Mathematics - 2016-04-13 (page 1 of 4)

12:01 AM

Wait. The complement isn't path connected. But it might still be connected.

I think it is connected, actually.

@MikeMiller My guess goes down to $1$. As for the equivalent problem with path components, they're still $3$ and $\infty$.

Mike Miller

12:19 AM

...really? One?

Samuel Yusim

I feel like I have no idea about how to compute the dimension of a riemann-roch space

i.e., $\mathcal{L}(D) := \{f \in k(X)^* : div(f) + D \geq 0\}$ for $X$ a smooth projective curve and $D$ a divisor

$\cup \{0\}$, that is

Akiva Weinberger

@MikeMiller Can you make it higher?

Mike Miller

12:36 AM

A small child would have trouble seeing how to make it only one.

Akiva Weinberger

…oh

*headdesk*

Ted Shifrin

1:10 AM

@Samuel: That's why you have the Riemann-Roch Theorem :P

Samuel Yusim

due to less-than-ideal circumstances I'm learning this for the first time for an exam tomorrow

Adeek

haha

Ted Shifrin

don't whine to me, @Samuel :P

You sound like Karim.

Adeek

I did in sometime of my undergrad it is stressful thing though

Ted Shifrin

I have news for you — grad school is more stressful.

Samuel Yusim

1:16 AM

I'm not whining, I figure I'm more just stating the facts as they are

Ted Shifrin

And life after that even more so.

Adeek

2

haha

Mike Miller

@Ted And when you're not stressed you feel bad because you know you're supposed to be... :)

Ted Shifrin

Precisely.

Oddly, even in retirement, I'm stressed. But not because of math.

PVAL

1:36 AM

I've never heard "Riemann-Roch space" before.

user19405892

hi

Let $Y$ be an unbased space and $f:S^{n-1} \to Y$ a free map. Then $f$ is homotopic to a constant function if and only if $f$ is extendable to a free map $\widetilde{f}:E^n \to Y$.

does the extension $\widetilde{f}$ if $\widetilde{f}(tx) = F(x,t)$ where $0 \leq t \leq 1$ work?

PVAL

Where is $x$ living?

user19405892

$x \in S^{n-1}$

PVAL

If $x$ is in $S^{n-1}$, then what is $f((1/2)x)$

anon

the homotopy to a constant "looks like" the n-ball, that seems right

PVAL

1:44 AM

half of something in $S^{n-1}$ surely isn't in $S^{n-1}$.

anon

what's your point PVAL?

wat, no

PVAL

Oh i see

anon

$f$ is defined on $S^{n-1}$, is being extended to $\tilde{f}$ on $R^n$

PVAL

That should work, you should check that it does.

user19405892

that's the part I'm trying to figure out

we know that $c \simeq_{\text{free}} f$ with homotopy $F$

Thus we must show that $\widetilde{f}i = f$ where $F(x,0) = c$ and $F(x,1) = f$

it follows that $\widetilde{f}(x) = f$?

ah which is what we wanted to show

@PVAL do you have any ideas how to prove the converse?

Mike Miller

1:56 AM

@PVAL Space of holomorphic functions with prescribed singularities.

PVAL

@MikeMiller I think that's called something different in the scheme world.

Ya usually I would just call it global sections of the associated line bundle.

I've never heard it called a Riemann-Roch space

Samuel Yusim

well regardless I'm finding it rather difficult to imagine how to apply the riemann-roch theorem to any particular variety (curves in particular) computationally

PVAL

2:13 AM

@MikeMiller Is there a proof that K5 isn't planar that doesn't implicitly use the Jordan curve theorem.

The proofs I see are just counting vertices edges and faces and if youre using Euler characteristic youre clearly implicitly using enough homology to build up Jordan anyway.

Samuel Yusim

I heard once that you can prove the JCT given that K5 isn't planar, but I have no idea how that would be a thing

PVAL

Well if you assume the classification of surfaces sure

David Wheeler

Not enough information

PVAL

since K5 embeds in every orientable surface with 1-component boundary besides the disk.

Mike Miller

No you need JCT. It's just slightly cleaner than the proof I would write.

PVAL

2:22 AM

I answered this question recently on main and now I'm questioning the cleanliness of my answer math.stackexchange.com/questions/1726844/…

Akiva Weinberger

Is there a name to the Jordan-like theorem that the complement of a homeomorphic image of $[0,1]$ ("Jordan arc"?) in the plane is connected?

PVAL

@MikeMiller I guess its probably easier to use the homological implications of the Jordan curve theorem instead of dealing with homeomorphisms.

Mike Miller

2:33 AM

If you can assume it's simplicially embedded jt's just an Euler characteristic statement. But that you can simplicially embed it is more or less a Schoenflies statement...

PVAL

@MikeMiller You can do something like "CW-complexes are ENR's" as in the appendix of Hatcher though without having to prove any homeomorphism statement.

Akiva Weinberger

3:29 AM

So apparently, this happened:

> "Let $\cal U$ be a nonprincipal ultrafilter."

> "No, let you be a nonprincipal ultrafilter!"

Ted Shifrin

3:45 AM

Wow, late night for you, DogAteMy!

Hi @EricS

Eric Stucky

nihao

mikeonly

Hi @TedShifrin! I want to thank you for your book on Multivariable Calculus. I suggested it to our library, and now it is there, so I am going to plough through it in Spring/Summer. It is exciting so far. :)

Ted Shifrin

You're welcome, so far, @mikeonly. Make sure you do good exercises.

mikeonly

@TedShifrin Sure thing.

mikeonly

4:40 AM

@TedShifrin What do you find fascinating in biochemistry, if you still do? (I found one of your old messages.)

Jester Tran

hi

what is the unit ball in M_22 (R)?

under the operator norm

||A|| = sup_{||x||_2 = 1} ||Ax||_2

Jester Tran

5:04 AM

Suppose (X,d) is a metric space and Y is a subset of X. What is the difference between the following statements:
1. Y is closed.
2. Y is closed in X?

TheProgrammer

5:46 AM

anon

@JesterTran they're the same

@TheProgrammer are you asking us?

TheProgrammer

why is the answer: $$n^m$$

yes

Jester Tran

@anon thanks

anon

@TheProgrammer suppose X has two elements. can you figure out why [X->Y] has n^2 elements?

Tobias Kildetoft

@TheProgrammer Because for each element in $X$ there are $m$ possible images

TheProgrammer

5:50 AM

can you give an example?

let's say: X={1, 2, 3} and y ={4, 5}

why would number of ([X->Y]) elements be 9? shouldn't it be 6?

{1->4, 1->5, 2->4, 1->5, 3->4, 3->5}

Tobias Kildetoft

@TheProgrammer what are the things you list there? They do not describe functions between those two sets

TheProgrammer

I am confused. then what should describe a function between those two sets?

Jester Tran

Is this correct way of thinking about the above problem: take each element in X={1,2,3}, there are two possible outputs in Y={4,5}. Hence answer is 2*2*2 = 2^3

Tobias Kildetoft

@TheProgrammer By saying where each element in the domain maps to

@JesterTran yes

Jester Tran

Take for example 1. It can map to 4 and 5 (2 choices)
Take for example 2. It can map to 4 and 5 (2 choices)
Take for example 3. It can map to 4 and 5 (2 choices)

Hence 2*2*2 total choices

TheProgrammer

5:57 AM

why are we multiplying them and not adding them?

Tobias Kildetoft

@TheProgrammer Because all combinations are possible

anon

@TheProgrammer how many outcomes are there if you roll a blue die and a red die? 6+6 or 6*6?

TheProgrammer

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh

nice

anon

such a scenario is essentially a map {red,blue}->{1,2,3,4,5,6}. generalize.

TheProgrammer

thanks :)

anon

6:00 AM

mmhmm

told you to work with |X|=2 first :)

Jester Tran

@anon that is an interesting way to think about it

TheProgrammer

because at any given time, we need to see all the outputs or states of the system

right?

(for all given number of input components)

anon

essentially

TheProgrammer

niceeeee

anon

now try to figure out how many one-to-one functions there are...

TheProgrammer

6:03 AM

thanks @anon @TobiasKildetoft @JesterTran

Jester Tran

@anon and onto and bijective :P

anon

bijective will turn out to be special case of injective, surjective doesn't really have a closed-form, just a different combinatorial interpretation

TheProgrammer

one to one will be n^m too?

anon

no

not every function is one-to-one

TheProgrammer

would it be n*m

anon

6:16 AM

no

TheProgrammer

oh wait

it will be just n

anon

no

TheProgrammer

:/

m?

but wait that would be incorrect as well

because if n<m

anon

let's do |X|=3. say Alice, Bob and Cindy go to play and pull out a toybox Y. first Alice picks a toy, then Bob picks a toy, then Cindy picks a toy. if |Y|=n, how many outcomes are there? (such a selection is essentially a one-to-one function {Alice,Bob,Cindy}->Toybox)

TheProgrammer

is toybox X, here?

oh

i was confused with that X =2

anon

6:20 AM

typo

TheProgrammer

*|X|=2

haha yeah

3 outcomes?

anon

nope

TheProgrammer

oh wait

anon

(a) how many toys does Alice get to pick from if she picks first?
(b) how many toys does Bob get to pick from if he picks second?
(c) how many toys does Cindy get to pick from if she picks third?

TheProgrammer

4

anon

6:22 AM

@TheProgrammer 4? wat.

remember there are n toys in the toybox

TheProgrammer

4*3*2

something factor - something factor

anon

where are you getting 4 from?

TheProgrammer

let me see

oh wait... I wrote down n=4

in my notebook

anon

answer my (a), (b), (c)

TheProgrammer

a- n

b- n-1

c- n-2

anon

6:23 AM

correct. so how many outcomes are there?

TheProgrammer

n*(n-1)*(n-2)

anon

right-o

in general, the number of one-to-one functions X->Y, where |X|=m and |Y|=n, is given by n(n-1)(n-2)... [with m factors]

you might have come across this as being called "permutations" in college algebra, denoted nPm

TheProgrammer

oh yeahhhhhhhhhhhh

Jester Tran

P(n,m)

TheProgrammer

this is some interesting stuff

is the answer (n!)/(m!)

anon

6:26 AM

actually n!/(n-m)! :)

TheProgrammer

oh right.... that was the formula for permutations

right?

anon

mmhmm

TheProgrammer

answer of what function would be the combinations?

Balarka Sen

Three stars on Ted's message on me being destructor of peace and quiet? Really!

anon

@TheProgrammer number of orbits of Perm(X) acting on the function space [X->Y]

getting group actions in the mix requires more exposition than I have available tonight. but it generalizes in combinatorics to "the twelvefold way"

Jester Tran

6:30 AM

@anon ah I recently learnt about orbits..

TheProgrammer

I don't know what orbits are

and didn't understand a single word of what you said after that

Balarka Sen

Hi @anon. Where's your alter-ego?

Jester Tran

@TheProgrammer dw about it. It's group theory stuff

anon

over there

Tobias Kildetoft

I still can't figure out if there is a nice description of what happens to Schur polynomials when one raises all the variables to some fixed power. But fortunately I could describe what happened when one also multiplied by some other Schur functions which was all I needed.

TheProgrammer

6:38 AM

I understand that it would simply be (A^2)*x

and g would also be linear because (A^2)*x=Bx which is of the same form

but what did it mean by definition? what is it that they are asking of

Jester Tran

g: ..... such that for all x in ... g(x) = ...

g: ? -> ? etc

TheProgrammer

basically, what is g? lol

Balarka Sen

I was thinking a bit about if there is an analogue of inverse function theorem for varieties with Zariski open sets (the exact analogue is false: take $\Bbb A^1 - 0 \to \Bbb A^1 - 0$, $x \mapsto x^2$ - the open sets are way too big). I was told to look at Milne's note for the etale version of the theorem. I don't know if it's surprising but it didn't excite me 'cause that's a tautology, seeing how they define etale neighborhoods.

TheProgrammer

y =z , but why?

Jester Tran

What does time-invariant mean?

independent of time?

TheProgrammer

6:48 AM

output doesn't depend on time

yeah

Jester Tran

no idea... I just keep thinking about matrices commuting

Aritra Das

7:19 AM

Hello guys! Does anyone here know (or have links to articles) on numbers expressible as sum of 2 squares in two distinct ways? I hear that it's a very well known problem, but my question about it hasn't been too active, to say the least.
http://math.stackexchange.com/questions/1738804/numbers-expressible-as-sum-of-2-squares-in-2-distinct-ways

Evinda

7:56 AM

Hello @robjohn

@robjohn We have $ E(t,x)=\frac{H(t)}{(2 \sqrt{\pi t})^n} e^{-\frac{|x|^2}{4t}}, x \in \mathbb{R}^n $, where H is the Heaviside-function.

So that we can write $ \langle E, \phi_t+\Delta \phi \rangle=\int_0^{+\infty} \int_{\mathbb{R}^n} E(t,x)(\phi_t+ \Delta \phi) dx dt $ , we have to show that $ E(t,x) $ is integrable.

$ \int_0^{+\infty} \int_{\mathbb{R}^n} E(t,x) dx dt=\frac{1}{(2 \sqrt{\pi t})^n} \int_0^{+\infty} \int_{\mathbb{R}^n} e^{-\frac{|x|^2}{4t}} dx dt= \frac{1}{(2 \sqrt{\pi })^n} \int_0^{+\infty} \int_{\mathbb{R}^n} e^{-|y|^2} dy dt=\frac{1}{(4 \pi)^{\frac{n}{2}}} \int_0^{+\inf…

robjohn

@Evinda why do you have to show that $E(t,x)$ is integrable simply to integrate it against a test function? If $\phi$ is a test function, we know that $\int_0^\infty1\cdot\phi(x)\,\mathrm{d}x$ exists even though $1$ is not integrable.

I thought you already did this problem.

2

Evinda

I hadn't shown that E is integrable... How else could we then show it? @robjohn

robjohn

@Evinda what is it actually that you are trying to show???

Evinda

Generally I want to show that $\frac{\partial{E}}{\partial{t}}-\Delta E=\delta(t,x)$. @robjohn

@robjohn I know how to show it, I just need to show that E(t,x) is integrable...

robjohn

8:15 AM

@Evinda WHY???

3

9 mins ago, by robjohn

@Evinda why do you have to show that $E(t,x)$ is integrable simply to integrate it against a test function? If $\phi$ is a test function, we know that $\int_0^\infty1\cdot\phi(x)\,\mathrm{d}x$ exists even though $1$ is not integrable.

Evinda

The prof told it to me... :/ @robjohn
So any function multiplied by a test function is integrable?

robjohn

The prof told you to show that $E(t,x)$ is integrable on $\mathbb{R}^{n+1}$? I don't see why they would ask that.

Evinda

Yes, he told so... In general is it right that any function multiplied by a test function is integrable? @robjohn

robjohn

@Evinda No, it is not. Can you think of a condition where it would be true? How about locally integrable?? why does it need to be integrable over ALL of $\mathbb{R}^{n+1}$?

You've already shown that it is NOT integrable over all of $\mathbb{R}^{n+1}$, so if you need to show that, then it must not be true.

Evinda

Do we maybe show that it is integrable for $t \in [0, m], m \in \mathbb{N}$ since the function is multipled by a test function whihc $\in C_C^{\infty}$ ? @robjohn

robjohn

8:24 AM

Locally integrable function

In mathematics, a locally integrable function (sometimes also called locally summable function) is a function which is integrable (so its integral is finite) on every compact subset of its domain of definition. The importance of such functions lies in the fact that their function space is similar to Lp spaces, but its members are not required to satisfy any growth restriction on their behavior at infinity: in other words, locally integrable functions can grow arbitrarily fast at infinity, but are still manageable in a way similar to ordinary integrable functions. == Definition == === Standard... ===

Evinda

@robjohn Do we just find the integral as for x and not the double integral, right?

robjohn

@Evinda well then how would you be showing the $t$ part of $\delta(t,x)$?

Evinda

And we show it also seperately for t? @robjohn

Evinda

9:06 AM

@robjohn How ca we find $\int_0^{+\infty} \frac{1}{(2 \sqrt{\pi t})^n} e^{-\frac{|x|^2}{4t}} dt$ ?

Jester Tran

@Evinda Suppose x \geq 0, then integrate. Suppose x < 0, then integrate?

It looks like you need to integrate by parts and make a recurrence relation

robjohn

@Evinda have you tried the substitution $t\mapsto\frac1t$?

Evinda

9:22 AM

@JesterTran @robjohn Ok, I will do later... Because I have to go to university now... Thanks for anwering :)

Algebra 2015

hI

How to find the basis of the space and the dimension of the space. Depending on the natural number n >= 3

Jester Tran

Is that real polynomials of degree n?

ADG

Good Afternoon everyone.

Algebra 2015

the solution depends on the PARITY of tne number n

which is >= 3

ADG

Does $|S|=|Y|^{|X|}$ for $S={f|f:x\to Y}$

Algebra 2015

9:27 AM

according to below, it is real polynomials

ADG

I'm talking about cardinality

Jester Tran

@ADG That is true. Do you know why?

ADG

yes, I guess

Jester Tran

@ADG Why?

ADG

because every element in X has |Y| choices

Jester Tran

9:30 AM

@ADG yes

ADG

@Algebra2015 I think you can assume $p(x)=ax^2+bx+c$ since $p'''(x)=0$ then $a+b+c=a-b+c$ so $b=0$ can you find dimension now

Jester Tran

An exact question (about cardinalities) was posted around 3-4 hours, and they extended to injective functions

ADG

@JesterTran I wass actually trying to find cardinality of finite polynomial over finite monomials

@JesterTran can you provide a link if possible/

Jester Tran

@ADG Timestamp 15:46

It's a short scroll

ADG

@JesterTran it's 15:02 here :P

anyways ill scroll

Jester Tran

9:32 AM

@ADG How do I link?

4 hours ago, by TheProgrammer

Algebra 2015

@ ADG what is dimension depennding "odd" and "even" parity

ADG

Oh thanks

Algebra 2015

@ADG what is then the BASIS for the space?

I am lost in this question

tnx

ADG

@Algebra2015 see clearly,basis is $\{1,x^2\}$ aint it since $b=0$

Algebra 2015

that IS or ISNT the basis ?

tnx

ADG

9:35 AM

it is

Algebra 2015

why u put "aint" then ? tnx

so, if n>= 3 and, n=odd the domension is n-1 ?

Am I wrong ?

ADG

it is

@Algebra2015 I'm not a native speaker. co-operate

Algebra 2015

OK

where are u from ?

germany?

ADG

India

Algebra 2015

ok

tnx

ADG

9:40 AM

btw does $\mathbb R_n[x]$ means polynomial of degree n over x?

Algebra 2015

yes

by the way

how to get private chat here? can u do it ?

tnx

ADG

create a room. btw if you're wishing to create a room with me for asking some doubts, it wouldn't be so good. I'm currently busy doing my discrete homework along.

click on a name and then you would then see the option

Algebra 2015

i did

click your name

then ? where is "create room" ?

Jester Tran

then "start a new room with this user"

under Actions

Algebra 2015

chat.stackexchange.com/users/118932/adg

where here?

ADG

9:45 AM

I can't see the same set of options as you since it's my own profile :P

Jester Tran

$Mathematics$ Test

I made the room

$Mathematics$ Test

ADG

9:56 AM

a monomail is a funtcion from a subset of countably finite set $\{x_1,x_2,...\}$ to $X=(\mathbb N\text{ or }\mathbb R)$. what does this mean?

user147690

Are you on a phone?

Balarka Sen

Hi @AlexClark.

user147690

Hey @BalarkaSen, what're you working on?

Balarka Sen

Algebraic geometry :)

user147690

Me too for the next 30 minutes :D

user147690

10:01 AM

Just quickly recapping some things and then looking at making proper sense of sheafs

Balarka Sen

Great. What're you working out in algebraic geometry?

user147690

(Or is it plural sheaves)

Balarka Sen

Sheaves is correct, yes.

user147690

Sheaves of regular functions for me of course

Balarka Sen

Did you prove that regular functions on an affine variety form a sheaf?

user147690

10:03 AM

I will

user147690

I will tonight after a little more recap

Balarka Sen

OK. It's not quite hard but not entirely trivial. Confused the hell out of me until I worked it out.

Anush

hi @robjohn

are you about to chat about your very interesting comment to my question?

user147690

10:20 AM

@BalarkaSen Do you know much representation theory?

robjohn

10:30 AM

@Anush Sorry, I have to drive someone to the airport. Maybe later

Anush

@robjohn no problem. That would be great. Just ping me

Soham Chowdhury

@AlexClark Alex, my man!

long time no see

user147690

Hey @SohamChowdhury! Where have you been :P

Soham Chowdhury

doing stuff.

user147690

Were you avoiding this part of the site due to procrastination :P?

Soham Chowdhury

10:42 AM

did parts of Tu's manifolds book.

user147690

Ahhh you surely know more than I do about manifolds then :P

user147690

Which for me is essentially nothing

Soham Chowdhury

I can calculate de Rham cohomology of the circle with M-V, that's all.

haven't learned to do torus and such right now

user147690

My knowledge of homology&manifolds&lie-groups&alg-top are all lacking

user147690

I am doing Alg geo atm though and stuff with the universal enveloping algebra of a lie algebra

Soham Chowdhury

10:44 AM

in a way, the whole purpose of going through Tu was to learn to do de Rham.

but I skipped the bits on Lie groups.

user147690

Ahh I see, why De Rham?

Soham Chowdhury

because it's a kind of (co)homolgy which I can learn right now.

user147690

Sure

Soham Chowdhury

Plus, M-V without ploughing through Hatcher.

I'll leave that to you serious people. :P

user147690

True true. I will likely be doing alg top next semester formally in class, so we will have much to talk about I imagine

Soham Chowdhury

10:46 AM

well, maybe.

user147690

Now is time for you to start on Humphreys introduction to lie algebras and rep theory

Soham Chowdhury

@AlexClark what book?

user147690

@SohamChowdhury Gathmann and Milnes notes

Soham Chowdhury

Gathmann's old ones or new ones?

old ones have a lot more buzzwords™ in them

user147690

New ones are where we are taking exercises, but two of us(of 7) are using the old ones

user147690

10:48 AM

Ahahaha what buzz words?

Soham Chowdhury

"Chern classes" and such. dunno what they are.

user147690

@SohamChowdhury Oh haha that's essentially at the end though

user147690

You have been using Gathmann's old notes too?

Soham Chowdhury

I notice the buzzwords™.

no, not studying AG right now.

I just took a printout of Vakil's sheaves chapter, so that I could interpret the ideas in Tu in sheafy language.

That's all.

user147690

Ahhh nice nice, I am going to be looking at sheaves again soon tonight

user147690

10:50 AM

Or maybe not, since I am meant to start working on my kac moody stuff now

Soham Chowdhury

wow, more buzzwords™

user147690

:P

user147690

Kac-Moody algebras are a class of infinite dimensional Lie algebras

user147690

So how is that secret community going?

Soham Chowdhury

apparently Balarka is recruiting more people.

(also, I'm apparently a member)

user147690

10:53 AM

I was apparently a member awhile back, but I didn't find out what that meant

user147690

I don't know what my benefits or commitments are...

Soham Chowdhury

be prepared to be assassinated in the near future for your lack of devotion to the Supreme Leader.

user147690

Oh god

user147690

Wait is Balarka the supreme leader?

Soham Chowdhury

repent while there still is time.

user147690

10:55 AM

:P

Soham Chowdhury

do you think he'd be satisfied with anything less?

user147690

It depends what the goal is :P

user147690

What are you working on tonight?

Soham Chowdhury

Galois theory is long overdue. Damn geometric things sidetracking me.

Actually, Mathcamp application + exams also got in the way.

user147690

What is mathcamp?

user147690

10:57 AM

Galois theory is very nice

Soham Chowdhury

a summer camp in the US.

user147690

I am intending to look back at it after doing some serious topology, but that'll be awhile from now

user147690

@SohamChowdhury Is the camp going to be filled with 'I am amazing at CALCULUS!!11' students?

Soham Chowdhury

don't think so

there was a spectral sequences class one year.

(Balarka went mad with rage when I told him that)

user147690

Wow nice nice nice

user147690

10:59 AM

Why was he mad? Because he missed out?

Soham Chowdhury

no.

because he thinks you can't learn serious math like that.

user147690

?

$Mathematics$ Test

$Mathematics$ Test

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$Mathematics$ Mathematics