Mathematics - 2020-03-28 (page 2 of 3)

Knight

16:01

@anakhro I could do just 4 questions

anakhro

@Knight so did you do those questions quickly? Or did they take too long?

Knight

@anakhro They took long

anakhro

So you need practice, and you need to work on test taking skills.

Knight

I tried 8 But was able to do only 5

We had 30 questions

anakhro

What is the average, do you know?

Like pass average.

Knight

16:05

12 questions correct

Something like that

Balarka Sen

@TobiasKildetoft Amazing

anakhro

@BalarkaSen thanks that was a good album.

Balarka Sen

Glad you liked it

anakhro

@Knight In terms for these exams, I think the general idea is doing more practice questions is the best policy.

The next thing to do after practicing a lot is doing actual practice exams. That is, you take an exam of which you have never taken before, and you do a "mock exam": pretend you are taking the exam, time limit and everything.

And see how you do.

That way, you get a feeling for progress.

"Am I better able to do this exam now that I have done all those practice questions?"

Knight

@anakhro I’m struggling with inner conflict, whether to study real things or work for exams

anakhro

16:08

@Knight well what do you want to do with your life?

Knight

@anakhro I want to invent some elegant idea

You know elegant

anakhro

Elegant is subjective.

Knight

The way Cauchy defined Limits

Maxwell refined Electromagnetism

Lord Kelvin made Physics more mathematical

anakhro

These are very prolific figures.

Knight

Something like that you know

anakhro

16:12

I am not sure if hoping to come up with something like they did is a healthy goal.

Knight

I don’t want that much recognition

feynhat

Literally unwatchable.

@TobiasKildetoft gmail doesn't allow underscores in email addresses.

Knight

I like some elegant work. I don’t have any desire to become like Von Neumann, Einstein or Gauss. All I want to do is that I should be know for my elegant idea, no matter in how small group but the idea should be real

anakhro

i.e. you want to do research?

Knight

Yes

anakhro

16:18

Okay, perfectly reasonable.

Research isn't terribly difficult of a goal.

I suppose you want to get into academia, then.

Knight

I think academia isn’t necessary but can be helpful

anakhro

Definitely not necessary, but it's difficult to get academic recognition from outside of academia.

Just ask that fellow on here who is frantically emailing professors, trying to get on the arxiv.

Knight

@anakhro Is that you?

anakhro

lol

No.

I am on the arxiv already. :P

Not for anything too special, though.

Bob

Hi

Knight

16:21

You’re very genial

anakhro

First, I want to make it clear to you that you can both do your own learning and research on your own, while studying at school stuff that you have to study.

2

Many people on this chat do just that.

But sometimes it takes a balance.

Knight

@anakhro Really? How?

Bob

If I have a system of two equations and two unknowns and I believe there is only one solution, how do I prove or disprove that there is a unique solution

anakhro

Because the school stuff that you have to do helps you get where you want to be.

Knight

@anakhro But they require a different kind of thinking, you know, we have to defeat others not to win ourselves

anakhro

16:24

I don't quite follow what you mean by that.

Knight

@anakhro Entrance tests or competitive exams require us to score more than the others

Just scoring enough is not enough others have to have scores lower than yours

anakhro

Indeed.

But if an entrance exam gets you closer to your dreams, you should study for it.

Balarka Sen

Nice motivationals, anakhro

You should open up a youtube channel

anakhro

lol

Knight

@anakhro It makes me to get closer and closer to reputation not the good ideas, at least in my area

My area:- my residential area

anakhro

16:28

@Knight you get at least 4 years of learning after you pass the exams.

Bob

If I have a system of two equations and two unknowns and I believe there is only one solution, how do I prove or disprove that there is a unique solution?

anakhro

Learning from world professionals, nonetheless.

For a trade off of what, a few months of learning independently from books you struggle to understand and can't find help for?

@Bob linear system?

Knight

@Bob You know determinants ?

anakhro

$ax + by = c$ and $dx + ey = f$, unknowns $x,y$?

Bob

it is not linear

0

Q: Black-Scholes and solving for both $r$ and $\sigma$ ; Do I have a unqiue solution?

Below is a problem that I am working on. I believe that my incomplete solution is correct as far as it goes. I would like to know if my solution is incorrect. I plan to solve the system of two equations by using a computer program such as SciLab. What I am wondering is, does this system of equati...

I do know demerminants.

Knight

16:31

@anakhro Some people are far better than me in studying for exams. But I can say it by standing in Pulpit that they have no, NO, thinking ability.

anakhro

@Knight stop comparing yourself to other people.

@Bob Two non-linear equations can have any number of solutions you want.

So this is more of a question about your particular equations.

Bob

okay

Knight

@anakhro What can I do? Should I act like a man ?

Bob

I am going to have lunch

thanks for the commets

have a nice day

anakhro

@Knight it has nothing to do with masculinity.

If you want to do research, you should find a way to do research.

If you aren't going to try to get into a college program to do so, then first you need to find a job which you can sustain yourself with while you do your own learning/reading.

But if you are going to try to get into a college program, you should at least try to do well on the entrance exams, since that's the first step.

Mind you, you don't have to get into the top university/college to do research.

So be open to possibilities.

Knight

16:38

How to console my parents? They want (want it badly) me to go into top universities

anakhro

That's up to you and your parents to figure out.

You do not have to go to Cambridge to succeed.

skullpatrol

Do your parents know you failed three times?

Knight

Yes

They know it

skullpatrol

What do they recommend?

Knight

Try it again

skullpatrol

16:42

Then go for it, pal.

Knight

@anakhro I know that, success doesn’t have to do anything with University

skullpatrol

The tests are a race against time.

Knight

@skullpatrol Pal i Sade It’s hard to make up your mind to work for those tests, but I would do it

anakhro

@skullpatrol I don't know that taking their advice for the 4th time would work.

Knight

@anakhro I agree

skullpatrol

16:45

Concentrate on getting faster, without losing accuracy.

anakhro

Especially since your work ethic is suffering as a result, and you are becoming contemptuous for others who succeed.

feynhat

@ManolisLyviakis What do you mean by breaking the counting into parts?

Knight

@anakhro Yes I hate those who succeed, it’s because of them people insults and disrespect me.

anakhro

Hating people doesn't help your case.

feynhat

@ManolisLyviakis Btw, there is a general expression for counting the permutations of a given cycle-structure, if that's what you're asking.

Knight

16:49

@anakhro It’s involuntary

anakhro

Hate is always voluntary.

Mats Granvik

“Never hate your enemies. It affects your judgment.”
― Mario Puzo, The Godfather

Knight

Have you been into my situation? (I’m really asking it no sarcasm intended)

skullpatrol

Focus all your attention on developing your test taking skills. @Knight

Knight

@skullpatrol I will, Pal

skullpatrol

16:52

Skills take time to develop

Tobias Kildetoft

@feynhat Ehh, I am not sure what that has got to do with anything

anakhro

Most people have been in situations where they don't want to do the work to do something, and then blame people who actually do the work.

feynhat

@TobiasKildetoft uhh... the video?

skullpatrol

The video is a gag

Knight

@anakhro Maybe

Tobias Kildetoft

16:54

@feynhat What does the video have to do with gmail adresses?

skullpatrol

no need for realism

Balarka Sen

some guy mentions a gmail id with an underscore in the video somewhere

feynhat

@TobiasKildetoft Its literally unwatchable because the guy says [email protected]

skullpatrol

how does that make it "unwatchable"?

Knight

@anakhro Thank you so much for today.

@skullpatrol Meet ya later Pal i sade

skullpatrol

16:57

cya, buddy

anakhro

@Knight just focus on your own improvement, rather than others.

10

skullpatrol

^

feynhat

I am joking people. There is a trope among gamers, that a game is literally unplayable if it misses some trivial detail from real-life. I was hoping run around with that trope. Turns out I can't.

@TobiasKildetoft @skullpatrol

anakhro

@BalarkaSen Hatcher has another exercise about covering space actions. If $G$ acts on $M$ as a covering space action via orientation preserving homemorphisms, then $M/G$ is orientable if $M$ is.

I assume that the point is that we are inducing the orientation along the projection $p\colon M\to M/G$

And you get the local consistency from the orientation preserving homemorphisms $g\colon M\to M$

Balarka Sen

Correct ideas. What is the orientation, explicitly?

anakhro

17:05

I am not sure. I am still struggling with understanding how to show orientability with this definition.

So I have a map $\mu\colon M\to \{\pm 1\}$ where $x\mapsto \mu_x$ has the local consistency condition.

I really just want to induce it with $p_*$ or something

Balarka Sen

:)

anakhro

That's just a gut feeling, I have no mathematical reasoning behind it.

Abhas Kumar Sinha

okay

anakhro

Just how do I get a generator on $M/G$? Push it forward.

Abhas Kumar Sinha

Hello

Everyone...

Balarka Sen

17:07

I will make some comments before you proceed. I like your notation, but it's a little misleading in the following way

Alessandro Codenotti

@Balarka Can you give me an understandable idea of what the Novikov conjecture says?

Balarka Sen

I wouldn't describe the orientation as a map $\mu : M \to \{\pm 1\}$ because there's no canonical identification of the generators of $H_n(M, M-x)$ togather a-priori. I would describe it as a collection $\{\mu_x\}$ of generators of $H_n(M, M-x)$ for each $x \in M$.

Alessandro Codenotti

I mean ok, the understandable idea is that some quantities are homotopy invariant, but what kind of information are those quantities encoding? Where do they come from?

Balarka Sen

But yes, essentially you want to give me a collection $\{\nu_x\}$ of generators of $H_n(M/G, M/G - x)$ from there which satisfies the local consistency. Thinking of it in the map way, given a map $\mu : M \to \{\pm 1\}$ you want a map $\nu : M/G \to \{\pm 1\}$. When do you have such an induced map? What do you know about $\mu$?

anakhro

So kind of like a continuously varying family of generators?

Balarka Sen

17:10

That's correct. There's a very convenient way to say it using sheaves but I'll hold off on that for now

anakhro

Heh

Balarka Sen

@AlessandroCodenotti Oh shit let's see.

Alessandro Codenotti

I'm asking because apparently Yu proved the conjecture for manifolds whose $\pi_1$ has finite asymptotic dimension (so for example for all manifolds with hyperbolic $\pi_1$), and that really surprised people and got them interested in asymptotic dimension, which Gromov had defined years earlier but had been kind of ignored

Balarka Sen

Ah I see

I think it's a classical result for hyperbolic manifolds, because the stronger conjecture known as Borel conjecture is true for hyperbolic manifolds

That's Mostow rigidity

Let me look around and try to see if there's an readable source for Novikov, I vaguely knew this but forgot about it

Alessandro Codenotti

Well there's a lot of non hyperbolic groups with finite dimension though :P

Balarka Sen

17:15

Yeah

This looks technical :S

Alessandro Codenotti

Don't waste time on it, I was asking assuming you were already familiar with it!

Balarka Sen

Nah I don't know much about this. @MikeMiller might

What is this assembly map garbage lol

I never understood what people in L-theory do

Alessandro Codenotti

I have no idea

My advisor actually needed all of this Higson machinery to prove some result about a K-theoretic assembly map, but I couldn't understand a word lol

Balarka Sen

Dang

anakhro

@BalarkaSen so what I am guessing is that $\nu_x := p_*\mu_x$, and this is well-defined because if $Gx = Gy$, there exists a $g\in G\,.\, gx=y$, and since $g_*$ preserves the generators, $\nu_x = p_*\mu_x = p_*g_*\mu_x = p_*\mu_y =\nu_y$.

Balarka Sen

17:25

I think the Hirzebruch $L$-class for a $4n$-manifold $M$ is an element $L \in H^{4n}(M; \Bbb Q)$ such that $L \frown [M]$ is the signature of $M$.

Mike Miller

Catch me up

Balarka Sen

Alessandro wants to understand the statement of Novikov conjecture

Alessandro Codenotti

Hi Mike

Balarka Sen

What is the relation with Borel conjecture?

I want to understand that

Mike Miller

Remind me statement of Novikov

Balarka Sen

17:28

I dunno man look at wikipedia

It's too hard for me

@anakhro Right, the "map" $\mu : M \to \{\pm 1\}$ is $G$-equivariant since $G$ acts by orientation-preserving homeomorphisms on $M$, so descends down to a map $\nu : M/G \to \{\pm 1\}$.

Mike Miller

What map mi

Balarka Sen

And that is your orientation on $M/G$. You should write everything down explicitly once, of course, instead of philosophizing; if $x \in M$ then you have an isomorphism $H_n(M, M - x) \to H_n(M/G, M/G - Gx)$ because $M \to M/G$ is a local homeomorphism near $x$ (it's a covering map). The image of $\mu_x$ you call $\nu_{Gx}$.

And then $\{\nu_{Gx}\}$ is your candidate orientation for $M/G$. You need to check local compatibility, which you can, etc.

$\nu_{Gx}$ only depends on the class $Gx$ and not on $x$ because $G$ acts transitively on the fiber over $Gx$, so you have well-definedness; essentially what you wrote.

Namely, if $x$ and $y$ such that $Gx = Gy = p$, there is a commutative triangle involving $H_n(M, M - x) \to H_n(M/G, M/G - p) \leftarrow H_n(M, M - y)$ and $H_n(M, M - x) \to H_n(M, M - y)$ where the last map is induced by the deck transformation permuting $x$ and $y$ on the fiber over $p$ in $M \to M/G$.

Therefore image of $\mu_x$ and $\mu_y$ are the same.

Mike Miller

What is mu

What are you doing

anakhro

@MikeMiller Cover map action $G$ on $M$ by orientation preserving homeomorphisms induces orientation from $M$ onto $M/G$.

Balarka Sen

35 mins ago, by anakhro

@BalarkaSen Hatcher has another exercise about covering space actions. If $G$ acts on $M$ as a covering space action via orientation preserving homemorphisms, then $M/G$ is orientable if $M$ is.

Mike Miller

17:38

I don't like that you are making a map to +-1 so I want you to tell me what my I

mu is!

Balarka Sen

I told him that earlier

Please don't sheafify and scare anakhro

anakhro

I know what a sheaf is, tbh

Mike Miller

I don't care what a sheaf is

anakhro

let's all put on our glasses and look very concerned about this map, $\mu$

and then we can conclude that we all understand it was a grave error

Balarka Sen

I didn't set it up as a map in my last proof

anakhro

17:41

and move on and forget about it :(

Balarka Sen

@MikeMiller Hey why is signature a cobordism invariant

I forgot

Suppose $M$ is a $4n$-manifold bounded by a $(4n+1)$-manifold $W$. The image of $H^{2n}(W) \to H^{2n}(M)$ is half-dimensional by half-lives half-dies.

And it's also isotropic under the pairing, clearly. I think that's enough to conclude signature of $M$ is zero.

anakhro

half lives, half dies sounds like a good summary for the Lefschetz hyperplane theorem.

Balarka Sen

A symmetric bilinear form with a half-dimensional isotropic subspace has zero signature. Right? Right??

Yes, by uh that theorem

inertia

Mike Miller

I wasn't picking on anyone, I just saw something I didn't understand and wanted to be caught up

Balarka Sen

$\mu$ was just an orientation on $M$. For convenience we were setting it up as a map to $\Bbb Z_2$, although that's abusive.

Mike Miller

17:52

You can make that work

anakhro

Yeah, especially if you introduce as axioms $CH$ and $\neg CH$.

Balarka Sen

Off-topic again: Is any cobordism invariant class a polynomial on the Pontryagin classes? Something like this was true but I forget what is the precise statement

Mike Miller

@anakhro I'm writing something, no need for smarm

anakhro

That's a word that I have never heard before.

Alessandro Codenotti

Did someone say CH?

anakhro

17:57

@AlessandroCodenotti I made a bad joke.

Mike Miller

Given $p \in M$, $M$ a connected manifold, there is a map $\Omega_p M \to \Bbb Z/2$ sending $\gamma$ to the neutral element iff $\gamma^*TM$ is an orientable bundle over the circle. Then we have the following properties.
1) This is a group map: if you compose two loops, then $(\gamma_1 \ast \gamma_2)^* TM$ is orientable iff $\gamma_1^*TM$ and $\gamma_2^*TM$ are either both oriented or both unorientable. This requires a little understanding of bundles over the circle. In particular, the map on $\pi_0$ is a homomorphism $w: \pi_1 M \to \Bbb Z/2$.

anakhro

How do you define the tangent space on a topological manifold?

Mike Miller

You're doing a topological manifold? Christ

Fine

Balarka Sen

You can define it similarly but you have to use the orientation double cover here

Mike Miller

Yeah I'm just going to keep telling the smooth story

Balarka Sen

18:06

Right, just wanted to make the point it basically doesn't matter

Continue

Mike Miller

Unfortunately this isn't the right way to think about orientations to be well-behaved for group quotients. The issue is that non-null loops downstairs lift upstairs to paths between different points. So the above description of orientability (it's a manifold for which loops don't flip orientation) isn't the right phrasing

But it's close to one!

Instead we're going to have to thicken this up so that it allows paths between different points --- instead of the fundamental group, this is one of those rare places where it's smart to talk about the fundamental groupoid

Balarka Sen

Ronnie disagrees in the background

Mike Miller

$\Pi(M)$ is a category whose objects are the points of $M$, and $$\text{Mor}(x,y) = \pi_0 \text{Map}([0,1],0,1) \to (X,x,y),$$ aka, the set of paths from $x$ to $y$ up to homotopy rel endpoints.

The composition operation just being concatenation of paths; the fact that we make this up to homotopy rel endpoints means that this is a category (composition is actually associative on the nose, and you have inverses).

What is the role of $\pm 1$ here, as we vary between points? Well, $\pm 1$ is really $\pi_0 GL(n)$, right? We check whether a self-map of $T_p M$ is homotopic to the identity ornot.

Set $O(M)$ to be the category whose objects are the points of $M$, and $$\text{Mor}_O(x,y) = \pi_0 \text{Iso}(T_x M, T_y M).$$

anakhro

Does the orientation double cover topologically replace the tangent bundle in some sense?

Mike Miller

I mean, what you're seeing above is that I'm really chucking out most of the data of the tangent bundle to talk about orientations

I don't care about the whole group $GL(n)$, just the path components

Leaky Nun

18:13

@BalarkaSen Magnus Carlsen is streaming here

Balarka Sen

On it

Fuck

Why would you do that to me?

Leaky Nun

ngl it's actually good

I like it

Balarka Sen

I'm real pissed

Mike Miller

Claim: there is a functor $\Pi(M) \to O(M)$, which is the identity on objects. It's a little irritating to write down. The point is that $\gamma^*TM$ is trivial over $[0,1]$, so that there are sections $e_1(t), \cdots, e_n(t)$ over the interval which restrict to a basis for $\gamma^*TM$ at each point. Then the map $T_x M \to T_y M$ sends $e_i(0)$ to $e_i(1)$.

Leaky Nun

i'm sorry

Balarka Sen

18:15

I'm going to forward it to some chess nerds

Mike Miller

Meh you won't like this, I'm going to stop

Balarka Sen

I was following, @MikeM

Alessandro Codenotti

I'm discovering so much LaTeX weirdness while typing my thesis. I might be doing something wrong

Leaky Nun

eg?

@MikeMiller so it's kinda "flowing" a basis from $x$ to $y$

Mike Miller

An orientation, really

Alessandro Codenotti

18:18

@LeakyNun So I suppose you're familiar with amsthm

Balarka Sen

Orientation class of the basis, rather

Sniped

Leaky Nun

@AlessandroCodenotti a bit yeah

Mike Miller

Now $M$ is orientable iff the map on $\Pi(M)(x,x) \to O(M)(x,x)$ is trivial.

Alessandro Codenotti

All of the environments you define have an optional parameter, if you have the environment thm for theorems and you want a named theorem you can do \begin{thm}[the name of your theorem goes here] the statement goes here \end{thm}, ok?

Leaky Nun

wow that's kinda surprising since $\Pi(M)$ only detects the first-dimensional issues right

ok

Balarka Sen

18:20

Orientation is a 1-skeletal issue, is the point.

Leaky Nun

really :o

Alessandro Codenotti

Now using a \cite{reference} inside the name of a theorem works fine. Using \cite[optional parameter]{reference}inside the name of a theorem breaks things and cannot be compiled

So you need to wrap the \cite with a \protect because of black wizardry

Mike Miller

a rank $n$ vector bundle (with connection) gives a homomorphism $\Omega M \to GL(n)$

Leaky Nun

solution: don't :P

Mike Miller

there is an induced map $\Omega M \to O(1)$ studying orientation

this corresponds to a map $M \to BO(1)$

the latter is a $K(G,1)$, and $\pi_0 \text{Map}(M, K(G,1)) = \text{Hom}(\pi_1 M, G)/\text{conj}$, which you can compute by hand

Alessandro Codenotti

18:23

This is the effect I wanted to obtain, rather than only having the Eng89 reference

Another funny one I stumbled upon is that in a figure environment you need to put the \caption before the \label or you can't actually reference that label elsewhere in the document

Balarka Sen

Dually, orientation is determined by triviality of the determinant bundle over $M$, which is a real line bundle. Give $M$ a CW-structure and look at the $1$-skeleton; it suffices to understand if the bundle is trivial over the 1-skeleton or not to determine global triviality of a line bundle.

Leaky Nun

interesting

Mike Miller

orientations are determined by elements of $H^1(M;\Bbb Z/2)$, in other words

that's really 2-skeletal since you need to know the relations in $\pi_1 M$ in the end

If $G$ acts on $M$, then we also have $G$ acting on $\Pi(M), O(M)$. On $\Pi(M)$ the action is obvious; on $O(M)$, if we have a homotopy class of isomorphism $T_x M \to T_y M$, it is sent to the composite $$T_{gx} M \xrightarrow{g^{-1}} T_x M \to T_y M \xrightarrow{g} T_{gy} M$$

And the functor $F: \Pi(M) \to O(M)$ is equivariant under this group action; $F(gx) = gF(x)$ and $F(g[x \to y]) = gF(x \to y)$

If $M/G$ is to be orientable, then necessarily $M$ must be as well. So suppose $M$ is orientable. Why must $G$ act by orientation preserving transformations?

Suppose $g: T_x M \to T_{gx} M$ is not orientation-preserving. In the above language, this means that $g_* \in O(M)(x,gx)$ disagrees with $F(x \to gx)$.

But that means precisely that if we have a loop $\gamma$ in $M/G$ which is induced from a loop $x \to gx$ in $M$, then the induced map $F(\gamma) \in O(M)([x],[x])$ is orientation-reversing, the non-identity element.

I hate the language I used here. This is a straightforward idea that I have obfuscated.

Oh well.

Balarka Sen

4 hours ago, by Balarka Sen

That's why you don't do orientation, you just say, well, stuff flips if you loopy

Mike Miller

The point of this stupid language is: so long as the map $w: \pi_1 M \to \Bbb Z/2$ is trivial, then "pushing along a path" lets you take an orientation at any one $T_x M$ and take it to an orientation at any other $T_y M$.

That gives you the global orientation. If $M/G$ is not orientable, then that means that the orientation on some two $T_x M$ and $T_{gx} M$ don't agree, where we're comparing them using the fact above.

And that means that $G$ acted by transformations which didn't preserve the orientation

anakhro

18:49

@BalarkaSen I am stuck on showing the local consistency condition.

Any hint?

I find it difficult to even phrase the local consistency condition in the definition (as in Hatcher).

Is it all supposed to follow from the isomorphism $p_*$?

Leaky Nun

@BalarkaSen a video showing police brutality on August 31 (every Hongkonger remembers this day) is now being circulated in India as a video showing Chinese police arresting COVID patients?

Balarka Sen

@LeakyNun Oh god.

I did not know this.

Leaky Nun

Boom live, a company in India to verify videos, has issued the following statement:
FALSE: चीन में संदिग्ध कोविड -19 मरीज़ों पर पुलिस की कार्यवाही दिखाने वाला वीडियो फ़र्ज़ी है

Balarka Sen

Ugh

Leaky Nun

@BalarkaSen I guess that means it isn't that widely circulated yet?

Balarka Sen

18:54

It's possible that it is; I am out of the COVID gossip group loop

Leaky Nun

do you read Hindi?

anakhro

He reads math, he can read anything.

Balarka Sen

I can read Hindi but it takes effort

Leaky Nun

can you read the sentence above?

Balarka Sen

Yeah it states that the video showing police violence against covid-19 patients is fake

@anakhro Yes, essentially.

Draw a similar commutative diagram

anakhro

18:57

But this time with the ball B around a point x in M?

Balarka Sen

Correct, let $B$ be an evenly covered open set in $M/G$ containing $x$ and $y$ and consider a slice $\widetilde{B}$ above it containing two fibers $\widetilde{x}$ and $\widetilde{x}$ above $x$ and $y$ in $M$

Mats Granvik

Someone who wants to help me here?

Balarka Sen

Then draw the commutative diagram interpolating between $H_n(M/G, M/G - x) \to H_n(M/G, M/G - B) \leftarrow H_n(M/G, M/G - y)$ and $H_n(M, M - \widetilde{x}) \to H_n(M, M - \widetilde{x}) \leftarrow H_n(M, M - \widetilde{y})$.

You should try practicing these arguments more, anakhro.

anakhro

Just so I am clear on your notation here, x,y are cosets?

Balarka Sen

Yes, they are points in $M/G$.

Leaky Nun

19:01

@BalarkaSen boomlive.in/fake-news/…

so it really exists

anakhro

@BalarkaSen thanks, I appreciate all the help.

Hatcher is a little bit better than before, in my eyes.

user21820

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A: Is $2^n$ always a sum of exactly $n$ primes for $n \geq 2$?

Actually, it is not a very difficult problem. We just need a stronger form of Bertrand's postulate, that for every positive integer $n ≥ 8$ there is a prime between $n$ and $n·3/2$. This fact has been known for a long time, for example based on a theorem of Jitsuro Nagura that if $n ≥ 25$ then th...

Can someone help to check my solution? I make mistakes now and then, but recently some trolls have been downvoting some of my posts, so I'm not sure whether this is one of them or not. Thanks!

By the way, in case anyone is interested in the coronavirus and haven't seen these yet:

Coronavirus SARS-CoV-2 structure

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anakhro

@BalarkaSen a problem I am seeing is that I see a lot of these things for the first time, and then they don't show up again---or if they do show up again, I don't notice that they are there again. Perhaps this is indicative that I am moving too quickly through Hatcher and I am not taking my time to relate things back to what I saw before.

Or simply that I am not actively thinking as much as I should be!

Probably a mixture of these and more.

Granted I regret not going through Hatcher last summer when you told me to.

Manolis Lyviakis

19:33

@feynhat thanks i did find it

Poline Sandra

19:46

Bonsoir @Astyx

Astyx

bonsoir

Poline Sandra

20:00

@Astyx je veux confirmer avec vous s'il vous plait, je cherche a trouver a de $[0,\pi]$ tel que cos(a)=cos(b), b dans $[-\pi/2,3\pi/2]$

alors a=b si $b\in [0,\pi]$, est ce que je rajoute un tour ?

@Astyx je veux confirmer avec vous s'il vous plait, je cherche a trouver a de $[0,\pi]$ tel que cos(a)=cos(b), b dans $[-\pi/2,3\pi/2]$

je veux dire est ce que je dit a=b ou $a=b+2k\pi $ ?

Poline Sandra

20:16

can you help me @TobiasKildetoft ?

Balarka Sen

@anakhro Yeah I mean you just need to practice more

You don't really ever understand math, you just get used to it. I think von Neumann said it, and it's partially true.

skullpatrol

20:53

if not completely true during certain stages of development

user76284

What's an example of a parametric family of distributions that is closed under translation, scaling, and maximum? This family, for example, is closed under maximum.

This would correspond to the underlying CDFs being closed under translation, scaling, and pointwise product.

Thorgott

21:47

Dirac distributions

user76284

21:58

@Thorgott Nontrivial :P

Posted the question here: stats.stackexchange.com/questions/457255/…

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$Mathematics$ Mathematics