the only subspace of $\Bbb C$ that is biholomorphic to it is $\Bbb C$, yes

@MikeMiller thanks

I think little picard + basic homotopy are enough to prove it, right

yes

could somebody here look at my post: math.stackexchange.com/questions/2955619/…

@LeakyNun this is a nice argument i hadn't thought of.

observe also that this implies the only subsets that $\Bbb C$ covers are once-punctured planes

indeed

and every map from $\Bbb C$ onto $\Bbb C \setminus \{a\}$ has the form $\exp(g)+a$ for an entire $g$

If you have two punctures it has universal cover $\Bbb D$, in which case you're in trouble

(will this process always terminate giving us a surjective entire function?)

Same argument if you have a finite number of punctures more than 2

i don't know what this process means

Given an entire function $f:\Bbb C \to \Bbb C$ that misses one point, express it as $\exp(g)+a$ for some entire $g$

now do that process on $g$ to get a new function

and then again to get yet another function

If $g$ misses $b$ then $f$ misses both $a$ and $e^b + a$

I don't think that's true, because if $g$ misses $b$ then it hits $2\pi i + b$

o fair

man the proof of the thing I stated is so cool in Freitag

(that means, the thing I stated is proved in Freitag, and the proof in Freitag is cool)

(the thing I stated = every map from C onto C\{a} has the form exp(g) + a)

That's not hard, it's just holomorphic map lifting lemma

Also, of course $\exp(\exp(z) - 1) - 2$ is an example for your process which requires a two-fold unfurling to get a surjective entire map.

I was being dumb

Good question, though.

@BalarkaSen and the cool thing is that they didn't use the lifting lemma

should I tell you the proof?

I think if you do an $n$-fold unfurling of $f : \Bbb C \to \Bbb C$ to get $g : \Bbb C \to \Bbb C$, then the order of growth of $g$ is $n$-fold logarithm of the order of growth of $f$. If $\rho_f$ is finite, then for some large enough $n$, you can make $\rho_g$ so small that $g$ is polynomial, or some such. I think that's the ideal proof strategy for this particular special case

@LeakyNun Ya go ahead

@BalarkaSen so let $f: \Bbb C \to \Bbb C$ miss $0$. Then $f'/f$ is still entire. So it has an antideritive $F$, i.e. $F' = f'/f$. Then let $G = \exp(F)/f$ and we have $G' = (\exp(F) (F' f - f'))/f^2 = 0$, so $\exp(F) = kf = \exp(c) = f$, so $f = \exp(F-c)$

Secretly that's still lifting. You're using that winding number of loops of the form $\gamma \circ f$ for loops $\gamma$ going around $0$ in $\Bbb C$ are zero.

I think

you don't need that to prove "entire -> has antiderivative"

but thats what it "means"

I don't need every loop, I just need triangles that have the origin as one of its vertices

i don't really care about minimal machinery :p

I see

the point is you're hiding the geometric meaning in symbols

fair enough

[Exercise 15, Section II.3, Freitag] Show that if $f$ and $g$ are entire functions with $f^2 + g^2 = 1$, then there is $h$ entire such that $f = \sin h$ and $g = \cos h$.

@BalarkaSen are you interested?

That is immediate from lifting again

how?

is $\sin$ also a universal cover?

(I don't really know, I'm asking you)

(f,g): C -> S^1, lift to R -> S^1

I'm not thinking about the details, but after that idea it just seems like a matter of writing it out

that's not S^1 though?

Maybe, maybe not. Think about how to fix it, shouldn't be a long walk from what I said

I mean, $f^2 + g^2 = 1$ doesn't restrict at all the range of $f$ and $g$

Sure, they can't both be 0 simultaneously for one :p

It restricts it to C^* very much

I don't understand what you mean by C -> S^1 lift to R -> S^1

Think

ah you mean the function $\Bbb C \to S^1_\Bbb C$ that sends $z$ to $(f(z),g(z))$

so $\Bbb R \to S^1_\Bbb R : x \mapsto (\cos x, \sin x)$ is a universal cover

and you're saying that $\Bbb C \to S^1_\Bbb C : z \mapsto (\cos z, \sin z)$ is still a universal cover

and then we're done

I'm not saying much. I wrote down an idea, making it into a proof seems like a good exercise for you

is that what you mean?

yeah but I need to understand what your idea is first

What you wrote is a reiteration of what my most immediate thought was, yes

it took me quite some time to realize that (f,g) is the product function, not just a pair of functions

nice

It's not clear to me how to get an entire $h$ yet, but also I am not thinking much about it

is $S^1_\Bbb C$ a complex manifold?

I think it's a variety so it must be a manifold

the preimage of each point is discrete

it looks like a nice map

Is Lehmer's theorem the best way to test primality?

I mean, it would seem like $\Bbb C/\Bbb Z$

I think it's definitely evenly covered

I'm just trying to understand the how to prove the $a^p\equiv a(\textrm{mod }p)$ version of Fermat's Little theorem, and come across this term

I think all those are worth computing/checking. Why z^2 + w^2 = 1 in C^2 cuts out a manifold, why C -> M, z -> (cos z, sin z) gives a covering map, ...

I haven't checked anything, to be clear

fair enough

I know how to prove the version $(a,p)=1, a^{p-1}\equiv 1(\textrm{mod }p)$ but is the proof similar to that of $a^p\equiv a(\textrm{mod }p)$

@IsanaYashiro the latter just makes you check one additional case

@TobiasKildetoft I only know that if it happen then it means $p | a(a^{p-1}-1)$, is this correct approach?

@IsanaYashiro If what happens?

$a^p\equiv a(\textrm{mod }p)$ happen, I have to prove it so I assume it may not happen, lol, I don't know i'm talking about

you already know it happens for all $a$ except possibly one. So check that one value of $a$.

@TobiasKildetoft I'm not sure s/he knows that

@LeakyNun They already mentioned Fermat's little theorem, and wanted to know whether the proof of this was similar.

if $(a,p)\not=1$ then $p|a$, then $p|a(a^{p-1}-1)$, and if $(a,p)=1$ then it's the version I know how to prove, is this correct?

@IsanaYashiro　にほんじんですか。

@Dalamar Yeah, all of them I think. And yeah, it's mostly authentic copies about I think.

@TobiasKildetoft they might not realize that a^(p-1)=1 (mod p) implies a^p=a (mod p)

@TobiasKildetoft how to prove that z^2+w^2-1 in C[z,w] is irreducible?

@LeakyNun It may be hilarious but I don't read Japanese.

@IsanaYashiro kimi nihonjin desuka?

@LeakyNun It it a sum of two homogeneous polynomials, which make sit easy to check what a factorization would look like if it existed

I see

@TobiasKildetoft (az+bw+c)(dz+ew+f) isn't the easiest thing to deal with..

or maybe I'm not getting you

I don't know much about sum of two homogeneous polynomials

@LeakyNun So clearly each factor will have constant term $\pm 1$, right?

I'm not dealing with integer polynomials

non-zero then

sure

I can normalize them so that each has constant term 1

aha

well, not really, since you need product $-1$

(az+bw+1)(dz+ew-1) = z^2+w^2-1 so -b+e=0 and -a+d=0

so (az+bw+1)(az+bw-1) = z^2+w^2-1

so (az+bw)^2 = z^2+w^2

so 2ab=0

thanks

so (z^2-w^2-1) is a prime ideal

so C[z,w]/(z^2-w^2-1) is irreducible and reduced

@TobiasKildetoft is this enough to conclude that it cuts out a manifold?

@LeakyNun I don't recall actually.

ok

Varieties can be singular.

You want to use inverse function theorem to prove that {(z, w) \in C^2 : z^2 + w^2 = 1} is a manifold.

I see

or maybe the two charts on $S^1_\Bbb R$ will work if we change the scalars to $\Bbb C$

hi @CaptainAmerica16

ok Freitag claims that writing $s = \sigma + it$ is the "Riemann-Landau convention" (the context is when working with prime number theorem)

When I google "Riemann-Landau convention", there are exactly 5 hits, all of which come from exactly Freitag

so is it a name if nobody else uses it?

(I know that it's a convention when dealing with prime number theory, I just am not aware of this convention having a name)

@Balarka @Leaky: Implicit, not inverse.

I just call it the preimage theorem

@TedShifrin ^^ context

If you know about $\partial\bar$, you can use the real implicit function theorem to prove a holomorphic version.

interesting

@Balarka: But have you proved that in the holomorphic category? Hmm?

Can $S^1_\Bbb C := \{ (z,w) \in \Bbb C^2 \mid z^2+w^2=1 \}$ be covered by two charts? @TedShifrin

Proof of inverse function theorem is easy in the holomorphic category once you know it in the real category

Ugh, no one writes $S^1_{\Bbb C}$. Please don't do that.

No, surely not, as you don't have global branches of square roots. I'll have to think about this.

what should I write?

@Semiclassical okay, so I looked into what you said. Made sense to work with polynomials, since the hinted (1+x)^n by definition is a polynomial. I discovered that going from $$\sum_{k=0}^n a_k\sum_{j=0}^m b_j$$ and $$\sum_{l=0}^{n+m}\sum_{q=0}^{l} {a_{q}\cdot b_{l-q}\cdot x^{q}}$$ is just a matter of terms ordering (not). In later form, for $${l}\gt {\frac{n+m}{2}}$$, we get extra terms that don't fit.Should I account for these extra terms? Am I even in the right direction?

What Leaky is writing down is two sheets of C cut along [-1, 1], then pasted sideways

The cut-and-paste description doesn't give you holomorphic charts, though.

True

I'd write $C$ for conic or $Q$ for quadric, @Leaky.

ok

I can cover it by four holomorphic charts

man, complex analysis is so beautiful

(ok this is the beautiful side of complex analysis)

@TobiasKildetoft Is it correct to say gcd(0,0) is infinite?

(I guess the algebraic sides are more beautiful then)

I think most of complex analysis is quite beautiful, but I like analysis far more than most in this room.

(so it isn't really complex analysis)

@IsanaYashiro No, it is $0$

more like, algebraic geometry

@Tobias: I'd say undefined. Doesn't every integer divide $0$?

but 0 only divides 0

Nonsense, Leaky.

what, 0*x=0

@TedShifrin Sure, and $0$ is thus the greatest one in the "divides" ordering

So $1$ divides $0$?

yes, as you said

@TobiasKildetoft Happy to know that, but my book only define gcd(a,b) for a,b not both zero, so I don't know it.

Huh? @Tobias

the partial ordering where a<b := a|b

@IsanaYashiro "undefined" is also fine. But definitely not infinite

@Isana: It is safest.

@LeakyNun You have some of those backwards

gcd(0,0) needs to divide 0

I definitely said "not both 0" in my algebra book.

that doesn't put any restriction

@TobiasKildetoft Could you provide a simple explanation for me?

for any m, if m divides 0 then m divides gcd(0,0)

Since $\langle 0,m\rangle = \langle m\rangle$, the gcd of $0$ and $m$ is $m$.

0 divides 0, so 0 divides gcd(0,0)

so gcd(0,0) = 0

@TedShifrin I think most books do. But once one then gets to the general notion in gcd domains, the definition changes to the one I used above where gcd(0,0) = 0 automatically

@TedShifrin now put m=0

Ted has an algebra book? :o

Right, I agree that it's consistent to put $m=0$. The ideal-theoretic version comes way later in my book.

amazon.co.uk/Abstract-Algebra-Geometric-Theodore-Shifrin/dp/…

I see

a geometric approach

It was my first one, Leaky.

smacks @ÍgjøgnumMeg

so it's algebraic geometry?

hahaha

@IsanaYashiro It could never be infinite since it has to be a number. But defining it as the largest common divisor (which is a good definition in this context), you need to only define it when at least one number is non-zero, since otherwise there it no largest common divisor.

No, no, but there's a chapter that gets to some algebraic geometry (and also affine geometry and hyperbolic/elliptic)

@IsanaYashiro you shouldn't care about gcd(0,0). We're just talking to ourselves.

@Balarka: So, for a projective conic, it's clear that you can cover with $2$ charts. So we can cover the affine part of it with $2$ — or perhaps $1$ — charts.

@TedShifrin does it define P^n? :D

For $n=1,2,3$, @Leaky. NOT using commutative algebra. Just equivalence classes. This is an undergraduate text.

:)

@Semiclassical sorry, this is the general form. Our particular case where $$a_k = \binom{n}{k}$$ works just great if we define negative factorials as zero. But I still don't know how to make a proof of this contraption

I mostly did it over $\Bbb R$, Leaky, although there's some discussion of $\Bbb P^1_{\Bbb C}$.

@TedShifrin Simply because it's projectively equivalent to something standard?

@Balarka: No, using the usual stereographic projection to $\Bbb P^1$.

Oh

$\Bbb C/\Bbb Z$ is a complex manifold right

where $\Bbb Z$ acts on $\Bbb C$ by addition

isn't that just the complex torus

Hmm, that must be biholomorphic to $\Bbb C^*$, I guess.

No, cylinder

What do you mean by complex torus?

It's certainly not compact.

I don't really know what I'm doing at this point. If I say I know very much, Ted would smack me in the face; if I say I know very little, all of my classmates will smack me in the face.

@TobiasKildetoft Because nothing to divide? Sorry for my keep asking but I really like gcd and thanks for your kind explanation!

ah right.

Leaky, best to keep quiet.

^

@IsanaYashiro Because all integers will be common divisors of $0$ and $0$

Can't I evaluate myself.

No, just do math

(and there is no largest integer)

@TobiasKildetoft 🆎

@TobiasKildetoft So we don't want that result, then we don't define it right?

@Eric and I were discussing $\Bbb C^2/\Bbb Z$ with a very different action. Very famous complex manifold.

@IsanaYashiro Right

@TedShifrin because the map $z \mapsto (\cos z, \sin z) : \Bbb C \to C$ is surjective with "kernel" $2\pi\Bbb Z$ right

@TobiasKildetoft Zero: "It's unfair!!!!!!"

Interesting, @Balarka. Is it biholomorphic to $\Delta^*$ or to $\Bbb C^*$?

@TobiasKildetoft anyway thanks for your idea, I understand it now~

@Leaky: What does kernel mean?

yeah I placed it in quotes now

I would say it in the language of fibration if I knew what it is

I don't understand. You're mapping to our conic $x^2+y^2=1$?

yes

Thanks

Right, of course. My brain isn't working well.

@TedShifrin what I mean is that each point has fibre that is an affine translation of $2\pi\Bbb Z$

I think what I really mean is that it is a universal cover

that is regular

How do you know it's even surjective?

ah!

Does anyone know the book A First Course in Abstract Algebra, by John Fraleigh? It looks nice and I'm about to read it

@IsanaYashiro that book gives me nostalgia

It's not my favorite style, @Isana, but it's popular. There are something like 12 editions of it.

@TedShifrin it isn't surjective is it

@TedShifrin What is the interesting action of $\Bbb Z$ on $\Bbb C^2$?

I honestly don't know, @Leaky. See above. My brain is not functioning. It has some good excuses.

Oops, $\Bbb C^2-\{0\}$, @Balarka.

Multiplication by $2$.

German word order! :P

@TedShifrin to be clear, that means multiplication by $2^n$ right

Yup. One could do that in French, too, but most people probably wouldn't.

but why do you need two complex variables?

Not $2^n$?

One last question of today: Consider the happiness of learning math, which realm you like the most?

@TedShifrin So it's interesting in the sense that the geometry is interesting?

You can do $\Bbb C^n-\{0\}$, indeed, @Leaky.

@TedShifrin $n \in \Bbb Z$ sends $(z,w) \in \Bbb C^2 -\{0\}$ to $(2^n z, 2^n w)$ right?

@Balarka: It's probably the easiest example of a non-Kähler compact complex manifold.

Oh, sure, @Leaky. I was just giving the equivalence relation.

aha

Not action of $\Bbb Z$. My fault.

nah it isn't your fault

Topologically you should figure it out very easily, @Balarka. Eric and I were discussing a generalization of this, too.

Balarka just wrote "(f,g): C -> S^1, lift to R -> S^1" and expect me to know what S^1 means

I stan limits.

@LeakyNun I want you to think instead of talking so much

@CaptainAmerica16 how are you doing

@Ultradark I'm doing pretty alright. I've only just understood the power of limits. I think I...love them.

@TedShifrin Yeah topologically it's just $S^3 \times S^1$, which is why I guessed the geometry must be important (you're forcefully identifying the inner and outer $S^3$'s by scaling)

@CaptainAmerica16 nice. Limits are great. I remember liking them too when I first learned about them

@TedShifrin do you like Freitag?

or do you like Rudin more?

heya @CaptainAmerica

Hey Ted

Yup @Balarka ... note the cohomology ring is totally not something you can have with compact Kähler. Don't even need ring in this case.

@TedShifrin Wassup :D

Leaky, I don't know Freitag, and I'm not a huge Rudin fan.

so that'll do much the same thing

Hi Ultra.

@CaptainAmerica16 appearantly 15 users thought the same :D

also, where did Will Hunting go?

I saw him a few days ago, but I dunno.

@Alucard About limits?

@TedShifrin Does the symplectic 2-form give nontrivial $H^2_{dR}$ for any compact Kahler guy?

right: Kähler means that the $2$-form you build out of the metric is closed.

@CaptainAmerica16 about having CaptainAmerica as a nickname :) (or, is it your age?)

@TedShifrin I mean, complex analysis is beautiful until you use 5 pages to set up the estimates for $\zeta$ and $\zeta'$ and $1/\zeta$ to show that the rest of the integral goes to $0$

@Leaky: I have no idea what you're babbling about, but I love estimates.

@LeakyNun Dude shut up and just do it

LOL, @Balarka is now playing the rôle of Mike.

@BalarkaSen channeling your inner shia i see

except shia never shuts up

what happened to Mike tho...

he keeps shouting "do it" at me

@TedShifrin Got it.

@Alucard Both, I thought it was clever.

Let $z^2+w^2=1$. Then, $(z+iw)(z-iw)=1$, so $z+iw \ne 0$, so $z+iw = \exp (i \xi)$ for some $\xi \in \Bbb C$ (this is the same method by which I proved the original theorem), and $z-iw = \exp(-i\xi)$. And then $z=\cos \xi$ and $w = \sin \xi$.

So indeed the map is surjective.

and it's clear that the fiber of each point is an affine transformation of $\Bbb Z \subseteq \Bbb C$

@Leaky: There is an obvious error in what you just wrote.