Mathematics

@Rob: talking about probability - when I was doing the presentation on Monday in Paris, one professor told me that the word supermartingale is so cool that he wants now to include it at least in one of his papers. I promised him that if write the paper together, this word will appear there.

ymar

Hello!

Asaf Karagila

@robjohn I am going to leave soon.

robjohn

@AsafKaragila No! where are you going?

Ilya

9:16 AM

@robjohn he is cooking

@ymar: hi

Asaf Karagila

@ymar On Monday I met with Haim and we discussed the answer. He was a bit annoyed that he couldn't have shown that there are $2^\kappa$ many non-isomorphic groups directly.

robjohn

@Ilya Ah, that is excusable :-)

Asaf Karagila

@Ilya Not today. Today we're going to a restaurant.

Ilya

@AsafKaragila so you lied

Asaf Karagila

@Ilya I made dinner last night.

ymar

9:17 AM

@AsafKaragila That's a pity... I don't think I will understand his answer soon. But at least I know what the answer is! :)

Ilya

@AsafKaragila Matt will agree that then you should use "I cook the meals" rather than "I am cooking the meals"

robjohn

@ymar which answer is this?

ymar

5

A: How many non-isomorphic abelian groups of order $\kappa$ are there for $\kappa$ infinite?

It's known by Fisher, Eklof and Shelah (see theorem 2.1 in [1]) that there are abelian groups which are stable but not superstable. Another well known result of Shelah (see [2]) is that, roughly speaking (again, check the paper below for details), if $T$ is not superstable, then $T$ has $2^{\lamb...

Asaf Karagila

@Ilya Matt isn't here.

Ilya

@AsafKaragila btw, do you know, why? I hope nothing serious has happened

Asaf Karagila

9:22 AM

@Ilya No idea. I went to bed around 4am last night and woke up only an hour or so ago.

Probably needed time without distractions for studying stuff which is not set theory.

Now I am going to trim my beard and take a shower. Then I am going to the university to work on my Whitehead thingie and then I am coming back home.

Ilya

@AsafKaragila Report accepted.

t.b.

@AsafKaragila still here?

Asaf Karagila

Yeah.

t.b.

What exactly is that theorem of Brunner you want to apply?

Ilya

@tb: good morning! haven't seen you for a while

t.b.

9:30 AM

Morning Ilya. I saw you liked Paris :)

Ilya

@tb yet another time - but I think I only like to visit it, not to live there

t.b.

It's hard not to.

Ilya

@tb I wasn't pretending I am original in that :) To be honest, I was in Paris only Sunday's and Monday's night (after 9pm) and in the morning - the whole Monday I spent in INRIA.

Asaf Karagila

@tb In Solovay's model every Banach space is dreamy.

t.b.

So every Banach space is isomorphic to $l^1(D)$ with $P(D)$ Dedekind finite? I don't follow.

Ilya

9:34 AM

But then with my supervisor we had a 2-hours walk on Monday's night, visited Seine and walk around the city center. That was so cool that even in a weekday, after 11pm there were still a lot of people there. So unusual for the Netherlands

Infinite compact discrete spaces?

t.b.

@AsafKaragila oh, silly me. You don't necessarily have a Hamel basis.... :/

Asaf Karagila

@tb I'm not following you... :-)

Ilya

@AsafKaragila Follow the White Rabbit

Asaf Karagila

@Ilya I am the White Rabbit.

Ilya

@AsafKaragila O RLY? have you already trimmed your beard?

t.b.

9:40 AM

@Ilya I am not the white rabbit if anything then I'm the white teddy!

@AsafKaragila I messed up completely. So the real point is that you want to argue that $\ell^1$ is reflexive but not isomorpic to its dual?

Asaf Karagila

@Ilya No, not yet. My girlfriend is about to leave and she goes in and out of the bathroom... it would make showering and trimming one's beard quite less comfortable.

@tb No, I want to argue that $\ell^1$ is algebraically reflexive.

robjohn

@Ilya trimming is allowed. I haven't shaved mine for 35 years.

Ilya

@AsafKaragila I won't comment

@robjohn trimming is allowed for rabbits?

@robjohn or, you have a beard! nice :) me and teddy we're beardless

Asaf Karagila

@Ilya You shouldn't because you don't know how our apartment looks like and what is the dynamics of our relationship in that sense.

robjohn

@Ilya with greens and some vegetables :-)

t.b.

9:43 AM

@AsafKaragila okay, but won't every $\ell^p$ do?

Ilya

@robjohn for Asaf that should sound quite evil :D

Asaf Karagila

@tb Because in Solovay's model topological duals are algebraic duals?

t.b.

If that is true then yes. DC is certainly enough to identify the topological dual of $\ell^p$ with $\ell^q$.

Asaf Karagila

Holy Scheiße.

So $\ell^2$ is isomorphic to its algebraic dual.

And that is what I wanted all along!

t.b.

Provided that it is true that Banach => dreamy you get a continuum of mutually non-isomorphic Banach spaces that are algebraically reflexive and I would bet that you can combine them using $\ell^p \oplus \ell^q$ to get a continuum of non-isomorphic examples that are isomorphic to their algebraic duals.

Asaf Karagila

9:51 AM

@tb Yes, that is quite the idea!

Now, why did you write Provided in such way?

t.b.

Because that's the proviso for me (I don't understand that statement -- that is I just believe you this point). You'd probably need the open mapping theorem from DC which I believe should be true because it only relies on the Baire category theorem.

Asaf Karagila

Why would I need the OMT?

t.b.

One of the ways I know how to tell those spaces apart would be to derive a contradiction from Pitt's theorem and the OMT.

There's certainly some checking to do that I didn't use more than DC anywhere, but I'm quite confident that something should be doable.

Asaf Karagila

Well you would only need that for asserting there is a fora of interesting examples.

t.b.

Yes.

Asaf Karagila

9:58 AM

My arguments for $\ell^2$ being self-dual and $\ell^\infty$ having a "smaller" dual stand.

t.b.

Yes. It should work for all $\ell^p$'s.

Asaf Karagila

Well, in the case of $p\neq 2$ we have the algebraically reflexive property but not necessarily the duality property. Right?

I have never seen this before, although all the needed axioms were out there already.

t.b.

Duality property? Isomorphic to the dual?

Asaf Karagila

Yes, that's what I meant.

I'm just excited because I've been trying to find this sort of space for almost a year now.

t.b.

I'm pretty certain that you can exclude the duality property for $p \neq 2$, but that would need some checking, as I said.

Asaf Karagila

10:02 AM

@tb Of course. :-)

Well, I am going to take a shower and all that and head out to the university.

t.b.

Okay, have a nice day!

I'll probably be here tonight.

Asaf Karagila

I will probably be away.

t.b.

Okay, then see you another time :)

Asaf Karagila

We're celebrating a year together today, so I'm probably gonna be preoccupied.

Ilya

@AsafKaragila congratulations! I think I've known you for more that a year and you always had this girlfriend

Asaf Karagila

10:05 AM

Thanks... :-)

t.b.

Okay, I should go, too. See y'all!

Ilya

oh no

@tb: I was waiting for you to finish with Asaf to ask you to take a look on this question of mine

maybe later now - if you have time, of course

t.b.

@Ilya I just got your ping. What exactly do you want to know from me?

Ilya

@tb: 1. what do you think about the method I've described there
2. how to resolve the issue there
3. maybe you can advise some references on more neat methods

but that might take some time - even the question is not too short. The answer I received is quite ok, but I also wonder about your opinion. Just in the case you have time - I can also wait, that's not urgent at all

t.b.

Then I suggest we discuss this at another point in time because I really should be gone in a few minutes. Two very useful things in this context are Dynkin's $\pi$-$\lambda$-theorem and the monotone class theorem. That is what immediately comes to mind given your question. But probably you look for something else.

Ilya

10:20 AM

@tb: I thought of them already - and I was looking for some exercises on that topic, just to understand if it would help or not. OK, let's discuss it another time, good luck with things you need to do

t.b.

I think I learned that stuff from Durrett's book, available here

I hope this helps for the moment. Sorry for not having more time right now.

See you soon!

Benjamin Lim

10:54 AM

$\ubsetneq$

$\subsetneqq$

Kannappan Sampath

Hi @Ben,

Benjamin Lim

Hey Kannappan I'm kinda busy typing an assignment now. I'll try to talk to you later.

Kannappan Sampath

Sure!

Zhen Lin

The fact that we even have a symbol $\subsetneq$ says that something has gone horribly wrong with $\subset$...

Ilya

11:11 AM

@ZhenLin yes, it annoys me - I'm so used for $\subset$ as just a weak inclusion

Benjamin Lim

@ZhenLin I am trying to see why the degree of $\Bbb{Q}(\sqrt{3} + \sqrt[2]{3})$ is exactly 6, I know it is at most 6.

Zhen Lin

It's contained in an extension of degree 6, so you just need to show it's not 2 or 3. I think you can do this by brute force.

anon

What's the difference between $\sqrt{3}$ and $\sqrt[2]{3}$?

Zhen Lin

He means $\sqrt[3]{3}$, I'm sure. :p

anon

oh okay

Benjamin Lim

11:23 AM

Sorry, typo I meant why $\Bbb{Q}(\sqrt{3} +\sqrt[3]{2})$

square root of 3 plus cube root of 2

N3buchadnezzar

Heya

Benjamin Lim

I know it is not 2

because the quadratic formula cannot produce cube roots

Skullpatrol

Hi @ZhenLin could I get get your opinion on the mathematical content of the book "The Road to Reality" by Roger Penrose please?

ymar

@BenjaminLim Take a look at this question of mine.

Skullpatrol

Hi @N3buchadnezzar What's up?

Benjamin Lim

11:26 AM

@ymar Thanks,

N3buchadnezzar

@Skullpatrol Integrals as usual. You?

ymar

@BenjaminLim Don't miss André Nicolas' comment!

Benjamin Lim

@ymar Not so easy to see why $\Bbb{Q}(\sqrt{3})$ is a subfield of what I wrote above

besides our questions are not exactly the same

Zhen Lin

@Benjamin: Then show it's not 3. Show that no non-trivial linear combination of $1, \alpha, \alpha^2, \alpha^3$ can be zero, where $\alpha = \sqrt{3} + \sqrt[3]{2}$.

Skullpatrol

@N3buchadnezzar Not much, just trying to get Zhen's opinion on the mathematical content of the book "The Road to Reality" by Roger Penrose.

anon

11:29 AM

What about the math content?

Benjamin Lim

@ZhenLin A lot of algebra crunching, I'll do that

ymar

@BenjaminLim They are very similiar. Try taking the same path and it should work.

Skullpatrol

@anon Is it pedagogically sound?

anon

Well, it gives you a fair feel for what math says and why, but it can't be used as a textbook even though it has "exercises." I only read/skimmed in a few hundred pages though, iirc.

Benjamin Lim

@ymar Thanks.

Skullpatrol

11:35 AM

@anon That is what I wanted Zhen's opinion on, can it be used as a textbook for an independent learner.

ymar

@BenjaminLim There is a theorem of Abel which says something to the effect that you'd have to be very unlucky for that not to work.

Matt N.

11:50 AM

@tb ...with $n \in \mathbb Z$, you meant. And for a finite cyclic group $G$ I don't get a nice duality, do I? bbl

Benjamin Lim

@ZhenLin Hey you told me to show that no non-trivial linear combination of $1,\alpha...\alpha^3$ can every equal zero. So when I write out this linear combination, is it valid to say that the coefficients of $\sqrt{3}$ must be zero, the coefficients of $\sqrt[3]{2}$ must be zero and so on? Can we compare coefficients like that?

robjohn

@ZhenLin you mean rational linear combination?

Zhen Lin

of course.

N3buchadnezzar

@Skullpatrol I would say no, but I think you can learn quite a bit from it

Zhen Lin

@BenjaminLim: Yes, provided you are willing to assume that $\sqrt{3}$ and $\sqrt[3]{2}$ are linearly independent. (They are, of course.)

N3buchadnezzar

11:59 AM

@Skullpatrol The problem I see with it is that the content is very diverse, and only skimms large parths of the math. Eg, the study on manifolds.

Skullpatrol

@N3buchadnezzar Thank you for being polite and replying to my question.

hhh

Is there any general definition for the term "real function"? Is it $f: \mathbb R\mapsto \mathbb R$? $f: \mathbb R\mapsto \mathbb R^n$? $f: \mathbb R^m\mapsto \mathbb R$? $f: \mathbb R^m\mapsto \mathbb R^m$?

1

Q: Harmonic real function, $\triangle f =0$, but not in origin?

Problem in English (original problem 7 on page 813 here) Suppose $f(|\bar{x}|)=\sqrt{x_{1}^2+...+x_{n}^{2}}$. For what kind of real $f$ it holds that $f$ is harmonic everywhere but not in origin? If $f$ is harmonic, then $\triangle f=0$. Definitions The "real function" apparently here...

Benjamin Lim

Well in the expansion I have $1,\sqrt{3}, \sqrt[3]{2}, \sqrt{2}\sqrt[3]{2}, \sqrt[3]{2}^2, \sqrt[3]{2}^2\sqrt{3}$

Skullpatrol

@robjohn Hi, how did the blood tests go?

Benjamin Lim

@ZhenLin There are exactly six of them. Now are they a basis for $\Bbb{Q}(\sqrt{3},\sqrt[3]{2})$?

If I know that they are then linear independence is immediate.

Zhen Lin

12:03 PM

They are, but again, this requires a little bit of work.

anon

@hhh What does $|\vec{x}|$ mean? If that's the same as $\sqrt{x_1^2+\cdots+x_n^2}$, then doesn't the supposition say that $f$ is the identity on $[0,\infty)$?

Benjamin Lim

@ZhenLin as in??

anon

hhh: Oh, nevermind, there's a typo in your problem. You forgot to put an f() on the outside, but you wrote the correct thing in the scan.

Zhen Lin

Well, showing that they're linearly independent!

Benjamin Lim

i know, but as in not in the usual way of crunching stuff

hhh

12:24 PM

@anon yes in this case $f: \mathbb R^{n}\to\mathbb R$ but generally, "real function" is an open term or? Is this actually correct domain $\mathbb R^n$?

Zhen Lin

@hhh: Don't use $\mapsto$ for that, use $\to$.

hhh

@ZhenLin Thanks, fixed.

anon

\mapsto is for explicitly defining a map, e.g. $x\mapsto 2x$. anyway a real function is anything $A\to \mathbb{R}$ I think.

Zhen Lin

A real-valued function is something with codomain $\mathbb{R}$. A function of a real variable is something with domain $\mathbb{R}$.

Kannappan Sampath

I had a downvoter. Random question. Very Very elementary but for all that is good, I don't see anywhere I can be a bit more explanatory. sigh But this time, I am going to wait for the next instance and bring it to the notice of mods.

hhh

12:36 PM

@KannappanSampath I did not downvote you if you inferred that. I would care less about downvoters, they may have some meaning later but it is better to concentrate on one's own working and try to do the best. Do not let the gamification to steer your attention from math...

Kannappan Sampath

@hhh Hmm. Not even in my dream, did I guess that some name for who possibly would have downvoted. So, this strikes me as strange confession to make. But, anyway, thanks forr your advice, there. : )