Mathematics - 2016-10-14 (page 3 of 5)

DHMO

16:00

my axioms are your axioms

none of your axioms mention anything like that

a and b and c are generic

Secret

Here's a breakdown

DHMO

@Secret you haven't proved the fourth diagram

Secret

We are given a=b and = is an equivalence relation

what else is missiing?

DHMO

what does an equivalence relation mean?

Secret

its transitive, reflexive and symmetric

DHMO

16:03

none of those properties include a=b <=> a+c=b+c

or a=b -> a+c=b+c

Secret

In algebra for a given equivalence relation, if you can go from left to right after applying a series of axioms, then you already proved $\rightarrow$
Unless the proof in logic require some more rigorous treatment which is normally omitted in an algebraic context (that I am unfamilar of)

DHMO

@Secret which axioms did you apply?

Secret

S is nonempty, definition of the addition operator and equality as an equivalence relation, I guess?

DHMO

5 mins ago, by DHMO

none of those properties include a=b <=> a+c=b+c

Secret

I don't understand, if you add c both sides on a=b, you get a+c=b+c thus showing $a=b \Rightarrow a+c=b+c$? because you have shown you can go from left to right?

DHMO

16:10

what is the axiom allowing you to add c on both sides of a=b?

user116211

@Balarka, I think I comprehended your statement earlier; if $f(k+1)\leq f(k),$ then $x = f(k+1)$ would be in the set $\{f(0), f(1), \ldots, f(k)\}$ which is absurd as $f(k+1)\in S\setminus \{f(0), f(1), \ldots, f(k)\}$ and thus from the Trichotomy Law, discarding the other two cases, it can be concluded $f(k+1)\gt f(k)\,.$

DHMO

@MAFIA36790 Could I look at what you are referring to?

Balarka Sen

that's right

user116211

@BalarkaSen \o/

user116211

@DHMO sure.

user116211

16:13

Go to transcripts.

DHMO

@MAFIA36790 but he/she said many statements earlier

user116211

7 hours ago, by Balarka Sen

Namely, $f(n+1)$ is recursively the least element in the complement of $f(\{0, 1, \cdots, n\})$.

DHMO

@MAFIA36790 thanks

Secret

Uh...

physicsforums.com/threads/prove-that-if-ab-ac-then-b-c.484638

Even the first step of this prove is adding -ac both sides,

why we cannot just add c??

DHMO

@Secret did i see the word "physics"?

arctic tern

16:17

@Secret you can add c to both sides. not sure why you'd want to do that though. you'd get ab+c = ac+c.

user116211

Can I conclude @Balarka $\{f(0)\lt f(1)\lt \ldots \lt f(n)\}~~\forall n$ then?

DHMO

@arctictern he's referring to "prove that a=b <=> a+c=b+c" when he's saying "add c"

Balarka Sen

Yes.

user116211

okay.

DHMO

@Secret it's interesting... the field axioms also do not contain that axiom

@BalarkaSen are you familiar with algebraic structures?

Secret

16:20

I am guessing it is part of the definition of +

Balarka Sen

I don't know what algebraic structures mean

DHMO

@BalarkaSen just something like fields

Balarka Sen

sure

DHMO

if you like, we can just talk about fields

@BalarkaSen is this an axiom? $a=b\ \iff\ a+c=b+c$

user116211

@BalarkaSen You mean a set with one or more binary operations?

Secret

16:22

+ is a binary relation such that $\{\forall a,b \in S, (a,b) \in + \iff a+b\in S\}$

user116211

@DHMO Cancellable operation.

Mary Star

How could we factorize the polynomial $x^4-2x^2-1$ so that at the factorization there is the term $\sqrt{3}$ ?
For example, if we wanted $\sqrt{2}$ instead of $\sqrt{2}$ then we would have $(x^2-1+\sqrt{2})(x^2-1-\sqrt{2})$.

DHMO

@MaryStar Why do you think that such a factorization exists?

Balarka Sen

Fields are cancellative, yes.

DHMO

@Secret what is $(a,b)\in+$?

proofwiki.org/wiki/Definition:Cancellable_Operation

Mary Star

16:23

@DHMO So it does not exist?

DHMO

@MaryStar no, it doesn't

Secret

When writing a binary relations R, we write the ordered pair (a,b) where a,b are elements of some set, $aRb =\{(a,b)\in R\textrm{ some contraints relating a and B}\}$

Balarka Sen

Also, it's not an axiom, but a theorem. The additive counterpart of fields is actually a group, so has inverses.

DHMO

@BalarkaSen how to prove that fields are cancellative?

Balarka Sen

it's enough to prove that groups are cancellative, because $(F, +)$ is a group.

there it's obvious because you have inverses

DHMO

16:26

@BalarkaSen it's not obvious to me, could you please explain?

Mary Star

We have that $\rho$ is a root of the polyomial $x^4-2x^2-1\in \mathbb{Q}[x]$. I want to check if $\sqrt{3}\in \mathbb{Q}[\rho]$ or not? How could we do that? @DHMO

Balarka Sen

$ab = ac$. multiply both sides with $a^{-1}$. $a^{-1}ab = a^{-1}ac$. hence, $b = c$.

DHMO

@BalarkaSen who allowed you to multiply both sides with $a^{-1}$?

Balarka Sen

the binary operator?

DHMO

so you're saying it's left cancellative because it's left cancellative?

Balarka Sen

16:27

huh no

where did I even say that

DHMO

is this an axiom? $b=c \implies ab=ac$

Balarka Sen

it's not an axiom, it's just what being equal and being able to multiply means. understand them clearly first.

DHMO

hmm... could you explain to me please?

Balarka Sen

if it didn't hold the binary operator wouldn't even be a map.

$\bullet : G \times G \to G$ is a well-defined map. if you had a relation $f : X \to Y$ which sent same elements to two different things, that's not a map.

Secret

ok, so that's the point I am struggling to explain, I really need to sharpen up my intuition on maps

DHMO

16:34

@BalarkaSen so it's an axiom?

Balarka Sen

if you want to call it that, sure. it's what a binary operator is, by the very definition.

I would distinguish between definitions and axioms, but I don't really want to talk semantics right now

Secret

35

Q: What is exactly the difference between a definition and an axiom?

I am wondering what the difference between a definition and an axiom. Isn't an axiom something what we define to be true? For example, one of the axioms of Peano Arithmetic states that $\forall n:0\neq S(n)$, or in English, that zero isn't the successor of any natural number. Why can't we defin...

In short, definitions have no truth values, it basically describes how a mathematical object acts

whereas axioms are propositions that are assigned true, from which lemmas and theorems follows from them

One can invalidate any axiom if they want to explore other mathematical structures

e.g. commutativity does not hold for matrix multiplication

and associativity does not hold for lie algebras

(At least to me) one can challenge an axiom, or throw it away, but one can never challenge a definition

DHMO

Alright

Anubhav

17:16

@MikeMiller I wonder, if there is any correspondence between $M$ bundle over $N$ and $N$ bundle ofver $M$ for some non-contractible manifold $M\neq N$.

Mike Miller

Nope. Take $S^1, S^2$. There are two $S^2$-bundles over $S^1$, one orientable, one not. The $S^1$ bundles over $S^2$, on the other hand, are the lens spaces $L(n,1)$.

I think it's true of 3-manifolds that if a manifold fibers over $S^1$ as a $\Sigma$-bundle and over $\Sigma$ as an $S^1$-bundle then it's $\Sigma \times S^1$.

Anubhav

Last statment sounds interesting

Balarka Sen

Oh, I misinterpreted.

Anubhav

I didnt your comment @BalarkaSen

what did you say?

Balarka Sen

Something irrelevant.

Anubhav

17:22

@MikeMiller Is there exists any spacial kind of manifold $M,N$ non-contractible, for which we can say that any $M$ bundle over $N$ is homeomorphich with some $N$ bundle over $M$?

Mike Miller

$M \times N$ :P

Anubhav

No, but I mean for any $M$ bundle, not just trivial

DHMO

@BalarkaSen If I twist a strip by $360^\circ$ and then glue the ends together, and then identify the two edges, would I have obtained a genus 0 surface?

Balarka Sen

He said "any" though.

@DHMO You'd obtain a standard strip.

Just embedded differently in R^3.

Mike Miller

I would guess there's no general theory of such manifolds.

DHMO

17:23

@BalarkaSen How to convert it to the standard strip?

Mike Miller

I basically would expect this to happen only when there are very few bundles of each type.

Balarka Sen

@DHMO You can cut-paste. I can't draw you pictures right now, but you can google that.

Anubhav

Do you have any immidiate example of such kind?

Balarka Sen

That thing is called the "double moebius strip"

Mike Miller

Uh, let me think.

There's a slight problem that fiber bundles are classified by maps into BHomeo(N), and Homeo(N) is really nasty for high-dimensional N.

Anubhav

17:26

Yes...

But I will be amazed if there exists any such pair of manifolds

Mike Miller

Oh, just any pair? I can do that pretty easily. Pick two non-isometric hyperbolic manifolds that each have trivial isometry group. Mostow rigidity implies that $\text{Diff}(M_i)$ is contractible, and hence that all $M_i$-bundles over anything at all are trivial. In particular, the only $M_i$ bundle over $M_j$ is the trivial bundle.

Balarka Sen

Cool.

Mike Miller

If you want to demand that there's a nontrivial bundle, it becomes harder, but probably still possible.

Anubhav

That is actually cool

Balarka Sen

I am going to try to use and abuse this example.

Anubhav

17:30

How does Mostow rigidity implies $Diff(M_i)$ is contractible.

The version I know is a bit different

DHMO

@BalarkaSen What would I produce if I identify the two edges of the strip?

Balarka Sen

Diff M is probably diffeom to Isom M for hyperbolic M

@DHMO Of which strip? The standard one?

Or the moebius (180 degree) version?

Mike Miller

@BalarkaSen I'm sure you know that's not true...

Anubhav

I'm not sure about it @BalarkaSen

DHMO

@BalarkaSen no, the full-twist (360 degree) one

Balarka Sen

17:32

Full twisted strip is the same things as no twisted stip

So you'd just get the torus.

@MikeMiller Sorry, homotopy equivalent, I meant.

Mike Miller

Good.

DHMO

@BalarkaSen but I have no idea how to identify the edges together in 3-space

Mike Miller

@Anubhav There's a version that says that the inclusion $\text{Isom}(M) \hookrightarrow \text{Diff}(M)$ is a homotopy equivalence. I don't know a reference off the top of my head, but google should get you there eventually. It only works for $\dim M > 2$, as always.

Anubhav

Ohk, that sounds interesting too

What is an easy example of a hyperbolic manifold with trivial isometry group?

Mike Miller

I don't know one off the top of my head... there are papers that construct them with given isometry group

Balarka Sen

17:34

@DHMO You probably can't see it without some difficulty. But the full twisted strip has two boundary circles (what you're calling 'edges'). Gluing them is equivalent to glue a tube (no twists!) to it, with gluing each bd circle of the tube to the other.

You'd get something with self intersections in R^3 probably

DHMO

@BalarkaSen why is R^3 so troublesome!

Mike Miller

I think usually you can do that by taking a hyperbolic knot in $S^3$ and then taking surgery on it with a sufficiently large coefficient

Balarka Sen

@DHMO Not enough dimension/degree of freedom.

Anubhav

It sounds horrible now

DHMO

@BalarkaSen is there a way to embed anything on R^2?

Balarka Sen

17:36

Only 1-manifolds (curves) embed in R^2

That is to say, circles and lines (and disjoint unions of them).

DHMO

@BalarkaSen but 2-manifolds would need to have R^4 for the full representation, right?

Balarka Sen

Yes, every 2-manifold embeds in R^4.

Mike Miller

Let $Y$ be a 3-manifold that fibers over $\Sigma$ with circle fiber and over the circle with $\Sigma$ fiber; suppose $\Sigma$ is not the torus. By prop 1.11 in Hatcher's 3-manifold notes we may take one of the $\Sigma$ fibers to be transverse to the foliation by circles. Now consider the map $\phi: \Sigma \to \Sigma$ given by taking a point on $\Sigma$ and following the unique circle passing through it in the positive direction until you hit $\Sigma$ again; let the point you hit $\Sigma$ at be

DHMO

@BalarkaSen do I say that a klein bottle embeds in R^3?

Balarka Sen

It doesn't.

Mike Miller

17:40

$\phi(x)$. Then $\phi$ is a periodic diffeomorphism of $\Sigma$, and $Y$ is diffeomorphic to the mapping torus of $\phi$. There is thus a $d$-fold cover of it diffeomorphism to $\Sigma \times S^1$. I think this is the best you can do.

DHMO

what would happen if I forcibly embed a 2-manifold in R^2?

deostroll

Hi could anyone point me to literature to the history of triangles.

Anubhav

@MikeMiller Let me post this problem in MSE...then you give your answer there...That would be good....Other wise I'll forget everything :p

Mike Miller

The question of which 3-manifolds fiber over both a surface and a circle?

Balarka Sen

I don't know what you mean by "forcibly embed". You can't do it.

DHMO

17:42

@BalarkaSen well, if I forcibly embed a klein bottle in R^3, it would self-intersect

n-manifold embeds in R^(2n)?

Balarka Sen

Yes, Whitney proved that.

Anubhav

Not Whitney actually

Mike Miller

Didn't he?

Anubhav

Whitney said 2n+1

Mike Miller

He later improved it to 2n.

Or maybe immediately. I don't know the history.

Anubhav

17:43

Is that by Whitney, I'm not sure

Balarka Sen

It's called the (strong?) Whitney embedding theorem nonetheless.

DHMO

What would a 3-manifold look like?

Anubhav

No, but the point is, who proved it first?

Balarka Sen

Shrug, @Anubhav.

Anubhav

OHK

DHMO

17:44

Also, some 2-manifolds can embed in R^3

is there any similar statements?

Balarka Sen

Orientable ones do.

DHMO

like, generalization?

Balarka Sen

There are generalizations of Whitney out there, I think.

Anubhav

@DHMO LOok at this link mathoverflow.net/questions/227435/…

Balarka Sen

Letting the dimension of R^k vary with the manifold.

Mike Miller

17:46

I think every oriented k-manifold embeds in R^(2k-1), as does every k-manifold when k is not a power of 2.

Anubhav

@BalarkaSen have a look at that link.

Yes....

DHMO

20

A: Can an oriented closed $n(\geq 2)$-dimensional manifold be embedded in $\mathbb{R^{2n-1}}$

A closed smooth $n$-manifold embeds into $\mathbb R^{2n-1}$ is and only if the normal $(n-1)$th Stiefel-Whitney class vanishes. This is due to Hirsch-Haefliger in dimensions $\neq 4$ and to Fang in dimension $4$. Massey showed that if the normal $(n-1)$th Stiefel-Whitney class is nonzero, then $M...

I only understand one word: "A"

Balarka Sen

Cute result.

Mike Miller

You will learn all of this one day if you want to.

abenthy

@MikeMiller Thank you! I see, so for a compact Lie group G with dim(G) > 0 to show that it contains a subgroup isomorphic to S^1 we go as follows: Since dim(G) > 0, the Lie algebra g = Lie(G) has dimension >0 and thus contains a nontrivial vector v. Now the map f : R -> G, t |-> exp(tv) is a Lie group homomorphism. Its image f(R) is a Lie subgroup of G which is isomorphic to R, hence f(R) is a connected and abelian subgroup of G and its closure is a torus in G.

But I think f needs to be injective, right? Is this the case? Sorry, my Lie group knowledge is still basic :/

Mike Miller

17:49

@abenthy Either it's injective or it's $S^1$, and we're already done!

If it's injective, we take its closure to get a higher-dimensional torus.

Anubhav

@MikeMiller do you think I should post that question to see whether someone actually had some answer ?

DHMO

@BalarkaSen can a double-twisted moebius strip be untwisted in R^3?

Mike Miller

Which question?

Anubhav

M bundle over N and N bundle over M

Mike Miller

I think it's probably too general to get anything interesting other than what I just said.

Balarka Sen

17:50

@DHMO Not by an ambient isotopy of R^3.

Alessandro

good evening

Anubhav

Yes..that is true in some sense though

DHMO

@BalarkaSen ok

Mike Miller

I should probably write my talk today.

Anubhav

What talk?

Balarka Sen

17:51

This should be a consequence of the Hopf link being nontrivial (that's what the boundary of the double twisted strip is in R^3)

Mike Miller

@DHMO That means you can't move it around in $\Bbb R^3$ without it intersecting itself to remove those twists.

DHMO

@MikeMiller yay, something in English

Mike Miller

@BalarkaSen DHMO is not a topologist - it's better to avoid terminology. Know your audience!

Balarka Sen

Sorry about that. I actually had to think for a bit on that question, so didn't remember to explain myself :P

Mike Miller

@Anubhav I'm giving a talk on this to a non-gauge theory specialist audience. The point will be to talk about how it could prove 4CT, the ideas involved in its construction, and how one might proceed from here. It's in 5 days, so I should get off my ass and write it.

Anubhav

17:53

Where? In UCLA?

Mike Miller

There's a series of colloquia called LATop where speakers from UCLA, USC, and CalTech talk about research. Students have their own variant of this, which died off for a bit but we're rebooting. It's going to be at CalTech this time.

Anubhav

Sounds great

Balarka Sen

For some reason I'm feeling a bit feverish today. Weird, huh.

Anubhav

Tell me about a day when you were not ill @BalarkaSen

Balarka Sen

Last sunday.

... of the year before the previous year.

Anubhav

17:58

you should try to become a doctor in future :P

Balarka Sen

Can't happen anymore. Dropped biology.

Anubhav

Dropped a year, and re join with a new ambition of becoming a doctor :P

Any way, take rest...I going to study...

Mike Miller

Title: The instanton approach to the 4-color theorem

Abstract: The 4-color theorem is famous for its lengthy and necessarily computer-aided proof. Recently, Kronheimer and Mrowka debuted a new type of Floer homology: one for embedded graphs in \Bbb R^3, which they hope can be used to give a new proof of the 4-color theorem. I will explain the idea behind this approach and summarize what we know, what we don't, and why it seems promising. No previous knowledge of gauge theory, Floer homology, colors, or the number four will be required.

Balarka Sen

lol

DHMO

@MikeMiller is it published on April 1st?

Mike Miller

18:02

No.

Anubhav

So then what is required ? :P

Mike Miller

Not much. The point of the talk is that it'll be accessible to a general audience of topology grad students.

I do expect most of them will have prior knowledge of the number four.

4

DHMO

@MikeMiller Not many do.

abenthy

@MikeMiller Thank you so much. I do see that for v nontrivial in Lie(G), the Lie group homomorphism f : R -> G, t |-> exp(tv), is nontrivial. But why is f(R) the circle subgroup if f is not injective?

Mike Miller

@abenthy There are only two 1-dimensional Lie groups, $\Bbb R$ and $S^1$.

Note that the homomorphism can't be trivial, or the Lie algebra homomorphism will also be trivial.

So if it has kernel, it's a closed proper subgroup of $\Bbb R$, which is readily seen to be cyclic.

Andrew Thompson

18:07

Someone changed the language of our washing machines to French. That was a challenge.

Mike Miller

Really? That's pretty much english.

Andrew Thompson

I seldom remember the degrees of which I'm washing and only the word used for the kind of wash I want. So I didn't know which one to press.

I think my clothes will still exist when I go get them in 30mins.

Mike Miller

I hope not.

Balarka Sen

18:23

a spanish washing machine would have been more exciting

Andrew Thompson

ohh our alggeo prof has put out exercises. Perhaps this friday night will have purpose.

Mike Miller

solve a major open problem

quickly

Andrew Thompson

One of the mandatory exercises for our calc1-kids is of the form "Solve the integral blablabla in less than 15 seconds."

Balarka Sen

I wanted to print Ted's notes out because that'd have been easier for me to read but my printer is dead

Enjoys Math

19:03

:^) math.stackexchange.com/questions/1968636/…

soft question

how does an amateur mathematician, being disconnected from academia, find what they need on the internet? Also in comments: looking for someone to write a paper with.

So far, no article that I am able to find has spotted this observation about the problem. It's as if they gave up or something or don't like abstract algebra, lol.

It would nice if I had someone to work with. Safety of numbers!

Andrew Thompson

Well, its tough for us to give an answer as you're asking if a buzzword "triggers" peer-reviewed work.

Enjoys Math

Yes, as I have finally found something that I don't want to share with this community, until it is published.

Mike Miller

are you triggered andrewt

@EnjoysMath you should know that nobody is going to steal your work. otoh, if you've found something you find interesting enough to put a ton of time and energy into it, awesome :)

Andrew Thompson

Slightly.

Enjoys Math

I also get booed off the stage and negative support. It's: ask something known or don't ask it all, which doesn't quite make sense.

I am paranoid that I will die, or someone will kill me, all the time (mentally ill; plus I've gotten real live death threats (not psychosis) about 30 times over the past 2 years, from people I knew and strangers to me, then community).

Anyway, that's not really of concern here, so I'll stop there.

Mike Miller

19:12

Sorry to hear that.

Enjoys Math

Yeah, moving soon to San Diego. Where I live in Arizona and have for 12 years, the "hippie culture" is more conservative than the far right. Uncool Gangster Hippies = UGH!

Andrew Thompson

Right. In any case, it is difficult to answer any of the given questions due to the lack of information. You may add more details to your question without fear of your work being stolen, as you could always change your name on math.SE to your actual name.

Mike Miller

San Diego is a lovely place. I hope you enjoy it there.

Enjoys Math

No, I will not share this on math.stackexchange. Too much can go wrong. Not all good advice applies to me.

It will either be brushed under the rug, stolen, voted down into oblivion, shrugged off, said to be obvious, demonstrated by some insane math I won't know for 10 years, and so on... keeping this one secret.

Andrew Thompson

Okay. In any event I do believe that it will be difficult finding a coauthor unless you give slightly more details, but you might get lucky. Good luck!

Enjoys Math

19:19

I will trust that gmail will deliver my letters when I write them, to the journal editors, and I will trust the journal editors to be honest; I will not trust them to be devastatingly overly opinionated against the ideas, as I have never published before. I am not sure how it will turn out. Though, to any one who has wondered for more than a week about the certain open problem, the article should be succinct, elegant & interesting.

If any one is interested in the details, they can check my post history and deduce with probability what I am most likely referring to. That's the most open I can be right now.

abenthy

Can i also argue like this: If $G$ is a compact Lie group with dim(G) > 0, and $g = Lie(G)$ its Lie algebra which has dim(g) > 0. Take v nontrivial in g. Then f : R -> G, f(t) = exp(tv) is a Lie group homomorphism. Then f(R) is a Lie subgroup of G of dimension <= dim(R) = 1. So since v is nontrivial, f is too and it follows dim(f(R)) = 1, thus f(R) = S^1 or R. Does this work?

Mike Miller

@abenthy Yup. I think it's really easiest to say that the image has dimension 1, its closure has dimension at least 1 and is a compact abelian group, so is a positive-dimensional torus.

abenthy

Ah,I see... and the closure of an abelian connected Lie subgroup is again abelian and connected, right?

Mike Miller

Yup.

abenthy

Wow, If I google for "closure of abelian" I get only three search results.

But yeah, I think this all works out :)

Mike Miller

19:27

It's not so hard you need to google it :) Suppose $g_n \to g$ and $h_n \to h$. We know that $g_nh_n g_n^{-1} h_n^{-1} = 1$. Now take a limit of the LHS.

abenthy

@MikeMiller Thank you so much for your invaluable help. Really helped me a lot.

Mike Miller

Glad to. I have to remind myself how to do this once every couple months, so it eventually got ingrained.

Balarka Sen

Thierno M. Sow got the Nobel prize 2016 in mathematics! Well received.

Semiclassical

hi chat

lol @BalarkaSen

Semiclassical

19:47

oh. a simple question that I'm for some reason having a hard time dredging up enough imagination for

Obliv

Anyone know why $\nabla f = \frac{\partial f}{\partial \rho}\vec{\rho} + \frac{1}{\rho}\frac{\partial f}{\partial \phi}\vec{\phi} + \frac{\partial f}{\partial z}\vec{z}$ in cylindrical coordinates? Specifically, where does the 1/radius in the second term come from

I know in cylindrical coordinates it's usually $f(\rho,\phi,z)$ where $\rho$ is the radius, $\phi$ is the angle b/t reference direction & the projected line P on the plane. Z is the height

Mike Miller

@BalarkaSen Your friend MM posted a good paper today.

Balarka Sen

ah, indeed?

Obliv

nvm I figured it out

Semiclassical

consider the surface $z=y^2+(x^2-1)^2$. the level sets for $-1<z<0$ will be concentric circles centered at $x=\pm 1$, and for $z>0$ it'll be concentric ovals

Obliv

19:52

$ds^2 = d\rho^2 + \rho^2 d\phi^2+ dz^2 \to ds = d\rho + \rho d\phi + dz$ then $\frac{df}{ds} = ...$ the thing I put up before

Semiclassical

that's obviously symmetric under a half-rotation, i.e. $(x,y)\to (-x,-y)$

Obliv

wait then isn't $\rho d\phi$ an arc length? it's not really an angle as I thought.

Semiclassical

How would one create a simple example of a surface that looks similar to this, but instead is symmetric through a 120 degree rotation?

Balarka Sen

Maybe something with the monkey saddle?

Semiclassical

never heard of that before. lemme google

ah, hrm. not really what I want. it having global minima is important

Balarka Sen

19:58

I see.

I probably wouldn't be able to write down such a surface, so I'm chickening out.

Semiclassical

heh

I thought initially to just do something like $(x - h)^2 + (y-k)^2$ for three different points $(h,k)$ and add them together. but the sum of three quadratic functions is itself quadratic, so just one global min

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