Anyone can help with my question above?

put one new vertex in middle with 6 edges coming out

can do so that you cut tiangles

oh but 3 new

I don't think that would work.

Hey guys!

hey

h

Oh wow a squad is forming very quickly, it seems

.

+

@Balarka thanks I keep forgetting to put a period at the end of my sentences, I know you're always there when that happens. :)

..

!

Oh oh this is getting intense

.

Heh, time-traveling punctuation correction ^

> How does NASA organise a party? They planet.

...

!

I can't

(And thus starts Beethoven's fifth)

...!

no symphony before ninth

no number before 9

the first meme song ever

I have no idea how the 9th goes

FREUDE

FREUDE FREUDE FREUDE

FEIGHELEIN FEIGHLEIN FEIGHELIN

Wait so this is a symphony so it's like an hour long

What part is the good pa–

oh wait I know this part don't I

Oh this is Ode to Joy

yup

Random question of the day: Why does movies like to depict alien objects as Platonic solids such as cubes?

Easy to draw maybe

I suppose it probably depends on what era of movies

i always thought it was reductive to think UFO's look like round discs

If I were an alien building space craft, what should it be shaped as?

Yeah but apparently farmers saw weird disks in the air or something

It was flying and they couldn't identify it, so they came up with the super-descriptive name

aka frisbees

does the function log(log(log(x))) ever crosses the x axis?

@Trey yes

@Trey Solve log(log(log(x)))=0 and solve

Set it equal to zero

@user202729 No idea

First thought is to throw V-E+F=1 at it

ok, gotta go

fast

I tried, but I know that only proves that there exists an vertex with degree < 6, and nothing more.

@user202729 There exists a vertex with degree < 6 other than one of the corners?

That's what I'm trying to prove/disprove. But I can't.

Hm... Well, I guess one problem is, technically, you have one solution, which is to draw no new vertices at all

You just have the original three vertices and edges

Technically, "all vertices other than the corners" have degree >= 6... because there are no such vertices.

one way to approach this, perhaps, is that in the usual V-E+F=1 statement, you've got no restrictions beyond it being planar. here you have constraints on the 'external' vertices

No, that is not what I want. A nontrivial example.

Right

The naive approach would detect that trivial solution, and wouldn't realize it's not a valid solution

which is why it didn't work

what I'm wondering is if one can break V-E+F up as (V-E+F)_internal + (V-E+F)_external

i'm not convinced that works, though

Is $\mathbb{R}^2$ - x-axis homeomorphic to 2 copies of $\mathbb{R}^2$?

Yes.

@KevinDriscoll

Similarly, $\Bbb R$ minus a point is homeomorphic to two copies of $\Bbb R$.

You just need to show that an open ray (such as the interval $(0,\infty)$) is homeomorphic to $\Bbb R$, and the map $f:\Bbb R\mapsto(0,\infty)$ defined by $f(x)=e^x$ provides that homeomorphism.

Is it right that I can cover $\mathbb{RP}^2$ with 2 open sets, each of which sits completely within one of the usual charts of $\mathbb{RP}^2$, ie the set $x_i \neq 0$ for some $i$? And then letting those 2 sets be $U$ and $V$, $U \cap V$ is $ \{ [x_1, x_2, x_3] : x_2, x_3 \neq 0 \}$ (choosing the open sets $x_2 \neq 0$ and $x_3 \neq 0$?

And then, $U \cap V$ is $\mathbb{R}^2$ with one axis removed

Ooooooh wait, there's a mistake there. My 2 open sets don't contain the line $[x_1, 0, 0]$ so I missed something

Your algorithm covers it with three charts

The standard "affine opens"

@BalarkaSen Ok so I have 2 ideas. 1) Take 2 of my chart open sets and union them and the leave the third as it normally is 2) OR make the opens sets all the $[x1, x2, x3]$ with $x1, x2,$ or $x3 < \epsilon$ and then everything else

Do you have any advice about how I should choose between my ideas

I like your second idea. $U = \{(x : y : z) | x < \varepsilon\}$ and $V = \{(x : y : z) | x \neq 0\}$.

Ok ya that is an easier way of doing essentially the same thing

Mhm

So you have to figure out how these look like, and what $U \cap V$ looks like

Ya it doesnt line up quite so neatly with the charts I know

so I have to think a bit

Okay

There's clearly some art to this

True :)

But it actually reminds me a lot of doing some calculations in physics

The full calculation is too hard. And there's LOTS of ways of splitting it up into smaller calculations. But an arbitrary way of splitting it won't be very helpful. One has to think about about what a good way is

Quite correct. In fact there's a general procedure to this

But it doesn't quite get apparent in the de Rham context

After you're done with this schtick I might tell you what cohomology is with $\Bbb Z$ coefficients, in the singular sense

Where does it get more obvious? Singular Cohomology?

Precisely so

The kind of decomposition you want your space to have gets quite apparent as you work with singular cohomology. The thing is called a "CW decomposition"

And once you have it, the cohomology calculation becomes a basic busywork. You can program a computer to do it for you

That's why it's a useful tool; it can be computed purely algorithmically

Unlike homotopy groups or something

it's closer to PDE stuff in that respect

@BalarkaSen This $U$ is real weird. Its an annulus around the equator of a sphere, but antipodal points are identified. So for example rather than rotating by $2 \pi$ to get back where you starter, you can also rotate by $\pi$ and then go down through the equator. Which means if I cut along the equator, and rotate one of the annuli by $\pi$ then glue it back along the cut equator..... that makes it seem like its just $S^1 \times I$?

damn you're close

@KevinDriscoll that actually sounds itself like something that'd show up in physics

It almost sounds like the spin group

Here's a thought experiment. Take the annulus around the equator of a sphere, mark antipodal points $x, -x$ on the equator.

o wait, spin group requires rotating by 360 deg, my bad

@Secret $4 \pi$ to get back where you started

Take a small vertical line (an arc on the sphere) at $x$

start with a nematic liquid crystal (order parameter space is a director in RP^2) then find some reason to forbid the director field from pointing too much away from the equator

and a small vertical line at $-x$

if you go from $x$ to $-x$, what happens to those arcs?

and what happens after quotienting by antipodal action?

hmm, can there exists wavefunction which flip sign when we rotate it by half a phase?

@BalarkaSen Is it a Mobius Band? Because there you can go all the way round, but you can also go part-way, cross through the equator and be back where you started?

Yup

Well shit. I was just thinking earlier, "I don't know how to do de Rham cohomology for non-orientable stuff"

lol

Whats the letter for a mobius strip?

$\mathbb{M}$?

ugh

Oh well if its not thats what it is now

jesus

your notation sucks

Maybe for people who already have something for that letter

I do not!

:thonk: is the best notation

$\mathbb{MOBIUS}$

@BalarkaSen Actually did you mean $x < \epsilon$ because I think Ive been thnking about $\lvert x \rvert < \epsilon$

ya that's what i meant

Ok, good. Didnt think the other one was open

basically all this is saying is that the normal bundle of RP^1 in RP^2 is a moebius strip

and an epsilon-neighborhood is diffeomorphic to the normal bundle

so

And a Mobius strip with the equator removed is just... $S^1$? Because now you can't go through the middle

it deformation retracts to the boundary $S^1$ of the strip, yes

but it's diffeomorphic to $S^1 \times \Bbb R$

Right, ya I meant $S^1 \times I$

mhm

@Semiclassical im going to use this

lol

@BalarkaSen So there are no never-zero 2-forms on the Mobius Strip. So that means you need at least 2 2-forms to generate all of them. Because the first will be zero somewhere, so you need a second thats not zero there.

Oh and this is actually an open mobius strip, so its not compact either

Ok we're gonna have to do things recursively. One can split the Mobius strip into another mobius strip and an $S^1 \times I$

This is neat.

Actually that may be a bad idea. Instead I can split it into.... 2 squares

@KevinDriscoll Right, you can't have a nonvanishing 2-form on it

@KevinDriscoll And do M-V on that? Also good idea

Yep thats the plan

Brilliant Twilight Zone episode

[7:56]

always loved Twilight Zone

$H^0_{DR}(\mathbb{R}^2 \cup \mathbb{R}^2) = \mathbb{R} \oplus \mathbb{R}$?

@BalarkaSen How do you feel about $H^k_{DR}(Mobius) = \mathbb{R}, \ k=0,1$ and $0$ otherwise?

That's correct

Ok, I'm getting better at the diagram chase

:)

Also its really easy when you can cover the space with 2 open sets homotopic to $\mathbb{R}^n$

ya

@BalarkaSen Is it right that since $H^1_{DR}(\mathbb{R}^2) \cong 0$ and $H^1_{DR}(Mobius) \cong \mathbb{R}$, this map $i^* - j^*$ seems like it has rank 0 since a 1-form on the Mobius loop has to have whatever covector you put at $(0, x)$ the same as what you put at $(2 \pi, -x)$ but $S^1 \times I$ has no such restriction?

So you're basically looking at the map $H^1(U) \to H^1(U \cap V)$ because the map is zero on the other factor, right?

just the restriction of a 1-form on U to U \cap V

Yea thats right

Ok yea, and $U \cap V$ just cuts out the equator here

right so $U \cap V$ is Moebius minus equator

Can we do the same loopy trick again?

I remember that I saw on some users profile a link to his own text about real analysis. But I cannot find it now. (I should have saved it back than.) Does anybody know who this might have been?

Im not sure if this is right but the 1-form that integrates to $2 \pi$ on $S^1$, this guy could be the pullback of a 1-form on the Mobius strip by the inclusion map. But then we can't homotope that $S^1$ to the an $S^1$ in $U \cap V$ because it would have to cross the equator somewhere it seems

Right!

I have tried searching "real analysis" site:math.stackexchange.com/users in Google, but it mostly returns user profiles of users who have the (real-analysis) tag among their top tags.

have you thought about $2S^1$ though?

by that i mean

the loop which goes twice around the equator

Oh, better idea was to look at my browser history and look there for pages containing "math.stackexchange.com/users". After a lot of scrolling I remembered which name rings the bell - it was Aloizio Macedo.

Sorry for disturbing, I hsould have figured this one myself.

@BalarkaSen To me it still seems like that guy has to make an 'x' somewhere so it will have to cross the equator twice

Oh wait maybe not....

@KevinDriscoll :)

draw a pic

@BalarkaSen Oh okay actually no, you can push the right half of the first go-round up, then connect it to the beginning of the 2nd go-round on the bottom, then keep the 2nd go-round on the bottom and push the beginning of the first up of the left to complete the loop

yuppers

I'm not sure what I should conclude from that though.

so take a form $\omega$ that integrates to $2\pi$ around the center circle of the moebius strip $U$

that generates $H^1(U) \cong \Bbb R$

now, $H^1(U \cap V) \cong \Bbb R$ too, 'cuz $U \cap V$ is a cylinder

if you pullback $\omega$ to $U \cap V$, and integrate it along the center circle of $U \cap V$, what do you get?

$4 \pi$

bingo

?

so if you choose the basis for $H^1(U \cap V)$ to be

the form $\omega'$ which integrates to $2\pi$ along the center circle of the cylinder $U \cap V$

then your map is basically $\omega \mapsto 2\omega'$

basechange; that's $\Bbb R \to \Bbb R$, multiplication by 2

an isomorphism

Ok, so the general idea is we want to start with closed forms that integrate to non-zero values on $U$ and $V$ because these guys form a basis for the de Rham cohomology there. And then we want to try to homotope the paths you integrate those guys over to paths that have closed forms that integrate to non-zero values on $U \cap V$.

yup

basically this is the homology-cohomology duality

to every homology class (in this case homotopy classes of loops) there is a dual cohomology class (form (upto an exact form) which integrates to 2pi over a certain loop)

@BalarkaSen This seems like its a problem..... because then I get a sequence $0 \oplus \mathbb{R} \to \mathbb{R} \to H^2_{DR}(\mathbb{RP}^2) \to 0$

yes, so the first map is an isomorphism.

and if the image of the first map is 1-dim, then then kernel of the 2nd is 1-dim, but the 2nd map is also onto, so the $H^2_{DR}(\mathbb{RP}^2) \cong 0$ but its compact so it should be $\mathbb{R}$

no, no

compact orientable manifolds have top cohomology R

Oh ballsacks

what you computed was exactly right

So yeah $H^{\text{top dim}}_{dR}(M)$ is a good invariant for the orientability of $M$

You have now proved completely cohomologically that RP^2 is not orientable

Yep says right here in my notes, "closed connected oriented"

dagnabbit

It's cool; you see this happening in principles rather than in theory

@AkivaWeinberger hello

@KevinDriscoll Do you know for which $n$ is $\Bbb{RP}^n$ orientable?

@BalarkaSen Ya it has to be that since the antipodal map is orientation-preserving for odd $n$ then $\mathbb{RP}^{2m+1}$ is orientable

yup

try to prove it cohomologically

@BalarkaSen Is $H^1_{DR}(\mathbb{RP}^2) \cong 0$?

yep

a pity, that's what you get for working in R coefficients

if you were working in singular with Z coefficients, you'd get Z/2Z as the answer

the cyclic group of order 2

but R kills all torsion

so

Very sad

much sadness

So what are the conditions where you get the orthogonality sum rule for the de rham cohomologies?

Sorry, what's that?

I thought there was some realtion like $\sum_{i=1}^n (-1)^{i-1} dim(H^i_{DR}) = 0$ or something like that

oh no haha

that's very not 0

Modulo small errors, it's the Euler characteristic

More precisely, $\sum_{i = 0}^n (-1)^i \dim(H^i_{\text{dR}}(M)) = \chi(M)$

Oh ok right, so its $0$ for odd-dimensional guys

and it ahppened to be $0$ for the Torus because its genus 1

Yeah very true

It's an interesting question to ask which manifolds have $\chi = 0$

Precisely the ones which admit a nowhere zero vector field, by Poincare-Hopf theorem, you could say (assuming your manifold is closed orientable)

@KevinDriscoll I don't know how to prove this without Poincare duality though

@BalarkaSen Youre in luck, Poincare duality was the last thing we covered. Which is why right after I wrote what I wrote i was like "o wait thats wrong"

I was reading some lecture notes earlier and I mustve just misread something

Ahhhh

Nice

My brains in 'absorption' mode for this final, so sometimes facts that are not actually facts get scooped up somehow

When's your final?

Dec 14

Aha

Mayer-Vietoris is the alst thing thats actually on the final. And we didnt have a homework assignment on it, which is why I spent todays tiem doing this stuff

Coolio

I was actually going to suggest you to compute cohomology groups of the surface of genus $g$

But maybe not today?

Yea I think I am done for today. I am going to do a complex prokective space tomorrow because thats much harder for me to visualize

and then I might be done with cohology via M-V for now

but I may come back to it

okie dokie

I'll tell you a story after you compute H^* of complex projective space

especially CP^2

In more seriousness though:

How will induction work in an ultrafinite universe?

hi

Any suggestions how to show that one can find a continuous projection onto a subspace with finite codimension of a banach space?

So finite codimension of $U \le X$ means $dim(X/U) < \infty$

And I can continuously project onto $X/U$

Oh, and $U$ is closed. Forgot to mention.

Hello! Could someone of you take a look at my question about teh convergence of a double series: math.stackexchange.com/questions/2559736/… ?

why are most people on this chat talk about cohomology stuff that I don't understand

PS I still know nothing about functional analysis

Can't you project on every closed convex set $A$ (linear subspaces are convex) by sending $x$ to $w\in A$ such that $||x-w||=\inf\{||x-v||:v\in A\}$? @brot

@Secret Because cohomology is spicy af.

@Alessandro I have that theorem for closed subspaces of Hilbert spaces. Not for Banach spaces though. Looking at the proof, it uses the parallelogram identity.

@Narcissusjewel Hey

You were learning some sheafy-flavored complex analysis right?

I was the guy who knew no complex analysis a few days ago :P. But yes, I would like to see this sheaf theoretically very soon. Mainly I just want to consider sheaves of smooth, holomorphic and meromorphic functions, and relevant duals hopefully.

I've just started Forster Riemann surfaces today too.

Ahh ok. I was wondering if you'd be up for learning Forster with me

Oh lol

That I am :D.

Fantastic. I have a chunk of the first chapter of Forster wrapped up

But didn't make much progress afterwards

Very nice. I was hoping to do the first chapter within the next month or so, but it's not my sole focus.

I think 10 and 11 might rek my plans though.

9 might also. I was thinking I'd look at Bott and Tu also.

I've been told that the best way to see why Serre duality morally holds is by this approach.

(Among other results, like the vanishing of the space of global sections for a coherent sheaf on a projective space)

I think you said something (approximately) about the Galois group of $\Bbb C(z)$ being covered here. What did you mean by that?

Is the difference between two divergent series divergent?

@MaryStar Depending on how you mean to take a difference, how about the difference of a divergent series with itself

@Narcissusjewel Ah I see.

I mean this difference: $\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}\left (\frac{1}{2k}-\frac{1}{2(k-1)}\right )$ @Narcissusjewel

Well that's also another thing I wanted to understand better. Basically if $X$ is a Riemann surface, you can realize it as a branched cover $p : X \to \Bbb{CP}^1$. Now, that gives an inclusion of the field of meromorphic functions $\mathscr{M}(\Bbb{CP}^1) \hookrightarrow \mathscr{M}(X)$

But $\mathscr{\Bbb{CP}^1}$ is really just $\Bbb C(z)$. So that's basically an extension of $\Bbb C(z)$

And if you think about the "holomorphic deck transformations" of this branched cover, that gives a topological way to interpret $\text{Gal}(\mathscr{M}(X)/\Bbb C(z))$

Hey everyone!

Hey

This uniquely defines the finite natural numbers without nonstandard elements

@BalarkaSen Thank you for this primer. I have to go in a moment, but did you say you were also following Szamuely's notes?

There were some sketchy plans regarding that which did not materialize into being because nobody seemed particularly interested

I can't get myself motivated enough to do algebra

I need peer pressure

I was hoping to read this also: websites.math.leidenuniv.nl/algebra/GSchemes.pdf

But I'm not ready quite yet in many areas. I thought I'd attack my ultra weak point of complex analysis first, and Riemann surfaces seemed a nice excuse to do this.

maths chat nowadays: cohomology, or CW complex, or algebraic geometry, or differential geometry, none of which I understand in finite time

@BalarkaSen I can pressure you into doing commutative algebra if you want to :P

@Narcissusjewel Hey. That's pretty good.

Bookmarked for future (ab)use

@BalarkaSen ;').

Apparently galois orbits are the only way to understand some parts of scheme theory intuitively. shrug

That's why I was considering it. But I must leave now. Talk later :).

@Alessandro but being a sneaky boi, i will most likely squeeze through the peer pressure by interpreting the algebra in terms of varieties

Isn't that the point of commutative algebra?

Nah. The point of commutative algebra is to overcomplicate and overobscenize beautiful pictures into symbolic pieces of nonsense

/jk

Does the series $\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}\left (\frac{1}{2k}-\frac{1}{2(k-1)}\right )$ converge?

Hello chat!

Hi!

I'm still dealing with this exercise: Let $\mathcal{F}$ be a sheaf on $X$ and define $\mathcal{F}^w(U):=\prod_{x\in U}\mathcal{F}_x$. Show that this defines a flasque sheaf $\mathcal{F}^w$ on $X$.

It is clear to me that $\mathcal{F}^w$ defines a sheaf on $X$. I am struggling with the flasqueness of it, i.e. to show that the restriction maps $\mathcal{F}^w(X)\to\mathcal{F}^w(U)$ are all surjective.

Given some $s_U=(s_x)_{x\in U}\in\mathcal{F}^w(U)$, I would need to "extend" $s_U$ to some $s_X\in\mathcal{F}^w(X)$ simply by choosing ANY $s_x\in\mathcal{F}_x$ for all $x\in X\setminus U$.

This would be sufficient, because then $s_X$ clearly maps to $s_U$.

However, I do not see why I can always choose such $s_x\in\mathcal{F}_x$ (for all $x\in X\setminus U$). Why can't $\mathcal{F}_x$ be empty in some case?

Or wait... maybe it doesn't matter, cause $\mathcal{F}^w(X) = \prod_{x\in X,\mathcal{F}_x\neq\emptyset}\mathcal{F}_x$.

@BalarkaSen every commutative ring is a finitely generated reduced algebra over an algebraically closed field, right?

yes, affine algebras, and no more

I see Zariski topology and schemes being useful, but if you're seriously using varieties in proofs, then for most of the results you're just proving special cases

and making commutative algebra useless for number theory

I'm fine with looking at what results imply for varieties

@MatheinBoulomenos that's okay for me :)

you're also making it useless for modern algebraic geometry

also okay

then why bother doing commutative algebra? you're missing the point

who's doing commutative algebra? :D

not me

i am reading Forster bro

plain vanilla $\Bbb C$

alg geo over $\Bbb F_p$ has more applications

applications where?

real-world stuff

why should i care?

coding theory

i don't care :P

Any suggestions how to show that one can find a continuous projection from a Banach space onto a closed subspace with finite codimension?

coding theory is what keeps a CD working if it has scratches or you're phone working if you have a shitty signal

good

let it keep working

but that's geometry over $\Bbb F_p$ for you

doesn't mean i have to learn it bro

that useless stuff over "shitty fields"

yep

total bullshit

Trolling aside, $\Bbb R$ and $\Bbb C$ is much much more important in "real world" than "geometry over $\Bbb F_p$" :P Literally all of analysis is built over the former two fields

Hence, all of physics is built over them. A wavefunction doesn't take values in a finite field.

It is true there are applications of the finite field theory in a lot of computer scientific worlds, however, true. I don't know much about coding theory but I know encryption

Stuff like ECM/ECPP uses elliptic curves over finite fields

BCH codes are a common example of applying finite fields

what's a BCH

Bose–Chaudhuri–Hocquenghem

Ah

Basically, they're error-correcting codes where you work with polynomials over a finite field

I see

Hello, can someone help me : math.stackexchange.com/questions/2559858/…

@BalarkaSen p-adic analysis is a thing

@LeakyNun Hello