Mathematics - 2014-06-04 (page 1 of 4)

AndrewG

00:29

Quick check: let $V_n$ be a complex representation of a finite abelian group $G$. Then every element $g \in G$ maps to an operator on $V$ -- and all of these operators are diagonalizable, and share the same set of eigenvectors? Is that last part right? I understand how, by Schur's lemma, V breaks down into 1-dimensional irreducibles. That's the same thing, right?

Ted Shifrin

00:46

Hi @Fernando @Andrew

Fernando Martin

Hi @Ted

AndrewG

Heyo Ted

Ted Shifrin

Quiet night on the ranch ....

Pedro Tamaroff

@FernandoMartin Hey.

Fernando Martin

sup @Pedro

Pedro Tamaroff

00:54

@FernandoMartin Remember that Ass M and Supp M exercise with minimal primes?

Fernando Martin

yeah

Pedro Tamaroff

All we need to assume is minimal primes in Supp M exist right?

I mean, for it to make sense.

Fernando Martin

01:06

which exercise?

Pedro Tamaroff

@FernandoMartin That the minimal primes coincide. I think we need A noetherian and M finitely generated.

At least I could prove it under those hypothesis.

Fernando Martin

yes, those are the ones Matsumura uses

Pedro Tamaroff

Cool bananas.

Fernando Martin

I haven't done it yet

Alicia added noetherian

but forgot f.g.

Pedro Tamaroff

Lel.

http://tabs.ultimate-guitar.com/f/foo_fighters/another_round_crd.htm
D = C9 wtf

Oh, lel.

The text is moved a little.

Kevin Driscoll

01:34

@Pedro Let's see if we can do some 'Black Magic' TM to make that equivalence more plausible. So C9 generally has C E G D which we can rearrange with D in the root as D G C E, which I will choose to read as D9sus4

BAM

I have a very 'soft' question for chat. I've heard of the ideas of homotopy and it seems like an active area of research. Can anyone describe how old this field is, how active a field it is, and compare it to other more familiar fields in terms of number of important results?

Mike Miller

@fernando some guy wrote vv for the dot product question. half credit

@Kevin what do you mean by "the field of homotopy"? can you be explicit about what you mean?

Pedro Tamaroff

@MikeMiller You correcting things?

Mike Miller

unfortunately yes

Pedro Tamaroff

What topic?

Kevin Driscoll

@MikeMiller I'm not sure I know enough to specify precisely, but the kind of application that would be relevant to me is that idea that we can solve certain easier problem and then somehow continuously deform the solutions to that problem into the solution of some more difficult problem of hte same type

Mike Miller

01:42

linear algebra

Kevin Driscoll

I guess its not really a field by itself but bundled up in algebraic topology or something

Mike Miller

@Kevin the idea of a homotopy between maps is a lot simpler than what could be called homotopy theory in the modem sense, which to my understanding is generally interested in stable homotopy groups of topological spaces, and more abstractly weird category theory stuff

I would guess that what's known as homotopy theory wouldn't be helpful for what you're interested in

Kevin Driscoll

@MikeMiller I see, so would you say that the mathematical work that would underlie the kind of application Im talking about has already been done?

Mike Miller

I dunno - probably, and what you say sounds like a simple enough idea that you can probably do it ad hoc

Kevin Driscoll

@Mike Yeah that was my impression. And I think people do do it ad hoc, but it kinda makes me nervous.

I was kind of hoping that there was still mathematics to be done that might turn out to be really useful for computing solutions of difficult problems

but alas.....

Mike Miller

01:50

@Kevin I'm doing something similar (and similarly ad hoc) for a problem of mine; if mainly amounts to keeping careful track of the inequalities involved

Kevin Driscoll

@Mike I think my issue is that I wasnt a math major and so I lack a lot of intuition about when things converge and when they converge in the correct space etc etc. My advisor proposed some approximation method for ordinary functions in terms of generalized functions that he'd constructed and it sent me into a tizzy several weeks ago

I didnt know which way way 'up'

Sawarnik

@r9m math.stackexchange.com/questions/487250/how-find-this-nice-limt/…

Mike Miller

02:13

@Kevin You should try reading an intro topology text's chapter on homotopy. Chapter 0 of Hatcher might help you.

Kevin Driscoll

@Mike How much background in topology is required? I know the basics but that's about it.

Mike Miller

@Kevin Do you know what a continuous map is? Then you're golden.

If you don't like the abstraction it should be pretty easy to restrict to metric spaces in your head.

Kevin Driscoll

@Mike a continuous map is one where the pre-image of an open set is an open set?

hope I remember that right

Mike Miller

Yes. But if metric spaces are easier for you you lose nothing by thinking in terms of metric spaces, @Kevin

Kevin Driscoll

That's good. I don't know anything about non-metric spaces. Everything I work with has a norm.

Mike Miller

02:21

A homotopy between $f: X \rightarrow Y$ and $g: X \rightarrow Y$ is a continuous map $X \times I \rightarrow Y$ such that restricting to $X \times 0$ gives you $f$ and restricting to $X \times 1$ gives you $g$

And that's perfectly doable in a metric space!

Kevin Driscoll

Yeah it seems so simple but..... like frighteningly so

because there are so many such maps

and its just like 'oh we turn this knob and now we have the solution of a different problem'

just feels os weird to me

Mike Miller

03:11

@Kevin But it's not that simple. Not any two maps are homotopic. Do you know about the fundamental group?

Kevin Driscoll

@Mike Not really. Something about [0,1] with a boundary and.....

Mike Miller

@Kevin Well, as a set, it's homotopy classes of maps from $S^1$ to your pace $X$ (keeping a basepoint fixed in your homotopy). But the point is that this is a useful tool; spaces have lots of non-homotopic maps from $S^1$.

So you can't just go from anything to anything else.

Kevin Driscoll

@Mike Yes this makes some kind of sense. If you could do from anything ot anything else it suggests a kind of triviality to what you're looking at. I need to figure out how this applies more concretely

skullpatrol

03:26

Hi guys :-)

AndrewG

@Kevin think about loops on the surface of a donut. Can you come up with two that aren't in the same homotopy class? :) (Hint: use the big hole, i.e. the topology of the donut.)

Mike Miller

@Kevin So for the application I did I needed all the maps in my homotopy to be "nondegenerate" in a sense and the hard part is figuring out when such a nondegenerate homotopy is actually supported. (Any two of the maps I was considering were homotopic; but they weren't necessarily homotopic in a useful way...)

mixedmath

hi all

Mike Miller

But when I could find such a nondegenerate homotopy I could reduce my problem to a trivial case.

hi @mixedmath

mixedmath

how goes it?

Mike Miller

03:30

it's alright

I intend to write something for our blog eventually, btw, but I haven't found the time yet

AndrewG

You guys are doing a math blog?

mixedmath

@AndrewG Yep! It just started, and the first post was just put up

AndrewG

Spiffy. I wish I knew enough to do stuff like that.

Mike Miller

@AndrewG I guarantee there's a topic you're both passionate and knowledgeable enough about to write on :)

(When I say "our blog", I mean the new Math.SE community blog - open to everyone!)

AndrewG

oh wow

that's an awesome idea

look forward to reading it

Awal Garg

03:33

Reset the Net Post this everywhere if you care for your PRIVACY!!!

mixedmath

oh yeah - community blog - anyone and everyone can write!

I thought we weren't resetting the Net until Thursday?

Awal Garg

@mixedmath The more people get to join in, the more will be the effect. Everyone doesn't knows of it. So spread the word :)

Mike Miller

@mixedmath Do you have a link handy? For a xommunity blog it's hard to find...

mixedmath

math.blogoverflow.com

@MikeMiller We aren't going to put it on the masthead until it's a bit older and slightly more developed, so right now the only way to know of it is to read meta

Kevin Driscoll

@Mike this sounds very similar to what happens in the area of nonlinear dynamics

where each trajectory actually corresponds to a set of trajectories due to some symmerty of the system

and so you need a way of modding out the symmetry to pick exactly 1 representative from each family to find you rpoincare sections

Mike Miller

03:39

@mixedmath do you intend to write anything for it?

mixedmath

@MikeMiller I probably will sometime. I'd like it to succeed

Mike Miller

@kevin don't frighten me with nonlinear stuff

quadratic is already bad enough

Kevin Driscoll

@Mike It scares me too. I do few-body QM. Everything is linear!

Mike Miller

@Kevin you should see if you can get a job working on the dynamics of sub linear polynomials

@mixedmath cool, glad to hear it. any idea what on?

mixedmath

something I've been thinking a lot about recently is pretty data visualization with matplotlib, actually

Kevin Driscoll

03:45

@Mike sub....linear...polynomial......WAT!?

Mike Miller

pretty things have wide appeal!

@Kevin I hear they behave particularly nicely :)

mixedmath

so one thing might be to just look at data about MSE and present it nicely and explicitly

Kevin Driscoll

those.....those just sounds like constants @Mike

Mike Miller

;)

@mixedmath by the way, you wrote me a long email a while back about life at Brown - I ended up accepting an offer somewhere else, but I want to say again I really appreciate it!

mixedmath

@MikeMiller oh yeah, that's right. So where are you going?

Kevin Driscoll

03:53

Did @mixedmath go to Brown?

mixedmath

@KevinDriscoll I'm there now. I'm a 4th year grad student

Kevin Driscoll

At Duke, Brown was always the ivy that we picked on. It frequently gets called 'clown college.'

Ted Shifrin

Howdy @Kevin @Mike @mixed ....

Kevin Driscoll

Ah okay. Grad program.... totally unrelated.

nabla blah

Hi @TedShifrin

Kevin Driscoll

03:53

@Ted howdy

mixedmath

Hi @Ted

Ted Shifrin

Hi @nabla

Mike Miller

@mixedmath I'll be heading to UCLA in the fall

hi @TedShifrin

mixedmath

@MikeMiller Way cool. There's some good number theory there (and I'm all about the number theory)

Mike Miller

@mixedmath Lots of good stuff all around. Glad to be heading there!

@Ted Too many margaritas?

Ted Shifrin

03:57

No, iPad typo. No booze whatsoever :(

nabla blah

lol

Mike Miller

Too few margaritas, then

Kevin Driscoll

Ive never understood the widespread love of margaritas

I think tequila is nasty

Also, I'm hardcore procrastinating on writing my thesis proposal

Ted Shifrin

I'm a gin fanatic, but many dislike that, too. :) At the moment I'm trying to lose weight, so have cut back on booze.

@Kevin: Imagine how much you'll procrastinate on the real thing.

AndrewG

Margaritas + seafood = win. Tis a theorem.

Kevin Driscoll

04:06

@Ted I've decided that the major obstacle to every PhD is the students fight with his own self-motivation

external factors are oftne neglegible

Ted Shifrin

Not if it's French or Italian food, as opposed to Mexican @Andrew

Kevin Driscoll

Also< i quite like gin

Mike Miller

@Ted Killing time with gin and lime?

Ted Shifrin

I was very motivated in grad school, but hit plenty of brick walls regardless.

Successive approximations @Mike?

Kevin Driscoll

Is there a well-known tiering list for schools int erms of math faculty positions? I hear people talking about tier 2 and tier 3 schools and I wonder where the 'official list' is

AndrewG

04:08

@TedShifrin Yes, I suppose in that case a good wine would be the necessary condition.

Ted Shifrin

There are AMS rankings, mostly based on department heads' opinions, which are often biased and way dated.

Kevin Driscoll

@Ted well rats you've disproved my hypothesis

Okay, so its very similar to how physics works. I really dislike this system. What happens is the department head says 'who do I know from X?' and their knowledge is of course skewed toward hteir field

Mike Miller

Well, USNews gets the top 5 right. Not necessarily in order.

Kevin Driscoll

so they artifically rank schools lower/higher based on a single field

at leas thtats how it usually works for us

Ted Shifrin

How do they do it? @Mike

I'm not sure any of us have an order. The order is random and/or field-dependent if there is one.

Mike Miller

04:11

@Ted I don't remember how they order them but their top 5 is MIT, Stanford, Berkeley, Princeton, Harvard

Ted Shifrin

Ah, Stanford has replaced Chicago?

Kevin Driscoll

If they do it like their undergrad rankings its a compound thing. Part is survey of department heads. part is post-grad survey. Part is work outcomes of students. Part is number of publications and what journals theyre in.....

Ted Shifrin

Otherwise, the list hasn't changed in 30 years.

nabla blah

I use ARWU

Kevin Driscoll

Man I don't care about the top 5 at this point. I'll take a tenure track position at a small/medium sized state school and be quite happy. Thank you very much.

Mike Miller

04:13

@Ted They might have it all as a top 6 now with Chicago tied in there, but maybe not.

Ted Shifrin

I am surprised Stanford is that high, but it's not silly. Chicago is no longer what it was in the 60s.

Mike Miller

It doesn't help that any such list is necessarily going to resist change

Kevin Driscoll

I think the top 6 for physics is identical

although you also nee dto include Cal tech

Mike Miller

I think those are generally just six of the best funded, most renowned universities I'm the nation, so you'll probably see the "top 5" for most fields in general just be 5 of some set of 10 schools.

Studentmath

Morning..

Ted Shifrin

04:17

Ah, done slumbering, @Studentmath?

Studentmath

Yeah, it flows well once again

No idea what I had, was so odd, everything went wrong :P

Ted Shifrin

LOL ... Not a catastrophe, nor a dogmatic failure. :)

Studentmath

Fortunately - or rather hopefully :)

AndrewG

hey @Ted, random topology question. Suppose I wanted to construct a surface, embeddable in $\mathbb{R}^3$, with fundamental group $\mathbb{Z}_2$. Only thing I can think of is the real projective plane but that's not embeddable.

Ted Shifrin

Every compact surface without boundary in $\Bbb R^3$ must be orientable.

Can't happen, by classification of surfaces. @Mike can crank universal coeff theorems:

AndrewG

04:21

So it's not possible

K

Studentmath

Also seems what I thought to be conditional yesterday is actually multipication

Bananarama

Here is the inequality I told you about

Ted Shifrin

Multiplication? @Studentmath ... You mean "and"?

Studentmath

Yes

Ted Shifrin

Get back to me when I'm teaching probability in a few months :)

Bananarama

04:25

let $n$ be a positive integer and let $a_1,a_2\dots a_n\in \mathbb R$ with $a_1+a_2\dots a_k\leq k$ for $k=1,2\dots n$ prove $\frac{a_1}{1}+\frac{a_2}{2}\dots\frac{a_n}{n}\leq \frac{1}{1}+\frac{1}{2}\dots \frac{1}{n}$

Studentmath

The probability of $Y_{(1)}<3Y_{(2)}$ requires two simultaneous conditions

When will that be?

Ted Shifrin

Ok, night, all ...

Mike Miller

@TedShifrin It's obviously not the case for closed surfaces by classification. The question for non-closed surfaces isn't obvious to me, though don't spoil it.

AndrewG

Later @Ted

skullpatrol

Later Professor

Studentmath

04:26

Good night Prof. @Ted

Parth Kohli

@skullpatrol Hi.

AndrewG

I love this place. Ask a question, and you get plenty to google and read about.

Mike Miller

I think I might have an answer @AndrewG but it's probably needlessly high-powered... let me fiddle with it a little more

Kevin Driscoll

This chat was made as a battfield to test the effects of nuclear mathematics on mosquito sized problems

4

Mike Miller

nope, argument doesn't work.

AndrewG

04:40

Darn :| what were you trying though? curious but probably won't understand

I've never taken topology beyond point-set, so I'm reading about classification of surfaces right now.

Mike Miller

Classification of surfaces is awesome! It's one of the things that convinced me combinatorics wasn't so bad after all, since it's essentially a combinatorial argument.

@AndrewG I wanted to start with a surface embedded into $S^3$ and use Alexander duality to show that its complement would have first homology = $\mathbb Z_2$ (this is the first place it fails) and then say that the complement would be an open subset of $S^3$ (also not true) and then that the first homology of an orientable manifold has no torsion in $H_1$ (also not true).

So I wanted to use one bogus argument and three false facts.

AndrewG

@MikeMiller That's the best way to prove things, though.

Yep, way over my head, but more to look up, at least.

Kevin Driscoll

Haha if you removed your admissions that these things are untrue you couldve successfully pulled off a proof by intimidation

Mike Miller

Anyway, one can have closed orientable manifolds that have fundamental group $\mathbb Z_2$. They just can't embed into $\mathbb R^3$.

I just don't know how to show they can't embed into $\mathbb R^3$ yet :p

AndrewG

Heh

Studentmath

04:55

Is there a way to put a cross on something written on the forums?

I.e. a line through

David Kirby

There's no markdown for it, but the HTML tags are allowed, like <strike>this</strike> or <s>this</s> or <del>this</del>

371

Q: What HTML tags are allowed on Stack Exchange sites?

The Stack Overflow site engine, as you know, uses Markdown for questions and answers. Per the Markdown spec, you are allowed to freely intermix HTML and Markdown tags. Not all HTML tags are allowed, as that would be an XSS paradise. Which HTML tags have been whitelisted and are allowed to be use...

Mike Miller

@AndrewG to prove Ted's assertion: by classification of surfaces, $\mathbb{RP}^2$ is the only compact surface without boundary with fundamental group $\mathbb Z_2$. (see this by representing every surface as a quotient of a polygon.) so one only needs to show $\mathbb{RP}^2$ can't embed in $\mathbb R^3$, which can be done with alexander duality

AndrewG

representing a surface by a polygon = generalization of building a torus by modding the plane by a lattice, etc?

identifying sides and so-on?

Mike Miller

yah

Studentmath

Thanks @David, eventually just deleted it..

Mike Miller

05:06

@AndrewG page 5

AndrewG

ah k

Studentmath

I still don't get that question though..

sigh statistics

David Kirby

@Studentmath, same question from last night regarding distributions?

AndrewG

Does he have a section on Alexander duality? I don't see it in the table of contents.

Studentmath

Not sure which one you speak of - the one regarding ordered statistics. I managed the other one with Gamma Distribution, went out nicely

David Kirby

05:10

Ah, awesome, yeah I was wondering about the Gamma distribution.. Glad to hear that you got it figured out

Studentmath

This oddie one now:
http://math.stackexchange.com/questions/819269/probability-that-order-statistic-is-larger-than-the-other

Yeah, I was over-complicating it. Should've just went straight on with the gut feeling, reasonable explanations and statements and you can use the gamma distribution to finish it off nicely :)

AndrewG

Also, I'm wondering now if there might be some crazy Seifert surface with the right properties.

Studentmath

At least I went from not understanding it to understanding it yet failing to solve

David Kirby

@Student, yeah, it seemed to me that you had figured it out but hadn't yet settled on an interpretation.

Mike Miller

@AndrewG It's at the veeeery end of chapter 3. It says, in particular, that $\tilde H_0(S^3 \setminus \mathbb{RP}^2, \mathbb Z) \cong H^2(\mathbb{RP}^2, \mathbb Z)$; but the first thing is always free abelian, and the second thing is $\mathbb Z_2$, which is a contradiction.

AndrewG

05:20

Nice.

Mike Miller

I feel like it should be easier. (There's almost certainly an easier way to do it if you want your embedding to be smooth.)

AndrewG

Is this a typical use of (co)homology groups? I've never really understood why one should bother with them, but showing that one space is not embeddable in another seems like a pretty practical thing.

Studentmath

@David Baby-steps. It's considered a tough question, hence my focus on it

AndrewG

I think I better back up and read this whole chapter :|

Mike Miller

@AndrewG (co)homology is incredibly useful! Some things you might like, that I know how to do most easily with them: $\mathbb R^m \cong \mathbb R^n$ iff $m=n$; every map from the closed unit ball to itself has a fixed point; no compact manifold retracts onto its boundary; the Euler characteristic of any manifold of odd dimension is 0.

David Kirby

05:33

@Student, you are wise to look deeper. it's an important question -- one that gets asked in different ways by countless disciplines.

Mario Carneiro

Is it possible to derive cauchy schwarz on $\Bbb R^n$ as a consequence of the triangle inequality?

Mike Miller

Okay that was silly, of course $A'$ has a fixed point. Ignore me.

AndrewG

Awww lol

Studentmath

I should do my research better :/

math.stackexchange.com/questions/735272/…

Mike Miller

My topology professor gave me that as an exercise, and I just realized it's trivial! He's got hell to pay, @AndrewG

AndrewG

05:38

I was still reading it when you yoinked it, so I dunno what it was exactly or how it was trivial :P but the Lefschetz fixed-point theorem does look awesome.

Mike Miller

@AndrewG Because $A(0) = 0$...

skullpatrol

Hi pal @Parth Kohli

AndrewG

Oh, haha, yes.

This Lefschetz thing is crazy.

Mike Miller

@AndrewG Chapter 2 of Hatcher doesn't require anything from chapter 1, so if you're interested in this, that's a good place to start. (Read chapter 0 first, though.)

AndrewG

There's so much beautiful math to learn :O

Mike Miller

05:45

yase

Parth Kohli

@AndrewG If it's beautiful, it's not math.

Mike Miller

how wrong you are

Parth Kohli

(Joke.)

Mike Miller

@AndrewG I did some googling, and it turns out that the fundamental group of any submanifold of $S^3$ is torsion-free... but it seems like it uses a lot of really hard stuff I don't know about yet.

Nevermind, it says any submanifold with connected complement.

AndrewG

@MikeMiller OK, so, I take a surface, cut it up and lay it out as a polygon, triangulate it, and start calculating the simplicial homology. I've seen this before (a bit) but should practice doing it. One thing, though: there are dozens of homology and cohomology theories, right? But I've often heard of "the" homology/cohomology of a space, as if they're all equivalent? Do all roads lead to Rome or something?

Also, that's cool, at least you're narrowing down the possibilities lol

Mike Miller

06:00

@AndrewG Yes, there are lots, but don't get lost in that right now. They don't necessarily even have to do with topological spaces. The ones you'll learn from Hatcher are called simplicial homology, singular (co)homology, and cellular homology... and those three are the same whenever they're both defined.

When one says the homology of a space they mean the singular homology unless otherwise specified.

AndrewG

K

Mike Miller

To clarify: simplicial/singular/cellular homology are all the same (on spaces where they're defined). Singular homology and singular cohomology are very different.

AndrewG

Right. But they have dualities relating them.

I've heard of Poincare duality but don't really know what it is except it relates those.

Mike Miller

For manifolds there are dualities relating them.

AndrewG

Ah

Mike Miller

06:03

As groups, cohomology is entirely determined from homology by the Universal Coefficient Theorem. Despite that, cohomology has a lot of its own stuff to say that homology doesn't.

AndrewG

Cool

Mike Miller

In particular, cohomology has a natural ring structure in addition to the group structure - and that ring structure can tell spaces apart that the group structure can't

AndrewG

So non-isomorphic spaces can have the same homology groups, but the cohomology groups can distinguish them through that extra info?

Mike Miller

Sure. (Not necessarily, though. For instance, neither homology nor cohomology can tell the Mobius strip apart from the cylinder - both are homotopy equivalent to the circle, and both homology and cohomology - including the ring structure - are homotopy invariants.)

bolbteppa

As I understand it, the fundamental meaning of cohomology is to be found in differential forms, which you can think of as the thing you integrate when doing a line integral (for my example, e.g. Pdx + Qdy say). Think of R^n with a hole at the origin, how do we detect it? We only have functions that are defined inside the space, if we go off to zero we don't know we're going to a hole at the origin, so we need some way to test for it.

All we know is that line integrals depend only on their endpoints for nice functions, so we try to find a differential form that, when we integrate it (in, say, a circle), we do not end up with zero. This means we went around a hole in the space. Furthermore, we know that integrals of crazy paths around the hole will give the same values as circles, and also that a load of different functions will give non-zero values as we go around the hole, so we need a way to simplify working with all these things... Based on this simple geometric picture, illustrated here:

we see that all that crazy algebra, group theory, ring theory etc... falls upon us naturally when we try to group all the differential forms with these properties into sets. You have to categorize the set of forms into equivalence classes using the only tools you have, the exterior derivative operator, and because this operator has some freedom in it you end up with group structure and all this stuff
http://en.wikipedia.org/wiki/De_Rham_cohomology
Hopefully that will give some geometric intuition and motivation behind the crazy algebra definitions

Maybe there's a better way to think about it, but the only other way I've seen is unmotivated algebra which I'm too stupid to understand :p

@MikeMiller how does one tell them apart? The only way I've seen is using fiber bundles, is there a way using homology or homotopy?

Mike Miller

06:24

@bolbteppa One is orientable, the other is not (and orientable can be defined completely homologically). but that's probably what you mean by "using fiber bundles".

AndrewG

Yeah, I don't understand differential forms well enough yet for deRham cohomology

Maybe soon

That's cool though

bolbteppa

@AndrewG if you read the link I gave you, and read the paper I linked to in that article, you'll understand differential forms very quickly, and with lovely pictures :p The idea is that a differential form sets up a field of sheets, and integrating a differential form along a curve is just a way of counting how many sheets the curve passes through.

Mike Miller

@bolbteppa You're new around these parts (or at least new to me). What's your story?

AndrewG

@MikeMiller am I understanding right that the original question is still open, so long as the manifold has a boundary and its complement is disconnected? lol

bolbteppa

That explains why line integrals are path-independent, depend on end-points, closed-loops give zero etc... (all those theorems in multivariable calculus), you just go forward through the sheets then back against them so you ultimately go through no sheets... Sometimes though, the field of sheets all spring out from one point and give a cone/fan shape which represents a singularity (see the answer to my question in that post), so when you go around in a circle you go around 2pi's worth of sheets

Mike Miller

06:30

@AndrewG i'm sure one can answer it from existing theorems, but I don't know the answer.

AndrewG

Well I'm certainly finding a lot to learn along the way, which is really better anyways.

bolbteppa

My story right now is that it's 7am, I've been awake for a day, and I have to now go and read a paper on the Ising Model and Algebraic Bethe Ansatz's work for quantum integrable models, all the while barely knowing elliptic integrals or how to geometrically derive ellipsoidal coordinates :p What's yours?

David Kirby

it's 2:34 am, woke up at 11 pm and brewed a pot of coffee -- supposed to be writing software but procrastinating and playing with math instead.

bolbteppa

Yeah I am supposed to learn mathematica on top of all this, keep getting distracted by math and coffee too

Do you think mathematica is the kind of thing you can pick up in about 2 hours, or is it insanely complicated?

David Kirby

eh, well ... you can do basic symbolic manipulation right out of the gate, but anything remotely non-trivial is going to hang your computer processor ;)

AndrewG

06:37

@MikeMiller what's it mean to say that a manifold is torsion-free? that its fundamental group has no elements of finite order?

bolbteppa

Great stuff

Torsion tensor

In differential geometry, the notion of torsion is a manner of characterizing a twist or screw of a moving frame around a curve. The torsion of a curve, as it appears in the Frenet–Serret formulas, for instance, quantifies the twist of a curve about its tangent vector as the curve evolves (or rather the rotation of the Frenet–Serret frame about the tangent vector). In the geometry of surfaces, the geodesic torsion describes how a surface twists about a curve on the surface. The companion notion of curvature measures how moving frames "roll" along a curve "without twisting". More gener...

AndrewG

Oh, sorry, I misread you earlier: you said the fundamental group was torsion-free. Nevermind.

I need more coffee.

David Kirby

good idea... (off to grab a cup and a smoke)

Balarka Sen

07:20

Hullo @bolbteppa

AndrewG

07:38

So, I bought Kolmogorov and Fomin used through Amazon. And apparently it used to belong to some lady professor. And she put a Hello Kitty sticker inside the front. I don't know why, but this makes me happy.

Balarka Sen

Bought?

AndrewG

It's not the same if you can't hold it in your hand.

Balarka Sen

@AndrewG Nonsense. I heard many people say that. I find no difference between holding and reading from the computer.

AndrewG

To each his own :P if one is just as good as the other for you, so be it. I can't get enough of paper.

Besides, you can't put Hello Kitty stickers inside the covers of PDF files. That alone is reason to buy books!

Balarka Sen

My fundamental reason to get PDF files is to prevent someone putting stickers in the math books.

AndrewG

07:46

Party pooper.

Balarka Sen

Someday, you'll find books that is not available in hard copy anywhere. Someday.

AndrewG

When that day comes, I will find a PDF of it, print it out, bind it myself, and put a Hello Kitty sticker in it, just to annoy you. :P

Balarka Sen

=D

AndrewG

@MikeMiller am I correct in the following understanding? Trying to get the point of relative homology groups and all this diagram chasing/homological algebra in Ch. 2. So, take for example the Klein four-group v.s. the cyclic group $\mathbb{Z}_4$. They both contain $\mathbb{Z}_2$ and quotient to $\mathbb{Z}_2$. I can represent both of these situations with a short exact sequence. If, however, I just decided to extend $Z_2$ by $Z_2$, there are two ways to do it. I have to decide which one.

@MikeMiller am I understanding correctly that all this diagram chasing stuff is just a fancy way of picking the 'right' extension?

Balarka Sen

sigh I never studied homology, even though I want to read it.

AndrewG

08:02

@BalarkaSen It looks pretty cool. Mike has me reading Hatcher :)

Balarka Sen

@AndrewG Yeah, someone recommended me that for homotopy instead.

I like homotopy, even though I am just scratching the surface with covering space stuffs.

MrWho

Hi folks

My Question On SE : math.stackexchange.com/questions/820200/…

Any clue would be appreciated, specially something with extreme validity!

@BalarkaSen @robjohn @nablablah @skullpatrol (With respect)

Balarka Sen

You're applying $x = \sin(\theta)$. But that restricts the whole domain $D$ (indefinite) to the some bounded domain $[0, 1]$, @MrWho.

MrWho

@BalarkaSen Well, how does it affect the final answer to the integral?

@BalarkaSen If you give me mathematical step by step reason for my question I absolutely understand what you mean then.

Balarka Sen

@MrWho My guess is that it takes up the "remains" and adds it to the constant term, i.e., pick up all the $x$ which is not in $[0, 1]$ and integrate up. I am not sure though, this is just a wild guess.

MrWho

08:20

@BalarkaSen I think my question is asking about something happening in the back-end of the method, very nature of this technique, just looking for robust answer.

Balarka Sen

@MrWho Yes, that's why I am saying it has something to do with the "crumpling" of the domain through that u-sub.

MrWho

@BalarkaSen Seriously, I myself don't really know anything about the effects of domain movement via substitution, anyway, I stay tuned looking for an answer.

Balarka Sen

Both of the answers seem to take the question to he theoretical level, instead of looking at the method, sadly.

Studentmath

I like his name math.stackexchange.com/questions/820210/…

MrWho

@BalarkaSen Yeah :(

Studentmath

08:23

Shame he shown no effort

Balarka Sen

@MrWho Maybe you should make it clear what you're asking.

MrWho

@BalarkaSen The answer is yes, it seems I didn't drive my question to the right destination!

@BalarkaSen Surprisingly, you are the only one who understood it!they think I don't know the constant would be the constant and there is infinite answer for the integral.

Balarka Sen

@MrWho I have - forgive me - an ability to think out of the box, yes, I wouldn't deny that.

At least we can call @robjohn. @robjohn Do you see what's intended to be asked?

Studentmath

Oh my god.. christianity.stackexchange.com/questions/952/…

MrWho

@BalarkaSen Help me! I don't really know how to paraphrase the question, seriously, I think I crafted the question in the artistic way!

Balarka Sen

08:31

@MrWho As for that, I am not sure either how to get the question in the right track, sorry.

MrWho

@BalarkaSen You see, I don't ask lots of questions, and even if I ask one (once in a season) the question would be convicted to be completely vague or intractable to be controlled!

@robjohn @BalarkaSen Well, why would $\frac {-1}{4}$ rise up when we use the trigonometric substitution rather than the general formula?For the sake of god,DO NOT remind me that constants are the same, and I certainly know that $\frac {-1}{4}$ is the part of the constants which can be applied to the question.But why $\frac {-1}{4}$ would be detached from the general constant and join the integration answer?

Balarka Sen

@MrWho You might want to edit our question. Someone changed that $\sin(2\theta)$s into $\sin^2(\theta)$, making it confusing.

MrWho

@robjohn @BalarkaSen Why $C$ which I call it the general constant is converted to $C'+\frac {-1}{4}$ when trig sub is applied? (The simplest way to represent what I mean) - Sorry for calling you guys several times.

r9m

@Sawarnik AH ... thanks a lot !! ;) :D

MrWho

08:46

@BalarkaSen Ah, I see, clumsy editor, damn it!

robjohn

@MrWho two anti-derivatives are equivalent if they differ by a constant. This is the purpose of the constant of integration, to account for this fact.

Balarka Sen

@robjohn I believe he much wants to know in which step $-1/4$ comes into play. Can you kindly point that out?

MrWho

@robjohn I know what you've just said and my question is not asking anything about this fact.

robjohn

If the constant of integration were going to be the same, we wouldn't need to leave it as $C$, we could simply leave it as $13$, or whatever.

MrWho

@BalarkaSen Your last statement is what I mean :|

Balarka Sen

08:50

@MrWho let's keep our temper a little down, ok? and pinpoint what we want to know to @robjohn.

robjohn

@BalarkaSen $\frac{\mathrm{d}}{\mathrm{d}x}\left(-\frac14\right)=0$

Balarka Sen

@robjohn That is understood. But I think what @MrWho wants to know is where does it appear in his derivation. My guess is that there is some domain inconsistency when applying the sub $x = \sin(\theta)$, i suspect this is the root of the problem.

But it's likely I am mistaken

MrWho

@robjohn @BalarkaSen Why $\frac {-1}{4}$ show itself when we use trigonometric substitution instead of general formual?I know it won't hurt integration answer but why is $\frac {-1}{4}$ detached?

robjohn

@MrWho This is simply an artifact of the method of integration. As you say, it does not affect the answer. One constant of integration is as good as any other.

MrWho

@robjohn With all due respect, I humbly disagree with you.There must be a clue to clarify the transpiration of the artifact you spot to.

@BalarkaSen @robjohn By mentioning artifact we're just escaping from finding the rationale behind it.

Balarka Sen

08:58

@MrWho How are you getting this equality? : $$\dfrac{-1}{4}(\cos^2\theta-\sin^2\theta)+C=\dfrac{-1}{4}+(\dfrac{-1}{4}+\frac{‌x^2}{2})+C'$$

MrWho

@BalarkaSen I'm not getting this?!

@BalarkaSen Think that the editor ruined my question!

Balarka Sen

That's what is written in the question.

MrWho

@BalarkaSen Editor! :(

Transcript for

$Mathematics$ Mathematics