Mathematics - 2016-09-30 (page 3 of 4)

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5:04 PM

anyone can help on on understandiing the above proof?

@Danu Does it?

@Secret I really don't like his syntax with the $:$ after the quantifiers

apparently $\forall x : A : B$ means $\forall x, A \implies B$

which is a pretty bad notation

Still, I don't see how $\{x:\forall x, x\ni B\}$ and $\{y:\forall y, y\ni f[B]\}$ will mean f[B] and B are empty unless their complement are whatever universal set that contains f[B] and B themselves, but nowhere in the proof seemed to suggest that?

um

I have no idea what those sets mean

oops sorry, those $\ni$ should be $\not \in$

5:17 PM

still, $\{x : \forall x, $ blablabla $ \}$ can't possibly be what you want to write

I am not sure if I understood this line properly in English $\langle \forall y :: y \not\in f[B] \rangle$ "For all y such that y is not in the set f[B]"

no it's "forall $y$, $y$ is not in the set $f(B)$"

@MikeMiller Yeah

"Floer homology" makes me go sigh and get all starry-eyed

Also I want to know equivariant (co)homology

You're easily excited :). Not a bad thing, however.

ok, even if y is not in f(B), it does not necessary mean f(B) is empty as y might be in the complement of f(B) that is not necessary the universal set that contains f(B)?

5:24 PM

no he means forall $y$

ever

Ah I see...

@Danu Personally, I'm terrified.

...and you're like a wizard in my eyes already

Can you imagine how I feel?!

on that note, what does Floer homology do?

hello, i need help why this $$\theta=\{\mathbb{R},\emptyset\}\cup \{A_q=]q,+\infty[\}$$ is not a topology on $\mathbb{R}$

we have that $\mathbb{R}, \emptyset\in \theta$

5:30 PM

it doesn't satisfy the axioms of a topology so it's not a topology

well

I'm assuming that you take all those $A_q$ qwith $q \in \Bbb Q$

so I'm cheating

yes

but since you didn't say what $q$ was...

$q\in \mathbb{Q}^*$ sorry

That matters

if i take $q_1, q_2$ then $A_{q_1}\cap A_{q_2}=]\min{q_1,q_2},+\infty[ $

5:32 PM

With $\Bbb Q$, I don't see why it shouldn't be a topology...

yes

it's stable by intersection

Consider infinite unions @Vrouvrou

(countable will suffice for your considerations)

@Danu Floer homology is one thing, but all those infinity categories

Question: Are there any nice, simple almost-topologies, i.e. things that closed under countable union (and finite intersection), but only fail when looking at uncountable unions?

$\bigcup_{n\in \mathbb{N}} A_{q_n}=\bigcup_{n} ]q_n,+\infty[ =]\min{q_n},+\infty[ $

5:34 PM

wait wait

@Vrouvrou $\max$, really?

what's $A_1 \cap A_2$ again ?

$\min$ is also incorrect :P

@Danu Take the open sets of R to be it's countable subsets?

why ?

5:35 PM

@BalarkaSen hehe

I know the authors learned infty catevories to write that paper. They thank the nlab.

@Vrouvrou Think about infinite sequences

because $\min$ of infinitely many things doesn't always exist

@MikeMiller lol

@MikeMiller eww

5:35 PM

7

Q: Is there a difference between allowing only countable unions/intersections, and allowing arbitrary (possibly uncountable) unions/intersections?

As in the title, I am asking if there is a difference between allowing set-theoretic operations over arbitrarily many sets, and restricting to only countably many sets. For example, the standard definition of an topology on a set $X$ requires that arbitrary unions of open sets are open. Do I lo...

general-topology analysis logic set-theory

I'm going to ask Sucharit this afternoon if we should hold a seminar to learn it or if I should just power through.

@mercio HEY don't spill the beans so easily

nlab sucks

@mercio we can say that it is equal to $]\min, +\infty[$ or $\mathbb{R}$

$\min$ of nothing ?

$]\min{q_n},+\infty[$

5:38 PM

Can anyone tell a good source for learning probaility, especially the expectated values part
I'm stuck whenever there is even small variation in general problems, like unequal probabilities in coupon collector's problem

@Vrouvrou No, wrong. Think about infinite sequences!

@Danu why d'ya laugh

@BalarkaSen trivial example is trivial :P

@Danu i don't know

@BalarkaSen It's a neat trick

@Vrouvrou What don't you know?

5:39 PM

you said nice and simple :)

you fulfilled all my desires* :P

@Danu how to think about sequences

no that's weird

@Vrouvrou So consider a sequence of numbers $\geq 0$ and $\leq 1$. Does it always have a minimum or maximum?

@BalarkaSen <3

@Danu yes they are numbers the max is 1 and the min is 0

5:44 PM

@Vrouvrou Incorrect

Perhaps you didn't understand my question right

i don't know

Consider a sequence of numbers such that each number is larger or equal to zero, and smaller or equal to 1

Must the sequence always have a max or min?

God fucking damnit Huybrechts...

inverse image sheaf

FUCK YOU SHEAF

Why are the corollaries so hard to prove :'(

??????

for example (\frac1n)

Those last few message were not aimed at you

but at my textbook

for me $0\leq \frac1n\leq 1, \forall n$ then the sup is 1 and inf 0

5:47 PM

your textbook is offended and flags your comments

seriously though---I just can't get through any proof without getting stuck for fucking hours

urrrrrgh

that's pretty standard @danu

I agree

@Vrouvrou Hey! Sup and inf? Those are new words you're using! I like them

Do you understand how sup and inf are different from min and max?

@mercio Neh

on the corollary of a proposition

not a serious theorem

yes if sup is in the set then it is equal to max

5:49 PM

I was fine with being stuck on understanding the Euler sequence etc

I usually skip ahead if I can't get through something. Nonlinear learning helps

I get stuck on simple corollaries too

@Vrouvrou So does that sequence you gave have a $\min$?

no

But it has...?

inf

5:50 PM

Great

In fact, it's funny and good that you picked that exact sequence

Now, consider the open sets in that "supposed topology" of yours, given by $(q_n,\infty)$ where the $q_n$ are the elements of your sequence.

What is their union?

$\bigcup]q_n,+\infty[=]\inf{q_n},+\infty[$

or $\mathbb{R}$

Sure, but consider the sequence you gave

So $q_n=1/n$

ok it is equal to ]0,+\infty[

...which solves your question, right?

and it is not in $\theta$

yes thank you

5:54 PM

WHAT

um

nevermind

but if we have $q\in mathbb{R}^*$

@mercio lol

(forgot it was $\Bbb Q^*$)

hehe @mercio

that's the whole point of the exercise

now do the same but with $\Bbb Q$

5:56 PM

if $q\in \mathbb{Q}$ it is a topology

Is it though? What if you take a sequence converging to $\sqrt 2$ (from above)?

how?

$\sqrt 2+1/n$

$1/n$ is your friend :)

yes

the inf is $]\sqrt{2}$ and sup is $\sqrt{2}+1$

We don't really care about the sup, I guess, but yeah that's right.

Then the union would be $(\sqrt 2,\infty)$

6:02 PM

yes and $\sqrt{2}$ is not in $\mathbb{Q}$

but also

$\sqrt{2}+\frac1n$ is not in $\mathbb{Q}

Great!

Good discovery

So now one should have the suspicion that it might work for $\Bbb Q$

To do so, you need to prove that $\inf q_n$ where $q_n$ are all rationals, is always rational.

so if $q\in \mathbb{R}^*$ it is not a topology ?

Well that's a bit of a moot question because then the definition of $\theta$ has to be changed

i think that if all $q_n$ are rational then the inf of then is rational

In fact, that is of course not true

Think about $\sqrt 2=1.4142135...$

Then take the sequence $q_1=1$, $q_2=1.4$, $q_3=1.41$ etc

This is a rational sequence converging to $\sqrt 2$ (in fact this is a way to define the real numbers :P)

6:07 PM

(That's not yet a counterexample though.)

But that doesn't prove that your proposed topology with $\Bbb Q$ isn't a topology

Because that sequence is not monotonously decreasing (its inf will just be 1)

I'm wondering if we can still make a counterexample out of this somehow...

hi @quid

hi @Danu!

What kind of mathematics do you study, @quid?

@Danu Look at the k-th decimal approximation + 1/10^k, say.

@BalarkaSen oh that's sweet

@Vrouvrou so what Balarka just said is the following: Take $q_1=1+1=2$, $q_2=1.4+0.1=1.5$, $q_3=1.41+0.01=1.42$ etc

This will still converge to $\sqrt 2$, but now it's monotonously decreasing (you should prove this!)

This answer is epic, @quid!

55

A: Unique candidate that fails

For a positive real number $x$ consider $$x^{x^{x^{\dots}}}$$ or formally (the limit of) the sequence $a_n= x^{a_{n-1}}$ (and $a_0= 1$). Determine the value $x$ (if it exists) such that $$x^{x^{x^{\dots}}}=4.$$ Uniqueness can be proved like this: since for such an $x$ we ought to have $x^4 = 4...

The answer below is also great:

55

A: Unique candidate that fails

The obvious example I immediately thought of is that, if the divergent geometric series $$1 + 2 + 4 + 8 + \dotsb = \sum_{k=0}^\infty 2^k$$ converged, it would converge to $-1$. Proof: If the series converged to some number $x = 1 + 2 + 4 + 8 + \dotsb$, then clearly this number $x$ would have to ...

6:19 PM

@Danu ok , and what about this topology : $\{\mathbb{R},\emptyset}\cup \{E_a=]a,+\infty[\}_{a\in \mathbb{R}^*$

@Danu mostly number theory (in a broad sense). Thanks for the feedback on my answer.

@quid I'm always curious about what motivates number theorists. Is it the number theory for its own sake, or the connections to other fields?

I never had much appreciation for numbers... :P I guess it comes with the physics background to be more geometric.

AFAIK, there's a relation with algebraic geometry, ideas of some of which are borrowed from topology (I have Weil conjectures in mind). I think number theory connects with A LOT of other fields.

But quid can tell you specific things.

Of course, there are many relations. I know that

@Danu as Jacobi said " le but unique de la science, c’est l’honneur de l’esprit humain, et [...] sous ce titre, une question de nombres vaut autant qu’une question du système du monde" :-)

6:27 PM

@quid heh

numbers are nice. I like them.

So are you working on Jacobi forms (or theta functions) @quid? :D Those come up a lot in strings-related physics

No, not much relation to Jacobi specifically. I just like the quote.

Yeah, fair enough :)

I think Tate once summed up the study of number theory as "study of $\text{Gal}(\overline{\Bbb Q}/\Bbb Q)$". I like that, but maybe there's more analytic number theory out there which doesn't revolve around that group (@quid is that right?).

6:37 PM

@BalarkaSen I know that quote but never saw it attributed. I think this is a very narrow view of number theory. For example, how does the Goldbach conjecture fit in their or the question whether the Euler-Mascheroni constant is rational.

Fair enough, so I thought.

I guess I am more topologically biased too, and the only place I see a shadow of topological ideas is there.

Yes, this makes sense. It is also an important part of NT but I like to see it as broader. I think both Weil and Erdős were great number theorists.

Yeah, I have great respect for Erdős too.

Also his drug use? :)

of course

7:08 PM

@Danu

please can you tel me

if this is a topology : $\{\mathbb{R},\emptyset}\cup \{E_a=]a,+\infty[\}_{a\in \mathbb{R}^*$

i think that no , if we take the same example as in $\mathbb{Q}^*

$

but in the exercise there is prove that it is a topology

hi @BalarkaSen

if we consider a manifold of dimension 0 then their is only 1 right ?

Hi.

their is only 1 smooth structure I mean

If you mean connected manifold, then yes. Just a point.

Otherwise a bunch of points is a fine 0-dimensional manifold (which give infinitely many examples)

I see

just a sec because I am confused little bit about so the maximal atlas determines the smooth structure on a manifold. If we have the manifold being 0 dimensional then it is discrete space, i.e singletons are open. So the map that sends the each point to the single point determine a chart, and collecting all of those points determine the maximal smooth atlas on M ?

right ?

7:14 PM

That's correct.

ok I see

7:28 PM

will it matter which smooth structure we define on manifold

does things change if we use different smooth structure ?

@BalarkaSen ?

It darn well will.

@Adeek It's a fact that in S^7 there exists multiple smooth structures quipping S^7 with which gives non-diffeomorphic manifolds.

I see

This is a nontrivial and surprising theorem by Milnor, and is a counterexample (but even stronger than that!) to smooth Poincare conjecture in dimension 7

Since then there has been more examples of homeomorphic but not diffeomorphic manifolds.

I see that is quite cool

geometry is actually pretty cool

It's still an active area of research, especially in dimension 4. Nobody knows if S^4 admits non-diffeomorphic smooth structures.

This is topology, not geometry though

7:31 PM

I see

yeah topology is pretty cool..

8:03 PM

@BalarkaSen am I correct in saying that given $f:X\to Y$ I have a map $f^*\Omega_Y\to \Omega_X$, $(x,\alpha)\mapsto (x,f^*\alpha)$?

I am disturbed by the fact that $f^*$ appears on the right hand instead of left hand side :p

What's $\Omega_X$?

forms

1-forms

So cotangent bundle

That's right.

aight

Dualizing is contravariant. Get used to it :)

8:05 PM

Yeah, that's the justification I also had in my head

but I'm just weirded out that $f^*$ goes out of $f^*\Omega$ :P

Does O(2) act freely on any sphere?

@MikeMiller By something else than rotations(+reflection)?

Literally any free action on a sphere.

I have no idea how to approach a question like that

SO(2) does, of course. But O(2)...

8:07 PM

Ah. It breaks my heart but it cannot.

Why not?

Z/2 x Z/2 is a subgroup and that can't act freely on any sphere. This is due to Smith originally but I don't know how his proof worked.

Good point.

Is that hard to prove? I have a proof on my other question.

@MikeMiller I know it for... odd? or even? spheres :D :P

Only Z/2 can act freely on even spheres.

8:11 PM

even, then :D

that was nice

The fancy theorem is that a group that acts freely on a homology $n$-sphere has $(n+1)$-periodic cohomology, which is not true of (Z/p)^2.

Some spectral sequence argument.

Yes, which I did prove here.

Nope, just build a K(G, 1) out of the quotient.

Coolbeans

The periodic cohomology statement is a spectral sequence argument.

I should learn chapters 4,5 of Hatcher at some point

8:14 PM

Ah, I see, I misread. But only having non-finite cyclic (n+1)-dimensional cohomology suffices.

Of course, the double cover Pin(2) of O(2) does act freely on a sphere. Instead of having "a reflection squares to 1", you have "an element of the non-identity component squares to -1", so you don't have a (Z/2)^2 subgroup.

Ah, now I understand why it's called Pin haha

I don't though. It's probably a Z/2-extension of Spin.

hi chat

@BalarkaSen It's $(S)O\Leftrightarrow (S)pin$

8:17 PM

@BalarkaSen Got a Riemann surface question for you, which I came up with in response to a question from my advisor

I'm listening in

Also lol @Balarka how you used Poincare conjecture in that question :D

what's the genus of $(y-1)^2(y+1)^2=x^4-1$?

isn't there a genus-degree formula

yup

^

8:18 PM

if it's smooth it's 3

Riemann-Hurwitz formula

heh. forgot about that.

christ that was fast

yeah

damn @mercio

btw, this implies then that SO(3) also doesn't act freely on any sphere, which also breaks my heart

8:18 PM

(but is it smooth ???)

nor ofc SO(n) for any n>2

Is that Riemann surface an acquaintance of yours? :P

it's a research thing, yeah

no but I had a friend who was also of degree $4$ and genus $3$

hehe

8:19 PM

I suspect mercio to be a version of anon/arctic

arctic tern has multiple accounts?

more generally, what would be the topology of a Riemann surface like $p^4+ap^2+P(q)=0$ where $P(q)$ is a quartic polynomial

Good evening

i'm guessing that the same logic gives it as $3$

@Semiclassical You have to count branch points

8:20 PM

good evening

@Danu yes, but I think mercio is a version who is not a different account.

just check out the Riemann-Hurwitz formula

$p\to y$ and $q\to x$, btw.

@BalarkaSen doubt it

@Danu Did I? Don't berember. Maybe I did that to show free groups acting on S^3 are precisely the finite 3-manifold groups.

8:21 PM

Ouch... I started drinking whisky but my mouth got burnt earlier

I think I've seen arctic tern in chat once, I can't be him

@BalarkaSen Yup.

i'm interested in an integral that looks like $\oint p(q)\,dq$ where $p(q)$ satisfies the above polynomial

so I wanted to know what that integral looks like on the Riemann surface

@MikeMiller mercio's mathematical skills seems oddly similar to anon, from the little I have talked to him. I don't believe they're the same person though, I was joking about that.

sad that SO(3) doesn't act freely on any sphere though. I'd think it does.

@BalarkaSen If you're going to invoke Poincare, you may as well invoke elliptization that they're subgroups of SO(4) acting freely.

lol get banned danu

8:24 PM

@MikeMiller isn't the genus-degree formula for curves in $\Bbb{P}^3$? what i've got is just $\Bbb{C}^2$, though projectivizing it is easy enough

it's for curves in P^2

if you have it in C^2, you can projectivize, yeah

that'd just be adding a bunch of points, so genus would be the same

were there any interesting questions earlier

@MikeMiller I only pretend to know elliptization. what's the statement of it?

Thurston elliptization conjecture

William Thurston's elliptization conjecture states that a closed 3-manifold with finite fundamental group is spherical, i.e. has a Riemannian metric of constant positive sectional curvature. A 3-manifold with such a metric is covered by the 3-sphere, moreover the group of covering transformations are isometries of the 3-sphere. Note that this means that if the original 3-manifold had in fact a trivial fundamental group, then it is homeomorphic to the 3-sphere (via the covering map). Thus, proving the elliptization conjecture would prove the Poincaré conjecture as a corollary. In fact, the e...

Right, P^2

8:27 PM

@MikeMiller heh

Though I should check if the example I suggested is a smooth curve

I apologize for my terrible language

I'm kind've suspicious of whether that example has any degenerate points

Not really, but formally

@MikeMiller Thanks. So it seems it literally says finite 3-manifold groups are subgroup of SO(4) acting freely by isometries.

Which is what you said, that is.

8:32 PM

:32648554 tsk!

@robjohn :'(

@Danu watch your [add appropriate swear words]-ing language.

@Danu I used to think Pin(2) was isomorphic to O(2) and was confused why people always used the name Pin(2).

@MikeMiller ah

I forget these groups. Spin(n) is the double cover of SO(n), and Pin(n) is the Z/2-extension of Spin(n), isn't it?

aka fits in the exact sequence 0 --> Spin(n) --> Pin(n) --> Z/2 --> 0?

8:41 PM

So the idea is just that spin is the universal covering of $SO$ and Pin of $O$

But it's a bit tricky because $O$ is not connected

ah, ok

> For a disconnected topological group, there is a unique universal cover of the identity component of the group, and one can take the same cover as topological spaces on the other components (which are principal homogeneous spaces for the identity component) but the group structure on other components is not uniquely determined in general.

The Pin and Spin groups are particular topological groups associated to the orthogonal and special orthogonal groups, coming from Clifford algebras: there are other similar groups, corresponding to other double covers or to other group structures on the…

From wikipedia

Huybrechts is confusing me again...

user image

"smooth" ?!

They are not the universal cover for n=2.

More like regular?

@MikeMiller Ah, okay.

But in other cases they are, no?

yeah, odd word

8:48 PM

Did I mess this up? :P

it's a submersion at x.

For n>2 yes.

Yeah right---just preimage of regular value theorem

I'm super weirded out by it because this comes at the end of chapter 2, after all that hard stuff

corollary 2.6.4 is the complex version of regular value theorem, yeah

I'm almost convinced it can't just be the definition of submersion

I feel like I'm missing something :P

8:49 PM

it is though

Super strange "timing" to put such a basic diffgeo-type result so late in the book

Also the terminology smooth is abominable

@MikeMiller Phew

I'd love to learn more about Spin structures

Hopefully soon!

unrelated: so clutching construction give a bijection between complex k-plane bundles on $S^n$ and $[S^{n-1}, U(k)]$. on the other hand there's also a bijection between complex k-plane bundles on $S^n$ and $[S^n, BU(k)]$ (because $BU(k)$ is the complex Grassmannian of complex k-dim subspaces in $\Bbb C^\infty$). I suppose the bijection $[S^{n-1}, U(k)] \to [S^n, BU(k)]$ is just the isomorphism $\pi_{n-1}(U(k)) \cong \pi_n(BU(k))$, then?

what is BU?

BG is the classifying space of G

when G = U(n) (unitary group), BU(n) is the "infinite Grassmannian" of n-dimensional subspaces inside $\Bbb C^\infty$

ok

And you just define that with a CW construction?

I saw it for the projective spaces

8:55 PM

yeah, it's an infinite union of the "finite" Grassmannians of n-dimensional subspaces of C^(n+k) for all k > 0

Generally, are the finite dimensional Grassmannians also built out of the lower-dimensional ones just are the projective spaces are?

@BalarkaSen Prove a version of the clutching construction correspondence for principal bundles over $\Sigma X$, $X$ a pointed space. Then prove that $\Omega BG \simeq G$.

@MikeMiller Thanks; I have bookmarked this. I'll do it tomorrow.

@Danu I don't actually know how to write down a cell structure of the Grassmannian off the top of my head. One generally does it in the manifold category by writing down the Stiefel manifold, and defining Grassmannian as a quotient of that manifold.

But one can moreover show it's an algebraic variety by explicitly embedding it inside some projective space and writing down equations.

That's the two ways to define the Grassmannian I know of.

I haven't played around much with them. Hatcher's VBKT has a chapter, so I guess I can fix that soon if I want.

I am off to bed, g'night all.

9:12 PM

@BalarkaSen It's in Milnor & Stasheff---I'll let you know once we cover it in my seminar :P

Bye @Balarka

9:23 PM

@danu yes, spin geometry is neat

Hey everyone, I have an open bounty on this question ('Why do we treat differentials as infinitesmals, even when it's not rigorous') here if anyone wants to take a look : math.stackexchange.com/questions/1943243/…

Why do we treat differentials as infinitesmals, even when it's not rigorous

@SimpleArt

Thanks!

9:42 PM

@Perturbative LoL, I wasn't sure if it worked in chat

XD cheers man $\ddot\smile$

9:56 PM

@Perturbative You're asking for a lot in your bounty statement ;)

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