Mathematics - 2011-10-28 (page 4 of 5)

J. M.

16:00

:D

t.b.

@Jack: you'll back there in no time!

Gortaur

terrytao.wordpress.com/2010/02/10/245a-notes-5-free-probability why here is written in 3. about algebras of RV just with values in R or C? I thought RV can have values in any measurable space (though clearly without algebraic structure)

Jack Schmidt

anyone know how to make a ghost or spider or something in tikz?

Gortaur

@JackSchmidt kill him brutally. oh no, sorry, read: ghost of spider

Jack Schmidt

hehe

J. M.

16:06

Spiders have souls? That's new for me... :)

Gortaur

no, but if you kill him in a really weird way you never know

it's a bad luck in Russia to kill a spider in your home (not a joke)

J. M.

It's never nice to kill anything. Unless it's about to bite you...

QED

I wonder if spiders are conscious

Gortaur

not all of them

QED

are some of them?

(which species?)

J. M.

16:12

We can't even claim a full understanding of the arthropod brain...

Gortaur

@QED I don't know, but some of then proved me to be unconscious

QED

ahh

Arthropod head problem

The arthropod head problem is a long-standing zoological dispute concerning the segmental composition of the heads of the various arthropod groups, and how they are evolutionarily related to each other. While the dispute has historically centered on the exact make-up of the insect head, it has been widened to include other living arthropods such as the crustaceans and chelicerates; and fossil forms, such as the many arthropods known from exceptionally preserved Cambrian faunas. While the topic has classically been based on insect embryology, in recent years a great deal of developmental mo...

this is very interesting

Gortaur

indeed ) unfortunately I have to leave

have a nice weekend (at least this wish refers here to all time zones)

QED

bye

J. M.

See you!

Srivatsan Narayanan

16:15

Bye, Gortaur.

Gortaur

wow, there are lots of you here, guys ))

don't forget to switch to winter time for whom it's still actual

QED

So what's the resolution of the folding a plane into a sphere problem?

It's impossible because it would crinkle and Gauss theorem is the formal version of this, but how do we know that we can apply that theorem?

J. M.

They're both surfaces. Gaussian curvature is an invariant of sorts...

e.g. the plane, the cylinder, and the cone all have zero Gaussian curvature, so wrapping cones and cylinders is a snap.

QED

I see it's about isometries which are distance preserving maps

so it's just exactly what we need

J. M.

16:30

Well, I must end here. See you guys!

QED

bye

Srivatsan Narayanan

See you J.M. I guess I will also get going for lunch.

Srivatsan Narayanan

17:35

There's always an enigma about non-mathematicians doing mathematics: math.stackexchange.com/questions/76716/….

Hi Nguyễn, I liked your question.

Nguyễn Duy Khánh

Thanks, Srivatsan

Alexei Averchenko

18:18

man, is algebraic geometry HARD!

Srivatsan Narayanan

That hard, eh?

Asaf Karagila

@JackSchmidt Funny how that was just when I went back to sleep :-)

Alexei Averchenko

i even suspect that that schemes business is easier than working with these projective spaces

i was wondering if a transformation from PGL_n exists that takes an arbitrary line into a line that lies entirely in some affine chart (except for one point). figuring out how to even write all this formally with homogeneous coordinates was hard

in fact, check it out:

consider a line L

L = P(W), where W \subset K^{n+1} is a 2-subspace

pick two vectors x, y \in W

we want to construct a basis e_1, \ldots, e_{n+1} of W such that all points of L except <x> lie in the chart Z_1 \neq 0

now, if i'm not mistaken...

e_1 = y

e_2 = x

pick e_3, \ldots, e_{n+1} in any way you please

the chart Z_1 \neq 0 is the complement of the hypersurface Z_1 = 0

Asaf Karagila

@Srivatsan: How did you sleep? :-)

Alexei Averchenko

so a point in L lies in the chart iff Z_1 \neq 0

since in our basis for each point in L we have Z_3 = \ldots = Z_{n+1} = 0, the only point that does not lie in the chart is [0, a, 0, \ldots, 0]

a \in K

which is exactly <x>

Srivatsan Narayanan

18:34

@Asaf It was surprisingly good, thank you! My alarm clock woke me up at 6 unnecessarily. I switched it off and jumped back to bed, getting up only at 10. ;)

Alexei Averchenko

now it is trivial to see that the inhomogeneous coordinates of points of L\<x> in the chart will be of the form (1 : t : 0 : \ldots : 0)

Asaf Karagila

Nice!

Alexei Averchenko

so they form a line

so we can apply Bezout

Asaf Karagila

I slept six hours, and just woke up with the +3 :-)

@Alexei: What's your home address? I'll send you some paper and pencils...

Alexei Averchenko

what? :D

Asaf Karagila

18:35

It looks like the chat is your notepad for solving questions.

Alexei Averchenko

i like peer review :D

Asaf Karagila

No one is reviewing... I think that me and Srivatsan are the only one following the chat right now, and I doubt anyone will read through this later.

Alexei Averchenko

don't be too harsh on me, i doubt anybody else in my entire town is into AG :)

could you review what i wrote, btw? please? :))

Srivatsan Narayanan

Unfortunately, the knob of my alarm clock that sets the alarm time has stopped working; I am stuck with something that screams everyday at 6.25 or so AM. ;)

Asaf Karagila

It's not even remotely close to my stuff.

Yikes.

Break it!

Srivatsan Narayanan

18:41

Well, grown rather fond of it that I do not want to replace it. Not when it's actually working at least. Also it was so cheap when I bought it that repairing it is just a waste of money.

Funny thing is, I am not even at home some days at that time. Wonder what the neighbors think. :=)

But yes, one of these days, I am gonna break it. :)

Asaf Karagila

Reminds me of the episode of Seinfeld where Elaine's neighbour went to Europe...

Srivatsan Narayanan

:) There was also the episode where a dog keeps barking, so they decide to kidnap it.

Are we speaking of the same episode?

Asaf Karagila

No.

I'm talking about the episode where Kramer had a pastrami slicer.

Srivatsan Narayanan

Not sure which one, but I will research it later.

Asaf Karagila

"The Slicer" is the 163rd episode of the NBC sitcom Seinfeld. This was the 7th episode of the 9th and final season.

Srivatsan Narayanan

18:50

Nope, definitely not seen it. Thanks for pointing it out.

Asaf Karagila

:-)

What's your mathematical inclination, @Srivatsan?

Srivatsan Narayanan

I work in theoretical CS. Some coding theory and complexity theory...

So I am generally interested in combinatorics.

Asaf Karagila

I see, nice. I have a good friend from my undergrad that did math & CS, and on our freshman year he came up with some clever algorithm to solve SATs and began working as a research assistant. He wrote a paper by the end of his undergrad.

I can tell you that I almost "proved" that P=NP on this very site the other day :P

Srivatsan Narayanan

I don't have much of a math background, but I am trying to learn some of those things slowly.

I mean, like calculus, general topology, functional analysis and the like.

Asaf Karagila

The last two years my university changed a bit the CS program and reduced the level of mathematics.

Srivatsan Narayanan

19:01

Who is this freshman friend who wrote the SAT paper?

Which is your univ, btw?

Asaf Karagila

arxiv.org/abs/1007.4935 my friend is Yoav, while the main author was the teacher in that freshman course.

Srivatsan Narayanan

Thanks. Not seen it before.

Asaf Karagila

I would have been surprised if you'd said otherwise :-D

Srivatsan Narayanan

:) You too in Ben Gurion University?

Asaf Karagila

Indeed.

Srivatsan Narayanan

19:05

In one of their lemmas, the number zeta^{-1}(2) figures. Isn't that cool?!

Actually, I haven't heard of your university before either.

Asaf Karagila

lol :-D

@tb: If you have nothing to do, help me instead of translating old papers :-)

Srivatsan Narayanan

Which papers is tb translating?

Asaf Karagila

His answer here

t.b.

Well, interestingly enough, people don't even see that JDH only answers half the question...

Asaf Karagila

Interestingly enough, you went Dubuque and used \mathrm :-D

Jonas Teuwen

19:20

Why does Dubuque do that?

t.b.

@Jonas: "I find the default italic fonts far too ugly. Alas, they messed up the roman fonts too with the recent font changes."

Srivatsan Narayanan

tb: How could you locate the comment so quickly?

Jonas Teuwen

Hmm, I find it more ugly if some people use a different font, but oh well.

t.b.

@Srivatsan: I knew what to look for: rahul roman dubuque site:math.stackexchange.com

Asaf Karagila

I know how he found it so quick. We discussed that a few days ago.

t.b.

19:25

Well, then it didn't take me much longer (and I forgot about that already). And who's that silly guy who thinks that \omega_1+1 is not compact?

Srivatsan Narayanan

Ok, although I can never seem to pull out things so quickly.

Asaf Karagila

@tb: Took you slightly longer, since at first you thought it was me :-D

Jonas Teuwen

Boss above boss.

t.b.

Oh, yes. Now I remember...

Heh, Joel's lecture at Harvard is finally up!

Asaf Karagila

Interesting.

The lectures appearing on that page have all been given, or this is a schedule for 2011-2012?

t.b.

19:43

Well, it's an ongoing workshop.

And I don't understand such questions open any book on C^* and you'll find the result!

Michel Kogan

hello there

t.b.

Hi there

Michel Kogan

i have a question: one of my friends found and prove some equations and theorems about Fibonacci numbers. The most important was extend fibonacci numbers into R. but today we found that most of there equations that he found was proved some years ago according to this link: mathworld.wolfram.com/FibonacciNumber.html ...

now i want to know that ... if he want to approve in some university in USA , is there any way for him ?

t.b.

That's tough luck... What do you mean by approve in some university?

Michel Kogan

in fact

he want to approve for Ph.D in math major

t.b.

19:49

Ah, you mean apply

Michel Kogan

and believe me .... he is the smartest person i have ever know ( in math )

yeah

in that link ... he found numbers: 50, 62, 63, 64, 65 and more ...

t.b.

Hm. I'm not too familiar with the usual requirements for a Ph.D.-position. But the discovery of his probably won't help him much, I'm afraid, but I'm sure he would get good recommendation letters. I'm sorry, I don't think I can help you. Maybe you want to drop by a bit later, when there are more US-people around.

Michel Kogan

hmm.. yeah ... really thanks

Asaf Karagila

@tb Interesting. I'm gonna take another yearly course with Magidor, so it's good to know that some classes will be canceled around April :-)

t.b.

@Asaf: good for you... The semester hasn't even started, and you're already thinking about the days off. Correct attitude, I must say !

Michel Kogan

19:55

he is really sad :( ... thanks anyway ... goodbye :)

t.b.

@Michel: good bye, and good luck for your friend!

Asaf Karagila

@tb Well, I took a course with him last year as well, and around that same time there were a lot of vacations and he went abroad a couple of times. So a proof that would take us a month took us three. It was really annoying.

t.b.

(and I can understand that feeling, it's very disappointing...)

Asaf Karagila

I wonder who was sitting in the crowd. I understood that Harvard is not exactly known for its set theory...

t.b.

@Asaf: Well, I believe that if even I know half of the names, then it can't be too bad.

Asaf Karagila

20:01

Rabin is also in Jerusalem :-)

t.b.

Anyway, as far as I understood it's quite a big collaboration.

Asaf Karagila

Impressive.

t.b.

But to be honest, the recordings aren't too great...

Asaf Karagila

Yeah, I started watching Woodin's lecture but stopped. I'll try again some other time when I can use a louder volume (my girlfriend is asleep right now...)

Jonas Teuwen

@MichelKogan If he's good enough he will figure out something new, don't worry :).

And hearing what they pay PhD students in the US, I wouldn't want to go there...

Asaf Karagila

20:09

Haha

t.b.

And off he goes...

So much for "the kindness of strangers"

Asaf Karagila

For an infinite dimensional space V, given a subspace U, there is a unique W such that U+W=V, and U\cap W={0}, right?

Jonas Teuwen

If it is Hilbert then you take U^\perp.

Asaf Karagila

Yeah, right. Hilbert.

Jonas Teuwen

Or do you prefer an UMD space?

Asaf Karagila

20:15

Anyway, the review of Lauchli's original paper seems to verify my suspicion.

Jonas Teuwen

(my favorite)

Asaf Karagila

There are no "big" subspaces in this kind of constructions, which I shall hereby name "Lauchli spaces" in his honor.

Jonas Teuwen

Why not Karagila spaces?

Asaf Karagila

Never name stuff after yourself. Also he invented them.

Jonas Teuwen

And you're the one naming them :-).

Asaf Karagila

20:17

@tb: If you can help me find and translate MR0143705, I'll be most thankful.

Jonas Teuwen

Didn't Porton name some things after himself, or am I confusing him with somebody else?

Asaf Karagila

I think you're confusing someone. Not sure.

Jonas Teuwen

In Triebel's book he names the Triebel-Lizorkin spaces "F-spaces".

t.b.

@Asaf: Well, here it is

Asaf Karagila

You're good :-)

Jonas Teuwen

20:19

If you use a dictionary, you will get quite far, @Asaf.

t.b.

To download it, click on the pdf icon on the left of the title

Asaf Karagila

Wunderbar!

Jonas Teuwen

Töll!

t.b.

Töll?

Toll?

Jonas Teuwen

Yes, but I like to add umlauts to German words.

Asaf Karagila

20:21

@tb: Which part deals with vector spaces without a basis?

I recognize page 13 as two bases with different cardinalities.

t.b.

Part III

Nr. 1.2 states "v has no basis"

Nr. 1.3 "v doesn't have proper complementary subspaces"

Nr. 1.4 "the dual space v* is trivial"

Asaf Karagila

Let me take a look.

"besitzt keine Basis."?

t.b.

" v possesses no basis"

Asaf Karagila

Does he specify anything on the characteristics of the field?

t.b.

besitzen: to own, to posses

Jonas Teuwen

20:25

A vector space with a "trivial" dual as in the dual is {0}?

(bezitten in Dutch if you care)

Asaf Karagila

Yes.

Jonas Teuwen

Hmm, odd!

t.b.

@Asaf: The field is assumed to be countable

@Jonas: you're dealing with the man without choice!

Asaf Karagila

:P

Jonas Teuwen

That is quite perverse, actually.

Asaf Karagila

20:26

What is the proof for the trivial dual? I am not sure I follow it right.

Jonas Teuwen

(A vector space with a trivial dual)

t.b.

First he proves that every proper subspace is finite-dimensional

That's 1.1.

Asaf Karagila

Hmm. Yes, as I came to suspect last night this would be the correct approach. Why didn't I check his paper earlier? :-D

t.b.

Now if you had a basis, this would be infinite, since the vector space is infinite-dimensional. Dropping some basis elements would then give a proper subspace.

which is infinite-dimensinal in contradiction to 1.1.

The real pervesity is 3, imo. Every endomorphism is a scalar multiple of the identity

Asaf Karagila

That's why there is no basis, but why are there no functionals?

t.b.

20:32

Well, the kernel would be a proper subspace of codimension one, I suppose. let me check

Asaf Karagila

@tb Yeah. This is what I thought I proved back then, which I now have to prove again: this sort of construction is consistent with bounded choice principles.

t.b.

Yes, that's exactly his argument of 1.4

Asaf Karagila

Hmmm... I'm not sure I could use this argument in my case. I'll have to check it.

t.b.

What would go wrong?

Asaf Karagila

In Lauchli's case the original space was of countable dimension, and then he shattered it so every linearly independent subset is finite. So finite+1=finite, which is a blunt contradiction.

In my case, every proper subspace has dimension \kappa, while the original dimension (which was shattered) was \kappa^+, but \kappa+1=\kappa.

So being of co-dimension 1 may or may not work here. I will have to check this more carefully.

t.b.

20:38

I forgot what you wanted to prove

Asaf Karagila

Although, if all proper subspaces are of dimension \kappa and the full space is not... Hmmm...

t.b.

You want trivial automorphism group?

Asaf Karagila

In essence I want either a trivial automorphism group, or no of functionals.

The former seems somewhat stronger, but I believe they can be equivalent.

Jonas Teuwen

Bye guys, I'm going to watch a German movie (Das Leben der Anderen).

Asaf Karagila

Can you help me with the proof of 1.1?

t.b.

20:41

@Jonas: have fun!, see you

@Asaf: I don't fully understand the notation, I don't know what exactly he means with "Feld", let me check

Asaf Karagila

Possibly support?

t.b.

Hm. It must be some model-theoretic thing. He has his model and he has the "underlying"(?) set of an object, so k is his field in the model, and t(k) is the set of its elements of whatever else wherever else. I don't know much about model theory.

Asaf Karagila

Is Feld a word in German?

t.b.

Feld = field, meadow

Maybe dict.leo.org helps

Asaf Karagila

Korperaxiomen geniigen?

t.b.

20:47

Körperaxiomen genügen, satisfy the field axioms

So you're on page 3: Ein Kôrper k ist beispielsweise ein Trippel k <h oc, (à} wobei oc (die Addition) und fi (die Multiplikation) Abbildungen von x X x auf x sind, welche den Korperaxiomen geniigen; x ist die Menge der «Korperelemente», auch das «Feld» von k genannt.

(I'm to lazy to fix that copy paste error

Asaf Karagila

Yeah, this is the first instance of "Feld" in the paper.

Next year, if all goes well, I'll take a course in German.

t.b.

Have fun!

Asaf Karagila

Ah. I think I got it.

t.b.

Well, anyway, he seems to build this set a_0 in the previous sections and he builds a vector space whose "set of elements" (I think that's the field) is that set a_0

Asaf Karagila

He says that the triplet is the formal object of the field and <x> generates the field or something?

A set which generates the field, that is.

Or the set of elements in the field.

Getting back to 1.1, assuming Feld indeed means field, what more can we extract from it?

(p. 9)

t.b.

20:55

So first he has 1. that states: Every subset of t(v) is either a subset of the Feld of a finite-dimensional subspace or else, the complement (wrt t(v)) of such a set.

Asaf Karagila

What is t(v)?

t.b.

The Feld of v

:)

a_0, the base set of the model, if I understand correctly

Asaf Karagila

The atoms, perhaps.

t.b.

He defines a_0 in the beginning of I on page 2

(or rather, axiomatizes it)

He assumes Fundierungsaxiom bezuglich einer Basismenge a_0

Axiom of foundation with respect to a base set a_0

Asaf Karagila

Yeah, also known as regularity in some places.

I think he gives the needed info at the beginning of III

t.b.

20:59

His axiom a) sounds like: The class A of all sets that contain themselves as only elment A = {x | x = {x}} shall be represented by a set a_0, the base set a_0

Asaf Karagila

Yeah, it's a nice way of getting ZF+atoms from ill-founded models of set theory.

t.b.

Do you want to assume 1 and proceed with the proof of 1.1 or do you want to do the proof of 1. first?

Asaf Karagila

What is the statement in 1?

t.b.

I gave it above "Every subset of t(v) is either a subset of the Feld of a finite-dimensional subspace or else, the complement (wrt t(v)) of such a set."

Asaf Karagila

Oh.

Well, let's give it a go. Couldn't hurt, I guess.

t.b.

21:03

Okay, I'm now top of page 9, Beweis (proof)

Let x \subset t(v) be a model set(?).

There exists a finite subset e' of a_0=t(v) such that...

x is transformed into itself by the group \tilde{h}(e'), (see I.6.).

Asaf Karagila

Wait wait wait... translate that last line before 1, I think it will give out what all the notations are.

t.b.

The automorphism group g of v is a permutation group of a_0. v is in the normal model N(g) an infinite dimensional vector space over k with the following properties:

then comes 1.

Asaf Karagila

Okay. I think I got it. He uses some very very arcane notations I think.

t.b.

Ah, I'm glad that even set theorists develop their notation :)

Asaf Karagila

He says that we have a_0 which is the set of atoms, v is a vector space and its underlying set is a_0.

t.b.

21:09

Okay, where do you want me to continue *fetches a beer*

@Asaf: yes, that seems to be the case.

Asaf Karagila

Now he says that every set of atoms is other in a finitely dimensional subspace or in a complement of such.

Okay, let's continue from where we left in 1

t.b.

Let e be the subspace spanned by e' then x is transformed by every automorphism of v rel. e into itself.

Asaf Karagila

rel.?

t.b.

relative to

(modulo?)

Asaf Karagila

Ah. I thought so.

Yeah, modulo.

Now he takes elements in t(v)\t(e) and defines a linear permutation, right?

t.b.

21:15

Since for any two elements s and t of t(v)-t(e) there is an automorphism of v rel e transforming s into t, footnote here it follows that either x \subset t(e) or x \supset t(v) - t(e) QED.

Asaf Karagila

So we have that x is a subset of e, or that x is a subset of the complement of e.

Yes, I know what the footnote will say.

t.b.

Sounds like that.

Asaf Karagila

Very well, so far I seem to be on the right track both with the paper and with my result.

t.b.

1.1 now?

Asaf Karagila

Yes, please :-)

t.b.

21:18

Without the axiom of choice we can prove that a subspace of a finite dimensional space is finite dimensional.

Asaf Karagila

Also, no need to use the t(x) notation, just use X instead.

It'll be clearer :-)

t.b.

pheew

Asaf Karagila

And we know that V=a_0 too, so no need to insist on that either :-)

t.b.

Because of 1. the feld U of an infinite dimensional subspace of v must contain the set theoretic complement of E of a finite dimensional subspace e of v.

From this we conclude that U = V

Asaf Karagila

Also, cut with the "feld" :-)

t.b.

21:21

yessir

Asaf Karagila

Wait, what? Your previous line seems like there's a beginning missing.

Oh right, it's above :-)

t.b.

Continue?

Let x in V.

V-E is non-empty because v is not finite dimensional.

Let y in V-E, so y in U

If x + y in E then x \notin E because y \notin E;

so x \in V-E \subset U.

If x + y \notin E then x+y in U since y \notin E;

whoops messed up.

If x + y \notin E then x + y \in U

and because y \in U it follows that x \in U.

Thus U=V, so u=v. QED

(scratch that line before the italics)

Asaf Karagila

Excellent.

He does better job than me, for obvious reasons :-)

t.b.

Well, but his writing is ghastly.

Asaf Karagila

I can show you the reason why.

t.b.

21:32

He formed the logic/set theory group at ETH together with Specker, but he died right before I enrolled.

Asaf Karagila

jmilne.org/math/tips.html - see the last one.

t.b.

Well, it's not only that! But it definitely is part of it.

I think one main point is number 9.

Asaf Karagila

:P

t.b.

Wanna do 1.2 and 1.3 as well?

Asaf Karagila

I know 1.2, and it's pretty obvious too.

1.3 is the codimension one, no?

t.b.

21:36

No, 1.3 is: There are no proper complementary subspaces.

Asaf Karagila

It is also pretty obvious from the above.

Srivatsan Narayanan

@tb Would like a quick check w.r.t. my comments 1 and 2 in math.stackexchange.com/questions/76777/…... Are they correct?

Asaf Karagila

One would have to be of infinite dimension :-)

t.b.

Exactly.

@SrivatsanNarayanan Undoubtedly correct. z ->z^2 is continuous, hence Borel, etc.

@Asaf: So 1.4 is fine as well?

Again the kernel would have to be infinite-dimensional...

Asaf Karagila

It's just a codimension argument, right?

t.b.

21:39

yes.

Asaf Karagila

So no need.

t.b.

So 1.5 says you can't extend a nonzero functional from a subspace u to all of v.

That's clear from 1.4

Asaf Karagila

Yeah.

Srivatsan Narayanan

Ok, I was just in the mood of ignoring measure theoretic complications (thanks to this answer: math.stackexchange.com/questions/76776/…). Only when I was posting an answer to the other question did I realize that this might be an exercise from a (rigorous) probability course.

Thanks, tb.

t.b.

Yes, but there's no problem. You have your random variable X -> R and squaring R->R. The preimage of a Borel under the square map is a Borel set since it is continuous and the preimage of a Borel set under X is measurable by hypothesis that X be a random variable.

No problem, @Srivatsan!

@Asaf: 2. is about vector space topologies (the only one is the trivial one, assuming k has a non-discrete topology).

So I hope we can skip that, too.

Asaf Karagila

21:44

I think I have what I need from that paper :-)

Many thanks.

If you want to see how Urysohn's lemma fails skip on to 6 :-)

Okay. So I think I have the underlying proof with atoms running. Now let's see if I can think of a quick reason why I can do that by a similar forcing argument.

First, let's find my treeware copy of Pincus' paper, the answer may be hidden there :-)

t.b.

Very good! Good luck!

Srivatsan Narayanan

I didn't know the word "treeware". ;) Now, I do, thanks to Urban Dictionary.

Asaf Karagila

It's from the free-on-web book's site: Practical Common Lisp.

t.b.

*pfuuh, reboots choice segments to feel whole again.*

Asaf Karagila

The about says something about treeware.

Srivatsan Narayanan

21:54

"I suppose this is an extremely general question, so I apologize, and perhaps it should be deleted. On the other hand it's an awesome question."

I like the starting lines... math.stackexchange.com/questions/76793/brave-new-number-theory

t.b.

Hm. No comment on the question itself, but such posts from people who asked such a question two weeks ago make me a bit doubtful...

I guess that was still a comment :)

Asaf Karagila

He went to a conference about large cardinals and homotopy theory, or at least was seeking references for the topic.

t.b.

Yeah, and then he asked a question like the one I just linked to.

Asaf Karagila

Before the conference I had coffee with Magidor (after a seminar, and everyone else left earlier) and he said he'll give a lecture there. I wonder how his talk was. He's a wonderful speaker.

Srivatsan Narayanan

I was speaking only about the starting lines though.

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