@ZhenLin How is it surprising that R^n / Z^n is flat?

@tb: No, but only if you think of a torus that way!

Of course, maybe our intuition for curvature is just not very good. After all, the cylinder S^1 x (0, 1) is flat even as an embedded submanifold of Euclidean space with the induced Riemannian metric.

@ZhenLin maybe? The curvature tensor is about the most unintuitive gadget I can think of!

the unit axiom doesn't work for either T t or t T

:(

one of the unit axioms works for T t, and another works for t T

@tb: I dunno, I like to think of the Riemann tensor as a measure of holonomy...

but the respective others do not work :(

@Alexei: That's unfortunate. But as I said, if you want to see monads in geometry, look into descent theory. ;)

but it's so close!

is there some weaker sort of monad that would do?

also, if we can somehow call T t and t T equivalent, we could use each for unit axioms where it works :D

that s looks like a possible route, but i don't know really

it's kind of like 'homotoping' natural transformations, which is insane in itself

2-monads in a 3-category...

this answer is a bit esoteric...

yes, i thought about that

is there a handy paper/book i could read for those?

I'm a lowly 1-categorist.

I only consider objects.

(at least at the moment)

really it's an appealing thought that one could treat the heart of the differential geometry in the essentially algebraic manner

even up to a 'homotopy'

Well, SDG is very algebraic too, if you're willing to give up classical logic and such...

@tb didn't you use derived categories in that paper where you did crazy resolutions?

at any rate, i'm satisfied with what i learned today :D

@AlexeiAverchenko Crazy resolutions?

i learned that the tangent bundle is best considered as a functor T: Diff \to Diff, and its projection is then a natural transformation

Yes, I used to use derived categories.

I've been scrawling some nonsense about dimension theory all day. I don't feel I've learned anything...

also, i learned about adjunctions and monads with some examples and non-examples

@tb did you stop using them? why? :)

@Alexei: Do more algebra, and you'll see loooots of examples.

@AlexeiAverchenko Have you seen the Kleisli an Eilenberg-Moore categories of a monad?

one thing i still don't know (and learning it would complete my day) is how all this allows us to build up objects from generators and relations

would be nice to apply this to locales (of which I heard that it's possible)

@tb maybe i did, maybe i didn't :)

Well, the standard free presentation of a (abelian) group or module gives you (the first step of) a comonadic resolution...

i'm at catsters adjunctions 6 now :)

or at least that's what my lecturer tried to tell me.

@ZhenLin That's nicely explained in Weibel's Introduction to homological algebra, Chapter 8, if I remember correctly.

Ah, thanks. But I think my homological algebra needs are considerably more basic at the moment.

If LHF = Low Hanging Fruit, does that mean LHS = Low Hanging Scrotum, or something like that? :-)

You do know that one always hangs lower than the other, no?

@ZhenLin Oh, but all bar-resolutions you know arise from monads (that's why monads were called "standard constructions" before some people insisted on that terrible "triple" thing). For instance, the standard resolution in sheaf theory.

@JM :-D

Weibel explains that very nicely.

(not the scrotum thing, of course)

does rotman explain this?

@AlexeiAverchenko Don't know, I suspect not.

It appears after the chapter about spectral sequences. I'm not reassured.

@ZhenLin You don't need spectral sequences for that.

Oh good. But I should probably learn spectral sequences too, at some point.

@ZhenLin At some point there's no way around them. Nasty beasts...

Right now though, my struggles are with dimension theory. So homology has to be put off...

if Grothendieck feared X, X is got to be very intimidating :)

But you know: You could have invented spectral sequences (a flattering lie, as George Elencwajg put it somewhere on MO).

Indeed. Is that paper a good introduction?

Why don't people look for the most obvious counterexamples themselves

@ZhenLin I don't know -- I found it nice, but I already knew about spectral sequences before. I learned about them in Bott-Tu and Hilton-Stammbach. But many people are put off by exact couples.

But Chow's article is definitely worth having a look at.

Actually, I dread to think what you guys have been saying about my questions before I started to hang out here...

: )

FWIW: we rip on questions, not on people. :)

(usually)

Corrected.

Cut down on your paranoia, Matt :)

Clearly, I've not done enough point-set topology to see the obvious counterexample...

A = B = X dense in Y, f = identity.

OK, now I do.

@tb: I think I can either be paranoid or naive. And I think paranoid is playing it safe.

Where's the hole in my argument: Let R be an integral domain (k-algebra of finite type, k algebraically closed), m a maximal ideal of R, dim R_m \ge dim R; but dim R_m = ht m \le dim R, so dim R = dim R_m.

This seems too easy...

Today I learned that cover spaces of pointed spaces correspond to subgroups of the fundamental group of the space. I went to ask if there's any related to Galois theory, and not surprising there is a deep relation which connected most of the advanced non-set theoretic courses together.

It seems to be an obvious question to ask for anyone who's ever studied both.

Indeed.

The relation was through alg. number theory and commutative and homological algebra (the latter which I am taking now as well).

There's now a subject called "categorical Galois theory" dedicated to the study of this phenomenon.

I don't like impatient people

@Asaf: Actually, I would have said the connection is via arithmetic and algebraic geometry.

@tb We do have quite a lot of people in a rush these days...

I don't like censoring words

if you don't want to say a word don't say it

@JM Well, leaving such comments won't help.

@ZhenLin Yeah, this is what was explained to me. However the words "alg. geo." were not said. I just know what is behind everything and it was apparent now how all that is connected.

@t.b.: I know; I think I've said before that I actively ignore questions with words like "hurry" in them...

@Asaf: The most amusing thing about it, really, is that classical Galois theory, from the geometric point of view, appears to be the study of covering spaces of a one-point space!

@AsafKaragila Have a look at Galois connections

@JM I think so, too, and I follow that rule. Interestingly, it's mostly people asking crappy questions who do that.

How do you choose what mathematics to study?

I was always good with set theory, and mediocre at best with the rest of the math... it was obvious for me. I was always enjoying large cardinalities. Infinite better than finite, etc etc.

Dinner time, bbl.

I seem to have done a negative amount of work in the last two hours. sigh There must be some way of understanding this dimension theory stuff...

I haven't done any maths for a long time

@ZhenLin You can always quit. :-P

Well, I've put it off as long as I could, and it seems that the time has come to learn it.

Nonsense. People put off anything they want for a lifetime.

I'm still putting off proper social conduct.

Well, I'm calling it quits for today anyway. :p Time to go do some other things (like eat).

Eating is for chumps! :-)

@QED It's admittedly inefficient, but I went with trial and error...

I've got to find something I can study

As long as it works

Set theory?

You have expressed interest in that before.

(though I've always liked geometry even as a kid, before I realized it was actually math)

@QED: Why don't you look at stuff you liked before and see if you can find some leads for further study? Good books almost always contain some pointers for exploring a subject further and deeper.

@QED Stein - Harmonic Analysis.

That reminds me, I never found out when a fourier series exists

Sometimes.

Almost everywhere.

Time for Aberlour!

@JM: I think that if the fifth closing vote is the executioner, the three deleting votes correspond to "Ready, aim, fire!"

Sounds good to me. :)

These flag alerts on chat are seriously annoying.

Whew. I thought I was seeing things...

I think that this is because you're a room owner.

I mean, they let me judge on flags but they won't let me delete stuff from the room I own. Talk about Koenigship.

Now it is time to pop some popcorn and go watch Total Recall with my lady.

@Asaf Have fun.

...and I better get to bed. Later.

Good night, @JM

Good night!

@JM Good night (if it isn't too late already).

Oops, it looks as if it is too late :-(

I was busy fixing up an answer.

Now I have to go to get groceries. It is raining out, of course...

@AsafKaragila: Funny you should watch that. We watched it just the other day and it was Total Awesome : D

@tb is too quick =). Within 20 seconds...

: D

@tb If you see this in time, could you correct the norm notation "||" to "\|" in that post?

Thanks

@Srivatsan: I see that you already have taken care of this :) I was already fixing David's answer...

No problems. I edited it after you ran out of your 5-minute edit window...

And here Davide has beaten me to it by a few seconds :)

What on earth could this mean?

We'll soon find out, he's going to edit his question.

This looks like a duplicate: math.stackexchange.com/questions/84011/…

The question or the user? : )

Changed, sorry =)

Does f_n converges strongly to f (in l^1) really mean lim_{n \rightarrow \infty} \sum_{k=0}^\infty | f_n(k) - f(k)| = 0?

Hello @tb, you around?

And similarly, does f_n converges weakly to f really mean \lim_{n \rightarrow \infty} \sup _{\| f_n - f\| \leq 1} |\phi(f_n) - \phi(f)| for all \phi \in (l^1)^\ast?

I need to know what does this sentence mean: In der Genievorstellung der Romantik waren große Dichter über allem stehende Persönlichkeiten.

@Gigili: Loose translation: Using the conception of a genius during romanticism the poets were above everyone else.

@Matt Phi(f_n) -> phi(f) for all phi in your dual.

@JonasTeuwen: Yes and now I'm trying to figure out what the arrow exactly stands for. Is what I wrote correct?

@Matt <f_n, phi> -> <f, phi> if that makes things clearer.

@Matt no, it means |\phi(f_n) - \phi(f)| -> 0 (note that this is a sequence of real numbers).

@JonasTeuwen: No that makes things less clear. My phi is already in the dual. If I'm not mistaken you should have a g in the product?

And this should hold for all \phi from the dual of course.

@JonasTeuwen: actually, I don't know. You tell me: what's your phi there?

An element from your dual.

@tb Thank you, but I don't understand it =\

@Gigili: The sentence asserts that poets were considered the greatest geniuses of all.

@tb: Confusing. The norm in the dual is sup \{|\phi(f)| \mid \| f \| \leq 1 \} but convergence in the dual is just convergence like in R instead of being sup \{|\phi(f) - \phi(f_n)| \mid \| f - f_n\| \leq 1 \}.

I should call it a day, probably.

@Matt But we're not talking about convergence with respect to any norm, we're talking about convergence of real (or complex) numbers. If you evaluate a linear functional at a point in the Banach space, you get such a number. Now weak convergence f_n -> f means that for all continuous linear functionals \phi we have |\phi(f_n) - \phi(f)| -> 0

@tb: Thank you.

@tb Got it, thank you.

At least no one can see you blush here when you realise your mistake : ) I like.

Given the color of your gravatar it is hard to imagine to see it blushing :)

: ) I was talking about my real face. I hate when I blush but I can't control it.

Maybe you should stop hating that. Less trouble for you!

Yes and hating it makes it actually more likely to happen : D

Exactly.

It's like a vicious circle.

@JonasTeuwen: Do you have it too?

@Matt have what too? Tendency to blush?

yes

@Matt Not anymore!

Curious question. May I know how old you are, Matt? =)

@JonasTeuwen: What did you do? Just stop trying to stop it and not care and then it stopped happening?

@Srivatsan: No. Sorry.

That's cool.

Well... Basically yes. (easier said than done). Plus I got older and noted that most people don't really care about you being wrong or whatever.

Personally, I don't worry about it much either.

Yes, the only one caring is: you.

@Srivatsan: I imagined you to look Indian so blushing wouldn't be an issue?

@Matt Really, I never gave it so much thought.

I feel you are overthinking this, if I may say so. =)

You may : )

Yes, what Jonas says is true. No one really has the time to keep track of who is wrong and so on.

@JonasTeuwen: Yeah, I guess I should try to grow up.

Oh, it happens in all sorts of situations, not just when I'm wrong...

Something else: Did anyone read The Cauchy - Schwarz Master Class by J. Steele? It's a great book.

Yes, sure, but the same holds there :D.

No, I've not read that book.

@JonasTeuwen Yes, it's very nice!

@Matt Then I suggest you do!

Yes it is!

It is not something you learn this systematically, at least not in the Dutch education system.

fedja's being blunt...

@JonasTeuwen: Thank you for the recommendation! I added it to my want to read list.

@tb Oh well :D.

Oh... :D. Strange.

lulu.com/product/ebook/xq-solution/…

Wow.

And 24 more books like it.

*le what? : D "bridging west and east through the medium of mathematics, the two great traditions of western science (cf boundaries of science) and eastern meditation (cf buddhist prescription)."

- what are the two traditions of western science?

Error and....

I am trying to get through some of the excerpts, but it's really hard.

Lulu also publishes Frelmin's books...

@tb Do you have hardcopies of Fremlin's measure theory books?

@tb But I didn't get the two answers. (a) sometimes, b) a crank.)

@Srivatsan: Where did you find the question?

@Srivatsan You need to look up the original questions

Oh, that's interesting.

I think Milne has used something called Booksurge. Appears to be owned by Amazon.

@JonasTeuwen No I don't. My flat is big, but not that big.

:D.

I wonder if I have a small budget to buy some books. My contract only consists of two pages.

Are these the two questions?

Ouch.

@JonasTeuwen: by the way (I haven't tried or researched this): Do you know whether hyperref also works with plain TeX? If not is there a substitute?

Well, I'm pretty sure hyperref uses some LaTeX specific things. I'm not sure about a substitute but I would think not as it would work with LaTeX as well. I'll check it out. One minute :-).

@tb hyperref does not work with plain TeX but you could write your own macros using tug.org/eplain.

@JonasTeuwen Thanks, I was fearing such an answer. I've been thinking a long time now how tremendously convenient hyperlinks in Fremlin would be.

That's plain TeX? Wow.

@Srivatsan Yes, of course.

@tb I figured that a bit after I typed it out. [I was reading the questions and answers in the opposite order, so they did not make sense first =)]

@JonasTeuwen I'm pretty sure it is, but I haven't looked at the source very closely.

(I was more interested in the contents of the book :))

I have tried to compile it with LaTeX and most of them just worked.

Unless I was drunk at the moment and have ran TeX instead of LaTeX.

Put I did use latex -> ps2pdf but hyperref also will work that way.

@tb arxiv.org/hypertex Check this out!

(not sure if it is useful ;-))

@JonasTeuwen Thanks, I'll try that!

But if you can compile it using latex, then the \usepackage{hyperref} should work...

As I said, I'll try that --- as soon as I find time... Too busy procrastinating here :)

Sure, I can try it now. One moment.

@tb Too bad :-).

There was a thick mist here all day! The world is going to end!

(it didn't work)

Thanks for trying! The world's about to end :)

I can't edit it anymore!

The world? Are we in the Matrix?

(going -> about)

Ah :) relief

: )

Yay, I have to watch House : )

(I'm being told to "close my bloody laptop", so I guess I'll see you shortly ; ) )

Have fun :)

=) math.stackexchange.com/questions/84046/…

@Srivatsan Whatever. I'd be curious to know what Carl Mummert has to say about it...

(never heard of that field)

Good quality questions and all that.

Well, I heard of type theory, but I wasn't aware that it's also called intuitionistic type theory.

That might have gone over better at CSTheory.

it's also called type theory - which is?

Type theory is essentially two different things.

FWIW, I voted to close it, tb. =)

See here: en.wikipedia.org/wiki/Intuitionistic_type_theory

One "type theory" is Whitehead & Russell's proposed foundation of mathematics in Principia Mathematica -- a higher-order set theory with particular syntactic constraints.

Another "type theory" is part of the abstract theory of programming language design in computer science.

There are genealogical connections between the two.

The "intuitionistic type theory" seems to touch upon both viewpoints.

So I linked to the second one you talk about, here's the page of the first one. Anyway, that question would have been better off with a just a little bit of information and background.

Thanks for the explanations, Henning.

Your links are the same.

en.wikipedia.org/wiki/Type_theory covers both of the directions I spoke about, though it does not go very much in depth about computer-science development.

Thanks, copy-paste error. That was the one I was intending to link to :)

@Srivatsan: To answer your previous question, I'm somewhere between 25 and 30. I did something else before I started my BSc in maths.

Ok.

@Srivatsan: But I don't believe in age. Every time I have to drive somewhere by car I get this "Oh. I'm actually all grown up."-feeling. That's the only time when I realise that I'm a grown up. : )

I didn't even know you are doing your BSc in math.

Oh. Well, never mind : ) It doesn't really matter!

@tb: I should work through your answer but now I'm too tired.

@Matt: no worries, take your time. Sleep well!

@tb: Thanks. You too.

Good night everyone!

@Matt: hello; good night :-)

@tb: your gravatar and Srivatsan's look almost the same color, though yours is a bit warmer (redder)

@robjohn - why are you making this observation now? Neither has changed for a few weeks now. =)

@Srivatsan I just happened to notice them next to each other, and I couldn't tell whose was who's. Perhaps I just had nothing better to do at the moment. I am not sure exactly why.

=) I thought it was a response to a previous comment...

No; was there some comment about colors or avatars recently?

something about Matt's gravatar. // chat.stackexchange.com/transcript/message/2483309#2483309

Ah, it does look a bit fuzzier.

or perhaps one would call it anti-aliased.

@Srivatsan Yes, it is already blushing, it would seem

@robjohn which one? Mine?

@Srivatsan No Matt's gravatar is not as sharp as it was before.

Actually, it is still sharp, but the rendering of the small size is anti-aliased.

The small sized ones look fuzzier. Even mine does.

@Srivatsan Yes, it has to do with the reduction algorithm that's used.

I just assumed they simply scaled it down. Sorry, not that knowledgeable about images.

I worked on graphics software at Apple in the 90's, so I tend to notice things like that.

@robjohn Oh, cool.

There must be a simple trick escaping me at the moment, but how do you evaluate the very last integral in this answer painlessly?

@tb Are you wondering about the +it/2?

or the integral of e^{-t^2}?

In fact, I think that the answer is not sqrt{pi} for all t.

Take int_{- INF}^{+INF} exp(-(x+it/2)^2) dx and differentiate w.r.t. t.

I expected the derivative to vanish, but it doesn't.

One second, I think I messed up something.

@Srivatsan the integral of e^{-x^2} along the reals +it/2 is always sqrt(pi)

@robjohn You mean e^{-x^2}, right?

I am getting mixed up in what we are talking about here...

@Srivatsan Yes

Ali's answer makes the chain of inequalities:

int e^{-(x+it/2)^2} dx = ... = int_{-infty + it/2}^{+infty + it/2} e^{-x^2} dx = sqrt(pi).

I am talking about the first integral because I am more comfortable with it than the one tb asked about (in the penultimate step).

@Srivatsan what you have there is good. It is just change of variables.

Ok, let me take the first integral and differentiate by t.

@Srivatsan You should get 0 (an odd function integrated over R).

@robjohn Precisely. Checking that... (Messed up something in the differentiation.)

No, not an odd function, @robjohn.

@Srivatsan well, equivalent to an odd function by contour integration, I believe.

int e^{-(x+it/2)^2} (- 2(x+it/2) (i/2)) dx // Oh, sorry. I just do not know how to compute this new integral either.

@Srivatsan which is int e^{-x^2}(-2x)(i/2) dx = 0 by contour integration

@robjohn Ok, I thought that the integral came out to be nonzero, and I was mistaken.

Thanks

Okay, now I'm happy. But this does require an argument, and after all I don't think it's easier than what I suggested in the comments there. Thanks guys!

The integral along the x+it/2 path and the integral along the x path are the same since there are no singularities (thus no residue) between them and the pieces at infinity drop out.

Exactly.

@robjohn Yes, that makes sense now. The answer requires some explanation though as tb says.

This is just silly, but the OP accepted it :)

@Srivatsan That depends on the intended audience, of course.

(who would ever consider the infinity measure as an interesting and useful example?)

@tb I guess I never even knew that was a valid measure.

@robjohn Well, it is a valid measure, but of no use whatsoever.

(except for building counterexamples)

@tb Yes, I was just thinking that there is no reason why it wouldn't be a valid measure.

@robjohn Well, the original question seems not much more complicated than this explanation, it seems. =)

@tb some counterexamples are just like that.

not useful for anything else but being a counterexample.

@tb - or counter-counter-examples :)

@robjohn I don't think measure theory would lose a single application when you add the requirement that all measures are semi-finite (every set of infinite measure contains a set of finite measure).

@Srivatsan right, exactly.

A counterexample like this just shows that the definitions include silly stuff. A good counterexample should involve things you could actually encounter.

@tb Ideal world. Welcome to MSE... =)

I didn't like the question in the first place. It was a patch on top of a good question just to make it a non-duplicate.

(well, OP doesn't like me because I rejected too many of his edit suggestions and once yelled at him that he should leave the answers he gets alone)

Hmmm.

@tb well... :)

B.t.w., I am wondering about this question: math.stackexchange.com/questions/84062/…. Make anything of it, @robjohn?

@tb Did you consider the possibility that the op did not understand your answer?

His is simpler. In a way, too simple... =)

I am not saying that you should expand your answer. That might not make any difference after all =)

@Srivatsan That may well be the case (and I suspect he didn't understand the answer I gave to the "duplicate" thread either). Anyway, I think the totally atomic case is the only one where you have uniqueness. And no, I wasn't planning on expanding on it, plug in the definitions and basta.

(remember that it's the same guy with the a[A] + b[B] question.)

@tb What's the significance of that observation?

Er, I think you are being a bit harsh on L. =) Just 2 cents.

@Srivatsan okay, just in time, sorry about that

Well, the way I see it is: his first question was good but turned out to be a duplicate. In an effort to non-duplicate it, he modified it to a "counter-* example" question. (I feel it's not that natural and all that, but leave that aside.) So I wouldn't blame that on him.

About the a[A]+b[B] one, that one suffered from a trivial counter-example, but if we cut that out by restricting 0<a<b, the resulting question is not half-bad.

I can see why you might think the op is yet to learn to walk. But I don't see where he is trying to run. =)

</monologue>

<dinner time>

Sorry, got to run. Bye @tb.

Bye @Srivatsan

:)