Mathematics - 2015-11-30 (page 2 of 3)

Frank Science

09:21

Every ideal is finitely generated: Noetherian.

I'm mostly just an idiot

@FrankScience Yes, I am working on understanding the equivalence of three definitions.

Frank Science

@I'mmostlyjustanidiot The equivalence of two of them is formal, so called well-founded relation.

Balarka Sen

Ascending chain condition <=> f.g. ideal, in particular.

Frank Science

Not that one.

a.c.c. $\iff$ existence of maximal

Balarka Sen

Ah, ok, I forgot this condition.

The one I mentioned can be proved as follows: if you have a.c.c, then ideals cannot be non-f.g. because if you had a non f.g. ideal, you can keep adding infinitely many generators to get an ascending chain which does not stabilize. If every ideal is an f.g. ideal, it is clear why you have a.c.c.

Frank Science

09:28

The one you mentioned isn't formal. The one I mentioned is formal, not only applied in ring theory.

Balarka Sen

ok, by formal you mean can be done purely set theoretically. I see. I completely forgot about that definition anyway.

Weird, I don't remember ever using that definition.

I'm mostly just an idiot

@BalarkaSen Yes I have proved the finitely generated $\implies $ ascending chain becomes constant, but not yet the converse.

Balarka Sen

Yeah, that direction is easier.

I'm mostly just an idiot

@FrankScience Thanks, I'll try this out using what you linked.

Balarka Sen

The other direction is slightly nonobvious, but I gave you a proof above.

I'm mostly just an idiot

09:32

@BalarkaSen I'll see if I can get it working (for me) now.

Balarka Sen

Sure.

Frank Science

The point is that the collection of finitely-generated ideals has a maximum.

That's so-called Noether induction.

Exercise (Cohen): Suppose $A$ is a commutative unitary ring, and every prime ideal $P$ of $A$ is finitely-generated, then $A$ is Noetherian.

Balarka Sen

You can Zorn the heck out of it, I believe?

(trying to recall the proof)

@FrankScience Oh, cool fact.

Frank Science

We need a hint for this exercise, otherwise it's a bit difficult.

I'm mostly just an idiot

So essentially we assume there is an ideal that is not finitely generated, with generators $\{x_1,x_2,\cdots\}$, and form the chain:

$$\langle x_1 \rangle \subseteq \langle x_1,x_2\rangle \subseteq \cdots$$

Which is a contradiction to having every ascending chain become constant.

Frank Science

09:42

Hint: If $a\in A$ such that $I+(a)$ and $(I:a)=\{x\in A\colon ax\in I\}$ are finitely generated, then $I$ is finitely generated.

Balarka Sen

@I'mmostlyjustanidiot You mean $(x_1) \subset (x_1, x_2) \subset \cdots$. None of those are equal.

I mean, you have to choose a set of generators that way.

Paradox 101

@BalarkaSen can I ask you a question?

I'm mostly just an idiot

Sure, that makes it more clear, but out of habit I use $\subseteq$(still correct, but less obvious what I mean).

Balarka Sen

@Paradox101 You can, but if it's about analysis, it's unlikely that I can answer.

Frank Science

You can use $\subsetneq$.

Balarka Sen

09:44

Hello @iwriteonbananas

@FrankScience Hmm.

Paradox 101

@BalarkaSen it's not about analysis it's to do with differential equations.

Balarka Sen

Oh, then I am definitely not the right person :)

iwriteonbananas

Good morning.

Frank Science

@I'mmostlyjustanidiot Have a try on Cohen's theorem. It's very challenging.

Paradox 101

Oh ok then thanks anyway :)

I'm mostly just an idiot

09:45

@FrankScience I was, it seems interesting, I'll try it now.

Frank Science

@Paradox101 Don't ask to ask; Just ask.

Balarka Sen

Ah, @FrankScience here knows diff eqns, of course :)

Frank Science

@I'mmostlyjustanidiot If you fail to solve it. don't worry. Spend days or even weeks on that. Don't expect to solve it in a moment.

Paradox 101

@FrankScience ok. if we have to solve a Legendre's equation about points $x=1$ and $x=-1$ then in the power series solution do we assume that $x_o=1$ and $x_o=-1$?

Frank Science

No, I don't know diff eqns.

I'm mostly just an idiot

09:47

@FrankScience Truly?

Balarka Sen

You were taking a course on them, weren't you?

I remember you mentioned.

Frank Science

Yes, but most of them are not about solving equations.

I'm mostly just an idiot

@BalarkaSen Is your exercise truly that difficult?

Balarka Sen

Ah.

Frank Science

And up until now, it's related to properties of pseudodifferential operators.

Balarka Sen

09:48

@I'mmostlyjustanidiot You pinged the wrong person.

I'm mostly just an idiot

Oh oops

Your images are so similar, I thought you posted that one.

Frank Science

@I'mmostlyjustanidiot Whether it's difficult depends on your experience, etc.

It's meaningless to say whether it's difficult, just like saying $1/2+1/3=5/6$ is easy.

I'm mostly just an idiot

I mostly agree, but definitely an open problem is more difficult compared to a standard first week commutative algebra tutorial question @FrankScience.

Balarka Sen

I have no intuition for why that statement is true, myself. It seems nontrivial, if not hard.

Frank Science

@I'mmostlyjustanidiot Cohen's theorem was a tutorial question for me (I mean, in Algebra 2 which I took).

@Paradox101 I don't understand the notations.

Paradox 101

10:03

@FrankScience we're supposed to solve this by assuming that the solution is in the form of a power series about a point $x_0$. So in this case we'll take $x_0$ to be $1$ and $-1$?

Frank Science

I don't know, but did you prove that any solution is analytic around $1$ and $-1$?

For ODEs like $\dot x=v(x)$ for analytic $v\colon\mathbb R^n\to\mathbb R^n$, maybe the solutions are analytic.

Balarka Sen

Is there a nice counterexample to the statement when prime ideals are replaced by maximal ideals? I am trying to think of a good variety/smooth manifold with stalk of structure sheaf having finitely generated maximal ideal. Don't think it's going to work.

I'm mostly just an idiot

So "Every ideal in $R$ is finitely generated" is equivalent to "Every non-empty set of ideals in $R$ has a maximal element".

Probably I am overlooking something simple, but why isn't $\{\langle x_1,x_2\rangle,\langle x_3,x_4\rangle\}$ not a counter example?

Where all of those generators are distinct.

Balarka Sen

I don't understand what your ring is.

Frank Science

@BalarkaSen It seems to me that from Gelfand representation, the maximal ideals of the ring of continuous functions are principal, just like Hilbert's Nullstellensatz.

Balarka Sen

10:12

Ah, @Frank? Is that really true? Interesting.

Frank Science

@I'mmostlyjustanidiot $a$ is maximal in $S$ means that if $b\ge a$, then $b=a$. So there might be more than one maximal elements.

I'm mostly just an idiot

@FrankScience Oh, of course!

Frank Science

@BalarkaSen For compact Hausdorff topological spaces.

@BalarkaSen Exercise 26 of Chapter 1 of Atiyah and Macdonald.

That's the start point of Gelfand representation.

Balarka Sen

I only remember the exercise $\text{mSpec} C(X) \cong X$ from chapter 1.

Frank Science

Sorry, I said something wrong.

Continuous isn't enough.

Balarka Sen

10:15

I suspected, because I think I have a counterexample.

Frank Science

But for smooth functions, if $f(x_0)=0$, then it's divisible by $x-x_0$ right?

Balarka Sen

@FrankScience You mean analytic functions. Yes.

Frank Science

No, only for smooth functions on $[0,1]$.

Balarka Sen

There are smooth functions which are not analytic.

Consider $f(x) = e^{-1/x}$ on $x > 0$ and $0$ elsewhere. This is smooth, but has no convergent Taylor series at $0$.

What do you mean by "divisible by $x - x_0$" in this case?

Frank Science

Divisible in the ring of smooth function.

Balarka Sen

10:20

So $f(x)/(x-x_0)$ is $C^\infty$, yeah? Is that true for this function?

Frank Science

Hadamard's lemma

In mathematics, Hadamard's lemma, named after Jacques Hadamard, is essentially a first-order form of Taylor's theorem, in which we can express a smooth, real-valued function exactly in a convenient manner. == Statement == Let ƒ be a smooth, real-valued function defined on an open, star-convex neighborhood U of a point a in n-dimensional Euclidean space. Then ƒ(x) can be expressed, for all x in U, in the form: where each gi is a smooth function on U, a = (a1,...,an), and x = (x1,...,xn). == Proof == Let x be in U. Let h be the map from [0,1] to the real numbers defined by Then since we have ...

Balarka Sen

Nice, that does the trick. Thanks.

Frank Science

And another point is interesting.

Suppose $(A,m)$ is the ring of germs of smooth functions at a point.

$\hat A$ is the completion of $A$ w.r.t. $m$.

You can see that $\hat A$ is the ring of formal power series.

Balarka Sen

By completion of $A$ w.r.t $m$ you mean $\varprojlim A/m^k$?

Frank Science

Yes.

The interesting thing is that the natural map $A\to\hat A$ is surjective.

Balarka Sen

10:31

Whoa, really?

Frank Science

Okay, there's one minor point: the example you given before, with convergent Taylor series, but not convergent to the original function at $0$?

Balarka Sen

mhm. what about it?

@FrankScience That is so false in general for arbitrary rings $A$. I wonder what makes it true for stalks of smooth manifolds.

Frank Science

I mean, you said something like $\exp(1/x^2)$.

Balarka Sen

$e^{-1/x}$, yes.

The Taylor series at $0$ does not converge to $0$.

Frank Science

That lies in the kernel of the natural morphism.

The point is really nothing to do with ring completion, so apparently nothing to do with general ring.

It's in fact an analytic property of the ring of germs of smooth functions.

Balarka Sen

10:40

Hmm, ok.

That makes sense, thanks.

Frank Science

And it's interesting to compare with Krull's Principal Theorem.

Well, in order to prove that property (so-called Borel's theorem), you need only to prove that $A/\bigcap_nm^n$ is complete.

Balarka Sen

I need to log out from the chat now. Thank you once again for this conversation, these facts are certainly very interesting.

Frank Science

The trick is that when you're given a serious $a_n\in m^n$, you need to sum up $\sum_na_n$.

Apparently, it would diverge.

Balarka Sen

ok, I see.

Frank Science

However, there exists $b_n$ such that $a_n\equiv b_n\pmod{\bigcap_n m^n}$ and that $\sum_n b_n$ converges in the canonical topology of smooth functions, which means that every partial derivative converges uniformly in a neighborhood.

I'm mostly just an idiot

10:48

**Question 7:** I am trying to prove that if every ideal of $A$ is finitely generated, then any nonempty set of ideals of $A$ has a maximal element. This is my 'intuition form proof', is there anything that is incorrect?

We have the set of ideals $\{I_1,I_2,\cdots\}$ If $I_i$ is not comparable to any other ideal in the set, then it is maximal, else it is comparable. If there is a simple non-comparable ideal we are done, assume all ideals are comparable, then we have numerous partially ordered chains consisting of a finite number of ideals. These chains terminate at a maximal element.

I'm mostly just an idiot

11:23

My question 7 above was discussed more in the Commutative Algebra room, it required zorn's lemma

Huy

"A subspace $X$ of a locally compact Hausdorff space $Y$ is locally compact if and only if X can be written as the set-theoretic difference of two closed subsets of $Y$."

I'm allowed to take one of the two closed subsets to be $X$ itself and the other to be the empty set, i.e. closed subsets of locally compact Hausdorff spaces are also locally compact, right?

Huy

12:17

I want to find a discontinuous, bijective homomorphism from $(\mathbb{R},+)$ to itself. Is this possible without using AC?

Karl Kronenfeld

12:32

@Huy Yes, a closed subset of a compact set is also compact. (So, your conclusion follows using the definition of locally compact)

Balarka Sen

13:01

@MikeMiller [re](http://chat.stackexchange.com/transcript/36?m=25759300#25759300): Pseudoisotopy theorem, as I understand it, says diffeo $f$ of $M$ (1-connected manifold without bd of dim >= 5) is isotopic to the identity iff there is a self-diffeo of $M \times [0, 1]$ which restricts to identity on one end and $f$ on the other.
That said, if $f: S^n \to S^n$ is self-diffeo such that $D^{n+1} \cup_f D^{n+1}$ is diffeomorphic to the standard $S^{n+1}$, then we can obtain an extension of $f$ to $D^{n+1}$ by removing a hemisphere. Call this $\tilde{f} : D^{n+1} \to D^{n+1}$. Now delete two sm…

Assuming I have proved that $\pi_0 \text{Diff}(S^n) \to \Theta_{n+1}$ is a group homomorphism of course. I haven't yet proved $\Theta_{n+1}$ is a group, but let me figure out why this is a group homomorphism assuming that.

(Everything I have done above is for $n \geq 5$)

Ok, doesn't seem too complicated. All I have to do is to prove that $D^{n+1} \cup_f S^n \times I \cup_g D^{n+1}$ is diffeomorphic to $D^{n+1} \cup_{fg} D^{n+1}$. Intuitively, I can see why, but not clear to how to prove it.

Paradox 101

Given the above question, the value of the coefficient $a_n$ comes out to be $0$ for all values of $n$ which implies that the final series will not contain the cosine terms. But given that the question asks for both sine and cosine series is $a_n=0$ possible?

Balarka Sen

13:18

@MikeMiller Whoops, I skipped some subtlety up there. What guarantees that chucking out small disk can be done in a way such that $\tilde{f} : D^{n+1} \to D^{n+1}$ gives a pseudoisotopy $S^n \times I \to S^n \times I$? In particular, if $B^{n+1}$ is a small disk in $D^{n+1}$ around the center, we want $f|_{\partial B^{n+1}}$ to be the identity diffeomorphism of $S^n$. This certainly might not happen.

Working on fixing it.

Ah, I can isotope $\tilde{f}|_{B^{n+1}}$ so that image is precisely $B^{n+1}$. I mean, obviously, as there is only 1 isotopy class of +-orientation embeddings of a disk. Now extend to an ambient isotopy by isotopy extension theorem. You can obtain a diffeomorphism of $D^{n+1}$ which fixes $B^{n+1}$ altogether. Now remove and continue with the procedure.

Um.

OK, no, I think that's fine.

Huy

13:59

@DanielFischer: I'd like to verify that $$m(B) := \lambda_{n^2} (\{tg| \, g \in B, t \in [0,1]\})$$ defines a Haar measure on $SL_n(\mathbb{R})$, where $\lambda$ denotes the usual Lebesgue measure and $B$ is Borel. I'm a bit confused because usually, I am given a Haar measure with a formula like $dm(x) = \dots d \lambda_{n^2}(x)$ and then I'd show that the integral of Borel sets is preserved.

@DanielFischer: My TA on the other hand uses compactly supported functions because any Borel measure on $\mathbb{R}^n$ is uniquely determined by its values on the compactly supported functions. Is there an advantage in doing that? Anyways, how do I find $dm(x)$ for the given formula in this case?

Private Pansy

During the revision lecture, this question came up. The lecturer took one look and said "Oh, this question. The year this question came out, zero out of 170 students managed to complete it. It broke my heart. Let's just skip over it."

Task is to test the series for convergence

BumbleBee

Anyone up for answering one question ?

You are given a 2N*2N matrix, A and B are playing a game where A chooses a cell randomly and fills a number there, and then B does the same. They do this until all the cells are filled. At the end, if the determinant is 0, B wins. Otherwise, A wins. Does B have a winning strategy?

user174558

@PrivatePansy They cannot complete it because they have been told to ignore it.

Private Pansy

Pffft :P

BumbleBee

@JasperLoy Can you answer the question that I mentioned above ?

Daniel Fischer

14:26

@Huy In this case, since $SL_n(\mathbb{R})$ is $n^2-1$-dimensional, a Haar measure on it can't be given by a density with respect to $\lambda_{n^2}$. But with the given formula, it's easy to verify the translation-invariance.

@PrivatePansy Testing it for convergence is easy with the given hint. Finding the value would be difficult, I believe.

Private Pansy

@DanielFischer I can't make head or tail of the hint

Wait, let me see...

Should compare to the series of 1/n I think.

y is obviously n, so we're looking at n^ln(ln(n-sqrt(n))), which is, uh...

Daniel Fischer

@PrivatePansy Look at the exponent. What do you know about the exponents?

Private Pansy

I'm thinking, since ln(ln(n-sqrt(n)) is an unbounded function, that is for large n it will go to infinity, so we can say that for large n it is smaller than the series of 1/n, thus the series should converge

Mike Miller

The sum of 1/n does not converge.

@Balarka: Seems reasonable.

Daniel Fischer

@PrivatePansy Right idea, but lacks rigour. It's not enough that the function is unbounded, that would still allow lots of small values. But the function is increasing, so that doesn't happen. And smaller than 1/n isn't sufficient, we need more. It's smaller than 1/n^a for some a > 1.

Private Pansy

14:45

Yes, I'm aware of the test for series 1/n^a, a > 1. I do need to go to sleep though - the test is tomorrow, and the questions that'll come out will probably be easier than this one :P

Mary Star

Hello!! Is there a condition so that a polynomial has at least one nontrivial root?

Private Pansy

@MikeMiller I mean, it is the boundary of where series will converge. 1/n^a where a > 1 will converge. So my thinking was at some point ln(ln(n)) would be larger than 1

Mike Miller

Sure, you seem to have gotten the idea with Daniel above. To drive the point home: 1/(n ln(n)) does not converge, nor does 1/(n ln(n) ln(ln(n))) etc.

Balarka Sen

@MikeMiller Thanks for checking.

Mary Star

Is there a condition so that a polynomial has at least one nontrivial root? @robjohn

Private Pansy

15:00

Right, in those cases n is the dominant term

Mike Miller

I'm not really sure what that means, but ok.

Keep in mind that 1/(n ln(n)^a), a>1, converges.

Balarka Sen

The pseudoisotopy theorem seems like a high-dimensional phenomenon. I'm curious why it fails in low dimensions. Is the proof involved?

Mike Miller

Yes. It's a cubic refinement of Morse theory. I don't at all remember the details since I did not make a careful study of it but it should fail for the same reason h-cob does.

Balarka Sen

oh, fun.

Mike Miller

Well, in some sense.

r9m

15:21

@robjohn could you check if this is a correct approach, I feel I might have nuked a mosquito for some reason! :-)

Dave

15:55

Hi, I'm wondering, is there a way to smooth the sinus at the top of the curve ? Is there a way to extends its time at the top ?

I'm using $y = sin (x/20*pi) * 500$ to simulate the number of bananas produced by a tree at the age x. It's okay but I'd like to produces maximum bananas for a couple years ("staying at the top of the sinus for a while") if that makes any sense....

evinda

Hi @DanielFischer @anon
Could I ask you something?
In order to show that $Y=\{ x \in \ell^2(\mathbb{N}) | \exists n \in \mathbb{N} \text{ such that } x_j=0 \forall j>n \}$ do we have to show that it is non-empty and $\lambda x+ \mu y \in Y$ given that $\lambda, \mu \in \mathbb{R}, x,y \in Y$ ?

Dave

Looks like calling it amplitude will make me look less dumb

Daniel Fischer

16:25

@evinda Looks like you forgot to type some words. It seems like you want to show that $Y$ is a linear subspace of $\ell^2(\mathbb{N})$?

evinda

No... a subspace... @DanielFischer
In this case don't we have to show anything further?
(I am looking at the exercise: Find in $\ell^2(\mathbb{N})$ a subspace Y and a $x \in \ell^2(\mathbb{N})$ such that d(x,Y) is not attained)

Daniel Fischer

16:41

@evinda Well. In context, subspace most likely means linear subspace. I would not be surprised if the verification that $Y$ is a linear subspace was not expected, since it may be assumed that students already have learned linear algebra. Then it remains to find an $x$ such that the distance is not attained, and show that for that $x$ the distance isn't attained.

evinda

We just show that Y is dense in $\ell^2(\mathbb{N})$ and thus $||x-y|| \to 0$ for all $y \in Y$ and thus $\inf \{ ||x-y||: y \in Y\}=0$..
Then we suppose that there is a $y \in Y$ such that $||x-y||=0 \Leftrightarrow x=y$ and that is a contradiction since $x \notin Y$, in general. Right? @DanielFischer

copper.hat

Has anyone here ever used/written answers for a site called slader.com?

Daniel Fischer

@evinda That doesn't really make sense as written. What is "$\lVert x-y\rVert \to 0$ for all $y\in Y$" supposed to mean?

evinda

17:00

I meant that the distance between any x and y will tend to zero... Isn't it right? @DanielFischer

Huy

@DanielFischer: And do you know of the reason why my TA keeps using compactly supported functions for showing left/right-invariance?

Daniel Fischer

@evinda The distance tends to zero as what varies in which way?

@Huy Personal preference, probably.

Huy

@DanielFischer: No advantage in technique or higher generality or something?

evinda

@DanielFischer for any x,y such that $x \in \ell^2(\mathbb{N}), y \in Y$. Or not?

Daniel Fischer

@Huy Not that I know. Sometimes one way is more convenient, sometimes another.

@evinda If $x$ and $y$ are fixed, nothing varies. You need to let (at least) one thing vary. [But I have to go now, so won't be available for further questions for a while.]

evinda

17:07

Ok, I will think about it and I will tell you later... :) @DanielFischer

PerplexedGuest

Hello all! :)

copper.hat

Excuse the repetition, has anyone here ever used/written answers for a site called slader.com?

hi!

Mike Miller

Morning.

copper.hat

hello!

Mike Miller

@Huy Might be that you defined your function spaces as completions of $C^\infty_c$, and your integrals as continuous extensions of this, &c.

In that case the only way to prove invariance would be to prove it for $C^\infty_c$ which then extends to everything.

Andrew Thompson

17:24

I came across a statement which kind of baffles me: if $R$ is any ring with finitely many ideals $I_i$ such that $R/I_i$ are Noetherian as rings, then their direct sum is Noetherian as $R$-modules. Is this true with no assumption on $R$?

Mike Miller

Why not try to prove it and find out?

Andrew Thompson

Been trying to find a counterexample, but might as well give it a go.

Overly Excessive

Can someone help me simplify and explain $(a-b)(a-b)$ ?

robjohn

@MaryStar Yes. If the polynomial is of odd degree, it has at least one real root. If a polynomial has at least degree one, it has at least one complex root.

Andrew Thompson

Ohh, I'm so silly. The $R$-module structure on $R/I$ is induced by the canonical surjection $R \to R/I$ and so I might as well consider $R/I$ as a module over itself, after which it is simple to work out.

Mike Miller

17:32

Hooray!

L33ter

hi

hi @MikeMiller

Balarka Sen

phew, schoolwork is a mess.

L33ter

@BalarkaSen

:D

I finished borsuk ulam thing

its so cool

Balarka Sen

Good to know. Got a proof sketch for me?

L33ter

now I think I am prepared for alg top class next semester

yeah I will write everything now and I will show you sketch of how I will present it after I am done.

Balarka Sen

17:43

Did you prove that a map $S^1 \to S^1$ which extends over a disk is nullhomotopic?

L33ter

yeah

iwriteonbananas

Are you guys talking about the Borsuk-Ulam in dimension 2 or in arbitrary dimension?

L33ter

lol no

Balarka Sen

@iwriteonbananas 2

L33ter

dimension 2

iwriteonbananas

17:46

Alright.

L33ter

I don't think I have enough knowledge of dimension n

to be able to do dimension n

Frank Science

You proved Borsuk-Ulam on your own?

L33ter

no

some results though

that I needed I did it on my own

actually the proof is not that bad if you rely on your intuition for some part

Balarka Sen

He's reading Munkres's proof.

iwriteonbananas

@L33ter It becomes easy once you know about the cohomology ring structure on $\Bbb RP^n$. But nevermind what I'm saying.

Mike Miller

17:48

lol

Balarka Sen

The whole deal is proving that odd maps have odd degree. It's not hard to see why B-U follows from this.

@iwriteonbananas You don't need that much.

Just the transfer sequence does the job nice and easy.

iwriteonbananas

@BalarkaSen Transfer sequence one way to do it (and in fact a cool way), but I wouldn't exactly call it easy. The cohomology way is more concise imo.

Balarka Sen

Ok, that is true. But computing cohomology ring of RP^n in the 1st place is a mess.

iwriteonbananas

Very true.

Unless you know of the Gysin sequence.

Balarka Sen

Speaking of, I am surprised you're alive.

iwriteonbananas

17:53

I am barely.

Balarka Sen

@iwriteonbananas Or the Poincare duality ...

L33ter

why ?

Balarka Sen

@iwriteonbananas I figured.

L33ter

what is happening with you @iwriteonbananas

Frank Science

I remember the hard time that there was an exercise to show that automorphisms of $\mathbb C\mathbb P^2$ are orientation-preserving.

Balarka Sen

17:54

he's going through the computation of H^*(RP^n) twice

that's what's happening to him.

iwriteonbananas

@BalarkaSen On the bright side, I think I understand all the details now and can write down the proof without looking at the notes.

Balarka Sen

Good to hear.

L33ter

btw again @BalarkaSen the homotopy is an equivalence relation on the set of maps from (I,X) right?

Balarka Sen

@FrankScience Um. It's not true that any homeom of CP^2 is orientation preserving, right?

Frank Science

And this year there is an exercise that $\mathbb R\mathbb P^2$ isn't a boundary of a smooth manifold.

Balarka Sen

17:56

@L33ter Yes.

Ah, @FrankScience. Good exercise.

Euler char of any closed cpt orientable 3-fold is 0. That's what is needed.

L33ter

I guess they use the notation Hom(I,X)

hm I am gonna give also quick introduction to fundmental group

Frank Science

I guess that computing Stiefel-Whitney classes is the ultimate weapon.

L33ter

so people with no background be able to understand the proof.

Balarka Sen

that's not needed, right?

(i dunno S-W classes)

Mike Miller

@Balarka: Re: CP2. Yes.

Frank Science

17:59

Sorry, there was no orientable condition. Euler characteristic should be done in $\mathbb Z/2\mathbb Z$.

Balarka Sen

Yes.

Also, you don't need the smooth condition.

@MikeMiller Thanks.

robjohn

@r9m I've added my own answer (hopefully simple enough). I am reading yours, but I need to look up the Theorem you mention.

Agawa001

hello

Balarka Sen

Fact I don't know how to prove but plan to know someday : any 3-fold bounds a 4-fold.

Frank Science

There's a characterization whether a manifold is a boundary by Stiefel-Whitney classes.

Mike Miller

18:03

@Balarka: Wrong. Want to fix the statement?

Balarka Sen

Closed, compact.

Mike Miller

I wouldn't pick on you for that.

Agawa001

@r9m hey

r9m

@robjohn Awesome!! (+1) :D .. I knew I was missing something simple! So keeping that $\frac{1}{x(1-x)}$ in the denominator really helps! :-)

Balarka Sen

:)

r9m

18:05

@Agawa001 hello :)

Mike Miller

You haven't fixed the statement. I mean that I wouldn't call it wrong if that was the error.

Because those are obvious from context.

Balarka Sen

Ah, too bad. Hmm.

robjohn

@r9m I didn't combine the fractions, just looked at each fraction separately

L33ter

hey @BalarkaSen

Mike Miller

Aw shit ignore me.

Balarka Sen

18:06

Do you need orientable?

Agawa001

@r9m have you already seen chriss she appears to be most of time absent

Balarka Sen

lol

copper.hat

If one writes a remark in chat to @whoever, will that person get notified in some way?

L33ter

intuitively the reason the fund group of the circle is the integers intuitively is of we say the positive integers is the number of times we go around the circle counter clock wise

Balarka Sen

yes.

L33ter

18:07

and the negative are the number of times we go around the circle clockwise

r9m

@robjohn 'kay .. got it!!

L33ter

ok ok good

r9m

@Agawa001 she is possibly busy ..

Balarka Sen

but you'd be hard pushed to turn it into a rigorous proof.

L33ter

nah that is enough for me :D I will just put it as a result and explain intuitively why that is the case

next semester I can do all that stuff

rigorously

Mike Miller

18:09

@copper.hat: Only if they were recently in chat. Say, the last week.

Balarka Sen

as you wish.

copper.hat

@MikeMiller: Thanks!

Balarka Sen

@FrankScience Heard of it, yeah. Is there an explanation of S-W classes without using too much machinery? I don't even know what they are.

Agawa001

take care

Mike Miller

They take a vector bundle and spit out a cohomology class. Any construction uses machinery.

Frank Science

18:15

@BalarkaSen I haven't systematically learned this subject. Maybe youcan consult @MikeMiller

Mike Miller

It is reasonable to learn them axiomatically and believe that such a thing exists. That's how Milnor starts his book.

Frank Science

I learned Pontrjagin classes and Chern classes from Hirzebruch.

Sadly, I didn't digest things well.

Balarka Sen

(By the way, I just finished eating, so here's a proof of the RP^2 fact: if M bounds RP^2, take double by gluing two M's along the boundary to get a closed compact 2-manifold. It's Z/2 $\chi$ is odd, as $\chi(M \cup M) = 2\chi(M) - 1$. But that's a contradiction, as Poincare duality implies $\chi = 0$. Just thought I should write down the proof.)

Mike Miller

I think Hirzebruch will do that to you.

Frank Science

Do what?

Mike Miller

18:16

Make it easy to not digest things.

Balarka Sen

@MikeMiller Hrm.

Frank Science

I didn't mean I learn directly from Hirzebruch.

It was just abbreviation for Hirzebruch's book.

Topological Methods in Algebraic Geometry.

Mike Miller

I know what you meant.

I tried to read that and didn't get much out of it when I did.

Frank Science

Okay, what did you mean? That's a bad book?

Mike Miller

I don't know if it's bad, but I think it requires a lot of effort and maturity, that I didn't have when I tried to read it.

Frank Science

18:19

Thanks for the comment. Since before I was told that it's a book good for undergraduates, so I read it. Yes, hard to digest.

Mike Miller

I guess undergraduates in your country are exponentially better than ours. I didn't even try til I was a grad student and had trouble.

r9m

@robjohn I forgot where, but I have seen an old answer of yours using/proving hermite-hadamard inequality for continuous functions (not assuming differentiability)

Frank Science

There are some great students, but most students aren't. (the undergraduates in the previous sentence is specific, not general)

In fact, after reading a book on Riemann-Roch theorem, I was eager to learn Hirzebruch-Riemann-Roch, so asked a teacher where I got such remark, and read it.

robjohn

@r9m Oh, no. I just looked at the envelopes in my answer, and realized it looks like something. One might caption the image as "The Crack of Doom"...

@r9m really? I have not had a chance to look up what that is :-)

copper.hat

@robjohn: You have some awesome graphics in your answers.

Mike Miller

18:24

I see. I think most native English speakers would interpret that to mean a general undergraduate student.

robjohn

@copper.hat I hope they add to the understanding of what is being explained (it's nice that they are appreciated, as well)

PVAL

Developing SW-classes (or Chern classes) without being able to compute things with Steenrod squares is pointless.

r9m

@robjohn That's really neat!! :D (+1) I was reading that before you pinged me the link! :)

Frank Science

General undergraduate students in my country won't continue to learn mathematics in my country. Most of them will go to areas like computer science and statistics, say.

PVAL

I think this is the reason Hatcher's book is terminally unfinished

copper.hat

18:26

@robjohn: For me, visuals often clarify instantaneously!

Mike Miller

I think I would have trouble computing chern classes with Steenrod squares.

r9m

@robjohn must have been a long time ago .. I remember faintly! but I do remember reading it with the [mean square] avatar below the answer! :)

iwriteonbananas

@FrankScience What country are you in?

PVAL

@MikeMiller Well at least all of the algebraic relations you need to prove things (after computing them) somehow boil down to Steenrod squares.

Agawa001

@evinda have you seen my response or not ?

Frank Science

18:29

@iwriteonbananas I was in China.

iwriteonbananas

I see.

Mike Miller

After thinking for literally a second I no longer wonder.

Dumb.

PVAL

@MikeMiller Have you read Wu's paper?

Or looked at it

It's in French I believe.

Mike Miller

No, I probably won't in the near future.

PVAL

Perhaps I should learn to read some French, though the papers I would really like to read I have heard are undigestable anyway.

Mike Miller

18:31

The only french I've really read is papers by Cerf, who is mostly a pretty good author.

Frank Science

It seems to me that it's not very hard to read mathematics in French.

Mike Miller

He occasionally abbreviates stuff which is very annoying but can be deciphered from context.

Frank Science

Cerf, you mean that guy in probability theory?

PVAL

I think Milnor describes the theorem of Cerf I cared about as "very hard" in his h-cobordism notes. I wasn't sure if understanding French would make it possible for me to read the "nullity of $\Gamma_?$" paper.

Is that what you've read?

Frank Science

Sorry, seems not the same one.

Agawa001

18:38

forget it

Frank Science

Maybe algebraic Riemann-Roch is easier than analytic Riemann-Roch.

Mike Miller

@PVAL: That's actually a book, not a paper. It was doable as a weekend project to read, though I certainly couldn't reproduce the technical details.

robjohn

@r9m I have looked unsuccessfully. Let me know if you find it, it would be nice to keep it in my log of interesting answers.

Mike Miller

He also proved that given Smale's conjecture the inclusion $\text{Diff} \hookrightarrow \text{Homeo}$ is a homotopy equiv for 3-folds. This is short and readable.

r9m

@robjohn sure! :-) I'll try and find it too :-) (gimmie some time)

Mike Miller

18:41

At some point I'm going to go back and really understand that first book, since while I could read it in a weekend, I was only able to walk away with a sketch.

@Frank: Riemann Roch either way is a corollary of serre duality. Whether you think algebraic is easier or not depends on how easy you find Hodge theory, which is how one proves serre.

Frank Science

With Hodge theory, Serre's duality is easy. However it's not easy to build Hodge theory from scratch to me.

Link

I have a stat question for an assignment I was working out, can someone just confirm that I did it correctly please? "If a team wins at a 15% rate, what's the chance they win at least one of their next 5 games?" I did this as $.85^5 = 0.443705312$ Chance to lose all 5 games, so winning at least one is at least $100-0.443705312 = 99.5562947?$

Did I do that right or am I missing something?

copper.hat

@Link: If p=0.15, then the chance that they win at least one of the next five (assuming the wins are independent) is $1-(1-p)^5$.

Frank Science

Another point is that, for analytic version, the geometric genus = 2 topological genus, is a corollary of Hodge decomposition theorem.

r9m

@robjohn AHA!!! Told you I have seen it :D (infact I seem to have upvoted it in past .. :) .. got it from my upvoting list)

copper.hat

18:46

@Link: you ar emixing percentages an d probabilites

Link

@copper.hat Okay, I see where I went wrong. It should be $1-0.44 = 0.56$

copper.hat

$1-(1-p)^5$ is about .556.

Link

Thank you!

copper.hat

you're welcome

robjohn

@r9m I searched on the tag convexity and on the term Jensen, but I should have sifted through a search on the term convex (though there might be a lot of matches).

robjohn

19:25

@r9m when I searched on the term convex, it was the seventh hit. Better searching wins.

r9m

@robjohn :-) 'kay! I just remembered having read it. Btw, have you tried this integral $\displaystyle \int_0^1 \log (1-x+x^2)\log (1+x+x^2)\,dx$? :)

robjohn

@r9m I remember that was posted on chat a week or so ago. I haven't tried it yet.

@r9m I believe I had an idea... I should look again.

r9m

@robjohn 'kay! :D

Transcript for

$Mathematics$ Mathematics