Taubes is baffling as usual.

@TedShifrin Me too, or I upvote someone else's violent objections in the comments before I downvote.

I mean, I get why I'm getting downvotes on that very subtle second partial derivative question, although I did work hard on what I wrote. But anyhow ...

speaking of bring radicals, this was on wikipedia for about a month. the edit summary was "adding an image to help explain the usefulness of Bring radicals and how they're distinct from radicals."

the comic?

yup

lol

what.

Ah, I remember that thing. Cute.

I'd rather take them than bring them ...

i mean, as an example of ultradicals in 'pop culture' it might warrant a mention. but placing it right smack-dab at the top?

that's probably not pop culture

yeah.

@MikeMiller It's not really "pop", and I'm not sure how much "culture" it actually has.

:)

i don't necessarily mind mentioning that joke somewhere on the page as an example of a mnemonic, but as a picture it just seems dumb

the comic isn't the joke, putting it at the top of the page is the joke

@arctictern and that it got away with being there for a rather long time.

(:

well, every article needs a picture

gotta illustrate it ysee

It's been so peaceful here this afternoon ... Balarka took a vacation?

guess so

Now, Ted... :)

Yes, J.M.? :)

Was that too orthodox on my part?

...tsk, fine. Carry on. :P

LOL.

how do you align expressions after equals signs in mathjax? e.g. 5 = 3 + 2 = 1 + 1 + 1 + 1 + 1, but having each equal sign sitting on top of each other, if that makes sense

you use \begin{align*}

thanks!

and then put &= \\ (for end of line) ... and then \end{align*} at the end of it all.

It's // and not \\?

Not that I've typed any TeX in years.

oops, sorry

hmm, wait.

no, I'm right ... I'll bet you a dinner

I won't take that bet.

have any of you met other math.se users in person (whom you previously had only known through the site)

This is overwhelmingly my favourite SE site.

The tone and spirit is highbrow and generous.

Thanks everyone.

and sometimes rude, @BenjaminR. :)

yes, @rorty ... Sadly, @MikeM and I know each other suchly.

Is that a pun? Because my surname is Rood?

LOL, no, @Benjamin, I saw only the R.

But I try to pun (as badly as possible) whenever I can.

@TedShifrin We do make an effort to tamp down on the rude, no?

I've never seen any rudeness in Answers to Questions, which is the most impt thing.

I spoke only for myself, @J.M. :)

I try never to be rude, but occasionally I get accosted rudely in comments and then my dander is up.

@TedShifrin O GOD. I just realized you wrote the textbook of the freshman honors math course at my university (which I did not take, though I do own your book).

Should I apologize, @rorty? Did you go to Yale?

@TedShifrin Oh certainly, and you then hope the dander makes their eyes water. ;)

At the very least, @J.M.

XD

Do you know each other?

To whom is that addressed?

You and mr. initials.

@MikeMiller Very likely not. Hint: it's morning where I am.

@rorty, ironically, I taught an individual Spivak calculus course to a high school student in Athens, GA, who then went to Yale to be a math major, only for the person teaching the Honors course that fall to stop using my book :P

I know quite well a mathematician (geometer, in fact) whose name is J.M. ... But they be different persons.

If two solutions to the equation $z''(x) + B(x)z' = 0$ have a zero at the same point does that imply that they are dependent?

Sure, @MaryStar.

My roommate last years first initials were JM.

i have to point out with some small amusement that my initials are M.J.

Don't confuzle me, @Semiclassic.

hah

Oh, so much for our peace and quiet.

@arctictern Are there uncountably many continuous maps $\prod_{\Bbb N} \Bbb Z/2 \to \Bbb Z/2$?

@TedShifrin And is one of the $xI$ and $I$ a subset of the other one?

Depends whether $x\in I$, doesn't it, @MaryStar?

okay, time to do grading

a set $S$ with a linear/total order relation is denoted $(S, \prec)$, right?

If you like, sure.

@MikeMiller believe so

since when were initials commutative, anyway? :)

(if the relation is $\prec$)

When $x\in I$, we have $xI=I$, @MarySTar, don't we? And why should the second one follow?

@MikeMiller such continuous maps are in bijection with clopen subsets. an open subset may be specified by picking a finite set J of indices and a subset of prod_J, and these are all clopen.

there's some redundancy but not enoguh to affect cardinality

Oh yes... So, at the second case we cannot say anything about that, right? @TedShifrin

sorry, continuous homomorphisms.

i agree there's plenty of continuous maps

i've been alerted it's clear from galois theory there are only countably many

@MaryStar: Seems like you'd need $x$ to be a unit, in which case $xI = R$.

Why does this hold? @TedShifrin

If $a\in I$, how do you write $a=xb$ for some $b\in I$, @MaryStar?

Hi @TedShifrin, @MikeMiller.

Hi @Balarka. My peace and quiet is over.

@MikeMiller I think every such continuous homomorphism is a dot product with an element from $\bigoplus_{\Bbb N} Z_2$

@arctictern it's a cool comic.

yep, they are all dot products

When we take $a=1$ then we have that $1=xb$, therefore, $x$ must the inverse of $b$, right? @TedShifrin

@MaryStar: For what ideal $I$ is it the case that $1\in I$?

@arctictern Me too

@Ted I decided against walking to Borscht today.

Lazy bum, @MikeM. So are you walking here for stuffed cabbage?

Probably not.

Cliff T demands my attention.

@AkivaWeinberger Now given a knot compute the fundamental group of its complement.

Use $\pi_1^{ab} \cong H_1$ to derive its first homology.

@Balarka: There's that pesky apostrophe again.

With multiplicity two.

(Some year you'll thank me for being obnoxious.)

You can't edit it away

WE ALL SAW WHAT YOU DID

smiles @DogAteMy.

@TedShifrin If $I=R$, if it is not a proper ideal, or not?

Nope, @MaryStar.

@MikeMiller I asked that question!

If you're using arrows I'm on my phone.

I'm on my phone too

long knot retract @MikeM

I can see the arrows

I was referring to retracting R^3 to the long trefoil.

I hate chatting here on my phone, so I don't.

They're just annoying to click on 'cause they're so small

I don't use the "new" interface.

I think I asked whether solid torus retracts into trefoil inside it, but yeah.

They give you the option of not using it?

@Balarka: If you have some good questions to contribute to my final exam, please send them.

Remember they should be at an elementary level.

Hmm. I have a couple, I think.

$\pi_1$ and covering spaces, right?

Starting with countability axioms, regularity, normality, Urysohn (metrization), Tietze, and Munkres's $\pi_1$ and covering spaces, yes.

@TedShifrin But?

@TedShifrin Q: How bad are non-Hausdorff spaces? A) Not that bad B) Horrible C) AAAH

A proper subset can be neither empty nor the whole set, @MaryStar. But?

Thanks for reminding me, DogAteMy. I have asked them for a non-Hausdorff quotient space (leaf space for foliations) before, and they didn't get it right. I'll put that on the final.

The empty set isn't considered a proper subset? I though it just mean subset-but-not-equal-to.

@TedShifrin Did you tell them about the Hawaiian earring?

Why it's not homotopy equivalent to the infinite wedge of circles, say.

I assume they'll read it in Munkres soon, @Balarka.

I am with akiva on "proper"

Ah, OK.

I was taught that the empty set wasn't a proper subset. Granted, that was a century ago.

@TedShifrin I don't know what a leaf space is, but I hope I didn't inadvertently lower your students' scores on that exam just now

indeed, why the xountavle wedge of circles doesn't embed in R^n for any n...

Wait, it doesn't?

Do you know what a foliation is, DogAteMy?

@AkivaWeinberger no.

Why not have the circles have radius $(1+1/n)$ instead of $(1/n)$?

@TedShifrin I learned that only fifteen or so years ago, with the caveat that I was reading old books. :)

I'm with Connes that the topological space one calls the leaf space is not actually what a leaf space is

@TedShifrin It contains at least one element of the set but not all of them.

Just a little bit of the structure

@TedShifrin Vaguely. I might have heard it a few times. You'd have to remind me

Right, by my definition, @MaryStar.

Same for fibration

DogAteMy, without pedantic details, it's filling up a big space with smaller-dimensional spaces, such that locally you have a product structure.

@Akiva See if you can prove it

@TedShifrin So what's a fibration?

@TedShifrin A rather hard problem is to figure out what the $\pi_1$ of a complement of a discrete set is in R^n.

Depends on the ambient space, @Balarka?

I just woke up :)

A fibration is a similar thing, @Akiva, but all the leaves are (homeomorphic to) the same space. Needn't be true for a foliation.

Ah

Foliations are one of my favorite geometric structures

Actually, that's a fiber bundle. A fibration is something only up to homotopy. Don't go there.

Me too, @MikeM.

Fibration is a generalization of fiber bundles. Has homotopy lifting, I think?

I believe you're correct, @Balarka.

I trust you already set them up to do Wirtinger presentations, for at least baby cases like Hopf link, trefoil, etc?

No, @Balarka. Remember how little Munkres does, although he does do classification of surfaces and a proof of Jordan Curve. We only have a few weeks left, and so I'm not sure how far they'll get.

Right, @MarySTar.

So, back to my original question, if $I\subset xI$, why is $x$ a unit?

of course once nice question is "when is a fibration fiber homotopy equivalent to a fiber bundle?"...

Ugh.

@TedShifrin Yeah. Jordan curves really do not belong in the $\pi_1$ chapter, I think :(

I'm retired. I don't have to think about these things.

Well, @Balarka, it's one of the few readable modern sources for it. I've actually not gone through the proof. I don't know if they will or not.

It's actually easier to see it using homology.

@TedShifrin are there some hypotheses on that?

I don't remember the answer to my question.

I prefer to prove it with smoothness. Done.

The answer is either interesting or yes.

@MikeMiller what do you mean by fiber homotopy equivalent?

I don't know, anon. :)

What could I possibly mean?

I think the answer is yes if you're allowed to change the fiber to something homotopy equivalent (you're not forced to use one of the fivers you have)

of course if you demand the fiber be some specific space the question becomes interesting; it's about the map BHomeo -> BAut

I am not sure what you mean.

It's quite a bit of sledge-hammer to get it with homology, @Balarka, as I recall.

oh well

I still don't know how to prove Schoenflies

Riemann Mapping?

@TedShifrin Well, one has to go through some inductive argument.

Is Schoenflies the one that says you can't have horned circles?

But most proofs use fancy duality stuff, @Balarka.

Like, the interior of a curve is a disk?

Yes, DogAteMy.

I don't think that will work without some sort of tameness assumption about the hounding curve?

(As opposed to a sphere with an Alexander-like thing pointing inwards)

LOL, hounding curve goes well with Dog.

:P

meow

Schoenflies fails in higher dimension, no? I can modify the Horned sphere so that both my components are not balls.

I think.

Don't be catty, DogAteMy.

maybe it works for rectifiable curves and clearly it works for piecewise analytic one's

well, not clearly, but there's little difficulty

Oh, now @MikeM has Balarka's apostrophe disease

sorry about the autocorrected apostrophe

LOL

I need to turn off autocorrect on my phone

@Balarka Schoenflies fails because it's just the wrong statement in higher dimensions...

So, we take $a,b\in I$, and since $I\subset xI$ we have that $a=xb$. Since $I$ is a proper subset of $xI$, so also of $R$, $I$ doesn't contain the identity element, right? @TedShifrin

What would be the correct statement? Replace embedding by smooth/PL embedding?

locally flat

there are no local pathologies in our pitifully small dimension 1&2; the interesting part of Schoenflies is that there are no global pathologies

hmm.

Why are you so preoccupied with the identity element, @MaryStar? Maybe $a=1$, maybe not. Think not.

But when $a\neq 1$, how do cocnlude that $x$ is unit? @TedShifrin

Well, do you see that if $x$ is a unit, then the inclusion holds?

What conditions on $R$ will make that have to be true?

ok, so the previous question is correct; a fibration of finite simplicial complexes is fiber homotopy equivalent to a fiber bundle, and you can in addition demand that the fiber is a compact PL manifold, possibly with boundary

but this theorem appears highly nontrivial

@TedShifrin Rather pathological, but fun: take the "expanding Hawaiian earring", i.e., union of circles of radius $n$ and center $(n, 0)$ (hence all touching at origin) for all $n$. This space is homotopy equivalent to infinite wedge of circles, but not homeomorphic to it.

So upto homotopy the infinite wedge of circles does embed in R^2.

why is that homotopy equivalent to the infinite wedge of circles?

You'd have to give me a minute to recall but the map $X \to \vee S^1$ which takes circles to circles and the opposite map I think constitutes a homotopy eq.

do you really believe that?

Wouldn't Pi end once it gets to atom level of accuracy?

Would this be an appropriate question to ask on one of the sites?

not sure if troll

not trolling

As I said, I need a minute to recall :) I believe they should be homotopy equivalent because $\pi_1$'s of both agree.

Planck says: no more pie!

pi is talking about mathematical circles, not arrangements of atoms

so you believe there's a continuous bijection in both directions, which are clearly seen to be inverse to one another? hmm...

@arctictern ok so isn't there an enigneering limit or universe limit or something

no. we're talking about abstract points in an abstract space modelled using coordinates and abstract number systems (the real numbers). atoms aren't anywhere in the discussion.

@William No. Math, in general, has nothing to do with the physical laws of the universe. $\pi$ refers to a mathematically perfect circle, which can't exist in the real world

the engineering limit is about 3.2, the engineers tell me

@MikeMiller Actually one of the maps I described is not continuous.

@AkivaWeinberger your joking a circle can't exist in the real world

Yup

Because of the atoms

Drat.

Its not continuous at the bad point.

A circle is connected. Anything physical is made of disconnected atoms, @William

Plus, they tend to move around a bit

So, no perfect circles

@AkivaWeinberger couldn't an atom it self be a perfect circle

…It's made of smaller things

can you prove that 1,1/2,...,0 is not homotopy equivalent to a CW complex?

I suppose it's possible that some really elementary thing is a perfect circle

but I personally doubt it. And, in any case, it doesn't change anything about math

$\pi$ is $\pi$ no matter if you can't find a perfect circle to measure it with

thank you!

The "is-ness" of mathematical objects is a different kind of "is-ness" than of the physical world.

@MikeMiller I'd start by trying to prove that if that (call it X) is homotopy eq to a CW complex then it has to be homotopy eq to a subcomplex of [0, 1]. Then use the quotient [0, 1]/X which is the earring, to make way somehow (the earring is not a CW complex).

doesn't make a whole lot of sense but hey why not

@William It's kind of like how we have no problem talking about numbers like $10^{5000}$, despite not really existing in the universe

or Graham's number

or, hell, plain old infinity.

We have lots of infinities. It's weird.

@MikeMiller oh wait

that's compact. compact cw complexes cannot have infinitely many connected components.

thanks I am reminded that I am not particularly good at mathematics. Especially considering most of my questions are considered to be trolls.

eek. upto homotopy.

@William some people get "ardent" about insisting some mathematical objects must actually exist, like tables and chairs do

perhaps we need to send you back to @Ted's topology class

There's a lot of troll-y people ranting against the use of infinity in mathematics

or irrational numbers

I thought Ted was retired?

Yes that was my other question regarded as a troll does .999 = 1

not related

@MikeMiller Yep. Doesn't matter: what I said was right.

@AkivaWeinberger I understand, say the "standard proof" about why $\sqrt{p}$ is irrational-but I must confess, some aspects of the real number system (and uncountable sets in general) leave me a bit uneasy.

what you said about what?

@DavidWheeler Have you learned about Dedekind cuts?

Sure. It's a clever construction.

david wheeler was around when they were invented

If $X$ is homotopy equivalent to some CW complex $Y$, then let $f : X \to Y$ be a homotopy equivalence. $X$ is compact, hence so is $f(X)$, thus $f(X)$ hits only finitely many components of $Y$. But # of components is homotopy equivalence invariant.

@MikeMiller heh

GPhys and I have been working with what are called "unlimited numbers" in something called IST. This isn't the best way to think about them, but you can sort of think of them as "things that are larger than anything else in this proof".

(That's actually a pretty horrible way to think about them. But not too horrible.)

@Balarka Sure. Now do it for an arbitrary totally disconnected Hausdorff space that's not discrete.

I guess you probably don't need Hausdorff.

@MikeMiller that just sounds like profanity to me :P

(not that it should. limits of imagination etc.)

They're not homotopy equivalent to CW complexes, so of course it's profane to speak of them.

mmkay

@AkivaWeinberger You'd have to be more specific for that to be meaningful to me

Anyway, the maps I described from the expanding earring to wedge of circles is not continuous because it's given by shrinking each circle of radius $n$ by a factor of $n$. However if I take preimage of a small nbhd of the bad pt of the wedge of circles then I get union of arcs of bigger and bigger arclengths $\cup A_n$ with $A_n$ an arc around $0$ on each $n$-th circle in the expanding earring. That's not open under the subspace topology.

I thought through this carefully at some point of time, but I apparently I forgot.

@MikeMiller Certainly any arbitrary totally disconnected space has to have either finite or infinitely many path components = points. If it has finitely many points, then it becomes discrete, so it has infinitely many points thus path components. Same argument goes through.

@TedShifrin Do we maybe take $a=b$ ? Then we have that $x$ must be the identity element, which is a unit...

@MaryStar What problem are you trying to solve, Mary?

Can I somehow rescale the coefficients of the first fundamental form to be constant?

@MikeMiller Here's why $\vee S^1$ and the expanding earring $E$ are homotopy equivalent. Take the map $\vee S^1 \to E$ which sends circles to circles. Define the inverse map $E \to \vee S^1$ by taking a small neighborhood of the origin in $E$ (which looks like a double infinite broom) and quotienting it.

These constitute a homotopy equivalence.

I worked all of these out a couple months ago. Needed some time to recall.

My question might not make too much sense. I guess I'm asking how much/in what cases can I control the value of the coefficients of the first fundamental form.

@MaryStar it's not true that when $x\in I$ we have $xI=I$. what exactly does Ted say forces $x$ to be a unit and $xI=R$?

@MaryStar Is $I$ the ONLY right ideal?

@arctictern Why isn't that true?

@DavidWheeler $I$ is the only maximum right ideal.

@MaryStar reverse your question: why would it be true? don't ask "why isn't X true" when you don't have any reason to believe X in the first place! anyway, obvious counterexample: the ideal 2Z in Z, when multiplied by 2, becomes 4Z, which does not equal the original ideal 2Z.

if you want an example where I is the only maximum right ideal, take 2Z/4Z inside Z/4Z

So, it must be $xI\subseteq I$, and not equality, or not? @arctictern

And what happens when $x\notin I$ ? @arctictern

@BalarkaSen I have no idae why you think the previous argument goes through.

Ah. Not compact anymore. Apologies.

I was thinking of too many things at once.

@BalarkaSen I dunno why that's a homotopy equivalence. I agree both maps are continuous.

@MaryStar in general if $I$ is a right ideal and $x\not\in I$, there's no guarantee that $xI\subseteq I$. but if $I$ is the unique maximal right ideal, then by virtue of $xI$ itself being a proper right ideal it must itself be contained in $I$. (every ideal is contained in a maximal ideal.)

How do we know that $xI$ is a proper right ideal? @arctictern

good question :) if it weren't, then it would contain 1, which means I contains a right inverse for x. we can't in general say it's a left inverse for x though, so not sure.

@MikeMiller Good point. I believe this is because a neighborhood of $0$ (call it $U$) in the expanding earring $E$ is contractible, and $(E, U)$ is a good pair.

I am actually not quite sure. Doesn't $U$ have a mapping cylinder neighborhood in $E$?

Seems so to me.

(note that $U$ is contractible because it's something like a deleted double broom space. the limiting line is not there, so we're safe and everything is path connected.)

Why would that mean that I contains a right inverse for x? I got stuck right now... @arctictern

i'll take your word for it

If $xI$ is not proper then $xI=R$, then $1\in xI$ implies $1=xy$ (i.e. $y$ is a right inverse of $x$) for some $y\in I$. But again, I'm not sure I actually can say that $xI$ is proper.

Actually, I am trying to prove that $I$ is an ideal. Am I on the right way by using the other right ideal $xI$ ? @arctictern

I thought you said we were assuming $I$ is a right ideal.

Yes, $I$ is the only maximum right ideal. I want to show that $I$ is also an ideal. @arctictern

you want to show I is also a left ideal?

Yes. Is it correct to use $xI$ ? Or is this way wrong? @arctictern

@MikeMiller Mind checking what I asked above? It might not make much sense, but I tried to strip the problem down to what my trouble was.

I don't know what you mean by "control" the first fundamental form. That's intrinsic to the surface.

@MikeMiller If $X$ is your totally disconnected space then whenever $X$ is homotopy equivalent to a CW complex it in fact has to be homotopy equivalent to a discrete set of cardinality $|X|$, actually. I just realized.

*weak homotopy equivalent, but who cares, $X$ has homotopy type of a CW complex.

Mhm.

What do you mean weak? I'm confused.

@MikeMiller Yes, that's true. But suppose the coefficients were constant, could I change the parametrization of the surface so that they become 1?

In an orthogonal parametrization, at least.

@SanicHodgeheg Warning that I've never thought about differential geometry from the lens of parameterized surfaces. But that should change the curvature, which would be bad.

@MikeMiller apparently that is the way to solve the question, but I'm having the same trouble. I'll check the hypotheses again to see if I'm missing some key information.

Hello all, vaguely mathematical question: I'm a math major going into summertime, and I'm curious what kind of mathematics communities exist outside the local university during summer? Is there way I can stay involved?

dis be math community

anyway, there are various REUs and whatnot you can go to, but it might be late signing up for most of them

I suppose I should add the word "physical" in there somewhere.

@MikeMiller Suppose $X$ is htpy eq to CW complex $Y$. This CW complex has $\pi_n$'s all $0$ except for $n = 0$ where it is the number of path components $|X|$. So take a bijective map $D \to Y$ where $D$ is a discrete set of card $|X|$. This is obviously continuous, and induces isom on $\pi_n$. So $Y$ is weak homotopy equivalent to a discrete set, hence is homotopy equivalent.

I was trying to be precise, since we're applying Whitehead here.

In any case, I don't think it's possible for a non discrete thing to be homotopy equivalent to a discrete thing. Let me check.

@arctictern There's a subtle difference between community members and homeboys.

@MikeMiller OK, so I have $X$ - totally disconnected - homotopy equivalent to discrete set $D$. Let $f : X \to D$ and $g : D \to X$ be the pair of maps that constitute a homotopy equivalence. Then $fg : D \to D$ is homotopic to identity. That means $fg$ has to be identity itself because verticals (in $D \times I$) of the homotopy form paths $I \to D$, which must be constant since $D$ is totally disconnected. $gf$ is similarly identity because $X$ is totally disconnected.

So $X$ is homeomorphic to $D$, contradicting hypothesis.

yup

a bit of a cannon to kill a fly though. stumbled upon this while thinking about a whitehead way to prove that $\vee S^1$ is htpy eq to $E$, but there should definitely be something easier out there.

I wouldn't call it a fly. The only way I know to see that something is not of the homotopy type of aCW complex is to see that it doesn't satisfy Whitehead.

Hmm. I haven't thought about it this way.

Perhaps because most spaces I have dealt with which do not have the homotopy type of a CW complex are extremely bad in the sense that they are not locally contractible, etc.

But even then it is easy to find spaces that are not locally contractible but still have the homotopy type of a CW complex.

crap I am thinking of homeomorphism type again

Boy it's hard to find answerable questions in the algebraic geometry tag

@Balarka Homeomorphism type is even worse; if your space doesn't have any of the obvious local pathologies I have no idea how one would prove its not a CW complex.

it's

Say, a compact manifold...

Ah, right.

I don't really expect an answer to whether all 4-manifolds have CW structures anytime soon.

Do you believe it?

No idea.

Are there 4-manifolds which are not homeomorphic to simplicial complexes?

I probably believe that if you can find a single non-triangulable 4-manifold that has a CW structure probably they all do.

Yes

Non-smoothable is probably the same as non-triangulable here though I would have to think about it

Weird.

Yes, usually it's much easier to be triangulable than smoothable.

Hello! Got some functional analysis questions.

Please take a look if you can help a cat out. Thx

I know when handwriting vectors we have to put arrow on top to indicate that it's a vector. Do do that for $\displaystyle \nabla $ as well, the gradient, since that's a vector too?

@BalarkaSen The open condition for $S$ is necessary here, right? "Suppose $f:A\rightarrow\mathbb{R}$ and $g:B\rightarrow\mathbb{R}$ are functions and $S\subseteq A\cap B$ is open. If $f$ and $g$ are both continuous at $c\in S$, then the function $h:S\rightarrow\mathbb{R}$ defined by $h(x)=f(x)+g(x)$ is continuous at $c$." (wording is my own)

mmm no

I think >.<

yeah I think it's fine for any $S\subseteq A\cap B$