Mathematics - 2016-02-05 (page 1 of 2)

12:21 AM

@Tobias let us consider the concepts presented in the three references. Can we form function between $A$ and $A^*$ (considering both infinite and finite $A$) that show relations amongst orders like wqo between $A$ and $A^*$?

Ted Shifrin

@mreyeglasses: You done did it. :)

hi @Julian.

mr eyeglasses

lol

Ted Shifrin

Not much goin' on here today ...

guest

...the calm before the storm :)

Julian Rachman

@Ted hello

Ropstah

12:55 AM

I'm watching a video about Leibniz calculus and at some point a derivative is taken of a function with a power ($3x+2x^2$), he also noted somewhere that he was 'taking down the power and lessening it with 1' resulting in a derivative of $3+4x$

does that mean that the power (2) is added to the 2 before the second $x$?

guest

No, it is multiplied.

Ropstah

ah ok

and does the $3x$ transform in a $3$ because that $x$ is also 'taken down'?

guest

Yes.

Ropstah

ok, so I tried to come up with some other examples, if i understand it correctly then $4+6x$ is the derivative of $4x+3x^2$

guest

Correct.

Ropstah

1:00 AM

and what happens to $5x+2x^4$?

$5+8x^3$? or should I bring down the cubic factor?

or is it perfectly fine to have a power in a derivative?

guest

It is fine.

Ropstah

ok cool thanks

guest

np

thanks for asking :)

Ropstah

do you know about Leibniz derivative notation?

guest

Yes, but it's better to go to Wikipedia first.

Ropstah

1:04 AM

oh lol, i guess that is the standard

I was into the history of Leibniz and ran into a video about "Leibniz derivative notation" thinking it was an alternative to some Newtonian method or something ;)

guest

Try the khan academy on youtube.

Ropstah

Yes i've done a lot there

But I'm working on math as a hobby next to my job as a programmer

It's hard to start from scratch as I have a lot of knowledge which is just named differently

And sometimes that's a bit troublesome as most of the times it's basic algebra rules or definitions

guest

I would suggest getting a "standard textbook".

Ropstah

anyway, I had a little moment of enlightenment (but was off a bit due to multiplication instead of addition)

I'm doing fine, and with your sharp comment I was corrected

Books don't work with me, I'm very visual

I have my programming environments, plotting capabilities and my pocket calculator

guest

There are some good ones with a lot of diagrams and sketches :-)

Ropstah

1:10 AM

I create 3D interfaces for my geometric 'explorations' finding the most beautiful theorems as 'crossing lines' etc.

But I just can't read a textbook :)

I need to work with the numbers and plots and get the concepts right

guest

I can relate.

Words are tough :)

Ropstah

very much

guest

I can spend literally hours on just one sentence.

Ropstah

I've spent the last two years on a set ;)

guest

What kind of set?

Ropstah

1:14 AM

which is another programmatic approach to a more scientific field

a set of concepts

/ words

each element represents a certain concept and can be related to other concepts

but it's not very mathematical, there are no mathematical properties to the set

guest

There are so many logical links between sentences :-/

Ropstah

more ontological, it is creating a knowledge base as a graph

yes I'm not working on sentences as of yet, all concepts are separate entities. But stories are the next part, which will include a rule engine for logical inferences

but at that part the complexity will go up exponentially

guest

Along with the amount of time spent on it.

Ropstah

Yes, but before that it should be online and crowdmanaged ;)

But before that I need to get my "prime" concepts correct

I'm trying to find good work about Leibniz' Characteristica universalis

guest

Ask on the main site.

Ropstah

1:24 AM

Oh that's not a question

It was a share ;)

guest

:-)

thnx

Ropstah

very interesting stuff, probably (too) complex

but having a 'knowledge' system or definition system for 'concepts' or thought should produce interesting objects

there was a company once who said their system was 'based on' Characteristica universalis

too bad that Leibniz never finished the work

lvella

2:45 AM

if I scale an object by a factor of x, the new volume will be old volume times x³, right?

Julian Rachman

3:35 AM

Can anyone tell me if we can form an order-perserving bijection between a wqo set X and a well-ordered (or suborder ordered) free monoid X^* over X?

Forever Mozart

4:53 AM

mjhvcxzsdfghjkl;'

guest

mjhvcxzsdfghjkl;' + 1

Forever Mozart

very good :) now we are on the same page

guest

:)

6

Q: Meaning of "on the same page"

Where on the scale from We are talking about the same thing to We are in agreement / We see it the same way is the phrase We are on the same page?

phrases

PVAL

5:09 AM

@MikeMiller I am sending you an email with quite a dense description of my recent work over the last few days, if you are interested.

Forever Mozart

yes

@PVAL in topology?

yes

manifolds?

yes

oh well I like topology but I don't do manifolds

I prefer set-theoretic topology

I'm a narcissist

Mike Miller

5:16 AM

@PVAL: I would appreciate that, yea.

guest

Who are you @MikeMiller cheering for Sunday?

PVAL

@MikeMiller Alright, sent.I added about a paragraph of context for you. A lot of it is another correspondence, so it might be difficult for you to grok.

Mike Miller

5:36 AM

@PVAL Thanks, I'll read it tomorrow. I'm a little distant from contact topology so will probably be a little slow.

@skull: Whoever the party's host is rooting for, lest I get no beer.

Aditya Dev

7:19 AM

Help. Someone is upcoming all my posts in math.SE. I don't want to be banned again.

Sorry. It's *upvoting and not upcoming

My reputation on math.SE just increased by 50.

Tobias Kildetoft

8:25 AM

@JulianRachman I am back in chat now. Well, the fact that $A^*$ is a wqo using the subword order when $A$ is finite seems like a bit of a miracle, so I am not sure if there is any obvious way to map from $A$ to $A^*$ in that case (though I suspect one can probably embed $A$ in $A^*$ in this case just because $A^*$ is large enough to allow a lot of things).

user153330

9:08 AM

@Hippalectryon (i mean by affordable that the ratio of acceptance of overseas students is the most high)

lucad93

9:40 AM

Hello!

Julian Rachman

@TobiasKildetoft I think embedding is an option

Tobias Kildetoft

This is quite interesting. So, if we have a finite group $G$ acting transitively on a set $X$, then this obviously gives a representation of $G$ by linearizing. And this representation has a unique (up to scalar) vector on which $G$ acts trivially.

But it turns out that if we start with a representation with a basis on which $G$ acts positively and "transitively up to summands" then this still holds. Here the requirement is more precisely that the matrix corresponding to the action of each element from the group has non-negative real entries, and that the matrix of the sum of the elements from the group is strictly positive.

I wonder how easy this is to show directly.

@JulianRachman What do you mean by an option?

Huy

@MikeMiller that's actually surprisingly obvious. -_-

Julian Rachman

10:00 AM

@Tobias though I suspect one can probably embed $A$ in $A^*$ in this case just because $A^*$ is large enough to allow a lot of things"

Tobias Kildetoft

@JulianRachman Right, but an option for what?

Julian Rachman

An option to embed

Like have the order on A^* to be a string-embedding preorder

@Tobias

Tobias Kildetoft

@JulianRachman you mean the subword order?

Julian Rachman

Is that essentially the same?

Tobias Kildetoft

Well, I am not sure what you mean by string-embedding preorder

Julian Rachman

10:07 AM

Page 8, Defn 3.1: cis.upenn.edu/~jean/kruskal.pdf

@Tobias

Tobias Kildetoft

@JulianRachman That is what you called the domination order earlier

robjohn

@I'manartist: Check out this blast from the past.

Huy

@TobiasKildetoft: surely you know of Jordan's curve theorem

Tobias Kildetoft

@Huy I know of it, sure.

@Huy Could you try to prove the statement I mentioned above? I think it might be a nice exercise (or turn out to be highly non-trivial)

Huy

@TobiasKildetoft: I'm thinking about something geometrical atm, just a second

@TobiasKildetoft: so if I have some simple closed curve $\gamma$ on $\mathbb{R}^2$ bounding a region $D$ and a homeomorphism $\phi$ on $\mathbb{R}^2$ and I know that $\phi(\gamma) \cap \gamma = \emptyset$, by connectedness, $\phi(\gamma)$ is either completely inside or outside of $\gamma$. $\phi(D)$ will always be bounded by $\phi(\gamma)$, yes? if $\phi(\gamma)$ is outside of $\gamma$, then $\phi^{-1}(\gamma)$ must be inside?

I have a strong feeling those are true but fail to make a rigorous argument

Tobias Kildetoft

10:21 AM

@Huy is $D$ open or closed?

Huy

open

just think unit circle and open unit disk

Tobias Kildetoft

@Huy so $\phi(D)$ might become the outside, right?

Huy

@TobiasKildetoft: that's what I thought and then it wouldn't be bounded anymore

Tobias Kildetoft

@Huy "bounded" is not a topological property

GBeau

lmao I screwed that up

Tobias Kildetoft

10:24 AM

Ahh, but the closure of $D$ is compact, right? And this would no longer be true for the imageif it became the outside

Huy

yes

GBeau

Am I misunderstanding this question and answer? math.stackexchange.com/questions/494979/primitive-root-theorem The way I read it, the answer calculates $$\frac{\#\{a \in \mathbb{F}_{p}^\times : a^{\frac{p−1}{q}} = 1\}}{\#\mathbb{F}_{p}^\times}$$ instead of $$\frac{\#\{a \in \mathbb{F}_{p}^\times : a^{\frac{p−1}{q}} \neq 1\}}{\#\mathbb{F}_{p}^\times}$$

Huy

so it can't be all of the outside

but will be bounded still since compact and in R^2

but is it precisely bounded by $\phi(\gamma)$ again?

it must be

take some point in $\phi(\gamma)$ and an arbitrary open neighbourhood $U$. the preimage will be some open nbhd of $\gamma$ so it will intersect $D$. since it's a homeo, that means $U$ must intersect $\phi(D)$

robjohn

@AdityaDev If that was one person, the votes will probably be invalidated by an automatic script that is run around 0300 UTC.

Huy

so indeed the "situation of the Jordan curve theorem" stays exactly the same if I apply a homeomorphism on the simple closed curve, interior and exterior

now if $\phi(\gamma)$ is outside of $\gamma$ I want to show that $\phi^{-1}(\gamma)$ is inside. @TobiasKildetoft: any idea? I'm trying to find a contradiction if it was outside but I'm not very successful

Tobias Kildetoft

10:36 AM

@Huy No idea, sorry

GBeau

I'm guessing @AndreNicolas just made a silly mistake there

Valentin Tihomirov

Could you imagine multiple dimensions and higher order direvatives if you lived in only 2D?

Huy

yes

@DanielFischer: do you see how to prove this rigorously? take the unit circle $\gamma$. let $\phi$ be some homeomorphism of $R^2$ such that $\phi(\gamma) \cap \gamma = \emptyset$. assume $\phi(\gamma)$ lies in the exterior of $\gamma$. I think that then $\phi^{-1}(\gamma)$ must lie inside it.

user189740

Hey Guys. I say an interesting equation today. I saw my textbook assert that $(1-x)/(1+x) = (2/(1-x))-1$. I can't for the life of me figure out how they got the left to be the right.

user189740

Any ideas?

GBeau

10:45 AM

I added my question as a comment to @Andre's answer as well math.stackexchange.com/questions/494979/primitive-root-theorem

Tobias Kildetoft

@byteofthat Write $1$ as $\frac{1+x}{1+x}$

user189740

@tob

user189740

@TobiasKildetoft I had tried multiplying by the conjugate, but I didn't really get anywhere

user189740

I suppose I should mention that I was only given the left hand side, and I was supposed to figure out I could turn it into the right hand side, so that I could use it for something else

user189740

so I am trying to figure out how just looking at the left, I could massage it to be the right

Tobias Kildetoft

10:50 AM

@byteofthat I agree that it is tricky to spot that it can be turned into the right hand side, but once you see it, it should be clear that they are equal

user189740

@TobiasKildetoft what is the first step I might take starting with the left hand side?

GBeau

oh well I'm done pulling my hair out trying to see if I misunderstood andre

hopefully he replies

nothing else to do today :<

Tobias Kildetoft

@byteofthat Well, you want to have a $-(1+x)$ in the numerator

user189740

@TobiasKildetoft but wouldn't that produce $-1-x$, which isn't the same thing that's already on the numberator

Tobias Kildetoft

@byteofthat right, you don't want the numerator to become that. You want it to include it

user189740

11:02 AM

@TobiasKildetoft I am sorry, I am not following

Tobias Kildetoft

@byteofthat Well, I am assuming you are looking to put in on a certain form specified beforehand?

user189740

I know I want to get it into the form of something that is similar to $1/(1-x)$

Tobias Kildetoft

@byteofthat Actually, the stated equality is not right. It should be $1+x$ in the denominator on the right

user189740

sorry, yes, you are right

user189740

Id like to get it in the form of something similar to $1/(1+x)$

Balarka Sen

11:17 AM

Hi @TobiasKildetoft.

Tobias Kildetoft

@BalarkaSen Hi

Balarka Sen

How's things?

Tobias Kildetoft

@BalarkaSen Good. Trying to understand transitive representations of finite groups (with the transitive part in the form I mentioned earlier)

Balarka Sen

I don't understand half the things you do :) Lots to learn.

Tobias Kildetoft

I am not sure I understand half the things I do

Balarka Sen

11:22 AM

:)

I was talking about the other half.

Tobias Kildetoft

@BalarkaSen You mean you understand the half I don't? Could you explain it to me? :)

Balarka Sen

That I don't understand the half the things you understand of the things you do doesn't mean I do understand the half the things you don't of the things you do.

GBeau

if $a$ is a quadratic residue and $p\equiv 3\pmod 4$, then $b=a^{\frac{p+1}{4}}$ is a square root of $a$ according to Stein (along with proof); is the other root $b=a^{\frac{3p-1}{4}}$?

Tobias Kildetoft

@GBeau Well, $-b$ is also one

GBeau

(I'm on page 87, it's open source: wstein.org/ent/ent.pdf )

@TobiasKildetoft oooooooooh.

Daniel Fischer

11:39 AM

@Huy Without further constraints on $\phi$, that needn't hold. Think of translations.

Huy

@DanielFischer: true, I forgot another condition. there are $x, y$ in the unit disk with $x = \phi(y)$.

does that suffice?

GBeau

@TobiasKildetoft why must $a^{\frac{p-1}{2}}\equiv 1\mod p$ in this case?

Tobias Kildetoft

@GBeau I get math processing errors on most of what you write

GBeau

@TobiasKildetoft Interesting, my mathjax is processing it

Tobias Kildetoft

@GBeau Hmm, seems like mine doesn't like pmod

GBeau

11:45 AM

is \mod better?

Tobias Kildetoft

apparently not

anyway, $a^{\frac{p-1}{2}}$ certainly need not be $1$ mod $p$

It could also be $-1$

GBeau

for a quadratic nonresidue, right?

the proof in Stein concludes it's the legendre symbol and equal to $1$ in the case above

Tobias Kildetoft

@GBeau Right

If $a$ is a quadratic residue then it is indeed $1$

(I had forgotten that)

GBeau

oh oh oh oh

I assumed that

okay that resolves it then, I just forgot my assumptions

Tobias Kildetoft

It is easy to see that it is $1$ by plugging in $a = b^2$

GBeau

11:52 AM

good point

Huy

@DanielFischer I think I got it

Daniel Fischer

@Huy Yes, then you have $\overline{\mathbb{D}}$ contained in the interior of $\phi(\gamma)$.

Huy

@DanielFischer: using the additional condition that there exist some $x, y \in D$ such that $x = \phi(y)$, say $\phi(\partial D)$ maps to the exterior of $\partial D$. this must be some simple closed curve "around" $D$ now. call $\phi(\partial D) = \partial C$. since $\phi$ is a homeomorphism, $\phi^{-1} (C) = D$. since $\partial D \subset C$, the claim follows.

Daniel Fischer

You should probably say explicitly what $C$ is.

Huy

the interior bounded by $\phi(\partial D)$

Daniel Fischer

11:58 AM

I know. I meant, you should probably explicitly say it when you write down a proof.

Huy

yeah, I don't have to write down a proof

I'm just filling the gaps of some proof that just writes "... follows from Jordan curve theorem"

(for myself)

Huy

1:20 PM

@DanielFischer: take some simple closed curve $\alpha$ on a hyperbolic surface. as we have established some other day, we can lift it to $\tilde{S}$ starting at $x_0$. let $x_1 = \tilde{\alpha}(1)$ be the lift's endpoint. then there is a unique deck transformation such that $\phi(x_0) = x_1$. (with some more work, this yields isomorphism of the fundamental group of $S$ and the deck transformations of the universal cover)

is there any reason why this particular deck transformation should be a hyperbolic isometry of $\tilde{S} \cong \mathbb{H}^2$? the text I'm reading repeatedly uses this fact but I don't see why it can't be parabolic or even elliptic. if I take some geodesic in the disk model, I could just rotate to move one endpoint to the other and that would also be a deck transformation, an elliptic isometry, no?

Julian Rachman

@Tobias so lemma 1.7 works in this situation right for my order-perserving bijection (isomorphism).

Tobias Kildetoft

@JulianRachman Between which objects?

Julian Rachman

A and A^*?

Tobias Kildetoft

@JulianRachman I thought we concluded there was never such a bijection

Julian Rachman

Nor isomorphism?

Even with the domination order

Tobias Kildetoft

1:33 PM

@JulianRachman Well, not necessarily never I suppose. But at least there was not always one, as the example of $\omega + 1$ showed.

Julian Rachman

On then... Ugh.

I need to find a relation!!!!!!!!!!!!

(@Tobias)

Daniel Fischer

@Huy "hyperbolic isometry" means "isometry with respect to the hyperbolic metric".

Tobias Kildetoft

@JulianRachman Well, I think the best you can hope for is an injective morphism

Huy

@DanielFischer: are you sure? it seems very clear in this text that they classify isometries of $\mathbb{H}^2$ in terms of fixpoints of the extension to the compactification. if there is one fixpoint in $\mathbb{H}^2$, that's an elliptic isometry, if exactly one on the boundary, it's a parabolic and if there's two fixpoints, it's a hyperbolic isometry. later in those proofs where they use "hyperbolic isometry", they also assume two fixed points in the boundary.

they also repeatedly use "the axis on which the hyperbolic isometry acts by translation", which also only makes sense with that notion

Daniel Fischer

1:50 PM

@Huy Okay, if they do that: an elliptic automorphism cannot be a deck transformation (unless it's the identity) because it has a fixed point. I don't see off-hand what rules out parabolic automorphisms as deck transformations.

Huy

oh, right

apparently it's only parabolic if the given curve can be freely homotoped into a nbhd of a puncture

Daniel Fischer

And, hmm, $z \mapsto z+2$ is a deck transformation for the modular function $\lambda \colon \mathbb{H}\to \mathbb{C}\setminus \{0,1\}$.

Julian Rachman

@Tobias what can I do with an injective morphism in terms of preserving wqo'ness?

mr eyeglasses

2:20 PM

"well-quasi-ordering" sounds like a cool term

privetDruzia

Hello everybody! :)

Julian Rachman

@mreyeglasses yep. It is a cool concept too.

@Tobias what can I do with an injective morphism in terms of preserving wqo'ness?

mr eyeglasses

Too bad I can't understand the concept yet

Albas

2:41 PM

@mreyeglasses Were you morphic before your new name?

mr eyeglasses

@Albas This is my old name I went back to, morphic was my new name

Albas

Okay.

GBeau

does the metric $d(x,y)=\min \{1,\lvert x-y\rvert\}$ on $\mathbb{R}$ have a name?

Albas

@GBeau The books I have read have no mention of that metric . Would you care to tell where you found it?

GBeau

an oddly motivated exercise in a book I was reading asks a question about it and I was deciding if it was worth my time to solve

@Albas says it has bounded sequence in $(\mathbb{R},d)$ without a convergent subsequence but sequences converge in $(\mathbb{R},d)$ iff they converge in the standard metric on $\mathbb{R}$

Mike Miller

3:15 PM

"A 4-manifold is constructed with some curious metric properties; or maybe it is many 4-manifolds masquerading as one, which would explain why it looks curious. Anyway, knots in the 3-sphere with complete finite volume metrics on their complements play a role in this story."

Semiclassical

3:28 PM

morning

robjohn

@Semiclassical Indeed it is :-)

mr eyeglasses

What is a volume metric

s.harp

Is it true in general if you have a bounded operator on a Banach space with eigenvectors spanning a dense subset, that the operator norm is equal to the suprememum of the eigenvalues?

Mike Miller

3:54 PM

@mreyeglasses: (finite volume) metric, not finite (volume metric). With a metric I can talk about the volume of the space.

I mostly just found the phrasing funny

iwriteonbananas

4:45 PM

Good evening

Mike Miller

morning

iwriteonbananas

Just learned about Jacobi fields

robjohn

@iwriteonbananas how often do they need to be watered?

iwriteonbananas

hahah

Mike Miller

@robjohn: They're the critical points for the watering functional

iwriteonbananas

4:48 PM

They didn't tell us that, unfortunatley

Mike Miller

Any good class on landscaped geometry should talk about the watering functional!

robjohn

@MikeMiller Does that play an important role in the Irrigation Theorem?

Mike Miller

At least in modern treatments, @robjohn; before sprinkler techniques, methods were much more ad hoc

robjohn

@MikeMiller filters and sieves were used to clean up the flow of arguments.

Semiclassical

I'd make a pun about ultrafilters, but my well of wit seems to have run dry.

guest

4:59 PM

We accept "dry" humor ;)

Mike Miller

hm, Thierno M Sow is emailing the department again

robjohn

@guest A lot of the humor this morning seems to be watered down

Semiclassical

@MikeMiller emailing the department?

Alessandro

Good evening everybody

Mike Miller

@Semiclassical: my department with his solution to the Kakeya problem, using his solution to the riemann hypothesis

Semiclassical

5:05 PM

ah. how pleasant.

is that sufficiently 'dry'?

though i'll confess, i have no idea what the kakeya problem is

Mike Miller

Me neither. I still suspect he hasn't solved it.

Semiclassical

that's a pretty good bet, giving the tool he purports to solve it with

i do recognize the kakeya problem after glancing at wikipedia, though; i read a math article about it at one point

the planar version, anyways. no idea about higher dimensional versions

alan2here

hello

"a" and "b" are constants.

Can I rearrange this "f[n] = (f[n-1] * a) mod b", To find f[n-1] in terms of f[n]? f[n-1] = ... f[n] ...

Alessandro

I have a question that has been hurting my mind for a while, can I have an uncountable, well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$?

guest

5:26 PM

@robjohn can watered down humor be dry?

Semiclassical

@MikeMiller btw, did you want me to continue rambling on about berry phase stuff at some point? (the conversation you asked me to link for reference)

guest

Seems like a contradiction to me :-)

Something can not be both watered down and dry at the same time.

Hippalectryon

@user153330 I don't know which ones accept the most students from overseas. The best will probably accept you provided you have excellent grades.

Tobias Kildetoft

@JulianRachman Not really anything I think, unfortunately

Imago

How is this general room used? Can anyone here just ask a quick question and look, if someone reacts on it? For example likes to read up on a certain topic?

Mike Miller

5:38 PM

@Semiclassical: Not today, I'm afraid; I'm both sick and busy

Semiclassical

oof

hope you weather it well

Mike Miller

And on top of that there's a new paper up today by my hero to read ;)

Semiclassical

ahh. which, out of curiosity?

Mike Miller

arxiv.org/abs/1602.01687

Semiclassical

yeah, that's about what i'd have expected :)

Mike Miller

5:41 PM

I am probably not going to read all 150 pages but I'd like to get the idea

Ted Shifrin

hi méchant @Hippa, @Tobias, @MikeM, @Semiclassic

hi ted

\o @TedShifrin

@TedShifrin Hi

good evening @ted

5:42 PM

Hi @Alessandro! Haven't seen you in a long time. Everything going well?

Semiclassical

right now my decision is between trying to read a paper or trying to grade homework

Ted Shifrin

is @guest a reincarnation of skull? Why does everyone insist on name changing?

Alessandro

very well! I've been very busy studying for my exams, but I passed all of them @Ted

how are you?

Semiclassical

though both are competing with the urge to go take a nap :P

Ted Shifrin

Very good, @Alessandro. I'm not surprised you did well :)

Mike Miller

5:44 PM

@Semi nap

Ted Shifrin

Stop that, @Semiclassic :D

guest

Yes professor @TedShifrin

Semiclassical

i see you're both in complete agreement :P

Ted Shifrin

@Semiclassic: I make a point of never being in complete agreement.

Semiclassical

there's a grad student talk i'm heading to in half an hour, so i should probably not bother

Ted Shifrin

5:45 PM

@Alessandro: So what are you studying these days?

Semiclassical

@TedShifrin For me, I seem to make a point of never being in complete agreement with myself.

Mike Miller

ah, 90 of the pages are an appendix.

Semiclassical

that's a hell of an appendix

Ted Shifrin

That's pretty much absurd, @MikeM.

Alessandro

right now nothing, the second semester begins on the 15th of February @Ted

Mike Miller

5:47 PM

@Ted: It's the proof of a theorem he needs but considers less important to the exposition.

Ted Shifrin

Ah, so you're on holiday, then, @Alessandro. Congratulations!

Mike Miller

I think he can write however he wants.

Alessandro

I'm reading some introductory things to set theory on my own because I find it interesting @Ted (actually I posted a question earlier in the chat, do you have time to look at it?)

Ted Shifrin

Of course he can, @MikeM, but a paper where the appendix is twice the length of the paper itself is somewhat strange.

@Alessandro: I'm not much good at set theory stuff. What is it?

Semiclassical

not to pester, @ted, but did you take a glance at that B. Simon paper I referenced? (aside from the abstract, i mean)

Alessandro

5:48 PM

it's here chat.stackexchange.com/transcript/message/27369972#27369972

Ted Shifrin

No, @Semiclassic, not yet.

Semiclassical

mmkay. it's here

Ted Shifrin

@Alessandro: My immediate answer to your question is no.

Semiclassical

which, judging from the title of that link, looks to have been used in a Diff-Geometry course

Balarka Sen

Hi @TedShifrin.

Semiclassical

5:51 PM

albeit probably a more physics-oriented one

Ted Shifrin

hi @Balarka

Semiclassical

but i do like that it's a Barry Simon paper on Berry phases which ends up citing Bott and Chern

Ted Shifrin

Well, perhaps, @Semiclassic, but then physicists need to learn about connections on hermitian vector bundles :P

Balarka Sen

@TedShifrin Did good on history today :)

Semiclassical

heh, yeah, or at least be able to tell what a mathematician means when they reference such

Ted Shifrin

5:52 PM

@Alessandro: It seems to me any ordered chain gives rise to a subset of $\Bbb N$, hence must be at most countable.

good, @Balarka.

Semiclassical

i'll confess, my knowledge of connections is pretty poor :/

Ted Shifrin

You could say you're feeling disconnected, @Semiclassic.

Semiclassical

nah, my spaces have $\pi_0 =0$

Balarka Sen

@TedShifrin Long pun no see.

Ted Shifrin

We'll see if that's improved you, @Balarka.

Balarka Sen

5:55 PM

After exams, we will. (8]

Semiclassical

i mean, my knowledge of bundles is pretty loose as well. i know them structurally--- a bundle space with a projection map into a base space

Alessandro

there can be uncountable chains, as I discovered reading this question, basically you map integers to rationals and pick Dedekind's cut, I don't know about well ordered chains @Ted

s.harp

Are there places in physics where axiom of choice or continuum hypothesis find use? I was wondering about experimental justifications for unprovable statements :)

Ted Shifrin

Oh, interesting, @Alessandro.

guest

Talking about "at risk students" Professor @TedShifrin I found an astrophysicist currently at Princeton who is apart of their prison teaching initiative that goes into state prisons and teaches them high school math and physics.

Semiclassical

5:57 PM

@s.harp i may just be lacking imagination, but I would say no

Kudos to him, skull!

So @guest is skull.

Yup

Just to be sure, @Alessandro, what do you mean by a well-ordered chain?

Balarka Sen