@psie that's obvious

how can one even compare the "number" of meromorphic function on a compact Riemann surface and a smooth functions onto $\Bbb CP^1$ on a compact Riemann surface

I know both are infinitely many and I should say "ratio" for that.

potato this is not my field but i am dimly aware that when people do that sort of thing they often are not "counting" meromorphic functions as such, but doing things like referring to dimensions of vector spaces of meromorphic functions. and not some full space of all of them, but just ones associated with a 'divisor' of the surface

e.g. the polynomials of degree <= n on C are the meromorphic functions on the riemann sphere having a pole of order at most n at infinity (the divisor there would be the single point infinity tagged with the integer n)

and thats a perfectly fine complex vector space with a well defined dimension (n+1)

and with partial fractions type stuff you can see what the dimensions of other spaces of meromorphic functions might be for other divisors on the riemann sphere (by choosing finitely many points where the poles get to be, and bounds on what order of pole you allow at each point)

so once you impose stuff like "finite group of points and bounds on orders of poles" and then do "vector space dimension" instead of literal counting, you do get some finite integer valued thing that you can literally count

and compare to whatever else

sadly our chat's once-foremost complex geometer has gone into a state of occultation

but i think there are some AG adjacent dorks who know all of this and more around here

leslie: ah yeah that divisor thing comes out when people talk about the Riemann-Roch. I was wondering some kind of "Dolbeault cohomology" type stuff comparing the ratio of "space" of smooth functions onto $\hat{\Bbb C}$ and "space" of meromorphic functions (I don't know if this makes sense). The problem with Riemann-Roch to me is that it only gives "at most" sense of poles and I need to prescribe the position of possible poles.

But it seems I can say something using the Beltrami differentials which precisely measure the failure of a smooth map between Riemann surfaces being holomorphic.

I recently left a comment here on why the holomorphic version of the Whiteny approximation theorem fails.

@leslietownes I miss him.

@onepotatotwopotato hash tag me too

hashtag not all chatters

Hello!

Quick question, am I doing something wrong here? math.stackexchange.com/questions/5037429/…

I don't understand the comment, I don't think such points are Lebesgue points (the measure $\mu$ outside is not necessarily Lebesgue)

We say two abelian groups $G, G'$ have the same type of acyclicity if either both $G$ and $G'$ are torsion or neither is, and the sets of primes at which each group is uniquely divisible agree. I'm looking at a claim that in this case $A \otimes G = 0$ iff $A \otimes G' = 0$ and $\mathrm{Tor}(A, G) = 0$ iff $\mathrm{Tor}(A, G') = 0$, does anybody happen to see why?

@Ben $A$ is any abelian group or f.g.?

any

I'm confused about why this should even hold when $A$ is, say, $\mathbb{Z} / p$. $\mathbb{Z} / p \otimes G = 0$ iff $G \xrightarrow{\cdot p} G$ is surjective, but unless it also acts injectively I don't see why it would also have to act surjectively on $G'$