Seems like the dead have come alive form the coffins like the Barlaka and ABeau

@BalarkaSen hey!

Everyone now calls me abeau, how nice.

@BalarkaSen Want a problem?

Hey @r9m. @Pedro Depends. What is it about?

@PedroTamaroff so you're doing Bsc?Msc?What do they teach?Is it interesting?SHould I go IIT? Learn CSE? or MAThs? But I don't think I'll et a good job then?from Maths so I think of doing CSE

@BalarkaSen Exhibit a flat $\Bbb Z$-module that is not free.

and ^that summarises how helpless the average indian student can sound ..

@ADG Sorry that's too many questions.

@r9m would _you_ help

@PedroTamaroff I am in topology mode right now, unfortunately.

@PedroTamaroff try answering them (please #beg!)

That's a silly excuse.

@ADG hmm .. nope

@BalarkaSen what does that mean topology? are you mathematician or cartographer? or geographer? do you map ?

Well you wouldn't find it silly if you were struggling with computing homology of cell complexes @Pedro

@r9m so blunt .I'm hurt , I think of filing a case against you (#cry!!!)

Definitely troll.

I think I have another person to ignore.

@BalarkaSen Yes, it is silly.

oh are you all 4 agianst me? think it's time to become seroius!

don't you people get bored or try to relax or enjoy or have fun?

@PedroTamaroff Right, OK. The real reason is that I don't know what a flat module is.

@ADG well for starters I can't be someone's career guide as mine is not particularly a stellar one .. and besides those Q's you are asking are too broad/lacking context .. etc .. :|

No.

I never got through much of chapter 2 in Atiyah-MacDonald.

Do any of you have a great Integration problem on hand?

@MikeMiller No to whom? to what?

You.

I never get bored or try to relax or enjoy or have fun.

@BalarkaSen OK. Then don't worry about it.

anyways see the great commotion at here

do we also support Math 101?

@MikeMiller So what do you do then the day long?

@PedroTamaroff Maybe I should start studying Chapter 2.

That's my business.

@MikeMiller sorry if that hurt you

@BalarkaSen If you're studying something else, just continue with that.

#Bye

Yeah, I was kind of thinking if it's a good idea to study two things at once.

Give it a shot.

@TedShifrin Hello!

howdy @Pedro @Balarka @Mike

Hi @Ted

oh, good night @Mike

Hello @ted

Morning @Ted

@Pedro Thanks for the encouragement.

hi again, Jasper

Hmm, @Pedro ... $\Bbb Q$ ?

@TedShifrin Yep.

Wow, I'm not a 100% idiot yet :P

@Ted I don't get the prism operator.

What was the quesrion?

That's why I told you to think about the boundary of a chain $\sigma$ cross $I$.

@Balarka wanted to understand why homology is a homotopy invariant, @Mike.

Why does that involve me, @Ted?

You asked what the question was ...

Oh, were you referring to Pedro's question?

He's still ignoring me @Ted

Oh, I see. Yeah.

sighs

@Balarka: You should have learned from your experience with me that stressing over it does no good.

But I could probably reverse engineer a quesrion whose answer is $\Bbb Q$.

Pedro's question was to find a flat module that isn't free, @Mike.

@TedShifrin OK, I used your hint and understood as much as this :

You could reverse engineer uncountably many questions whose answers are all $\Bbb Q$, @Mike.

You have a homotopy $F : X \times [0, 1] \to Y$ between $f, g : X \to Y$

Yes, "Find a field of characteristic 0" was one I was considering.

And you want this homotopy to descend all the way down to a homotopy of chain complexes.

@TedShifrin when $(x,y)\mapsto \phi(x,y)$ is some parametrisation of some area then $\phi_x \times \phi_y$ is a surface normal right?

Well, sure, @Balarka, but I was trying to give you geometric intuition, rather than formal intuition :P

Yes, bananas.

There is this map $F \circ \sigma : \Delta^n \times [0, 1] \to Y$

I never imagined Think would make such a badass combination with a shoot-out scene :P lol

if we do the product in reverse order, the sign changes?

That's why you need an orientation on surfaces in order to define flux, bananas.

@TedShifrin I am trying to get the geometric intuition, Ted. I don't understand chain homotopies.

okok

So imagine that $f_*\sigma$ is at the $0$ level and $g_*\sigma$ is at the $1$ level, @Balarka. What does the boundary of that prism give you?

@MikeMiller Note that any finitely generated flat $\Bbb Z$-module is free. So this shows $\Bbb Q$ is not finitely generated as a $\Bbb Z$-module.

I imagine we could give a direct argument for that, @Pedro :)

OH. The boundary is $g_*(\sigma) - f_*(\sigma)$ plus something.

And that something might just be $-P \partial$

Right, @Balarka.

The other part is $\partial\sigma\times I$, not worrying about signs.

Wait wait wait let me visualize this.

Draw pictures, for sure.

@Pedro: Remind me of the proof that finitely generated + flat = free.

@Ted: Maybe you consider it cheating, but you could just use the classification.

That doesn't make me happy.

@TedShifrin One uses that torsion-free f.g. modules over PIDs are free.

Figured.

Well, so you're giving me @Mike's answer.

aka, classification

And flat modules are always torsionfree. In particular, over a domain where every finitely generated ideal is principal, torsionfree and flat are equivalent.

Of course. If there's torsion, there's an obvious module to tensor with which destroys exactness.

@TedShifrin Right.

How does one use PID (don't say classification)?

One uses submodules of free modules are free.

You can show any torsion free module embeds in a free module.

Ah, submodule of free is free is false without PID? Example?

Of course you need that $A$ is a PID.

@TedShifrin $\Bbb Z[x,y]$, $(x,y)$.

$A=\Bbb Z[x,y]$.

@TedShifrin It is an iff, in fact.

Hmm, why isn't $(x,y)$ free? Oh, 'cuz $xy=yx$. Drat.

Cool. When I'm retired you can reteach me all the stuff I've forgotten, @Pedro :P

more obviously: there's a non-free sub module of $\Bbb Z_4$...

This sort of stuff leads to the Koszul complex, in fact ...

@TedShifrin Yes, more generally if $I$ is an ideal inside a ring that is not principal it cannot be free: if it is free it has a basis of more than two elements say $a_1,a_2,\rm stuff$, but $a_1a_2-a_2a_1=0$.

Huh, @Mike? Any quotient of a PID is a PID.

Right, @Pedro. That's the Koszul complex starting ...

uh...

Oh, any quotient of a PID is a principal ideal ring.

@TedShifrin Stops, fears the Koszul complex.

and domain is essential here

Where? @Mike

It's homological algebra, @Pedro.

@MikeMiller Well, no. If $A$ is a ring such that every submodule is free, then it is a domain.

That was, I believe, the point @Mike was trying to make to me, @Pedro.

@TedShifrin I've bumped into it in Jacobson's BAII. Yes.

@TedShifrin Oh. Sorry @MikeMiller.

No fun when you guys gang up on me :D

Balarka sure got quiet. Maybe he's understood his picture.

The fact that it's a domain is essential to the fact that submodules are free modules are free, @Ted. That's all I was saying. Nothing intelligent.

OK, so, $f_\#, g_\#$ are maps from $C_\bullet(X) \to C_\bullet(Y)$ sending a simplex $\sigma : \Delta^n \to X$ to $\Delta^n \to X \stackrel{\text{f or g}}{\to} Y$. So you have the kind of diagram like this

That wasn't the picture I wanted, @Balarka. I wanted a picture of $\sigma\times I$ :)

But, yes, what you have is germane.

so $F_\#(\sigma\times I) = $?

Yeah, yeah I'm thinking about it. Don't give hints.

Don't worry. I'm not.

Awww, damn, I forgot my pen at home.

Your own whom, @Mike?

LOL oh

I turn off autocorrect.

The minor bizarre errors are worth the convenience.

@MaryStar: Is this an assumption about a particular integral domain $K$? Or are you asserting that it's true for every integral domain?

I guess I'll just do some reading instead of some writing.

@Mike: You'd probably forget your head if it weren't screwed on.

I take it you're at one of the coffee shops, not your office, @Mike?

Out to lunch, but close enough.

Well, I should be working on clearing out closets ... agh, I hate this.

@MaryStar Principal means it is generated by a specific number of elements...

@TedShifrin Does this stand for each integral domain when the ideals are principal??

Think about $\Bbb Z$, @MaryStar. How many ideals are there?

I'll defer to @Pedro for algebra. :) I can go do hateful stuff.

@Ted: in algebraic geometry, is a blowdown just undoing a blowup?

Yes, @Mike, although not necessarily of a point, of course.

Sure. Thanks.

hello guys! I'm sorry to promote my question but I have an exam the day after tomorrow and I would like to understand it well :)

Can someone take a look at math.stackexchange.com/questions/1127735/… ?

@PedroTamaroff So when the ideals of a $K$ are principal, they are finitely generated, right??

I'd appreciate it :)

@MaryStar What do you think?

@Ant It is rigorous if you think about it as $$\lim_{M\to\infty}\int_0^M e^{-ts}dt=\lim_{M\to\infty}\left.-\frac{e^{-ts}}s\right|_0^M$$

Well $F_\#$ is a map from $C_\bullet(X \times [0, 1])$ to $C_\bullet(Y)$

That doesn't look too promising.

I think that it stands that when the ideals of a $K$ are principal, they are finitely generated. Is this correct?? @PedroTamaroff

@MaryStar A principal ideal is generated by how many elements?

@Balarka: You've distracted yourself from my specific task.

A finitely generated ideal is generated by $n$ elements for $n$ a nonnegative integer.

By one, @PedroTamaroff right??

@MaryStar Yes.

$\sigma \times [0, 1]$ is the map $X \times [0, 1] \to Y \times [0, 1]$.

Not sure what's special about that.

So it is generated by finitely many elements, namely $1$.

@PedroTamaroff Yes I don't understand why I can calculate the integral as the difference of the antiderivative if I have a complex function of real variable.. Can I apply the fundamental theorem of calculus in that case?

Hey guys

Please help me with my question

But maybe I should think more.

Ok, thank you!! :-) @PedroTamaroff

Just think of $\sigma$ as a $k$-simplex, @Balarka.

@BalarkaSen Me too..

Oh gah \sigma was a singular simplex

I was confusing maps

@Ant If you take a complex function $f$ and $F$ such that $F'=f$, the for a linea integral $$\int_\gamma f=F(\gamma(1))-F(\gamma(0))$$

In this case you have an infinite path.

$\sigma \times I $ is the map $\Delta^n \times [0, 1] \to X \times [0, 1]$, sure.

You can take the limit of the paths $0\to M$ as $M\to\infty$.

That's what I told you, essentially.

And wait wait wait.

@PedroTamaroff Yes but doesn't that theorem hold for holomorphic function? As a function of real variable it is not differntiable in the complex sense, is it?

@Balarka: For concreteness, take $\sigma$ to be an actual geometric $k$-simplex, e.g., with $k=2$.

@Ant If $F'=f$, then $f$ is certainly holomorphic.

@PedroTamaroff But if the function does not depend on y (the imaginary part) how can the cauchy riemann conditions hold? I would need that the real part of my function derived with respect to x to be equal to 0, but that is not the case, is it?

@Ant Here $s$ is complex.

The function $f(z)=e^{-sz}$ is holomorphic.

@PedroTamaroff yes but I am integrating over t.. I should treat s as a constant thorugh the calculations, right?

@Ant $s$ is constant and you're integrating over the real line.

$$\int_{\Bbb R}f(z)dz=\int_0^\infty f(t)dt$$

@PedroTamaroff I am sorry.. I don't understand who is f(z) on the left integral.. I only defined f(t) for a real t

@Ant $f(z)=e^{sz}$.

What I said has 7 stars? LOL.

@MaryStar I don't understand your graph. The height of a function of both $x$ and $y$ cannot be represented solely in the plane. You need a 3D graph.

@PedroTamaroff ah I see. Basically I have a function of a real variable $t$, $f(z)$ is its analytic continuation over $\mathbb C$ , hence it is holomorphic and I apply that theorem above; then of course integrating over the real line yields the same answer as integrating out "real variable" function f(t) over $(0, \infty)$. One should prove that exists the analytic continuation but of course $f(z)$ works so here it is

@PedroTamaroff that correct?

Yes.

@PedroTamaroff Okay thank you very much! I'm sorry but I was a little confused :) If you want to take the time to write a short answer to my question I would gladly accept it :)

@Kevin: It's the level set $f=0$.

Kevin Theorem: The analytic continuation always exists, even if it doesn't

@TedShifrin That was my guess as well

@Ant You can post an answer yourself.

But then you're looking at the intersection of a plane with your surface and so it looks like there's a cusp at $(0,0)$ even though there isn't

I don't really understand whence the worry about analytic continuation. The fundamental theorem of calculus works fine with maps $\Bbb R \to \Bbb R^n$, with the same proof.

@MikeMiller Ah, I see. I didn't know that.. And I am worried about the "easy generalizations", because there's a good chance I don't really understand what's going on if I apply them blindly

@PedroTamaroff I will then! Thanks again :)

@MikeMiller Aha.

How can I do a 3D graph?? @KevinDriscoll

@Ants You might see how the proof translates yourself.

@MaryStar Are you familiar with wolfram alpha?

@TedShifrin I think I am seeing it. $F_\#$ maps the top of the singular simplex $\sigma \times [0, 1]$ to $f_\#$ and the bottom to $g_\#$

@KevinDriscoll wolframalpha.com/input/?i=f%28x%2C+y%29%3Dy%5E2-x%5E3

@MikeMiller I surely should! If I have some time I will :) I am studying engineering and we don't really go into details (and I can't devote all my time to analysis, sadly :-/ )

@MaryStar I prefer to look at it with a 1:1 aspect ratio

well, probably reversed, @Balarka, but that's no big deal. But what is $\partial(F_\#(\sigma\times I))$?

AspectRatio->Automatic [ironically] @Kevin @MaryStar ... I don't know if you can do that on Wolfram Alpha

It's a question of whether the curve has a tangent line, @MaryStar. Try graphing $f(x,y) = y^2-x^2(x+1) = 0$.

continues to work on clean-up duty

Perhaps I have misunderstood the question

@TedShifrin Surely you are not implying that $F_\#$ is something like a chain homotopy? I mean it looks a lot like it but surely isn't the correct analogue?

The derivatives of the surface certainly do exist at $(0,0)$, although the derivative of the curve $f(x,y) = 0$ certainly does not

It is involved in one, yes, @Balarka. Remember that a chain homotopy goes from $C_k()$ to $C_{k-1}()$? or something like that

Yeah, which is quite confusing.

Well, not entirely. Because isn't it $C_k(X\times I)\to C_{k-1}(X)$?

I'm rusty on this, plus when I've taught such things recently it's been with differential forms (which is cohomology).

@KevinDriscoll So, the derivatives that I found are the derivatives of the surface ?? Which is the difference between the derivatives of the surface and the derivatives of the curve??

I'm going to ponder on your hints @Ted

@Kevin: You'd better make it extremely explicit what you mean by surface.

Still, @Balarka, I wanted you to realize that most of what's going on comes from the formula for $\partial(\sigma\times I)$, which I think you sort of got an hour ago.

Yeah, I kind of saw it when you told me to identify the top face with f_# and bottom one with g_#

My professor told me to think about this chain homotopy idea whenever I get time. He also said that correctly getting the idea and the full power of the analogy might take years, which is frightening.

@MaryStar @TedShifrin What I believe i mean is that at the point $(0,0)$, the surface $z = f(x,y)$ has a well-defined tangent plane, it is the plane $z=0$. However, if we consider the curve $f(x,y) = 0$ then there is no single line which is tangent to the curve at the point $(0,0)$.

@Ted: Apparently Pedro gets to watch me mumble about smooth 4D Poincare in March.

Oh, you invited him and not me, @Mike? :(

Well, I'll catch you on reruns next year :P

He's in the area, @Ted.

Ah... this is a one-time event.

I think usually the topology seminars are more technically challenging, but this and last quarter were quite accessible.

As I told you, I gave my first seminar talk my first year, @Mike. Wonderful little paper of Atiyah-Hirzebruch on non-multiplicativity of signature of manifolds. Nice algebro-geometric stuff plus topology. That was one of the dozens of folders I tossed into the recycling bin trying to get my office cleared out a bit.

:(

Yes, lots of memories being thrown away ...

That's a beautiful little paper. I assume it's easily available electronically these days.

I had Griffiths, Kobayashi, and Wu-chung Hsiang (who happened to be visiting from Princeton) and all sorts of people trying to help me puzzle part of it out for the second talk. We never quite figured it all out. But the seminar was after wine and cheese hour, so most of the 60 people didn't fuss at me :P

I am too tired to think about it. Maybe I should sleep and restart my thinking tomorrow morning.

Yes, @Balarka. Math doesn't work well when you're sleep-deprived.

Thanks for all the great hints @Ted.

You're welcome, @Balarka.

Yeah @Ted, but chatting sure does.

:P

This is the definition of a singular point, after that there is the example that we are talking about..

You had regular wine and cheese hours?!

No, this was a once a quarter type thing, @Mike. Just happened to be Friday at 3 PM, the hour before the geometry seminar (Fridays at 4 PM).

Unrelated : Hopf fibrations are interesting stuff. I figured how to visualize them a few weeks earlier.

We don't have wine and cheese hours once a quarter, either :P

We have cookies and coffee before seminars. No wine or cheese, sadly.

MAybe that paper was just Atiyah. Chern-Hirzebruch-Serre had proved the theorem that signature is multiplicative when there's no monodromy.

Hey, that was the 70s, @Mike @Kevin. We were all hedonists then :P

Essentially, the visualization consists of showing that S^3 is union of two solid torii joined across the boundary.

Maybe I should leave UGA's math department money for such a thing. Oh, no booze allowed on campus in GA.

Good night, @balarka!

@TedShifrin Whaaaaaaaaaaat? UGA is a DRY CAMPUS!?

Does this not sound weird that S^3 can be decomposed into two genus 1 surfaces?

Yes, @Kevin. And bars or liquor stores have to be several hundred feet from the boundary.

@TedShifrin You can't dismiss me like that!

Wanna bet @Balarka?

Think about Van Kampen, then, @Balarka :P

I'm still a hedonist, @Ted. That's why I want wine and cheese hours.

What, you dismissed Van Kampen too?

No, @Balarka: What you said is wrong. It is decomposed into two 3-manifolds with boundary; the boundaries are genus-1 surfaces.

@Ted There's no way that rule could possibly be followed or enforced!WORLD'S LARGEST OUTDOOR COCKTAIL PARTY and all

@TedShifrin yes, i use surfaces and n-manifolds interchangeably.

I don't know how football drunkards get dispensation, @Kevin. Mine is not to wonder.

@Balarka: "genus 1 surfaces" refers specifically to surfaces, not 3-manifolds with boundary. Genus isn't defined ...

@Ted Very interesting. There must be some kind of approval procedure. The school of physics has alcohol at its homecoming every year

Anyhow, @Mike: Here's the reference. Very beautiful little paper. M. F. Atiyah: “The sig­na­ture of fibre-bundles,” pp. 73–​84 in Glob­al ana­lys­is: Pa­pers in hon­or of K. Kodaira. Edi­ted by S. Iy­anaga and D. C. Spen­cer. Prin­ceton Math­em­at­ic­al Series 29. Uni­versity of Tokyo Press, 1969. MR 0254864 Zbl 0193.​52302

oh right. two 3-manifolds pasted across a genus 1 surface then

glued along, yes, @Balarka.

actually i asked this question to somebody in the uni and he said that it's a property of 3-manifolds that it can be decomposed into two 3-manifolds with genus g boundaries, glued along each other by a diffeomorphism for any g. weird stuff.

This is handlebody decompositions. All sorts of stuff special for $3$-manifolds.

i mean it's easy to see this for S^3. any genus g solid surface in R^3 has complement that looks like a R^3 minus huge ball plus g pillars, one-point cmpcfication of which is just a 3-manifold with g holes on the boundary

so the diffeomorphism is the identity map here

@TedShifrin ooh cool name

i am tempted to google that

@Mike: Not that you're remotely interested, although you might be: Here is the paper.

Yes, @Ted, I'm interested. What kind of 4-manifolds fellow doesn't like the signature?

Mike is obsessed about 4-manifolds.

There's some interesting topology and Riemann surface stuff. I never did get to where I could see the monodromy explicitly (it has to be there by non-multiplicatively). That's where all the pros got stuck, too.

I should look at the Kodaira paper he references before I give away my collected works of Kodaira :P

Hm... I won't stake a claim on that one.

@BalarkaSen Give two nonisomorphic groups with isomorphic automorphism groups.

LOL, if you're staking claims, you'd better send someone to pack and ship to you :P

You know I'm broke, @Ted. :(

Oh, cool exercise for you, @Pedro. I don't think I gave you this before. What is $\text{Aut}(\mathscr Q)$?

@Pedro Assigning trivialities now, eh?

By $\mathscr Q$ of course I mean the order-8 quaternion group.

Other good exercise: Find the smallest non-isomorphic groups with the same number of elements of each order.

Right. It is $S_4$.

Yup, @Pedro. I think that's very cool.

@TedShifrin Why?

Of course, I interpret $S_4$ as the symmetries of the cube.

Because there are (although I've forgotten) geometric ways to see it.

@PedroTamaroff Well, Z_2 and 0 :P

That was way too easy.

I think @Balarka won that one.

But it took some time, haha

@BalarkaSen OK. I want one to be infinite and the other finite.

@Balarka: This one might be good for you too.

Wrong, @Balarka.

Once you have products you get lots of automorphisms.

figured that @Ted

hmm

actually semidirecting with Z_2 leaves the automorphism group same

oh

Stop inventing words Balarka.

Z and Z_3

Yes.

there, i haven't forgotten everything