Mathematics - 2019-07-15

Akiva Weinberger

04:08

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

Martin Sleziak

04:59

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

05:16

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

Karl Kronenfeld

05:30

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

Leaky Nun

05:56

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

Martin Sleziak

06:51

Thanks for the responses!

Julius

08:43

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

09:02

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

@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

@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

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

lol

Rithaniel

12:32

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

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

13:05

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

13:14

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

13:24

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

13:40

@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

14:26

@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

14:47

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

15:04

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

15:25

@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

15:45

@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

16:08

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

16:30

My construction must be wrong then

Balarka Sen

16:46

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

17:02

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

Ted Shifrin

Hi, a @Balarka

Mike Miller

There must be some way to relax this assumption on boundary monodromy

Don't know anything else

H

Semiclassical

hi @ted

I applied for a physics instructor position in the fall at an undergraduate liberal arts college nearby yesterday

bit of a crapshoot but we'll see how it goes

Ted Shifrin

17:18

Hi @Semiclassic: It would probably be wiser to cast a wider net.

Semiclassical

yeah

this is one which just happened to show up on my radar

(i.e. they need someone for a specific course in the fall)

Ted Shifrin

Sounds like hardly any money.

Drew Brady

17:32

Ted do you have .a minute to be bothered by a differential geometry question?

Ted Shifrin

We'll see what the question is. I'm half packing, half paying attention.

user131753

web.archive.org/web/20020308033648/http://…

Drew Brady

[More details here, but sadly not receiving much attention: math.stackexchange.com/questions/3290962/…

Here's the setup. We're on the n-sphere, and considering convex functions on the n-sphere

Mike Miller

Oof too hard for me

Drew Brady

What does this mean? It means for a map f: S^{n-1} -> R any geodesic gamam on the n-sphere must yield a convex map f \circ gamma.

*gamma

Wow. Let me try that again: for every geodesic $\gamma: [0, 1] \to S^{n-1}$, the composite map $f \circ \gamma$ should be convex.

Ted Shifrin

17:35

You might as well change coordinates to diagonalize $F$ in the first place.

Drew Brady

Absolutely.

(my partial response does this)

in which case f looks like a weighted sum of squares.

Ted Shifrin

Since geodesics are great circles, it should just suffice to study $F$ on a $2$-plane through the origin.

Drew Brady

yes that's right. (that's what my answer does too)

So you can restrict to span{e_i, e_j} intersected with S^{n-1}

Ted Shifrin

But if $F$ isn't positive-definite, can't you see it won't be convex on a $2$-plane with mixed sign (or even on which it's negative definite)?

Drew Brady

Yielding a great circle, and e_i, e_j picked to be assocaited with eigen values which are distinct.

Leaky Nun

17:36

@TedShifrin good luck with the packing!

Ted Shifrin

Oh, I didn't see you'd answered your own question.

I hate it, but thanks, @Leaky.

Drew Brady

Really? Does positive definiteness matter here? I think distinctness of eigenvalues is all one needs.

Tobias Kildetoft

@TedShifrin Moving or going on holidays?

Ted Shifrin

I doubt distinctness matters.

@Tobias: Moving 3 miles.

But that isn't so much simpler than moving 3000 miles.

Drew Brady

Well what I think I showed is that if even two eigenvalues are distinct then you're hosed.

Tobias Kildetoft

17:38

well, at least it is not that far. But moving always sucks

Drew Brady

But let me know if something is not convincing in my argument. Basically, we can restrict to the great circle in the plane spanned by the two eigenvectors associated to the two distinct eigenvalues.

Ted Shifrin

Interesting. So $x^2+2y^2$ is not convex on the unit circle?

Drew Brady

Parameterizing the function in this space, we get f(theta) = (lambda_1 - lambda_2) cos^2(theta) + constant.

Ted Shifrin

That's $1+y^2$.

Hmm ...

Drew Brady

Yes, that is true...

Ted Shifrin

17:41

Yeah, I guess you've convinced me. So you don't need me!

P.S. This isn't differential geometry :P

Drew Brady

I'm confused, 1 +y^2 is convex, though

What am I misunderstanding.

Ted Shifrin

You argued in your proof that $y^2$ is not convex. The $1$ is certainly irrelephant.

Mike Miller

I never understood questions without a question mark

Ted Shifrin

If we take the points $(0,\pm 1)$, it's pretty clear we violate convexity on the semicircle, isn't it?

LOL, @MikeM?

Balarka Sen

@MikeMiller Is that so

Mike Miller

17:47

@BalarkaSen hm?

Balarka Sen

lmao

Secret

@AkivaWeinberger I will say it this way: There are so many infinities such that the whole list is an infinity larger than any of them

(that's really what a proper class of cardinals is really...)

Ted Shifrin

I agree with you, a @Balarka?

Albas

@MikeMiller Lol yeah. Your first reaction is to read it as a statement lol

Drew Brady

Just to be concrete, you're saying we should essentially consider f(t) = sin^2(t) + 2cos^2(t) for t in [0, pi]

Ted Shifrin

17:48

Oh, wait, I got that precisely wrong, @Drew. Let's try the points $(\pm 1,0)$.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Nah, let's do $\sin^2 t$ on $[0,\pi]$.

hi, demonic @Alessandro

Drew Brady

well yes its obvious that as a map [0, \pi] -> R it is not convex

Wait, I'm confused are you saying you agree with my sketch or that you don't agree with what I'd sketched @TedShifrin?

Ted Shifrin

I'm thinking only about this discussion. So doesn't that function violate your definition of convexity on the circle?

Drew Brady

Yes, this means that y^2 as map from the circle to R is not convex.

(when I said earlier that $1 + y^2$ is convex, I thought you meant as a function from $\Bbb R^n \to \Bbb R$ !!)

Ted Shifrin

17:56

So this says your argument must be right. I had never thought about this before.

Drew Brady

Bizarre that in the tangent space this function is convex but not once on the manifold.

Ultradark

Why do musicians stick to one genre?

Drew Brady

Anyways, thanks.

Btw, I agree this isn't serious differential geometry. It did reduce to polar coordinates after all ;)

Balarka Sen

@ultradark lol they don't

Ted Shifrin

@Drew: I've never thought about this before, as I said, but it's the curvature of the circle/sphere that's causing problems.

s.harp

17:59

@BalarkaSen one could say they stick to music

Balarka Sen

good point

i am become defeat

Ultradark

I wonder if there's a famous artist (meaning musician) that's also a famous mathematician

s.harp

little known fact: homer's last name was pythagoras

Ted Shifrin

Raoul Bott was a somewhat accomplished pianist. Lots of mathematicians play instruments pretty seriously.

Balarka Sen

Huh I didn't know that

Ted Shifrin

18:04

Well, I have to know a few things you don't.

Balarka Sen

lol

Érico Melo Silva

18:17

@TedShifrin im really good at being bad at the mandolin

Ultradark

hahaha

Albas

The linking number for two smooth manifolds $M,N$ is defined as the degree of the map $l(x,y) = \frac{x-y}{||x-y||}$. If $M$ is the boundary of an oriented manifold say $\Sigma$ disjoint from $N$, the map $l$ can be extended to a map $L : \Sigma \times N \to S^k$ which when restricted to $\partial \Sigma$ is just $l$. This is smooth and also well defined because $\Sigma$ is disjoint fro $N$. Using the boundary theorem, it implies that $\operatorname{deg}(l) = 0$. Is this right?

Tobias Kildetoft

@Ultradark Tom Lehrer was probably more known for his music. But he was also a mathematician.

Ted Shifrin

Yes, he even taught at MIT and Harvard.

I think he ended up in the Econ departments, though.

@Albas: You needed to say that $M,N\subset\Bbb R^{k+1}$, dimensions, etc. But yes.

Balarka Sen

Linking number is very good

You can actually say that if $\partial \Sigma = M$, then the linking number is the number of points of intersection between $\Sigma$ and $N$, once made transverse.

(Try to prove it @Albas)

Albas

18:28

Cool. Will try to.

I heard it today from my project guide. Apparently they are used in TQFT's somehow.

Ted Shifrin

each one counted with sign @Balarka @Albas

Balarka Sen

Yes ^ intersection number, rather.

Ultradark

** Star this if you are right handed **

** Star this if you are left handed **

Ryan Unger

@Ultradark I lost both my hands in a chemical accident.

s.harp

Lefschetz actually did lose both his hands

Ultradark

18:42

@RyanUnger before the accident were you right handed or left handed?

Ryan Unger

@s.harp It was a reference to him

Leaky Nun

18:55

cool til @RyanUnger

user193319

19:14

Does anyone else find it strange that the only numbers needed in analysis are the naturals and numbers of the form $\frac{1}{n}$ or $\frac{1}{2^n}$ for $n \in \Bbb{N}$?

Tobias Kildetoft

@user193319 How would those numbers suffice for analysis?

user193319

At least those are the only numbers needed to prove certain "foundational" results in real analysis.

Ryan Unger

@user193319 having all the real numbers is kind of important

complex numbers too...

user193319

Well, e.g., whenever I do a "$\epsilon$-$\delta$ proof, I really only use natural numbers (don't need any irrational epsilons).

As another example, I only need naturals to describe what $\lim_{n \to \infty} f_n(x) = \infty$ means, where $f_n$ is, say, a sequence of continuous functions.

Tobias Kildetoft

that is just because those always just require you to pick a small number

(or a large one)

user193319

19:18

In fact, I needed the fact that to show that $\{x \in \Bbb{R} \mid \lim_{n \to \infty} f_n(x) = \infty \}$ is a countable intersection of $F_\sigma$ sets.

Just need the naturals--that's all.

I suppose that should be expected, though, since all other numbers can be constructed from them.

It's still strange though.

Ryan Unger

I think you've just discovered what an Archimedian field is

user193319

I guess so

So is anyone here a weeb?

Alessandro Codenotti

Hi @Ryan

I have a functional analysis question which is either easy or a typo in my notes

Ryan Unger

@AlessandroCodenotti ok let's hear it

Alessandro Codenotti

I have a densely defined operator $D$ on a $H$ separable Hilbert such that $D$ is self-adjoint and $(D\pm i)^{-1}$ are compact ($D$ comes from a spectral triple essentially)

Now I can show that the bounded transform of $D$, namely $D(1+D^2)^{-1/2}$, is Fredholm

I want to show that $D$ itself is Fredholm

Ryan Unger

19:33

is $(D\pm i)^{-1}$ compact automatic or a hypothesis

Ramit

Hi guys. How to represent conjugate of z and mod of z in mathJax? Thanks.

Alessandro Codenotti

@RyanUnger hypothesis

In my notes I just wrote $D=D(1+D^2)^{-1/2}(1+D^2)^{1/2}$, where the first part is the bounded transform so it's Fredholm, while the second half is unitary, so $D$ is Fredholm

However I don't see why the second part is unitary now

Ryan Unger

you mean why is $(1+D^2)^{1/2}$ unitary?

Alessandro Codenotti

yes

I don't see it

Ryan Unger

I don't buy it...look in finite dimensions and take D to be the identity

the compactness is automatic

the matrix is diagonal $\sqrt 2$ right

not unitary

Alessandro Codenotti

19:39

@RyanUnger Looks right to me

Ultradark

how many distinct ways can you torus fibrate a sphere

Ryan Unger

also unitary implies bounded and no way is that bounded right

ÍgjøgnumMeg

waddup yo

Alessandro Codenotti

Uhm wait so $1+D^2$ is invertible, right?

Ryan Unger

we decided that last time, right

how else do you define the bounded transform

Alessandro Codenotti

19:43

Yes

True

Doesn't have $0$ in its spectrum, I remember

Leaky Nun

@AlessandroCodenotti studierst du nun in Deutschland?

Alessandro Codenotti

Ja, seit Oktober

Ryan Unger

so uhhh, idk. I haven't studied unbounded Fredholm operators

Leaky Nun

@AlessandroCodenotti aber studierst du auf Deutsch oder auf Englisch?

Alessandro Codenotti

Auf Englisch

@RyanUnger I don't actually buy that $D$ is Fredholm now

ÍgjøgnumMeg

19:48

@Alessandro muss glaub ich in Heidelberg auf Deutsch studieren :(

Alessandro Codenotti

Because we want to use the bounded transform to get a Fredholm module from a spectral triple, why not use $D$ directly if it is Fredholm?

Bonn is very international, so we have all classes in English

Almost 40% of the students in the maths master program are foreigners

Ryan Unger

I actually don't know what an unbounded Fredholm operator is

ÍgjøgnumMeg

Fair, I think most of Heidelberg is German

Alessandro Codenotti

Can't you use the same definition? finite dimensional kernel and cokernel?

Ryan Unger

maybe, but there's a very nice theorem in Conway that gives 7 equivalent definitions for bounded ones

idk how much of this carries over

Alessandro Codenotti

19:54

Oh ok so the idea is that you have a closed operator $D\colon\mathrm{dom}(D)\to H$, and it is bounded if you put the graph norm on its domain, so you can use the usual definition

(In this case $D$ is closed because it is self-adjoint)

Ryan Unger

Oh my god this guy is computing $[\nabla,\star]=0$ in coordinates

Ultradark

20:09

Can $S^2$ be expressed as a 3D projection of a 4D shape?

Ryan Unger

22:24

@ÉricoMeloSilva did you sign up for healthcare

Transcript for

$Mathematics$ Mathematics