Mathematics - 2017-03-18 (page 1 of 4)

12:01 AM

RP^3 should admit such a self-diffeom. reflect along a hypersphere to get S^3 --> S^3. This is Z/2 equivariant so quotient down to get RP^3 --> RP^3. Isn't this degree -1?

Daminark

RIP

Mike Miller

Sorry, I'm wrong. Every lens space has one coming from the one on $S^3$.

But eg the Poincare sphere certainly doesn't admit one.

Balarka Sen

Oh, I guess reflection is still Z/p equivariant up there

so my example generalizes

@MikeMiller Interesting. Why is that fact about Poincare sphere true?

Mike Miller

I don't know if there's an easy proof off the top of my head. You could prove that the mapping class group is the same as the path components of the isometry group and work that out. (I think all elliptic manifolds have the same property by the same proof?) Alternatively Froyshov's h-invariant is 1 on P and is negated under orientation reversal.

The last argument is basically the same as using Donaldson's theorem to see that $P \# P$ can't bound a homology ball, but of course $P \# \bar P$ can.

Balarka Sen

Ah, I was thinking along those lines but didn't stop to look at homology cobordism instead of cobordism (P # P of course bounds a 4-manifold, as do all oriented closed 3-manifolds of the world, so that's boring).

Ted Shifrin

12:18 AM

@Balarka's sleep clock is just destroyed.

Whose, you mean?

I get $\infty$ typos on the iPad.

Balarka Sen

yes, even though my overused apostrophe's'ses' disagree

Ted Shifrin

Well, you iz wrong a.e.

Balarka Sen

:P So, how's your day?

Meow Mix

Hi @Ted

Ted Shifrin

martini time ;)

Meow Mix

12:22 AM

I'm having a fruit salad right now.

Ted Shifrin

Better than what you usually eat, Zach.

Balarka Sen

Enjoy

I think my eardrums died a bit while listening to Death Grips ...

Ted Shifrin

@Balarka: Of course you know this, but any submanifold with a hyperplane of symmetry has an orientation-reversing diffeo.

Balarka Sen

I agree.

Daminark

Hey everyone!

Meow Mix

12:27 AM

Hi @Dami

Ted Shifrin

Demonark returneth?

Meow Mix

-eth was the suffix for verbs conjugated in third person, right?

Daminark

Yeah

Ted Shifrin

i thinketh so.

Meow Mix

You thinks so?

Me thinks as well.

Daminark

12:29 AM

Wait... But how.. I... ???

Meow Mix

@Ted Have you heard Balarka's problem

Ted Shifrin

He has so many ...

Meow Mix

about lattice points and $n$-spheres

Ted Shifrin

That sounds like a classical Weyl-type problem.

Meow Mix

That sounds too complicated for my teen brain.

Balarka Sen

12:31 AM

i asked him to prove that for any $n$ there is a circle in the plane which contains exactly $n$ lattice points

Meow Mix

Anyways, like, in $\Bbb R^n$, there exists, for every $k \in \Bbb N$, a $n-1$-sphere containing strictly $k$ lattice points

Ted Shifrin

Oh, different. I doubt it.

Balarka Sen

generalization of that, yeah

@TedShifrin It's true :)

Ted Shifrin

I still doubt it.

Daminark

The real lattice? If so then definitely not

:P

Meow Mix

12:32 AM

no

like, $\Bbb Z^n$

Daminark

I know I know, I was just kidding

Meow Mix

Actually, if you choose any "lattice", that is, linear combinations of some set of linearly independent vectors, but only with integer coefficients, it still holds true

Ted Shifrin

So, if I change the lattice, you'll need ellipsoids instead of spheres?

Oh.

Meow Mix

Right @BalarkaSen?

Because the reasoning still appears to hold.

@TedShifrin If we allowed ellipsoids the problem would be trivial :P

Unless you just mean, for the lattice.

Ted Shifrin

Trivial? Indoubt it.

Meow Mix

12:36 AM

In $\Bbb Z^n$ I meant

Balarka Sen

@MeowMix It should for any countable subset of the plane, I think. It's really counter-intuitive when you wonder about $\Bbb Q^2$ in $\Bbb R^2$ but it should be true.

arctic tern

discrete subset?

Meow Mix

I'm not a topologist, but are all points in $\Bbb Q^2$ limit points?

Balarka Sen

I think you need that, yeah. Let me think for a second.

Right, if you try to blow a ball from a point stuff should keep coming

Meow Mix

The only requirement I had in my proof was countably many pairs of points

Balarka Sen

12:37 AM

So discrete is needed

@MeowMix The point is you needed to do a "minimal distance" argument

You can't do that if you have a circle with center $0$ say, and the points being $1, 1/2, 1/3, 1/4, \cdots$

@arctictern Actually, closed discrete.

arctic tern

given any two $\lambda_1,\lambda_2\in\Lambda$, the centers of spheres that contain both form the hyperplane halfway between them, which has codimension $1$. If we take the union of this for all such pairs in $\Lambda$, we get a subset of measure $0$, so we can pick a center not in that subset. Then the number of points in a ball around that center will only jump by $1$ at a time.

Meow Mix

Ah, alright. So it's kind of like you need to have no limit points at all in the subset?

Balarka Sen

Right, @arctictern.

Meow Mix

puts head down in shame

Balarka Sen

The point is to find a point from which you can order the lattice points by distance by a strictly increasing order

arctic tern

12:41 AM

@BalarkaSen discrete subsets are closed

Balarka Sen

Not always.

I think generally they are closed, though, yeah.

arctic tern

are you not talking about R^n?

Balarka Sen

I am. {1, 1/2, 1/3, ...} is sometimes considered discrete (every point is isolated) even though it's not closed. It's a matter of convention.

See eg this

arctic tern

I see

Meow Mix

@BalarkaSen do you have any ideas for the "on" instead of "contains" scenario?

Balarka Sen

12:52 AM

@MikeMiller Just to verify, would the Donaldson argument go by saying the double of the homology ball P#P bounds would have nondiagonalizable intersection form? That would only prove it's not smooth, though - is there anything forcing the homology ball to be smooth?

Adeek

I figured it out @BalarkaSen

p was just a fibration

Balarka Sen

@MeowMix Not particularly. I think you need to vary the center there. If you fix a center the number of lattice points lying on a circle of radius $r$ are number solutions to $x^2 + y^2 = r^2$ which seem to grow too fast relative to $r$.

Adeek

If $p : E \rightarrow B$ is a fibration where B is path connected then all fibers are homotopy equivalent

this makes sense geometrically too

like intuitively

Balarka Sen

Ok, great, then, @Adeek.

Adeek

hi @TedShifrin

Ted Shifrin

12:54 AM

Weird, I wasn't here.

Adeek

@TedShifrin I am enjoying both algebra and topology

gonna start reading Hartshrone soon btw @TedShifrin

Ted Shifrin

Glad to hear it, Karim :) ... Well, if you start Griffiths/Harris, you can talk to me. Otherwise, not.

Meow Mix

@Ted I looked at chapter 8 for the lulz. I'm just gonna stick with my linear algebra

Adeek

Oh okay I will try reading it along with Gritffitah

along with Hartshrone @TedShifrin

Ted Shifrin

Yes, Zach, proceeding linearly would do you well.

Balarka Sen

12:57 AM

G/H is complex algebraic geometry, whereas Hartshorne is algebraic algebraic algebraic geometry. Probably not a suitable companion?

Meow Mix

You've been around Balarka too much, @Ted

Ted Shifrin

Karim: It's more complex-analytic and differential-geometric :P And I can send you lots of exercises from when I taught a grad course ...

Meow Mix

You've become Balarkafied

Ted Shifrin

Zach, I've been around all of you too much.

Adeek

@TedShifrin that would be awesome

can you send it to me ?

Balarka Sen

12:58 AM

Plan to learn some actual geometry after exams, @Ted, so hopefully I'll get to talk to you.

Adeek

@BalarkaSen I like seeing both pictures

Balarka Sen

@Adeek pictures? it's all algebra man

Ted Shifrin

In Hartshorne's defense, he has lots of pictures ... Whereas Rudin has 0 in all his books put together.

Adeek

I like analytic and algebraic pictures @BalarkaSen

Balarka Sen

i refuse to put algebra and picture on the same sentence :D

Adeek

12:59 AM

our differential geometry prof mentioned you can study algebraic geometry using analytic method and algebraic method

Meow Mix

@BalarkaSen What if you use "not"?

Ted Shifrin

Balarka: You'd be amazed by how many pictures I put in my algebra book :)

Balarka Sen

@TedShifrin Yeah, Rudin is way terse

Ted Shifrin

And I taught with lots of colors.

Adeek

@BalarkaSen no that is not true algebra is influenced from the geometry and geometry influenced through algebra

Balarka Sen

1:00 AM

algebra is influenced through geometry. geometry is influenced through geometry, tee hee

I'm just pulling your leg :)

Meow Mix

My wrist is getting worse, to the point where using a fork hurts

Adeek

yeah @BalarkaSen I think that there was a mathematician maybe Sophus lie who said something like that

Balarka Sen

@TedShifrin Right, of course, your way of teaching things is way different (and, IMO, more efficient) from many people.

ALannister

Hey @arctictern, am I sniffing glue? math.stackexchange.com/questions/2191650/…

Meow Mix

Anyways, I'm off to play a word game. Good bye math.SE

Ted Shifrin

1:02 AM

Zach, stick to spoons. Good luck.

ALannister

Teaching math with colors is the only way to do it.

Meow Mix

How does one cut a chicken cutlet with a spoon?

Ted Shifrin

You have teeth, don't you?

arctic tern

@ALannister That's correct. More succinctly you could write $\ker = I_n+mM_n(\Bbb Z)$

Meow Mix

I suppose I could eat with my hands

ALannister

1:03 AM

@arctictern dang it. Thanks.

Meow Mix

Welp, bye!

Adeek

@TedShifrin btw you know ever since I quite facebook and youtube and those things I noticed my concentration level has sky rocketed

ALannister

You could cut a chicken cutlet with a spoon if it was freshly made, still hot, and the meat was especially tender.

Ted Shifrin

well, I still saw you on FB, Karim :P

Adeek

@TedShifrin but I don't go there often as I used too haha

ALannister

1:05 AM

Or if you threw it in a blender first.

Balarka Sen

Honestly, though, I really don't grok the algebraist's perspective on a lot of things - say commutative algebra. Us mortals prove the Hilbert's Nullstellensatz (an essentially too, especially if one wants to understand the sheaf theory of varieties) in the very beginning and slowly ventures through varieties and eventually proves Noether normalization through a geometric argument (it essentially says any variety admits a finite surjective map to some affine space)

Ted Shifrin

Reverting to babyhood is probably not Zach's desire, @ALannister.

ALannister

Well, to us he is a baby.

Ted Shifrin

@Ali: You're burning the 2 AM oil again?

Balarka Sen

But the algebraists do it right up backwards: prove Noether normalization using some algebraic tools first in a completely incomprehensible language after some fiddling with commutative rings, and then prove Nullstellensatz as a consequence

Really!

ALannister

1:07 AM

Anybody born after the internet was invented belongs in a crib wearing diapers.

Ali Caglayan

@TedShifrin I woke up earlier so I guess its happening

Ted Shifrin

Well, @ALannister, everyone is a baby here compared to me.

ALannister

True. You O.

Naw.

Ali Caglayan

I'm not sure I can wait till July for Game of Thrones

ALannister

I was so pissed when I heard that @AliCaglayan

Adeek

1:10 AM

@AliCaglayan hi

Ali Caglayan

Hi @Adeek

ALannister

Somebody should call up HBO and play the Reynes of Castermere.

Ali Caglayan

How is comm going?

Balarka Sen

Invention of internet was a very major thing which I am not sure I'd regret if it didn't exist.

Adeek

awesome @AliCaglayan we are doing DVR atm

Ali Caglayan

1:11 AM

Do you hate talking to us that much?

ALannister

Yes! Now I remember! @MeowMix does go by another name. He is @ZachHauk son of Bob, the Warden of New York City Maximum Security Penitentiary.

I must be getting O.

Forgetting things.

Balarka Sen

@AliCaglayan No. But I am not sure if what we do classifies as talking.

Ali Caglayan

I suppose

Balarka Sen

Writing a few disjoint sequence of messages in some internet chatspaces was never meant to be communication, was it? It's changing the very understanding of human communication everyday

ALannister

Escape from New York

Escape from New York is a 1981 American dystopian action film co-written, co-scored, and directed by John Carpenter. The film is set in the then near-future 1997 in a crime-ridden United States that has converted Manhattan Island in New York City into a maximum security prison. Ex-soldier Snake Plissken (Kurt Russell) is given 24 hours to find the President of the United States (Donald Pleasence), who has been captured by prisoners after the crash of Air Force One. Carpenter wrote the film in the mid-1970s as a reaction to the Watergate scandal. After the success of Halloween, he had enough influence...

Ali Caglayan

1:14 AM

@BalarkaSen At least it makes us aware of each others presence.

Ted Shifrin

absent

Ali Caglayan

hey @arctictern have you ever seen crystals in representation theory?

arctic tern

like, plethysm, or rep thry of crystallographic groups?

Balarka Sen

@AliCaglayan We are not just giving a cry "I'm here!" from moment-to-moment to our best buddies on the internet. We are also trying to convey our ideas and emotions, but it's completely unclear how much distortion it goes through upon reaching the receptor.

arctic tern

either way, no

Ted Shifrin

1:18 AM

thinks it's past Balarka's bedtime

Ali Caglayan

@arctictern yes plethysm

Balarka Sen

i feign being high by talking garbage philosophy when i'm way past my bedtime

arctic tern

"crystal pleth" would be a nice math blog name

play on words from "crystal meth"

Balarka Sen

lol

Ali Caglayan

nice

Ted Shifrin

1:25 AM

Just what I need, mr tern.

arctic tern

got 83% in paper io

proud of my algorithm

I think a root beer float is in order

@ALannister Oops, look like I misled you.

my bad

ALannister

1:55 AM

What do you mean you misled me?

@arctictern?

arctic tern

the $\ker$ is not all of $I_n+mM_n(\Bbb Z)$

ALannister

@arctictern What is it then?

arctic tern

probably a crazy set

ALannister

Well, it has to be answerable. It's a hw problem.

arctic tern

as Qiaochu said,

> It's not an easy question to answer, so it's quite possible that what you wrote is all that was intended.

ALannister

1:59 AM

So, what I say it's a subset of $I_{n}+mM_{n}(Z)$?

arctic tern

sure

ALannister

ugh.

arctic tern

it's actually defined by a nonlinear multivariable diophantine equation, so not simple

ALannister

Well, the question did say "describe" it.

@arctictern in this guy, Qyburn's answer: math.stackexchange.com/questions/1267777/… His proof doesn't explicitly use that $R$ must be a ring with unity, does he?

arctic tern

you mean qyburn's question?

ALannister

2:07 AM

Yes, in the proof that he gives.

Argh. Nevermind. He does mention $1$.

I'm trying to prove that the product of two principal ideals (x) and (y) of R is the principal ideal (xy) but all I'm assuming is that R is a commutative ring. Not that it has 1.

arctic tern

$\sum r_is_i$ may not cover every possible $r\in R$ if $R$ is not unital

you can't prove that, because it's wrong

ALannister

Interesting.

arctic tern

2Z is a commutative ring, and (4)(4)=(32) there

ALannister

So my prof meant to say that R has unity?

arctic tern

either that or they believe all rings have unity

ALannister

2:11 AM

No, they certainly don't

arctic tern

well, if it doesn't have unity I would call it a rng and not a ring

my rings have unity

Meow Mix

Hi

ALannister

I know, but he doesn't use that terminology

Hello, young Hauk.

arctic tern

(I guess I should call 2Z a rng in order to not be a hypocrite.)

ALannister

Lord of the Rngs.

One rng to rule them all.

Oh please star that last part.

It's cheesy, but it's sooo cool!

The one rng. My preeeecious.

Meow Mix

2:16 AM

Oh, I see the problem now :D

ALannister

Whatchu talkin' bout @MeowMix?

Meow Mix

I realize how to show this

The problem is "Prove that for any $k$, there exists a circle intersecting only $k$ lattice points"

Oh, shit.

Secret

rng
rings
semirings
nearrings
near semirings
pseudo rings
commutative rings
rg
division rings
noncomutative ring
nonassociative ring

domains

Daminark

Ideally we'd have more of an index for this terminology

Meow Mix

hi @Dami

arctic tern

2:27 AM

what's a rg?

Secret

a ring when you don't have an identity and additive inverses. I am pretty sure the maths community find these too unuseful to deserve its own name

Meow Mix

@arctictern Can you help me with something?

Daminark

Wait wha...?

Meow Mix

Actually, anyone

arctic tern

@Secret wouldn't you just call that a semirng?

@MeowMix what?

Daminark

2:29 AM

Depending on the topic, I might be able to help @Meow

Meow Mix

Prove that for any $k$, there exists a set of $k$ integer points which lie on a circle, such that all other points on the circle are not integers

Secret

@arctictern semirings has identities for both + and * (at least if the convention in wikipedia is used), they do sometimes jokingly referred as rigs

Daminark

Uh... The best argument I can immediately come up with is contained here: theproofistrivial.com

arctic tern

@MeowMix I don't actually have an idea for that

Meow Mix

try the integer one

it might be easier

Ok, I'll "Just biject it to a
continuous
manifold
whose elements are
exact
Hilbert spaces"

I proved it for $8 | k$

Secret

2:34 AM

O I almost forgot, some authors only need 0 or 1 in their semiring axioms, thus yes if you structure is only missing one type of identity, then it is still a semiring. "rg" takes this further by throwing away both identities

Another way to think about these things is take two types of algebraic structure (often semigroups) with one operator each, and then add a distributive law to relate them

In most cases you will end up with boolean algebra and lattices, but there are cases where you don't get a lattice

ALannister

A rg is that thing on Donald Trump's head.

Secret

2:49 AM

that should be a wig, not a rg

unless there is a hairstyle called rg

Mike Miller

@Balarka: No, the double is probably just $S^4$. $P$ bounds a manifold with intersection form $E8$, so if it also bounded a homology ball, you'd get a closed manifold violating Donaldson.

But homology cobordism is trivial if you want to work topologically instead of smoothly.

On the other hand every homeomorphism is isotopic to a diffeo in dim 3.

ALannister

@Secret a rug.

arctic tern

someone should invent a sport called homology ball

Mike Miller

i'd play

Secret

How would that be played, given how hard is to compute the homology of something?

and homology does not have as visual meaning as homotopy?

Ali Caglayan

3:03 AM

only when its cohomology and 2am

Secret

O wait a sec, both homology and homotopy can be used to characterise holes?

(I am not so sure about these objects anymore)

Ali Caglayan

3:20 AM

homology can be thought of as characterising 'holes'

Homotopy is about maps from the sphere to your space

for a fundamental group its about loops

in general homology groups are much easier to compute than homotopy groups

Forever Mozart

thought I found a mistake in my new result. almost had a heart attack

Ali Caglayan

and have nice properties like being 0 for higher dimensions

but homotopy groups are not neceserily 0

for example for the 2-sphere we still don't know all the homotopy groups

Forever Mozart

if it's wrong, I quit math

Ali Caglayan

I like to think of homology as the failure of a sequence to be exact

Usually these notions make a lot more sense when talking about cw complexes

which is basically lego for topologists

Forever Mozart

soul youtube.com/watch?v=HCMf66rZEXc

Topologicalife

3:50 AM

Hi.

If $a_n < b_n \, \forall n \in \mathbb{N}$ and $\lim a_n = A$, $\lim b_n = B$, is true that $A<B$ or $A \leq B$?

I think $A < B$

arctic tern

$a_n=-1/n$, $b_n=0$, $A=B=0$

Topologicalife

:D

Where is my fail here?:

arctic tern

$A\le B$ is true, $A<B$ is not necessarily true

Topologicalife

Suppose $A > B$. Then, for $n > n_1$ then $|a_n - A| < \epsilon$, for $n > n_2$ then $a_n - B < \epsilon$, let $n_3 = \max\{n_1,n_2\}$, then, since $a_n < b_n$ we have $A+ \epsilon < B + \epsilon$

Which is a contradiction

Ali Caglayan

An easy one to try is the same sequence but one is an index ahead

The are the same limit but if it is monotone then they have their own order

Topologicalife

4:00 AM

I meant $b_n - B < \epsilon$

Where is my fail there? Uhm

Oh, wait.

I had to suppose $A \geq B$

which is still a contradiction

arctic tern

@Topologicalife how are you deriving $A+\epsilon<B+\epsilon$ from $|a_n-A|<\epsilon$, $|b_n-B|<\epsilon$ and $a_n<b_n$?

Topologicalife

Since $|a_n-A| < \epsilon$ then $a_n < \epsilon + A$

Similarly, I get $b_n < \epsilon + B$

and because $a_n < b_n$ by hypothesis, then $\epsilon + A < \epsilon + B$, i.e: $A < B$

arctic tern

4:16 AM

do you know what an absolute value is?

Topologicalife

Yes.

arctic tern

I am derp.

Topologicalife

More in detail: $|a_n - A| < \epsilon \iff -\epsilon < a_n - A < \epsilon$

So, in particular, $a_n < \epsilon + A$

arctic tern

anyway, how do $a_n<b_n$ and $a_n<\epsilon+A$ and $b_n<\epsilon+B$ give $\epsilon+A<\epsilon+B$?

Topologicalife

if $a < b$, and $b < c$ then $a < c$

arctic tern

4:19 AM

what's your point?

$a<b$ and $a<x$ and $b<y$ does not imply $x<y$

Topologicalife

mm you are right

Thank you, I messed it up :)

So... $a_n > A - \epsilon$, $b_n < B + \epsilon$ and since $a_n < b_n$ then $A - \epsilon <B + \epsilon$

$A < B + 2 \epsilon$

If I let $\epsilon = \dfrac{A-B}{3}$ then...

I get $A-B$ < 0 which is false

now I'm right but I have the same problem.

arctic tern

you can't let $\epsilon=(A-B)/3$ if $A=B$

Topologicalife

right, ty :)

Mm and can I define $\epsilon = (A-B)/3 + 1$?

Yeah and then $\dfrac{A-B}{3}-2 < 0$ which is undefined

Nice, I love maths.

Thank you so much :)

2017

4:40 AM

If there are 6 players and each pair has to play with each other pair then for calculating the total number of games possible why is $$\dfrac{ \binom{6}{2} \binom {4}{2} } {2}$$ the wrong method?

Ah, wait....it seems my textbook was wrong

nvm, this is the correct method ^

Daminark

So there's this 2 week program I'm applying to over the summer called summer school in analysis, and they want a resume

So guess what's about to be created right now?

BAYMAX

You want to create a resume ?

Daminark

Yeah

Evan

7:22 AM

math.stackexchange.com/questions/166648/… Hey folks, why is it that when you move the points on Z, the map is continuous?

user243301

9:22 AM

All users, I have found in YouTube an interesting talk (in spanish) with slides about Cusanus and $\pi$, from professor Jesús Guillera. You can find it typing in YouTube Cusanus y el Número Pi, and see it from the official channel Jesús Guillera. Thus if you know spanish o have a spanish friend you can see it this morning or afternoon. Good weekend.

Secret

9:51 AM

What is an example of a space where it is impossible to place any charts and coordinate systems on it?

Secret

10:06 AM

(For details, I think I am looking for a manifold which cannot be covered by any number of charts)

Aayush Agrawal

10:17 AM

Anybody here know any good Highschool Level mathematics books?

Nothing in particular as long as the content is interesting and picks up from a highschool level

2017

@AayushAgrawal You are in which grade?

Alessandro Codenotti

@Secret what do you mean? every manifold is covered by charts

2017

@AayushAgrawal I don't know what you mean by "highschool level". If you are preparing for competitive mathematics (for example the olympiads) then the books you have to read will be of a considerably higher level than normal school textbooks.

Secret

Well I remember things like a sphere cannot be covered by one single chart, and that it need 2 different charts to be covered

Thus naturally I wonder whether there are manifolds that can only be covered by n charts where n is some arbitrary number

Alessandro Codenotti

Ah, you meant a finite amount of charts

I think a countable connected sum of tori should work then, not sure though, you should ask @Balarka

BAYMAX

10:25 AM

Hi chat , any idea where there is a nice introduction to Manifolds,I am finding this name used in several contexts , papers, directly and I always overead it!

Martin Sleziak

@BAYMAX Perhaps it could also be reasonable to ask this in Geometry & Topology room if you don't get any feedback here.

BAYMAX

ok @MartinSleziak

Martin Sleziak

Or you can simply browse a bit through the Wikipedia article. There is also a question on Quora: What is the best way to explain the concept of manifold to a novice?

BAYMAX

ok

Martin Sleziak

If you are looking for some introductory text, you might have a look a the posts tagged manifolds+book-recommendation.

But I hope that somebody who knows a bit about this topic will give you a better advice. (Basically all I said was: Try to look in the usual places.)

BAYMAX

10:37 AM

thanks@MartinSleziak , ok.

Martin Sleziak

BTW congrats on reaching 1k posts in the main chat room.

BAYMAX

ha ha :), I was unaware of that ,thanks!

Mike Miller

10:53 AM

@Secret @AlessandroCodenotti You only need at most $n+1$ charts for a closed $n$-manifold. They'll have complicated intersections with each other. I don't know what you can say about non-closed things off the top of my head.

I think it should probably also be true for arbitrary n-manifolds...

Balarka Sen

11:16 AM

@MikeMiller Ah, alright.

That makes sense.

@AlessandroCodenotti Like Mike said, you can cover every surface by at most 3 charts. It's interesting to construct it, even for the torus. Maybe take that as an exercise.

(For the torus you can't do it less than 3)

Once you do it for the torus the other surfaces shouldn't actually be hard. It's the same idea.

Alessandro Codenotti

If I cover the torus with 2 open sets at least one of them won't be simply connected, right?

Balarka Sen

Correct. Otherwise van Kampen gets you.

Alessandro Codenotti

They don't necessarily have a connected intersection though

Balarka Sen

Ah fair point. You can fix that; use Mayer Vietoris instead.

Give me a moment. M-V seems to say that there has to be two connected components in the intersection and the boundary map is an isomorphism. Trying to see if I can use that to contradict anything.

@Alessandro Actually it's easier than all that. Your open subsets of torus are both manifolds. Simply connected open 2-manifolds are exactly disks, so that's the same as asking them to be charts.

Alessandro Codenotti

11:34 AM

I should find out what Mayer-Vietoris is sooner or later

That makes sense though

Balarka Sen

I am not sure if my M-V argument works. I can prove that you can't have 2 charts by a cup product argument.

Mike Miller

@BalarkaSen You agree with me you shouldn't need more than n+1 even when you're noncompact, right?

Alessandro Codenotti

It's not hard to cover the torus with 4 open sets following the usual CW-structure, I hope my drawing skills are enough to cover it with 3

Balarka Sen

@MikeMiller If you're noncompact the top dimensional class is zero; so I think you can have less than n+1. But not more.

Eg the torus with a puncture can be covered by two charts I believe.

Alessandro Codenotti

It can, if my drawing is correct

Oh, wait, it's not, what I drew is a sphere with 4 punctures

Actually can't you just take an open disk as a noncompact 2-manifold that can be covered with less than 3 charts?

2017

11:51 AM

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is? My Attempt: I initially assumed card 1 is already placed in envelope 2. So the answer should come out to be derangement of 5 objects. D(5)=44. But ans is 53.

Which cases am I missing out?

Balarka Sen

@Alessandro I meant you can have generically less than n+1.

Alessandro Codenotti

I don't think I'm following

@2017 2 can go into any envelop

Balarka Sen

I mean I think for all noncompact 2-manifold you need 2 charts to cover it. I am a little skeptic about that now though; what about like infinitely torus connected summed up?

2017

Umm, I think card 2 doesn't need to be deranged

Alessandro Codenotti

I agree

2017

11:54 AM

@AlessandroCodenotti Right, it just now struck me :)

Alessandro Codenotti

@BalarkaSen aha, I see now, I have no idea about that though

Balarka Sen

Torus with a puncture (enlarge that to a hole, say) is $S^1 \times I$ with $S^1 \times 1$ glued to $S^1 \vee S^1$ by $aba^{-1}b^{-1}$. Cut along a path $\{p\} \times I$; that's a chart, no?

Or am I daydreaming

Secret

When thinking about the countably infinite connect sum of torii suggested by Alessendro, this is the picture that first came to mind on what the manifold look like:

Transcript for

$Mathematics$ Mathematics