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
@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.
[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 $g = \cos h$.
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, ...
[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 $g = \cos h$.
@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?
@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
@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.
@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.
@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 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.
@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)
@Michael.P typically you don't define negative factorials as zero. But you often do define the binomial coefficient $\binom{n}{k}$ to vanish unless $0\leq k\leq n$
@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 $0$
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 $\Bbb Z \subseteq \Bbb C$
