@r9m I looked, but trying to verify it numerically in Mathematica didn't work. I looked deeper and noticed that perhaps your notion of $\omega$ might be off.

I computed the value numerically so it would be easy to check if someone came up with a correct closed form :-)

@robjohn Mother of God! I just realized the silly mistake I have made! I am temporarily deleting my answer while I fix it .. thanks!!!!

@robjohn could you check if it checks numerically now? I guess I have fixed it now ..

@Alizter @BalarkaSen There is a pretty interesting post on mathoverflow about looking for when a graph is a cayley graph here.

There are also some cool remarks there too, like graphs which are not quasi-isometric to any groups

@robjohn remember that link to your pictures of things not on earth

Can you link it?

@skullpetrol hey! How are things going? :-)

Fine thanks @r9m how are you pal?

anybody here about some statistics?

Alex? :O

Welcome back pal

@AlexanderGruber

What y'all thought y'all wasn't gonna see me? \m/

thanks @skull.

:D

@Tien-ChengHuang quantum mechanics, electricity and magnetism, statistical mechanics / thermodynamics, machine learning, game theory

and throw some art and literature in there to grow you a soul, too

@AlexanderGruber how's the back feeling?

@skullpetrol Not great, they're gonna chop it open before long here.

but evidently it should be good after that

-_-

@skullpetrol I'm fine too :)

Did you get a second opinion? @AlexanderGruber

From a specialist, I mean.

@skullpetrol took me a lot of opinions just to get a diagnosis actually, i thought it had nothing to do with it for a while. I was having pain elsewhere and had all kinds of tests done there, eventually turned out to be displaced from back injury

I see.

I hope for only the best pal :-)

thanks. I should be alright.

One of my undergrad friends just got hit on his bike and broke his collarbone in like 20 places so I don't have too much room to complain by comparison

-_-

Ouch!

hit by another guy on a bike too which is really just ridiculous

20 places?

That's really strange.

He must need major therapy

Yeah, I guess he couldn't type or text for a long time afterwards, so I didn't even know about it until a month after the accident. (He lives in Colorado now.)

hi

hi

anyone know a good book about spherical harmonics?????

Wow! FIVE question marks, you must be desperate.

bye bye thanks for your help

How is the lifting correspondence for a covering map $p:E\to B$ well defined. I cannot understand the lifting correspondence because it is a map from the fundamental group of B to the inverse image of a point in B and can't understand how it is well defined.

Hi!

@AlexanderGruber Can you help me on the question above?

Hey I am trying to solve this question in a more direct manner.

just a sec first

let me thnk of something before posting it

@Rememberme ok what's up

You still around?

Yes

ok, so, let's see

You got loops in $pi(B,b_0)$

Path homotopy classes of loops.Yes

right

$p^{-1}(b_0)$ is some points in $E$

Yes.

hey hold on just one sec i have a phone call

Okay.

hm

Maybe you could help me with this problem

What is the question about Karim?

Suppose that X has the finite complement topology and Y is subset of X where X is infinite, then prove that the limit points of Y is X.

@Rememberme any path from x to itself in B lifts to a path from a chosen lift of x to another point in its fiber

well, we know finite complement topologies are $T_1$

so in particular they satisfy the following theorem Let A be a subset of X Then the point x is a limit point iff every neighborhood of x contains infinitely many points of A

@KarimMansour you want to specify Y is infinite too don't you?

yeah

where Y is infinite I meant

that was a typo

then any punctured nbhd of x is cofinite which must nontrivially intersect Y, so every x is a limit point

I don't understand what you mean is cofinite?

finite complement

indeed, I call your topology the cofinite topology

yes by theorem we have that every nbhd of x will intersect nontrivially with Y and it must be also of infinite cardinal, but why is it X ?

read what you just wrote again. every nbhd of what?

edited

where x $\in Y`$

where x in X

(also, you should say punctured nbhd, not just nbhd)

there's no need to invoke any theorems here

if x is an elt of X, then any punctured nbhd of it is cofinite, which must intersect Y nontrivially, hence every x in X is a limit point

well any punctured nbhd is cofinite, which means for a arbitrarily punctured nbhd U we have X - U is either finite or is all of X. we also have that $U-\{x\} \cap Y \neq \emptyset$

just a sec @anon I am a bit slow today because I am sick so trying to wrap my head around this.

we know x is a limit point of Y iff every nbhd of x intersect Y in some point other than x.

Yes.

so how did we show here that every nbhd of x intersect Y in some point other than x?

where $x \in X$

and x is arbitrarily

every punctured nbhd of x is cofinite

every cofinite subset of X nontrivially intersects any infinite subset of X

in particular, every punctured nbhd of x nontrivially intersects Y

@anon Whats the intuition behind the lifting correspondence?

lifting is

if you travel around in the base space, its lift travels around in the covering space

going in a loop in the base space may get you somewhere new in the covering space

that somewhere new must be a lift of the original position

i.e. in its fiber under the covering map

Ahh. I see.

@anon Do you know a good place to learn about representation of functors?

Hello @TedShifrin

Hello everyone

Does anyone know what happens if one defines filter on preordered sets which aren't partially ordered?

@Remember It's clear what the lifting correspondence is about. If you lift loops in basespace, you get paths in the top space. But endpts of the paths might not be the same.

In any case, the endpt would at least be in the fiber.

So you get a natural map $\pi_1(B, b_0) \to p^{-1}(b_0)$

You were asking why is this well defined. Prove that it is.

My question is not purely theoretic, I am dealing with multifilters which are "filters" on the preordered set $\mathcal{P}_0(\mathcal{P}_0(X))$ together with the finer relation given that $\alpha \preceq \beta$ if every element of $\alpha$ is subset of an element of $\beta$

@Remember You can also have a $\pi_1(B, b_0)$-action on $p^{-1}(b_0)$ given as follows : take point $\tilde{e_0}$ in the fiber over $b_0$, take a loop $\gamma$ based at $b_0$. lift to a loop with basept $\tilde{e_0}$. define the endpt to be $\gamma \cdot \tilde{e_0}$.

so you even have a map $\pi_1(B, b_0) \to Aut(p^{-1}(b_0))$

this is known as the monodromy.

these are fascinating concepts, because they neatly match up with many ideas in Galois theory.

one can actually recover the fundamental group $\pi_1(B, b_0)$ by all it's monodromy actions on all finite covers, modulo a bit lying.

Hey...so 1m is approx 3780 pixels...how did they arrive at this figure?

unitconversion.org/typography/…

that seems weird, a pixel is not a length unit in my opinion

If x <= y is inf{x} <= y true?

as inf{x}=x yes

@DominicMichaelis Given that I know nothing other than inf{x} exists and the sets are not empty.

Can you please explain why the above is true/false?

the infimum is the least upper bound, while y is an arbitrary upper bound ...

so, is it true?

yeah and furthermore inf{x}=x ;)

Thanks a lot!

you mean inf{y} = x?

if you have any partial order set you know that $x\leq x$, as partial orders are reflexive and furthermore you have via antisymmetry that $x \geq a$ together with $x\leq a$ implies $x=a$

no inf {x}=x

Thank you very much

Hi, in a 1-d ball physics problem, how do we model upward ascend of the ball after it bounces...?

I know $h = \frac{1}{2} gt^2$, but without any resistance, what do I assume the velocity of the ball be after it bounces?

I recently read a proof that [0,1] is compact in R. The key to the proof is to consider " sup{x in [0,1] : [0,x] has a finite subcover} " and finally prove this sup is 1. How did mathematicians think of this idea? I am so curious.

oh i made a false statement...! I will restate:

I recently read a proof that [0,1] is compact in R. The key to the proof is to consider " sup{x in [0,1] : [0,x] has a finite subcover} " and finally prove this sup is 1 AND the sup is in the set. How did mathematicians think of this idea? I am so curious.

Hi guys! I had a silly doubt. If the diagonals of any quadrilateral intersect at right angles, is the quadrilateral bound to be a rhombus?

This doubt came to mind while I was solving an Olympiad problem

@SwapnilDas draw a pair of intersecting mutually perpendicular line segments .. what makes you think they will form a rhombus?

it's monday morning where I try to solve as many interesting problems out of my analysis book as I can :D

I wish I hadn't learned about nonstandard analysis continuity definitions before trying to do these problems with epsilon-delta analysis

I think it's good in that the nonstandard analysis definition sometimes gives me faster insight into the answer of a problem before I try to prove it classically

@Balarka the monodromy idea seems very cool. I never thought we could think of automorphism group of $p^{-1}(b_0)$. But I had this question: for that map to work we have to have p to be a covering map. So what other maps 'p' are there for which the above map (that is the monodromy) works?

vector bundles

@GBeau: If you have the insight and intuition to prove things in analysis, the espilon-delta argument shouldn't be very far away.

I guess it just seems much more obvious in nonstandard..for if $f$ is classical, $f: \mathbb{R}\rightarrow \mathbb{R}$ and $x\in\mathbb{R}$, $f$ is classically continuous at a standard point $x$ if $x\simeq y\implies f(x)\simeq f(y)$

@GBeau You just need to replace $\simeq$ with $\rightarrow$

(well, essentially)

where $\simeq$ is "infinitely close" ($x-y\simeq 0\iff \forall ^\mathsf{S} c>0, |x-y|<c$)

where that means "for all standard"

@GBeau But if $f$ is classical, how do you evaluate $f$ at a non-standard point?

it is valid to evaluate classical functions at nonstandard points in the version of NSA I learned

(axiomatic NSA)

indeed, even the standard set $\mathbb{N}$ has nonclassical elements in this formalism to my dismay when attempting to use the word finite

@GBeau How? Given some arbitary function from $\mathbb{R}$ to $\mathbb{R}$, how can you decide where to send non-standard elements?

they points within $\mathbb{R}$ itself @TobiasKildetoft, so you can use the normal definitions

@GBeau No, now we have a completely classical thing, being a function from the classical reals to themselves. This is the starting point of classical analysis

the axioms it adds modifies ZFC to extend sets like $\mathbb{R}$ to include the points you need

the set of classical reals does not necessarily exist in this system

A big point of nonstandard analysis is that it is actually supposed to apply to the original problems

(I think you can actually derive a contradiction from assuming it)

Ohh, I thought you had studied the kind of nonstandard analysis that actually was able to deal with classical problems

the only one I've studied is "Internal Set Theory", which I gather is the less common variant

@GBeau If you have not studied one that can even speak about the classical problems, then it is probably better to try and forget all about it when studying those. There is just too much of a risk of something you "know" being wrong in the new context

let me give an example

I Just finished solving (with regular analysis), a continuity problem

f(x)=x if x is rational, f(x)=x^2 if x is irrational

I was working to show it continuous at $1$

since $f(x)$ is a classical function and $1$ is a classical point, you can prove that the definition I gave above for continuity is sufficient to show classical continuity

@GBeau But you said that the classical reals were not even an object in that nonstandard axiom system

indeed not, we don't use it in the proof

(I have it open in my text to make sure I'm not crazy)

@GBeau How could you not? You start out with a question about the standard reals

so how can you argue about something that does not even exist in the system?

a classical formula is true for all x as soon as it is true for all standard x (that's actually just one of the axioms itself)

the "T" axiom of IST ^

it took me a good 3 or 4 weeks before I wasn't tripping over myself with the axioms

@GBeau But then clearly the classical reals do exist in this system

but they cannot be the only reals

not sure what you mean by "only"

you can show from one of the axioms that if a set is infinite, it must have nonstandard elements

(any set)

@GBeau Hmm. This is sounding really weird

it is

hello

Especially because you still have the problem of getting a function defined on the "reals" in your system to work with, starting from the one defined in the problem

@TobiasKildetoft reals take the classical definition, transferred to all $x$ by an axiom

the words "standard" and "nonstandard" aren't necessarily set-forming in IST so statements that denote a set that uses them needs to be constructed axiomatically to show it even exists

@GBeau So there is an axiom that any function from $\mathbb{R}$ to itself has a unique "analogue" defined on the "reals" in this system?

yes, actually (standardization, or transfer depending on what you're interested in)

(the "S")

how is *f defined?

@Brennan.Tobias depends on context

that notation isn't used in the IST I learned

(this: en.wikipedia.org/wiki/Internal_set_theory )

but at any rate, with my example

the NSA definition is quite easy to understand, that for $1\simeq 1+\epsilon$, then $f(1)\simeq f(1+\epsilon)$

ncatlab.org/nlab/show/nonstandard+analysis I found one defintion here

"Then there is a nonstandard extension"

the above is obvious (just pick whether or not $\varepsilon$ is irrational, then take the standard part of both sides)

ie that $1=\mathrm{st}(1+O(\epsilon+\epsilon ^2))$

@GBeau The classical argument is pretty similar. You just take a max of two numbers

(and the RHS there is clearly $1$)

or, if ep was rational, it's just O(ep)

as I remember all this stuff applies to natural numbers as well as reals

then you get infinite natural numbers (but no infinitesimals)

it gets silly in IST since you modified set theory itself

so if you define finite as a bijection to some $\{1,2,\ldots n\}$ where $n\in\mathbb{N}$, then you run into the idea that $n$ can now be unlimited

so there's a finite set of numbers that contains all the standard natural numbers

(under IST)

so it's silly in the sense that it hardly seems "finite"

it's similarly easy to show $2$ is discontinuous (just pick a single irrational $\epsilon$, and the RHS easily fails)

@GBeau Same is true for the classical. Any open interval around $2$ will contain an irrational

yeah I was just looking at that, I pick (working with classical definition of continuity), $\epsilon = 1$ and an irrational $x$ in $(2,2+\delta )$

then $|f(x)-f(2)|>2>\epsilon$

what is it you're proving with nonstandard analysis?

f(x)=x if x is rational, f(x)=x^2 if x is irrational

found it

nothing I was just talking about how the problems I'm working are more obvious to me with NSA

(but I'm proving with classical definitions)

yes that's one of the best things about NSA

@Brennan.Tobias for that function, was showing continuous at 1 and discontinuous at 2

@GBeau Actually, it is probably a good exercise to try to understand how the nonstandard and classical approaches actually say "the same thing"

I think of it a bit like how complex numbers make some basic number theory proofs easier

hmmm

can I prove a sequence type statement about continuity? like if I can find a sequence where $\lim x_n=c$ but $\lim f(x_n)\neq c$ then, $f$ is not continuous at $c$

@GBeau Yes

mmmmmm actually this seems straightforward let me try to prove it

hmm this is interesting: if $f$, $g$ are continuous (both R->R) and $f(x)=g(x)$ for all rational x, show this implies equality for all $x$

@GBeau One of my favorite type of results

I tease out the same irrational number from the neighborhoods in the continuity definition I would guess?

@GBeau but come to think about it, I don't think I have ever proven it in the narrower context of analysis, only in the slightly broader formulation of topology

(knowing the ways that this generalizes gives a very neat proof of the Cayley-Hamilton theorem)

yeah if they agree on rational points then continuity mandates what values it takes everywhere else

@Brennan.Tobias I meant the generalization to varieties

what is that?

@Brennan.Tobias Replace the rationals by any dense subset and "continuous" by "morphism of varieties"

in case of Cayley-Hamilton, one uses that the set of matrices with all eigenvalues distinct is an open (and thus dense in the Zariski topology) subset, and that these clearly satisfies the required

still trying to show that using the classical continuity definition :3

@GBeau One way is to take some point $x$ and a sequence of rationals converging to $x$

then comparing the functions on the sequence

oh I think I can do that

actually I think the proof I already did earlier was pretty much sufficient then

(the sequence one)

Hello! I like this book's style: Katznelson's A (Terse) Introduction to Linear Algebra

Is there a book like "A (Terse) Introduction to Abstract Algebra"?

@Tien-ChengHuang Your mean Katznelsons' (there are two of them :) )

@TobiasKildetoft This one amazon.com/dp/0821844199

@Tien-ChengHuang Right, there are two authors, both called Katznelson

ok then Katznelson and Katznelson 's "A (Terse) Introduction to Abstract Algebra"...

Why is it so hard to find a basic example on how to calculate some divisor of a curve?

two functions nowhere continuous whose addition is continuous...opposite value assigned Dirichlet functions?

@GBeau good choice

@Krijn It is an unfortunate thing that as topics get more abstract, it becomes rarer for examples to be provided (annoyingly for anyone wanting to learn)

I've proved most of these book problems with the sequence lemma I mentioned

@GBeau it is often handy, yes

I dont see what I'm doing wrong...

Let $C$ be the curve given by $Y^2 = XZ$ in the projective plane. What is the divisor of the function $g = X/Z$?

I would say, $(X:Y:Z) = (X/Z : Y/Z : 1)$ if $Z \neq 0$ so we can look at $t^2 = u$ with $t = Y/Z$ and $u = X/Z$?

then $div(g) = div(u)$

Ah I figured it out, great!

@TobiasKildetoft $\Bbb Q(a, b) \cong \Bbb Q[a, b]$ implies $a, b$ are algebraic. Am I being silly or does this just follow from the fact that if $a$ is transcendental then $(a)$ is an ideal of $Q[a, b]$?

which is a contradiction, as fields have no proper ideal

if f is not continuous at c does there always exist a sequence x_n converging to c where {f(x_n)} does not converge?

(stronger than asking if f(x_n) merely doesn't converge to f(c))

oh I think I got it

Suppose that $k = \mathbb{F}_q$. What are the number of effective divisors of $C$ (a smooth projective curve) defined over $k$ and equivalent to $D$?

How do you approach this?

I guess if an effective divisor $D'$ is equivalent to $D$ it means that it differs by an element of $\operatorname{div}(k(C)^*$

@robjohn you are possibly right .. I guess there are issues in the integrals that need to be fixed .. it's better if I take it down for now! :-) Thanks for the numerical verifications! :D

Given an arbitrary independent set of a finite-dimensional vector space, how to expand the set to obtain a basis?

@r9m Numerically integrating and summing the series I got both agree, so I am inclined to believe those results.

(oh... possibly a newbie question...)

@robjohn I am probably making some kind of mistake and I need to identify it ,, thanks!

I.e. finding row reduced echelon form of a matrix.

well sorry i haven't work through the whole LA book...

Then you know for sure that you can add (0,0,1,0) and (0,0,0,1).

I.e., the ones are in the positions where the rref does not have pivots.

@Tien-ChengHuang Could you give an example which you are trying to solve?

Or some example similar to your problem.

I have two exercises giving spanning set and requiring trim them to obtain a basis. but have no exercise giving independent set and requiring expand them to obtain a basis.

so i can just learn form your example now

And you are working in $R^3$ or $R^4$ or some similar space?

$R^5$

Ok, let's do that then.

Suppose we are given vectors $(1,2,-1,3,2)$ and $(1,1,0,2,1)$.

We check whether they are linearly independent. (That should be easy right?)

We want to add some other vectors - we need to add 3 vectors.

Let us denote them by $\vec a-(1,2,-1,3,2)$ and $\vec b=(1,1,0,2,1)$.

Notice that $\vec a-\vec b=(0,1,-1,1,1)$

Would you agree that $[\vec a,\vec b]=[\vec a-\vec b,\vec b]$?

I.e. that $a-b$ and $b$ span the same subspace as $a$ and $b$?

@Tien-ChengHuang

you mean the same span? yes

ok

So we need some three vectors which are not in their span.

I write them under each other so thet we can see better what ve are doing.

That should be b and a-b, unless I have made a mistake.

Maybe I could use b-a instead (it does not matter that much).

Notice the first and the third column.

Those are columns where I have number one and all other entries in that column are zeroes.

This is the important thing.

Now if I ask whether I can get (0,1,0,0,0) as a linear combination of these two vectors, could you answer that?

I will write them again into a matrix, so that we can see them better.

We are asking whether the third row can be obtained as linear combination of the first two rows.

Can you answer this question @Tien-ChengHuang?

i think no

That's correct. They are linearly independent. Is there some easy argument how to see that the third row is not linear combination of the first two?

As I said, the first and the third column might be useful to find such an argument.

i am listening :)

ok

To make this shorter I will denote the first two rows by $\vec r_1$ and $\vec r_2$.

If I write a linear combination $c\vec r_1+d\vec r_2$ then I can see what will be on the first and the third coordinate.

The first coordinate must be equal to $c$. The third coordinate must be equal to $d$.

So if I want to get $(0,1,0,0,0)$, I must have $c=0$ and $d=0$, because of the first and third coordinate.

But this does not work on the second coordinate.

So I can't get the third row as linear combination of the first two.

Does this argument make sense?

yes i can follow you

Great.

And for the same reason I can add two more vectors.

Maybe this is not the only possibility, but for these vectors I can be sure that they are independent without checking it by manual computation.

Would you agree that in this way we extended {a,b-a} to the basis of $R^5$?

ok i see the second column can be made containing only one 1 in the 3rd row

the same as the 3rd and 4th column

4th an 5th*

Yes. Using row operations. So you have learned a bit about elementary row operations right?

yes so that is a use of row operation

ok

So we extented {a,b-a} by these three vectors.

This is the same as extending {a,b}.

But b-a was better, because I had a column containing zeroes on all places with one exception.

I simply guessed that I can use b-a. (For two vectors this was easy.)

But no guess work is needed.

so {a,b}U{the three vectors} are maximal independent set

which is spanning

Yes.

To avoid guesswork, I could have taken the original two vectors:

Use row operations until I get matrix in such form that some column has all zeroes with one exception.

This is what I meant by Gauss-Jordan elimination and rref. (I am not sure whether you have learned about this yet.)

If I get such matrix, I can add the vectors as described above. I.e., I take vectors from the standard basis, but avoid these columns.

thank you!! you really teach me something :)

I could also have done this.

Now I got pivots in different place (the first and the third columns).

So in this case I see I can take e_3, e_4, e_5 from the standard basis.

I have just added this to show that e_2, e_4, e_5 is not the only possibility. (Those were the vectors we took before.)

ok, I hope this helped a bit and that you would be able to solve something similar with three vectors

I wish I can add someone as a friend in MSE now.

one of my book problems was wrong :s

It happens.

I made the errata page ^^

congratulations

@anon: are you here?

yeah

the problem was: Let $c_1$ be a cluster point of $A \subset \mathbb{R}$ and $c_2$ be a cluster point of $B \subset \mathbb{R}$. Suppose that $f \colon A \rightarrow B$ and $g \colon B \rightarrow \mathbb{R}$ are functions such that $f(x) \rightarrow c_2$ as $x \rightarrow c_1$ and $g(y) \rightarrow L$ as $y \rightarrow c_2$. Let $h(x) := g\bigl(f(x)\bigr)$ and show $h(x) \rightarrow L$ as $x \rightarrow c_1$.

where that's a notation for a limit

@anon: do you by any chance know of some resources where I can see examples of computations of Haar measures? my measure theory is a bit weak and I need to confirm that some measures are indeed Haar measures

I spent quite some time on it before I realized the statement was false

not really

why there is a lot of people here, haha