Mathematics - 2017-03-24 (page 3 of 5)

Fawad

06:05

Wait,whose birthday is today?

Happy birthday @arctictern

Balarka Sen

Hi @AlexanderGruber

Alexander Gruber

Hi there @BalarkaSen

Balarka Sen

long time

Alexander Gruber

true

are you a grown up yet?

Balarka Sen

Not officially, but I like to think I grew up a little since we last talked.

Alexander Gruber

06:16

Well well well

Good going

Balarka Sen

:P

skill patrol

Ted has you on the accelerated plan :P

Alexander Gruber

haha

Balarka Sen

How's life?

Alexander Gruber

It's busy

research is pretty consuming atm

Balarka Sen

06:19

Ah. Are you still teaching machines how to learn?

Alexander Gruber

More or less.

I haven't gotten anything new written unfortunately though

been too much hustle and bustle with the life this year, have been working a lot but not being very productive

Balarka Sen

I know that feeling. Well, I'm sure you'll get things done :)

Alessandro Codenotti

Hi chat

Balarka Sen

Heya.

Alessandro Codenotti

How was chemistry?

Balarka Sen

06:24

Hasn't happened yet. It's going to be pretty bad

Alessandro Codenotti

Hm, I see and I doubt it

arctic tern

I am going to ask a homotopy question on main

hopefully it is not trivial

skill patrol

Why not go to their room?

Balarka Sen

Do link us when you're done.

Alexander Gruber

@arctictern What's with that automorphism group answer?

Balarka Sen

06:25

@skill Homotopy theory is a rather frightening chat.

skill patrol

True dat pal.

arctic tern

since I don't have the background to know if my question is trivial, I'll stick to MSE over MO. too much material in my motivation section to fit it into the homotopy room without hogging, although maybe I'll link there.

@AlexanderGruber what? I can think of two things you might be talking about, but neither is one I would guess you would reference without further context...

Balarka Sen

@AlessandroCodenotti Nah, I am forgetting everything I know and don't know. But I also don't care a lot since the rest of the things went good.

Mike Miller

@arctictern i'm watching for it

skill patrol

@BalarkaSen most of them are pretty nice guys though.

Balarka Sen

06:30

I agree. but it's scary nonetheless.

Alexander Gruber

@arctictern From about a month ago. Looks like good content, it'd be a shame if people couldn't see it

arctic tern

the one I deleted on the question with the bounty?

Alexander Gruber

Right.

arctic tern

I was going slowly insane alternating between thinking it was a nice answer and meaningless gibberish

and guilty the OP might dislike it and feel swindled out of their bounty

link it and I'll undelete

Alexander Gruber

@arctictern here you go

arctic tern

06:33

unbaleeted

Alexander Gruber

i'm all about those species man

We don't see enough of that on MSE

arctic tern

yeah, there's like one guy who regularly answers those questions

marco reidel or something approximately shaped like that

also, I never understood the discussion of the 15-puzzle in my group theory class, or after. it wasn't until I learned about groupoids that it suddenly made sense.

Alexander Gruber

I learned about them from Kerber's book Applied Finite Group Actions

alright alright, got my latest bounties up

Here they are if anybody's interested 1 2 3

3

I'm outy for the night y'all.

skill patrol

Cya pal, take care :-)

Tim The Enchanter

06:57

Happy birthday Arctic.

arctic tern

07:09

@MikeMiller @BalarkaSen @TedShifrin Okay, asked.

Balarka Sen

You can probably write S^7 as S^4 semidirect S^3 in different ways than the obvious Hopf bundle. Eg, Milnor's fibration in his exotic 7-sphere.

arctic tern

that'd be weird. I don't think it affects the question though since I'm only caring about the composition factors, not how they fit together necessarily.

Balarka Sen

Ah, alright.

arctic tern

wonder if there's a (numerology) tag...

nope

Balarka Sen

lol

I upvoted, btw.

Alessandro Codenotti

07:24

You could create it. But that doesn't mean you should :P

Balarka Sen

Huh. I just realized that my Hopf bundle comment, if works, can be made into an answer for this.

Danu

12:30

Does anyone here have access to e-books from the AMS? I'm trying to get a look at a chapter from "A celebration of the mathematical legacy of Raoul Bott"

Semiclassical

13:33

@Danu Our university only seems to have it in print at our math library.

Akiva Weinberger

14:00

@BalarkaSen How do you say "thank you" in Bangla?

(I'm thinking it would be nice to thank some local merchants in their native language)

nbro

any help for this question: math.stackexchange.com/questions/2200017/…?

Fawad

Q)Express 29 as a linear combination of4147 and 10672.

Answer is weird,can someone help

SteamyRoot

I'm going to guess that $29$ is the gcd of those numbers.

Fawad

@SteamyRoot yes

How should I come up with 551 and 9×58 ?

SteamyRoot

14:17

Extended Euclidean algorithm

In arithmetic and computer programming, the extended Euclidean algorithm is an extension to the Euclidean algorithm, which computes, besides the greatest common divisor of integers a and b, the coefficients of Bézout's identity, that is integers x and y such that a x + b y = gcd ( a , b ) . {\displaystyle ax+by=\gcd(a,b).} This is a certifying algorithm, because the gcd is the only number that can simultaneously satisfy this equation and divide the inputs. It...

Fawad

@SteamyRoot thanks!

GPhys

Hello @AkivaWeinberger

Secret

15:02

[A series question from the maths club I have been to]
The following is the original question:

Find the sum without using generating functions
$$\sum_{i+j+k=17}ijk$$

(NB I already looked at the answer, thus the following is a new version of it)

Find the sum without using generating functions nor the official answer
$$\sum_{i+j+k=17}ijk$$

Current thought in progress: trying to inscribe a sphere of radius 7 into a lattice of points

Almost forgot: i,j,k are integers

And of course, extended questions:

find the following sums:
$\sum_{\sum_{i=1}^n a_i=r}\prod_{j=1}^na_j$

$$\sum_{\sum_{\sum_{\dots _{\sum_{i=1}^3 a_i}}}}k$$

NB, I actually don't know if a sum with nested indices make sense

Maple SE - Area 51 Proposal

15:36

Hello guyz I am not going to say anything obvious

arctic tern

lol

skull petrol

lol

Secret

16:02

[to be done]Find $f$ such that $f(transcendental) \neq 0$ and $f(algebraic)=0$

Preliminary thought $\sum_{p \in \Bbb{C}[T] - \{0\}}[p(x)]^2$

However this is too complicated

thus more thought is needed

sorry typo: switch $\neq$ and = around in the criteria of $f$

SteamyRoot

What do you mean, find such $f$ ? Just define $f$ like that?

As in, take the indicator function of the algebraic numbers or so

Secret

find an $f$ in terms of polynomials $p$. The idea is to try to construct an indicator function for the algebraic numbers such that it is expressed in terms of polynomials

Alessandro Codenotti

I feel like I'm missing something obvious, any hint on how to prove that a finitely generated abelian group has a finite torsion subgroup?

SteamyRoot

Because such group is isomorphic to $\mathbb{Z}^n \oplus \mathbb{Z}_{p_1} \oplus \mathbb{Z}_{p_2} \oplus \dots \oplus \mathbb{Z}_{p_k} $ ?

with the $p_i$ prime powers

Alessandro Codenotti

That's the next thing to prove

SteamyRoot

16:11

Oh

Well, I guess idea is the same: it's abelian, so you just have to consider the order of the generators

torsion subgroup will be the subgroup generated by all the generators with finite order?

Alessandro Codenotti

Ah, no, wait, the next thing is that such a group $G$ is isomorphic to $\Bbb Z^n\oplus T(G)$, but that's the same thing modulo specifying the structure of $T(G)$

@SteamyRoot oh, duh, it's all abelian of course, product of finite order elements has finite order

Ok thanks, I should be able to fill in the details

BAYMAX

Any good resource for studying Principal Ideal Domain ?

like video lectures?

Sophie

16:50

Someone needs to teach Ukrainian farmers how tile circles

Akiva Weinberger

17:28

I just noticed something annoying about homology: the simplex $abc$ is not homologous to $bca$. (They don't even have the same boundary; $\partial(abc)=ab-ac+bc$, but $\partial(bca)=bc-ba+ca$. The problem is that $ab\ne-ba$ and $ac\ne-ca$.)

It turns out not to matter; the above problem can't happen with cycles. That is, if we have two cycles that differ only by even permutations of vertices of some simplices, then they are homologous. This means that the homology we'd get by "declaring samely-oriented simplices to be equal" is the same as the usual homology. It's still annoying, though.

Mike Miller

I don't find it to be, particularly

s.harp

I don't get it, abc is a triangle and bca is another orientation on that triangle (right?), these spaces should be homeomorphic?

Akiva Weinberger

They're homeomorphic, but not homologous @s.harp. Homologous means that their difference is the boundary of some sum of 3-dimensional simplices.

(where the boundary operation is defined so that each term in the boundary has the same order as the original simplex,

s.harp

I see, you are taking them to lie in $\Bbb R^3$ right

Akiva Weinberger

so for example $\partial(ab)=a-b$, $\partial(abc)=ab-ac+bc$, and $\partial(abcd)=abc-abd+acd-bcd$

IIRC

and you can check that the boundary of a boundary is zero.

Mike Miller

17:40

Of all of my complaints with the singularchain complex this is rarely one of them

Akiva Weinberger

@s.harp You can. In the more general case, these are singular simplices, or functions from $\{(x_0,\dots,x_n)\in\Bbb R^n:x_0+\dotsb+x_n=1\}$ to the topological space you're studying.

But you can think of this as formal sums of geometric simplices (with ordered vertices) in $\Bbb R^n$, I guess.

Mike Miller

I'm pretty sure s.harp knows what singular homology is but didn't understand your original phrasing

Akiva Weinberger

Oh, sorry

Right, yeah, didn't realize

s.harp

I wasn't thinking really about the context of singular complexes, thats why I was confused :)

s.harp

17:56

Sometimes I see a "homology version" of the Cauchy formula, meaning if $\gamma,\gamma'$ are $C^1$ curves in some open $U\subset\Bbb C$ that have the same homology class, then $\int_{\gamma} f dz = \int_{\gamma'}f dz$.

I'm a bit confused by that, I wonder if there are any $C^1$ curves that are not null homotopic but have zero homology class

Mike Miller

yeah, this already happens for the twice punctured plane

s.harp

because its homotopy equivalent to the "$8$", ok that was simple :)

Mike Miller

the point is that CS really just wants to say that bordant curves give the same integral - you're just using Stokes

s.harp

because if $f\equiv R(f) dx + I(f) dy$ then Cauchy Riemann implies $df=0$, so indeed $\int_{\partial D} f = \int_D df = 0$

and bordant curves really is the same concept as homologous curves

SoumyoB

18:17

I think I should join the stats stack exchange, though I really don't know much of stats

but I'll probably get more probabilists over there

An old man in the sea.

18:29

Could someone help me with this simple question? math.stackexchange.com/questions/2201628/…

I think I used to know it, but somehow I forgot how to do it.

Thanks

arctic tern

there seem to be more MSE-like questions getting upvoted on MO, judging from their front page atm

SoumyoB

19:01

mse?

arctic tern

the website you're on

SoumyoB

and what's MO?

Meow Mix

MathOverflow

LegionMammal978

19:33

Where do logarithms belong on the standard order of operations? My teacher is refusing to give me a straight-up answer

arctic tern

either log is only applied to the thing directly after log, or you use parentheses around whatever you're taking the log of

LegionMammal978

I'm, but what counts as a "thing"

arctic tern

what is your actual question? give the expression you're looking at or describe the one you're trying to write.

LegionMammal978

Because logs just seem to be like unary prefix operators

arctic tern

yes. same goes for trig functions too.

LegionMammal978

19:37

I'm thinking perhaps somewhere between multiplication/division and addition/subtraction

SoumyoB

I've never understood what's the difference between math overflow and math stack exchange

I mean why the need for another website

isn't it redundant to have the two

arctic tern

it is pretty obvious to me what the difference is. MSE is for lower level stuff. they used to actually be independent things, it was only recently that MO got absorbed into the SE network. (if I understand the situation correctly.)

it takes 2 seconds to make the personal decision to not write down things that are ambiguous, which is a decision that is independent of how exhaustive order of operations conventions are, and yet so much time is spent mulling over said conventions or lack there of

Meow Mix

Hi

Sha Vuklia

Show that for each $A\in L(\mathbb R^n,\mathbb R)$ there exists a unique $c\in\mathbb R^n$ such that $Ax=(x,c)$ $(x\in\mathbb R^n)$. Prove that $\Vert A\Vert=\vert c\vert$.
I don't understand the question, because $A$ is a linear map from $\mathbb R^n$ to $\mathbb R$, so how can $Ax$ be a 2-tuple? What does $(x,c)$ even mean in this sense? Is it some element in $\mathbb R$? Because it should be, given that the codomain of $A$ is $\mathbb R$.

Ted Shifrin

@SoumyoB: Most researchers spend their on-line time on MO. They don't want to waste their time with most of the questions on MSE.

arctic tern

19:50

@ShaVuklia they mean Ax is the dot product of x and c

Ted Shifrin

Ayup.

Meow Mix

Hi @Ted

arctic tern

often it's written using inner product notation, e.g. (x,c) or <x,c>

Ted Shifrin

hi Zach

arctic tern

also hi

Meow Mix

19:50

Should I be using only the tools in the chapter to do exercises?

Ted Shifrin

Hi tern. I gather I missed handing you birthday salutations. I most humbly apologize and so do now.

Of course yes, Zach.

Meow Mix

Alrighty.

arctic tern

I didn't mention it until the day after technically. I don't want to know I'm getting older.

Ted Shifrin

well, mr tern, you're still a baby in my eyes :P

Sha Vuklia

@arctictern ah thanks, and happy (belated?) birthday

arctic tern

19:51

@MeowMix you can also use outside tools for extra edification, but that's extra

Ted Shifrin

Zach, what hadst you in mind?

Akiva Weinberger

Do rotations of $\Bbb R^2$ fix any points of the $\Bbb C\rm P^2$ it sits in (when you extend them)?

arctic tern

the identity is the only one

Mike Miller

how do you want to extend them?

Meow Mix

@Ted It was an exercise from Section 3.

arctic tern

19:54

wait, you mean $\Bbb CP^1$?

Akiva Weinberger

Wait, I messed up the statement

Ted Shifrin

Section or chapter, Zach?

Meow Mix

Ugh, sorry.

Ted Shifrin

No, he's thinking of $\Bbb R^2\subset\Bbb C^2\subset\Bbb CP^2$, I think.

But maybe not.

G'night, @MikeM.

Ah, Zach, if you're thinking about the one I think you are thinking about, you can't give a proof of one belief without later notions. Notice I asked for a conjecture, not a proof.

arctic tern

(err, -I is another one besides I, regardless of which we're talking about)

Ted Shifrin

19:56

$I$ projectively, tern :P

Akiva Weinberger

@TedShifrin I am

arctic tern

although to be honest, I do think some books structure pedagogy in a morally wrong or at least suboptimal way. then I would feel compelled to build and use tools in a different order than the prescribed one. but you can't really have that reaction if you're new to something.

Akiva Weinberger

What are the eigenvectors of real rotations

Ted Shifrin

So you think of the linear map on $\Bbb R^3$ by putting a $1$ in the other diagonal entry, and then complexify and it induces a map on $\Bbb CP^2$.

Akiva Weinberger

(the eigenvectors being complex)

Ted Shifrin

19:56

$e^{i\theta}$, of course.

arctic tern

those are the eigenvalues @Ted :P

Akiva Weinberger

That's the eigenvalue

Ted Shifrin

Oh, I misread.

I'd have to compute.

Zach, while we're hypothetically discussing the hypothesis, do you think it's true without a finite dimensionality assumption?

Akiva Weinberger

Apparently it's $(\pm i,1)$

For all rotations. Weird.

Which means, in $\Bbb C\rm P^2$, the infinite point on the line through that and the origin, $[\pm i,1,0]$, is fixed for all rotations.

Ted Shifrin

Yeah, that's right. Well, you have the same $\Bbb C$ invariant subspace, DogAteMy.

Meow Mix

20:01

@Ted Which one were you thinking?

arctic tern

If a rotation of a real inner product space $V$ has an invariant plane $P$, then it acts as a rotation by an angle $\theta$ on $P$; if $P$ has orthonormal basis $\{p,q\}$, then $p+qi$ and $p-qi$ are eigenvectors with corresponding eigenvalues $e^{i\theta}$ and $e^{-i\theta}$.

Ted Shifrin

$V$ and $V^{\perp\perp}$, Zach.

arctic tern

(Given any isometry of $V$, we can decompose $V$ into an orthogonal direct sum of invariant lines and planes.)

Meow Mix

@Ted Yeah... that was the one.

Akiva Weinberger

This seems to work for complex rotations as well, though

arctic tern

20:04

elements of U(n) are all diagonalizable (to a diagonal matrix in U(n))

Akiva Weinberger

@arctictern You've lost me

Meow Mix

@Ted I think so...

arctic tern

@AkivaWeinberger I don't know which comment lost you.

Sha Vuklia

Would this be a fair proof then?:
Let $A\in L(\mathbb R^n,\mathbb R)$ and let $\{e_1,\dots,e_n\}$ be a basis for $\mathbb R^n$. We can identify the linear map $A$ with a unique matrix $c=(c_1\dots c_n)$, where $c^T\in\mathbb R^n$. If we multiply $c$ with $x=(x_1\dots x_n)\in\mathbb R^n$, we get $c_1x_1+\dots+c_nx_n$. This is the same as $(x,c)$.

Ted Shifrin

Zach: Do you want a hint?

Akiva Weinberger

20:05

@arctictern Oh, never mind, I think I get it

Meow Mix

@Ted Sure.

arctic tern

@ShaVuklia If you know that all linear maps can be represented by matrix multiplication, then yes that's fair. You can do it more abstractly though. If $V$ is a real inner product space and $A:V\to \Bbb R$ is not the zero map, then by rank-nullity $\ker A$ has codimension $1$, so it has a dimension $1$ orthogonal complement on which $A$ restricts to an isomorphism; pick $c$ to be the preimage of $1$ then.

Akiva Weinberger

I was trying to understand this answer:

Sha Vuklia

@arctictern Yea, I'm allowed to use that. Maybe if I mention that if we choose $\{1\}$ as the basis for $\mathbb R$, then the coordinate of $Ax$ is the same as the actual value of $x$.

Akiva Weinberger

5

A: Which one result in mathematics has surprised you the most?

When I was a freshman at the University of Rome I took a course in Geometry in which some projective geometry was covered. I was really amazed by two facts: a Moebius strip sits inside the real projective plane. the group of rotations of the real plane fixes two (complex) lines and all the circ...

He also mentions that the two points lie on all circles, which seems true

Ted Shifrin

20:08

Zach: Consider the set of sequences $\{a_n\}$ with $\sum a_n^2$ finite. There's an obvious inner product.

Akiva Weinberger

(the projective equations being $(X-aZ)^2+(Y-bZ)^2=r^2Z^2$)

which is surprising.

Ted Shifrin

Consider the subspace $V$ of sequences that are eventually $0$. What is its perp? What is its double-perp?

Sha Vuklia

@artic I'm not really familiar with the orthogonal stuff you mention later on:P so I guess I'll stick to the matrix representation then

Ted Shifrin

Zach: Just to be sure — do you know what the inner product of $a=\{a_n\}$ and $b=\{b_n\}$ will be?

Meow Mix

$\sum a_ib_i$?

Ted Shifrin

20:10

Yup. Taking the limit of Cauchy-Schwarz you can prove it makes sense.

And the usual properties are easy to check.

Akiva Weinberger

So, all circles in the complex projective plane contain $[i,1,0]=[1,-i,0]$ and $[-i,1,0]=[1,i,0]$, and these are fixed by all rotations. Wild.

That answer I linked to called them the "cyclic points".

Ted Shifrin

DogAteMy: I think I've mentioned this before, but you will find the book by Pedoe called Geometry: A Comprehensive Course very interesting.

2

arctic tern

since SO(2) is just exponentials of a single right-angle rotation, being fixed by all rotations corresponds to just being fixed by that particular one (or its opposite)

Astyx

Hi chat

Ted Shifrin

Yeah, tern, that's another way of phrasing my statement that there's always the same $\Bbb C$-invariant subspace.

Hi @Astyx

Astyx

20:14

How are you ?

Ted Shifrin

Busy cooking, Astyx :)

I'm about to disappear — my fish mousse is almost baked and I have to work on a chocolate hazelnut torte.

Akiva Weinberger

Also, fun fact, the arc in the bottom-right of this image is not a wild arc (there's an ambient isotopy taking it to a straight line)

Meow Mix

It looks like $U + U^\perp = \Bbb R^n$

Actually, that's most definitely true.

Akiva Weinberger

The one on the top-right is wild, but its complement is homeomorphic to $\Bbb R^3\setminus\{p\}$.

Astyx

Bon appétit @Ted

Ted Shifrin

20:21

@Astyx: It's not just for me. :P

Astyx

Yeah, I read you were having a party yesterday

Ted Shifrin

Yes, Zach ... You'll tie this into the story of projections (in general) later. ... Are you gonna think about the infinite-dimensional example I gave you?

Meow Mix

Yeah.

Ted Shifrin

Cool.

Astyx

Have a good time anyway

Ted Shifrin

20:22

It ended up as a challenge problem on one of my homeworks in Chapter 4, Zach.

LOL, @Astyx: Merci bien. :)

OK, I'm outta here for about an hour. Back later.

Astyx

See ya

Sha Vuklia

For the second part of this exercise: Show that for each $A\in L(\mathbb R^n,\mathbb R)$ there exists a unique $c\in\mathbb R^n$ such that $Ax=(x,c)$ $(x\in\mathbb R^n)$. Prove that $\Vert A\Vert=\vert c\vert$.
We know that $\Vert A\Vert=\sup_{\Vert x\Vert=1}\Vert Ax\Vert=\sup_{\Vert x\Vert=1}\vert (x,c)\vert=\sup_{(x_1^2+\dots+x_n^2)^{1/2}=1}\vert(x_1c_1+\dots x_nc_n)\vert\overset{?}=\dots=\vert c\vert$

What should I do in the last step?

Astyx

Use CS

Before that

And what is that $^{1/2}$ ?

Oh right

Sha Vuklia

oh...

of course

well the norm is the square root of the square of the elements, no?

Astyx

But you don't need the $^{1/2}$ anyway

Sha Vuklia

20:28

haha yea ok

:P

you are right

Astyx

That happens :p

Have you done the first part of the question ?

Sha Vuklia

haha yes, also here :P

maybe 10 min ago

Astyx

Ah right, I'll check that

Sha Vuklia

I've rewritten the proof

Let $A\in L(\mathbb R^n,\mathbb R)$, let $\{e_1,\dots,e_n\}$ be a basis for $\mathbb R^n$, and let $\{1\}$ be a basis for $\mathbb R$. We can identify the linear map $A$ with a unique matrix $c=(c_1\dots c_n)$, where $c^T\in\mathbb R^n$. We have that $cx=c_1x_1+\dots+c_nx_n=(c,x)$, which is the coordinate of $Ax$ w.r.t. the basis in $\mathbb R$. Since our basis is $\{1\}$, this coordinate equals the value of $Ax$.

this is my final proof

oh wait

i forgot to mention the inner product

fixed*

Astyx

Right

Meow Mix

20:40

Hi @Astyx

Astyx

Hi @Meow

How's it going ?

Meow Mix

Meh

Doesn't look like I'll be skipping any math as of now..

Astyx

Oh .. why is that ?

Meow Mix

District policy

Astyx

What do you mean by "skipping math" ?

Meow Mix

20:41

I want more room for other classes, because most of the math classes I need to take are superfluous (that is, I know them already)

Astyx

That's something I can understand

Meow Mix

I'm not allowed to take the AP test per district policy, I have to take the class as well

Astyx

If I were you I would still make the most out of math classes

Meow Mix

LOL sure...

I don't learn anything, sorry.

Astyx

It's not that much about learning

It's about practicing

Meow Mix

20:44

...seriously, the math is extremely easy.

Astyx

I guess so

Anyway, any progress on that lemma I told you about some days ago ? :)

Meow Mix

Sperner's?

Astyx

Yup

Meow Mix

Do you have any hints?

Astyx

I do

How big of a hint do you want ?

Meow Mix

20:48

Little.

Astyx

Count the edges that have one extremity red and the other green in two different manners

It's not actually enough, but maybe it'll help you find the solution

Akiva Weinberger

Remind me: that's the lemma that says that if you triangulate a triangle, and color it with three colors such that each of the three edges avoids a different color, then there's a small triangle with all three colors, right?

Astyx

Yup

Akiva Weinberger

Can I give a one-word hint?

Astyx

Well, that's for Zach to decide

:p

Meow Mix

20:52

Sure

Akiva Weinberger

@MeowMix "Parity"

Benjamin

What is a word for a plane-like curved surface?

Meow Mix

Manifold?

Akiva Weinberger

@Benjamin Specifically in two dimensions?

Euclidean, maybe?

Benjamin

@AkivaWeinberger Sort of. It curves in 3d space, but would itself appear locally flat. To explain, let's pretend that a plane can be any function and is not required to be flat. I mean a plane (see last sentence) formed perpendicular to the x-axis with z=sin(x).

Meow Mix

20:58

That sounds like a manifold.

Astyx

I would agree with Zach, that's a manifold

Akiva Weinberger

Aren't manifolds more general (they can be many-dimensional)?

I'd just say a "surface" (or a "smooth surface")

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$Mathematics$ Mathematics