Mathematics - 2012-03-10 (page 1 of 3)

t.b.

00:03

I can't decide which one I like the most. Those I looked at are essentially epsilon variations of each other.

Asaf Karagila

Because the principle's the same, and it's so compelling to use that argument that you can only disguise it so much.

Jonas Teuwen

Is "agoodguy" a troll?

t.b.

@AsafKaragila The annoying thing is that I was stuck at something seemingly unrelated and it took me a hell of a long time to see that I was just looking for those darn families...

Jonas Teuwen

I can't imagine that one comes with such a reasonably advanced material and then fails to answers that question.

Asaf Karagila

@tb Well. My email is not just for pdfs and personal problems you want to discuss! :-D

Jonas Teuwen

00:09

Can I e-mail you with my personal problems, Asaf?

Asaf Karagila

@JonasTeuwen You can try.

Jonas Teuwen

Ft. Let me get something.

t.b.

@AsafKaragila :) Well, the main problem was the re-casting of the question in those terms; after that was done, the problem was resolved instantly.

After the fact it looks soo obvious :s

Asaf Karagila

@tb I see. I know that often mathematicians prefer to try on their own and not ask other people for help. I can tell you that several times I began writing an email, which essentially made me put the question into exact details - and solve it on my own.

t.b.

I know that effect too well... That's why I ask so few questions on this site. While writing them up I solve them :)

@JonasTeuwen well, it really depends on how you're taught complex analysis. Our prof lulled us into sleep for about two months by massaging formal power series. When I started realizing that the action had begun, it was already over. :)

Jonas Teuwen

00:16

That is a very handy skill when computing integrals.

Asaf Karagila

@tb There are so many "in bed" jokes one can turn your message into...

t.b.

@JonasTeuwen don't forget that continuous functions are Riemann integrable :)

Jonas Teuwen

Can't I get a PhD by compiling some integral tables? It is quite entertaining.

@tb Yes, but then your norm is only defined on Riemann integrable functions right? What is the "extension"?

t.b.

@AsafKaragila thanks for spelling it out, I was aware of it.

@JonasTeuwen The Riemann integral gives you the $L^2$-norm, right? So the completion is $L^2$. Who wants to know that the completion consists of classes of measurable functions? :)

Jonas Teuwen

Right.

Asaf Karagila

00:20

@tb Jonas might be drunk, and did not notice. It's just my public service... :-)

Jonas Teuwen

I wonder if there is something circular in that definition, as if you take the $L^1$ that won't work I think.

But I forgot why.

Must be the Port Ellen substitute.

t.b.

@AsafKaragila I see: The König doing grassroot campaigning.

@JonasTeuwen It will. You probably have regulated functions in mind, but there the issue is that you take the uniform completion of the step functions in order to guarantee that the limit function is still integrable.

@JonasTeuwen Put differently: The Riemann integral gives you a functional on $C[0,1]$. So the measure's already there by Riesz-Markov.

Jonas Teuwen

Yes, I realised this some seconds ago. Oh well.

Asaf Karagila

Damn! For the first time ever I need to use colored markers and I can't find my girlfriend's markers!

Kannappan Sampath

@AsafKaragila Hmm, I thought you could always call people. : )

Asaf Karagila

00:31

She's asleep it's 2:30am.

Jonas Teuwen

It is important enough to wake her up.

What's a man without colored markers?

t.b.

What's a man with colored markers (if he's lucky) but without girlfriend?

Asaf Karagila

@JonasTeuwen Well... a straight man, I'd think.

Jonas Teuwen

I always carry at least one with me.

@tb Very happy.

Asaf Karagila

I can wake her up, sure. I could also procrastinate and continue playing chicken invaders pointlessly!

t.b.

00:33

^ sounds like a plan :)

Jonas Teuwen

Excellent choice, Sir.

I'll go for Lagavulin. Because I don't have any Port Ellen yet.

Kannappan Sampath

@AsafKaragila How I wish you give up Chicken Invaders.

Asaf Karagila

Why???

Kannappan Sampath

That's a lame game. :p

Asaf Karagila

Is not!!

Kannappan Sampath

00:35

Have you tried something like Counter strike, or may be GTA series or something .....?

Asaf Karagila

Hm. Another alarm.

I think I'll go to sleep and that's it.

Goodnight!

t.b.

Good night

Kannappan Sampath

@AsafKaragila Good night. : )

Jonas Teuwen

Alarm?

Kannappan Sampath

Grad alarm it seems?

They are bombing the militants?

t.b.

00:39

@JonasTeuwen here

Jonas Teuwen

A grad alarm? Is there a swarm of grad student zombies running in the streets?

t.b.

Okay, I should be going too. Good night!

Jonas Teuwen

Good night!

EISBÄR!

Kannappan Sampath

00:58

@tb Good night. : )

Kannappan Sampath

01:10

Contradiction in AC

Asaf is going to pounce, methinks.

Kannappan Sampath

02:28

@Mariano Can you assist me with GAP please?

I setup GAP but I just cannot query anything into it, it replies with some error message.

So, I'd like it if you can query your GAP system for those groups G such that G and Aut G are abelian?

Mariano Suárez-Alvarez

/me tries

Kannappan Sampath

@MarianoSuárezAlvarez Thanks for trying. : )

Mariano Suárez-Alvarez

There are 200 of size at most 200

The cyclic ones :)

Kannappan Sampath

@MarianoSuárezAlvarez Oh, so, the order of the groups are bounded by 200?

Mariano Suárez-Alvarez

No, I limited the search at 200

I guess the only such groups are the cyclic groups

Kannappan Sampath

02:36

@MarianoSuárezAlvarez I should attempt a proof. This question was asked in a paper in American Math. Monthly.

Mariano Suárez-Alvarez

a group G satisfies your conditions iff its primary components satisfy them

so it is enough to suppose that |G|=p^r for some prime p

Kannappan Sampath

(In fact, IRC they said this was not interesting or something. : ) )

Mariano Suárez-Alvarez

then the group is a direct sum of cyclic groups of the form Z_{p^k}

Kannappan Sampath

@MarianoSuárezAlvarez I am not getting this point. Can you explain why so, please?

Mariano Suárez-Alvarez

if there are two isomorphic sumands, then Aut(G) is not abelian

suppose G is an abelian finite group

and let G_p be the p-primary part

that is, the subgroup of elements killed by some power of p

Kannappan Sampath

02:39

OK.

Mariano Suárez-Alvarez

then G is the direct sum of the G_p

and Aut(G) is the direct product of the Aut(G_p), because an autom. of G must map each G_p into itself.

Kannappan Sampath

Needs a proof but I think I am seeing one. So, ok.

Mariano Suárez-Alvarez

So if G is finite abelian with abelian Aut, then each G_p satisfies the same conditions

of course it needs a proof :)

and conversely, if all the G_p satisfy the conditions, the observations made imply that G does too

Kannappan Sampath

Ah, yes. You put all of this under that carpet? : )

Mariano Suárez-Alvarez

no, I am trying not to ruin all the fun for you :)

Kannappan Sampath

02:42

@MarianoSuárezAlvarez Sure. Thank You. But, I did not think along these lines at all. : (

Mariano Suárez-Alvarez

this trick will come to you more naturally with time

I am just decomposing G into simpler parts

Kannappan Sampath

Yes. So, here we are also having the fact that Automorphism preserves the order of the element. So, this seems to be more natural. Is this correct line of thought for this choice of decomposition?

Mariano Suárez-Alvarez

yes

automorphisms do not mix the different G_p

(that is why Aut(G)=$\prod_p$Aut(G_p)

Kannappan Sampath

Yes. So, we have simplified the case to $|G|=p^n$ for some prime $p$.

Mariano Suárez-Alvarez

indeed

now a p-primary finite abelian group G is the direct sum of cyclic groups of the form $Z_{p^n}$.

You can easyly show that $Aut(\mathbb{Z}_{p^n}\oplus\mathbb{Z}_{p^n})$ is not abelian

and from that, that if in the decomposition of $G$ as a direct sum of cyclic groups there are repeated summands the autom. group is not abelian.

Kannappan Sampath

02:48

@MarianoSuárezAlvarez I'll attempt a proof. So, we have essentially solved the problem, right?

Mariano Suárez-Alvarez

so your group must be of the form $\mathbb{Z}_{p^{n_1}}\oplus\mathbb{Z}_{p^{n_2}}\oplus\cdots\mathbb{Z}_{p^{n_r}}$ with $n_1<n_2<\cdots<n_r$.

Kannappan Sampath

@MarianoSuárezAlvarez How are these cyclic?

Mariano Suárez-Alvarez

they aren't

unless r=1

you have to prove that r=1 :)

Kannappan Sampath

Yes, that's why I am wondering too.

@MarianoSuárezAlvarez Oh, sure. I'll prove. Thank you for taking it this far. : )

Mariano Suárez-Alvarez

np

Kannappan Sampath

02:51

I hit at this problem after reading a paper by Diaconescu. Let me fish out a link.

Sorry that's Deaconescu

@MarianoSuárezAlvarez Here's the link, at last. : )

Benjamin Lim

Hey

Kannappan Sampath

Hey Ben, How are you?

Benjamin Lim

@MarianoSuárezAlvarez I understand if $M$ is a $\mathbb{Z}$- module then we have an isomorphism $\operatorname{Hom}(\mathbb{Z}, M) \cong M$

@KannappanSampath Not bad

But I have been told the same is true if we replace $\mathbb{Z}$ with any commutative ring

I used the assumption that $M$ is an $A$- module then a homomorphism of $A$ - modules $f : A \longrightarrow M$ is completely determined by $f(1)$.

Mariano Suárez-Alvarez

It is true if A is a ring

Kannappan Sampath

It's breakfast time. Bye all of you. Have a good day.

Benjamin Lim

03:00

@MarianoSuárezAlvarez If $A = k[t]$ is a polynomial ring then shouldn't my homomorphism $f : A \rightarrow M$ be completely determined by say the image of the indeterminate $t$ in $M$?

@KannappanSampath Bye enjoy your dosa

Kannappan Sampath

@BenjaminLim Yes, my favourite. : )

Mariano Suárez-Alvarez

it is determined by the image of $1$

Benjamin Lim

@MarianoSuárezAlvarez I don't really see why.

Mariano Suárez-Alvarez

(the image of $t$ will be the image of $1$ times $t$)

well: try to prove it :D

Benjamin Lim

Well yeah that's obvious I tried to use that

but then if you want to know what is say the image of $t^2 + 1$ in the module

then it is $f(t)^2f(1)^2 + f(1)$

Mariano Suárez-Alvarez

03:02

I claim: given a ring $A$ whatsoever, and an $A$-module, two $A$-linear maps $f$, $g:A\to M$ which coincide on $1$ are in fact equal.

Benjamin Lim

but then we don't know what is $f(t)$ yet

Mariano Suárez-Alvarez

no

where do you get that from?!

f is a map of modules, not rings

moreover $f(t)^2$ simply does not mean anything

for $f$ takes values in $M$, which is a module.

Benjamin Lim

Oh sorry It should be $f(t^2 ) + f(1)$

whoops

Mariano Suárez-Alvarez

suppose $f:A\to M$ is a map of left $A$-modules such that $f(1)=0$.

then if $a\in A$, we have that $f(a)=f(a\cdot 1)=a\cdot f(1)=a\cdot0=0$.

Benjamin Lim

Do we need to talk about left/right modules can we assume $A$ is commutative?

Mariano Suárez-Alvarez

03:06

if you delete the left I said and pretend that $A$ is commutative, everything I said works

Benjamin Lim

Oh Mariano I got it

$f(t^2)$ is just $t^2f(1)$

Mariano Suárez-Alvarez

indeed

Benjamin Lim

Oh crap yes I was confused over the multiplication :D

Mariano Suárez-Alvarez

that is good

as soon as you unconfuse yourself about that, the better

Benjamin Lim

Yeah I forgot that now $A$ is viewed as an $A$ module not a ring :D

Mariano Suárez-Alvarez

03:07

and it seems to have happened quite soon, so everything is cool

:)

Benjamin Lim

One more thing

I have been told one does not simply define the product of two submodules

Mariano Suárez-Alvarez

well, it is not Mordor...

Benjamin Lim

but in $\mathbb{Z}$, given the submodules $(2)$ and $(3)$ we can define the product of submodules just like the product of two ideals?

Mariano Suárez-Alvarez

what do you mean by «define the product of two submodules»?

Benjamin Lim

sorry I am addicted to that phrase

well I have been told it is not possible to define it

But what if we define it like the product of two ideals

Mariano Suárez-Alvarez

03:09

the thing is, the question does not make sense really :D

Benjamin Lim

Where $M_1M_2 $ is the set of all finite sums $\sum x_iy_i$ where $x \in M_1,$ y \in M_2$

Mariano Suárez-Alvarez

if a module has a product, then you can multiply things in it

if not, not

Benjamin Lim

Oh crap

we cannot multiply two elements in a module

Mariano Suárez-Alvarez

indeed

Benjamin Lim

shit forgot again it's an abelian group under addition

i am going nuts: rings, modules, algebras......

Mariano Suárez-Alvarez

03:10

heh

it stabilizes after a while: don't worry :P

Benjamin Lim

Two many algebraic structures

I just started learning modules this morning :D

Mariano Suárez-Alvarez

well, you are allowed ot be overwhelmed by the multitude of definitions for about 6 months

so you have plenty of time

Benjamin Lim

I see at the end of chap 2 in AM I need to know homological algebra?

Mariano Suárez-Alvarez

you can simply skip that

no one reads AM from cover to cover

(I would suggest reading a different book, in fact...)

Benjamin Lim

how will I know which parts to skip and which not to? I here it is the canonical beginners text in AC

Mariano Suárez-Alvarez

03:14

yes, that's why most people end up hating commutative algebra

Benjamin Lim

ahahahahhahaha

@MarianoSuárezAlvarez I found exercise 26 in chap 1 to be super rad

Mariano Suárez-Alvarez

I confess not to know the problems by heart :)

Benjamin Lim

Well

If you have C(X)

X a compact Hausdorff space

C(X) the ring of all real valued continuous functions

then the exercise is to show that there is a homeomorphism between X and Max (C(X))

the rad part was that the topology on X was constructed out of nothing

Mariano Suárez-Alvarez

that's cool indeed

it was constructed out of the hard work of a few generations of mathematicians :P

Benjamin Lim

the best part was how we already had a basis for Max (C(X))

Then by the Urysohn Lemma constructed a basis for X out of nothing

I had never seen anything that rad before

Mariano Suárez-Alvarez

03:19

:)

Benjamin Lim

@MarianoSuárezAlvarez Do you know the history of a problem like that? For example how it arose?

Mariano Suárez-Alvarez

it comes from functional analysis

Benjamin Lim

oh ok

Mariano Suárez-Alvarez

its code name is Gelfand duality and/or Gelfand-Naimark theorem

Benjamin Lim

which areas? I would like to check it out

Mariano Suárez-Alvarez

03:23

WIkipedia surely talks about this

Benjamin Lim

yes checking it out

Mariano Suárez-Alvarez

Historically, this arises from Fourier analysis

Benjamin Lim

Thanks for the context. Kannapan has started a commutative algebra room where we discuss AM. I told him it is not possible to do AM without knowing general topology

Mariano Suárez-Alvarez

"knowing general topology" may mean many things

youcan do AM with very little general topology

Benjamin Lim

Well I think you need to be familiar with at least material of chapter 2 Munkres

@MarianoSuárezAlvarez I should get back to work. Been talking too much :D

Thanks again for your experience.

Mariano Suárez-Alvarez

03:31

np

robjohn

03:42

I have just thought too long about this problem, but I think my answer should be convincing, even if lacking some of the rigor that I usually like to give to answers. After having thought about it for a long time, I saw this and said, "of course!"

Rajesh D

04:15

Hi @rob

robjohn

@RajeshD Hey there. How is your day going?

Rajesh D

just started

after a break fast

An Oat meal with no sugar and a small pich of salt

I am low on power....i'll back after a while

Why 21 upvotes for it !......looks like there is a lot of change in voting patterns

robjohn

04:37

@RajeshD I think a lot of the voting frenzy was because the problem is hard to think about using normal tools. The problem is pretty simple, but you have to use simpler optimization techniques. I think that the simplicity of the answer will prevent the voting frenzy to continue to my answer :-)

Off to get dinner. BBL

Mariano Suárez-Alvarez

05:52

@robjohn, what you wrote is more an argument than a proof... :P

Jeff

06:15

hey folks

anyone still here?

Kannappan Sampath

@Jeff Hi!

Jeff

@K, hey

Kannappan Sampath

@Jeff You did not ping me. The first three letters are required by this @system for identifying someone to ping.

Jeff

@kan, hey :D

Kannappan Sampath

@Jeff You did! : )

Jeff

06:18

spose i received over 100 poiunts in a day and i look at my reputation and it lists the questions. but those questions aren't changed, or voted on, or anything...

robjohn

@MarianoSuárezAlvarez that's sort of what I was saying by lack of rigor. I am working on trying to add that.

Jeff

and you also get 100 point 'associated bonus'. where did all that come from?

also, spose some gave me a -2. what was it for?

Mariano Suárez-Alvarez

I wonder where he is getting the problems from...

Kannappan Sampath

Robert Israel and Gerry's post often are caught by the poor quality posts filter and I have no choice :/

Mariano Suárez-Alvarez

attribution would be nice

Kannappan Sampath

06:28

@MarianoSuárezAlvarez Who do you mean?

Mariano Suárez-Alvarez

Gingerjin

338 1 8

Kannappan Sampath

@Jeff Your answer was downvoted because it did not sound satisfying to the person who did it. And, they are expected to drop in a line as to why they downvoted the post. But most people don't.

@MarianoSuárezAlvarez I see. : )

Jeff

@kan how come the down vote doesn't appear on my screen?

Kannappan Sampath

@Jeff I mean, you should be able to see it on the rep page, no?

Jeff

@kan it's there now. last time i looked there were no up or down votes on my answer.

robjohn

06:31

@Jeff to whom are you speaking?

Kannappan Sampath

Spam?

Jeff

the OP is wrong though. my answer may not be a proof, but it's not wrong, either. sheesh

Kannappan Sampath

Conjugacy in statistics is beyond my comprehension though.

Jeff

@rob, talking to @KannappanSampath

Kannappan Sampath

@Jeff Can we start talking about something particular? Which question and which post of yours is giving you a head ache?

Jeff

06:33

@KannappanSampath, several of them are. How about the +100 'associated bonus' I have in my profile? I can't tell where that came from?

Kannappan Sampath

@Jeff Likely to have got it from the network when you crossed some fixed rep. points. I think even I got some such bonus.

robjohn

@Jeff okay :-) I haven't gotten a bounty for quite a while, so I was wondering (since your comment immediately followed mine)

Kannappan Sampath

@rob Is there a notion of conjugacy in statistics?

Jeff

okay, i guess i'm outtie here for now. have a weekend everyone

Kannappan Sampath

@Jeff Bye. Wishing you a good weekend.

Kannappan Sampath

06:53

Hey Ben.

Just did the first two sections of Chapter 2 and the exercises from Munkres.

@Mariano Would like a discussion about closing a question as exact duplicate. Are you here?

anon

what about it kanna?

Kannappan Sampath

@anon Should this be closed as exact duplicate of the question whose answer I have linked in my answer there?

anon

I would so that no, those are not quite the same question.

Kannappan Sampath

@anon In which case, does my answer really answer the question?

anon

It does not. Counting is not the same as listing.

Kannappan Sampath

07:04

@anon Well, the answer I had linked gives a way to list all the classes.

anon

Oh, okay.

@Kanna A partition of n is something like (a1,a2,...) with the entries' sum being n. The number of partitions of n is p(n). So "is equal to the partition of n" is bad English. Also, I could have swore there was a closed-form for p(n) in the news recently...

Kannappan Sampath

@anon I'll fix that. About closed-form, I am not sure...

Thinking of which, thanks for fixing the exponent. : )

anon

np. it's clearly the reason for the downvote.

Kannappan Sampath

I mean, people should allow small room for errors. I mean nobody deliberately makes such errors. And this is not gross as well. :/

Are you serious that $p(n)$ closed form remark?

anon

yes.

Kannappan Sampath

07:11

Can you please point me to some reference?

anon

5

Q: Closed-form Expression of the Partition Function $p(n)$

I feel like I have seen news that a paper was recently published, at most a few months ago, that solved the well-known problem of finding a closed-form expression for the partition function $p(n)$ which enumerates the number of integer partitions of $n$: does anybody have the reference of this pa...

number-theory reference-request partitions

Kannappan Sampath

@anon I removed my comment about that...

Thanks for the pointer.

Kannappan Sampath

07:29

Still I'm not convinced that it is not duplicate. : (

anon

@Kannappan: Re-reading, to me it looks like what the OP wants is set partitions of a set X of size n quotiented by the equivalence relation x~gx where g is the action of $S_X$ on partions of $X$.

Actually, scratch that, I have no idea what he is talking about.

Kannappan Sampath

Let me finish my lunch and see what this is.

His comment below my answer makes no sense whatsoever

Kannappan Sampath

08:05

I am going to sleep.

(nap, yeah.)

Rajesh D

08:39

@robjohn After reading your answer, I am impressed with the question too..(no complaints) !.........But i did not understand at one point...How did you get (3) from (1) and (2)...a small elaboration is required for me to appreciate

@KannappanSampath lucky to have a siesta !

Hey @anon : Watz up Myan ?

anon

nada rajey.

Rajesh D

How do you know that

anon

know what

Rajesh D

that people at home and close friend call me 'raje' (short form)

?

anon

I didn't.

Rajesh D

08:47

the what is nada rajey ???

anon

But it's rather predictable.

Rajesh D

darn.....

anon

thefreedictionary.com/nada

Rajesh D

I thought 'Raj' sounds stylish

so @anon : Whats your name If i may ask ?

anon

Rumpelstiltskin.

Rajesh D

08:51

r u kidding

anon

:)

Rajesh D

you are a tounge twister !

Rumpelstiltskin

Rumpelstiltskin is the eponymous character and antagonist of a fairy tale which originated in Germany (where he is known as Rumpelstilzchen). The tale was collected by the Brothers Grimm, who first published it in the 1812 edition of Children's and Household Tales. It was subsequently revised in later editions. Plot In order to make himself appear more important, a miller lied to a king, telling him that his daughter could spin straw into gold. The king called for the girl, shut her in a tower room with straw and a spinning wheel, and demanded that she spin the straw into gold by morn...

@tb tb tb !

Now you don't have any excuses

t.b.

I think this should be closed as duplicate of this

@robjohn It's interesting to see the difference of the reactions here and on MO. Why 22 upvotes and only $\pm 1$ for you, by the way?

Hi, Rajesh, what's up?

Rajesh D

@tb nothin...i just wanted to show you this It is simple but I made it look tedious

t.b.

Thanks, I will take a look later, but I'm not quite in the mood for reading it in detail right now, my brain hasn't really started up, yet.

Rajesh D

09:05

okay

t.b.

09:32

@RajeshD Are you still there? I've now looked at your argument, but I'm pretty confused by it, to be honest.

For instance, I do not understand this:

> The minimum value of $n$ = sum of, the number of times $f_1$ is differentiable at $\tau-x$ and the number of times $f_2$ id differentiable at $x$, $\forall x \in (0,1)$.

Asaf Karagila

Mornin'

t.b.

Hi, Asaf

Still on your quest for color markers?

Asaf Karagila

Good thing you remind me, I'll ask my girlfriend.

Couldn't you have let me loaf around for a while? I just woke up!

Now I have to work...

t.b.

This reminds me of a conversation I overheard in the lift to the library: A: do you use highlighters? B: Yes I started using them recently. A: Good, they're immensely practical. You can see immediately if you're reading something for the second time

@AsafKaragila Don't worry, you'll find plenty of ways to avoid that :p

Asaf Karagila

@tb Oh, of course I will. I'm a smart guy when it comes to procrastination and loafing around.

t.b.

09:45

exasperation

Asaf Karagila

Good. Now to email Magidor and ask him which of these theorems we already covered in class, because I cannot remember!

@tb It's not the first time he's been doing that.

Once he did that after his answered turned into CW. I flagged another answer which was CW-ified to be un-CW and mentioned his answer and that someone should look into that.

I was hoping to write an answer to that new question on AC, but I simply don't understand what on earth does that guy want me to answer.

t.b.

@AsafKaragila I don't understand this question at all.

Asaf Karagila

Which one?

t.b.

The self-contradictory AC

Asaf Karagila

Ah. I thought you meant my question, because I didn't ask one and it seemed weird.

t.b.

09:57

Interesting comments

Jonas Teuwen

8-).

So GEdgar was trolling me!

Matt N.

10:16

Morning everyone.

Asaf Karagila

Guten Morgen!

Matt N.

Guten Morgen, Asaf.

t.b.

morning

Matt N.

Hello Teddy. Look at all the sun!

t.b.

Yeah :)

Matt N.

10:19

Got any plans to go to the lakeside and play Hacky and drink beer with your buds?

(Or is that not your idea of "enjoying the sun"? : ) )

That seems to have left him speechless...

t.b.

I'm in serious want of dexterity when it comes to any kind of juggling. The only thing I really could handle was the diabolo. No I'm amply satisfied by just sitting in the sun (provided it's warm enough, which it currently isn't) :)

Matt N.

: )

Asaf Karagila

I am unfamiliar with the concept of a cold sun.

Matt N.

Never been skiing?

Asaf Karagila

Nope.

The only time I've been abroad was in Ireland and the sun was still warm (late September).

Matt N.

10:24

: O

t.b.

@AsafKaragila you'll find out what it is when you go visit your metal friends in Norway :)

Matt N.

Asaf has metal friends in Norway? Heh.

Asaf Karagila

@tb Well, I used to have several friends from Norway related to the metal scene. I've since cut ties with "everyone" several times so by now I'm not in touch with any of them.

Matt N.

My integral got starred : D

And? Did you compute it? : ) I saw Jeff had a go.

Asaf Karagila

Well there was another volley around 2:30 (one hit in the city somewhere) and another around 6...

And the problem is that I'm living right by the alarm horns, so I can't just ignore them and keep on sleeping.

t.b.

10:31

Notation question: given a cardinal $\kappa$ how do you usually denote the cardinality of the iterated powerset, that is $\kappa_0 = \kappa$ and $\kappa_{n+1} = 2^{\kappa_n}$?

Asaf Karagila

$\mathcal P^\alpha(\kappa)$, if you want $\alpha\ge\omega$ then you have to specify what happens at limit points though (direct limit usually works fine).

There's a variation on $\beth$ numbers in which you also note the starting point. That could come in handy too if you prefer it that way.

t.b.

That would be $\beth_{\alpha}(\kappa)$?

Asaf Karagila

I think so. I've only see that used on Wikipedia though, never in books or papers. It's best to also define that in a line or two.

t.b.

@AsafKaragila the problem with that is that I've seen people use this for the set of subsets of cardinality equal to $\alpha$

(even though it strikes me as slightly nonsensical)

Asaf Karagila

That would be a subscript index, usually.

Although that too is nonstandard since $\mathcal P_\kappa(\lambda)$ means all the subset of $\lambda$ of cardinality less than $\kappa$.

t.b.

10:38

Those guys would write $P^{\leq \kappa}(\lambda)$ for that

Asaf Karagila

How nonstandard of them.

I do know the notation $[\lambda]^\kappa$ for "all sets of size exactly $\kappa$"

t.b.

Thanks, I'm not running the applet, obviousy :)

Matt N.

You're funny : )

Asaf Karagila

I have very strong feelings that the Wikipedia article on Beth numbers is full of mistakes.

Matt N.

And mysterious. The plug in is just too convenient to not be run.

anon

10:41

Quick question on algebra: "Describe the set of equivariant maps $G/H\to G/K$." Would I be correct saying they are $xH \to xvK$ for each element in the set of $v$ such that $vHv^{-1}\subseteq K$ quotient'd by the action of $H\cap K$?

t.b.

@AsafKaragila Yes, that's what I encountered most frequently. Thanks, that's about what I wanted to hear, I'll just write down the iterative definition since I don't seem to need to use that systematically.

Asaf Karagila

Where are you using it? if I may ask.

t.b.

I've written up the isomorphic classification of some special types of Banach spaces and for one type there's a cardinal invariant of this form coming up.

Asaf Karagila

I see. Sounds like fun! :-D

I'm gonna go and kill some chickens now (not literally). I'll be back shortly...

t.b.

I see. Sounds like fun! :-D

Matt N.

10:45

: )

anon

(actually, I should have written $\to xv^{-1}K$ I guess.)

t.b.

You're right, anon.

anon

Cool, thanks.

Asaf Karagila

@tb It is fun.

Moreover I fear that my thesis might be in danger, they are releasing Chicken Invaders 5 soon!

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$Mathematics$ Mathematics