Mathematics - 2016-04-08 (page 1 of 3)

Akiva Weinberger

12:00 AM

@PVAL I think I heard someone say that algebra was algebraic topology with the soul sucked out of it

or something like that

Carry on Smiling

@AkivaWeinberger can you speak foodtongue?

Akiva Weinberger

No. But I know what it is…

Are you a Mathcamper?

Carry on Smiling

no, I have a friend who was

Akiva Weinberger

Oh

Joshua Lamusga

Does anybody know what this question asks? http://i.imgur.com/bAWmIW6.png
Not how to do it, but what it asks. I think it's asking 2 questions and the first looks like I should verify the given information (using the given information?)

Carry on Smiling

12:11 AM

I was too busy doing my military service

Akiva Weinberger

Who's your friend? I might know them

Carry on Smiling

Alex Desatnik

Akiva Weinberger

Oh. No, I don't know them

("him?")

Carry on Smiling

it's one person

Akiva Weinberger

Alex is a neutral name, so I don't know if it's a boy or a girl

Carry on Smiling

12:13 AM

o right

Akiva Weinberger

What made you think I'd know foodtongue?

Carry on Smiling

I thought it was likely given you went to mathcamp

Akiva Weinberger

How'd you know I went to Mathcamp?

Carry on Smiling

Sorry, I wanted to test the veracity of this statement:

" I'm not the only person with my name so Googling it won't tell you much."

Akiva Weinberger

@JoshuaLamusga Do you know what parametric equations are and how to graph them?

@CarryonSmiling …ah

Joshua Lamusga

12:15 AM

@Akiva yes. I just need to know what it's asking. Like I said, it looks like it's asking me to verify info using said info, which is a fallacy. Otherwise, why would it begin with "show that the..."

Carry on Smiling

@AkivaWeinberger, I don't think my name will get you anywhere

Jorge Fernández

Akiva Weinberger

@JoshuaLamusga Show that the graph of those equations is a line.

Joshua Lamusga

Ah, so it is only one question?

Carry on Smiling

and it contains the points

Akiva Weinberger

Yeah, just one question

Joshua Lamusga

12:17 AM

@CarryonSmiling do you happen to have a phd at Pennsylvania University?

Carry on Smiling

sadly no

I'm an undergrad

Akiva Weinberger

Yeah, your name is much more common than mine.

I think I share my name with just one other guy, actually :P

Joshua Lamusga

Darn. Well it seems there are a lot more people with your name.

My name is unique, I think. Especially if I include my middle name

Carry on Smiling

@JoshuaLamusga yeah, there's even a Mr Universe winner (although that might be myself, I'm not giving more info)

catch you guys later

Joshua Lamusga

Bye.

Akiva Weinberger

12:20 AM

@JoshuaLamusga i.imgur.com/rPsaesR.jpg

That you?

Joshua Lamusga

Of course. I also intentionally proliferate my name, which makes it easier :)

Joshua Lamusga

12:59 AM

@AkivaWeinberger mind taking my pic off imgur? lol

Also, I think I've stared at the question I posted not too long ago enough to realize I have no idea where to start

Akiva Weinberger

It's unlisted

Not sure how to delete it, actually…

Joshua Lamusga

Don't worry about it. It'll just sink into the never-ending hole that is imgur and be buried under countless memes.

Akiva Weinberger

It's unlisted. Pretty sure that means no one can see it unless they have the link

Joshua Lamusga

Alright. Could you point me in the right direction for this question? I'm following Webassign and it has a tendency to deviate from the text (Stewart Calc) enough that this question seems to have no relationship to anything I've touched before. I have no idea where to start.

Akiva Weinberger

Wait, no, I figured out how to delete it

Maybe find the $y=mx+b$ formula for that line, and see if it's equivalent to the parametric stuff?

somehow

Joshua Lamusga

2:48 AM

I've got the answer to my question, but I don't understand it. It's here: i.imgur.com/fAYRRG1.png

Ted Shifrin

3:15 AM

@JoshuaLamusga: Solve for $t$.

Joshua Lamusga

4:13 AM

Thanks @TedShifrin.

user21820

5:24 AM

Hello.. Someone disputed my flag of math.stackexchange.com/questions/1571961/… as low quality. Why?? It explicitly talks nonsense about Z (integers) being the only finite subset of Q (rationals) that is "closed to multiplication". I had been previously told to flag it that way at meta.math.stackexchange.com/q/17512/21820.

Jester Tran

hi

user21820

5:37 AM

@JesterTran: Hi

skill patrol

@JesterTran @user21820 hi

user21820

@skillpatrol: So what do you think of my flag?

skill patrol

I posted your message in the math mods' office i hope you don't mind

user21820

@skillpatrol: I certainly don't mind. I am concerned that many 'answers' on Math SE are nonsense, not just this one and definitely not just this user who has a history of nonsense answers. See math.stackexchange.com/questions/46833/… for one example which Asaf was so kind as to respond nicely to.

@skillpatrol: One would think that some of these people would learn not to do that, but it doesn't seem to happen and they repeat their behaviour soon enough.

skill patrol

Indeed.

Let the mods handle them :-)

user21820

5:50 AM

yea that's why i don't understand the review process either.. i'm told explicitly at the meta site that i shouldn't flag for moderation attention

so i didn't lol

last time i posted in the reopen/undelete/close/delete chat-room and nothing got done.. i guess now i should be posting in the math mods chat-room?

skill patrol

yup @user21820

user21820

@skillpatrol: Thank you for your help!

skill patrol

np

Huy

6:12 AM

morning @EricStucky (sorry just wanted a reason to ping you)

Tobias Kildetoft

6:35 AM

@AlexClark Never really studied any Lie theory, so I don't have any recommendations for the topic.

user147690

6:45 AM

What do you study again @TobiasKildetoft?

user147690

Representation theory?

Tobias Kildetoft

@AlexClark Yeah, and higher versions (well, 2-, but that is still higher)

user147690

@TobiasKildetoft How does higher rep theory differ from rep theory?

Tobias Kildetoft

@AlexClark it is the study of $2$-representations of $2$-categories (well, in the version I do)

so it specializes to ordinary rep theory, though not everything in ordinary rep theory will come from higher stuff

Huy

@AlexClark: why do you want to study Lie theory ?

user147690

6:49 AM

Any good textbooks for this @TobiasKildetoft?

Tobias Kildetoft

@AlexClark To compare the two, you can compare my paper with Mazorchuk called "Special modules for positively based algebras" with his paper with Miemietz called "Transitive $2$-representations of finitary $2$-categories"

user147690

@Huy I find it interesting, and I am doing a research project at the moment on a specific type of basis for the universal enveloping algebra of $\mathfrak{n_+}$

Jester Tran

jo

Tobias Kildetoft

@AlexClark I don't think there are any proper textbooks yet, as the topic is still rather new. Mazorchuk has written one on categorification, which could be seen as a start, but a lot of the theory has been developed since, and many of the ideas make more sense when seen in the light of the newer stuff

Jester Tran

hi

Tobias Kildetoft

6:51 AM

@AlexClark The main thing that got the topic starter was Khovanov's categorification of the Jones polynomial

Together with some work of Lauda

user147690

@TobiasKildetoft Fair enough, I definitely believe that with what I have been trying to learn haha. I searched for a little to try to find some textbook on KLR algebras, and realised there were none

Tobias Kildetoft

@AlexClark Yeah, those are probably also still too new to have proper textbooks yet

(not that I really know anything about them)

@AlexClark Will you get to Soergel bimodules in the course you are following?

Jester Tran

The probability of getting a head on any single toss is 1/2. Find the probability that we get 3 heads in a row (from the start) before there occurs 2 tails. Write answer in numerical form

any help?

Huy

what's the opposite of this

user147690

@TobiasKildetoft It's not a course, just the research program with Peter Mcnamara, so I am not sure

Tobias Kildetoft

6:54 AM

@AlexClark Ahh, I thought those talks were as part of a course

user147690

@TobiasKildetoft Oh no, I gave a talk for no credit :D

Tobias Kildetoft

@AlexClark I have given many of those

Huy

or just draw a big tree

user147690

smp.uq.edu.au/people/PeterMcNamara/qft/index.html

user147690

@TobiasKildetoft Do you think it would be contained in one of these? math.stanford.edu/~petermc/syllabus.pdf

Tobias Kildetoft

6:56 AM

@AlexClark Probably not, though they are related to the KL conjecture (or rather, to the recent algebraic proof)

It is quite funny that it is still called the KL conjecture, even though it was proven within like 2 years of being stated (so whenever it is mentioned to non-experts, one has to add the "now a theorem" part).

user147690

Hahaha

Tobias Kildetoft

Soergel bimodules are a category of bimodules which categorify the Hecke algebra. The category can also in a natural way be considered as a $2$-category with one object, and its $2$-representation theory has a lot of nice consequences

For example, the type $A$ case was used in the classification of parabolic projective functors on category $\mathcal{O}$ by Mazorchuk and myself

user147690

@TobiasKildetoft Oh I see, your 2015 June paper

Tobias Kildetoft

@AlexClark yeah

We are working on proving similar statements in other types, and I hope this can be used to better understand parabolic projective functors in other types as well, though there will be a long way to go before that.

Paradox 101

The normal group $S$ in $G$ has an index $20$ and $|S|=7$ then we have to show that if $g \in G$ such that $g^7=e$ then $g \in S$. So if I say $|gS|=|S|=7$ then $(gS)^7=g^7 S=S$ which implies that $g \in S$. Is this correct? I know it should be simple but I don't get it. Any help would be appreciated thanks!

Tobias Kildetoft

7:07 AM

@Paradox101 You don't really say how you use that the index is $20$.

Paradox 101

@TobiasKildetoft I haven't used that and I'm not sure how to in this case

Tobias Kildetoft

@Paradox101 Well, how do you conclude that $g\in S$ from $(gS)^7 = S$?

Paradox 101

@TobiasKildetoft because $g^7=e \in S$ although I don't think my reasoning is correct

Tobias Kildetoft

@Paradox101 no, that was what you used to conclude that $(gS)^7 = S$.

You need more to conclude from this that $gS = S$

@AlexClark It was quite interesting actually. In studying the dihedral case of this, the Catalan polynomials showed up, and we needed to use that these are irreducible for prime indices, which follows from the same property of the Fibonacci polynomials, which is a result I found in a '69 paper in the Fibonacci Quarterly (which I had never heard of before that).

Paradox 101

or if I say $|G|=140$ implies that $g^140=(g^7)^20=e$ then $((gS)^7)^20=S$ which implies that $gS=S$? @TobiasKildetoft

Tobias Kildetoft

7:19 AM

@Paradox101 Try again with more brackets

Paradox 101

Sorry I don't follow

Tobias Kildetoft

@Paradox101 You are missing a lot of {} in that, so it is very hard to read (the exponents are messed up)

Paradox 101

Oh sorry. $g^{140}=(g^{7})^{20}=e$ then $(gS)^{140}=S$ which implies that $gS=S$? @TobiasKildetoft

Tobias Kildetoft

@Paradox101 But why does it help to make the exponent larger?

The idea here is that you should be working in the quotient group $G/S$.

Paradox 101

@TobiasKildetoft I still don't get how I can arrive at $gS=S$

Tobias Kildetoft

7:27 AM

@Paradox101 What is the order of $G/S$?

Paradox 101

It's $20@

$20$

Tobias Kildetoft

ok, so if $g^7 = e$ in a group of order $20$, why must we have $g=e$?

Paradox 101

because $g^20=e$?

and gcd(7,20)=1?

Tobias Kildetoft

@Paradox101 No, that is not why

well, the second one is part of it

Second part plus Lagrange

Huy

@MikeMiller: is it just me being unfamiliar with the subject or are many of the proofs in Farb & Margalit just brilliant

Adeek

7:33 AM

@AkivaWeinberger hi. Sorry Akiva for not replying early been pretty busy with exams and projects and assignments as this is the end of the semester !

Paradox 101

so gcd(7,20)=1 and as the order of an element must divide the order of the group, $|g|=1$ and $g=e$? @TobiasKildetoft

Tobias Kildetoft

@Paradox101 Yes, precisely

Huy

is writing $|g|$ for the order of an element common?

Tobias Kildetoft

@Huy Yes

much more common that $o(g)$ certainly, as the latter would get too cumbersome

Huy

let alone $\operatorname{ord}(g)$

Paradox 101

7:42 AM

@TobiasKildetoft and we're using gcd here because $g^{|G/S|}=e$?

Tobias Kildetoft

@Paradox101 right

Paradox 101

Ok thanks a lot!

Are normal subgroups always abelian?

Huy

no

Paradox 101

why not? I mean in normal groups the right and left cosets are equal?

Tobias Kildetoft

@Paradox101 Yes, but that does not mean that elements inside the normal subgroup commute

Huy

7:53 AM

take any non-abelian group $G$ and $N = G$

Tobias Kildetoft

Probably better to think of normal subgroups in terms of conjugation

Paradox 101

Oh ok.

thanks

If a normal group and it's factor group is abelian, then the group itself is also abelian?

Tobias Kildetoft

@Paradox101 No (try some small examples)

Paradox 101

8:12 AM

@TobiasKildetoft I'm not sure I know of any but if I try to prove it by using the properties that the normal and factor groups are abelian wouldn't any two arbitrary elements in the group $G$ commute?

Tobias Kildetoft

@Paradox101 Try the smallest example of a non-abelian group you know

Paradox 101

so a dihedral group of order 6?

Tobias Kildetoft

@Paradox101 right

Paradox 101

so then should i consider all the normal subgroups or would just a single non-trivial suffice?

Tobias Kildetoft

@Paradox101 Any non-trivial one will work

Paradox 101

8:36 AM

@TobiasKildetoft so then the group itself is also a normal subgroup right?

Tobias Kildetoft

@Paradox101 yes, any group is a normal subgroup of itself (in the example I wanted you to look at you should of course take a proper subgroup)

Paradox 101

@TobiasKildetoft dihedral group is centerless so does'nt that mean it has no abelian subgroups?

Tobias Kildetoft

@Paradox101 No, it means that it has no non-trivial elements that commute with everything

all the proper subgroups are abelian (after all, it was the smallest non-abelian group, right?)

Paradox 101

@TobiasKildetoft yes

In cases where we need to prove or disprove, do we always need to write a proof or is an example sufficient to disprove the statement?

Tobias Kildetoft

@Paradox101 If the statement is false then an example suffices (assuming the statement is of the form "for all...")

Paradox 101

8:52 AM

@TobiasKildetoft what's the best way to approach such problems? When i tried to prove it I thought it was a true statement (in hindsight my proof must be wrong) so is it better to find a counterexample or to arrive at a proof?

Tobias Kildetoft

@Paradox101 Depends a lot on the concrete problem

Paradox 101

9:03 AM

Are there any infinite groups such that if an element of order $n$ is in its factor group, then an element of order $n$ is not in the infinite group itself?

Tobias Kildetoft

Not sure what you mean

Paradox 101

The question stated that if $S$ is a normal subgroup of a finite $G$ and if $G/S$ has an element of order $n$ show that $G$ must also have an element of order $n$. I've done this part but it also asks to show an example that proves that $G$ must be finite in order for the statement to hold. I can't think of any examples

Tobias Kildetoft

@Paradox101 Consider the integers

Paradox 101

9:21 AM

So if $G=(\mathbb Z, +)$ then $N=((\mathbb 2Z, +)$ and $G/N=((\mathbb Z_2, +)$ and then in $G/N$ there is an element $1$ with order $2$ but no element in $G$ has order $2$?

Tobias Kildetoft

right

barto

@J.M. I found a possible explanation: the 1/2 allows to write $\xi$ as $(s-1)\pi^{-s/2}\Gamma(\frac s2+1)\zeta(s)$, whereas otherwise there would have been a factor $2$ in that last expression

r9m

10:35 AM

@RandomVariable re the problem you posted in I&S .. shouldn't the range of $a$ be $(-\frac{\pi}{2}, \pi)$?

that way $e^{-ia}$ avoids the 3rd quadrant and we can safely choose the triangular contour that doesn't overlap with $z = ib$ the essential singularity of $e^{\frac{1}{z-ib}}$ and the principle branch of log?

the case $a = \pi$ needs to be handled separately (but we done that before .. with that semi circular arc contour .. )

sorry I meant the second quadrant ..

Tobias Kildetoft

11:47 AM

So if $f(x_1,x_2,\dots,x_n)$ is a symmetric polynomial (let's just take a Schur polynomial associated to some partition for simplicity), then also $f(x_1^k,x_2^k,\dots,x_n^k)$ is a symmetric polynomial. Is there a systematic way to see how to write it as a sum of Schur polynomials?

Balarka Sen

12:13 PM

Hi @TobiasKildetoft.

Tobias Kildetoft

@BalarkaSen Hi

Hmm, so one can of course subtract $f^k$ (which can be calculated) to get something that might be simpler, though it still seems tricky

user 1618033

1:03 PM

@r9m Hey. Tortured here by a top research problem I cannot share ... :|

Tobias Kildetoft

@user1618033 Why can't you share it?

r9m

@user1618033 you speak like Nazi Germany .. :|

user 1618033

@TobiasKildetoft This problem can produce a revolution in my area of interest if I understand that horrible symmetry that makes my life a nightmare.

@r9m lol :-))))

Tobias Kildetoft

@user1618033 But why does that mean you can't share the problem?

user 1618033

@TobiasKildetoft Well, one with a depth understanding of the problem, in case of getting the solution, is like getting a golden mine.

Tobias Kildetoft

1:06 PM

@user1618033 So you mean that the problem is such that anyone else could solve it as well, if only they thought of the problem?

user 1618033

@TobiasKildetoft Putting the right problem might be even more important than finding a solution.

@TobiasKildetoft There is a low probability though, but it's not about the problem itself, but about the doors that probem can open. I mean the problem also keeps the key for other similar problems unsolve so far.

Daniel Fischer

@user21820 A disputed low-quality flag is always the result of a mixed review - this case. If a moderator deals with it, it'll be either marked as helpful or declined. And if declined by a moderator, there's always a decline message. If it's declined without message, that's also from the review, when a sufficient majority of the reviewers selects "Looks OK" (don't know what exactly "sufficient" is).

Tobias Kildetoft

@user1618033 Ahh, nevermind, I had forgotten this was you under a new name (I should have remembered as soon as I saw someone mention not being able to disclose a research problem)

user 1618033

@TobiasKildetoft Not really. I bought the account for 1 buck from Chris's sis with the promise to behave like Chris's sis.

Tobias Kildetoft

Certainly a good imitation so far.

user 1618033

1:09 PM

Thanks! :-)

r9m

It must be Chris's bro this time .. >_> and God knows what he did with the sister .. -_- (probably locked her away in a basement)

user 1618033

@r9m :D

@TobiasKildetoft By the way, do you realize how important is to ask yourself the right questions? Even in your research?

Asking yourself the right problems is a golden mine!

@r9m :D

r9m

@user1618033 seriously though .. get that damn editor and rough him up a bit .. what the hell is leaving a contract midway?

user 1618033

@TobiasKildetoft I think you don't understand because you never created an important problem.

Tobias Kildetoft

@user1618033 As I have said before, I prefer to solve problems rather than create them.

user 1618033

1:15 PM

@TobiasKildetoft That's perfect, but creating a proper problem can lead you to amazing results in math.

r9m

land mine alert ..

user 1618033

I would be glad only to accept that.

Tobias Kildetoft

@user1618033 I don't agree

Balarka Sen

@r9m save them for future. you'd need them on a daily basis henceforth.

user 1618033

@TobiasKildetoft Pfffff (I cannot have other reaction)

@r9m Did you finish my inequality? :-)

r9m

1:18 PM

@user1618033 oh .. forgot about it .. lemme find it again!

user 1618033

Prove that $$\int_3^4\cos(\pi/x) \ dx<5-3\sqrt{2}$$

@r9m it's the one above.

r9m

ya ,, got that from message history (thanks!)

Jester Tran

hi

r9m

@user1618033 okay ,., that reduced to $5\sqrt{2} > 7$ (which is obviously true)

user 1618033

@r9m How?

Jester Tran

1:29 PM

why do people think {study math => can do all math}?

user 1618033

@r9m Something very elegant and simple would be so awesome!

r9m

@user1618033 $\cos (\pi/x)$ is increasing in the range .. thus the integral is less than $\cos (\pi/4) (4-3) = \frac{1}{\sqrt{2}}$

user147690

Hey @BalarkaSen

user 1618033

@r9m Yeah, it's not such a good problem. It's just the way I received it.

r9m

@user1618033 :) it's just a simple bound .. I suppose we can devise analytic means of improving upper bound ($\cos (\pi/x)$ being such a nice function and all .. )

user 1618033

1:37 PM

@r9m Yeah, it's annoyingly simple

r9m

I chill man .. no point in getting annoyed at simple things :)

anyway .., BBT time (bbl) :-)

Vrouvrou

What is the definition of a bilinear continuous function please

user 1618033

@r9m not sure about what editor you mentioned, about what man, what him, and so on. This account was sold ...(and Chris's sis is a she all day long)

Back to my research.

Tobias Kildetoft

@Vrouvrou Just combine the definitions of bilinear and continuous

Vrouvrou

@TobiasKildetoft , $\forall\varepsilon, \exists \delta>0, \forall (x,y)\in E\times F, ||(x,y)-(x_1,y_1)||\leq \delta\Rightarrow ||B(x,y)-B(x_1,y_1)||\leq \varepsilon$ ?

where $E,F, G$ are normed spaces

and $B: E\times F\rightarrow G$ is a bilinear function

Tobias Kildetoft

1:50 PM

@Vrouvrou Assuming the topology comes from a norm, yes.

user 1618033

@TobiasKildetoft some of the reasons I don't like to come here are: 1) lack of math maturity and 2) I feel some of you behave like the haters instead of doing serious math. I never heard you (or others - there are exceptions) here trying to do some interesting teaching lessons if the case.

If any start some teaching lessons in chat, let me know to come in. Too many unuseful contradictions. And then in my math I got top results - if the case to talk about results, but this is not the point.

Vrouvrou

@TobiasKildetoft if this is true how to find that $\exits M>0, ||B(x_1,y_1)||\leq M ||x_1|| ||y_1|| $

Tobias Kildetoft

@user1618033 Fortunately, I really don't care whether you consider me mature or a serious mathematician (I don't consider you either, so I guess we are even there).

user 1618033

@TobiasKildetoft At the beginnin I only chatted with robjohn, but then one after another you had stuff to add. Not only you, but more.

I suppose you are not used to the excellence idea in mathematics, maybe it's annoying to look to the others trying to reach very high peaks in mathematics.

Tobias Kildetoft

@Vrouvrou Hmm, been a while since I did this stuff

user 1618033

1:56 PM

Maybe we can add some fun then to make things seem less serious.

Well, the internet also allows you to talk with Obama, and with this I said all I had to say and closed the discussion.

Back to some serious stuff.

Vrouvrou

@TobiasKildetoft do you have an idea ?

Tobias Kildetoft

@Vrouvrou Hmm, I think you can just reduce this to vectors of unit length by using bilinearity

Vrouvrou

@TobiasKildetoft i don't understand

Tobias Kildetoft

@Vrouvrou divide each of the vectors by their length and use bilinearity to take those scalars outside

user 1618033

@TobiasKildetoft Then I don't care to consider you in any way, I'm not here to assess you or others. I'm here to talk about my math.

Tobias Kildetoft

2:04 PM

@user1618033 I thought you had closed the discussion.

@user1618033 But you never talk about your math, since it is all top secret stuff

user 1618033

@TobiasKildetoft lol, there are parts of my math I discussed. The point is that you're not funny with all you say, you just seem another hater to me.

user147690

You bring us into your "I do real math - it's beautiful" spiel every day...

user147690

I've added you to ignore to save myself some sanity

Tobias Kildetoft

@user1618033 I am not trying to be funny

user 1618033

@AlexClark aha, learn some quadratic equations first of all ...

Tobias Kildetoft

2:07 PM

@user1618033 You misremember. I was the one you accused of not knowing how to do quadratic equations because I didn't want to solve them for someone here on chat.

Or did Alex commit the same faux pas?

user 1618033

@TobiasKildetoft No, you asked us how you can solve a quadratic equation. Anyway, this discussion leads nowhere.

Tobias Kildetoft

@user1618033 I did? Did someone tell me?

user 1618033

@TobiasKildetoft One thing to remember from me: never ever a real mathematician won't be busy with what the others do.

Anyway, any discussion is a loss of time. I prefer to spend my time on useful research than give you the possibility to consider yourself overly important.

Tobias Kildetoft

Too many negations for me to parse that sentence.

user 1618033

@TobiasKildetoft I my class you wouldn't pass the class.

Joshua Lamusga

2:12 PM

Aha, so there is an opposite to parametrization, which is called implicitization. :)
I was looking for that yesterday.

user 1618033

Closed case (back to my research).

r9m

@user1618033 that's a mean thing to say .. -_-

user147690

I would love to see the IP connections of r9m=Chris

Joshua Lamusga

Lots of hate in chat today, eh?

Vrouvrou

@TobiasKildetoft i consider $\frac{(x,y)}{||(x,y)||}$ ?

Mike Miller

2:17 PM

@Huy I haven't read it in much detail but I probably couldn't disagree, geometric topology is charming.

@JoshuaLamusga It's not clear to me why people have discussions that make them unhappy so frequently.

Agawa001

@TobiasKildetoft for how long u r intending to keep this derisive tone about other people's language!

Tobias Kildetoft

@Vrouvrou no, divide each of the two vectors by their length

@Agawa001 As long as they have it towards me

@Agawa001 Ohh, and the comment about not being able to parse the sentence was sincere. I did not know what was meant.

Vrouvrou

@TobiasKildetoft $\frac{(x,y)}{||(x,y)-(x_1,y_1)||}$ ?

Joshua Lamusga

@MikeMiller I'm not sure, but that's why I stopped watching TV news, lol.

Tobias Kildetoft

@Vrouvrou No, $x$ and $y$ live in different spaces, right?

Mike Miller

2:20 PM

@PVAL If I recall the original page was a letter to someone, and then he later added the devil, saying that the mathematics was a waste of our time.

Vrouvrou

yes @TobiasKildetoft

user147690

Studying hard @Sodre?

Sodre

@AlexClark Extremely :P

Tobias Kildetoft

@Vrouvrou So you want to divide each of $x$ and $y$ by their length

user147690

@Sodre I got all of my textbooks btw, they weren't stolen :D

Vrouvrou

2:24 PM

wht is the length of x and y ?

Sodre

@AlexClark Very nice :) Although, who would steal textbooks like that xD

Tobias Kildetoft

@Vrouvrou The norm

user147690

@Sodre My seedy neighbours

Vrouvrou

@TobiasKildetoft ah so i conider (\frac{x}{||x||}, \frac{y}{||y||})

Tobias Kildetoft

@Vrouvrou Right

Vrouvrou

2:27 PM

@TobiasKildetoft but where in the definition of continuity ?

Tobias Kildetoft

@Vrouvrou So far you are just reducing this to showing that there is a bound on how large the norm of the image of a pair of unit vectors can be

Vrouvrou

but how this can help me in my question ?

Tobias Kildetoft

@Vrouvrou You will probably also need to know something about the possible distance between unit vectors

Mike Miller

@AlexClark There are always plenty of nice survey papers about things. You should ask your adviser about such things when you're looking for one.

user 1618033

@r9m I'm not mean at all, but the point is this: how should we receive the achievements in mathemtics? If I had considered the comments here I should have considered that it's a very bad thing to get very good results, or if you get them to hide them well, not to ever dare be proud of them, or share them.

Random Variable

2:36 PM

@r9m I integrated the function $f(z) = \frac{\exp \left(\frac{1}{z+ib} \right)}{z}$ around a sector of a circle that makes an angle of $a$ with the positive real axis.

user 1618033

I always congratulate people that manage to get amazing results! It's a time of joy for all, really!

Tobias Kildetoft

@MikeMiller Not about all things

user147690

@MikeMiller Which post of mine is this referring to?

Mike Miller

The discussion you were looking about references for.

@TobiasKildetoft At least in my field there's a pretty healthy supply... anything that was introduced, say a decade ago, and some things that are much more recent

user147690

There are no introductory survey papers in my opinion to KLR algebras or to my $\Bbb Q(q)$-algebra $\bf{f}$

Mike Miller

2:37 PM

Wow my first sentence there is mangled.

r9m

@RandomVariable yes .. and I used $f(z) = \frac{\exp \left(\frac{1}{z-ib} \right)}{z}$

user147690

The things I am looking at are 2011 and onwards

user 1618033

@AlexClark Yes, my results are amazing these days. Most of them are at the level of Ramanujan's problems. Is this bad?

Mike Miller

There appear to be about a billion nice talks upon googling for those. Anyway, it's harmless to ask your adviser; have you done so?

user147690

@MikeMiller Yep

Mike Miller

2:39 PM

Fair enough.

user 1618033

Yeah, I'm very proud of it! ;)

user147690

He gave me a run down on the board of what I need to learn to get into learning some specific references

Mike Miller

Interesting use of the word geometric here.

user 1618033

OK, back to my stuff.

r9m

@RandomVariable my confusion is with the choice of branch of logarithm thereafter .. wait .. I'll pm you my solution and lemme know if it's correct ..

user147690

2:40 PM

Lmao @MikeMiller

Random Variable

@r9m OK

Akiva Weinberger

@BalarkaSen All right, I have no idea how to compute the singular homology of a surface with genus $g$ without going through simplicial homology.

I'm not even sure how to do it for a torus.

Mike Miller

@AkivaWeinberger What tools do you have?

r9m

@RandomVariable there .. sent it.

user 1618033

Just to mention it: if any of you want to initiate some math lessons with me, let me know ...

I sacrifice my work for that!

If not, I don't wanna hear A WORD!!! @TobiasKildetoft do you understand?

Akiva Weinberger

2:45 PM

@MikeMiller Mayer–Vietoris, homotopy equivalence, long exact sequence for a pair

Mike Miller

Oh, you have plenty.

Maybe you could calculate the homology of the sphere as the simplest case.

r9m

@user1618033 perhaps I am just sensitive to the way you said you'll fail my class .. (you know I have issues with that kinda stuff .. )

Akiva Weinberger

I know the homology of a sphere

And I know how to do simplicial homology

Mike Miller

Forget simplicial. Can you calculate the homology of a sphere without that?

user 1618033

@r9m ooo, bad thing I returned in this chat full of negativity. I said absolutely nothing about you. I addressed myself to @TobiasKildetoft, not to you.

This chat is a way to fail for sure in research and in many other serious math activities. It's to me like that! In the past it was amazing when I had chats with robjohn only.

Akiva Weinberger

2:52 PM

@MikeMiller Yes

Mike Miller

How?

user 1618033

I'm out.

Akiva Weinberger

Two hemispheres, Mayer–Vietoris. Or do the long exact sequence of $H_n(D^n,\partial D^n)$

Mike Miller

At least one of those sounds promising for a torus.

Which is good, since those are your only tools.

What goes wrong with uour attempts?

Random Variable

@r9m My approach is actually a bit different. I don't have to worry about the branch cut of the logarithm. Also, you put the essential singularity in the upper half-plane, while I put in the lower half-plane.

Akiva Weinberger

2:57 PM

I know $0\to A\to B\to0$ and exactness gives $A\approx B$. But I can't find a way to do Mayer–Vietoris with zeroes like that.

On the sphere, the hemispheres are contractible, so I get 0s in the sequence.

Mike Miller

You won't be able to. Get your hands dirty.

r9m

@RandomVariable will I post that as an answer? ..

Vrouvrou

@TobiasKildetoft i still don't know how to do

Random Variable

@r9m It looks OK to me.

user 1618033

@robjohn you have my email for any case, feel free to use it anytime :-) I don't plan to return here for months, I just made this decision.

Maybe I'll visit the chat at the end of June, beginning of July when I finish another important project.

OK, that was all.

r9m

3:11 PM

@RandomVariable okay :) .. I'm posting it. Any progress with S's post?

Random Variable

@r9m I haven't been thinking about it.

r9m

@RandomVariable 'kay .. I was wondering if Flajolet's residue sum technique works .. but I'm not good at that ..

Mohamad

In my CS book, it shows that n! = (n + 1)! \ (n + 1). I'm trying to expand this, but I'm unsure how n(n + 1)!. Let's assume n = 3: 3! = (3 + 1)(3 + 2)(3 + 3) / (3 + 1) = 3! = (4)(5)(6) / 4. This is 30, which is not equal to 3!... where have I gone wrong?

r9m

@Mohamad $3! = 3 \times 2 \times 1 = 6$, the notation $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$

Mohamad

@r9m fair enough. But the book says that this is mathematically correct: "if n >= 0, then n! = (n + 1)! / (n + 1). When I expanded it, I found that this wasn't true. Which means I expanded the equation incorrectly. I'm confused about (n + 1)!. It seems to say that factorials can simulateneously be n * (n - 1) and (n + 1)(n + 1)/(n + 1)...

r9m

3:26 PM

if they are defining factorial recursively as $n! = \frac{(n+1)!}{n+1}$ .. it means $(n+1)! = (n+1) \times n!$ .. when you wrote 3! = (3 + 1)(3 + 2)(3 + 3) / (3 + 1) .. it's wrong

instead what is says is that $3! = \frac{4!}{4}$

Mohamad

ahhhh.... OK. I see now. Thank you!

Akiva Weinberger

@MikeMiller OK, I think I got it

Let $p\in T$ be a point. Let $U$ be a disk-shaped neighborhood of $p$ and let $V$ be $T\setminus\{p\}$.

($T$ is the torus)

Using the reduced Mayer–Vietoris, we get:$$H_1(U\cap V)\to H_1(V)\to H_1(T)\to\widetilde H_0(U\cap V)$$

Agawa001

@Mohamad wud kind of feakin logic are u followin ?

Akiva Weinberger

That's $\Bbb Z\xrightarrow f\Bbb Z\oplus\Bbb Z\to H_1(T)\to0$

Agawa001

n! =/= (n+1)(n+2)(n+3)...

its is the way backward

Akiva Weinberger

3:40 PM

But $f$ sends everything to $0$, because that loop in $V=T\setminus\{p\}$ is a boundary.

So $H_1(T)\approx\Bbb Z\oplus\Bbb Z$

(I don't know if I need to be more rigorous)

@MikeMiller

@BalarkaSen This works for tori of arbitrary genus because we know $M_g\setminus\{p\}$ deformation retracts onto the wedge sum of $2g$ circles

As for $H_2$, we get that from the $0\to H_2(T)\to\Bbb Z\xrightarrow f\Bbb Z\oplus\Bbb Z$ bit of the sequence

'cause we already know $f$ is the zero map

so $H_2(T)=\Bbb Z$, and this is true of $M_g$ as well by the same argument

Mike Miller

3:57 PM

@Akiva Sure, that works. The way I was thinking was to do Mayer-Vietoris with two cylinders.

Akiva Weinberger

I tried that, but I couldn't figure it out from the sequence

Transcript for

$Mathematics$ Mathematics