@anon Well, I know that searching wiki:1 on the search bar will filter these questions to me. I kinda see them linked somehow.

which was the cause of the confusion

@anon Yep. Maybe because I got another native languange.

wish I knew more than English

@anon Let's study another languange!

@anon Something not very usual,such as akkadian.

@anon, where are you from, if I may ask?

Midwest USA

@FortuonPaendrag And where are you from, if I may ask?

Colorado, USA.

@FortuonPaendrag Is that your real name?

No no. I would have been beaten up in school for that.

@FortuonPaendrag I was thinking you're greek, because of the name.

It is homage to a character in a book. "Tuon Paendrag"

@FortuonPaendrag Tuon Athaem Kore Paendrag

?

Formerly. Now she is Fortuona Athaem Devi Paendrag.

She became empress, you see.

@FortuonPaendrag I guess I've already read about WoT somewhere.

That's exciting. They are not very popular, unfortunately.

Are you into fantasy?

@anon, It is considered ok to discuss non-math stuff here,right? Or is there an upper bound to how much of that we can do?

it's fine if there's no other convo you're getting in the way of

and certainly the regulars do it even while others discuss math

@FortuonPaendrag I've heard about it, but I never read some formal definiton on it. Doing it now.

fortuon means like fiction, novels, stories

though fantasy is a specific genre I think

Yeah, it is a little more specific than that. It's more like things involving "magic" of some sort.

Although I hate to describe it like that.

@anon Yep. I imagined I knew it, but as I never researched into the topic, I feared to presume what it is.

@Gustavo Where are you from?

@FortuonPaendrag Brazil.

@FortuonPaendrag My difficulty in math is explained now. Haha

Ah. Then it must be very late for you?

@FortuonPaendrag Very early ;)

Are you in the same place, @N3bu?

@FortuonPaendrag 05:13 A.M.

@FortuonPaendrag I know the time difference from my place to various places..

So, where are you located?

@FortuonPaendrag "I am an unramified abelian extension looking to ramify." -> Does it mean you're searching for a wife?

You could guess, here it is 10:15 AM.

@N3buchadnezzar I guess Europe or Africa

@N3buchadnezzar Praha?

@FortuonPaendrag Norway!

Haha

@GustavoBandeira , No No. I am not looking for a wife. I tried to write something that would indicate my interest in Algebraic Number theory and still sound cool.

@N3buchadnezzar I like the extension of my guess: I've almost said: "Planet earth"?

@N3 Ah, very nice. We are such an international community.

@FortuonPaendrag The ramify thing, I imagined the biological relations, ramify reminded me specifically of plants.

Ahaha. I was talking of Ideals Ramifying in Extensions. Math terms are so weird.

@FortuonPaendrag Also, I'm no saying you're a plant. But I don't know if you are, are you? In Brazil, plants can talk.

@FortuonPaendrag btw, searching about this now.

@Gustavo Did you know there's a subfield of math called Surgery theory?

Are you searching about plants talking? Well, I am a palm tree and I can talk.

@FortuonPaendrag Indeed. Except foreigners, death to all foreigners!

@FortuonPaendrag Oh, last night I discovered about umbral calculus.

@FortuonPaendrag Haha. No, about ramification.

Oh dear. I think we have a situation here. @N3bu

Oh, what is your mathematical background @Gustavo?

@GustavoBandeira First time I read about it I thought it was umbilical calculus. Gave me nightmare for weeks.

@FortuonPaendrag Not much. I'm going for calculus 1 now. I'm reading mathematics for the nonmathematician and what is mathematics?.

Hanging around this site and trying to answer questions as you can will really broaden your horizons. And people here are usually very patient if you are trying to understand something, @Gustavo

@N3buchadnezzar I guess it was worse with me, "umbra" means shadow, it reminded me of ghost pokemons.

Umbreon is the best! @GustavoBandeira

After Jolteon, I mean.

@FortuonPaendrag Yep. Now i'm kinda losing the fear I had of asking. All the questions seemed to be so hardcore and my questions are so silly.

@GustavoBandeira Umbilical as in umbilical cord, the cord connecting the unborn fetus to its mother.

@N3buchadnezzar Are you acquainted with Pokemon?

@N3buchadnezzar Yep. but why the nightmares?

@GustavoBandeira mangarush.com/manga/uzumaki/11/p-1 Read at own risk.

@FortuonPaendrag Yellow ftw?

@FortuonPaendrag I've never evolved eevee to Umbreon. =/

@N3buchadnezzar Is this that kind of hardcore non-sense japanese porn?

@N3buchadnezzar Yes! Yellow, always. With pikachu trailing behind.

@GustavoBandeira Horror, no crazy porn.

@GustavoBandeira I have read guro as well, that shit is messed up yo.

@N3buchadnezzar Yes.

Turned a dead girl into a floating submarine, I laughed so hard...

@N3buchadnezzar This is everyday routine in Brazil.

@GustavoBandeira Can they fire rockets?

If oi want a good laugh, check out "Mai chans Daily life" at your own risk of course..

@N3buchadnezzar Rockets and various types of missiles. Haha

What is all this stuff?! I am so ignorant! Perhaps for the Best.

@FortuonPaendrag What stuff?

definitely for the best

All these links you are posting?

@anon, I was about to ping you.

My week has been perspective changing.

@anon lol

With 4chan and what not

@FortuonPaendrag Mine is just an horror story, no wierd stuff.

:) I will trust, you friend, norwegian. @N3buchadnezzar

@FortuonPaendrag Oh, when he suggested me that history,it reminded me of ero guro.

Ha!! - I have just been insulted after trying to help a poster with an answer :( WTF

@OldJohn Why?

here

@OldJohn, such rudeness. I am sorry for you.

does not implement a decent algorithm for computing the integral solutions = well-optimized algorithm for finding the integral points on an elliptic curve ?

11-year old, heh

@anon He has not looked at my profile, I think :)

But I think he needs to learn some manners if he expects help here :)

@OldJohn He doesn't know you have 12yo!

@GustavoBandeira times 5

@OldJohn Would you mind if I said something there?

@FortuonPaendrag Fire away

@FortuonPaendrag FIGHT! FIGHT! FIGHT! FIGHT!

@anon I am 11 and what is this? Elliptic curves and booties?

Hi there.

@FortuonPaendrag Thanks! Maybe he will learn some manners :)

@Matt Hi there

@OldJohn Anytime, OldJohn. I should have been a little more firm, perhaps.

I seem to have somehow lost the ability to think.

@FortuonPaendrag Nah - that was enough ... for now :)

Do arrows also look like this for you on the site:

@Matt It happens - and as you get older, it only gets worser and worser :)

@Matt Only when I drink. At the moment all arrows looks fine.

@OldJohn "Retired school teacher with a bit of experience in fine topology" - Last night I dreamed of stealing a book on topology.

@Matt I sometimes see things like that with arrows and long square-root signs - always assumed it was my browsers fault

@OldJohn Oh no, I had recently gained some. But I had an oral exam on Thursday and since then it seems that I can't make sense of anything anymore. Not sure how to get rid of this.

@GustavoBandeira a "fine" one, I hope

@Matt Ask Jonas.

@Matt some combination of sleep/alcohol/coffee/exercise...

@OldJohn Is it possible to explain the difference to someone who know only how to sum? I promise I'll not say that every 11yo kid knows that.

@GustavoBandeira you want an explanation of fine topology?

@OldJohn Yes.

When I was 11 I wrote a program implementing the Eratosthenes sieve algorithm. That made me very happy.

@GustavoBandeira in short, it is the coarsest topology on $\mathbb{R}^n$ which makes all the subharmonic functions (of classical potential theory) continuous.

@ZhenLin Really? You parents were math teachers or something related?

in the normal topology, subharmonic functions are only required to be upper semi-continuous - so the fine topology is a bit finer than the usual one

@OldJohn I guess I'll need more background for that - At the moment, I'm starting calculus 1.

@GustavoBandeira Nothing of the sort. It's a very simple algorithm. You just need to know how to write loops and work with arrays...

@ZhenLin When I had 11, I didn't even know that compilers exist. =/

@GustavoBandeira OK! - well, you are in the right place to get help if you have any problems :)

@GustavoBandeira When I was 11, I'm not sure any compilers existed :(

@ZhenLin Computer was primarilly for gaming. =/

@OldJohn Haha. Well, Zuse made some comercial computers in the forties, I guess.

@OldJohn ALGOL was developed in 1958, and it's pretty much the granddaddy of all modern programming languages.

@ZhenLin Yes - I was exaggerating ... slightly :)

@OldJohn You mean, not drink alcohol, I assume.

@Matt either way! - sometimes a little alcohol can lubricate the thinking process

I'm going to sleep now. Good morning to ya.

Goodbye, all of you folks!

BYe!

@FortuonPaendrag Bye!

@ZhenLin I just found out that $\textrm{Hom}(V,V) \cong V^\ast \otimes_F V$

At the moment I can see why it's true just simply by comparing dimension

$\hom_k(V,W)\cong V^*\otimes_k W$

consider $V^*\otimes W\to \hom(V,W):f\otimes w\mapsto(v\mapsto f(v)w)$ extended linearly

yes

was that an exercise? :<

@anon No it was just that in Serre today

He was talking about how some representations were isomorphic

It was just a hunch that the isomorphism came from the fact I stated above

@anon If you have a commutative diagram of vector spaces

can you apply $GL$ to the diagram (as a functor) ?

so, it'd have no effect on the objects, it'd take isomorphisms to themselves, and wouldn't be defined for other arrows? what do you want this functor to do?

just a random question really

because I'm now wondering what is $GL(Hom(V,W))$

there'd be some kind of inclusion functor that takes a category of vector spaces with only isomorphisms into Vect, not sure about adjoint..

ok

@anon I'm confused about GL(Hom(V,W))

It'd probably be the contragredient action of GL(V) combined with the usual action of GL(V)

woahhhhhhh

in the case of V=W, these actions would coincide

@anon If we already know that $Hom(V,W) \cong V^\ast \otimes W$

@anon Then by applying GL

maybe. look at basis effects on $V^*\otimes W$

Do we get that $\textrm{GL}(Hom(V,W)) \cong \textrm{GL}(V^\ast \otimes W)$

oh, yes

why?

if $A\cong B$ then $GL(A)\cong GL(B)$.

why is that so?

with $\alpha:A\xrightarrow{\sim} B$, consider $(x\mapsto Mx)\mapsto(\alpha(x)\mapsto \alpha(Mx))$

wait the A's are the not same yes?

oops

ok

that's what happens when the brain hemispheres are not on speaking terms with each other

or it's 5am

@anon Only today I thought about whether or not you can just apply GL to it

@anon I should not disturb you

@BenjaLim of course. GL(V) is isomorphic to the group of invertible matrices that are $\dim V\times \dim V$

yes

oh yes

because $V^\ast \otimes W$ and $\hom(V,W)$ have the same dimension

you can just apply gl

it must be isomorphic

@anon but wait

GL has no structure of a vector space

so?

you talked of $\dim V \times \dim W$?

no, I said $\dim V\times \dim V$

sorry yes

but why did you mention dimension?

meaning GL(V) depends only on the dimension of V, and any two vector spaces of the same dimension are isomorphic

ok I see.

@anon I am really tired now. I guess I should go. Thanks.

Hi folks

Mushrooms mushroooooms.

Toadstoooooools

Hi @skullpatrol - cool new avatar - does it mean "no entry to anyone with a chicken on a stick"?

@OldJohn Hah, I was wondering that as well.

Do not Feed the Troll.

Hi @OldJohn

images.google.com: don't feed the trolls

But the trolls also need to eat! 8-(.

@skullpatrol Great - must bookmark that one :)

Should we feed them?

@skullpatrol nah - let them eat hobbit

Don't break anyone's heart, they have only one. Break their bones, they have 206.

As adults we have approx. 206 bones.

hi

@JasperLoy Hi

@JasperLoy why delete them - can't you just not use them?

Hellllllllo

@anon I am working on this one here: math.stackexchange.com/questions/181277/…

why precisely 1? I will look more later. I have to wake up in three hours.

@OldJohn Specifically this comment motivated my avatar.

@anon Great. Continue like this and you will need no sleep. I have tried.

@skullpatrol got it :)

Cool, an American/Korean friend is in Delft! Meet him now.

It's raining rain.

Hallelujah.

:-D

@skullpatrol You don't make sense, dude.

@JasperLoy Unify them? Become the Brian Boru of SE?

@PeterTamaroff "It's raining rain." makes sense?

Dude.

well, I've always thought I

would be very worried if it rained cats and dogs

the curse of repetition :-x

exactly!

@mixedmath Random stranger, can I ask you a question about indentification topologies?

raining cats and dogs is possibly less worrying than "hailing taxis and buses" :)

@JayeshBadwaik But repetition is the mother of learning?

@mixedmath hey

are you familiar with representation theory of finite groups?

@BenjaLim I was first. Damn you!

huh?

@PeterTamaroff Sure, although I am not great at a lot of topology

@BenjaLim I know a bit of this and a bit of that. What's up?

@mixedmath Oh, OK. I miss Brian Scott.

See the edit to my answer here: math.stackexchange.com/a/181297/5783

@mixedmath Actually my worry can be simplified down to one thing

@mixedmath Let $S_5$ act on the three element set $\{1,2,3\}$ simply by permuting the numbers.

@mixedmath How would you represent say the action of $(15)(3)(4)(2)$ on the set using a matrix?

@mixedmath the matrix would be $3 \times 3$ yes?

@skullpatrol here, the problem is I have learnt all these series and sequences as undergrad but in the applied manner and now when studying analysis, there is a lot of repetition and bits of new material embedded within. The repetition dulls my senses and I frequently miss the new material, making me having to track back frequently.

@BenjaLim Yes.

but what would the matrix look like?

@mixedmath I mean the problem is in the cycle we have $(15)$ there....

ah, I see. Well, what would the action do?

as in?

you say that S5 acts by permuting the numbers. I'm unclear how that permutation would act on the set

@mixedmath For example the cycle $(12)$ sends $1$ to $2$.

yes, but what does (15) mean?

that's the problem now...

I tried to compensate for that in my answer by adding extra zeros in the matrix

I think that maybe there's more to this action than simply saying that it acts by permutation. I don't know what permutations with 4 or 5 would do to the 3 element set.

yes correct.

In general, how would S_n act on a set with say k elements where k is different from n?

well, it's easy to see how Sn might act when k is larger than n

yes

perhaps this is one of those things that I've never noticed, but that we don't act on smaller sets. I can't think of any times that I have right now.

you might have to do something like let S3 act on it, and choose a particular subgroup of S5 isomorphic to S3 (why not the obvious one?)... and then do something

the thing is if you say that

then the question is meaningless then

no, Sn can act on many sets, and it doesn't always have to be strictly by permutation.

really?

hmm, let's think about this. When I wrote that, I was thinking that it's just a restriction to some particularly-easy-to-phrase actions. But the permuting is a consequence of the fact that we represent Sn by permutations, so any action will carry this over

yes

to some degree, at least

but we could let S5 act by conjugation, pretending (123) to be a permutation itself, for instance

the result would be a permutation in the sense that it's a subgroup of a permutation group, but I wouldn't say that it acts by permutation

right.

@mixedmath the important fact I used in my answer was the fact that the trace of the permutation matrix was the number of elements fixed by it.

ok

@mixedmath I should go to bed now.

Thanks for the discussion.

ok - good night. I hope that something became clearer, or something

later

Hmm - I seem to be upsetting everybody today :(

I dislike anonymous downvotes >8(

are you getting downvoted for something?

@OldJohn You are not upsetting me, unless you downvoted my answer without commenting.

@robjohn which answer?

@robjohn Nope - never downvoted you, I am sure :)

Peoples. Quick question. Let $X_n=\{\{0,2\},\{\varnothing,\{0,2\}\}\}$ and define the topological product space $X=\prod_{n\in \mathbb N}X_n$.

@mixedmath Yes. I thought I did a passable job of not using homotopy in this answer. I just wish that if I did something wrong, they would say what.

Now, define $f:X\to [0,1]$. Should I assume the author means $[0,1]$ as a space itself, or as a subspace of the real line?

Depends on what he asks next, if there is a next part.

@PeterTamaroff There are two elements in each $X_n$?

@robjohn Looks OK to me

@robjohn Sorry, I mean $X_n$ as the topological space $\{0,2\}$ with discrete topology.

@PeterTamaroff Ah :-)

@PeterTamaroff Thta definition above looks more like the indiscrete topology, doesn't it?

@OldJohn Hahah yes, bleh. I meant that.

@PeterTamaroff OK

@PeterTamaroff How is one different from the other? (perhaps I should shut up and let more knowledgeable people answer)

@robjohn Well, let me think.

$(a,b)\subset [0,1]$ is relatively open iff it is open in $\Bbb R$.

@JasperLoy why are you deleting the accounts? Is there a reason against simply not visiting those sites?

@OldJohn I know.

@robjohn was that aimed at Jasper? (I'm not deleting anything)

@OldJohn That's my recollection, too.

Similarily, $[a,b]\subset [0,1]$ is relatively closed in $[0,1]$ iff it is closed in $\Bbb R$.

@OldJohn yes, the messages were moving under my mouse :-)

The only "picky" points are for $[0,a)$ or $(a,1]$ because they are open in $[0,1]$ but not in $\Bbb R$.

@robjohn happens to me all the time :(

@JasperLoy But deleting accounts removes rep from others... :-(

@robjohn never knew that

@OldJohn There have only been three times that I know of (since I've been here), but it always raises a small furor when someone deletes their account.

so ... if people start closing accounts, I could lose my hard-earned guru badge ? :(((

@OldJohn badges don't go away (afaik)

@robjohn not that I really care about rep - I come here to learn from real gurus :)

@JasperLoy - Nah - never consider myself in that category, despite the badge

@OldJohn Yes, I understand, but it still causes a fuss. I don't know if a deleted user's questions or answers remain.

@JasperLoy that is good.

changing to anonymous sounds like a sensible process

you end up going back to your original user##### name

I would end up becoming user9754 again - that'd be sad

@JasperLoy Dawg. Let $X_n=\{0,2\}$ with the discrete toplogy. Then $X=\prod_{n\in \mathbb N}X_n$ inherits the discrete topology too, correct?

@PeterTamaroff No, it does not. Follows easily from the Tychonoff theorem, but I'll look for an elementary explanation.

@MichaelGreinecker SORRY! I mean INDISCRETE!

Ah, here is an answer by a deleted user (Chandrasekhar).

BAH!

@MichaelGreinecker I have the words "discrete" and "indiscrete" short circuiting.

Sorry.

I will call the discrete topology the "powerset topology" for my own sake.

@PeterTamaroff Yes, the product of indiscrete topoogies is indiscrete.

@MichaelGreinecker Yes. OK. Just checking.

I'm doing a lil' exercise on the Cantor Set.

If you endow $\{0,1\}$ with the discrete topology, $\{0,1\}^\mathbb{N}$ is homeomorphic to the Cantor set.

@MichaelGreinecker Well, I'm actually doing this: "Endow" $\{0,2\}$ with the discrete topology. Set $X=\{0,2\}^{\omega}$

Define $f:X\to [0,1]$ by $f(x)=\sum_{n=1}^\infty x(n) 3^{-n}$

Yes, I know.

Then prove $f$ is one one and continuous.

$1-1$ follows from the equality of series.

Suppose $f(x)=f(x')$

Then $x'(n)=x(n)$ so $x=x'$

Now I'm proving that $f$ is continuous, but I had to check what the topologies were.

Should I assume $[0,1]$ is given the relative topology? @MichaelGreinecker

@PeterTamaroff Yes.

@MichaelGreinecker OK. So the open sets of $[0,1]$ are the two trivial ones, $(a,b)$ $0\leq a<b \leq 1$ and $[0,a)$,$(a,1]$,$0<a<1$

@PeterTamaroff ...and their arbitrary unions.

@MichaelGreinecker Well, yes.

Hm, I think know.

@PeterTamaroff The function is continuous if you endow $\{0,1\}$ with the discrete topology and $\{0,2\}^\omega$ with the product topology. The indiscrete topology is too coarse.

Guys, who does know explicit formulas of reconstruction of $x_1, \ldots, x_n$ by $x_1^k + \ldots + x_n^k$ known for any $k = 1, \ldots, n$ ?

@MichaelGreinecker Fuu, I misread.

In wiki I've found only Newton identities that allow to reconstruct elementary symmetric functions