Yeah actually I have proved $H^1(X, O) \cong \Bbb C^g$ assuming the Abel-Jacoi map gives an isomorphism $\text{Pic}_0(X) \cong (\Bbb C/\Bbb Z^2)^g$. I don't know how easy that is though.
I don't know what it means in terms of the Cousin problem, that's a good question. I think Ted's differential forms approach must be the right way to go.
I could never wrap my head around these various groups beyond the homological algebra.
But I didn't think of divisors for Riemann surface as the intuition for why the multiplicative cousin problem shouldn't be true in general. That's nice !
By the way, am I missing something or the fact $H^1(M, O) = \mathbb{C}^g$ follows directly from an easy application of Leray's theorem ?
So for example a RS with genus $g$, it's easy to see taht you can find an open covering such that each open set is biholomorphic to $\mathbb{C^*}$, and there are $g+1$ pairwise nonempty intersections which is also biholomorphic to $\mathbb{C^*}$
so for this covering, you get $H^1(U, O) = \mathbb{C}^g$
I thought you can argue similarly as in the case when the sheaf is constant sheaf, and since the only holomorphic function defined on the entire surface is the constant ones, but of course this don't make any sense.
there's this guy on this lecture that pretends to know what he's talking and is asking idiotic questions phrased vaguely so as not to give away that he doesn't know what he's talking about
He/she was talking about the inequality between sup @JoeShmo
@love_sodam The statement I gave is sufficient to show the inequality between infs. The statement you gave is sufficient to show the inequality between sups
If $A$ is an abelian group, let $\hat{A} := Hom_{\Bbb{Z}}(A, \Bbb{Q}/\Bbb{Z})$. I want to argue that $\hat{\Bbb{Z}_n}$ is a cyclic group of order $n$ (i.e., it is isomorphic to $\Bbb{Z}_n$). It's clear that any $f \in \hat{\Bbb{Z}_n}$ is determined by its image on $\overline{1}$. I conjecture that $f_0(\overline{1}) = \frac{1}{n} + \Bbb{Z}$ generates $\hat{\Bbb{Z}_n}$, but I'm having a little trouble verifying this.
Clearly $f_0$ has order $n$. If $f \in \hat{\Bbb{Z}_n}$ is nonzero, then $f(\overline{1}) = \frac{p}{q} + \Bbb{Z}$, where we can take $0 < p < q$ and $gcd(p,q) = 1$.
I want to argue that $q = n$, but I am not sure how to.
@Thorgott I'm not so sure my professor would like that answer. As it is, I don't think he'd like my way of proving it either, since he wants me to explicitly construct an isomorphism from $\Bbb{Z}_n$ to $\hat{\Bbb{Z}_n}$, but I don't see how to construct it.
I have something which I think is quite well-known but I don't know the name for it
I have a random variable with some probability distribution, and a function which maps these random variables to some another variables, and I'd like to find the probability density of the codomain of this function
Sorry to interject. I have some conflicting information written in the notes I've been accumulating, an $n\times n$ matrix is diagonalisable if and only if it has $n$ linearly independent eigenvectors, which necessarily means it doesn't have degenerate eigenvalues. But I also have it written down that degenerate eigenvalues don't imply a matrix can't be diagonalised.
The gravitational force exerted by a symmetric sphere of mass M on a particle external to itself is exactly the same as if the sphere were replaced by a particle of mass M located at the centre.
@Stupidquestioninc It is essentially the statement that if you add up all of the vectors representing the "force" from each piece of the shell it is the same as one big vector at the centre
If that's the case, then the matrix can't have $n$ linearly independent eigenvectors since the repeated eigenvalue must correspond to two linearly dependent eigenvectors, which contradicts the first statement that a matrix is diagonalisable if and only if it has $n$ linearly independent eigenvectors :(
Suppose $f: (X_1, d_1) \to (X_2, d_2)$ is a mapping of metric spaces. For any $\delta, \epsilon > 0$, let $$U_f^{\delta, \epsilon} = \{x \in X_1: \text{ there are }y, z \in B_{\epsilon}(x)\text{ with }d_2(f(y), f(z)) > \delta \}$$ How would use set operations with $U_f^{\delta, \epsilon}$ to obtain the points at which $f$ is continuous?
Thanks @Ted, I've given the concept a bit of a skim.
Early on in this semester, I set up a chat room for all of us students to interact during this two-semester measure theory + probability sequence. I thought perhaps my classmates weren't interacting with the chat room recently because they finished the assignment already. Turns out they haven't even started, and they're full-time students.
Compared to just taking classes on the side and working full-time, I prefer this over being a full-time student. Then again, I'm not aiming for a tenure-track position.
