Mathematics - 2018-05-24 (page 3 of 4)

I think L'H put in his book that he learned everything from Bernoulli but ppl attributed it to L'H anyway which made Bernoulli mad. something like that

Abcd

@mercio so in reality it is "Bernoulli's rule"?

GFauxPas

math history has a lot of drama

mercio

in reality it is "the definition of the derivative"

Ted Shifrin

6:18 PM

Well, no, it's not quite that.

GFauxPas

and a lot of mathematicians seem to have been petty, from the anecdotes I've seen

Ted Shifrin

It really takes the generalized MVT to prove it, so it's not the definition of the derivative.

GFauxPas

the secret formula for roots of a cubic that started a lifetime feud for spilling the beans

Ted Shifrin

I try to be as petty as possible, @GFauxPas.

loch

i feel that l'hopital is only taught because it's more algorithm-like compared to using taylor expansions (which results in some students misapplying l'hopital quite often)

(disclaimer - ive never taught calculus)

GFauxPas

6:19 PM

that way you'll make it into the history books Ted

Semiclassical

Cool math of the day: math.harvard.edu/~elkies/Misc/pi10.pdf

Ted Shifrin

@loch: L'Hôpital is typically taught months before one gets to Taylor polynomials.

Semiclassical

Nothing too shocking but I hadn’t actually seen that explanation before

GFauxPas

using Taylor's implies continuously differentiable, though, and LHR only needs differentiable? or am I misremembering

Daminark

Hmm, I guess the question then is whether there is a situation where L'Hôpital helps and Taylor doesn't?

Ted Shifrin

6:20 PM

Elkies is a smart dude, @Semiclassic.

@GFauxPas, wrong on both counts.

Semiclassical

@GFauxPas I gave a talk for one of my undergrad courses on the discovery of the cubic formula (Tartaglia, Ferrari, Cardano etc). It makes for a fun story with a good amount of drama

GFauxPas

good thing I added "or am I misremembering", otherwise my street cred would be ruined

Ted Shifrin

You need differentiable not at the point but nearby, yes.

For the $k$th degree T.P. at $a$, you need $k$ derivatives at $a$. So it is different.

GFauxPas

ah

so for a silly example like

Ted Shifrin

In most actual applications, you have plenty of derivatives at $a$ for the functions involved, but you may have repeated $0/0$ occurrences when you do L'Hôpital.

GFauxPas

6:23 PM

$\dfrac {|x|}{|x|}$ as $x \to 0$

Abcd

@TedShifrin In that question, whose series should I use?

Ted Shifrin

@Abcd, for me it's easiest to switch everything over to $0$, so I don't have to rethink Taylor polynomials at $\pi/2$.

But you can do the latter quite easily, too.

GFauxPas

Lhr would give $\dfrac {\operatorname{sgn} x}{\operatorname{sgn} x}$ at $x \to 0$?

Abcd

@TedShifrin So I should choose h = pi/2-x ?

GFauxPas

I know it's a silly example

Ted Shifrin

6:24 PM

@Abcd, or better yet $h=\pi/2 - x$.

@GFauxPas: An example where $f'(0)$ and $g'(0)$ don't exist but $\lim f'(x)/g'(x)$ exists?

Abcd

@TedShifrin yes sorry

Ted Shifrin

Either way works, @Abcd.

@GFauxPas: Good question for you. Is there an example where $f(x)/g(x)$ has a limit ($f(x)\to 0$, $g(x)\to 0$) but $f'(x)/g'(x)$ does not?

GFauxPas

I have to catch a train but I'll think about it

Abcd

@TedShifrin Okay, how to solve it using $x \to \pi/2$ only?

Ted Shifrin

Use Taylor polynomials at $\pi/2$, @Abcd.

Abcd

6:29 PM

@TedShifrin It doesnt help

Ted Shifrin

Of course it does.

GFauxPas

if Ted gives a tip it's generally helpful

2

Abcd

$x(x+ x^3/ 3 + 2x^5/(15)+... - \dfrac{\pi/2}{1- x^2/2 + x^4/ 4!}...$

This is the polynomial I get @Ted

Ted Shifrin

@Abcd: Do you know what a Taylor polynomial (or series) at $x=a$ means?

Abcd

@TedShifrin I am just supposed to substitute a in the expansions?

Ted Shifrin

6:31 PM

No.

You do a polynomial in $x-a$.

Coefficients are derivatives at $x=a$.

GFauxPas

what's a more common notation for initial segment of a woset? one book of mine uses $\mathcal I_x$ and one uses $S_x$

I guess "I" for initial and "S" for segment

Ted Shifrin

or S for section. I dunno.

Abcd

@TedShifrin Sorry, not getting how to do the problem exactly.

Ted Shifrin

Look up Taylor polynomials, @Abcd.

Abcd

@TedShifrin thats just f(0)+f'(0)x + f''(0)x^2/2++ f'''(0)x^3/3!....

Ted Shifrin

6:34 PM

You'll find something like $\sum\limits_{k=0}^n \dfrac{f^{(k)}(a)}{k!}(x-a)^k$.

No, @Abcd.

That's only for $a=0$.

Safder

Hello

GFauxPas

There is no standard notation or convention for this concept.

Welp

Ted Shifrin

Hello, @Safder.

Abcd

@TedShifrin So here you want me to use f(x)= xtan x - (pi/2) sec x instead of expanding them separately ?

Ted Shifrin

No, I want you to write it, as before, in terms of $\sin$ and $\cos$, but expand those in Taylor series at $a=\pi/2$.

user193319

6:42 PM

How does one show that $\Bbb{Z}_n$ is an Artinian $\Bbb{Z}$-module? The $\Bbb{Z}$-module structure on it is $z \cdot \overline{a} := \overline{za}$, but $\overline{za} = \overline{z} \cdot \overline{a}$, so it seems that the module structure is identical to the ring structure (this seems like a facile argument, though). So submodules will just be ideals, so I just need to consider descending chains of ideals.

But I'm not sure how to show that any descending sequence is eventually constant.

Ted Shifrin

Ideals in $R/I$ correspond to ideals in $R$ that contain $I$?

mercio

how many ideals does $\Bbb Z_n$ have ?

user193319

A finite number.

Ted Shifrin

OK, so that's two ways to approach the problem.

user193319

Is it possible to show that one of the ideals in the sequence is $(0)$?

That would certainly kill the problem.

Ted Shifrin

6:45 PM

No, that needn't happen.

user193319

Hmm...you're right.

Alessandro Codenotti

@GFauxPas $\downarrow\{x\}$

sup nerdoids

gwarn

ohai

6:48 PM

rwk

B. Mehta

Hi all, perhaps I'm missing something obvious, but I'm struggling to see why $X_n \to X$ almost surely implies $\mathbb{E}(\|X\| I_{\|X\| \geq M}) \leq \mathbb{E}(\liminf_{n \to \infty} \|X_n\|I_{\|X_n\|>M})$

Ted Shifrin

Hi nerd Eric.

Eric Silva

nERdIC

how goes it @Ted

philmcole

7:08 PM

Helloo

We defined the oscillation for a bounded function $f: X \to \Bbb R$ on a metric space $X$ as $$\omega(f,x) = \lim_{\delta \searrow 0} \left( \sup f\left(B_\delta(x) \right) - \inf f\left(B_\delta(x) \right)\right)$$

and I showed already that $f$ is continuous at $x \in X$ iff $\omega(f,x)=0$. Now we proved a lemma which says for every $\eta \ge 0$ the set $N_\eta = \left\{ x \in X \mid \omega(f,x) \ge \eta \right\}$ is closed.

My question is: Why is $M := \left\{ x \in X \mid f \text{ is continuous at } x \right\}$ not necessarily closed?

I tried and came up with a proof that shows $M=N_\eta$ for a specific chosen $\eta \gt 0$ and know that $N_\eta$ is closed...

(Possibly wrong proof:) Let $x \in M$. Then $\omega(f,x)=c$ for some $c \gt 0$. Let $\eta := \min\{c=\omega(f,x) \mid x \in M\}$. Then $x \in N_\eta$ since $\omega(f,x) \ge \eta$, and so $M \subseteq N_\eta$. Let $x \in N_\eta$ then we have also $x \in M$ since $f$ is not continuous at $x$ because $\omega(f,x) \gt 0$. So $N_\eta = M$ and since $N_\eta$ is closed by the lemma we conclude that $M$ is too.

Can somebody tell me where my mistake in the reasoning is?

Nûr

Do you know how to prove for all function $f : I \to \mathbb R$, there exist $g,h$ satisfying the intermediate value theorem such that $f = g + h$ ?

Alessandro Codenotti

@philmcole Isn't $\omega(f,x)=0$ rather than $c>0$ for $x\in M$?

philmcole

Omg I wrote "continuous" where it should say "not continuous". Sorry

$M := \left\{ x \in X \mid f \text{ is not continuous at } x \right\}$

Alessandro Codenotti

You have no guarantee that $\{\omega(f,x)\mid x\in M\}$ has a minimum

philmcole

mmh can you explain why?

It's certainly $\ge 0$

Alessandro Codenotti

7:22 PM

it has an inf, but not necessarily a minimum I think

mercio

@Nûr yes

philmcole

@AlessandroCodenotti And with inf instead of min it would not work?

Alessandro Codenotti

The inf could be $0$

philmcole

Ah yeah that would be a problem... So is there a simple way how I could show that the min doesn't exists?

Alessandro Codenotti

I don't think so, what you want to prove is false in general, but true for some functions

philmcole

7:28 PM

This means I would search for a counter example then

Nûr

@mercio What is the argument ?

Alessandro Codenotti

Thomae's function works

philmcole

Cool thx

Zee

What alcohol goes well with beef jerky ?

Alessandro Codenotti

What's true (and easy to prove from you already have) is that the set of points on which $f$ is continuous is a $G_\delta$ (and that the set of points on which it is discontinuous id an $F_\sigma$)

mercio

7:31 PM

well as you know, functions can be ugly

so not only we are oging to find $g,h$ that satisfy the intermediate value theorem

philmcole

What are $G_\delta,F_\sigma$?

mercio

but in fact we can have that $g(I) = h(I) = \Bbb R$ for any nonsingleton interval $I$

which would certainly be enough to get the IVT property

Alessandro Codenotti

$G_\delta$ is a set that can be written as a countable intersection of open sets, $F_\sigma$ is a set that can be written as a countable union of closed sets

Nûr

Yes certainly it is enough :D

mercio

to do this well you biject $\Bbb R / \Bbb Q$ with $\{0;1\} \times \Bbb R$

philmcole

7:34 PM

@AlessandroCodenotti Okay, thanks!

mercio

for $x$ in $\Bbb R$, if its class is in correspondance with $(0,y)$ then you decide $g(x)=y$ and $h(x) = f(x)-y$

and if its class is in correspondance with $(1,y)$ you decide $h(x)=y$ and $g(x) = f(x)-y$

since every class is dense in $\Bbb R$, $g$ and $h$ have the property I promised

Nûr

it's nice, thank you :)

mercio

also asa bonus they are quite easy to draw

so you can make nice illustrations

Nûr

sure :D

mercio

are you cheating in your exam ?

Eulb

7:42 PM

it's not even exam week right now is it?

@BalarkaSen sup

mercio

sometimes every week is exam week

Balarka Sen

Hi @Eulb

How's it going

Nûr

Who is cheating ?

Eulb

@BalarkaSen ded

Nûr

8:10 PM

Let $n \in \mathbb N$ and $ I \subset [| 1, n |]$. Find a sufficient and necessay condition on $I$ so that $\{M \in M_n(\mathbb R), \ rank(M) \in I \}$ is connex.

Paul Plummer

8:34 PM

@BalarkaSen Hello

Balarka Sen

Hey @Paul!

Long time

Paul Plummer

Yah it has been

Semiclassical

hmm. How active is the stats SE chat? My relative ignorance of the subject is showing b/c this question seems like it should have a simple answer

Balarka Sen

Last I checked, really inactive

Semiclassical

Figures

Alessandro Codenotti

8:38 PM

Hi @Paul

Paul Plummer

Any one know anything about $\eta$-invariants and what they tell you? (I don't)

@AlessandroCodenotti Hi

Semiclassical

More or less: I’m trying to decide whether there’s a nice realization of a particular 3-by-3 correlation matrix

MatheinBoulomenos

Hi everyone

ÍgjøgnumMeg

Hey man

MatheinBoulomenos

@BalarkaSen I attended a colloquium talk on fundamental groups today. You would've enjoyed that

Balarka Sen

8:42 PM

hah

MatheinBoulomenos

I didn't know that the relation between coverings and the fundamental group were discovered by Heidelberg topologists

He started with topology and then went on to algebraic geometry

Balarka Sen

I didn't know that either

Lozansky

@Semiclassical Got a second?

ÍgjøgnumMeg

@Mathein don't think I'll get into heidelberg for the master

Lozansky

I got a tiny (quantum) question :>

MatheinBoulomenos

8:45 PM

@ÍgjøgnumMeg why not?

ÍgjøgnumMeg

@Mathein I'm pretty sure the deadline for international applicants is like mid-June or smth and my A-Level grades still haven't arrived in the post

lol

so I might miss the deadline

MatheinBoulomenos

ah, damn

ÍgjøgnumMeg

bleh

I'm also looking at Frankfurt am Main tho

Paul Plummer

What is going on with you @BalarkaSen

Balarka Sen

Things

MatheinBoulomenos

8:48 PM

@BalarkaSen Grothendieck proved that if you have a category $C$ and a functor $F:C \to \mathbf{FinSet}$ satisfying some axioms, then there exists a unique profinite group $G$ and an equivalence of categories from $C$ to the category of finite discrete sets with a continuous $G$-action, such that under that equivalence, $F$ is just the functor which forgets the $G$-action. As a special case, one can obtain the relation between finite coverings of $X$ and $\hat{\pi_1(X,x)}$

Balarka Sen

For now, tryna remember a chunk of math I forgot

@MatheinBoulomenos Yeah I "know" this story

Ted Shifrin

hi a @Balarka

Balarka Sen

Profinite representability, etc

MatheinBoulomenos

this also works in algebraic geometry (unlike talking about paths)

Balarka Sen

Hi @Ted

MatheinBoulomenos

8:50 PM

but in finite characteristic, this doesn't give the "right" fundamental group, which was already noted by Grothendieck

Hi @Ted

Nûr

Consider a a billiards whose edge is $C^1$. Show that there exists a periodic trajectory with $n$ rebounds, for all $n≥2$

Ted Shifrin

Hi @Mathein

MatheinBoulomenos

If $k$ is algebraically closed of characteristic $0$, then $\pi^{ét}_1(k^\times) = \hat{\Bbb Z}$, the finite connected "coverings" of $k^\times$ are just $x \mapsto x^n$ for $n \in \Bbb Z \setminus \{0\}$. If $k$ is algebraically closed of characteristic $p$, then the $p$-th power map $k^\times \to k^\times$ is not étale. For this reason,$\pi^{ét}_1(k^\times) = \prod_{q \neq p} \Bbb Z_p$ (so this is like $\hat{\Bbb Z}$, but the factor of $\Bbb Z_p$ is missing)

this is like in topology we have $\pi_1(\Bbb C^\times)= \Bbb Z$ and the finite connected coverings are $z \mapsto z^n$

Balarka Sen

Yeah

Pretty cool

MatheinBoulomenos

we would like a notion of a fundamental group to also consider the map $z \mapsto z^n$ when $\operatorname{char}(k)$ divides $n$. The problem is that the "group" (actually non-reduced group scheme) that this corresponds to doesn't have enough points. When we consider $z \mapsto z^p$ in char p, then the group scheme this corresponds has only one point. It's impossible that this group scheme acts nontrivially on a finite set

loch

8:59 PM

it's pretty cool how etale topology works in alg geom -- by cool i mean i dont understand it very well and i would like to understand it more

MatheinBoulomenos

the way to fix this (due to Nori) is basically to consider linear representations instead of group actions. So to a variety over $k$, where $k$ has finite characteristic, we associate a category of "essentially finite vector bundles" (he didn't define this in the talk) and then we have a fibre functor that goes to the category of finite dimensional $k$-vector spaces

Balarka Sen

What's $\pi_1^{et}(\Bbb A^1_k \setminus \{0, 1\})$, now I wonder.

Affine line minus two points is a perfectly fine quasiprojective variety so that should make sense

($\pi_1^{top}(\Bbb C \setminus \{0, 1\})$ is free on two generators)

Ted Shifrin

I have the answer when $k=\Bbb Z_2$. :P

Balarka Sen

Lol

Let's take $k$ to be algebraically closed, char 0

Ted Shifrin

Oh.

MatheinBoulomenos

9:02 PM

So similarly to the case where you have a category $C$ together with a functor to the category of finite sets, there's also a kind of "representability results" when you have a category $C$ together with a functor to the category of finite-dimensional $k$-vector spaces, satisfying certain axioms, then there's a theorem on "Tannakian categories" that says this is equivalent to the category of finite-dimensional $k$-linear representations of some profinite group

if you use this, you get the "right" definition for the fundamental group in finite characteristic (and it agrees with the étale one in char 0)

@BalarkaSen for $k=\Bbb C$, this will be the profinite completion of the topological fundamental group, so free profinite of rank 2

Balarka Sen

I know that :P

I was asking for arbitrary fields, alg. closed and char 0

For complex quasiprojectives the etale fundamental group is just profinite completion of the topological dude

loch

$\mathbb{A}^1_{\mathbb{F}_2}$ minus 0 and 1 isn't empty though right.. - at least as a scheme (im assuming that's the joke lol)

Ted Shifrin

Really, @loch? What points does it have?

loch

For example the prime ideal $(x^2+x+1)$? When I say scheme I meant $\mathbb{A}^1_{\mathbb{F}_2} = \mathrm{Spec} \mathbb{F}_2[x]$ - so really just any irreducible polynomial since that defines a prime ideal.

Balarka Sen

@MatheinBoulomenos That sounds technical

Ted Shifrin

9:10 PM

Oh, I guess when the Nullstellensatz fails, I have no idea what's going on.

How does one algebraically remove a point, in terms of the coordinate ring?

Balarka Sen

@loch When I say $\Bbb A^1_k$ I mean $\text{maxSpec} k[x]$, not $\text{Spec} k[x]$, I think?

Yeah the affine line is much larger to a scheme theorist than a classical algebraic geometer I think

loch

@TedShifrin For example if you remove the point $(0,0)$ from $\mathbb{A}^2_{\mathbb{C}}$, you don't get an affine variety. So I guess in this sense you can't really see what happens to the coordinate ring (the plane minus the origin has the same coordinate ring as the plane)

Balarka Sen

It is a quasiprojective variety.

loch

@BalarkaSen the ideal there is a maximal ideal!

Balarka Sen

I think it's still a scheme, right?

loch

9:12 PM

yes

Ted Shifrin

sure ...

Now we know why I did all my algebraic geometry over $\Bbb C$. :)

loch

but affine varieties (schemes) are nice because they're really the same thing as their coordinate rings

MatheinBoulomenos

@TedShifrin that Tannakian category stuff was inspired by some result on the representation theory of compact groups

Ted Shifrin

And I actually did have a few scheme-theoretic things in a few papers. @Balarka: I think I mentioned the interesting Whitney umbrella to you once.

loch

but when it's not affine then you don't have a coordinate ring to work with - e.g. alg functions / holo functions on $\mathbb{P}^1_{\mathbb{C}}$ are all just constants :p

Balarka Sen

9:13 PM

@TedShifrin Yep you did

Ted Shifrin

@loch: My question remains. Algebraically, how do we work with $X-\{p\}$ if we understand $X$?

I keep thinking we need something with localization ... But I'm way rusty on algebra :)

Nah, that can't be right.

Balarka Sen

@loch I agree. I think for projective varieties the better thing to look at is the homogeneous coordinate ring

Ted Shifrin

Yes, of course.

MatheinBoulomenos

If we remove a closed point, then this an open subscheme

Ted Shifrin

Sure ... So now what?

MatheinBoulomenos

9:15 PM

all open subsets of $X - \{p\}$ are also open in $X$, so we know what the sections look like

I don't think we can say more in general

loch

For example if you take the ring (let's say a finitely generated $k-$algebra) - inverting $f$ gives you the open subscheme corresponding to the complement of $f=0$.

I guess if you're expecting a purely ring theoretic way of looking at what $X-\{p\}$ is in general you can't really do it (without resorting to schemes)

Ted Shifrin

By Hartogs' Theorem, in dimension $\ge 2$, any holomorphic function on $U-\{p\}$ extends across $p$. So we can't tell the difference by looking at the rings of holomorphic functions.

Balarka Sen

Good point

loch

Yes - which is kind of the problem here - because e.g. the ring of regular functions on $\mathbb{A}^2$ is the same as the ring of regular functions on $\mathbb{A}^2 - \{(0,0) \}$

Ted Shifrin

Yeah, I'm genuinely confuzled.

Right, @loch.

loch

9:18 PM

So the coordinate ring of $\mathbb{A}^2-\{(0,0)\}$ and $\mathbb{A}^2$ are the same - which is why if I interpret your question as saying how to look at $\mathbb{A}^2-\{(0,0)\}$ purely via their coordinate rings - I think the answer is you can't

Balarka Sen

So we can't distinguish them by looking at their structure sheaves?

loch

Yes you can

Balarka Sen

How? I think that's the question

loch

but when you talk about a sheaf you need a topological space to start with!

Ted Shifrin

Interestingly, then we're looking at $q\ne p$ ... and I still don't see the difference in $\mathscr O_q$ ...

Right ... so one space has $p$ and the other doesn't.

But I can't say that in an algebraic way :P

Balarka Sen

9:20 PM

@loch That's what Ted's saying. You're using the topological properties of the underlying spaces of the schemes to distinguish the schemes. What's a purely algebraic way to do it?

I guess you can compute A^2 - {(0, 0)}'s etale cohomology

(@Mathein?)

Ted Shifrin

That's not the issue.

MatheinBoulomenos

I think $H^1(\Bbb A^2 - \{(0,0)\},\mathcal O)$ is infinite-dimensional

you don't need étale cohomology

loch

I'm saying I don't think you can (whatever purely algebraic means - to me it means let's ignore the topological space) ! You only have an equivalence of categories between the category of rings and the category of affine schemes..

Ted Shifrin

If I know how to give $Y\subset X$ and $Z\subset X$ by equations, I know how to do $Y\cup Z$ and $Y\cap Z$. But can I do $X-Y$? Clearly not!!

loch

I mean if you want to talk about sheaf cohomology then there's also a topological space there too

Ted Shifrin

9:22 PM

As it's not closed.

Balarka Sen

@TedShifrin You can write down equations for an affine variety whose image under a projection map is $X - Y$, though.

loch

@BalarkaSen Oh and I wanted to mention this just now - it's not that nice for projective varieties in that you dont get an equivalence of categories as you do here for the affine guys (of course they are still useful though!)

Ted Shifrin

Yeah, we were talking about this last week. I need to pass to a higher space, @Balarka. I can't do it inside $X$.

Balarka Sen

Yeah

@loch Ah yes

Ted Shifrin

LOL ... I'm glad to know that projective varieties are still useful! /sarcasm

MatheinBoulomenos

9:24 PM

We know that the inclusion $\Bbb A^2 - \{(0,0)\} \to \Bbb A^2$ induces an isomorphism of global sections. If $\Bbb A^2 - \{(0,0)\}$ was affine, then this would imply that the inclusion $\Bbb A^2 - \{(0,0)\} \to \Bbb A^2$ is an isomorphism of schemes, but it's not even a bijection!

loch

Haha - no I meant the homogeneous coordinate ring

which when I said there's no equivalence of cat. - might suggest to some that it's not that great (or maybe not)

MatheinBoulomenos

this shows not only that $\Bbb A^2 - \{(0,0)\}$ is not the same as $\Bbb A^2$, but that it's not affine

Balarka Sen

Hm, I seem to remember there's a way to reconstruct the projective variety from it's homogeneous coordinate ring, using the projective spectrum construction or something

loch

Anyway - now that I think of it there is a purely algebraic way - the point being you can think of schemes as certain functors on the category of rings to the category of sets satisfying some properties

so I guess in that sense then you can distinguish them purely 'algebraically'

Balarka Sen

Am I incorrect?

Ted Shifrin

9:26 PM

That fails for removing codimension $1$, though, @Mathein.

loch

You are right
but you can do this with two different homog. coordinate rings

Balarka Sen

@TedShifrin In which case the coordinate rings distinguish them

Ted Shifrin

Right ... We're back to Hartogs in the analytic category.

Anyhow, I'm sorry I took us down this rabbit hole.

I'm leaving algebra land now. ... :D

Balarka Sen

Sometimes the algebra land is refreshing

... until I have to do the actual technical computations

loch

I like to think I suck at both geometry and algebra.. so I learn both at the same time to cover my weaknesses (or be twice as bad)

Ted Shifrin

9:28 PM

Well, you say that about every field, @Balarka :P

MatheinBoulomenos

I'll take algebra technical computations over diff geo technical computations any day

@TedShifrin is $\Bbb C^2 \setminus \{0\}$ Stein?

Ted Shifrin

@loch: It's convenient to be across two fields. I always claimed to be a complex geometer until someone asked me a question I couldn't answer, and then I said I was a differential geometer ... and vice versa :P

Balarka Sen

@TedShifrin At least in topology I can fail to do the computations, learn it from someone else, and build a mental model to evade the technicality with pictures!!

Ted Shifrin

Obviously, you and I differ in that, @Mathein.

Hmm, @Mathein. Offhand I'd guess no.

Balarka Sen

As they say, the true technical computations are those in which your hands flab around in the air at lightning speed

Ted Shifrin

9:30 PM

I've never heard that, @Balarka, so I don't know who "they" are.

MatheinBoulomenos

that implies that $\Bbb A^2\setminus \{0\}$ is not affine over $\Bbb C$ by GAGA

loch

@TedShifrin Same :) maybe replace complex with algebraic - but things that I'm interested in somehow are all happening over $\mathbb{C}$, so I should pick up my complex geom soon

Balarka Sen

@TedShifrin I'm referring to myself in gender-neutral sense, of course

geocalc33

Anyone know of a place where I can graph a function with both a complex variable in it and a real variable in it?

Ted Shifrin

Nope, @geocalc33. Even with just the complex variable is challenging to "graph."

geocalc33

9:32 PM

Yeah but there are online websites i can use to graph complex functions

Ted Shifrin

Oh, it follows from Hartogs, @Mathein.

You only think you're graphing them, @geocalc.

You're seeing slices of the graph.

geocalc33

oh arent they actually four dimensional?

Ted Shifrin

Yup.

And to do complex + real, you need a function of three real variables, and you're plum out of room.

MatheinBoulomenos

@TedShifrin I also think that $H^1(\Bbb C^2 \setminus \{0\}, \mathcal O)$ is non-zero which contradicts Cartan B

geocalc33

I can't figure out how to turn on math jax

Ted Shifrin

9:37 PM

I would have to sit down and think to compute that, @Mathein.

MatheinBoulomenos

I was thinking about computing the Cech cohomology of the covering by $\{(z,w) \mid z \neq 0\}$ and $\{(z,w) \mid w \neq 0\}$

loch

that would be the way to do it (or at least I don't know how else would you compute it..)

Ted Shifrin

9:51 PM

So we do double Laurent series on $(\Bbb C-\{0\})^2$. That seems right.

MatheinBoulomenos

yeah

Ted Shifrin

But it does follow immediately from Hartogs. You can't have a proper holomorphic embedding in $\Bbb C^N$, as any holomorphic function on a neighborhood of $0$ extends across $0$ and hence stays bounded.

MatheinBoulomenos

I see

Daminark

10:42 PM

Hey everyone!

MatheinBoulomenos

Hey @Daminark

geocalc33

Hey