This came up in algebra today. The prof was stumped. Somehow I knew you guys would be johny on the spot.

@AndrewG What was le problem?

@BalarkaSen My main motto would be learning the tools, as that would not just help me understand RH but also several other RHs, would it?

@BalarkaSen it's up not too far

@Swadhin There is only one RH, the last time I checked.

@BalarkaSen I mean other difficult conjectures such as RH... :)

I dunno.

That's the only million dollar problem I understand.

=P

@PedroTamaroff Does that not also have index 2?...

Oh, wait.

$\langle s,r^4\rangle=\{1,s,r^4,sr^4\}$, since $r^4=r^{-4}$?

Ok...bye, I will do some reading and get it touch with you, and hopefully solve the problem too.

$D_8$ has $16$ elements.

So it has index $4$.

And it is not normal since $sr^3\cdot s\cdot r^{-3}s=sr^6$ is not in it.

Hello

Hello Group Theory Master.

Derp, of course, I forgot how to count.

hello Andy

sup mike

not much

@Pedro @Andrew: Glad y'all are ignoring my suggestion :D

hi @Mike

hello @Ted

@Ted anytime! :)

i'm actually just slower than you guys and it takes me longer to process all this...

I'm slower than most of the young'uns here, @Andrew.

And @Pedro is totally ignoring me ... sulk

@TedShifrin I comfort you.

@Andrew: Here's a hint. Conjugation in $GL(n)$ means change-of-basis ... i.e., the conjugacy class is all different matrix representations of a given linear transformation.

How so, @Jasper?

@TedShifrin Simply by what I said.

@TedShifrin I'm reading some ComplEX Analysis from Remmert.

Kinda boring...

Complex analysis is one of the most beautiful (and broad) subjects. It makes me sad when you're bored.

@PedroTamaroff That book is not on my list of favourites. I don't know why. It looks weird.

@Jasper: We already know that your taste is suspect.

@TedShifrin Maybe it is the book.

I just woke up after 3 hours sleep, should go back to sleep later.

@PedroTamaroff Why did you choose Remmert?

@JasperLoy Recommended.

@PedroTamaroff By your professor?

@Ted If he's bored, he should read Hormander.

Freitag's Complex Analysis 1 and 2 is probably the most comprehensive treatise @PedroTamaroff

Hörmander is very terse, but very elegant. But all of one variable is a 15-page chapter, if I remember correctly.

I've never heard of Freitag, @Jasper.

@TedShifrin Translated from German as well.

You're correct, @Ted. I struggled on that chapter for a long time. I gave up someway into page 12 or 13, maybe.

@Pedro: Cartan's book is succinct and good. Ahlfors is a classic.

I'm blanking on the classic text that's German.

@TedShifrin libgen has the french version.

Learn French?

Ah, Hille is the one I was blanking on. Two volumes. Classic.

@TedShifrin Hrmmmm....alright. Let me think. enters the Chamber of Understanding

@TedShifrin There's one by Janich, not translated.

@Andrew: I'm doing my best to turn you into a geometer :D

@TedShifrin >:(

Yeah, I'm definitely giving up on Remmert.

LOL, pauvre @Pedro. :)

You need it for grad school anyhow, @Pedro.

@PedroTamaroff Have a look at Freitag Pedro. It is the best.

@TedShifrin Need Remmert?

Who is this guy? ;)

@Jasper discards all the significant classics with that unknowledgeable remark.

No, @Pedro. Need FRENCH.!!!!

@Pedro Read Cartan or Ahlfors.

@TedShifrin I live in the modern generation, lol.

Well, that's a big mistake, @Jasper.

@TedShifrin Oh. But Remmert is in English. Cartan is in French.

@PedroTamaroff Cartan's book has been translated, a Dover book.

But I'm telling you you need French/German/Russian for grad school, @Pedro.

I only have it in French, actually. :)

@Pedro I do not speak French but I can read Godement's book, which is in French. Just read it. You can too.

Godement is much harder than Cartan. Élie Cartan, a different matter.

@PedroTamaroff Cartan is in Eng!

Not free, apparently, @Jasper.

I.e., pilfered.

@MikeMiller Yeah, I guess I can roughly translate stuff to Spanish.

French is easier for you than German or Russian, @Pedro, I will surmise.

@Ted For sure. My point is that I know trivial amounts of French and I can read it nonetheless with minimal difficulty.

I'm not disputing you, @Mike.

When one knows math, one can glean a great deal from notation and context.

Subtleties may be missed, however.

True.

God knows how many subtleties I've already missed!

The Ahlfors book is so expensive.

Hell, @Mike, you've missed plenty in English :D

@MikeMiller Only if you believe in God.

I don't, @Jasper, but he/she/it still knows.

@TedShifrin OK, Hormander uses differential forms....

Differential forms are useful ...

and essential for where he's headed for several complex variables.

@Pedro He also uses measure theory for everything, which is what ended up making me give up.

I always introduce them in my graduate complex variables class, @Pedro, not formally, but I use them.

Pedro is not listening to my recommendation, sigh...

@Ted Can you reasonably do Riemann surfaces without them?

no @Mike

you can do nothing interesting on manifolds without them :D

I remember how Pedro insisted on using Mendelson for topology.

Thought not

(waits for lightning to strike)

@Ted Watch yourself, I will invite an army of topologists in.

I'm used to being hated, @Mike.

@TedShifrin Not a sure thing; depends on the school. My observations suggest foreign language exams being steadily dropped from PhD requirements in math departments around the U.S.

(Observations being visits to grad program pages of math departments.)

@Ted Question: why are smooth manifolds triangulable? I remember this is true but not why.

slowly, @Thursday, true ... but most of the top schools still have it ... serious computer skills replacing the second language

@Thursday I hope they are dropped too when I enter.

The language requirement is essentially a triviality.

with a Riemannian metric it's not too hard ... but it's a theorem, @Mike.

@TedShifrin Do you know what the baricentric division of a simplex is?

yes, @Pedro

So does @Mike

OK, explainz?

>not knowing barycentric subdivision

@MikeMiller Have you decided on your thesis topic?

haha

@Pedro: In 2D: Take a triangle. Take its centroid. Use the centroid to divide the triangle into 3 subtriangles.

@MikeMiller I have great expectations of you.

@TedShifrin OK.

@Ted You want six subtriangles. Only doing 3 means the diameters will not go to zero after repeatedly subdividing.

OH yeah, @Mike is right. I told you to ask him.

@Pedro Pick a centroid, and take the line through each vertex and that centroid. That subdivides the triangle into six subtriangles.

@MikeMiller OK.

Or rather take the centroid

And for a tetrahedron?

Or in general for $\Delta^n$?

That's the barycentric subdivision of a 2-simplex

In more generality pick an (n-2)-simplex in the boundary and take the hyper plane that contains it and the centroid

And all of those, together, provide the subdivision

Actually the better description is inductive

It is 6 am. I should go back to sleep.

@JasperLoy Yes, go and sleep.

Wow, that's the only thing you said to me. I am hurt.

It's for your own good.

Subdivide each simplex in the boundary . Then the subdivision of the simplex is [v_0, ..., v_n, C], where the first are the vertices of one of te subdivided simplices, and C is the centroid

There's a description of this process somewhere in the beginning of chapter 2 of Hatcher's book. It has pictures.

blue

Someone is hammering on the wall at 6 am, nuts.

I think the person living below me is nuts. Hammering is usually done at 12 am.

@blue a friend and I got a bound on that problem. we'll post it later.

cool

I almost responded to the @blue

my fault, @Jasper

I am the only true blue, lol.

well, anon is allowed to feel blue, too

In fact, this blue is called "blue".

well, I guess it's time to cook dinner ...

@blue for your favorite positive $\varepsilon$ there's an $f \in O(x^{4+\varepsilon})$ such that for any abelian group $G$, $|\text{Aut}(G \times G)| \leq f(|\text{Aut}(G)|)$.

the argument does not extend to even a little bit more generality

I saw what was in a casino the first time in my life a few weeks ago.

I did not know it was so huge.

@MikeMiller Cool, Mike.

Maybe I should try to make some money out of gambling.

@JasperLoy That's not smart.

That's cool. To prove an $n$-simplex has topological dimension $n$, one uses Sperner's lemma!

@Pedro Why do you care about simplices so much today?

Maybe he is studying algebraic topology today.

@MikeMiller I'm reading a book about Combinatorial and Differential topology.

Why?

I'm interested in it.

I killed 3 mosquitoes this year. They sucked my blood.

Good reason.

I am still itching from the last bite.

Oh, Ted has gone to cook dinner.

Did you guys know that xkcd also autoexpands in chat?

I am off to bed. Goodnight @PedroTamaroff, I will see you in my dreams.

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