Mathematics - 2016-05-07 (page 1 of 3)

hello

Hmm, so suppose $M$ has an orientation on it, that is a chart $(U_i, \psi_i)$ with $\psi_i : U_i \subset \Bbb R^n \to M$ being local parametrizations so that declaring $\{\partial \psi_i/\partial x_k\}$ to be a positively oriented basis on each $T_pM$ for $p \in \psi_i(U_i)$ is consistent, i.e., at the intersections of two charts the basis are equivalent in the sense that one can be basechanged to another by a positive det matrix.

Then each $\psi_i(U_i)$ admits a nonvanishing top dimensional form: $dx_1 \wedge \cdots \wedge dx_n$, $(x_1, \cdots, x_n)$ being the local coordinates. I have to partition of unity this thing out, do I not?

If I have a nonvanishing form $\omega_i$ on each chart $\mathcal{U}_i$ of $M$, I can take a partition of unity $\rho_i$ and take $\sum_i \rho_i \omega_i$. That's still nonvanishing, and is obviously a well-defined top dimensional form. That's it? @MikeMiller.

@Forever How's it cooking?

well I am trying to write everything down but its frustrating

because I never know how much detail to give

did you figure out if your ctrexample to erdos' thing worked?

yeah I'm pretty sure it does

no major flaws yet

1:42 AM

did you tell anyone?

yes a couple of people. they thought it was ok, but one was so disgusted by my lack of details that he quit talking to me

urk, heh

so now I am in panic mode trying to be very precise

what are you and mike always working on?

a bit off topic, but at some point of time after you're done with this i think you could try to learn some algebraic topology. i think some crazy mutants from your branch of topology uses them to try to do crazy things. who knows, you might get some good applications (like lefschetz did when he applied algebraic topology to a completely unrelated field of algebraic geometry). newer tools are always nice.

i have shape theory, etc in my mind when i say those, btw

i should try but my algebraic topology background is only very basic

1:48 AM

yeah, i mean it could be a long term project. pick things up as you go on, try to apply them to crazy spaces.

just a suggestion

but how crazy can they be? Manifold is always separable metric

does this stuff apply to nonmetric spaces?

sure, the mainstream algebraic topology always works with good spaces - cw complexes. but there are people - shape theorists and so on - who do try to find analogous tools in the nonmetric category

not that I am against working with separable metric spaces

but locally euclidean is the really strong part

yeah, i mean the point of my suggestion was to see if you can find a good algebraic invariant for nonmetric things.

analogous to fundamental group, homology groups, etc.

as i said, it's probably an ambitious and long term project. but it's worth the try, imo

what would that be like for the topologist sine curve?

1:53 AM

here.

i think its a branch which has not fully developed yet, so has a lot of opportunity for good research (and close to you line). you might want to look at it, that's all

en.wikipedia.org/wiki/Shape_theory_(mathematics)#/media/… why is this shape equivalent to a circle?

i dunno, i am not familiar with shape theory.

and so I guess topologist sine curve is shape equivalent to a line

the standard classical homotopy type of that space is horrible, by the way.

it essentially cannot distinguish that space from a point

homotopy is study of paths in a space, right? A path from $p$ to $q$ in $X$ is a continuous mapping $\gamma:[0,1]\to X$ such that $\gamma(0)=p$ and $\gamma(1)=q$.

1:58 AM

yeah.

well, paths modulo "perturbations" of the path.

yes you can define homotopy between two paths and put these into a single class

yep

and that forms a group if you restrict attention to paths with same endpoints.

so the set of equivalence classes is just the set of singletons if a space is totally disconnected, right?

correct.

thus the need for a different idea to define an analogue of $\pi_1$.

can there be only one homotopy class?

2:03 AM

well, if you fix a basepoint, then there is only one homotopy class in your totally disconnected example

a fundamental group is not of a space $X$, but a space with a choice of a basepoint $(X, x_0)$.

maybe I am confusing the set of equivalence classes with the fundamental group

and usually basepoint does not matter, correct?

no, you aren't. space of equivalence classes of loops based at $x_0$ is the fundamental group.

@ForeverMozart if you space is path-connected, then it doesn't matter.

but otherwise you can look at a singleton disjoint union a circle. the two components have different $\pi_1$ (the point has trivial pi_1 and the circle has Z). so it matters where the basepoint is.

so why is fundamental group of the warsaw circle trivial

because there is no loop, correct?

because no path can pass through the wiggly bit of the warsaw circle (otherwise it'd be discontinuous). so no loop can "wind" around it.

yes.

well, there are loops, like go ahead a bit and come back to the point to started. but no homotopically nontrivial ones.

this is a problem I would encounter a lot. So there needs to be a different definition of path

2:09 AM

yep, there needs to be some analogous thing for bad spaces.

i don't know what.

maybe use the cantor set instead of $[0,1]$

think about it :)

i will. and once I define something reasonable, I have lots of crazy spaces to test it on to see if it does anything

:P

it's also worth noting that just thinking about paths might not lead to a fruitful analogue. grothendieck was smart: he used a different definition of $\pi_1$ in topology which could be analogized in algebraic varieties (which are badly nonhausdorff, but varieties have more structure than just a bare topology). so you may have to think out of the box sometimes

it's just a long term project, is all.

Hey @EricStucky.

2:13 AM

eyu

Dreamhack has started :D :D

what's a dreamhack?

esports tournament

Starcraft, specifically

er ok

Well, not exclusively, but I don't care about the other games :P

So I just downloaded the notes from this number theory course

that I "took" this semester

501 pages O.O

the man basically wrote a book

he probably has revised/added to them many times over many years

lvella

2:26 AM

Hi, knowing $\frac{dx}{dt} = \frac{d^2y}{dt^2}$, how can I write $\frac{dx}{dy}$

?

number theory is good stuff

2:37 AM

@Balarka: For some reason previous messages aren't loading for me. Is there something you wanted me to look at?

Yeah, I proved that on a smooth manifold a choice of orientation and a choice of nonwhere vanishing top dimensional form are equivalent. Here's the summary of what I did:

nonvanishing top form => orientation : if $\omega$ is my form on $M$, then $\omega = f_i dx_1 \wedge \cdots \wedge dx_n$ in some local coordinate chart $U_i$ and $f_i > 0$ or $f_i < 0$ (globally) on $U_i$. $\omega(\partial/\partial x_1, \cdots, \partial/\partial x_n) = f_i$ so if $f_i < 0$, change local coordinates to $(-x_1, \cdots, x_n)$ and then look at $\{\partial/\partial x_1, \cdots, \partial/\partial x_n\}$.

So after a suitable change of coordinates these form a consistent choice of basis, hence an orientation.

orientation => nonvanishing top form : orientation implies there exists local parametrizations $g_i : U_i \to M$ so that defining $\{\partial g_i/\partial x_1, \cdots, \partial g/\partal x_n\}$ to be a positively oriented basis of each tangent space on $g_i(U_i)$ is consistent i.e., at intersections they are equivalent in the sense that one can basechange from one to another by a positive det matrix.

So if that's possible on each $U_i$ I have a nonvanishing top form $\omega_i = g_i dx_1 \wedge \cdots \wedge dx_n$. Take a partition of unity $\rho_i$ and consider $\sum \rho_i \omega_i$. That's a nonvanishing top form on $M$.

does that sound ok?

Yes. Good work with the partition of unity.

Was afraid for a second you were going to just try to define it chartwise.

Thanks, it came pretty naturally.

@ForeverMozart Some keywords are "shape homotopy theory", "Cech fundamental group". A name I know associated to all this is Jeremy Brazas. If you're interested in this, you mihjt Chase through his papers and references.

I had been told the idea about partition of unity is to glue local objects into global things, when I was told a bit about how to construct a Riemannian metric on manifolds. That stuck with me, so it doesn't seem too unnatural.

I think I was even told the names of the sheaves one can do partition unity on to glue local sections to global sections. "flasque sheaf" or "flabby sheaf". i forget, but who cares.

2:49 AM

Goddamn, you're a topology lover aren't you Balarka? xD

I like topology, yes.

Not poking fun, just slightly overwhelmed by what you're writing. xD

Fine sheaf.

You will more or less only ever encounter them when they're "xontinuous/smooth sections of some bundle".

@SAWblade I didn't think of it as poking fun. :)

Ah, fair enough. xD

2:52 AM

@MikeMiller aha, ok/

well now nothing stands between me and research

great :^)

I keep telling myself that before getting nothing done.

^

I have no research, so I have nothing to feel guilty about. Tee-hee.

Saaaaaaaaaame.

My school is paying me to research so extra guilt here. xD

To be fair my problem may very well be NP-complete. xD

2:59 AM

The dynamic I get into: "Hmm, I have both A and B to do. What should I do first? I can't make up my mind, so I'll do C until I figure it out."

and C usually involves wasting time on my laptop

Me too, in different contexts.

yup yup yup

I should really be reading more papers about my problem ...

@MikeM: Have you ever read the German drama The Physicists? Huy sent an audio drama to me ages ago, but I only watched in the previous week. It's probably one of the darkest and psychological things I have ever read/heard. I like it a lot.

Thought it might suit your tastes.

No, sorry. I would like to but I doubt I will soon,

It's ok, thought you wouldn't get the time.

3:09 AM

@Semiclassical I have been making a concerted effort lately to spend my time doing things I actually enjoy (work, video games, movies) and not much time mindlessly browsing the internet.

yeah, that's sensible

it's only procrastination if you think you should be doing something else, rather than actively deciding to take a break and pursue something

nah, sometimes it's still procrastination

i just try to optimize my time so that if i'm procrastinating it's in doing something i enjoy.

useful procrastination is good.

yes indeed

Wow, this TeX file I wrote has >1000 lines of text in it ... xD

and yet the very same projct done by my classmate has 2100 lines

clearly i am not working to my fullest potential

eh, i'd insist on labelling the former and not the latter as procrastination. the latter may or may not be a good decision, but it's a choice rather than a compulsion.

and hoo boy is procrastination compulsive for me.

3:14 AM

"Starman" is pretty nice. This man knows his songs.

yo are we talking about david bowie now

i love david bowie

@Semiclassical Maybe for you, personally, but it's definitely true that instead of working i'll pull out my 3ds and play a bit

yeah. i somehow developed a habit of listening to at least one of his songs a day.

what a Pure thing

Have you heard the live version of his song Heroes? :)

nope, not really.

3:17 AM

Ha, okay, I won't force you. xD

i just developed the habit a couple days ago, so... :)

Ah, fair enough. xD

stone love / she kneels before the grave, a brave son / who gave his life to see the slogan / that hovers between the headstone and her eyes / for they penetrate her grieving

i'm going to bed soon. was there anything else you needed?

rebel rebel

nope, nothing else. i'll try to understand how to integrate forms on manifolds a bit for now.

3:20 AM

Man. xD

it seems a bit unmotivated to me

:C

good night, btw.

I don't understand what could be more motivated

I suppose I don't entirely understand what it means geometrically. Not that I have any examples with me.

3:24 AM

You don't understand what integration means geometrically? ;)

I understand integrating forms over parametrized manifolds. But I don't understand what it means to glue a bunch of integrals over charts by a partition of unity. What does it do geometrically?

That's what I want to understand.

@MikeMiller of course i do; it's the gluing by partition of unity bit that's unclear to me.

What do you mean, what does it do geometrically? Do you see that it's the exact same thing over $\Bbb R^n$?

Like if you picked an open cover of $\Bbb R^n$ and then used a partition of unity and integrated

it's no more than saying "Integration is linear, $\omega = \sum \omega_i$, I know how to integrate $\omega_i$"

Hmm, no, I haven't thought about it. I don't see at a glance why they are the same.

Because integration is linear.

Oh yeah, fair. I was being dumb. If I integrate cptcly supported form $\omega$ over opencover $U_i$ over $\Bbb R^n$, I integrate $\rho_i \omega$ on each $U_i$ and sum up to $\sum \rho_i \omega$ and $\omega$ pops out.

so it's the same as integrating $\omega$.

4:08 AM

thanks @MikeMiller anything with "Cech" attached to it seems to be worth checking out

Cech homotopy/homology are pretty useful, afaik.

@BalarkaSen my paper for the result is 6 pages now, and counting...

I think I can do it in under 8

I think the idea there is you replace paths by "chains", i.e., a bunch of open sets enumerated by some countable set with two consecutive sets intersecting.

But I don't know.

@ForeverMozart nice

how do you say two chains are equivalent?

that's what I am not sure about

4:12 AM

you could assume we are working in a compact connected space

sure.

Cech homology on the other hand is quite simple.

cause there are lots of nice chainable spaces like that, that are not path connected

you say two chains are equivalent if one refines the other?

@ForeverMozart right.

@ForeverMozart i don't think that's viable. take R^2 and two paths going from (1, 0) to (-1, 0) along two arcs. open cover these two paths.

so would this make the warsaw circle just a circle right?

would you not say the corresponding chains are equivalent?

because so are the paths.

but none of them refines the other

4:15 AM

it may not reduce to usual homotopy in path-connected spaces, but it could be interesting anyway

i dunno

let's see topologist sine curve becomes a line

warsaw circle becomes a circle

i don't really think this refinment idea is going to pan out, to be honest. i think you need to do something like in cech homology. a basepoint is going to be like an open cover and a chain is going to be through open sets in that cover. and then "pertubation" has to be analogized. there's a lot to do.

the idea is that you build some simplicial model of your space. for each open cover, build the following geometric structure: for each open set, draw a dot. for each intersection of open sets, draw a line between dots. for each triple intersection, fill in the triangle. so on and so forth.

and then do algebraic topology over that geometric model. then take inverse/direct limit or some sort over all such open covers/models.

that's at least what cech cohomology does.

user116211

5:33 AM

@BalarkaSen: o/

user116211

@Cody: cortana?

user116211

Okay, I'm having difficulty with this line:

Cody

What is the least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.

anyone knows how to solve this

user116211

> The plane $z'= \textrm{constant}$ is described in the $x,y,z$ system by the plane $$z'= \mathbf r\cdot \mathbf{\hat{z}'}= \textrm{constant}\;.$$

user116211

where $\mathbf r= x\mathbf{\hat x}+ y\mathbf{\hat y}+ z\mathbf{\hat z}\;.$

user116211

5:38 AM

How did the author conclude that?

anon

what does z'=constant have to do with x,y,z if z and z' are completely different things?

@Cody chinese remainder theorem

split into prime powers first

Cody

@anon thanks let me try

N3buchadnezzar

9:31 AM

any suggestions? stackoverflow.com/questions/37086388/…

9:54 AM

@BalarkaSen the one by Dürrenmatt?

Yep, @quid.

@BalarkaSen maybe you will like the "The Visit" too.

By the same author?

Yes see en.wikipedia.org/wiki/The_Visit_%28play%29

@BalarkaSen good to hear you liked it

10:00 AM

@Huy yeah, I wanted to tell you but couldn't catch you around.

Thanks, @quid. I'll have a look.

Hmm, I don't really like wikipedia's synopsys of The Physicists. Makes it a science fiction story, which it is not.

10:15 AM

Why do I find so many variations of the same set of slides for a topic when I Google it?

I've found 3 variations of a set of slides with different theme and professor name but the same exact slide content.

On another note, if someone can help me with this again:

$a_{m-1}\ b_{m-1}$
$b_{m-1}\ a_{m}$

@BalarkaSen: I haven't read the English one, but the German one seemed fine when I checked it

That is a Matrix called B:

@Huy OK, I am talking about the English one here.

The Wilkinson shift is defined: $\mu_{w}^{(k)} = a_{m} - \frac{sign(\delta)b_{m-1}^{2}}{|\delta| + \sqrt{\delta^{2} + b_{m-1}^{2}}}$

That thing just has several different interpretations. I like to think insanity is the main theme (those 3 guys were hallucinating everything)

10:23 AM

The problem I have is that $b_{m-1}^{2}$ is not defined in the Matrix B

When B has two different values at the positions $b_{m-1}$....

10:37 AM

Hi everyone

I'm new to this section (maths)

@KennyLau hi!

@quid Hi!

@BalarkaSen: yeah, just compare the length of the English wiki to the German one. you should start learning German.

I should do a lot of things. But will I?

Why doesn't $$$$ work here?

10:49 AM

it does

oh

ok

Hello.

hi

I came here for discussing about matrix inversion, because I don't really know how exactly it works.

This is my Matlab code explaining a little bit more of inv(X): eval((1/det(X))*adjoint(sym(X)))

Was planning to program matrix inversion in Python without libraries.

For any kind of matrix size.

@AnastasiaDunbar This may help you: mathworld.wolfram.com/MatrixInverse.html

you could write a function for determinant first

10:58 AM

Currently doing it.

Is it multiplying each value diagonally?

no

@AnastasiaDunbar wiki, mathworld

it is a recursive function

Now I am confused, do they behave differently in sizes?

no

Is there a quick way of finding out which permutations would fix a polynomial?

11:11 AM

What does "fix" mean?

as in sigma(f)=f

well... what is that (sorry)?

An example is that all elements of $S_3$ fix $f=x_4^2 +x_1x_2x_3$ as any permutation such as (1,2) or (1,3,2) leaves the polynomial the same

Evinda

Hey @robjohn
I have a question...

I wanted to know that given any polynomial is there a method for checking which permutations fix it without case by case analysis. e.g. $f=x_1x_2+x_3x_4$. Other than checking which permutations fix it manually, is there a short cut?

Evinda

11:17 AM

@robjohn I want to show that $ H^1(\mathbb{R}) \subset L^{\infty}(\mathbb{R}) $ and that the embedding is continuous.

Let $ u \in H^1(\mathbb{R}) $.

Then it holds that $(1+|\xi|^2)^{\frac{1}{2}} \widehat{u}(\xi) \in L^2(\mathbb{R}^n)$.

$ u \in L^{\infty}(\mathbb{R}) $ means that $ ess \sup |u|<+\infty $. Right?
But how do we deduce this from $ u \in H^1(\mathbb{R}) $?

@GridleyQuayle what is $S_3$? What does (1,2) mean? I thought there are four parameters in $f$.

Oh, sorry. $S_3$ is a symmetric group, in particular the set of permutations from the set {1,2,3} to itself. (1,2) is a cycle notation for a permutation that sends 1 to 2 and 2 to 1. (1,3,2) Would send 1 to 3, 3 to 2 and 2 to 1. If the numbers are not shown in the cycle notation that means they are invariant. So (1,2) is an element of all $S_n$ for $n \geq 2$ as in $S_4$ for example (1,2) sends 1 to 2, 2 to 1, 3 to 3 and 4 to 4.

@GridleyQuayle Well, I thought (2,1),(4,3),(3,4,1,2),... would do it?

For $f=x_1x_2+x_3x_4$ I got (1,2) (3,4) (1,2)(3,4) [that is just the composition of the previous 2] (1,3,2,4). (3,4,1,2) would send f to $x_2x_3+x_4x_1$ so it is not invariant

I did these by checking cases which is impractical for polynomials with more parameters so I was wondering if there was a quicker, less computational way of doing it.

well I mean $x_3x_4+x_1x_2$, however that is notated

11:31 AM

ghostbin.com/paste/mo24s

How does a 4x4 look like.

Well, note that if a term consists of more than one thing then you can permutate them however within themselves

and if two terms have the same degree then you can also swap them

...is this good enough?

@AnastasiaDunbar youtube.com/watch?v=H9BWRYJNIv4

@Balarka What have you learned?

I thought matrix inversion was going to be easier than expected.

So if we had $f=x_5^3x_2^2+x_1+x_3+x_6^2x_4^3$ Then we would look for the permutations that send 1 to 3 (and vice versa) and 5 to 4 only if 2 goes to 6.

Thanks, I'll see how sticking to this quickens the process for more complex polynomials

Why is it recursive?

11:40 AM

@MikeMiller I am learning the proof of Stokes' theorem.

Good morning, by the way.

Another permutation question I had: Consider a 4x4 tiled grid where arranged like so:
1 2 3 4
5 6 7 8
9 10 11 12
13 15 14 [].

Where [] is a blank space. We can move the tiles around by moving 12 down and then 11 to the right for example.

1 2 3 4
5 6 7 8
9 10 [] 11
13 15 14 12.

Can you show that:

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 [].

is impossible to reach from the starting arrangement ?

Adjoint is easier than determinant?

Albas

What do you mean by easier @AnastasiaDunbar

Just look at this Python code for determinant:

def determinant(matrix):
n=len(matrix)
if (n>2):
i=1
t=0
sum=0
while t<=n-1:
d={}
t1=1
while t1<=n-1:
m=0
d[t1]=[]
while m<=n-1:
if (m==t):
u=0
else:
d[t1].append(matrix[t1][m])
m+=1
t1+=1
l1=[d[x] for x in d]
sum=sum+i*(matrix[0][t])*(determinant(l1))
i=i*(-1)
t+=1
return sum
else:
return (matrix[0][0]*matrix[1][1]-matrix[0][1]*matrix[1][0])

Albas

Oh. I do not know python.

11:50 AM

The code doesn't get fixed?

Well whatever.

It works fine.

Why not format it?

There's no indentation.

Just hit ctrl k before you send.

I'm afraid that it won't format.

Now I can't edit.

:o okay.

Here anyways

I need a live preview for chat.

@BalarkaSen Oh, but I think you pretty much already know it. :)

12:00 PM

Hello guys

$\frac{\omega}{Q} = 2+ \frac{1}{R_{3}}$
How do you get R_{3}?

In Matlab, can you make your formula more explained?

@MikeMiller Yeah, it looks fairly straightforward.

@trilolil what is that?

@KennyLau it is just a formula with a bunch of constants

It just asks if you know how to do it for little cubes (half-spaces, if you prefer), and you do.

12:02 PM

I just want to rearrange everything to get R3 @KennyLau

$$\frac\omega Q-2=\frac1{R_3}$$

Right.

omg....

$$\frac1{\frac\omega Q-2}=R_3$$

this was so stupid....

12:03 PM

$$\frac Q{\omega-2Q}=R_3$$

Great! So now that that's done, what's up?

Aren't we all stupid? :)

Just doing it on charts suffices. And on charts I can just look at cubes or half-cubes since my forms (after taking a partition of unity) are compactly supported (I can extend to cubes containing those).

Why can't I see LaTeX in chat?

@AnastasiaDunbar math.ucla.edu/~robjohn/math/mathjax.html

12:03 PM

@KennyLau I feel even more stupid since I entered university...

How do I run in console?

@trilolil because the more I know, the more I know I don't know

@MikeMiller Dunno, there's a bunch of examples next. The next chapter is physical interpretation (curl, grad, div) which I already know so probably going to skip. The chapter after that is topological applications of forms.

And then... I'm done with calculus?

Oh, you're almost done. Have you been doing the exercises?

Yeah, I have been sending them to Ted.

12:06 PM

This is what happens when you actually spend time and effort doing something ;)

It just feels too surprising that I will be done with calculus so soon. I thought I needed another year or so.