Mathematics - 2017-04-22 (page 1 of 3)

Benjamin

00:03

Would a question about math journals be welcome here?

Ted Shifrin

rehi @Semiclassic

Benjamin Gadoua

Sure @Benjamin

Ted Shifrin

Like what, @Benjamin?

I can't for the life of me understand that, @Semiclassic.

Benjamin Gadoua

aaaaahhh three's two of us noooooooooooooo

Ted Shifrin

It talks about a variable parabola and then we end up with the equation of a circle.

Benjamin Gadoua

00:05

I also can't grok that question

Ted Shifrin

Oh yeah, there are two Benjamins. Worse than having two Teds.

Benjamin

@TedShifrin Yes, I got buzzed yesterday about another Benjamin's post.

Ted Shifrin

Two Benjamins with very similar avatars, too.

Benjamin

What journal would be most appropriate for a paper introducing a new trigonometric function?

Ted Shifrin

Well, the pings goes on the first four or five letters, so you're screwed, Benjamin. One of you should change names :P

Benjamin Gadoua

00:08

can you change how your name displays? my MSE name is my full name, haha

Ted Shifrin

A "new" trigonometric function? This sounds dubious. Say a bit more.

Benjamin

@TedShifrin I am fine to just deal with it because the discussion I ran into actually turned out to be interesting.

Benjamin Gadoua

anyway, you had some sort of question about a paper or journal?

Ted Shifrin

But this why I always refer to MikeM and DanielF.

Benjamin

@TedShifrin Not wholly new, but a function that is combination of other trigonometric functions to get a new and interesting result in the pattern of circles and arcs.

Ted Shifrin

00:12

Depending on the mathematical content and depth, possible the MAA's Mathematics Magazine or American Math Monthly.

But I've certainly turned down articles for those.

Benjamin

@TedShifrin What is the average length for papers in those journals?

Ted Shifrin

I don't know for sure. I would say something like 5 printed pages.

There are shorter ones, for sure.

Best for you to take a look at them (probably can't see them on line without a membership) ...

Benjamin

@TedShifrin I can get them through the library.

I think.

@TedShifrin It is a nice short paper, probably about that length, if not shorter so it should fit that. I will definitely take a look. Thanks!

Ted Shifrin

At some point if you want to email it to me, I'll be happy to give you an opinion, Benjamin.

Benjamin

@TedShifrin Once it is a bit more polished, I'll look at that.

Ted Shifrin

00:15

Sure thing.

Benjamin

Do those journals allow preprint publication?

Ted Shifrin

What do you mean by "allow"?

Benjamin

@TedShifrin Do they permit authors to publish preprints on ArXiv?

Ted Shifrin

I'm not sure that ArXiv wants things of this level, but I'm not sure about that. I have no idea, actually, but most journals know that people post pre-publication research-level stuff on the ArXiv.

Benjamin

@TedShifrin I didn't mean ArXiv per se, just a major priority of mine is open access.

Ted Shifrin

00:22

I cannot speak with authority on this.

Simply Beautiful Art

The Story of Large Numbers, a dive into the unknown

Apr 1 at 0:14, 9 minutes total – 2 messages, 1 user, 0 stars

Bookmarked Apr 1 at 0:24 by Simply Beautiful Art

The Story of Large Numbers, a dive into the unknown: 2nd chapter

13 mins ago, 5 minutes total – 2 messages, 1 user, 0 stars

Bookmarked 7 mins ago by Simply Beautiful Art

Chapters 1 and 2 are available for those looking for a simple read

Ted Shifrin

Are you calling me a simpleton, @Simply?

Simply Beautiful Art

@TedShifrin No, the 'simple' was my trademark ;)

ramanujan_dirac

hi @Adeek

Simply Beautiful Art

Why does everyone name themselves ramanujan? I know he's awesome and all, but c'mon, I've seen so many of your unoriginal username lol

Ted Shifrin

00:37

I concur, @Simply.

I should have named myself Gauss or Riemann.

ramanujan_dirac

for example? :P dont know any other MSE user named ramanujan

Simply Beautiful Art

xD

Benjamin

@SimplyBeautifulArt These look interesting. What about $ω^ω^ω^ω^...$

Simply Beautiful Art

@ramanujan_dirac
https://math.stackexchange.com/users/419835/srinivasa-ramanujan
https://math.stackexchange.com/users/235721/ramanujan
https://math.stackexchange.com/users/276533/ramanujan

@Benjamin Not finite. At least not yet

ramanujan_dirac

in a world dominated by the work of grothendieck and serre, ramanujans work can never be overappreciated

Benjamin

00:39

@SimplyBeautifulArt Okay.

Simply Beautiful Art

@Benjamin Heard of the fast growing hierarchy?

Benjamin

@SimplyBeautifulArt Maybe, but not with that term. What is it?

Simply Beautiful Art

@Benjamin A way to classify function growth rate using ordinals

It works like this:

Benjamin

@SimplyBeautifulArt No, I haven't seen that.

Simply Beautiful Art

$f_0(n) = n+1$
$f_{\alpha+1}(n) = f_\alpha^n(n) = f_\alpha(f_\alpha(...f_\alpha(n)...))$
$f_\alpha(n) = f_{\alpha[n]}(n)$

where $\alpha[n]$ means to take the $n$th term in the fundamental sequence of $\alpha$ if $\alpha$ is a limit ordinal.

Ted Shifrin

00:42

@ramanujan_dirac: Suffice to say some of us appreciate his intellect but are not big fans of that mathematics.

Benjamin

@SimplyBeautifulArt That makes sense, I should go read more about this.

Simply Beautiful Art

@Benjamin It grows extremely rapidly. Everything I've written in chapter 2 of the above 'Story' eventually grows slower than $f_{\omega2}(n)$

ramanujan_dirac

@TedShifrin: thats exactly my point. i would be happy if there was more place in academia for that sort of math, and if journals were not dominated by category theorists (for example)

Benjamin

@SimplyBeautifulArt As would be expected given its finite nature.

Ted Shifrin

Journals are far from dominated by category theorists.

But how does what I say make you say "that's exactly my point"? I don't like Ramanujan-style math.

I don't like category-style math, either.

Simply Beautiful Art

00:45

@Benjamin Well, I've seen some python programs that have functions on the order of the Large Veblen Ordinal, if that makes any sense to you.

Benjamin

@SimplyBeautifulArt Don't know much about it as it is not an area I do much with, but I know enough to know it is sizeable.

Simply Beautiful Art

@Benjamin Well, if you wish, you can learn how to write large computable ordinals just as @Secret is learning.

TGIF btw guys

Benjamin

@SimplyBeautifulArt I wouldn't mind learning, I always can make some free time for math.

@SimplyBeautifulArt Yes, definitely.

Ted Shifrin

LOL ... someone just said "Shit. I see my mistake now." after I'd pointed out he hadn't read the question right and he insisted he was right. :P

Simply Beautiful Art

The definition of the one argument ordinal collapsing function is given as follows:
$\Omega$ is the first uncountable ordinal
$C(\alpha)_0=\{0,1,\omega,\Omega\}$
$C(\alpha)_{n+1}=C(\alpha)_n \cup\{\gamma+\delta,\gamma\delta,\gamma^\delta ,\psi(\eta):\gamma,\delta,\eta\in C(\alpha)_n,\eta<\alpha \}$
$C(\alpha)=\bigcup\limits_{n<\omega}C(\alpha)_n$
$\psi(\alpha)=\min\{\beta:\beta\notin C(\alpha)\}$

@Benjamin You will need some MathJax and a lot of brain space if you aren't familiar with this area of math

Benjamin

00:52

@SimplyBeautifulArt I am not great at MathJax, but I can do LaTex and I have brain space.

Simply Beautiful Art

@Benjamin Well, let me explain the first value: $\psi(0)$.

Since we can never have negative numbers, we don't need to worry about the condition $\psi(\eta)$.

$C(0)_0=\{0,1,\omega,\Omega\}$

Ted Shifrin

Benjamin: See the LaTeX in chat link up there >>>>> ^^^^

Simply Beautiful Art

Now add these, multiply them, and exponentiate them.
$C(0)_1 =\{0,1,\omega,\Omega,1+1 ,\omega+1,\omega +\omega,\omega\omega ,\omega^\omega,\Omega+1,\Omega+ \omega,\dots\}$

$=\{0,1,2,\omega,\omega+1,\omega2 ,\omega^2,\omega^\omega,\Omega\text{ stuff}\}$

We then do this again. Add, multiply, and exponentiate all of these to get $C(0)_2$

Benjamin

@SimplyBeautifulArt So, this is somewhat similar to $\aleph$ notation.

Simply Beautiful Art

$\psi(0)$ is the smallest ordinal you never can reach

@Benjamin Meh, I don't really know about that...

Benjamin

00:57

@TedShifrin I have it. Is MathJax basically LaTex?

Simply Beautiful Art

Yes, with a few tweaks

Ted Shifrin

Yup, Benjamin. I hardly ever tweak.

Simply Beautiful Art

So, as you might be able to notice, the smallest ordinal you can't reach is...
$$\psi(0)=\omega^{\omega^{\omega^{\omega^{\dots}}}}$$

20 mins ago, by Benjamin

@SimplyBeautifulArt These look interesting. What about $ω^ω^ω^ω^...$

The thing you hinted to earlier

Now, $\psi(1)$ is even larger...

$C(1)_0=\{0,1,\omega,\Omega\}$

Now, add these, multiply them, and exponentiate, but you may also take $\psi(\eta)$ under the following conditions:
$\eta\in C(1)_{n-1}$
$\eta<1\implies\eta=0$
Clearly, $0\in C(1)_0$, so...

$C(1)_1=\{0,1,\omega,\Omega, 1+1,\omega+1,\omega+\omega,\omega\omega,\omega^\omega ,\Omega+1,\dots,\Omega^\Omega,\psi(0)\}$

Benjamin

I am following with pencil and paper.

Simply Beautiful Art

So now everything to the left of $\Omega+1$ will end up producing everything less than $\psi(0)$. Likewise, we may now do things with $\psi(0)$ to get even higher. The smallest ordinal we can't reach is...
$$\psi(1)=\psi(0)^{\psi(0)^{\psi(0)^{\psi(0)^{\dots}}}}$$

Ted Shifrin

01:03

parenthetically remarks: in favor of pencil and paper.

BAYMAX

How do I visualize $|z1 -z2| \geq |z1| - |z2|$

Simply Beautiful Art

xD

Especially for math

@BAYMAX Triangle rule implies the use of triangles

@Benjamin Are you good so far?

Ted Shifrin

Yes, @BAYMAX, draw a triangle and think about lengths.

Benjamin

@SimplyBeautifulArt Is $$\psi(2)=\psi(1)^{\psi(1)^{\psi(1)^{\psi(1)^{\dots}}}}$$?

Simply Beautiful Art

Yes @Benjamin

And so it continues on like this...

Benjamin

01:05

@SimplyBeautifulArt But what would $$\psi(0)^{\psi(1)^{\psi(2)^{\psi(3)^{\dots}}}}$$ be?

Simply Beautiful Art

until we reach $\psi(\omega)$. Let's look at its construction:

BAYMAX

the length of a side of a triangle is less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides.

Yes good @SimplyBeautifulArt

Simply Beautiful Art

@Benjamin That would be $\psi(\omega)$, since every higher psi function is a fixed point of the previous

BAYMAX

yes@TedShifrin

Benjamin

@SimplyBeautifulArt Okay.

Simply Beautiful Art

01:06

@Benjamin That is, notice that $\psi(1)=\psi(0)^{\psi(1)}$ and so on...

BAYMAX

like how the third side is thought of as $|z1 -z2|$

Benjamin

@SimplyBeautifulArt That makes sense.

BAYMAX

if we treat $|z1-z2|$ as third side

we can apply the triangle rule

and we get the result

Ted Shifrin

Precisely, @BAYMAX, if two sides are $z_1$ and $z_2$, the third side will be $\pm(z_1-z_2)$.

Simply Beautiful Art

Now, $\psi(\omega)$ looks like this:
$C(\omega)_1=\{0,1\omega,\Omega,1+1,\omega+1,\dots,\Omega+1,\dots,\psi(0),\psi(1)\}$
Eventually, you will notice we will construct every natural number, and since every natural number is less than $\omega$, we may take $\psi(k)$ for each $k<\omega$. Likewise, each $\psi(k+1)$ is larger than anything you can reach from $\psi(k)$

Also, you will also be able to fill in all the gaps and construct all the in between ordinals

BAYMAX

01:09

consider the right angled triangle 3,4,5@TedShifrin

Benjamin

@SimplyBeautifulArt Just to note: I definitely see connections to $\aleph$ notation.

BAYMAX

oh

Simply Beautiful Art

Thus, the smallest ordinal we can't construct is...
$$\psi(\omega)=\sup\{\psi(1),\psi(2),\psi(3),\dots\}$$

Ted Shifrin

What am I supposed to consider, @BAYMAX?

Simply Beautiful Art

@Benjamin Well, it is set theory-ish

BAYMAX

01:09

got it

Benjamin

@SimplyBeautifulArt True.

BAYMAX

one of the vertices in origin

then one side islength is z1

Simply Beautiful Art

@Benjamin So what would be $\psi(\omega+1)$?

BAYMAX

other length z2

Ted Shifrin

Right triangles won't be anywhere near the problem.

BAYMAX

01:10

and vector rule

?

Benjamin

@SimplyBeautifulArt Can you have non-integer ks?

Ted Shifrin

Make a very flat triangle (with one angle almost 0).

Simply Beautiful Art

@Benjamin No

Adeek

@ramanujan_dirac hi

BAYMAX

Make a very flat triangle (with one angle almost 0).

here comes equality right?

Benjamin

01:12

@SimplyBeautifulArt $$\sup\{\psi(1),\psi(2),\psi(3),\dots\,\psi(\omega-3},\psi(\omega-2},\psi(\omega‌-1}, \psi(\omega}$$

Ted Shifrin

Right, that's where you see that you need slightly better than equality.

Simply Beautiful Art

@Benjamin $\omega-k$ doesn't make sense though

You also used } instead of )

Benjamin

@SimplyBeautifulArt True, but it is out of habit.

BAYMAX

nice1@TedShifrin

Benjamin

@SimplyBeautifulArt Thank you, I missed that.

BAYMAX

01:13

thanks@TedShifrin@SimplyBeautifulArt

Ted Shifrin

Of course, @BAYMAX.

Simply Beautiful Art

@Benjamin Well, $\psi(\omega+1)$ is just an infinite tower of $\psi(\omega)$

BAYMAX

:) >>>

Simply Beautiful Art

$\psi(\omega2)$ is just the limit of $\psi(\omega+k)$

@Benjamin You get the general idea? Or am I going too fast?

Benjamin

@SimplyBeautifulArt Can you have $$\psi({\psi({\omega}))}$$

@SimplyBeautifulArt I get the general idea and will probably review this later to be sure.

Simply Beautiful Art

01:15

@Benjamin Yes, you can have that

We then have this:
Let $\alpha$ be the following ordinal:
$$\alpha=\psi(\psi(\dots\psi(\omega)\dots))$$

It is extremely large, no doubt

BAYMAX

bye all.

Benjamin

@BAYMAX Bye.

Simply Beautiful Art

And we clearly have $\psi(\alpha)=\alpha$

@BAYMAX bye

But what is $\psi(\alpha+1)$?

BAYMAX

@Benjamin@SimplyBeautifulArt bye,good day.

Simply Beautiful Art

@Benjamin Well, let's look at its construction

Benjamin

01:18

@SimplyBeautifulArt I am sort of lost on this one.

Simply Beautiful Art

$C(\alpha+1)_0=\{0,1,\omega,\Omega\}$

$C(\alpha+1)_1=\{0,1,2,\omega,\omega+1 ,\omega2,\omega^2,\omega^\omega, \Omega\text{ stuff},\psi(0),\psi(1),\psi(\omega)\}$

Benjamin

Is it equal?

Simply Beautiful Art

$C(\alpha+1)_2=\{0,1,2,3,4,\omega+(0,1,2,3),\omega(2,3,4),\omega^{\omega+1} ,\dots,\Omega \text{ stuff}\\,\psi(0),\psi(1),\psi(\omega), \psi(0)+1,\psi(1)+1,\dots,\psi(\psi(0)),\psi(\psi(1)),\psi(\psi(\omega))\}$

Benjamin

Sorry, I have to go for now, but I will keep considering this and try to figure it out.

Simply Beautiful Art

@Benjamin Wait! Before you leave!

Notice that you can never truly do $\psi(\dots)$ an infinite amount of times. Therefore...
$$\psi(\alpha+1)=\alpha$$

Likewise, for bigger ordinals, $\psi(\beta)=\alpha$

Benjamin

01:23

@SimplyBeautifulArt Okay, this is sort of the lines along which I was thinking.

Simply Beautiful Art

Even $\psi(\Omega)=\alpha$

But $\psi(\Omega+1)$ is different

Benjamin

@SimplyBeautifulArt So, it forms a maximum. Can it ever be larger?

Simply Beautiful Art

@Benjamin Write down the rules of this function a few times and see if you can spot what's so diffferent about $\psi(\Omega+1)$

@Secret Can also read the above stuff

@Benjamin Every maximum shall be exceeded, except for the impossible. Which would be the set of uncomputable ordinals

Benjamin

@SimplyBeautifulArt Well, I will read it later and try to figure it out. Goodbye. Should I read the stuff you showed secret?

Simply Beautiful Art

@Benjamin Lol, I suppose it is good stuff to read under the bed sheets in candle light

and good night!

Sergey Dylda

01:33

Hello, people. Is there anyone who is familliar with tensor/exterior algebras?

Daminark

03:31

Hey @Adeek, sorry I was watching a play for my theater class at the time :P

BAYMAX

03:42

Is difference of two increasing functions increasing?

Semiclassical

Doubtful.

BAYMAX

Actually a montonically incresing function is differentiable almost everywhere

Semiclassical

As a simple counterexample, $x$ and $2x$ are both increasing but $x-2x$ isn't increasing.

BAYMAX

yes

Like how functions of bounded variation imply function to be differentiable almost everywhere

?

Semiclassical

Couldn't tell you.

BAYMAX

03:46

ohhh

see the amazing sequence

I hope you may appreciate it, here is the statement -

"Every absolutely continuous function is of bounded variation and hence is differentiable almost everywhere."

Now I got the first part

Daminark

A monotone function has only countably many discontinuities

BAYMAX

now I wanna show bounded variation implies differentiable almost everywhere

Daminark

Thus, so does one of bounded variation

BAYMAX

now a function of bounded variation can be written as ifference of two increasing functions

now the difference of two incresing functions nedd not be incresing in general but can be Monotone I guess ----- ".."

and now montone functions are differentiable almost everywhere

so functions of bounded variation implies differentiable almost everywhere.

.............

Daminark

Oh wait if you already know that monotone functions are differentiable almost everywhere you're done

Say $f = g-h$ where $g$ and $h$ are monotone

BAYMAX

03:51

is this true that difference of two incresing functions is a MONOTONE .

Daminark

Call the set of points where $g$ is not differentiable $E_g$

BAYMAX

@Semiclassical@Daminark

Daminark

And similarly for $E_h$

Then the set of points where $f$ is not differentiable is a subset of $E_g \cup E_h$, which is measure 0

Like, if two functions are differentiable at a point, so is their difference

Semiclassical

@BAYMAX still no. for a piecewise continuous case, take $f(x)=2x$ for $x\geq 0$ and $f(x)=x<0$.

Daminark

So even though functions of bounded variation are not necessarily monotone, it doesn't matter

Semiclassical

03:53

Then $f(x)-f(-x)=x$ for $x>0$ and $-x$ for $x<0$.

Both functions are increasing, but their difference is increasing for $x>0$ and decreasing for $x<0$.

BAYMAX

Ok NOW I SEE THAT THIS IS NOT TRUE FOR ALL FUNCTIONS

SO THIS MUST BE TRUE FOR A SPECIAL SET OF FUNCTIONS

AND HENCE HERE COMES THE ORIGINAL STATEMENT OF THEOREM - :)

"Every Absolutely continuous function is of bounded variation and hence is differentiable almost everywhere"

I think the piecewise continuous functions is not absolutely continuous function@Semiclassical

Semiclassical

Presumably not. But I imagine one can product a smoothed version of the above.

BAYMAX

ok

Semiclassical

@BAYMAX Yeah, take $f(x)=x\left(\dfrac{2-e^{x}}{1-e^x}\right)$.

For large positive $x$, that behaves as $x$. For large negative $x$, it behaves as $2x$. In between, it interpolates between the two.

BAYMAX

04:10

hmm

Semiclassical

And then one can take $f(x)$, $-f(-x)$.

Both are monotonically increasing, but the difference between the two is an even function.

Hence, can't be monotonic.

BAYMAX

so in general difference of two increasing imply monotone in general is not true.

Semiclassical

Right.

BAYMAX

I have to think other way around.

MMM

04:50

Hello everyone

Anyone interested in Maple software then come and join us here

67

Maple

Proposed Q&A site for students, teachers and researchers who are using Maple for symbolic and numerical computation.

Currently in commitment.

Vrouvrou

05:38

Hello

Secret

09:24

Is it possible to have a "short line"?

Studentmath

09:39

@Secret What do you mean?

I'm having some trouble with functional analysis problem - I think I am not understanding something, I have no idea how to proceed

Should be a rather simple question..

Secret

I am not sure, recall that a long line is $[0,1]\times \omega$, thus it seemed longer than the real line due to the topology effectively ballooning up each point in the interval into a countable set.

Therefore, a short line will be something like $[0,1]\times ?$ where ? will be some number or set such that the resulting line is still an uncountable set, but "shorter" because e.g. infinite many points in the interval [0,1] get identified into one single point still within the interval

Studentmath

Ah, hm

Studentmath

10:38

I'll pop my question:
I am defining $L$ as a linear operator in $l^2$, defined by the infinite matrix $(a_{ij})_{i,j=1}^\infty$.

We say that $\sum_{k=-\infty}^\infty sup_{i\in N} |a_{i, i+k}|< \infty $

(Where $a_{ij}=0$ if $j<=0$)

Balarka Sen

Hi @Danu.

Danu

Hi Balarka

Studentmath

I am asked to show that $L$ is bounded - and more so the bound is the above sum. How do I even approach such a proof?

Danu

Wanna check a dumb proof of a linear algebra fact that I wrote down?

Balarka Sen

Maybe. What's the fact?

Danu

10:41

I wanna show that any complex structure on a vector space $V^{2n}$ is compatible with (i.e. orthogonal wrt) an inner product

So I do induction: Pick $x_1\in V$ and set $x_2=Jx_1$. Those are linearly independent.

Then complete to a basis and declare it to be orthonormal to find $V=\operatorname{span}(x_1,Jx_1)\oplus V'$ where $V'$ has a basis of the form $x_2, Jx_2,\dots, x_n, Jx_n$ by induction assumption. THen declare that basis to be orthonormal to obtain the inner product I want

That works, right? Really all I need to show is that I can find a basis of the form $x_1, Jx_1,\dots x_n,J x_n$

@BalarkaSen Whaddaya think?

Balarka Sen

Ok, and how's the inner product defined using that basis?

Danu

That basis being orthonormal

Then $J$ is orthogonal

Balarka Sen

Ah, so you're defining <-, -> to be orthonormal on that basis and then extending (bi)linearly

Yah, that most certainly works

Danu

yeah

I always feel so dumb when doing linear algebra

There's so little to work with

So when I don't immediately see something it's weird: Don't know what to try :P

Balarka Sen

I almost always try to think of R^n and C^n when I want to figure out why something is true, and when I see it's obviously true there I try to figure out what made it true.

Danu

10:49

hmm

Balarka Sen

Eg your proof is pretty natural if you think about C :)

Danu

So now I can answer the question Ted linked me yesterday

Balarka Sen

Cool

I'm trying to remember some differential topology but failing.

Danu

What difftop?

Studentmath

I was thinking to show that $||Lx||\leq K \cdot ||x||$ for all $x\in l^2$, where $K$ is that infinite sum..

Balarka Sen

10:56

@Danu There's a slick proof that nondegenerate (Morse) critical points are isolated which does not invoke Morse lemma. I have been trying to remember that for the last 5 minutes; I know it's in G-P but I don't want to look.

Danu

@BalarkaSen Oh, really? Morse functions in G-P? I must've forgotten that too, haha

Ah, Lefschetz critical points

Right, I remember

Balarka Sen

ya it's a single section of the first chapter

Danu

The Lefschetz stuff is in ch 3

Balarka Sen

If $f$ is a smooth function, the critical points are zeroes of $Df$ and the nondegenerate ones are such that $DDf$ is nonvanishing. One has to do inverse function theorem with $Df$ to prove isolation.

Mike Miller

@Danu: Remind me the definition of omega-compatibility?

Balarka Sen

10:58

Or something like that.

Danu

@MikeMiller $\omega(-,J-)$ inner product

The proof that the space is conctractible that I nkow is the one in McDuff

it uses some annoying linear algebra stuff which is not very elegant

I'm probably not going to write that out in the answer I'm writing.

Though I probably should. But I don't want to spend time on it right now.

I already wrote it out once in my own notes

Secret

Hey guys, any example of a topology that is neither first nor second countable, thus making its limit points unable to be described by converging sequences?

Alessandro Codenotti

Disjoint union of a a space which is not first countable and a space which is not second countable

Or $\omega_1+1$ should also work

Secret

I see

But are there more extreme examples, where the uncountable set itself is neither first nor second countable, and has no subsets that is either first or second countable?

Alessandro Codenotti

11:15

Ah, my first suggestion is kinda pointless, since second countable implies first

No, singletons will always be second countable subspaces, as well as all finite subsets

Secret

So for general topological spaces, we can always talk about approaching limit points via converging sequences?

One reason I wonder about that is I seemed to rely on sequences too much when trying to find limit points given a topology, and wonder if I can always use this method, or have to get used to ways of finding limit points in the absence of converging sequences

I recall DHMO said finding limit points require taking arbitrary intersection of neighbourhoods

Danu

I guess I can't answer that question Ted linked after all

Mike Miller

what was the q

Danu

Show that the space of complex structure $\mathcal J(V)$ is homotopy equivalent to the space of symplectic forms.

Alessandro Codenotti

In a first countable space sequential closures coincide with the actual closures, that's not true in general

Danu

11:21

What I can say is that each $J$ lies in some $\mathcal J(V,\omega)$ and that that is contractible

Mike Miller

ah i see

Danu

But I don't know much about the space of symplectic forms with which $J$ is compatible. Maybe it can be easy since I have some kind of convexity going on

Yeah, right. That's nice.

So that space is contractible too

Secret

Hmm, I guess I need to figure out how to train my arbitrary intersection computation skills...

Danu

Because the space of inner products is convex

Secret

I really suck at computing arbitrary unions and intersections because they are not very visual

Mike Miller

11:24

@Danu And that's the proof, more or less

Danu

Right. I need to arrange things in my head, but I think that should be enough.

Right, so I associate to $J$ some $\omega$, unique up to homotopy

Mike Miller

Consider the space of pairs $\mathcal J \times_{comp} \mathcal S$ of a complex structure and a symplectic form such that $J$ is $\omega$-compatible

Danu

then use that $\mathcal J(V,\omega)$ is contractible

Mike Miller

There are projection maps from $\mathcal J \times_{comp} \mathcal S$ to each factor, and you just proved that their fibers are contractible. You need to know the bonus fact that the projection map is a fibration, but this is probably not that bad?

But in any case once you know that you can apply the homotopy LES to see that this bigger space is equiv to each of the other spaces.

Danu

Do I really need it to be a fiber bundle?

Mike Miller

11:27

fibration is weaker than fiber bundle

have to have homotopy extension for discs

fiber bundle is probably pretty hard to prove

Benjamin Gadoua

just a fibration in the sense of Serre then, I guess

Danu

I run into a little bit of trouble here since I know 0 homotopy theory

Balarka Sen

one probably doesn't need Serre fibration to have homotopy LES but i don't recall

Benjamin Gadoua

I think a Serre fibration is pretty close to the minimal requirement, you don't need a Hurewicz fibration

Mike Miller

That's the least you can get away with.

Balarka Sen

11:31

Ah I see, Serre says you have HLP wrt disks (hence CW complexes) and Hurewicz says for every space. not the other way round

Danu

So can any of you finish up the proof and write it out?

2

Q: Space of all complex structures on $V$ is homotopy equivalent to the space of (linear) symplectic forms on $V$

How can I prove that the space of all complex structures $I\colon V\to V$ is homotopy equivalent to the space of all non-degenerate skew-symmetric $2$-forms $\omega \in \Lambda^2V$? Skew-symmetry is seems to be related with the property $I^2=-\operatorname{id}_V$, but I can't formalize it to com...

'cause I don't know no fibrations

Benjamin Gadoua

That's what I thought, the proof I know passed through the LES of a pair & you definitely need that homotopies from disks lift

Danu

That $\mathcal J(V,\omega)$ contractible is kind of ugly and proven on pages 64-65 of McDuff and Salamon's intro to symplectic geometry

Balarka Sen

@Benjamin Yeah I agree now.

BAYMAX

can any one help m in searching out the source of this question

5

Q: Measure theory qualifying exam questions

The following question is a qualifying exam question, though I don't see how these two parts are related. (a)Let $(X, \mathcal F, \mu)$ be a measure space with $\mu(X)=1$ and suppose $F_1 , F_2, ...F_7$ are 7 measurable sets with $\mu(F_j) \geq 1/2$. Show that there exist indices $i_1<i_2<i_3<i_...

real-analysis measure-theory lebesgue-integral lebesgue-measure

Balarka Sen

11:33

Oops, there are two Benjamins. Sorry, other Benjamin.

Benjamin

@BalarkaSen It's fine.

Danu

Actually, that book also says it's "easy to see" that the projections are (locally trivial) fibrations in this setup

Benjamin Gadoua

@Danu there's a theorem of Serre's that's useful here, I'm trying to recall the source, but basically if you have local trivializations you have a Serre fibration

might be in switzer

Balarka Sen

Local trivialization means you're already a fiber bundle, or am I misunderstanding?

Danu

I'd assume so

Astyx

11:45

Hi chat

@Danu It seems I haven't recieved the notes :/ (not that it is urgent or anything)

Danu

Huh.

Did you check your spam folder?

Astyx

Good call

Cool, thanks again !

Danu

Beware of the many typos!

SBM

oh, hello

Secret

[Random] We have:

$$\sum_{i\in S} a_i$$

$$\prod_{i \in S}a_i$$

We should have:

SBM

11:53

wait what?

series

Secret

$$E_{i\in S} a_i=a_1^{a_2^{a_3}\cdots^{a_n}}$$

Studentmath

I'll mumble here since it might help me figure it out:
So I was thinking of writing $Lx=(\sum_{k=1}^\infty a_{1k}\cdot x_k, \sum_{k=1}^\infty a_{2k}\cdot x_k, ... )$. Then I should try to rearrange that so I can see the norm of $Lx$ as the norm of $x$ times some series of scalars - then somehow limit said series with that series of $sup$..

Simply Beautiful Art

@Secret lol

The Story of Large Numbers, a dive into the unknown

The Story of Large Numbers, a dive into the unknown: 2nd chapter

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$Mathematics$ Mathematics