Mathematics - 2016-10-08 (page 2 of 5)

Balarka Sen

11:01

In my impression the motivation comes more from number theory than algebraic geometry. People want to count solutions to diophantine equations mod p.

user228700

Hi again. For a frequency distribution of salaries of employees in a company, which would be the better choice for measure of central tendency? Median or mode?(The data is severely skewed) I feel that since, in the particular example that I'm working with, there are about 6 people who earn 90K a month, the mode would be the best option but does the median(which is 12K :/) work too?

Balarka Sen

So might as well look at the algebraic variety over F_p, try to generalize the techniques for characteristic 0 fields to understand it better, etc.

@Kaumudi I vote for mode.

user228700

@BalarkaSen Huh. Then why does my textbook think that the median would be better? >.<

user228700

It says "The data has outliers at 90K and 95K and so we should use median in place of mode". Does that make any sense..?

Balarka Sen

Shrug. I don't know statistics.

user228700

11:07

Oh, crap :/

Danu

Don't ask us to do your homework? :)

user228700

@Danu I don't understand why you feel that I've been asking you to do my homework. Did you read the question properly? I'm trying to understand why my textbook thinks that the median would be the better choice.

user228700

And that doesn't strike me as particularly homework-tsy.

Secret

"A subset S of M is said to be a generator of M if M is the smallest set containing S that is closed under the monoid operation, or equivalently M is the result of applying the finitary closure operator to S"

@MikeMiller Is it saying that $S$ closed under the monoid operation of $M$, or only $M$ is and applying the monoid operations on $S$ gives $M$?

Danu

Okay, fair enough.

Balarka Sen

11:16

I should read more about characteristic classes.

Danu

@BalarkaSen You started on the book without me? ;)

Balarka Sen

A bit.

Sorry :P

Danu

Meh, it's fine.

Balarka Sen

You're getting to know so many cool stuff, so I felt I should try to cover up my envy by studying cool stuff too. :P

Danu

By the way @BalarkaSen how does that Lefschetz theorem give any obstruction? Isn't that isomorphism there due to Poincaré duality already?

I guess PD gives

$H^k(M;R)\cong \operatorname{Hom}_R(H^{m-k}(M;R),R)$ if $R$ is a field*

Balarka Sen

11:24

You have real coefficients here though. Hm.

Danu

By universal coefficients---if there is no torsion you always get this

so if stuff is finite dimensional you get that isomorphism

Balarka Sen

I guess you're right.

Stuff would surely be finite dimensional.

Danu

Yeah, exactly

since compact manifolds are retracts of finite CW complexes

Balarka Sen

That's odd. Oh well.

Danu

So now I wonder how your initial statement connects to the theorem I'm learning about :P

I hate it when theorems are phrased in a way that doesn't make their utility clear :P

Balarka Sen

11:27

Yeah, maybe the weak Lefschetz theorem is the thing you want to understand.

That's the whole deal.

Danu

Does it follows from mine though?

Balarka Sen

I don't see it. Also I don't understand your version of weak Lefschetz.

Danu

I don't have anything called "weak Lefschetz"

I have "hard Lefschetz" and "Lefschetz on (1,1)-classes"

Balarka Sen

You said the (1,1) thing.

OK.

Danu

which is a statement about the map from the Picard group into $H^{1,1}$

namely that the map that assigns the first Chern class surjects onto a certain subset

namely $H^{1,1}(X)\cap \text{im} \iota$ where $\iota$ is the inclusion of $H^2(X,\Bbb Z)$ into $H^2(X,\Bbb C)$

Balarka Sen

11:30

I am out of ideas.

Danu

Where's Ted when complex geometry gets out of hand in the morning instead of evening?!

Balarka Sen

For some reason I didn't notice the real coefficients when I told you that obstruction thing. Ugh.

Ralf17

12:31

@Danu There was no price given. I just need to show it:

How can I show that it is impossible to add two mixtures of candy worth $4 per pound and $5 per pound to obtain a final mixture worth $6 per pound?

With no other additional information.

I thought you mean I represent it this way:

4a + 5b = 6(a + b)

where a represents the pounds for $4 and b represent the pounds for $5. a + b is the resulting pounds by mixing these to a mixture worth $6

@Danu How can I show that that is impossible?

Vrouvrou

Hello, is there an easy way to prove this: $$\forall x\in\mathbb{R}, \forall\varepsilon>0,\exists r\in \mathbb{Q}, |x-r|<\varepsilon $$

thank you

Danu

@Vrouvrou Think about $1/2^n$.

Vrouvrou

$\frac{1}{2^n}\in \mathbb{Q}$

and then ?

I think that $r$ must have a relation with $x$ and $\varespsilon$

@Danu?

someone here ?

user116211

12:48

@Vrouvrou You don't have to ask whether someone is here.

user116211

If someone is interested, they will help you now or later.

user116211

You have to wait.

Simple Art

To anyone who's interested:

5

Q: Did I derive a new form of the gamma function?

I wish to extend the factorial to non-integer arguments in a unique way, given the following conditions: $n!=n(n-1)!$ $1!=1$ $$f(x):=\ln(x!)$$ $$f(x)=\ln(x!)=\ln(x)+\ln((x-1)!)=\ln(x)+f(x-1)$$ $$f(x)=f(x-1)+\ln(x)$$ $$\frac d{dx}f(x)=\frac d{dx}f(x-1)+\ln(x)$$ $$f'(x)=f'(x-1)+\frac1...

It's a nice read I think.

iwriteonbananas

13:30

Suppose we have maps $f:Z \rightleftarrows X: g$ both of which are $\pi_*$-iso's and such that $f\circ g\simeq \operatorname{id}_X$. Suppose $Z$ is a CW complex. Do the maps form a homotopy equivalence?

Balarka Sen

13:42

Hi @iwriteonbananas

I doubt this. $X$ can be horrible.

iwriteonbananas

Hi Ballaka

Find an example

Fuck I wanna know if it's true or not....

Can u come up with a counterexample?

Balarka Sen

Hmm, the $f \circ g \cong \text{id}_X$ condition is interesting.

I'd have to think. You should post it in MSE.

iwriteonbananas

Yeah

Ok, fine

0celo7

Reminds me of a theorem from Whitehead's Combinatorical Homotopy.

I think the condition $fg\sim \mathrm{id}_X$ is called being "dominated" by a CW complex.

Balarka Sen

Yes.

iwriteonbananas

13:50

Hm ok

0celo7

maths.ed.ac.uk/~aar/papers/jhcwch1.pdf

page 215

iwriteonbananas

I see

0celo7

@BalarkaSen Do you know about Arzela-Ascoli type lemmas?

Balarka Sen

No.

0celo7

Damn.

My analysis prof keeps giving wrong hints for the homework, and it's pretty infuriating.

Wrong / takes way more work to fix

iwriteonbananas

13:56

I hate when that happens

It borders on disrespect to give half-assed/wrong hints

0celo7

To be fair, the thing you need to complete the suggested proof seems reasonable.

$X$ a metric space, $\{K_i\}$ a set of increasing compact subsets such that $\cup K_i=X$. Then any compact $C\subset X$ is contained in one $K_i$.

That's not true, somewhat to my surprise.

iwriteonbananas

Oh yeah, classic

One needs T1 or something

0celo7

@iwriteonbananas Huh? Metric spaces are automatically $T_1$.

Balarka Sen

@iwriteonbananas I tried to work out an example for your thing with the Warsaw cycle. Didn't work.

iwriteonbananas

Oh we got a metric space, nvm

@BalarkaSen Too bad :(

0celo7

14:00

@iwriteonbananas My space is even separable and it's still wrong.

Consider $X=\{0\}\cup\{1/n\mid n\in\Bbb N\}$ with the subspace top from $\Bbb R$.

And $K_n=\{0\}\cup\{1/i\mid i\le n\}$.

iwriteonbananas

Mhm

0celo7

Then $X$ is not contained in any of them.

iwriteonbananas

Yea

0celo7

This can be easily adjusted so that $X$ is noncompact

(the thing I'm trying to prove is only interesting in that case)

So, now the goal is to see if I can prove A-A directly on separable metric spaces.

I might pick up a copy of Bourbaki at the library. I hear they prove it there.

iwriteonbananas

What course is this for?

0celo7

14:08

Intro analysis

iwriteonbananas

Seems like a hardcore intro

0celo7

Agreed

@iwriteonbananas I'm glad I went against one prof's advice and took baby analysis in 1D first.

Balarka Sen

@iwriteonbananas Hm, I have an idea.

iwriteonbananas

Let's hear it

Balarka Sen

Oh, wait a second, you want $Z$ to be a CW complex, not just homotopy type of a CW complex? Then I'm out of ideas.

I wanted to set $X$ = double comb space, $Z$ = one of the comb components, $X \to Z$ to be pinching the other comb component, $Z \to X$ to be inclusion.

iwriteonbananas

14:13

$Z$ is an honest CW complex

Balarka Sen

Oh well.

iwriteonbananas

What's the double comb space again?

0celo7

Algebraic topology is too much for me, and I need to eat.

Bye

iwriteonbananas

Enjoy the food

Balarka Sen

This

Too hard to write down.

iwriteonbananas

14:16

Oh right

Balarka Sen

I don't berember any other examples of maps $X \to Y$ which is isom on $\pi_*$ but not homotopy eq

iwriteonbananas

I'm out of examples too

Balarka Sen

why do you need to find a counterexample/proof for such a horrendous theorem

i mean, yuck ;)

iwriteonbananas

Well, I am trying to show that if the functor $[X,-]:Top\to Set$ sends weak equivalences to bijections, then $X$ has the homotopy type of a CW complex.

Balarka Sen

huh

iwriteonbananas

14:21

So we take a CW approximation of $X$

I.e. a weak equivalence $f:Z\to X$

this gives a bijection $f_*:[X,Z]\to [X,X]$

a pre-image of $id_X$ should give us a homotopy inverse of $f$

but a priori we only know it is a right homotopy inverse

Balarka Sen

that kind of thing always makes me wonder about Yoneda

iwriteonbananas

and I wonder if in general it's automatic that it's also a left htp inverse

Balarka Sen

I see

iwriteonbananas

Question:

Mike Miller

@iwriteonbananas No. That would prove every space is homotopy rwquivslent to a CW coplex

iwriteonbananas

14:31

Oh heah

hehe

wait

why?

Mike Miller

Sometjing with a right homotopy inverse that's also a left homotopt inverse is a homotopy wquivalence?

oh I am so bad

nevermind

iwriteonbananas

Whew, ok

Balarka Sen

Morning, @MikeMiller.

Ramanujan

Hi nerds

Danu

Hey @MikeMiller

Ramanujan

14:37

Hi Danu,the nerd.

Danu

@Ramanujan ?? Is it some kind of joke I'm not getting?

Ramanujan

Nor really

0celo7

@BalarkaSen That's not even that horrible.

I was expecting some fractal.

@Danu No, you're just a nerd.

Danu

I don't perceive myself as a nerd.

iwriteonbananas

You're a nerd, end of debate.

Danu

14:40

What do you know about me?

iwriteonbananas

That you're a nerd

Mike Miller

I mean, iwritebananas did end the debate

0celo7

You're a wannabe physicist/mathematician. End of debate.

Mike Miller

You can't argue with that

Danu

OK. Cool story bros

Ramanujan

14:40

Is there any of you on fb? So that I can send links of my any asked question?

iwriteonbananas

Sorry Danu, I'm just letting my frustration out on you

Danu

You guys are doing great representing the shitty stereotype ;)

Balarka Sen

@Ramanujan No, we're nerds, we don't.

Mike Miller

If I was on FB "Can I friend you so I can send you links to my questions" would be the least convincing request I've ever heard.

Danu

^

Ramanujan

14:44

Is there any?

user116211

2 hours ago, by MAFIA36790

@Vrouvrou You don't have to ask whether someone is here.

user116211

2 hours ago, by MAFIA36790

If someone is interested, they will help you now or later.

user116211

._.

Ramanujan

OK ;) I will keep in mind

Danu

@BalarkaSen what is the index of the intersection pairing?

Balarka Sen

14:55

mathworld.wolfram.com/QuadraticFormIndex.html.

Danu

So the number of "negative directions"

But my index is given as $(a,b)$.

Balarka Sen

Not sure what you mean by direction.

Danu

@BalarkaSen Linear subspaces generated by a unit vector

Mike Miller

It's giving you both the dimension of the maximal-dimensional positive definite subspace and negative definite subspace.

$(b^+, b^-)$.

Danu

Ah, okay. Thanks.

Balarka Sen

14:59

I think that's called the signature.

Adeek

Yes @BalarkaSen

Mike Miller

At least in topology, signature means the difference.

Balarka Sen

Interesting.

Danu

That's how I learned it too

user227867

Anyone seen Chris's Sis around? Hasn't been to chat for long

Mike Miller

15:09

@BalarkaSen It's the difference that is an oriented cobordism invariant.

Danu

@MikeMiller This is also the only thing I learned about signature, haha :D

Balarka Sen

@MikeMiller Ah

user227867

@idomathart how is your book coming along?

0celo7

I don't think that person comes here any more.

Danu

Your ping won't work.

user227867

15:18

Thanks for telling me. The chat ping rules keep changing so I don't bother to find out the latest version, lol.

Mike Miller

If your ping autocompletes to a name, it will work. If it doesn't, it won't.,

user227867

There is a way for it to work which I shall demonstrate now, hang on...

Ramanujan

0

Q: Number of solutions in an equation

Why do we get $2$ solutions for a quadratic equation and $3$ solutions for a cubic equation and $4$ for biquadratic equation and so forth?

soft-question

user227867

@Idomathart How is your book coming along? I deleted my channel again but today I made a new one again with one video. =) youtube.com/watch?v=-aIm6kRPDz4

Mike Miller

Are you sure those work?

user227867

15:24

They will get it in the SE global inbox notification.

Mike Miller

Interesting.

Balarka Sen

I have never heard of this before.

user227867

Chat pinging is a difficult art to master.

user227867

If you feel very bored and have nothing better to do, you may listen to my masterpiece above in the link. =) Remember to subscribe and upvote if you want. =)

user227867

I am now eagerly awaiting the second edition of Riemannian Manifolds by John M Lee to be published next year. Just a few more months and my book list of 24 books is complete.

user227867

15:28

It now has 23 books and together the 24 will fill up exactly one row on my bookshelf.

0celo7

WHAT

He's publishing a second edition?

Proof?

Mike Miller

Jasper emails him.

user227867

You can visit his website to find info.

user227867

I also email him now and then but that is privileged information. =)

user227867

On his site he does say he is working on a second edition. =)

0celo7

15:30

@WillHunting I know that but he doesn't give a year

user227867

Of course, it might be delayed again. I have been waiting for the second edition for five years now. Also, I have been trying to get well for over a decade.

0celo7

A shame on both counts

user227867

In the words of Nash's wife in the movie, I need to believe that something extraordinary is possible.

0celo7

Did he say what will be in the new edition?

The first edition isn't anything special, it's a nice little book

I wouldn't mind getting the second edition when it comes out

user227867

Well, if there is anyone here interested in Buddhism, I would like to recommend these four books: Long/Middle Length/Connected/Numerical Discourses of the Buddha published by Wisdom Publications. I write it here because they are quite unknown in the world.

0celo7

15:34

(I think do Carmo covers way more material)

user227867

@0celo7 do Carmo does not contain modern machinery, which is a shame.

0celo7

What modern machinery?

user227867

@0celo7 There would not be any drastic differences. Just the usual improvements in a second edition.

user227867

I don't know. I haven't really seen do Carmo but I read many reviews on Amazon etc.

0celo7

I don't know of any "modern machinery" in basic Riemannian geometry

user227867

15:36

But I just wanna mention that Lee has the best treatment of Hopf's Umlaufsatz and the Gauss-Bonnet theorem I have seen in that book.

0celo7

That is true

user227867

In terms of the level of generality and the presentation of the theorems and proofs. He is a genius. But maybe by the time I can enter grad school he would have retired. Otherwise maybe I will try go Washington to work with him on Riemannian geometry.

0celo7

> This is probably what makes the book good to start with, but there is still going to be a somewhat difficult transition from this book to a typical differential/riemannian geometry book. Namely, the basic language of vector bundles, pull backs/push forwards, tensors and tensor fields are either covered in a very specific framework or not at all.

user227867

Well, I have hope because people like Serge Lang retire only at 75. =)

user227867

And then he died at 78, I think.

user227867

15:38

I wonder where he has been reborn. =)

user227867

OK, I am going to pee now, bye.

0celo7

...

Semiclassical

16:24

hi chat

0celo7

@Semiclassical Do you want to read a long-ish analysis proof

Semiclassical

not really

0celo7

@Semiclassical I think I have a fabled 5$\epsilon$ proof

Semiclassical

that sounds horrible

0celo7

yes it is horrible.

in the notes it's just a 3$\epsilon$ proof but I think the notes are wrong

you need two more estimates lol

unless we're implicitly using uniform equicontinuity, hmm

Semiclassical

16:39

epsilon management just seems kind've horrifying

0celo7

problem is, I know my 5 epsilon proof is correct

there's no doubt

but a 5 epsilon proof is horrifying

\begin{align*}||f_n(x)-f_m(x)||&\le ||f_n(x)-f_n(x_i)||+||f_n(x_i)-f_n(d_i)||+||f_n(d_i)-f_m(d_i)||\\
&\quad+||f_m(d_i)-f_m(x_i)||+||f_m(x_i)-f_m(x)||
\end{align*}

@Semiclassical how do you like that

Semiclassical

that's not so bad. essentially just writing the LHS as a telescoping sum and then applying the triangle inequality?

0celo7

yes

wait until I post the full proof in a minute lol

Let $(X,d)$ be a $\sigma$-compact metric space, $\mathscr F\subset C_E(X)$ a family of continuous functions with $E$ Banach. If $\mathscr F$ is equicontinuous at each point of $X$ and $\mathscr F(x)$ is precompact in $E$ for each $x\in X$, then any sequence of functions in $\mathscr F$ has a subsequence converging uniformly on compact sets.

*Proof.*
Since $X$ is $\sigma$-compact, it has a countable dense subset $\mathscr D$ (see problem 6). Let $(f_n)\subset\mathscr F$ be a sequence of functions. We enumerate $\mathscr D$ as $\{d_1,d_2,d_3,\dotsc\}$. The set $\{f_n(d_1)\mid n\in\Bbb N\}$ i…

Semiclassical

gag

0celo7

@Semiclassical pls read :)

Semiclassical

16:49

ugh

Missing a word in the first sentence: "To show that a sequence of continuous [?] converges uniformly on compact sets,..."

0celo7

that's not the first sentence

but thanks

Semiclassical

well, first sentence you posted

0celo7

uh

expand the window

or scroll

Semiclassical

oh, you're right. i already had scrolled a bit

yeah, I can't wade through that

0celo7

the main part is the 5-epsilon part

Semiclassical

16:52

my brain just spontaneously reacts as in this: viruscomix.com/page420.html

0celo7

my prof tried to be clever and choose the ball radii cleverly, but I don't think it works

:(

you gave me the finger

Semiclassical

no, I gave the proof the finger

0celo7

I think the proof is correct

Semiclassical

didn't say it wasn't

0celo7

maybe a little long

but tell that to my prof who gave it as homework

Semiclassical

16:57

just meant that it's the kind of proof (and probably the kind of problem) that my brain just won't read

Adeek

let us see @0celo7

0celo7

@Adeek see the huge text box above.

the proof is actually straightforward in a perverse sense

Transcript for

$Mathematics$ Mathematics