Mathematics - 2016-10-14 (page 1 of 5)

Adeek

00:21

hey @TedShifrin our U.S was telling us how she wants to move to Canada

permanently if trump wins haha

Ted Shifrin

00:42

She is far from alone, Karim. But I'm reasonably confident this won't happen. I just voted and I want to remove myself from the next month.

Mike Miller

I just want a landslide.

Ted Shifrin

Me too.

But there are a lot of haters out there, including a few who show up here, apparently. (Danu will censor me again, no doubt.)

I just added a comment on one of your comments, @MikeM. Things are a bit more subtle when you stay in the holomorphic category, re Chern classes.

Mike Miller

@Ted Oh, good call. The tautological bundle doesn't have many sections. I ignored the holomorphic bit.

I already contacted him privately, saying I don't think he's going to get much "from the exact sequence alone".

Ted Shifrin

Well, the exact sequence leads you to the right $(1,1)$-form, actually.

Mike Miller

I also don't think he wants the curvature story, since it kills torsion.

Ted Shifrin

00:51

For $c_1$ there isn't torsion.

You're too used to real stuff.

Adeek

oh your here @TedShifrin

Mike Miller

If the manifold isn't simply connected there can be.

Ted Shifrin

No I'm not.

Mike Miller

Take an Enriques surface.

Adeek

I wanted to ask you if you got the solution manual for your book ? I am planning to solve it and go systematically over it

Ted Shifrin

00:52

Hmm, how do you get torsion by pulling back cohomology from $\Bbb P^n$?

Adeek

but I wanted to check my my soln are correct

Mike Miller

The map goes in the other direction!

Ted Shifrin

Say what, @MikeM?

Mike Miller

It would just say that eg twice the generator dies in the pullback.

Adeek

the multi variable mathematics book @TedShifrin

Ted Shifrin

00:53

I know, Karim. There's no solutions manual freely available. If you want to send me occasional well-written things to critique, I'll be glad to do that for you.

Adeek

okay sounds good. yeah I can send my solutions to you just to get some critique over my soln

I didn't take a rigorous multi-variable course so that would be really nice

Ted Shifrin

Very few people did, Karim.

Adeek

it is such a shame

Ted Shifrin

OK, @MikeM, I sit corrected. I'll have to ponder this a bit.

Adeek

brb gonna go grab the bus

Ted Shifrin

00:56

Buses are awfully big to grab.

Mike Miller

@Ted: There should be a flat holomorphic line bundle on an Enriques surface.

Ted Shifrin

It's been years since I've thought about Enriques surfaces. :(

You mean flat but nontrivial, of course.

Mike Miller

Right.

I don't have many examples to work with... I like k3 and related objects.

Ted Shifrin

I haven't thought about surfaces in many years. I miss them. I wonder if the canonical bundle of an Enriques fits your bill.

Adeek

01:15

I am gonna attend this talk tomorrow

seems interesting but I don't think I will understand a lot

Title: Some new Fano fourfolds

Abstract: I will discuss some new Fano fourfolds found in quiver flag varieties as zero loci of sections of homogenous vector bundles. This is work with Tom Coates. The classification of Fano varieties up to deformation is known in dimensions 1, 2, and 3. The program of Corti, Galkin, Golsyhev, Kasprzyk, and others seeks to use mirror symmetry to find and classify Fano varieties in dimension 4 (and more). Küchle classified 4 dimensional Fano varieties of this form in Grassmannians; quiver flag varieties are a generalization of Grassmannians that include flag…

I think @MikeMiller you understand those stuff right ?

I can't wait for the day I am actually understanding seminar talk

Forever Mozart

What are the open problems in algebraic topology? I'm curious because everyone here seems interested in that vs general topology

Ted Shifrin

@Adeek: That's some serious algebraic geometry there.

But you definitely need to be learning Grassmannians soon :)

@Forever: There's more stuff going on in geometric topology of various sorts than in algebraic topology. There is some long-standing work in homotopy theory questions, but I honestly don't know what "pure" algebraic topologists are doing.

Adeek

I learned grassmannian from differential geometry class

the is the only word I know haha @TedShifrin

there is a lot of algebraic geometers here @TedShifrin

Ted Shifrin

01:31

Grassmannians were the key to my thesis work, Karim. I love them. :)

Adeek

I should buy ipad so I can read books while I am on the bus going home

Mike Miller

@Ted: Yeah, I think it does.

Karim: I don't know any of that stuff.

Ted Shifrin

I do too, @MikeM. I just don't berember that stuff.

Adeek

cool @TedShifrin yeah I need to learn more and more. So, I can have a clear idea when I start writing my thesis I know what I will write about or prove.

Ted Shifrin

That's a bit down the road, Karim. Just learn in the meantime.

Adeek

01:32

I think for now I should concentrate on subjects and learning in my free time and problem solving

yeah @TedShifrin

Mike Miller

@Ted: I would have to think about it. My favorite approach is to think of K3 as a resolution of T^4 mod pm 1.

Probably pretty easy to write down from this perspective.

Kodaira proved K3s are all deformation equivalent. Is that true for Enriques' too?

ffahim

05:31

If ylogx = 6 & log14x^8y =3 then how to get the value of x.....?

Forever Mozart

the first equation says $x^y=6$

wait in the second equation is y in the power ?

arctic tern

@ffahim do you mean log_14(x^(8y))? use parentheses.

use power rule for logarithms

and change of base formula so both logarithms have the same base

actually that's probably log(14x^(8y)), in which case forget change of base formula, apply product rule then power rule

ffahim

06:25

Can you pls kindly give the whole solution...pls @arctictern

@ForeverMozart yes the power is 8y

Pls can u pls give the full solution @ForeverMozart

Forever Mozart

let me see

ffahim

@ForeverMozart @arctictern I have tried to go towards....something like this Log_(14(x)^8y)= log_14 + log_(x)^8y= 3 ... then log_14+8log_x^y=3 then ...?:/

Forever Mozart

im confused. Is it ln(14*(x^(8y)))

ffahim

No its log(14(x^8y)).... is it okay now :-)?

user116211

Where is Balarka ._.

Forever Mozart

06:40

lol what is the difference?

so it is log base 10?

ffahim

Yeah..

Not e based

Forever Mozart

I do not think there is a solution

$y\ln x=6=2*3=2*\ln(14x^{8y})=16y \ln(14x)$

oh

so $\ln x=16\ln(14x)$

but then $x=14xe^{16}$

$1=14e^{16}$

no solution

user116211

My bad; sorry.

user116211

Anyways, @Balarka, you here?

Balarka Sen

Yes

Forever Mozart

06:46

i guess $y$ could be $0$ and then $x$ could be anything

so there's a solution

user116211

Well, last night I asked a question about countability of subset of $\mathbb N;$ got a pretty answer from Brian.

ffahim

Okay... tell me one thing @ForeverMozart is there any way or trick to find out the value of log2 ... without calculator.....?

Forever Mozart

yes

ffahim

?????

Pls

user116211

The question was basically about one statement in the proof $f(n)\geq n$ for each $n\in \mathbb N\,.$

Forever Mozart

06:48

log 2=a means 2=10^a

but that is not easy to solve

user116211

Where $f$ maps the subset to $\mathbb N\,.$

user116211

It is an injection.

user116211

The proof was showing then how it is surjective too.

user116211

Then the author invoked the condition above.

ffahim

Ooh I see... and why don't we take the base of log negative? @ForeverMozart

user116211

06:49

I asked how the author concluded the condition.

Balarka Sen

I don't understand. $f$ is just an injective map from a subset of $\Bbb N$ to $\Bbb N$?

Without more hypothesis, it need not be surjective.

user116211

@BalarkaSen The proof needed to show $f$ is bijective.

Balarka Sen

I am asking what are the hypothesis on $f$.

Forever Mozart

what is the question

user116211

@BalarkaSen Initially, it was just taken as a map.

user116211

06:53

$f: S\mapsto \mathbb N\,.$

user116211

where $S$ is the subset.

Forever Mozart

Q: What does a drowning number theorist say?

Balarka Sen

log log, log log log

Forever Mozart

yes :)

Balarka Sen

@MAFIA36790 What else?

user116211

06:54

Nothing more.

user116211

Anyways, coming to my question....

Tobias Kildetoft

@MAFIA36790 then you can conclude precisely nothing about the map

Balarka Sen

Well, that doesn't make any sense, precisely for the reason Tobias said

user116211

Wait, I'm showing the post....

user116211

Probably missed something or so; but check the complete excerpt:

user116211

06:56

0

Q: Problem in understanding the proof of existence of a bijective function between a subset of Natural number and set of Natural number itself.

I was reading this pdf where I came across this proposition and its proof: Proposition 2: Let $S \subset \mathbb N$ be an infinite subset. Then there exists a bijection between $S$ and $\mathbb N.$ Proof: You might think this is intuitively clear. But let’s prove it carefully. We define a...

elementary-set-theory

Balarka Sen

$f$ pretty darn sure isn't "just a map" there. They define it explicitly.

Not all maps $\Bbb N \to S$ are bijections.

(Which, speaking of, your domain and codomain are switched)

user116211

@BalarkaSen yeh, noticed that; typo.

user116211

Re-reading it again; I think I can sought out my confusion; otherwise I'll ask.

Balarka Sen

Sure

user116211

:)

Forever Mozart

07:03

Mmmm

user116211

BTW, if you don't know already, @Balarka, JD will not be returning for another year in h bar. So, the bar is pretty good now.

Balarka Sen

That's (good) news. How did that come to happen?

user116211

@BalarkaSen You know...

user116211

ban...

Balarka Sen

Funny, I never thought that guy wrote anything ban-able. But physics.SE mods are strict and impatient.

user116211

07:08

Anyways, will not talk about the man who is banned, but ~~except 0celo7~~ everyone is quite okay with it ;P

Balarka Sen

understandably

Forever Mozart

he is banned?

what a bunch of horseshit

user116211

He dared to talk physics in h-bar which is a punishable offence.

Forever Mozart

;)

user116211

H-bar is only for maths.

Forever Mozart

07:09

where is that?

Balarka Sen

I disagree with that.

Forever Mozart

link?

user116211

And sometimes maths-physics ;)

user116211

re-reading...

user116211

@ForeverMozart oh; you don't know h bar? Are you asking for the link of the room?

user116211

07:15

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...

10

32

Forever Mozart

so he's banned here too?

user116211

Don't know whether he comes here or not.

Forever Mozart

0celo7?

user116211

?_?

user116211

I was not talking about 0celo7.

Tobias Kildetoft

07:25

@BalarkaSen One of my collaborators has just sent in a proposal for us (as well as some others) to form a SQuaRE (except we will be 5 people, so it will really be a pentagon).

Balarka Sen

@TobiasKildetoft What's a SQuaRE?

Tobias Kildetoft

@BalarkaSen aimath.org/programs/squares

Balarka Sen

hah, that's pretty nice

Forever Mozart

i am not a collaborator

hey if I submitted a paper over 2 months ago and have not heard anything, should I ask what's going on?

Tobias Kildetoft

@ForeverMozart you mean not heard even a confirmation that it was received?

Forever Mozart

07:34

I think they said it was received

but I have no idea if anyone is reviewing it

Tobias Kildetoft

then 2 months is not that long

I think anything less than 6 months is too optimistic to hope for, but after 6 months it is usually considered ok to inquire about the status

(math reviewers are slow)

Forever Mozart

I would share the link to my paper on Arxiv

but then I would have to reveal my true identity ;)

Tobias Kildetoft

heh. If I really felt like it I could probably find it given the amount of information you have revealed about what the topic is. But that sounds like way too much work

Forever Mozart

yeah

i like to leave a margin of indefiniteness

Balarka Sen

whatever

Itachí Uchiha

07:41

How to solve this differential equation?

\frac{d^2 \theta}{dt^2}=-k\theta

Forever Mozart

your papers are very difficult

$\frac{d^2 \theta}{dt^2}=-k\theta$

Tobias Kildetoft

@ForeverMozart mine?

Forever Mozart

yes the symbols!!!

Tobias Kildetoft

symbols? What symbols?

user116211

@ItachíUchiha How?

user116211

07:50

Take the non-trivial trial solution as $\theta = Ce^{mt}\ne 0\,.$

Tobias Kildetoft

08:23

This is a weird combinatorial thing I am getting into now. So I am considering those Young diagrams which have the property that whenever an odd-numbered row has an odd length, then the next row has the same length. Out of these, how many (relatively) have the further property that whenever an even-numbered row has even length, then the next row has the same length?

Forever Mozart

you want to consider some type of density

Tobias Kildetoft

Well, ideally I would like a nice formula for the numbers of each type of diagram, but that seems unlikely. So I would settle for (even an approximation) of the ratio

Forever Mozart

oh

given the dimensions

how many have the property

Tobias Kildetoft

Ohh, right, these are corresponding to partitions of the same (even) number. I also actually forgot to mention a condition on them which is that they should be tileable using dominoes

I think the fact that they are tileable together with the condition on the odd-numbered rows mean that whenever an even-numbered row has an odd length, then the row above it has the same length (this is at least true up to partitions of 6), but I am not sure.

Forever Mozart

08:50

user116211

09:14

@ForeverMozart Was it from the Presidential campaign? Quite bold.

user116211

Okay, @balarka, I've read it again; but still I'm not getting the last point of Brian's answer:

user116211

1

A: Problem in understanding the proof of existence of a bijective function between a subset of Natural number and set of Natural number itself.

Let $B=\{n\in\Bbb N:f(n)<n\}$. If there is some $n\in\Bbb N$ such that $f(n)<n$, then $B\ne\varnothing$, so $B$ has a least element $m$. Clearly $m>0$ (why?), so $m=k+1$ for some $k\in\Bbb N$. Then $k\notin B$, so $f(k)\ge k$. But $f(m)<m$, so $f(m)\le k$, and therefore $f(m)\le f(k)$, contradict...

user116211

I got $f(m)\leq f(k)\,.$

user116211

But how is it "contradicting the construction of $f$"?

Tobias Kildetoft

@MAFIA36790 because by construction you have $f(n) \geq n$.

Balarka Sen

09:18

$m = k+1$, so that says $f(k+1) \leq f(k)$. But that's garbage due to how $f$ is defined.

user116211

@TobiasKildetoft yes.

Balarka Sen

Namely, $f(n+1)$ is recursively the least element in the complement of $f(\{0, 1, \cdots, n\})$.

If $x = f(k+1)$ was really smaller than $f(k)$, $f(k)$ would have have taken the value $x$.

Tobias Kildetoft

09:39

@BalarkaSen My mentor thinks what I have been doing recently ought to be worth writing up as a paper, so now I need to figure out what sort of thing goes into a paper in computational algebra

Balarka Sen

@TobiasKildetoft May I ask what you have been working on?

Tobias Kildetoft

@BalarkaSen computing the left-, right- and twosided Kazhdan-Lusztig cells in some small rank cases

as well as Hasse diagrams for the two-sided orders (still working on ones for the other orders)

Balarka Sen

ah, ok

Tobias Kildetoft

The diagram for type $B_6$ looks quite nice dropbox.com/s/3vaxzpfwbn0jhy9/B6-twosided.pdf?dl=0

The full twosided cells look less nice dropbox.com/s/wa1rj8h9pvem0st/B6-full.pdf?dl=0

Balarka Sen

cute, although I don't really know what those signify

Tobias Kildetoft

09:48

I have a computer working on type $B_7$ at the moment

The nodes in the graph are twosided KL cells which correspond to a certain partition of the representations of the Hecke algebra

and the arrows are the twosided order which correspond to twosided ideals generated by the corresponding elements

Balarka Sen

ok.

Tobias Kildetoft

These cells are indexed by those types of partitions I mentioned counting earlier (with the subset I wanted to count indexing those that are strongly regular), and the order is the dominance order on such partitions (at least for the cases I have calculated so far).

Vrouvrou

Hello. someone know a counter example to prove that $f(A)\cap f(A') \subset f(A\cap A')$ is not true where $f:E\rightarrow F$ is an application and $A,A'\subset E$

Tobias Kildetoft

@BalarkaSen I had to set the page size to 100x230 inches for that full pdf of the cells

Vrouvrou

if $f$ is injective the statment is true , but have a counter example where $f$ is not injective ?

Tobias Kildetoft

09:56

@Vrouvrou take $E$ to have two elements which are sent to the same element in $F$

(also, it is called a function in English)

Vrouvrou

@TobiasKildetoft thank you for example $f: \{a,b\}\rightarrow \{a,b\}$ whith $f(a)=b, f(b)=a$

Tobias Kildetoft

No

Vrouvrou

so how?

Tobias Kildetoft

that one does not satisfy what I said

Vrouvrou

$E$ has two elements

and F has the same elements as E

Tobias Kildetoft

10:01

I did not say that $F$ should have the same elements as $E$

Vrouvrou

ok

$f:\{a,b\}\rightarrow \{a,b,c\}$ and $f(a)=a, f(b)=c$ for example ?

Tobias Kildetoft

"which are sent to the same element in F" is the important part.

Vrouvrou

$f(b)=b$

you mean this this by "sent to the same element in F

$f(a)=a, f(b)=b

$

Tobias Kildetoft

those are not the same element.

Vrouvrou

you means $f(a)=f(b)$

Tobias Kildetoft

10:07

yes

Vrouvrou

ok so i take $f:\{a,b\}\rightarrow \{a,b,c\}$ and $f(a)=f(b)=c$

Tobias Kildetoft

good

Vrouvrou

and $A=\{a,b\}$ and $A'=\{b\}$

Tobias Kildetoft

try taking the sets disjoint

Vrouvrou

$A=\{a\}$ $A'=\{b\}$

Tobias Kildetoft

10:10

right

Vrouvrou

$f(A\cap A')= \emptyset$ and $f(A)\cap f(B)=\{c\}$

thank you

ypercubeᵀᴹ

10:41

Is it possible to search the site for question with specific equation? Example: $a^3+b^4=c^5$

The search yields nothing but I don't trust SO search too much

Tobias Kildetoft

@ypercubeᵀᴹ I think someone on meta mentioned a customized search engine which allows for such

namely, it also allows for the possibility that someone used a slightly different notation

66

Q: Announcing a third-party search engine for Math StackExchange.

Approach0 is a math-aware search engine. “Math-aware” means you can add math expression(s) as some of your keywords to have search engine help you find similar expressions. Check out here: https://approach0.xyz This is my side-project, hopefully it can be useful in some cases to help Math SE use...

discussion search

Secret

Is the following rather trivial looking algebaic structure self consistent (i.e. no two distinct elements can be shown to be equal). Can we really use the properties of the left zero subsemigroup formed by the + cayley table to simplify the distributive law and + associative law for this algebraic structure as follows?

Tobias Kildetoft

@Secret not sure what you mean.

ypercubeᵀᴹ

@TobiasKildetoft Thnx. Looks very good.

Secret

$a,b,c$ are any elements in the structure. The elements 0,1,q are the 3 distinct elements (and in no way should be treated as the zero or one as we knew, thus they are just abstract labels for the elements)

(i should fix the labels to emphasize they are just arbitrary elements, give me a sec..)

So... basically...

In this algebrac structure, we have 3 elements p,q,r

Their + cayley table is isomorphic to the left zero semigroup

while their * cayley table is isomorphic to the cyclic group of 3 elements $\mathbb{Z}/3\mathbb{Z}$

Now for a distributive law to hold:

a(b+c) need to be = ab+bc

From the * cayley table, we knew that * is closed thus any products of the form ab must be any of the 3 elements p,q,r

Now from the + cayley table, all 3 elements are left zeros thus $\exists z,\forall a, z+a=z$

Combining the above two reasonings we found that ab is also a left zero

Therefore ab+ac=ab, and a(b+c)=a(b)=ab. Thus the distributive law holds for all a,b,c

Is this a correct deduction?

Tobias Kildetoft

10:59

@Secret Yes, for left distributivity

Secret

We should be able to use simialr reasoning to show that (b+c)a=(b)a=ba, and ba+ca=ba, thus right distributivity should also holds?

Tobias Kildetoft

yeah

Secret

and similarly , to show that (a+b)+c=a+c=a, and a+(b+c)=a+b=a, thus + associativty also holds

Secret

Now the interesting question:

In group theory we knew that the fitting subgroup will control the sovability of a solvable group. Here

we also see a simialr phenomenon where nearly everything of this algebraic structure is controlled or dictated by the subsemigroup that is the + cayley table. Is there a general name for a subobject that has such controllign behaviour to the object in question?

Tobias Kildetoft

@Secret But this is not a subobject (it is the image of a forgetful functor instead).

Anyway, I need to go teach now.

Secret

11:07

ok, see ya

ffahim

12:13

(2^X-1)^1/2=y ...what's it's domain?

Anubhav

@MikeMiller I wonder if you know any easy explanation of the fact that " $S^1\times S^2$ is the only prime (3) manifold is a cover of a non-prime manifold $ \mathbb RP^3#\mathbb RP^3$"

This sounds a bit odd...so let me put in this way... The only example of a prime manifold which is a cover of a non prime manifold is $S^1\times S^2 \to RP^3 \# RP^3$

Secret

@ffahim in the reals or complex?

user228700

12:34

@Ramanujan: Do u have the time to solve a problem together? I dunno y I'm not getting it right :/

ffahim

As far I have learned in text book only real answers are granted...@Secret :-)

Secret

well, in that case the domain are the x where y is real

the squareroot is important here

Dartek12

hi guys

anyone who knows some circuit theory?

DHMO

12:50

Someone said that there is an "aleph fixed point", which means that a number for which the subscript of the aleph is equal to itself. Then, he proceeded to say that that number is an "endless cascade of aleph", meaning $\aleph_{\aleph_{\aleph_{\cdots}}}$. My question is: is this legit?

2. a torus surface (genus 1) can be formed from a square by identify the two pairs of opposite edges. is there a way to generate a sphere surface (genus 0) with a similar method?

Ramanujan

@Kaumudi yes

ffahim

hey...... does 0 belong to R+ ?

DHMO

@ffahim no

Ramanujan

@ffahim why 0 does not belong to R-? @DHMO

SteamyRoot

Who is "someone"? I don't dare say if it's right or not, but in any case using Aleph as a subscript like that seems fishy, and I've definitely neve come across it. I've only seen $0,1, 2, \dots$ and ordinals $\omega$ as indices for Aleph

The h Bar

Transcript for

$Mathematics$ Mathematics