g)$ be a compact Riem. manifold of dimension $n\ge 4$ which is strictly $1/4$ pinched in the pointwise sense. Then $M$ is diffeomorphic to a spherical space form.

Balarka Sen

I don't know what "strictly 1/4 pinched" and "spherical space form" means, admittedly

Well, neither do I know what any of Riemannian geometry means, so...

0celo7

Strictly $1/4$ pinched means the sectional curvature is in $(1/4,1]$ at each point

Balarka Sen

Ah.

0celo7

then there's a result about weak pinching, which is just $[1/4,1]$.

$\partial_t g(t)=-2\mathrm{Ric}_{g(t)}$

Such a simple PDE

Juan Fran

simple?

highly non trivial!

0celo7

14:05

Looks simple

Juan Fran

true dat

but Ric, Riem, etc are quite nontrivial

0celo7

I know

I've studied general relativity intensively

Juan Fran

nice

i never bothered too, i guess lack of time

and if you understand the math behind it already it is probably less interesting

i guess most people start learning differential geometry because they want to learn general relativity, and not conversly

0celo7

If you know Lorentzian geometry already, then you're a weird person :P

Lorentzian is so much different than Riemannian

For instance, there is no distance function to be obtained from the metric

Juan Fran

true, but you can make up a generalization of both (just a general Framework, of course the results are different)

0celo7

14:10

If you define the distance as an inf, you always get 0

If you define it as a sup, you generally get infinity

Studying when the distance function blows up is interesting

Juan Fran

is it physically relevant?

0celo7

Then there is the Geroch theorem, which is probably one of the coolest theorems

Juan Fran

i mean what does it mean in physics

that this happens?

0celo7

@JuanFran Eh

Our universe is thought to be pretty well-behaved

There is a conjecture which states the universe is globally hyperbolic

Which is a very boring topology

Juan Fran

when i look up geroch's Splitting theorem

they mention "cauchy surfaces"

not sure how they made up that terminology :D

but it looks like a very nice theorem

0celo7

14:14

It's from PDE theory

Juan Fran

basically sayinig that space time is a sort of fibre bundle

i see

0celo7

The usual way of thinking about it is that spacetime has a global time function

Juan Fran

well PDE's is not something i know a great deal about

0celo7

and that allows you to split it into a "time" and "space" in a topological sense

@JuanFran Basically, if you know the metric and its derivatives on a Cauchy surface + boundary conditions, you can solve the Einstein equations

Juan Fran

well that is something i actually find very interesting

it is the sort of physics i can understand (i.e. maths :D )

0celo7

14:16

+ stuff about matter fields

@JuanFran there are 3 kinds of GR books

books for physicists

books for mathematical physicists

books for mathematicians

My favorite book of all time falls in the first category.

Juan Fran

which one? dirac's ?

0celo7

The third category is pretty dry, honestly.

Juan Fran

:D

0celo7

@JuanFran Zee's

Juan Fran

hmm some Problem with the third category is

0celo7

14:18

10

A: Getting started self-studying general relativity

This list is extensive, but not exhaustive. I am aware that there are more standard GR books out there such as Hartle and Schutz, but I don’t think these are worth mentioning. Books with stars are, in my opinion, “must have” books. (I) denotes introductory, (IA) denotes advanced introductory, i.e...

I have a list of books I think are good

Juan Fran

that when i would look at their mathematical discussions, i would probably get very mad

0celo7

Why

Juan Fran

i like maths the bourbaki style

elegant, neat, coordinate free

but an interesting list you got there

0celo7

In differential geometry coordinate cannot be avoided, period

Juan Fran

it seems like no one reads the only GR book i own

i.e. dirac's :D

0celo7

14:20

@JuanFran The third category is written by mathematicians.

Juan Fran

it is very short

what you mean by that? it cannot be avoided?

0celo7

There are things which cannot be proven without coordinates in differential geometry

Juan Fran

obviously you cannot do calculations without them

give me an example!

0celo7

For instance, I'd be very surprised if you can show that the dimension of the tangent space is the dimension of the manifold without coordinates

Juan Fran

thats probably true

0celo7

14:22

:D

Juan Fran

but this is because the dimension of a manifold

is basically defined via coordinates

0celo7

You can probably define it via homology or something stupid

Juan Fran

yeah via H_n(X,X-{x},Z)

but only for closed manifolds

0celo7

There you go

Juan Fran

which is Kind of sad

0celo7

14:24

How about showing that geodesics always exist, locally

You need coordinates to write down the geodesic equation in a chart

Juan Fran

true

well the very nature of a manifold gives you coordinates

so you cannot avoid it entirely

but when you can, i find it preferable

0celo7

I don't see any shame in using coordinate for proofs

Juan Fran

how would you define the tangent space?

me neither, but i just don't find it appealing

0celo7

Oh, the tangent space is easy to define without coordinates

It's the vector space of derivations of germs of functions

But you need coordinates to show that this is spanned by the partial derivatives

Juan Fran

yeah, i was just interested which definitiion you would give me

i mean you can also define it via coordinates

0celo7

14:27

I like the algebraic one

Juan Fran

pretty ugly and cumbersome

0celo7

@JuanFran Yeah

Juan Fran

me too

it actually works for any locally ringed space :D

0celo7

But that's how Hirsch does (coordinates) it because it's the one that works on $C^r$ manifolds, $r<\infty$.

The algebraic definition fails in the non-smooth case

Which is probably my favorite factoid of all time

Juan Fran

it also Fails for complex manifolds i think

0celo7

14:29

I doubt that

The complex tangent space is the complexified tangent space of the underlying real manifold

I think

Juan Fran

let me find a reference

Balarka Sen

what are we talking about

0celo7

I'm right, according to Jost

Juan Fran

i think there is one Definition of tangent space

which Fails for complex manfiolds

0celo7

In my mind, the complex tangent space is the complexified real tangent space

So as long as the manifold is smooth, the algebraic definition works

Juan Fran

14:32

yeah, i think the Definition i had in mind is a derivations of C^infty(M) at that points

Balarka Sen

I don't understand the craze about coordinate freeness. Whenever coordinates are elegant, use them. Whenever coordinate free things are elegant, use them. Done.

@JuanFran I think that fails hard for C^k manifolds, no?

Juan Fran

when are coordinates elegant?

0celo7

I already told him that, @BalarkaSen

Juan Fran

you were quite right about that @0celo7

0celo7

Juan Fran

14:34

actually i admitted it before :D

Balarka Sen

@JuanFran Elegant for understanding, not for conciseness. I understand differential forms better coordinate-wise than sections of the exterior product bundle.

Juan Fran

i said the Definition works for any locally ringed space

and of course a complex manifold is a locally ringed space

0celo7

The most physicist proof of all time right there

Juan Fran

i understand them as the exterior Algebra over the 1 forms :D

Balarka Sen

@JuanFran That completely subsumes the geometric content for me.

Juan Fran

14:36

well the geometric content is contained in the 1 forms

covector fields

Balarka Sen

Depends on what you mean by that.

Juan Fran

sections of the cotangent bundle

Balarka Sen

No, I mean, what do you mean by "geometric content"?

Juan Fran

surely covector fields have a very vivid geometric intepretation?

or not?

Balarka Sen

Sure, but I am asking: what does differential form mean to you?

What do you mean when you say "geometric content"?

Juan Fran

14:38

something i have a geometric picture of. vector fields are just functions assigning arrows

in the most primitive sense

and differential forms are just induced by them via natural algebraic constructions

basically my Definition makes this clear

Balarka Sen

OK, I do not see how that is a useful geometric interpretation. But each to his own, maybe you think algebraically.

Naturally thinking algebraically is totally fine.

Juan Fran

so how is the coordinate stuff more useful as a geometric Interpretation?

i mean you think of a differential forms assinging to each point some mulitlinear alternating map

how is that different from what you say?

in your geometric interpretation

you can*

Balarka Sen

I think of differential forms as something which eats infinitesimally small parallelograms and spits out their signed area :) This is also easy to see from the section definition, but only becomes apparent - I repeat, to me - when I look at it via coordinates.

The coordinate-free definition is a nice and concise packaging of that to me.

Juan Fran

i see, yeah well i think our interpretations are quite similar

i just stated them more fancily

in the most primitive way i would say

differential forms is what you integrate over :D

not very convincing, but it is essential

Balarka Sen

@JuanFran Which distinguishes our thinking style, I think :)

Yes, that you can integrate over them becomes apparent to me from the parallelogram thing I said.

Juan Fran

14:44

might be very well

i think what is nice about my interpretations (and this is what i just noticed now) is how it puts the focus on vector fields

oftentimes students find differential forms harder to understand than vector fields

i basically say vector fields+natural algebraic construction = differential forms

i guess these students just have Problems with natural algebraic construction :P

Balarka Sen

Tensor fields are precisely parametrized version of "infinitisimally small parallelograms" :)

Juan Fran

so to speak yeah

Balarka Sen

Mathwise, given $n$ vector fields $X_1, \cdots, X_n$ on $M$, at each point $X_1(p), \cdots, X_n(p)$ span a parallelogram at $T_p M$.

Juan Fran

haha did you star my message?

you know 3 references, but how does the proof work? - 2d ago by Juan Fran ▼

:D

Balarka Sen

Nope.

Juan Fran

14:51

ok ok, i just thought. it somehow appeared on the board

i recently had an amusing discussion

with someone claiming that number theory is "by far" the hardest subject in math

turned out he did a phd in number theory, expected

not sure why people always think their own subject is the hardest

Krijn

To make themselves look smarty

Juan Fran

mostly this is just due to ignorance about other subjects

god forbid krijn!

Balarka Sen

@JuanFran Probably it's true.

From some point of view.

Juan Fran

well it seems that every subject is the hardest

Balarka Sen

"Hardest" is probably a wrong choice of word. "Vast" or some such would have been more appropriate.

Juan Fran

14:54

from some pov

imo both the hardest and most interconnected subject is algerbaic geometry :D

but i might be wrong haha

0celo7

@BalarkaSen what the hell

Balarka Sen

huh?

0celo7

Oh

Never mind

Balarka Sen

lol

0celo7

@JuanFran they do?

Balarka Sen

15:00

assign to each vector it's length (a number in general)

Juan Fran

what are you referring to?

0celo7

For higher forms what Balarka said makes sense

But I've never had a feeling for covectors

Juan Fran

but you have a feeling for the dual space?

0celo7

Not geometrically

Juan Fran

it is just a vector field

in a parallel universe :D

Balarka Sen

15:05

it's easier to visualize covectors than the whole dual space

0celo7

What's a covector

ELI5

Balarka Sen

a thing which assigns a number to a vector.

yikes i should get back to studying

procrastinating for too long

0celo7

@BalarkaSen so?

Is that supposed to be a geometric interpretation?

The only one I've heard is the "egg crate" one

Which is terrible

Anubhav

15:38

vote for reopen this question math.stackexchange.com/questions/1883064/…

@BalarkaSen help me

0celo7

Not even a please?

Anubhav

@0celo7 I was in a hurry at that time, so missed it

yes please vote for re-open that qes

Akiva Weinberger

15:55

@Agawa001 It's not plagiarism. They both evolved from a common ancestor.

Aritra Das

17:54

Guys, I'm doing calculus (or analysis, I don't know which) and I've encountered the "Cauchy Criterion"

And whenever there's a problem based on it, my profs use the "contractive criterion" to solve the problem instead

And I've seldom been able to come up with solution, using the statement of the Cauchy criterion, for such problems, myself.

So, do you know of some problems, solutions or links where the cauchy criterion can be directly used to examine convergence?

Or is it totally useless?

Thanks in advance

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