@TedShifrin No, in fact I do. I'm out of beers :(

@Krijn Are you Dutch?

LOL

...your profile suggests so

which means we had the same(-ish) high school education ahah

I guess you're Dutch

I guess I am

You guess?

I'm from below the rivers :o

made some slight organizational/writing edits on the post

What do Dutch people think about people calling the Netherlands "Holland"?

I love it.

I always call it Holland.

Don't care, really

But I do call it the Netherlands

Also, I've had 10+ people ask me "what do you think about this"

nobody really cares

...but I do call it Holland

Huh

4 syllables is worse than 2

I'm guessing you live in one of the Hollands

(also, I'm from Amsterdam)

There you go

^^^yes

When our football team is playing, everybody agrees on Holland

That is the real measure of nationalist feelings

Today was the real measure

Lebesgue is the real measure

@Krijn ?

@AkivaWeinberger King's Day

King's day today

Ah

(which is my excuse for drinking)

...I think

What's a king's day?

Does he have any real power?

@AkivaWeinberger Neh

A king's birthday

@BalarkaSen Birthday

We partayyyy

Can I just say, as an American, that sounds so bizarre

I didn't know Holland had a king.

I have missed King's/Queen's Day for the past 2 years now :(

Although only when the king's birthday is in spring or summer

@BalarkaSen It's in Europe. So…

@BalarkaSen And a beautiful queen, too!

When it's too cold we just change the date

Mucho caliente

Meh

...for a queen!

I like the guipure lace. The woman... well, the picture does not show her in the best way.

Just Google "Maxima". I dare you to find a more beautiful queen!

@Danu Jordan's

"Beautiful queen" is a nonexistent object.

Arguably.

I frequently think it would be faster for people to learn [X] than it takes for people to ask questions about how to learn [X].

@AkivaWeinberger Hmm... comparable.

eg math.stackexchange.com/questions/1761757/…

@MikeMiller True

Especially on Physics and Mathematics

...but I think your link is not a great example.

@MikeMiller sadly, the speed of finding answers to X is frequently dependent on background knowledge in the area of X

@MikeMiller It's probably for the greater good, in a way; now, whenever someone Googles "books to learn [X] from," they can easily find a nicely compiled list

guys how can I prove that there exists $ab = nq$ for all $a,n,b,q \in \mathbb{Z}$?, assuming $a,n$ are fixed integers and $1 \leq a \leq n$ and $1 < n$?

Picking a good textbook is a delicate matter!

hahaha @Obliv, gotten desperate? ;)

:(

@Danu You perhaps don't see how much energy some people spend on picking textbooks.

He's drunk. Ignore him. @Obliv

With all that discussion, they could have just read three of them instead.

it seems so intuitive I'm having trouble finding words to establish a proof

@MikeMiller I do spend quite a lot of time on it too

@AkivaWeinberger I've started on the whisky collection now :3

Slainte!

Tobermory 10 it is, tonight

@Obliv I'm not 100% sure what you're asking, but what if you let $b=n$ and $q=a$?

(I think you made a typo in the question)

oh sorry I forgot to put bounds on $b$

$1 \leq b < n$

there are no bounds on $q$

pantachiquement

So, I'm confused. If we fix $a=3$ and $n=5$, say, are you claiming that there are $b$ and $q$ satisfying those bounds where $ab=nq$? @Obliv

what does that word mean?

what does that mean?

pointwise?

It means no worries for the rest of your days @ForeverMozart

lol

ok so I can forget this paper

no, that was hakuna matata, @Akiva. you confused.

I'd say what The Hound said about the king, but this is a family show...

yes but it seems there is no such $b$ for that case. I see, it makes sense that they prefaced it by saying $(a,n) \ne 1$ I thought it wouldn't matter if they had a common factor or not :p

@akiva

Yeahh

with those bounds that is. I was thinking it wouldn't matter because you could just multiply by the opposite number like you said

So, you're saying the thing you wrote is false, unless you have $(a,n)\ne1$?

yes with the bounds correct

Arright

Speaking of word meaning, I think this is something everyone should see. Not sure if I posted it before.

Well, I guess if you do $q=a/(a,n)$ and $b=n/(a,n)$, it should work. @Obliv

@Ted: Maybe you can write a nice answer for this guy.

They're integers because $(a,n)$ is a common divisor

you would get the least common multiple that way, no?

Uh, yeah, I think so

ah yes. $ab = lg$ where $l$ is the lcm and $g$ is the gcd between the two integers $a,b$ (sorry for using different variables)

$q$ also equals $\lcm(a,n)/n$, and $b$ is $\lcm(a,n)/a$

Wait, that's not a proper LaTeX code? What.

i didn't know \lcm existed as tex

It doesn't, apparently.

$\gcd$ does, though

I like to denote lcm as $\ell$ in my proofs for some reason

\newcommand{\lcm}{\operatorname{lcm}}$\newcommand{\lcm}{\operatorname{lcm}}$

There we go

$\lcm$

oh nice i didn't know people had that kind of power here

It doesn't work forever, I don't think. If it's off the page it won't work

\newcommand{\sin}{\operatorname{cos}} MWAHAHAHA

(I didn't actually do it)

The level of evil is unspeakable.

Wait. What happens if I do \newcommand{\newcommand}{}? Would I not be able to change it back?

@BalarkaSen Tell me about math.

You're off probation until the room goes back to math again, so whatever you want.

Hey, TeX is math…

…sorta

Any suggestions on how to improve the looks of this diagram?

Nope.

^cool story bro

@AkivaWeinberger

I have a question

@MikeMiller If I start talking about $e^{{\rm d/d}x}f(x)$, does that count?

Nah.

It's equal to $f(x+1)$

Is there a way to calculate get the elements of GL(3,GF(2)) which fixes (1,1,1)^T?

Think about it

@AkivaWeinberger?

@Adeek I have no idea what that means O_O

@MikeMiller I got exposed into some basic Riemannian geometry during my stay in TIFR. Am I allowed to talk about that?

I mean, maybe a vague idea?

Why not?

GF(2) is just $Z_2$

and GL(3,Z_2) is just the set of invertible matrices with coeffients from Z_2

So, what am I supposed to talk about? I know a few basic definitions, a few basic facts. I don't know what I can do with it.

@Adeek I have no idea why you tagged me

Tell me a story!

@MikeMiller Once upon a time, there was a linear operator from the set of polynomials to itself…

Not you.

So, a Riemannian metric on a manifold is a smooth choice of inner products at each tangent space $TM_p$ where $p$ is point of $M$, smooth in the sense that given any two smooth vector fields $X, Y$ on $M$, my metric eats $X(p), Y(p)$, spits a scalar in $\Bbb R$ and this gives me a smooth function $M \to \Bbb R$.

What happened to the linear operator?

And every smooth manifold admits a Riemannian metric, because, well, partition of unity.

@rumtscho Um, it got eaten by the big bad wolf. The end.

@BalarkaSen Note that this is the same as "smooth section of a certain bundle"

Or perhaps someone took the exponential map of it, and noticed it looked similar to the Taylor series…

@MikeMiller Oh. Hmm.

The end? Now I want to know what happened to the wolf's stomach.

…and, uh, got that $e^{{\rm d/d}x}f(x)=f(x+1)$

Maybe the operator transformed it to a lamb's stomach so the wolf had to eat grass from then on.

I guess a smooth section of $TM \otimes TM$.

Wrong bundle, but close enough. Anyway.

@rumtscho He got a bad stomach ache, and the doctors had to perform an operation on him

Sorry, should have been cotangent bundle instead of tangent.

So, Riemannian metric actually induces an actual metric on the manifold.

Suppose $\gamma : [0, 1] \to M$, be a path between $p, p' \in M$.

Um, I think you take the inf of something or other?

Then I can define it's arclength $\ell(\gamma)$ to be $\int_0^1 \sqrt{\langle \gamma'(t), \gamma'(t) \rangle} dt$.

Yeah, $d(p, p')$ is defined to be the inf of arclengths of all the paths between $p, p'$.

You need to explain what that means

I mean, you don't, but I'd like to know

(The integral, I mean)

It's just the standard definition of the arclength for $\Bbb R^n$, if you think about it.

A smoothly varying inner product on the tangent spaces is all we need to define it.

"all we need"

:D

Oh, it's the sum of infinitely many infinitesimal segments, and an infinitesimal segment is small enough to fit in the tangent space? Or something

(Pre-rigorously, of course)

You should prove that infinitely many infinitesimal thing agrees with $\int_0^1 \|\gamma'(t)\| dt$ for paths in $\Bbb R^n$.

This is just easy calculus.

Hello @BalarkaSen

Hi @Ali.

Never mind, misunderstood

What have you been up to recently @BalarkaSen

@AliCaglayan Studying differential forms.

@BalarkaSen Yeah, I've seen that before

@BalarkaSen Very slightly maybe related

I am studying diff geo

Specifically Riemann manifolds and pseudo-riemann

Nice, you should talk to Ted.

More of a ricci calculus thing tho

I did

a bit ago

I don't know anything about those :)

It's essentially the mathematics needed to understand GR

I guess I'll stop telling my story because @MikeM is probably bored since he knows this.

All I know about Ricci is that it was used to prove Poincare's thingy

@BalarkaSen No, no, keep going

I'm thinking about going to a summer school on geometric flows this summer

Does anyone have any input on whether that's a good idea, or not?

(I don't know anything about geometric flows)

@Danu Looks interesting

@AkivaWeinberger I don't feel comfortable telling stories about things I don't know well to someone who isn't familiar with that topic, because I may tell them my personal wrong point of views about things (resultant of not knowing the story well) or even worse, wrong facts about things.

Sorry, I've been distracted

If you're busy, it's fine. I am not keen on telling this story since I don't know it well :)

what did I miss

I defined arclengths on Riemannian manifolds and proposed a metric on it, but I didn't prove it worked.

Also, a Riemannian metric is a section of cotangent bundle tensor cotangent bundle, not tangent. Mistype.

how common is it for math texts to ask you to create computer programs in the exercises?

@Danu why not first try to learn it on your own?

dummit & foote has been asking for them once a section from my experience

@BalarkaSen A special kind of that

Yes, of course.

@Obliv even high school textbooks do that.

It needs to be symmetric and positive definite.

Sure. What else you got?

So, I don't remember how to prove that the metric induced by the Riemannian metric is actually a metric. I think $d(p, p') > 0$ for $p \neq p'$ is the hardest part.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I never encountered any math txtbook asking me to create computer programs before. What is the point of that? Perhaps coding something procedurally gives more insight on what they are trying to understand. What about functional programming languages like haskell then? I don't know much about them but they seem to take out a lot of the 'effort'.

$d(p, p'') = d(p, p') + d(p', p'')$ is more or less obvious. So is symmetry.

Yes. You need to work in a special coordinate system to prove that.

Yup, @Obliv they take a lot of time and effort. There are also "calculator key-in" exercises.

It helps to learn number crunching ;)

@BalarkaSen At a guess, is this the only one of the three where you need the smoothness of the inner product things?

I think so, yes.

yeah, though of course you only need C^1

It's not clear to me what sort of coordinate systems I should choose. Should I be able to figure it out if I think a bit harder?

nah

alright. so are you going to tell me how to do it, or should i move on?

move on

@BalarkaSen Should be an inequality sign there, I think

Yes, apologies.

Of course.

OK, so next on the list is affine connections.

work in "normal coordinates", and use these to prove that the shortest path in these coordinates is an actual geodesic (as opposed to a broken one); see that any geodesic from the center of the coords to your point x that leaves the chart must have length at least, say, C; and therefore any geodesic is at least the length of the unique geodesic from the center to the point in your normal coords

something like this

only works for close points, but if the points aren't close... :)

Makes sense. If the points aren't close, I can just cover my path with such normal charts, nope?

if the points aren't close the distance couldn't possibly be zero :)

oh, sure, fair enough.

I need to work this out explicitly at some point of time.

start with some more basic manifold theory

OK.

and then before that, calculus :)

but now we're back to math, and that was the point

Right you are.

@AkivaWeinberger You should work out why inf of sum of all piecewise linear partitions of a curve $\gamma$ in $\Bbb R^n$ is the same as $\int_0^1 \|\gamma'(t)\| dt$ at some point of time. not really complicated, just a proof-worthy fact you didn't seem to know.

Alright, so I guess I will get to sleep today instead of more math. Will start on serious math tomorrow.

No, I did know that. I was just confused about something else @BalarkaSen

Ah, ok.

Though, um, how do we define $\gamma'(t)$ here?

Where?

Oh, probably some magic involving the definition of a tangent space

Oh for smooth manifolds

(I don't actually know what the formal definition of that is)

yes, it's the image of the differential

there's nothing magic. if you could define a tangent space, you could define the derivative of a curve

Yeah, I think that's essentially what I was confused about.

@MikeMiller Well, yeah. I can't define a tangent space

oh well

I haven't seen him for at least a couple of months @Agawa001 he got really quiet for awhile and then left without a word.

I mean, if these are subsets of Euclidean space, then probably

but that's not necessarily the case, is it?

Start with submanifolds of R^n

Which brings you to multivariable calculus ;)

I know some.

I have no idea what a differential form is, though. (Something about matrices?)

so you may as well.

every smooth manifold is submanifold of a euclidean space

Oh, OK. That's part of the definition?

No.

sure

What

you can take it as a definition, but manifolds defined in the abstract sense can be embedded in some euclidean space is nontrivial fact.

but the point is it doesn't matter much

also that it's better to do algebra with rings before studying higher category theory over artin stacks

I am skeptical that it is a nontrivial fact :)

nice to see you back @quid echo echo :D

clarification: nontrivial to me, because I haven't studied the proof of Whitney embedding theorem.

it's weaker than the whitney embedding theorem, which restricts the dimension

Hmm, fair point. Admittedly I don't know how to prove it in either case.

Good night, all

n

I feel guilty about not doing much work today, but on the other I hand I don't feel like it.

that's fine, as long as you feel bad about it

I should probably stop discussing fairy tale math and head to bed. That should make me less guilty, because discussing fairy tales while not doing the real work is probably more sinful.

seems like a promising idea. so, g'night.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ :-) I am not sure what the echo tolerance of this room is.