Mathematics - 2014-06-21 (page 1 of 5)

Ted Shifrin

12:00 AM

Or go to sleep and do it tomorrow :)

Alizter

Sleep before that. However I will do Z__I_5 thingsy

$\Bbb Z_5[i]^*$

Mike Miller

12:19 AM

"We apologize for any inconvenience, but an unexpected error occurred while you were browsing our site. It’s not you, it’s us. This is our fault. Detailed information about this error has automatically been recorded and we have been notified."

Alizter

$\Bbb Z_{29}[i]^*$ click to see better

Pedro Tamaroff

Fuck Hypnotoad. All hail Hypnocayley.

Ted Shifrin

No comment.

G'night, @Alizter

Alizter

@TedShifrin Good night! I shall go too!

Good night every body!

Enjoys Math

dats a mesh

@PedroTamaroff what are you studying this week?

Mike Miller

12:28 AM

Why are these damn sites always read only lately?\

Pedro Tamaroff

@EnjoysMath Combinatorics.

Enjoys Math

@PedroTamaroff specifically?

Pedro Tamaroff

At the moment graph stuff.

Enjoys Math

@ComTruise neet name, that's a band?

algebraic graph theory?

Com Truise

hehe yes

Studentmath

12:29 AM

Gave up on sleep. Can anyone explain the 3+3+3+6(v(G)-3) step in this paper: math.uwo.ca/~srankin/courses/2156/2012/ass10.pdf

Pedro Tamaroff

@Studentmath I think I have the proof now.

I am writing it down.

Enjoys Math

don't write it down, write it up!

down sounds negative, bro

Studentmath

Let me know how it goes @Pedro

Pedro Tamaroff

My first step was showing that $\# N(v)\leqslant n$ for any $v\in G$.

Enjoys Math

@Studentmath you are aware your paper is called ass10.pdf?

XD

Pedro Tamaroff

12:32 AM

Assignment 10.

Enjoys Math

They abbreviated to a synonym for butt to be funny

Studentmath

@enjoys that's nice. It's less nice someone in the world gets this as the sole question in the assignment, but still.

Enjoys Math

►

@Studentmath what's a simple planar graph?

Studentmath

It's a graph without loops/multiple edges that can be drawn in a plane without two edges crossing each other

Graph theory wise, it's a simple graph that has no $K_5$ or $K_{3,3}$ subgraphs, not even inuced by elementary subdivision operation

Enjoys Math

I get everything except the last part after the comma

Mike Miller

12:42 AM

@EnjoysMath A planar graph is one that can't be drawn in the plane without self-intersection

robjohn

we should be back in read write mode!!

Studentmath

The subdivision operation is an operation where you take a uv edge and make it a u,w,v path, or vice versa.

Enjoys Math

add vertices ?

Studentmath

Now I am off to sleep. @Pedro ping me once you have it done, will check it tomorrow, interested in how you did it in the end!

@Enjoys add a vertex and two edges, yes.

Or 'eat up' a vertex

Mike Miller

one might note that this characterization of planar graphs is not a definition, it's a nontrivial theorem

Ted Shifrin

12:45 AM

Night ... AGAIN ... @Studentmath

@Mike ... Your statement is a negation off.

Mike Miller

kuratowski's theorem is trivial?

Pedro Tamaroff

@Studentmath I have it,.

It's a bit informal, I think, but I will ask my professor and see what he thinks.

@MikeMiller Prof said it has a topological proof, and that it should be understandable to undergrads.

Mike Miller

ah, that makes it trivial

Ted Shifrin

@Mike: Read chat.stackexchange.com/transcript/message/16194898#16194898 again!

Mike Miller

oops

Ted Shifrin

12:58 AM

Uh huh. Apology accepted.

Pedro Tamaroff

@TedShifrin What is he apologizing for?

Ted Shifrin

For being snippy at me? ;)

Mike Miller

I would never be snippy

Ted Shifrin

Between snippy and Hippa's memes, I should skedaddle.

Mike Miller

:(

Pedro Tamaroff

1:16 AM

@TedShifrin Aw, Ted.

Please!

Antonio Vargas

Am I reading it wrong or is the quote from Ahlfors in the answer to this question still saying that the number of singularities must be finite?

robjohn

I really dislike these nonsensical downvotes. At the request of a comment, I add some explanation to this answer, and immediately get a downvote.

Antonio Vargas

@robjohn That's nuts.

robjohn

@AntonioVargas I didn't understand the first downvote, but the second one is just beyond my comprehension.

Com Truise

maybe someone hit the wrong button

robjohn

1:30 AM

@ComTruise There have been enough downvoting of good answers going on that I generally discount that idea

off to the park... BBL

Com Truise

@CHELL

user97303

Hey @ComTruise

user97303

(I don't know how to put tex)

Com Truise

I dont understand what is going on in that matrix

user97303

the natural numbers

user97303

1:43 AM

the first row is 1, 2, 3,... up to n

user97303

the second line is n+1, n+2, etc

Com Truise

oh ok

Pedro Tamaroff

It goes up to n^2?

user97303

yes

user97303

maybe even more

user97303

1:43 AM

if m>n

Com Truise

up to n*m

user97303

yes

user97303

now, is it true that the rank of this matrix is min(m,n)?

user97303

intuitively I've been trying to figure this out from modular arithmetic

user97303

but my background isn't very strong in it

Com Truise

1:47 AM

maybe a good question for the site

it doesn't look obvious

user97303

I don't know, I don't like posting questions because they get flagged and marked for deletion

Pedro Tamaroff

Post your question. If it is well written and you show your work, it won't be closed.

user97303

alright, @PedroTamaroff, I did

user97303

0

Q: The rank of a matrix with the natural numbers in ascending order.

Say you have a matrix, $\begin{bmatrix}1& 2& 3& \dots& n\\(n+1) &(n+2) &(n+3)& \dots &(2n)\\(2n+1) &(2n+2) &(2n+3)& \dots &(3n)\\ \dots &\dots &\dots &\dots &\dots \\(m(n-1)+1) &(m(n-1)+2) &(m(n-1)+3)& \dots &(mn) \end{bmatrix}$. Is it true that for any such matrix $A(m,n)$, $rank(A)= min(m,n)$?...

Ted Shifrin

I'm confused @chell: it seems to me the rank is $2$ whenever $n\ge 2$.

user97303

1:56 AM

wait, nvm , I don't know what I was thinking

Ted Shifrin

Did you mean something else?

user97303

I was thinking of something else, but I didn't write it down, so I guess I remembered it wrong

user97303

well, whatever

blue

@chell subtract the 2nd from the 1st, then 1st from the 2nd columns. then the first two columns are (0 n 2n ...) and (1 1 ... 1) respectively, and everything else is a linear combination of them

Ted Shifrin

I've been known to do that.

user97303

1:57 AM

yeah, I realised it after I typeset it

blue

(that was the answer I was writing before you deleted)

Ted Shifrin

Similar argument with rows ...

user97303

sorry, I'd delete these posts if I could

Mike Miller

Hi @Ted

Ted Shifrin

Boo

Karl Kronenfeld

2:00 AM

ew

Ted Shifrin

Heya @Karl

Mike Miller

that's what I say when I see @Karl

Ted Shifrin

@blue, are you the new incarnation of turtles?

blue

it's turtles all the way down

Ted Shifrin

Down to where?

blue

2:05 AM

http://en.wikipedia.org/wiki/Turtles_all_the_way_down

Mike Miller

@TedShifrin it is, yes

Pedro Tamaroff

@FernandoMartin

Ted Shifrin

Ah ... Ignorance be bliss.

Karl Kronenfeld

@TedShifrin Somewhere below my location.

I'll give you more information as it comes in.

Ted Shifrin

Hi @Fernando

Gee thanks @Karl

You're more anonymous than all of turtles ...

Karl Kronenfeld

2:07 AM

@Ted Breaking news! It's at least ten feet below!

Fernando Martin

hey there @Pedro, @Ted

Karl Kronenfeld

@TedShifrin What's this measure of anonymity?

Pedro Tamaroff

LEL

16

Q: Why didn't Britain's nuclear weapons deter Argentina from invading the Falklands?

The United Kingdom has nuclear weapons. All the [major] political parties support this policy. A common justification heard is that they are a deterrent against foreign aggression. For example: UK nuclear deterrence policy consists of 5 main principles: preventing attack - the UK’s nu...

united-kingdom falklands-war nuclear-weapons

Karl Kronenfeld

I guess I have deleted system32 more times than anon has

Mike Miller

@KarlKronenfeld If someone worked hard enough they could probably figure out who anon is

Karl Kronenfeld

2:08 AM

@MikeMiller I think Jayesh Badwaik has worked sufficiently hard. So you are correct.

Ted Shifrin

I've talked enough to anon that I know more than I expected to.

But @Karl stays opaque.

Com Truise

Argentina invaded the Falklands?

Karl Kronenfeld

@TedShifrin True, skull and I are probably twins.

No idea who the hell that guy is.

Mike Miller

just a troll like you

Ted Shifrin

Yeah, but @Karl knows math.

Pedro Tamaroff

2:13 AM

Could someone help me count?

Ted Shifrin

No. I cannot.

Mike Miller

@PedroTamaroff I can get up to seven, tops

Karl Kronenfeld

@PedroTamaroff My computer can.

Mike Miller

"You haven't voted on questions in a while!"

piss off, I'll vote what I want

Pedro Tamaroff

I need to count triangles in a graph. Prove there are at least $\frac E{3V}\, (4E-V^2)$ triangles in a simple graph.

Ted Shifrin

2:15 AM

Now someone posts asking to translate a problem from German to English. What is this place?

Pedro Tamaroff

@TedShifrin That's fine!

Ted Shifrin

I will answer a specific question, but I'm not going to translate a page.

Mike Miller

@TedShifrin Voted to close as not math.

Ted Shifrin

Being multilingual is a curse :)

But one needs to know math to translate, @Mike ...

Pedro Tamaroff

@FernandoMartin

Mike Miller

2:18 AM

Ok, voted to close as not math as defined in the help center.

Pedro Tamaroff

I have already proven that given an edge $\{vw\}$ there are at least $\deg v+\deg w-V$ triangles including that edge.

Now I need to sum some stuff and use $\sum_{v\in G}\deg v=2E$.

Ted Shifrin

Planar or not, @Pedro?

Pedro Tamaroff

It says "simple graph".

That's it.

Mike Miller

@TedShifrin I have 25 bucks to spend on a book.

Pedro Tamaroff

I'm working on it, anyways.

Mike Miller

2:21 AM

Positivity in Algebraic Geometry II is $26... :D

Karl Kronenfeld

Here's a dollar

Ted Shifrin

I'm amazed there are any books for less than $60.

Karl Kronenfeld

Is it published by dover @MikeMiller?

Mike Miller

@TedShifrin Well, I'd probably need to know algebraic geometry and have read Positivity in Algebraic Geometry I.

@KarlKronenfeld You won't believe who it's published by

Karl Kronenfeld

@MikeMiller Cambridge University Press?!

Mike Miller

2:23 AM

Well, Springer, actually

CUP wouldn't put something for so reasonable a price.

Karl Kronenfeld

Still surprising

Mike Miller

trying to decide what I'd like to read about

I'm open to suggestions

Pedro Tamaroff

Thanks guys. =/

Mike Miller

welcome

Karl Kronenfeld

@PedroTamaroff I have been thinking about it pal. :D

Just haven't made any real progress.

Ted Shifrin

2:26 AM

@Pedro: For reasons I don't understand yet, for regular polyhedra we get equality.

Pedro Tamaroff

@KarlKronenfeld Let me tell you what I've done.

The number of triangles containing edge $\{vw\}$ is $|N(w)\cap N(v)|$.

Mike Miller

I think I had this exercise in my graph theory course

I vaguely remember it being gross

Pedro Tamaroff

By the usual inclusion exlusion this is $\geqslant \deg v+\deg w-V$.

Since $|N(v)\cup N(w)|\leqslant V$.

Karl Kronenfeld

yeah

Pedro Tamaroff

So given an edge $e=\{vw\}$ there are at least that many triangles in $G$ that contain it.

Now we need to sum that thing appropriately.

Karl Kronenfeld

2:28 AM

That's where I'm stuck

Not sure how to tell how much I am overcounting

Pedro Tamaroff

We're counting each edge three times, as I see it.

Mike Miller

I could get Kirby-Siebenmann for $25 even

hmm that's +shipping

Ted Shifrin

@Mike: Did you win the lottery?

Karl Kronenfeld

lol

Com Truise

dover books?

Mike Miller

2:31 AM

I graduated, @Ted

I wonder if this would be a well-accepted MSE question

Karl Kronenfeld

@PedroTamaroff We're counting each edge three times for each triangle it belongs to.

Pedro Tamaroff

@KarlKronenfeld Yes, that's what I am trying to say.

ftp.bolyai.hu/~p_erdos/1964-21.pdf

Karl Kronenfeld

4E-V^2 looks like the discriminate of a quadratic, so I wouldn't be surprised if you end up doing some algebraic manipulation here, getting a quadratic polynomial.

Ted Shifrin

Huh? @Karl

Pedro Tamaroff

@TedShifrin I've seen that in graph theory, I have.

I once attended a course in Rosario.

Ted Shifrin

2:34 AM

No, I meant counting each edge 3 times ...

Pedro Tamaroff

@TedShifrin What do you mean?

Ted Shifrin

If we have a closed surface, then each edge is in two faces, but that needn't be true.

Karl Kronenfeld

Yeah, what did I mean there.

Ted Shifrin

Each face has 3 edges ?

Pedro Tamaroff

Face of what?

Ted Shifrin

2:36 AM

Face = triangle

Pedro Tamaroff

Well, yes that is what a triangle is...? =P

Mike Miller

@TedShifrin Face is the wrong word here I think, we're not assuming planarity

"Lectures on symplectic manifolds" is $17

Ted Shifrin

Nor am I ...

r9m

@G.T.R :D .. if I'm not making a silly mistake ... me thinks (as of now) the bound is $\displaystyle \forall x\in [0,1], |f(x)|^2 \leq 0.707107 \cdot \int_0^\pi {f'}^2$, when $\displaystyle \int_0^\pi f=0$ ... is the bound $\pi/3$ sharp ? (this is btw the WA estimate of $\min(\frac{t^2}{8}+1)\frac{\coth {\pi t}}{t}$ = 0.707107, it can be improved further I guess .. I've most likely complicating something easy(?) :P

Ted Shifrin

Weinstein? @Mike

Mike Miller

2:37 AM

Yup, @Ted, and it looks about 10x beyond my level

Ted Shifrin

It's a good book.

Mike Miller

The course after Riemannian at LA is symplectic anyway, so I can wait.

I think sometimes they use that as a course on contact geometry instead, though.

Ted Shifrin

Complex manifolds somewhere too?

Mike Miller

Unfortunately I don't think that has its own course.

Ted Shifrin

Too bad.

I taught it as a topics in geometry at MIT ...

Com Truise

2:39 AM

@MikeMiller amazon.com/…

do it

Mike Miller

no thanks @Com

There's a 60s book on 3-manifolds (by Fort and Silver) that's $16...

Com Truise

so price is a factor?

Mike Miller

$25 tops, and that includes shipping and tax

@Ted I could get Chern's complex manifolds without potential theory for $27, but alas, that's too much

Karl Kronenfeld

@TedShifrin It does indeed boil down to that. If you loop through the edges, counting the triangles that contain the given edge, you will have counted each triangle three times. (As I see it)

blue

@Pedro Can you get $\sum_{v\in V}(\deg v)^2\ge 2E^2/V$?

Com Truise

2:42 AM

@MikeMiller you want differential geometry books?

Pedro Tamaroff

@blue What do you mean by that?

Mike Miller

Ah, I found it for $22... @Ted, have you read that book? Is it good? Will I get anything out of it?

Pedro Tamaroff

If I can prove it?

blue

yes

Mike Miller

@com geometric or topological stuff is preferable, and I like manifolds of all flavors

Pedro Tamaroff

2:44 AM

SIGH I think we did evaluate that sum in class, but I lost my binder, so I'm literally on my own at the moment. I should go to my university and ask for it, it should be there. FUUUUUUU

Ted Shifrin

What book, @Mike?

blue

@PedroTamaroff you probs want to sum $\deg(v)+\deg(w)-V$ over all edges $vw$, then divide by $3$

Mike Miller

Chern's "Complex manifolds without potential theory", @Ted

Pedro Tamaroff

@blue Yes, that I know. But I don't know what to do with that sum. I think I am getting the following.

Ted Shifrin

Oh, I know it intimately. I helped with the revision for 2nd edition and have an autographed copy :)

Mike Miller

2:46 AM

"with an appendix on the geometry of characteristic classes"... sounds like it's for older kids than me.

Ted Shifrin

You know some of the stuff in there ...

Pedro Tamaroff

If I sum over all edges without worrying about order, I get $3\Delta\geqslant 2\sum_{E}\deg v-2EV$ where the sum runs through all the edges of $G$ without repeating.

Mike Miller

Certainly not well enough and not intimately, @Ted

Pedro Tamaroff

That looks wrong.

Ted Shifrin

Well, get it. Unless you expect me to give you that, too.

Mike Miller

2:48 AM

LOL @Ted

I don't expect anything

blue

@PedroTamaroff $$\sum_{vw}(\deg v+\deg w-V)=2(\sum_{vw}\deg v)-EV=2(\sum_v(\deg v)^2)-EV $$

Ted Shifrin

I have to keep some books. Maybe I'll find an apartment with room for a big library.

Mike Miller

The only thing you promised me was Gunning and Kodaira...

But I don't hold people to their promises if they decide they don't like me anymore

Pedro Tamaroff

@blue Ah, OK.

Ted Shifrin

Is that a threat or a promise?

Com Truise

2:51 AM

you should look through dover editions if you want under $25 new

I just got a magazine from them

Pedro Tamaroff

@blue When you write $vw$ you mean sum through edges without repeating yes?

blue

@PedroTamaroff yes

Mike Miller

@Ted Neither, it's a joke.

Ted Shifrin

Hard to find grad level Dover.

Com Truise

oh

Mike Miller

2:52 AM

One of my fave series for odd topics is Universitext.

they published the nevanlinna theory book I have

which is I suppose not an odd topic

but it's not a "standard curriculum" topic, rather

Ted Shifrin

It's dated and not popular ...

blue

@Pedro I might have an extra factor of 2

Pedro Tamaroff

@blue Where?

Oh, yes.

Mike Miller

@Ted luckily for me, I'm not looking to do research in it or use it :P

I even have enough money left for a coffee.

Ted Shifrin

Chern's book has a little bit of value distribution theory in it ...

blue

3:00 AM

@PedroTamaroff $V\sum_{v\in V}(\deg v)^2$ counts the tuples $(u,v,w,w')\in V^4$ such that $vw,vw'\in E$ and $v\ne w,w'$, while $4E^2$ counts the tuples $(a,b,c,d)$ such that $ab,cd\in E$ and $a\ne b$, $c\ne d$. Is this useful?

Mike Miller

I mostly find it to be a good testament to the incredible rigidity of (mero/holo)morphic functions, @Ted

Though my impression was that Vojta's analogy between value distribution theory and diophantine approximation was fairly successful

Pedro Tamaroff

@blue Yes.

Damn sun my counting is so bad.

@robjohn

How can I contact the "big fish" moderators here?

To report an issue that is rustling my jimmies.

You already know the issue.

Ted Shifrin

"Rustling my jimmies"?

What kind of cowboy is @Pedro?

Mike Miller

@TedShifrin That phrase is pop culture now

3:11 AM

I'm 80 years behind the times ... My students have never said any such phrase!

Mike Miller

Not in front of you :P

Ted Shifrin

Nor to the side of me, in hundreds of hours of office hours.

Pedro Tamaroff

SIGH at serial downvoting.

Ted Shifrin

Serial? In this case the OP wanted you specifically to address his question. That's legitimate.

Mike Miller

3:18 AM

Anyway, @Ted, you can't be 80 years behind the times. You don't look a day over 75.

Pedro Tamaroff

@MikeMiller Ted is 50 something?

blue

@TedShifrin the OP has three downvotes in 2.5 years of MSEing, I doubt they used it on Pedro. It's possible OP's comments influenced someone's voting decision though.

Pedro Tamaroff

@TedShifrin Nah, the point is lately I've been receiving daily downvotes in waves of 3.

Some were reversed.

blue

and I do not find the downvote legitimate; failing to address a narrow question by answering the broader question in a different way that's good to know is useful information

Pedro Tamaroff

But it is just bothersome some guy is taking its time downvote me everyday.

Mike Miller

3:24 AM

@Pedro It's a joke.

Ted Shifrin

Let's face it, @Pedro: Neither of us is universally beloved.

Pedro Tamaroff

@MikeMiller What?

Com Truise

can you see who votes?

Pedro Tamaroff

No, votes are anonymous.

Mike Miller

Not like I am, at least, @Ted

Com Truise

3:43 AM

►

Vibhav Pant

4:47 AM

Hello

Com Truise

hi

who goes there

eXtremiity

5:06 AM

I am unsure why the other conditions are necessary. Isn't (a) enough to prove the question :s ?

Unless it's because that "infinite" condition on (a) doesn't provide enough certainty.

Mike Miller

I'm trying to think of a counterexample but I can't so far

blue

@eXtremiity no, ${\Bbb C}_p$ is not spherically complete even though it is complete

Mike Miller

there's got to be something better than that

blue

it's the only one I had in my memory

blue

5:31 AM

@Mike, @eXtremiity: let $S$ be the space of functions $f:(-\infty,\infty)\to\{0,1\}$ with discrete support. if $f,g\in S$, then define $d(f,g)$ to be $e^{-r}$, where $r$ is the first real in which $\{f(r),g(r)\}=\{0,1\}$. let $f_n$ be the indicator function of $\{-1/k:1\le k\le n\}$, and $A_n$ the ball of radius $e^{1/n}$ around $f_n$.

(that's my remake of the $p$-adic example)

Mike Miller

I'll look at that when I get home

blue

*discrete support bounded below

Mike Miller

@extremiity Tip for your problem: pick $x_i \in A_i$. Show that this sequence converges. Show its limit is in the intersection.

eXtremiity

Thanks @blue, at least there is a counter-example. Too hard for me to understand, but it exists >_>. @MikeMiller. Yep, I'm on that way :). Thanks.

Mike Miller

simple counterexample

let $X=\Bbb R$, $A_k= (-\infty, -k] \cup [k,\infty)$