Mathematics

Akiva Weinberger

4:08 AM

@Ultradark Homeomorphism doesn't care about "pinches" so those are all homeomorphic

Martin Sleziak

4:59 AM

Is Lebegue an alternative spelling of Lebesgue, or is it simply a typo/misspelling?

I have noticed Lebegue/Lebesgue is not in this list on meta.

Martin Sleziak

5:16 AM

(But I asked mainly because I saw it in several posts on the main site.)

Karl Kronenfeld

5:30 AM

Heh. Henri Leb\'egue (1856-1938) was a French palaeographer.

Leaky Nun

5:56 AM

@MartinSleziak it looks like Lebesgue always signed his letters to Borel with "H. Lebesgue"

Martin Sleziak

6:51 AM

Thanks for the responses!

Julius

8:43 AM

Hi. If $E[X \mid Y, Z]=E[X \mid Y]$ and $E[W \mid Y, Z]=E[W \mid Y]$, is it true that also $E[f(X,W) \mid Y, Z]=E[f(X,W) \mid Y]$ for measurable $f$?

Leaky Nun

@Julius does LOTUS help?

Albas

9:02 AM

Can you classify all possible Circle actions on spheres? Since $S^1$ is a Lie Group do all these actions generate global vector fields?

Julius

@LeakyNun, if it does, I don't see how. I don't think so, since those are conditional expectations. The result would be true for a single random variable since $\sigma(f(X))\subset \sigma(X)$. I think it's also true for $f(X,W)$ since, I think, each $A\in \sigma(f(X,W))$ can be written as a combination of unions and intersections of $B_i\in\sigma(X)$ and $C_i\in\sigma(W)$.

Albas

I mean they obviously do because you can always define the vector field as the one generated by the lie algebra of S^1 but how do these vector fields look like?

Balarka Sen

10:34 AM

@RyanUnger Too many Homs

@Albas Linear actions correspond to $S^1$ subgroups of $SO(n)$. Classifying them is tantamount to understanding the maximal torus of $SO(n)$, which is the block diagonal matrix consisting of $2\times 2$ rotation matrix blocks (with a $1$ somewhere depending on the parity of $n$).

In that torus you can have many $S^1$-subgroups, by taking an appropriate rational slope curve.

For example, consider $S^1$ acting on $S^3$ by $\theta \cdot (z, w) = (e^{i\theta} z, e^{i\theta} w)$. This is such an action, it's the Hopf fibration explicitly.

But this corresponds to the $(1, 1)$ curve in the 2-dimensional maximal torus of $SO(4)$. You can take, I dunno, a $(p, q)$ curve. Explicitly, $\theta \cdot (z, w) = (e^{i\theta} z, e^{i\theta p/q} w)$, something like that.

I don't know about nonlinear actions.

You can probably do crazy shit. Take a homology $3$-sphere $X$ and look $S^2 X$. By the double suspension theorem, $S^2 X \cong S^5$. There's always a circle action on a suspension, you can rotate along the meridians. I bet this is a nonlinear $S^1$-action on $S^5$.

No, the above was nonsense. Scratch.

ÍgjøgnumMeg

11:04 AM

@Mathein hab mich jetzt fuer die Studentenwohnheime beworben lol, mal schauen obs noch was gibt

MatheinBoulomenos

@ÍgjøgnumMeg cool!

ÍgjøgnumMeg

Die moderneren Studentenwohnheime in der Bahnstadt schauen mal fein aus

lol

Balarka Sen

If you take various complex line bundles on $S^2$, the total spaces of their unit circle bundles are linear quotients of $S^3$, I think, but I don't have an argument off the top of my head

For example $O(2)$ is the unit tangent bundle, and the total space of it's unit circle bundle is $\Bbb{RP}^3 = S^3/\Bbb Z_2$.

I think for $O(4)$ it's $S^3/Q_8$.

MatheinBoulomenos

@ÍgjøgnumMeg Bahnstadt ist generell ein edles, modernes Viertel

Balarka Sen

Yeah alright. Circle bundles over $S^2$ are classified by the linking number of the fibers. So they are all quotients of the Hopf fibration $S^3 \to S^2$, which is $\Bbb Z_n$-equivariant, by $\Bbb Z_n$.

Wait

Yeah that seems right. So total space of unit bundle of $O(4)$ is... $S^3/\Bbb Z_4$? They are all lens space!

Alright, so this gives some picture of how some linear $S^1$-actions on $S^3$ behave. Look at $S^3 \to S^3/\Bbb Z_n \to S^2$, where the last map is the bundle projection of $O(n)$. Let $S^1$ act on $S^3/\Bbb Z_n$ by acting principally on the circle fibers of $S^3/\Bbb Z_n \to S^2$.

This lifts to an $S^1$-action on $S^3$, right? Use the $n$-fold covering map $S^1 \to S^1/\Bbb Z_n = S^1$ and a big commutative diagram.

Hm, so now to fix the nonlinear nonsense. I feel like I can cook up a nonlinear circle action on spheres someway, somehow.

ÍgjøgnumMeg

11:28 AM

@Mathein schaut cool aus, ist wahrscheinlich besser als privates Leben

lol

Rithaniel

12:32 PM

A quick question: Does $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ have uncountable cardinality?

I want to say yes because allow $f:\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\mathbb{R}$ to be defined such that $f(x)=\text{lim}_{n\to\infty}\frac{x_{2n}}{x_{2n+1}}$

Where $x=(x_1,x_2,\cdots ,x_i,\cdots)$

Since this includes all rational numbers as well as the accumulation points of rational numbers, I believe the function is onto, but I feel like there might be some oversight on my part, because $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ looks like it would be a countable union of countable sets.

(Actually, verified it by just doing a google search. It's the set of sequences of integers. Easy enough.)

Mathphile

12:50 PM

11

Q: Conjecture: "For every prime $k$ there will be at least one prime of the form $n! \pm k$" true?

Using PARI/GP, I searched for primes of the form $n!\pm k$ where $k \ne 2$ is prime and $n\in \Bbb{N}$. With the help of user Peter, we covered a range of $k \le 17028457$ and couldn't find a prime $k$ for which $n!+k$ has no primes. Observations: $(1)$ When $n \ge k$, $n! \pm k$ cannot be ...

Any ideas, heuristics or guesses?

Alessandro Codenotti

@Rithaniel For every $x\in\bigoplus\Bbb Z$ there is $N\in\Bbb N$ with $x_k=0$ for all $k\geq N$

So in particular $|\bigoplus_{n\in\Bbb N}\Bbb Z|=\sum_{n\in\Bbb N}|\Bbb Z|^n=\sum_{n\in\Bbb N}|\Bbb Z|=\aleph_0\cdot|\Bbb Z|=\aleph_0$

Rithaniel

Is this a syntax error on my part? Because I know what you're referring to, and I meant to be specifying the group where infinitely many elements can be nonzero.

Alessandro Codenotti

Then you're talking about $\prod_{n\in\Bbb N}\Bbb Z$, a very different object

Rithaniel

Okay, then yeah, I was using the wrong syntax.

Alessandro Codenotti

1:05 PM

That one is uncountable, in fact it has the same cardinality as the reals

Rithaniel

I suspected that might be the case, and was trying to formulate a surjection in either direction to prove there existed a bijection, but the other direction wasn't coming to me.

Alessandro Codenotti

That's $|\Bbb Z|^{|\Bbb N|}=2^{|\Bbb N|}=2^{\aleph_0}=|\Bbb R|$ just by cardinal arithmetic

Rithaniel

I've not had any course which introduced me to cardinal arithmetic, unfortunately.

Martin Sleziak

Or, in other words, $\aleph_0^{\aleph_0}=2^{\aleph_0}$.

Main site: What is $\aleph_0$ powered to $\aleph_0$? and What's the cardinality of all sequences with coefficients in an infinite set? Found using Approach0.

Alessandro Codenotti

@Rithaniel Then you can inject $[0,1]$ into $\prod\Bbb Z$ by sending $0.a_1a_2a_3a_4\cdots$ to $(a_1,a_2,a_3,\cdots)$, it doesn't matter what you do with number with two decimal expansions, just pick of the them

Except with $1$ for which you want to use $0.\bar{9}$. Or inject $[0,1)$ instead

Rithaniel

1:14 PM

But what if what I want is to surject $[0,1]$ or $[0,1)$ or $\mathbb{R}$ onto $\prod\mathbb{Z}$?

Alessandro Codenotti

I think it's easier to inject $\prod\Bbb Z$ into $\Bbb R$

Actually the easiest is probably to inject $\prod\Bbb N$ into $\Bbb R$ with something like $a=(a_0,a_1,a_2,\cdots)$ goes to $0.a_0a_1a_2\cdots$ if $a$ is not eventually constantly $9$ and it goes to $3.a_0a_1a_2$ if $a$ is eventually constantly $9$

(and $\prod\Bbb Z$ and $\prod\Bbb N$ have the same cardinality because you can pick a bijection $\Bbb N\to\Bbb Z$ and use that on each coordinate to get a bijection)

Martin Sleziak

Hm... I don't really follow the last one; if you meant decimal expansion then you cannot have arbitrarily large digits.

Alessandro Codenotti

Oh lol right, there's stuff bigger than $9$ in $\Bbb N$

I guess it's time for continued fractions then

Martin Sleziak

Perhaps something similar would work with continued fractions, but that's probably overcomplicating things. Something like that is here: Baire space homeomorphic to irrationals.

Rithaniel

So, the theorem is that if there exists a surjection from A to B and a surjection from B to A, then there exists a bijection between then, but having an injection from A to B doesn't mean you have a surjection from B to A, right? Like, I can inject from $\mathbb{R}$ to $\mathbb{N}$ by taking every real number to it's first digit.

Martin Sleziak

1:24 PM

Anyway, cannot we simply use that $\mathbb Z$ can be embedded into $\{0,1\}^{\mathbb N}$ and then use that $\{0,1\}^{\mathbb N}$ has the same cardinality as $(\{0,1\}^{\mathbb N})^{\mathbb N} \cong \{0,1\}^{\mathbb N\times\mathbb N}$?

Alessandro Codenotti

The theorem is that if there is an injection $A\to B$ and an injection $B\to A$ then there is a bijection $A\to B$. With choice we can use surjections instead. If there is an injection $A\to B$ then there is a surjection $B\to A$. With choice if there is a surjection $A\to B$ then there is an injection $B\to A$

Martin Sleziak

@Rithaniel Did you mean to write about surjection from $\mathbb R$ to $\mathbb N$ rather than injection?

Rithaniel

Ah, right, that wouldn't be an injection, my mistake.

My brain for some reason had me thinking injection meant "surjects onto a subset."

Alright, yeah, thinking this stuff through with the proper definition in my head again, the world makes sense again.

Mike Miller

1:40 PM

@BalarkaSen @Albas Fibered knots allow you to construct nonlinear circle actions on $S^3$ and $S^4$. I think these were first found in dimension 4 by Pao.

Albas

Are non linear circle actions not possible on $S^2$?

This brings me to another question whose answer I believe is in the negative but given a smooth manifold, is it true that every global vector field possible on the smooth manifold can be generated by the action of some Lie group?

Ryan Unger

2:26 PM

@BalarkaSen do you understand the point of these Lie theory papers

there's one on math.dg every day

I have no idea what their goal is

Akiva Weinberger

2:47 PM

Speaking of injections

My arm hurts

but at least I won't get chickenpox

By the way

If you're telling someone about Cantor stuff, and you say "There's more than one infinity"

Mathphile

3

Q: Does there exist some prime $k$ for which there will be exactly two primes of the form $n!+k$?

This is a question related to my recent question Conjecture: “For every prime $k$ there will be at least one prime of the form $n!\pm k$” true? Using PARI/GP I searched for the number of primes of the form $n!+k$ possible for each prime $k\le 2300$. I observed that for some prime $k$, all number...

elementary-number-theory prime-numbers conjectures

any ideas about this?

Akiva Weinberger

and they ask "How many infinities are there?"

How would you respond?

If you say "Infinitely many" the natural response, given what you've just told them, is "Which infinity"

In this imaginary conversation I'm imagining myself muttering something about Russell's paradox and losing them

In any case that's the vaccines done, now to go to the glasses place to get my eyes checked

ÍgjøgnumMeg

3:04 PM

Alan Turing revealed as the new face of the £50 note!

Mike Miller

@Albas No. You can prove this by hand by studying what the possible fixed point sets are

anakhro

Hi Mike.

Mike Miller

@Albas If the vector field has a flow for all time then flowing defines an action of R whose derivative at zero is the vector field. If you don't have a flow for all time then that's not possible.

Hi. I'm just responding to a question.

anakhro

@AkivaWeinberger "many" would be my response.

But I think Russell's paradox is pretty accessible to the layman.

Albas

3:25 PM

@MikeMiller Ah, yes the one parameter way of talking about flows. Thanks.

Thorgott

Given a set $\Omega$, a ring of sets $\mathcal{R}$ on $\Omega$ and a pre-measure $\mu$ on $\mathcal{R}$, is there a good criterion to affirm the existence of non-$\mu$-measurable sets (assuming AC)? I'm particularly interested in the case $\Omega=\mathbb{R}$, $\mathcal{R}$ the ring of finite unions of half-open intervals and $\mu$ being absolutely continuous w.r.t. the Lebesgue measure.

Balarka Sen

3:45 PM

@MikeMiller Huh I see.

@Albas Any complete vector field can be thought as a homomorphism R -> Diffeo(M), like Mike said.

Albas

@BalarkaSen Yea, I saw it was written in my notes. Must have forgotten

Watcha upto @BalarkaSen?

Balarka Sen

Hm, if I let $S^1$ act on $S^2$ by rotation, and then I take suspension of this I get an $S^1$-action on $S^3$ which fixes a circle, right?

@Albas Nothing much, freewheeling.

Can I get $S^1$-actions on $S^3$ which fixes a knot?

Then I can suspend to get an $S^1$-action on $S^4$ which fixes a locally non-flat submanifold. That cannot be conjugate to a smooth action.

I dunno, weird questions maybe

Ultradark

4:08 PM

The following is the metric tensor for the $2$-sphere -----> $ds^2 = d\theta^2+\sin^2(\theta)d\phi^2$

Balarka Sen

There are no smooth $\Bbb Z/n$-actions on $S^3$ which fixes a knot.

Smith conjecture

In mathematics, the Smith conjecture states that if f is a diffeomorphism of the 3-sphere of finite order, then the fixed point set of f cannot be a nontrivial knot. Paul A. Smith (1939, remark after theorem 4) showed that a non-trivial orientation-preserving diffeomorphism of finite order with fixed points must have fixed point set equal to a circle, and asked in (Eilenberg 1949, Problem 36) if the fixed point set can be knotted. Friedhelm Waldhausen (1969) proved the Smith conjecture for the special case of diffeomorphisms of order 2 (and hence any even order). The proof of the general case...

This is too complicated for me lol

Ryan Unger

@BalarkaSen doesn't that say it can't fix only a knot

Balarka Sen

That's what I mean.

I'm not a precise person

I mean, otherwise the question is dumb lol.

Take the involution on $S^3$ which switches the two solid torii in it's genus 1 Heegaard decomposition. Take a $(p, q)$-torus knot on the fixed torus :P

Mike Miller

4:30 PM

My construction must be wrong then

Balarka Sen

4:46 PM

Wait, this was your strategy?

Alessandro Codenotti

@AkivaWeinberger at least 4

Karl Kronenfeld

over 9000 ofc

Mike Miller

@BalarkaSen I understand my failure now. I tried to get a circle action from a fibered knot. This is fine if the knot complement has a circle action giving the fibering.

But that's only going to be possible if the knot / fibering is trivial.

Apparently this strategy will actually work one dimension up.

Balarka Sen

Ah I see. So the only way to get an open book decomposition on the knot complement which fibers over S^1 is the trivial knot

Is that correct?

You wanted to act by S^1 on the total space by acting on the base of S^1 (it's like a family of Seifert surfaces parametrized over S^1)

Ultradark

5:02 PM

Any theorems out there for counting closed orbits on surface homeomorphic to $S^2$?

Clairut's relation seems appropriate here

Mike Miller

@BalarkaSen Right. You would need the monodromy f to be a periodic diffeomorphism to cook up a circle action at all. But the monodromy is assumed the identity near the boundary and if it was periodic it would globally by the identity.

Balarka Sen

Aha, gotcha.

Very nice ideas

Hi @Ted

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$Mathematics$ Mathematics