doesn't $e^(e^z)$ contradict Picard's theorem since it is entire but does not have the values 0,1 in its range?

it does have 1 in its range

let $w$ have $e^w=2\pi i$...

ah

Given that soap films minimize area given a boundary, how do bubbles (spheres) fit into the conversation? In one case the film is pulling tight trying to cling to the boundary. In another case, the film is pulling tight encapsulating a fixed folume of air.

However soap films are minimal and spheres are not

@Prototank: They minimize the surface area of all surfaces that enclose a fixed volume. Surely there's some physical explanation for why this is what's ultimately obtained, but I don't know it

@AlexClark Yes

@AlexClark The $m$ is in reference to $\Bbb Z / d_i \Bbb Z$ at the begining, where $i \leq m$, so if there is, for example, and element that lies only in one of the finite groups

(where every other coordinate is 0)

and finite group elements have finite order, but $\Bbb Q$ does not have elements of finite order other than the identity

Which leads to the contradiction

Does that make sense? You have a nonzero element that is multiplied by a nonzero element, which ends up being zero.

There are multiple ways to represent the elements

You can represent them as linearly independent sums, or tuples

morning @AlexW

Evening @Mike. How's it going?

It is definitely not uncommon. Its the same idea to you can look at $\Bbb R^n$ as a vector space of $n$-tuples or the sum of linearly independent parts

It's alright. Making coffee.

You never implemented my IV idea?

Yourself?

Oh, no. I told you, there's more to coffee than the caffeine!

Damn kids, injecting everything they see.

Damn, that sounds good, I am going to make some coffee

Next thing you'll be telling me there's more to beer than getting drunk, @Mike.

Why would I ever say that, @AlexW?

That's also, of course, why people drink whiskey

Hehe. I actually do love me a good cup of coffee, and I'm not much for alcohol. So it would seem I've poorly represented myself entirely, here. =P

I'm good though, @Mike. On Thursday, I realized an issue with a problem I'd solved a while back.

What problem? What issue?

A stupid, silly thing. $fg$ primitive iff $f$ and $g$ are both primitive.

The issue was (partially) borne out of misreading the problem. I read $f$ is primitive if the gcd of its coefficients is $1$.

So to prove $f,g$ "primitive" implies $fg$ primitive, I went by contrapositive. If $fg$ is not primitive (under the definition I gave), then there's some $d$ such that the every coefficient of $fg$ lives in $(d)$. Here was the dumb mistake: I argued that $(d)$ is contained in some maximal ideal $(p)$, whence $fg \in (p)$ implies $f \in (p)$ or $g \in (p)$, so we're done. Of course, that would be assuming that the maximal ideal containing $(d)$ was principal, so that's obviously too strong.

What's the correct definition?

The ideal generated by the coefficients is $(1)$. Of course, in a PID, these definitions are coincident.

Oh, I see. What's the context? This is in A-M somewhere?

Yeah, one of the very first problems. I don't even know why it came to me.

Of course, under the actual definition, the problem is very easy. I was fiddling and fiddling until it hit me that gcds aren't necessarily sensible outside of an integral domain, and that I should probably look at the problem again.

I don't see how to do it with the right definition. What's your solution?

Oh dear, I hope I haven't misstated something or made a mistake. Let's see if I have this right.

@AlexClark That is sort of funny and strange.

Let $f = a_{0} + \cdots + a_{n}x^{n}, g = b_{0} + \cdots b_{m}x^{m}$.

This will take forever to write out explicitly, but the idea is this. Say $fg$ is primitive. The coefficients of $fg$ are sums of products of $a_{i},b_{j}$. Each coefficient appears in the product at least once. So if there's an $A$-linear combination of the coefficients of $fg$ that equals $1$, then you can just group the $a_{i}$'s to show $f$ is primitive. Same thing for $g$.

@AlexClark Glasses seem to work okay for me, not sure if it will help with skipping over words though :P

Maybe you are trying to hard @AlexClark

To read

Its like, if you read each letter carefully, instead of the words and ideas you won't "see" the sentence @AlexClark

Maybe you do need your eyes checked, idk

What does this say:

$ \Huge {E}$

Hmm, interesting

Just so you know my vision is even worse than you are pretending you vision is

Yes, couple years ago I was being shown that eye vision card, and the doc just showed the giant E, and I thought it was three lines of letters

Oh, I don't think he was surprised, I am sure they see that kind of stuff all the time. My mother was surprised though

I was a little surprised

@AlexWertheim I thought we were trying to prove that $f,g$ primitive implies $fg$ primitive? It seems like you proved the converse.

Its sort of weird how you mind tries to force a picture, so you end up seeing three lines and letters where there are none @AlexClark

What does this say? $\dfrac{\bar{\Xi}}{\Xi}$

Iff actually, @Mike. But now I'm running into issues with the second part - I don't think my idea works as well as I thought. Let me get back to you :)

Were you thinking of going into finite group theory @pjs36

?

@PaulPlummer Thanks for asking, I don't think so. Truthfully, I don't have the slightest clue about what I'd like to get into, besides "everything"

Haha, I guess that is sort of how I feel about things, I like just about everything @pjs36

Nevermind, missed context

@MikeMiller Indeed, whoever the story was about/told by, with "This notation sucks"

right, I just missed Paul's joke about the E above

Ah, gotcha

@PaulPlummer Yeah, basically! I worked on some out-there stuff for my thesis; building combinatorial polytopes out of posets. It has its roots in algebraic geometry, so I'm trying to see if I can figure any of that out

Cool. I wish I remembered more detail, at the JMM this year I saw something like constructing certain combinatorial geometric objects out of things like posets, and it also had its roots in algebraic geometry. Sadly that is about as good as my memory gets... @pjs36

If you look at $a_2x_2=0$, that is the contradiction

Because they are both nonzero, but multiply to be zero, which does not happen in $\Bbb Q$

(as it is a field)

@AlexC: Here's the point. If $\mathbb{Q} \cong \mathbb{Z}_{d_{1}} \times \mathbb{Z}^{n}$, then some element of $\mathbb{Q}$ looks like $(a, 0, 0, \ldots, 0), a \in \mathbb{Z}_{d_{1}}$. This element has finite order, since every element of $\mathbb{Z}_{d_{1}}$ has finite order. No nonzero element of $\mathbb{Q}$ has finite order.

Another way of saying this is that if $\mathbb{Q} \cong \mathbb{Z}_{d_{1}} \times \mathbb{Z}^{n}$, then $\mathbb{Q}$ has a subgroup which is isomorphic to $\mathbb{Z}_{d_{1}}$, which is torsion. But $\mathbb{Q}$ is torsion free.

Cool, glad to hear it! :D

@PaulPlummer Interesting, I'm having a look now... Thanks for the heads up

Oh you found it off that incredibly vague description? @pjs36

Well, I found the JMM 2015 website; I hadn't heard of JMM before, heh

Looking at the schedule now

And that is a HUGE schedule! I'm like 10% through it...

it's the largest math conference in the US

unlike more specialized conferences, where one might expect to see every talk, that's not even close to feasible here, where there are 20 or so at any given time

nor is it desirable

Definitely; it was crazy big! I think I found the one that would be at all related, @PaulPlummer, it would be this talk

Although there are a few other candidates

Maybe, although maybe it wasn't posets.... it could have been just something that made me think off them, or the speaker mentioned them.

@pjs36

@MikeMiller Did you go to JMM?

went once a couple years ago to present a poster

@Mike: What do you think of this? Suppose $fg$ is not primitive, i.e. the ideal $\mathfrak{a} \subset A$ generated by the coefficients of $fg$ is not $(1)$. Then $fg = 0$ in $A/\mathfrak{a}[x]$. If $f = 0$, say, then the coefficients of $f$ all lie in $\mathfrak{a}$, and we're done. Suppose $f \neq 0$ and $g \neq 0$.

Then $f$ is a zero divisor in $A/\mathfrak{a}[x]$, so there's some $\alpha \neq 0$ in $A/\mathfrak{a}$ such that $\alpha f = 0$. Then every coefficient of $f$ is contained in the ideal quotient $(\mathfrak{a}:\alpha) \subset A$, which can't contain $1$ since $\alpha \notin \mathfrak{a}$.

gotta type out the whole mathfrak i guess

Sorry, bad habits from my own TeX macros. =P

seems reasonable. the following seems a little more elementary: if $fg$ is not primitive, pick a maximal ideal $\alpha$ (too lazy to type out mathfrak every time) that contains the coefficients of $fg$. then $fg = 0$ in $(A/\alpha)[x]$; but because $A/\alpha$ is a field, $(A/\alpha)[x]$ is an integral domain, and one of $f$ or $g$ must have its coeffs in $\alpha$, and thus not be primitive.

I made a macro \sl := \mathfrak{sl}, turns out that \sl is some old deprecated command, still used by some of the packages I had loaded, caused a lot of problems and me scratching my head at math mode errors, when I had no math mode in document

it's essentially what you said, except i avoided the ideal quotient, since it's not something I'm used to thinking about. :)

Nice! :) That's actually what I wanted to do originally. For some reason, I got held up with something stupid, for reasons I can't quite understand now. Oh well, very nice. :)

@AlexW: I think it's probably good to think about the ideal quotient. I read part of Cassels-Frohlich's algebraic number theory book once, and had a lot of trouble understanding it precisely because I wasn't comfortable with lots of the ideal operations they did.

@Mike: I guess it is good practice. I suppose it's also nice that it avoids the implicit appeal to Zorn's lemma... ;)

Why should you avoid appealing to the most well-known result in mathematics?

I don't believe in invoking the Axiom of Choice unless I absolutely have to. Like, say, if I only have one orange left in my fridge, and I'd really like two.

:D

lol'd @AlexWertheim

Howdy @Cbjork

How goes it @Kaj? Your last day of classes must be coming up soon.

Hey @KajHansen

(If it hasn't happened already)

hello @KajHansen

Already happened @AlexWertheim. I have a topology and French final on Monday, then complex on Tuesday

Hey there @Rememberme . do we know each other?

@AlexClark Did you press the button?

Well that's exciting. Good luck! How do you feel about topology?

Indeed @AlexClark. But I don't have a Reddit account.

@AlexClark I pressed it the first day thinking that it would be a one day thing. Silly me.

@AlexClark, do we know each other?

@KajHansen Do we know each other?

@AlexClark I love that 60s flair

@AlexClark You changed your picture

Oh cool. What happened to Incurrence? @AlexClark

Murdered....

he is enjoying algebraic topology @KajHansen

Everything makes sense now.

@AlexClark You are like the spy in team fortress 2, always changing form

If every point on my surface is umbilic, how do I know that I have a (portion of) a sphere or a plane?

@PaulPlummer you have played gods of asgard?

@AlexClark Hahaha

@Rememberme No

I saw something from my PDE class of 6 or so posted on MSE. Necer figured out who it was but didn't put much effort into it

I see

Let me think @Cbjork

The proof is pretty ugly, @Kaj

Well, Gaussian curvature would be constant and necessarily $\geq 0$.

(At least the proof I have on hand, given in the awful book by Oprea)

I thought I remember classifying surfaces with constant non-negative Gaussian curvature was fairly easy, but I could be wrong?

It's been a long time.

@KajHansen I'll think abut it. @AlexWertheim The proof isn't too bad, I hope, because it's a hw problem

from a long time ago

How is umbilic being defined? The definition I have is that the principal curvatures are equal at every point. It's part of the theorem to prove that this means that the curvature is constant everywhere.

Once you do that, I agree it's not bad.

@AlexWertheim that definition is the one I have. I've proved that part

I see geometrically why it's true

It's a pretty common trick, I think. You have to use the constant curvature and the normal vector to show that every point on the surface is equidistant from some point.

If you take the derivative of some fairly well cooked up expression, it should fall out. I can't quite remember what it is at the moment though.

I should clarify that when I said ugly, I didn't mean difficult, but just that I didn't feel that the ideas involved were very elegant.

Ok more thoughts: Every direction is principal, and so $II_p(V, V) = \sqrt{K}$ (in particular, it is constant for all $p$ in the surface. Then let's see...

Oh wait, @Cbjork, combine constant Gaussian curvature with Theorem 3.6 from the text.

I think I got it

@KajHansen yes that definitely works, but it can be simpler

If a surface has only circles as its geodesic curves, then must it be a sphere? @Cbjork, @AlexWertheim

@KajHansen hmmm... let me type out my proof

I suck at differential geometry. Ugh.

That's almost certainly untrue, @Kaj :)

I can't remember. That's a reasonable enough sounding statement, but I have no idea how the proof goes, if it's true. I need a refresher on my differential geometry.

I'd love to see a counterexample @AlexWertheim. I want to post on main, but it looks too much like a homework problem, and I have little to contribute to my question as far as work of my own goes.

I just realized I have shown that the normal is in the opposite direction of the surface without proving that the surface has constant magnitude.

@TedShifrin I am sure those are not standard basis vectors and i am not working with change of basis formula...so is my method right now...due to the following condition ?

a surface in $\Bbb R^3$?

or just an abstract surface with a metric?

Definitely $ \mathbb{R}^3 $

because $\Bbb{RP}^2$, with the metric it inherits from $S^2$, also only has circles for its geodesics

And @kaj 3.6 uses the hw problem I'm doing

but there are lines as geodesics for any constant curvature metric on any other compact surface

if you demand it be embedded in $\Bbb R^3$ I think I might believe your claim

It's in R3.

@MikeMiller, do you think that'd be an unreasonable question to ask on main, given that I don't really have the background/memory to provide any personal insight into the question?

Nevermind, I think I was being not too bright. $ \textbf{n} $ is a unit vector so I have it for free.

nah, it'd be a good question

@Kaj: if every geodesic is a plane curve, then the surface is part of a plane or sphere.

@AlexWertheim, oooh. Can you give me a source?

Ah, found by googling.

don't see how that helps

@MikeMiller, circles are planar curves?

@Kaj yes

Yeah, I know. I'm saying doesn't that help...

no

define planar curve and you'll run into trouble

@Mike: for the problem? It probably doesn't. I just was trying to remember the statement that I knew that concerned geodesics which characterized planes/spheres, that's all. I found it, so I sent it to Kaj.

@MikeMiller, a curve whose torsion is zero?

because if your definition says every circle in $\Bbb R^3$ is a "planar curve", as is everything homeomorphic to a line, then every surface is part of a plane or a sphere

which seems problematic

@KajHansen look at prop 2.4 on page 15 of the text

k, one sec

Yeah, that's what I was referencing above @Cbjork

@Kaj: Is it clear what I'm talking about now?

I'm not sure @MikeMiller. We take "planar" as equivalent to zero torsion in this course.

If we add the assumption of constant curvature, then every zero torsion curve is part of a circle.

@Kaj: Why does every circle, embedded in $\Bbb R^3$, have zero torsion?

@MikeMiller, every such circle (centered at the origin, say) can be parametrized by $v_1\cos^2(t) + v_2\sin^2(t)$, where $v_1, v_2$ are appropriately chosen orthogonal vectors of equal magnitude. We can work out the Frenet frame from there?

@AlexWertheim, to answer your question from a long while ago, I'm nervous about topology. There's an intimidating amount of terminology to remember. The exam is cumulative, and we've covered a lot of material as well. For example, we blasted through filters, ultrafilters, and applications of them in the last 2-3 lectures or so amidst doing other things.

I'm not too fast with LaTeX, so I just wrote out my proof of all umbilic points means you have a sphere (or plane), starting with knowing that the shape operator is always a scalar multiple of the identity. Here is my proof (imgur link): i.imgur.com/hnrBD3B.jpg

@Kaj: OK, to be more explicit: you want to say "If a surface has its geodesics all planar curves, we know it's part of a sphere or plane. By hypothesis, our geodesics are circles, and circles are planar." But it's not true that every closed curve (i.e., circle) in $\Bbb R^3$ is planar.

For instance, pick one that's not contained in a single plane. :P

@MikeMiller, I agree that not all closed curves in $\mathbb{R}^3$ are planar. I do say that all such circles are planar, and in particular they lie in the plane spanned by $v_1$ and $v_2$ from the parametrization above.

What does all 'such' circles mean? Circles that are geodesic on some surface?

@MikeMiller, circles in $\mathbb{R}^3$ I mean to say.

Do you not consider the trefoil knot a circle?

I think... knot...

Then your definition of circle is stupid.

Wait, let me look up how Shifrin defines "circle" in his geometry text.

No. It doesn't explicitly define "circle", but it strongly implies that a circle is a closed curve with nonzero constant curvature and zero torsion, where "curvature" and "torsion" are defined here: en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas

So just because something is topologically a circle doesn't mean it's a circle in this context.

If that's your definition of circle, then I agree your proof is sound. It gives rise to a much more interesting question, the one I thought you were asking: if every simple closed geodesic on a surface in $\Bbb R^3$ is (homeomorphic to) a circle, is the surface necessarily the sphere? Or less generally, if every closed geodesic is periodic, is the surface a sphere?

I've got no idea how to answer that. I think it's true.

Oh very nice. I hadn't thought of that version until now, now that I understand where you are coming from.

How do you define "periodic" @MikeMiller ?

Whoops, yeah. I think Kaj and I have similar ideas of what was meant by a circle. Not that that means my suggestion was necessarily helpful.

@AlexWertheim: Your suggestion gives a full proof :)

Oh, neat, so it does! (sorry, had to catch up on the intervening text)

@KajHansen: Just that every geodesic, considered as a map from $\Bbb R$, descends to a map from $S^1$.

@Kaj: that sounds tough. I have every confidence you'll do quite well though. :)

So that we're demanding that every geodesic, not just the closed ones, are the images of circles, not lines.

can someone discuss this problem?

I want begin this $F(x)=\int_{1}^{x}f(t)dt$, then $F'^3(x)\ge F(x)$

this inequality I can't deal it? can you someone following How to do?

@robjohn,I hope you can see it

@MikeMiller Shouldn't that be obvious by enumerating how many Riemannian 2-manifolds there are? (especially after knowing that every 2-manifolds can be equipped with a Riemannian metric, and knowing all the 2-manifolds upto homeo/diffeomorphisms).

The geodesics depend on the metric. The space of conformal classes of metrics on any given compact surface is finite dimensional, but is far from discrete. And (I think) geodesics depend on more than just the conformal class of the metric!

oh, right. the metric is the problem.

Right. There's no such thing as doing 2-dimensional geometry just by enumerating the cases.

Well, at least outside of lucky cases. I guess it works for some questions. But you see the point.

right, right. i can, like, equip the any surface with a buckload of different metrics. topologically, they're the same thing, but can be vastly different analytically.

i haven't really thought about geodesic metric spaces for a while.

no one technically needs analysis to know about geodesic metric spaces, @AlexClark. :) I had learned them in the context of geometric group theory, something totally algebraic/geometric.

@Mike i think the correct topological idea to do that problem would be to see if every loop in this weird manifold of mine can be homotoped through the circle geodesics. Then I know that this manifold is simply connected, and there is only one such 2-manifold.

Weird idea, though, probably won't work.

I have been doing algebraic topology lately, from Hatcher.

I didn't "finish" D-F, it has a lot more than I can digest one at a time. Done the algebra I need, I guess.

I don't know what you're suggesting. I guess you're talking about the problem that "all geodesics are circles" implies sphere.

@MikeMiller Yeah.

someone discuss this problem?

@MikeMiller Well, recall the homotoping-through-geodesic thingy I was talking about when I was doing basic fundamental group stuff. I don't know how to formalize it, but if one can prove that any loop $\gamma : S^1 \to M$ in the Riemannian 2-manifold $M$ can be homotoped to a point via sliding along the geodesics, then you're done.

Think about the sphere.

Also, even if you proved that it was simply connected, it's not clear why you're done. Perhaps the plane could be given a wacky metric under which all geodesics are circles.

Huh.

Perhaps compactness is needed somewhere.

No, I don't think so.

@MikeMiller What is the loop in $S^2$ you have in mind which one cannot homotope through the geodesics (i picked the standard metric).

Any geodesic? Geodesics are great circles.

You're going to have a piss-poor time if you want to contract a loop through great circles.

Yes, they are. Now if you're basepoint is on one of the intersection of all the great-circles, then you can clearly slide through the great circles so that it contracts, right?

i.e., you slide each point of the loop individually to the basepoint via sliding along the great circle.

I think we had this discussion when you were first talking about it, and I didn't buy it.

ok, i guess i am not seeing your argument. it's too many things to do anyway.

What you're talking about now is not what I was suggesting a counterexample for; I misunderstood what you meant by homotoping through geodesics, but now I remember.

The problem with your new argument is that, a) I don't think you can always homotope through geodesics. b) I don't see why doing this would show you're simply connected.

i was thinking about something like straightline-homotopy, yeah. that's kind of homotoping through geodesics in the plane.

@MikeMiller fair, that's why i said "too many things to do".

Right, I understand the motivation. I don't think you'll be able to do it in general.

But yeah. I'm not thinking about it much now.

me neither

hello

hi

what's happening here?

ok.

@AlexClark you an undergrad?

where?

You can delete the comment once I've seen it.

Ah ok. You should have Massoud as one of your lecturers yea

@AlexClark Wonderful guy. I really like Massoud.

He's also really really like pro.

There's also Peter McNamara.

@AlexClark You should come for this: sites.google.com/site/masoudkomi/mooloolaba

I asked Tony for funding, atm he said it's just for within Australia.

Well you can just show up right

Why don't you go ask Massoud tomorrow

Yea. FYI, I did my undergrad just down the road at ANU :D