so define a map $u: Bil (bla) \rightarrow \hom(V \otimes W, U)$

(I am willing to pay reasonable amount of money, though.) @ymar

but it's not really what he's asking for

that sends $f$ to $f'$

uniqueness does follow from your proof, but it's not really direct

but it shows at the end that the dual of the space of bilinear forms is isomorphic to the tensor product

@KannappanSampath I don't know, maybe they'll make an electronic copy if you pay them enough money.

does this not mean that that the dual has the universal property of the tensor product?

basically you're giving an alternative proof to what he's trying to prove

(they're all in the US)

@ymar Crap monkies.

@AlexanderAmenta so the proof is ok??

OK. What are they? Does any of them have an e-copy itself?

your proof is fine, but it doesn't answer the question as stated

why?

I have no idea. I just found them via a website.

he didn't ask for a proof that Bil(V,W;F)* was isomorphic to the tensor product

If it is

does that not mean it satisfies automatically the universal property?

he asked how to show the $\tilde{\alpha}$ was unique

link Typing "London" in "location" worked for me.

your proof does imply that, but it's not a direct proof

ah ok

I went round the back door

exactly

and the back door IS there, but it's not the point

hahaahahahahaha

completing the question as stated would establish the existence of the tensor product

this is exciting me more than that damn analysis assignment

your proof doesn't work if the existence hasn't been proved yet

pffffff

=p

what existence

????

you have to prove that such a vector space/module/whatever exists

@KannappanSampath Here there seems to be a digital copy. But I don't see what you need to do to obtain it.

@ymar Madison Library does have an electronic version--so, someone from Madison should help.

you can't write your first isomorphism without V tensor W existing

JINX.

@AlexanderAmenta but he already knows that no?

possibly, possibly not

ah ok

but that way is very messy

the way the OP suggested

as is any construction of the tensor product (in my opinion)

hahahaha

but this is exciting me more

I am procrastinating by answering this

@BenjaminLim I think your answer misses the point of the exercise. You can prove that the tensor product of vector spaces exists by identifying it with a subspace of the dual space of the space of bilinear forms.

@ymar Do you happen to know anyone from Winconsin Madison?

@KannappanSampath no, sorry...

@tb The pro is here! What do you mean?

than the analysis assignment

@BenjaminLim best kind of procrastination

haahahahhahahahahahaha!!!!!!!

anyway i have to go home and sleep

'night all

ok night alex

@AlexanderAmenta nite!

@tb What should the point of that exercise be? To show directly ?

@BenjaminLim 4:30 tomorrow, i probably won't be around earlier

ah ok

@KannappanSampath Just send them an email. they'll tell you everthing probably.

Good night Alex.

I have two people on my mind: Brian -- the resident topologist .

@KannappanSampath math.stackexchange.com/questions/138715/…

@BenjaminLim $\DeclareMathOperator{Bil}{Bil}$ For $(u,v) \in U \times V$ let $f_{(u,v)}(b) = b(u,v)$. This defines a bilinear map $U \times V \to \Bil(U,V,k)^\ast$ which has the universal property of the tensor product map.

@BenjaminLim Oooh, the tensor products. o_0!

@tb My proof solves the OPs problem no?

@BenjaminLim I don't think so.

But I showed that the tensor product is isomorphic to the dual

provided you know that it exists.

OK, see you guys.

And, math.stackexchange.com/users/15649/tyson-williams

the tensor product?

Later @ymar. Take care.

But that is like assumed knowledge no? @tb

no.

haiiii

Check what I said and get the existence of the tensor product for vector spaces for free.

ah ok

I shall edit my answer :D

I am procrastinating by doing this and not typing up an assignment :D :D

(of course you take the subspace generated by the $f_{(u,v)}$, not all of the dual space).

@tb Shall I ask you for an advice?

(and a clarification.)

If I say you may, I guess you shall.

: D

OK. So, Brian is from Winconsin-Madison, right?

Somewhere in Ohio, yes.

And, is it advisable to write to a random stranger asking for favour? (of course, some reference material.)

I wouldn't call him a random stranger, but it heavily depends on the favor.

OK. The language problem again. I'll call him a stranger graduate student. But, I would like to ask him for a copy of thesis from the library , available online.

@KannappanSampath I don't understand: the thesis is available online but you don't have access and you hope he has?

(if he has, I don't think it would be too much to ask but I'd ask him here first before sending him an email).

@tb Yes. His home institute library has a soft copy of the thesis.

(And, I was not referring to Brian as random stranger, if that was not clear.)

I was referring to: math.stackexchange.com/users/15649/tyson-williams as one.

@KannappanSampath I wouldn't ask that guy. You can certainly ask Brian if he's here, but I'd ask him if you meet him here, not send him an email. I myself always found getting unsolicited email from people here intrusive and bothersome.

(see also Arturo's profile)

@tb OK. So, I'll keep my fingers crossed. I hope he'd come.

@tb I have seen and I know it is irritating. But, well, you might understand that some references are simply not some things you can miss.

I would like to read that to understand more about the collineation groups of projective planes. It is conjectured that they are trivial.

Sure, I understand. I think you had enough exchanges with Brian that it might be okay, but bear in mind that people are willing to help here does not mean that they will be pleased to do random favors to people. After all, here you can still choose what you want to spend your time with.

Yes, I agree.

Sheesh how does this piece of imbecility get two upvotes?

sympathy + incomprehension

The Chocolate Teapot...Hello @JM

Hah! I guessed right!

Hi Kan.

@tb ayt?

@MattN well, now you got your tautological non-answer I predicted...

@tb Haven't had time to read anything, been running around until this very second.

@tb I assume you knew that already. But I thought just in case you didn't.

Now I need to sit here for a minute and recuperate before I read through all the stuff in response to that question.

@JM Is that a chocolate magic lamp? : )

@MattN Hi! Well, I was going for something that looked like a cow's hide, but that works too.

What do you use for these graphics @JM

@KannappanSampath Mathematica.

I see.

@JM What's a cow's hide?

I am no good with them anyway.

@MattN Its skin, may be?

@KannappanSampath yeah... coat, hair+skin... that sort.

I assumed the skin of a cow is hairy by definition :D

(This habit of mathematical precision is getting to me... :D )

@JM Never mind, I did not tell you what my definitions were. So, I AM a bad mathematician-will-be, not you.

Hm... I guess it would help if I looked up what local convexity means. I think I've looked it up before and if I remember correctly, it means the topology on the space is induced by a bunch of seminorms.

Exactly.

Then I don't understand why that's a necessary assumption to ask when the weak and the topology on $X$ coincide.

I'm not sure it's ok to call the other thing strong topology because I think strong means the same as norm topology.

@tb ...and hi to you.

@MattN A linear functional $\varphi$ is continuous if and only if the seminorm $x \mapsto |\varphi(x)|$ is continuous. Since the weak topology is the initial topology induced by the linear functionals it is also the initial topology induced by those seminorms, so it is locally convex.

@MattN strong topology has many meanings...

@JM and hi to you, too!

Byee!

Bye Kannappan

@tb I see. I didn't know the thing about the seminorms. I think I should look at some more basic stuff first because I even struggle to understand everything your 'prediction' : )

@MattN I think what you should understand right now is that on an infinite-dimensional space the weak topology is never metrizable (it's not even first-countable), hence it is weaker than the topology you start with in most situations where you don't already know that you start with the weak topology.

@tb Ok, I'll do. I assume you assume that it's obvious to me but it's not.

^ this is not a request for further explanation.

(Looks as if Martin's comment answers my question.)

@MattN Do you know that a Hausdorff locally convex space is metrizable if and only if it is first countable? If it is first countable, there are countably many seminorms $\{|\cdot|_n\}_{n=1}^\infty$ inducing the topology. Then the metric $d(x,y) = \sum_{n=1}^\infty 2^{-n} \min{\{|x-y|_n,1\}}$ is compatible with the topology.

This just scared the hell out of me.

I've seen better cardboard-box robots, yes... :)

@tb No, of course I don't. This is all new to me. But thanks.

@JM May I ask for your vote here?

It's done.

Thanks.

bbl

@tb I've posted the link to the article on weak topologies an an answer - as you suggested. I made my answer CW - just in case you (or someone else) would like to elaborate more on it.

@MartinSleziak Thanks, Martin. By the way: the AMS journals are freely available on the ams homepage (except the last five years), so I usually prefer to link there instead of jstor which isn't accessible to everyone.

Thanks.

I did not know that.

This morning was hilarious: the dude supposed to give the talk just didn't show up.

It turned out to be the best thing that could've possibly happened since I got some awesome advice regarding many things among which my thesis : )

Now I need to find something cool to write about. And a way to ask this as a question on SE without getting closed.

Time to cook. Thai noms today, yesterday was Korean : ) I think I'll get the "boyfriend of the week award" if I keep this up.

Hi @Martin. Long time no see.

Hi @Matt and others.......

Hi

Happy Queen's day everybody!

@MattN Look a monkey! spikedmath.com/508.html

I'd totally give bonus points for that.

Heh.

@JonasTeuwen I'd tell some of your students that. :)

Taking a test will take some time, if they start using flower/giraffe and such things for variables.

@KannappanSampath Yes.

@MartinSleziak These days, people are smart. So, they might have facsimile stamps prepared for a monkey and Jonas would have to give more points.

@MartinSleziak Even more bonus points if they use several apes and monkeys!

Yeah something, okay, your integral is wrong but you have a monkey so you will get 1/4 anyway.

But, may I ask, why this obsession towards monkeys?

I don't know. Probably because they look a lot like humans but are still wild animals.

So they remind me of something still "pure". If monkey no like, monkey will sure let you notice.

Plus they are so cute when they eat.

May be.

But you live in India and they seem to be quite a plague there? 8-).

Yes, I am annoyed on several occasions by them.

They just broke into my room and robbed all my biscuits.

However, they left me alone, thank SF.

:D.

I would think that's great.

Maybe not if they do it all the time.

@JonasTeuwen You! What are you? :-)

@KannappanSampath A hoomin?

Whatever. :)

@KannappanSampath Do you have picture of those criminals? 8-).

@JonasTeuwen No, I did not have a camera with me then. I brought it from my house only last month. :-(

I will never understand why people insist on this distinction between induction and strong induction. There is none and no wonder people get confused all the time...

@tb Hmm. Isn't it just that you also assume that everything before $P(n)$ is true and not just $P(n)$ itself?

@tb Nice beer!

Guys if you are asked in an exam to show that a matrix is diagonalizable is it enough to show that the geometric multiplicity is equal to the algebraic multiplicty of its eigenvectors,?

@JimCS Depends on the theorems in your course book!

Right, they'd be slightly shocked with you using Chapter 13 theorems for a Chapter 3 problem...

@JM If he would say that it is a theorem from Ch. 13 on the exam and it is correct I would still give full marks :-).

But if it is a theorem in some other book which he cites without any source -> not full marks.

I see. :)

@JM You would not?

@JonasTeuwen Well, if I was testing only the first five chapters, and there's a theorem from chapter 13 with no explanation of how it got there...

I'm torn. :) On the other hand I'd be delighted that the kid (apparently) reads ahead.

@JM Hmm, yes, but if he says: Use theorem 13.1 or whatever, that is just a correct argument, right?

Yeah... it's not like I'm above nuking mosquitoes myself. :D

@JonasTeuwen Haha, funny. Sorry, was afk.

@MattN 8-).

I'm stuffed.

@MattN Like the monkey that devoured @KannappanSampath's biscuits?

I'd love to meet that criminal.

@JonasTeuwen No : ) Like myself after eating a mountain of stir fried noodles.

Not achieved anything today : (

I'm gonna have to step out. You kids have fun, okay?

Okay : ) Be seeing you!

Look at that.

So I'm not the only one who thinks lectures are torture / boring to death.

Although I think telly is much worse.

Day 4 could be a "bad day".

bbl

Hi @Brian. I have not seen you in a while!

yay! Рајко's back

@KannappanSampath Likewise. First you were away, and then I was out of town for a while.

@BrianMScott So, how are you?

Not too bad. A bit sore from too much driving. My back doesn't like that much.

Not nice to hear that... Hope you'll feel better soon.

I'm trying to help a friend with her complex homework. Can someone provide a hint towards the first step to $\int_0^\infty \frac 1{x^5+a^5}dx$.

@KannappanSampath Sleeping in my own bed instead of on a friend's couch helps a great deal. :-)

Hello everybody. Give me please some reference on Riemann function in mathematical physics if you have.

@tb Why "For the sake of having an answer"?

@MattN because the question was already answered in the comments.

(I think the main confusion stemmed from the fact that newbie had a wrong definition of the basic neighborhoods).

Yes but in the comments you merely give a hint. Am I missing anything?

I haven't read and understood all of your answer but it looks like a full answer to the question.

I don't like "for the sake of having an answer".

Why not omit it?

@tb Nice : )

There I removed it. What's the problem with it?

Don't know but it puts me off somehow. It's unnecessary. No?

But now I'm happy : )

Thanks.

@tb who's Рајко?

@robjohn A crank who is especially fond of the identity $\pi = \lim_{m\to\infty} m \sin{180/m}$ (in degrees).

@tb Is that the unknown user?

@robjohn No, I think it's this one.

@MattN Yes, but I was looking at tb's earlier comment

@robjohn That's what I thought.

which, after looking, has no dates more recent than last summer, so I am way confused.

@robjohn no, the phantastic contribution was removed as spam, apparently. It's basically the same post as this deleted one

Has no one been nominated for moderator yet, or is this phase anonymous?

@ZhenLin no, the banner popped up only a few minutes ago. People have to nominate themselves and I think it's visible as soon as someone does.

Ah. That explains things.

(at least last time it was)

@ZhenLin there are moderator nominations?

boy, I feel out of it :-)

@tb ah, that would explain my cluelessness

@robjohn - Could you answer me a question? What foreign language you took as your phd requirement?

@Victor German and French

@robjohn Heh, das gemeine Quadrat spricht Deutsch?

@MattN nur ein bischen

: )

@robjohn - i knew that most college only give you the choice of Russian, German, french as requirement, is it true at least in US?

I took a few classes as undergrad and my latin pushed me through the french

@Victor I think those were three, we may have had one or two others.

@robjohn - I think it make no sense that majority of medical doctor and lawyer make more money than scientist and mathmatician, but they may speak no foreign language

Wasn't this question asked before?

@tb never saw a banner. I guess only some people saw the banner?

@tb Oh, I see it is on the main site :-)

@robjohn with link here

(oh, I see you found it :))

Also, quick question: a sequence space equipped with the metric $d(x, y) = \inf \{ 2^{-n} : x_i = y_i \text{ for all } i < n \}$ is compact, right?

@tb Thanks. I was just reading that post :-)

@ZhenLin are your sequences from $\{0,1\}$ or a finite set? Then yes, you get a homeomorph of the Cantor set.

Ah, yes, finite set.

Hm, I thought it would be homeomorphic to $X^\mathbb{N}$ equipped with the product topology.

Ah... they're the same. Good. Intuition hasn't failed me!

@ZhenLin that seems right to me.

@ZhenLin This is the Cantor set (or a version of it) if $X$ is finite. (if $X = \{0,1\}$ send a sequence to the corresponding triadic number in $[0,1]$.

@robjohn - i think this is funny that the graduate student major in commuication: cla.purdue.edu/communication/graduate doesn't require foreign language as requirement but most of the graduate student of mathmatics does.

I proved for myself today that a function $f : X^\mathbb{N} \times Y \to Z$ that depends only on finite initial segment of the sequence is continuous (for each fixed $y \in Y$), when $Y$ and $Z$ are given the discrete topology. It seems like there's something deeper going on to do with profiniteness, but I don't know enough.

@ZhenLin What you write is the usual way people write down the metric on $\prod \mathbb{Z}/(p^n)$ in the explicit construction of the $p$-adic integers as limit.

@tb Ah, yes, I remember! Thanks for reminding me.

@ZhenLin Actually, I think it has more to do with trees and descriptive set theory. Using such observations is the way one embeds Cantor sets into various other sets (e.g. uncountable Borel sets in Polish spaces).

Hmmm. Possibly. I was working with oracle-computable functions. But it reminded me of things like Galois groups acting on finite sets.

Well, I don't know enough about the effective side of that stuff, but I think it might have to do with Willie's question here. I link to a pdf version of Moschowakis's book in a comment to Asaf's answer where there should be more explanations on the relations and differences.

Oh, already two nominees... and a meta thread.

Let $p_i$ be -1 or 1 with 50% probability, does $\sum_n p_n/n^{1/2+e}$ converge for all e>0, but not e=0?

What do you think? The first part is as trivial as trivial can be!

oh duh

the last one ?

Noo, now the teddy has disappeared and I missed saying good night.

Good night!

@MattN Bye 8-).

@Holowitz Well if you have finitely many times $1$ or $-1$ it still diverges.

@MattN He might have just lost connection. He usually says goodbye

@JonasTeuwen but that almost never happens.

Yes, so you can remove that situation 8-).

I'm sure the convergence sought is "almost always"

So then the question is if we have $a_n$ which is $1$ infinitely many times and otherwise $0$, does $\sum a_n n^{-1/2}$ diverge?

(for all such sequences of course)

Or am I talking out of my ass? 8-).

@JonasTeuwen How does that apply?

Oh, it doesn't...

It was just trying to make it simpler, and failed 8-).

As it can diverge but our original series can still converge.

The question is probabilistically based on the expected random walk of length $n$ ending about $\sqrt{n}$ away from the origin.

At least that is how I see it.

He no wants analysis? 8-).

Fourier Analysis may be brought to bear :-)

Yes.

I was trying to think of something really elementary...

@MattN Night. Sorry about that, I felt like disappearing for a moment...

Hmm... If you're ever in the neighborhood, I'll show you some good places for Feuerwasser and the like, @tb 8-).

(of course, if you'd want that...)

Thanks, of course I'd love to do that, but I think it won't be too soon that I end up to the north of my current location. Still shivering from this year's winter...

:-). It is 20 degrees here!

I'm almost melting.

And now I need a fridge to cool my beer...

@robjohn Could we apply PUB? 8-).

Well since I consume a lot of milk I always need a fridge :)

Yes, but you need a different temperature for many beers!

@robjohn So say we have the operator $S_N$ from the bounded sequences to $\mathbf R$. Which is the partial sum of our thingie there! So, then the operator norm is either bounded in supremum or we have a sequence where it blows up 8-).

Okay, that doesn't make any sense.

@JonasTeuwen: I'll let you know if ever I should be nearby :) Do the same if you come around here. But I should go now. Good night!

@tb I will! Good night!

@JasperLoy Bloody monkey! So blistering hot!

Ack! Now I've missed tb. Teach me for dealing with things afk.

@robjohn He will return 8-).

There is a quantifier difference, Jasper...

@JonasTeuwen tomorrow, yes.

Something like "for all" and "there exists".

@robjohn Could we show this using Banach-Steinhaus, btw?

@JonasTeuwen I don't see it right off, but it might be

hi all

can we nominate other person to the moderator elections?

I certainly don't hope so! Who you had in mind @leo?

@JonasTeuwen I was thinking in what persons full fill the description of a moderator. Two names comes to my mind: @tb and @robjohn. I don't know the current nominees.

I think @tb and @robjohn are very capable of nominating themselves when they would want that.

It can be felt as some kind of harassment if you do this unsolicited...

@JonasTeuwen yes. You are right. After all, the moderators are voluntary.

you've got to be kidding me

@anon That's an interesting nickname.

@anon :-|

It would be much better if he would ask some very deep questions 8-).

better would be to consistently write deep questions and answers and gain a reputation on the site, all the while having "ilikesaggytits" ROT13'd as his username.

@anon Evidently, they were.

@anon That'd be nice.

@leo I've seen a number of posts from Chris, chessmath, and Peter.

@anon vyvxrfnttlgvgf?

@robjohn Yes 8-).

@BrianMScott what do you think?

see you all

@leo I'm not going to give it any real thought for a few days yet; it's too early. I will say that Chris's nominating statement sounds very reasonable.

@BrianMScott Seems that he really wants to do so

see you later

@JasperLoy It seems that the account was deleted

Also see the url of the link.

@JasperLoy They may have talked to the person behind the curtain and nudged things that way, but I don't think they would remove the account, though I could be mistaken.

it is quite strange that nominees, so far, are users with relatively little participation in the site

@MarianoSuárezAlvarez I noticed that too. I don't remember any of them having 10k powers, which I thought would be a prerequisite.

do we know how many new mods are to be elected?

@MarianoSuárezAlvarez Chris Taylor has been reasonably active, though he's not yet at 10k. Peter Tamaroff has been very active recently but has been around for only three months.

@JasperLoy Not strictly speaking. But it seems like a bit of a jump.

@BrianMScott I have seen a lot of activity from those users recently, as well.

How often do they have elections, or is it when the current moderator load is too much?

the current load is not too large

so it is probably a yearly thing :)

@MarianoSuárezAlvarez On the right of this page it says two.

ah

thanks!

Kind of a weird place to put information you want people to read.

Is there a clever way to get a good-looking \pm with the plus and minus two different colors?