Mathematics - 2023-06-09

anak

02:05

Balarka!

Balarka Sen

Hi @anak!

anak

I came across your papers on stratified spaces the other day and thought of you.

Balarka Sen

Oh lol glad you remembered me

anak

I was trying to find something about symplectic reduction and it turns out you did a stratified symplectic reduction.

Hope you have been well and having fun with math still.

Balarka Sen

Sort of. Our version of stratified symplectic spaces do not quite gel well with symplectic reduction, but we state a relationship.

I was very tired after thinking about stratified spaces for a long time, so I switched gears and now I am trying to learn contact topology.

anak

02:14

Contact topology is thrilling.

I was going to try to do some stuff with contact geometry, but I don't think anything will come out of it. I have been doing mostly just Riemannian stuff recently, so I strayed from my beginnings.

What sort of stuff are you looking at in contact topology?`

Balarka Sen

Oh cool. Maybe we can read together again, finally.

@anak My MSc thesis, which I am currently writing, is about loose Legendrians. I am thinking about Honda-Huang a lot lately.

anak

Have you been in contact with Emmy at all?

Balarka Sen

Nope, but that'd be fantastic lol

Maybe I'll have an occasion someday

I'm a big fan of her work

anak

Yeah, you were really into the Eliashberg-Murphy paper.

Balarka Sen

What are you doing in Riemannian geometry?

(In the sense of, what the hell are you doing over there, chief?)

anak

02:22

flopping around like a fish trying to find an interesting enough result for a thesis and failing

the usual

Balarka Sen

Average grad student life

anak

Pretty much. Found any good music lately?

Balarka Sen

I think I went through a couple of musical phases since we last spoke.

anak

A couple?

Balarka Sen

In the middle I was listening to Burial a lot. Future garage/UK garage/dubstep.

The umbrella term for what I was listening to at that time is "hauntology", you can look that up.

anak

02:28

british deconstruction movement, intriguing

Balarka Sen

yeah lol

anak

UK garage is also british...I am sensing a theme

shintuku

is figuring out whether $I_? = \{h/h' \in K[x,y]_{(x,y)}[1/y]: h \notin \langle y \rangle \frac{K[x,y]}{(x,y)} \}$ is an ideal of $K[x,y]_{(x,y)}[1/y]$ messy or am i doing it wrong

anak

the notation is sadistic in the least

Balarka Sen

@anak indeed

anak

02:36

I think the only british house artist i know well is Fatboy Slim

but he's like big beat/acid

shintuku

$\langle y \rangle$ powers of $y$; $(x,y)$ ideal at which we localize, $[1/y]$ is what we're extending by.

anak

Any current rec.s for what you are listening to right now?

Balarka Sen

I really enjoyed "Engine of Hell" by Emma Ruth Rundle, which I heard a couple weeks ago.

anak

but she isn't from the UK!!!

Balarka Sen

Yeah, I got over the UK phase :P

anak

02:39

American post folk now? Or this is a one off?

Balarka Sen

One off, do not have any coherent themes ongoing.

anak

Sometimes it's good to not have a musical commitment.

Balarka Sen

True

anak

I like those seasons where I can jam to pretty much any tune.

Balarka Sen

02:59

Fill tool is a bit strange in Inkscape but this will do I think

Sourav Ghosh

03:13

Does there exists any set $A$ containing the Cantor set $C$ such that $D(A) =A-A$ doesn't contain any non empty open interval?

shintuku

i used a frac instead of a set minus

we call that optimal notation

Sourav Ghosh

Q.2) Let $A_n\subset \Bbb{R}$ be such that $D(A_n) $ doesn't contain any interval. Can $\cup_{n\in\Bbb N} A_n$ contain an interval?

@shintuku This is not setminus.It is Minkowski difference or algebraic difference.

We call the "difference set"

Balarka Sen

Doesn't $A - A$ contain $C - C$? Isn't $C - C$ an interval?

shintuku

sorry i was talking about a notation mistake i made above, not your comment

Sourav Ghosh

@BalarkaSen Done!

Ignore :)

Both problems are done:)

So $S=\{A\subset \Bbb R : D(A) contains an interval }$ is not a sigma ideal.

Countable sets, measure 0 sets, meager sets, sets having the property of Baire are sigma ideals.

I am trying to find more interesting sigma ideals in \Bbb R

*Correction: $D(A) $ doesn't contain any interval.

Thomas Finley

03:41

What are the various ways/strategies to prove that a set G (G\subset R) is dense in R ?

shintuku

choose a point in R and make an $\epsilon$ neighborhood, and find at least one element of $G$ in that neighborhood

Thomas Finley

Now, I know the elementary defn that a set G is dense in R, iff for every x,y in real set, there exists a g in G, s.t x<g<y

@shintuku is that a definition ?

Balarka Sen

shintuku

it's equivalent @ThomasFinley

Balarka Sen

improved

Sourav Ghosh

03:43

@ThomasFinley cl(G) =R

Any non empty open set intersect G

Thomas Finley

64

Q: Proving that $m+n\sqrt{2}$ is dense in $\mathbb R$

I am having trouble proving the statement: Let $$S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$$ Prove that for every $\epsilon > 0$, the intersection of $S$ and $(0, \epsilon)$ is nonempty.

Say for eg in this post

The user suggests a point in OP to prove the set is dense

It says for every epsilon>0, (0,epsilon)\cap S is non-empty

Is that a criteria for being dense in R ?

I mean an equiv defn or something?

Sourav Ghosh

Side note: Any additive subgroup of $\Bbb R$ is either discrete or dense.

Thomas Finley

I am not quite sure about it?

Sourav Ghosh

G\subset R is dense iff every basic open set ( i.e open interval) intersect G

Thomas Finley

@SouravGhosh yes, but is :"for every epsilon>0, (0,epsilon)\cap S is non-empty" a debn for being dense in R ?

Sourav Ghosh

03:47

No. One way is true.

Thomas Finley

Because this criteria/defn takes only sets (0,\epsilon) into consideration

@SouravGhosh So what about the link of the post I posted here ?

Sourav Ghosh

If $S$ is dense then $S\cap (0, \epsilon) $ non empty for all \epsilon >0

Thomas Finley

@SouravGhosh Oh, but then in the post, the user is trying to prove, S is dense by the other way

Sourav Ghosh

For an example take [0, \infty)

Thomas Finley

That's erroneous, isn't it?

Anyways, I am trying to prove, that the set $\{m+n\sqrt 2:m,n \in\Bbb Z\}$ is dense in R. I dont know where to start ?

Sourav Ghosh

03:51

@ThomasFinley The additive structure will be crucial here.

Thomas Finley

My knowledge about dense sets is that: a set G is dense in R, iff for every x,y in real set, there exists a g in G, s.t x<g<y or equivalently, a set G is dense in R, iff, every point of R, is a limit point of G.

@SouravGhosh You mean that the set is a group under addition?

Sourav Ghosh

$S\cap (0, \epsilon) $ non empty implies $S\cap (-\epsilon, 0) $ also non empty for every $\epsilon>0$

Thomas Finley

@SouravGhosh but why will it be useful here ? Has this something to do with a set being dense in R ?

Sourav Ghosh

See the second answer math.stackexchange.com/a/73272/977780

A clean proof proofwiki.org/wiki/…

AMRIT KUMAR PATEL

04:19

I just got the ability to chat here :)

Thomas Finley

@SouravGhosh this is indeed helpful but here in this case, S is an additive subgroup, so what if it's discrete ?

Koro

04:40

@ThomasFinley Forget m for a moment and consider only the fractional part of $n\sqrt 2$.

then show that it is dense in [0,1).

@ThomasFinley think of discrete as without any limit points.

if you know groups then you can use the fact that additive subgps of R are either dense in R or cyclic.

but that's not necessary really. You can prove this using pigeon hole principle (which is roughly - if you have n>m objects, m placeholders for those objects, then atleast one of placeholders will have more than 1 object in them).

closure of totally bounded set is totally bounded?

cl A= \cup G, where G is open in cl A. Take any e>0. Take a p in cl A.
I want to show that B(p, e) sits inside some G.
B(p,e)= {a\in cl A: d(a,p)<e}
if p is in A \subset cl A, then it does sit inside some G by total boundedness of A so let's suppose that p is in cl A\A.

A\subset \cup G. Given any e>0, for every a\in A, there exists a G such that B(a, 2e)\subset G. closed ball B(a,e)\subset B(a,2e)\subset G for some G.
So B(a,e) (this ball is in cl A) should also be in G.

(first attempt, it should probably work.)

Sourav Ghosh

It looks fine to me.

Koro

thanks.

2

Sourav Ghosh

04:58

$\forall \epsilon >0$ there exists $a_1, a_2, \cdots, a_n\in A$ such that $A\subset \cup_{k=1}^{n} B(a_k, \frac{\epsilon}{2}) $
Then $cl A \subset \cup_{k=1}^{n} B[a_k, \frac{\epsilon}{2}] \subset \cup_{k=1}^{n} B(a_k, \epsilon) $

Koro

I think this is what I should have done.

my answer is not correct. I confused between Lebesgue covering and total boundedness somehow.

I started with cl A = union of some open family

and then tried to find e>0 such for every x in cl A, B(x,e) is in some member of the open family.

this was not required at all.

Thomas Finley

@Koro can you please elaborate the meaning of discrete, i.e some definition? I only know that intutively?

Koro

@SouravGhosh indeed, this is how one would do it. thanks.

@ThomasFinley $A\subset R$ is said to be discrete if $A$ has no limit points in R. e.g. the set of all integers Z

Sourav Ghosh

@ThomasFinley Derived set of A i.e $A'=\emptyset$

Thomas Finley

05:17

@Koro I tried to prove that $n\sqrt 2$ is dense in $[0,1)$, using the fact that irrationals are dense in $\Bbb R$ and so in [0,1) hence, for any x,y in [0,1) we have, an irrational i s.t x<i<y ...but then I am stuck?

Sourav Ghosh

An interesting fact: $Z, √2Z $ both are discrete but their algebraic sum/Minkowski sum Z[√2] is dense.

@ThomasFinley $\{ √2 n : n\in\Bbb Z\}=√2Z $ . Isn't √2Z discrete? Does it have any limit point?

Koro

05:52

@ThomasFinley I meant fractional part of $n\sqrt 2$ not $\sqrt 2 n$. Anyways, here is one simpler way (it doesn't use showing fp of $\sqrt 2 n$ dense in [0,1) first): step 1: show that $0$ is a limit point of {m+ n\sqrt 2: m,n integers}=T. step 2: take any interval (a,b), a<b and show that this interval has an element of T. This uses step 1 and also completeness of R.

shintuku

$A \subseteq B \subseteq C$ with $A,C$ local rings and $B$ a ring does not imply $B$ local

this is a sad moment for $K[x,y]_{(x,y)}[1/y]$

Koro

@ThomasFinley Denoting $\sqrt 2$ by $b$, note that for any integers $n\ne m$, $\{nb \}\ne \{mb\}$ (here {,} denotes fractional part) because if $\{nb \}= \{mb\}$, then $b=\frac{[nb]-[mb]}{n-m}$, which is rational but b is irrational so this is not possible.
Step 1 from the last message: Take any $n\in N$. Divide [0,1) into $n$ parts: [0,1/n), [1/n, 2/n),... Consider n+1 numbers {b},{2b},...,{(n+1)b}. The subintervals are only n but the numbers are n+1 (distinct by the remarks in the first part of this message) so some subinterval must contain at least two of the numbers-say {i b} and {jb}. …

now, you try step 2.

PM 2Ring

@Thomas Dirichlet devised the pigeon-hole principle for precisely this stuff. See math.stackexchange.com/a/56054/207316

It's a beautiful result. And it's probably a good idea to play with a few simple examples until it "clicks".

On a (somewhat) related note, take a look at Beatty sequences. en.wikipedia.org/wiki/Beatty_sequence

robjohn

06:39

@TedShifrin: good evening.

Ted Shifrin

Hi, Mr. sheriff!

robjohn

it was a beautiful day, here. Mid 70s, sunny. There is a chance of rain on Sunday.

Ted Shifrin

06:53

@robjohn Well, maybe here, too.

robjohn

We have a bunny living in our backyard. We just hope it's successful at avoiding the coyotes that come through the neighborhood.

Balarka Sen

07:09

Consider the first Reidemeister move, but take an axis and spin it to get a "Reidemeister move" for 2-knots in R^4 (in terms of knot diagrams, immersed surfaces in R^3).

Iterate to get it for $n$-knots in $\Bbb R^{n+2}$. What should you call this thing? "Torus symmetric Reidemeister move"?

There's different ways to generalize the first Reidemeister moves, hence the question.

Ignore the fact that there are no knotted $S^n$'s in $\Bbb R^{n+2}$ for $n \geq 3$. I am working with knots with more structure, so this is no longer true in my setup.

Torus-symmetric... is a lot of words to say. Toric would be better, if not for that word already being taken. Toral sounds pretentious.

Toroidal? lol

Hopeful Whitepiller