Looks good @feynhat

have you an idea @NicholasRoberts ?

Hello

hello

should one put parentheses on integrands that are sums?

I used to think yes but now I'm rethinking it

example please

@AlessandroCodenotti Thanks.

@feynhat you should give an epsilon-delta proof for metric spaces

it's a good exercise

I didn't even think that was a sum, but i think that if it's clear from context then probably no parentheses

@0celo7 proof of what: Transitivity of desnsity relation or that C[a,b] is separable?

@0celo7 putting $\mathrm{d}$whatever at the end is not an option?

Hi,

Weiertrass new

@AlessandroCodenotti ew, why would I do that

I omit measures as much as possible

@Daminark wot, there's a +

@feynhat $A\subset B\subset C$: if $A$ is dense in $B$, $B$ dense in $C$, then $A$ dense in $C$.

I like to write the measures explicitely

do this for subsets of metric spaces

@0celo7 Oh ok. Thanks for the suggestion. Will try later.

I don't write that many integrals however

salut, Gérard

@AlessandroCodenotti oh lord. Let's do a search for \int

tu fais des heures sup ?

my tex editor won't tell me how many occurrances

I can tell you! More than three

tu as des nouvelles de CC ?

3 is a lower bound, yes

if I actually wanted to find out how many, I'd probably search "\int_" rather than just "\int"

100 is as well

just in case there's other \int-stuff

@0celo7 ugh there should be a law regulating the maximum number of integrals in the same document

aller je te laisse, et tu me vois ravies pour ta promotion...lol

tchuss

the string \int occurs only 102 times in the current draft of my thesis

Allowing no more than 10 to be precise

the string zeta, on the other hand, occurs almost 300 times

\int occurs 0 times in the current draft of mine

Which is one of the very first drafts, but I can assure you that number won't grow from now to the final draft!

I can't figure out how to do this

and I haven't even written the chapter on local zeta functions, yet

@XanderHenderson you use mac, how are you doing it

I think i've got about 50 total \int's

I know how to do this on my PC

@0celo7 it has nothing to do with the OS, you need a better text editor

I use BBEdit

bah, some hipster shit

fyi, in TexWorks it only searches one at a time by default

but you can do a checkbox to 'search for all occurences'

Jokes apart I would write the brackets

and then it'll list them all + the total number

so in your editor it may be a similar thing

@0celo7 it is more a matter of inertia

I've been using BBEdit for almost as long as I can remember

it used to be free, even :\

back in the 90s, it was either BBEdit, or SimpleText

Am I the only one who picked the easiest to install editor? Which happened to be Kile

and SimpleText was less feature rich than Notepad

@AlessandroCodenotti I prefer a text editor and the command line

I use TeXShop

it was the first thing I saw on Google years ago

I don't like the integrated TeX dohickies

seems to work except for stupid shit like this

i used texcentre on my previous laptop, but on this one it always crashed

so I went back to plain TexWorks

749 occurrences

loool

not surprising

>tfw sums register as $\Sigma$ to me and I forgot how + works

@XanderHenderson that's painful when you have to do multiple steps for index/internal references, biber, the other index etc.

@AlessandroCodenotti that's what makefiles are for

What would be funny is the percentage of pages that integrals appear on

i finally gave up on doing everything in bibtex

believe it or not there are some

since all the papers I'm adapting already had the sources in bibitem form

and BBEdit has a macro system which allows me to run a makefile with a keyboard shortcut, if I really want to

I just use sharelatex

yeah

@XanderHenderson I know nothing about those. I only used TeX from command line once because I had to generate some big tables so I made a Lua script that wrote and compiled a .tex file

for getting stuff working initially I was using Overleaf

It's the easiest to install since you don't install anything

with a bunch of these things, 749 is totally reasonable i.gyazo.com/613cd179144cbef8e70cf3fb29c80745.png

Sharelatex and Overleaf are merging apparently?

not sure wth that actually means

@Semiclassical it means all of your projects on there will break

I thought they were gonna stay separate but just have a partnership or something

maybe that's what it was. I don't remember the details at all

Overleaf is annoyinf because of the automatic preview (which can be turned off), the impossibility to zoom on the pdf and the internal references not working in browser unless you download the file

the zoom is very annoying, yes.

does sharelatex not suffer from that?

But we still use it for the group assignment in algebraic number theory because it's the easiest way to habe many people working on the same file

I never tried sharelatex, I don't know

kk

I tried to use github for latex

it was awful

You can zoom in sharelatex

I don't mind the auto previewing when I'm trying to figure out something simple

I like zooming to see the beautiful vector graphics

I just like being able to see text at a reasonable size

except integrals are significantly lower rez than other things

you can make the Overleaf output window bigger and so make the output easier to read, but you do so by making the input window smaller

which is...less than ideal

I used overleaf once when helping a friend proofread a paper, the auto preview stuff kinda annoyed me and felt like it made it a bit slower

With the internal reference thing I mean that if you usehyperref they seem to be clickable in the preview, but clicking on them will be interpreted by overleaf as "find the point in the source code corresponding to this" rather than "take me to where this reference is pointing"

oh, it defeinitely makes it slower

I don't actually zoom too much so I haven't benefitted from that side of sharelatex

so if i'm generating latex writing at length then it is a good idea to turn off auto preview

but it is convenient at times being able to just change a few characters and not have to tell it to re-compile

so sometimes it's nice. but I don't understand why it's the default

and I always turn off the rich text option for the input

@Semiclassical did you see Sam's hilarious Penrose diagram notation

no

hahaha

christ

the fact that I can actually read that to some extent is the best part of that

That looks like an attempt at the three utilities puzzle

$A_{\mu\nu}^{ij}(\partial_i u^\mu) (\partial_j u^\nu)$

Probably that should be $\nabla$ not $\partial$

Not sure what the last bit is supposed to be

Well it could be $\partial$ but...fuck...why did you have to remind me of that

this is probably all wrong now that I think about it

oops

lol

The problem with doing things in ridiculous generality is that the derivatives should be covariant in the bundle

and now I need another symbol

all of the usual ones are taken, crap

another symbol for?

$\nabla =d+A$, where $A$ is the connection form

but $A$ is taken

oof

$\aleph$ :P

I might do $V$

that $A$ is the potential in Yang-Mills

and $V$ is used for potentials

so why not

yeah

could in principle do $U$ but I'm guessing that's taken

hmm, no

not in this section at least

(since U is potential energy)

yeah

I don't really like how people use V for potential energy

if you're doing potential, sure

but writing the hamiltonian for instance as $H=T+V$ just confuses me

(and I have no idea why T gets used for kinetic energy by some authors)

kineTic energy

:/

@Semiclassical I think this has been around for 100s of years

yeah

google says it's French

makes sense

Hey everyone

I have this problem, "Let $x$ be a symbol and define $f(x) = x^2 + x + 1 \in \mathbb{Z}_2[x]$, prove that $f(x)$ is irreducible in $\mathbb{Z}_2[x]$"

@Perturbative If $f(x)$ is reducible, then $f(x)$ has a root. Does $f(x)$ have a root?

Shouldn't $f(x)$ be written as $f(x) = 1_{\mathbb{Z}_2[x]}x^2 + 1_{\mathbb{Z}_2[x]}x + 1_{\mathbb{Z}_2[x]}$ though?

Because $1$ is not an element of $\mathbb{Z}_2[x]$ but the equivalence class $[1]_2$ would be in $\mathbb{Z}_2[x]$ however

Or if you go by the coset definition of quotienting the ring then you'd have a coset $1 + 2\mathbb{Z}$ instead of $1$

maybe, but uh

that's really s***** notation

Hiiiii

@Semiclassical feel free to write this diagramatically i.gyazo.com/9fd37401e166142c74af7369b09dd97b.png

also integral count + 10 since last measured

Aren't there usually different symbols for different kind of potential energies ?

@0celo7 I could, but uh

no

I have a grid of 3x8 points and I want to choose 8 grid-points out of in it so that no two points are adjacent horizontally or vertically. How would one go about doing this?

I could, let's say, write it as a 8-string word with letters U, D, M, C and O corresponding to picking a point up, picking a point down, picking a point both up and down, picking a point in the middle, or picking no points at all in a 3-point column.

The constraints are that I can't write a U or M after U, a D or M after D, neither of U, D, M after M and C after C. Also everything except M contributes to 1 on that column, which contributes a 2, so those contributions must sum up to an 8. This seems like a nightmare please kill me now.

(@AkivaWeinberger I am going to ping you because you're the only person I would subject this horror to)

Oh, and O contributes to a 0 of course

So for example a string which produces a solution is MCMCMOOO (2+1+2+1+2+0+0+0 = 8)

MC Moo is my new rapper name now

@Balarka the progression in that sequence of messages was gold

Who's Balarka? Call me MC M$\!\infty$

Lmao

this is MC Moo / I be counting all night till I fuck up your poo / deranged angels in a derangement like / factorial n factored by n factorials slides off / to one over constant e, just got a Phdee / in comb-in-atorics / up my sleeve I have lots of trix /

Oh my Lord what has gotten into you

We need to not let you do combinatorics

Talk to me about fibrations

Another problem, let $f(x) = x^2 + x + 1 \in \mathbb{Z}_2[x]$. "Write down all elements of $\mathbb{Z}_2[x]/f(x)\mathbb{Z}_2[x]$ and prove that your list is complete". But like for example $(x^{999}) + f(x) \mathbb{Z}_2[x] \in \mathbb{Z}_2[x]/f(x)\mathbb{Z}_2[x]$ and so is $(x^{47393} + x^{884} + x^1) + f(x) \mathbb{Z}_2[x] \in \mathbb{Z}_2[x]/f(x)\mathbb{Z}_2[x]$, I don't see how I could write down all elements in any sane way

@Perturbative those two examples are in the same class as much much smaller polynomials though

@AlessandroCodenotti hello

Hi

@BalarkaSen are you wanting to count all the ways to do that?

coming up with individual examples seems simple enough

@AlessandroCodenotti I'm not sure what class (something to do with irriducibility maybe?) that is though, all I can say at this point is that any polynomial in $\mathbb{Z}_2[x]$ has co-efficients either $0$ or $1$

I don't have time to think about this now, sorry

@Vrouvrou What is adherent value?

@Semiclassical Of course, sorry I wasn't clear about that.

np

@Perturbative What do you know about $x^2+x+1$? What does that tell you about the ideal it generates? What does that tell you about the quotient?

I want to count the number of ways I can do it.

we say that $x$ is an adheret value of $u_n$ iff $\forall V\in\mathcal{V}_x,\rm card\{n\in\mathbb{N}, u_n\in V\}$@AlessandroCodenotti

I guess what I'd be inclined to do is start with the 3x1 case, in which case there's 5 possible ways to put in dots without violating adjacency

I know what an adherent value is, but I don't want to think about that question now, it looks like an uglier than I'm willing to do exercise :/

@Daminark The ghost of MC Ride

so... ooo, oox, oxo, xoo, xox

@Semiclassical That's what O, D, C, U, M stands for

derp

Well, in different order

$x^2 + x + 1$ is irreducible in Z_2[x], so the ideal it generates is maximal proper in Z_2[x] and the quotient is a field because of that @AlessandroCodenotti

with certain adjacency constraints, as you noted

@Perturbative Right, more specifically a field of which characteristic? And with how many elements?

hmmm, a thought

Lord

you can represent the adjacency constraints by writing a graph with O,D,C,U,M as nodes and with edges between elements that are allowed to be adjacent

@Daminark Disjointed Houdini baby

with the nodes carrying a weights 0,1,2,1,1 respectively

@Semiclassical That's a good idea, yes.

you then want all 8-vertex paths for which the weight of the vertices is 8

I want paths in that graph which add the weight up to 8

Rip but yes

That's not something I'm so versed in but it seems like a good formulation

C stands for Central, so that's 1, M stands for "mixed" (so D and U both), so it's 2

Sorry for not being clear with terminology

ah, gotcha

I figured it was 'middle'

Sobolev

@AlessandroCodenotti what is the pushforward filter?

and, actually

suppose you've got something like SUUD

UU can't happen

Can't it?

oh, woops

can SUCC happen?

@BalarkaSen well I'm still in Z for the time being

SOOD

@AlessandroCodenotti Z_2 has characteristic 2, not sure about the characteristic of Z_2[x] or the quotient though

@0celo7 Nope, very unfortunate

@LeakyNun you have a filter $F\subseteq\mathcal{P}(X)$ and a function $f:X\to Y$, the pusforward filter $f^*F$ is the set $\{Z\subseteq Y\mid f^{-1}(Z)\in F\}$

@Semiclassical Ok.

@AlessandroCodenotti and what is the meaning behind those symbols?

Which symbols?

my thinking is that OO has the same weight as O itself, so any 8-path which includes OO can be mapped to a 7-path which has O instead, and this won't change the weight of the path

$X,Y$ are just sets, $f$ is just a function

I mean, I can understand which set it is, I just don't get it @AlessandroCodenotti

the intuition behind. I can see the symbols.

I don't even get the intuition behind filter

The main reason I like this is that it would allow one to ignore the possibility of OO, at the cost of worrying about shorter paths

people say "it's telling you which sets are big", but I don't see that at all

@BalarkaSen CUCC?

the cure may be worse than the disease for that, though

I think about filters has saying which sets are bigs, kinda like a $\sigma$-algebra (filters and measure are somewhat related)

@Semiclassical yeah I'm worried about that

well, let's get the graph first. I'll ignore O for the moment since it'll connect to everything (including itself)

@0celo7 Nope, only two O's can consecutively appear

so just D,C,U,M

so...

C - U
| X |
D - M

@AlessandroCodenotti and I already said I can't understand that intuition

@LeakyNun It's just a way of saying that a set "is big" if it is in the filter. That makes sense in the context of ultraproducts, because you identify two sequences if the set of coordinates on which they agree "is big" (is in the filter you're quotienting by)

I mean, a singleton can generate a filter

is a singleton big?

it is according to the filter

just like a singleton is big according to some measures

so my intuition fails

Is that right?

@Semiclassical M doesn't connect to D

Or to U, for that matter

Yeah

at least in the epsilon-delta definition I can visualize an ever-shrinking rectangle near the point

Just C

I can't visualize anything for the filters

@AlessandroCodenotti and you know the filter around infinity right

M
|
C - D
| /
U

That should work.

Yeah this looks right.

and then U connects to everything, with a loop on it itself as well

which is annoying

Actually consider the measure $\mu:2^{\Bbb R}\to [0,+\infty]$ defined by fixing an $x_0\in\Bbb R$ and declaring that a set has measure $1$ iff it contains $x_0$ and measure $0$ otherwise. I claim that $\{X\subseteq\Bbb R\mid \mu(X)=1\}$ is the principal ultrafilter generated by $\{x_0\}$

part of the issue is I'm more familiar with graphs where the edges have weights

not vice versa

That measure considers big a singleton, just like that filter does

@AlessandroCodenotti you mean the point mass at $x_0$

@Semiclassical Maybe take the dual graph?

@0celo7 I suppose

it's a thought

@AlessandroCodenotti or the delta function

That's how the physicists call it!

I don't see a systematic way to count the paths, I guess.

that's how most people call it

@0celo7 for god's sake

I suppose you can "do it", give you have written out the graph

@AlessandroCodenotti right, but I can visualize a measure as some kind of notion of "length"

I just can't visualize anything when it comes to filters

@LeakyNun ?

@LeakyNun Not really, I guess it's the one generated by the open rays $(a,+\infty)$?

/-- `at_top` is the filter representing the limit `→ ∞` on an ordered set.
  It is generated by the collection of up-sets `{b | a ≤ b}`.
  (The preorder need not have a top element for this to be well defined,
  and indeed is trivial when a top element exists.) -/
def at_top [preorder α] : filter α := ⨅ a, principal {b | a ≤ b}

(NB: ⨅ is indexed infimum of filters, under reverse inclusion)

I wish I had a collection of lots of creative combinatorics problems but which simultaneously didn't make me feel like shit for not getting it

There's no mechanic by which I can enjoy these problems. It's like, lots of "Aw fuck yeah"s when I get them, and lots of frustration on the contrary

@LeakyNun I don't think about filters visually, just like I don't think about $\sigma$-algebras visually, I just think about them as families of sets with nice closure properties

@AlessandroCodenotti :(

@Alessandro Sad reacts only.

@AlessandroCodenotti do you think about topologies visually

Could someone explain why this happens $(-1)^n(l+in\pi)+(-1)^{-n}(l-in\pi)=2l(-1)^n$ ?

Also because the interesting filters are the nonprincipal (ultra)filters, those are really strange

@BalarkaSen if it weren't for the weight constraint I'd say to write down the adjacency matrix and take the 8th power

that would probably be overkill but it'd work

@Isa note that $(-1)^n = (-1)^{-n}$

Oh right. I remember that.

So the adjacency matrix is the $n\times n$ matrix with $a_{ij}$ corresponding to $1$ if there's an edge between the $i$-th and the $j$-th vertex and $0$ if there's none?

$n$ = number of vertices

@0celo7 I think about the resulting topological space visually, but I don't really think about the collection of sets satisfying a bunch of properties that we call open when thinking about a torus

right

which in this case would be...

that explains everything, thanks @LeakyNun

@BalarkaSen If a 2-surface in $R^3$ contains a line, is $K=0$ on the line? Google suggests I can only say $K\le 0$ but I'm pretty sure it should be $K=0$.

I don't remember why the $ij$-th entry of $A^k$ counts the number of $k$-paths on the graph between the $i$-th and $j$-th vertices.

I see it for $k = 2$, though.

Oh

@BalarkaSen it's induction time

Lol, yes

Ok, I see it

@0celo7 Think about the hyperboloid of one sheet

I'm pretty sure $K \neq 0$ along the rulings

Yeah it isn't

 11111
 10110
 11011
 11100
 10100

fixed

Nice

in the order O,U,C,D,M

So if you had weights on the edges, you'd change those 1 by the weights

Man the dual graph should do it

does that actually work? I forget

I think it should. Hm.

Wikipedia says that's the adjacency matrix of a weighted graph

hmm

link?

@BalarkaSen Hmm I have a different picture in mind, need to write down exactly what's happening

I mean actually it's obvious, I think. If you have weight k on the edge, replace it by k weight 1 edges.

hm

yeah, ok

@Semiclassical Here, right below the section

thx

So, what's the dual graph here

@BalarkaSen Oh, what if it's also a geodesic of the surface. I don't think the ruling there is a geodesic.

@0celo7 Oh, then I think in an appropriate frame one of the principal curvatures becomes 1

1? Not 0?

The other one should be 0 if your claim is true

$K = k_1 k_2$, and $k_1 = 1$