@TedShifrin I'm taking a seminar on abelian varieties. Of course, we're working over general base fields, but if I want to get some intuition from the complex case, what would you recommend?
Mumford has a classic book. Lots of stuff, too, in Griffiths/Harris (e.g., conditions for a torus to be algebraic) and even Griffiths’ little book on curves. Also Gunning has stuff in the Princeton series.
there used to be a restaurant on hwy 29 on the way to napa called Tonelli's. i always wanted to open up Fubini's next door and run them into the ground.
okay I think I see how one might get a map. We have the Abel-Jacobi map $X \to J(X)$ (assuming we fixed a base point). If we think of $\mathrm{Pic}^0(X)$ in terms of Weil divisors, then because the group of Weil divisors is freely generated by the points of $X$ and because $J(X)$ is an abelian group, the Abel-Jacobi map extends to a map $\mathrm{Div}(X) \to J(X)$, which we can restrict to $\mathrm{Div}^0(X) \to J(X)$
I suppose the difficult parts are showing that this is surjective and the kernel consists of principal divisors
I was searching for some clever sequence of sheaves that gives the map, but it's easier to think about Weil divisors here
If I ever happen to contribute anything in mathematics, I think I'm going to name things formally with zoomer terminology solely because I think it would be humorous to see in titles of formal papers and their contents.
"Let a member of this set be formally defined as a bruh moment. If the value is real, then it is a real bruh moment."
I don't know why I think these things sometimes :)
I see that lots of probabilistic proofs start with simple uniform distribution for their random variable at first. Any hint or clue please as to why they usually make that assumptions besides to start simple :/
Here is the problem: Suppose that one has proven the proposition that if $A \subseteq B$ and $C \subseteq D$, then $A \cup C \subseteq B \subseteq D$. Prove that for any integer $n \geq 2$ that if sets $A_1, A_2,...,A_n$ and $B_1, B_2,...B_n$ are sets that satisfy $A_j \subseteq B_j$ for $j = 1, 2, ..., n$ then $$\bigcup_{j=1}^n A_j\subseteq \bigcup_{j=1}^n B_j."$$
@Avra Ok, I see it! So, in assuming $P(k)$, I can automatically say that $\bigcup_{j=1}^k A_j \subseteq \bigcup_{j=1}^k B_j$ without having to prove it like the base case?
And then just show how this assumption leads to $P(k+1)$, like in your answer?
people were talking about approximation in the sup norm the other day. that is a good example of a norm that is not induced by an inner product but is still of high interest.
a good deal of work in normed spaces involves approximate replacements for stuff that you get from an inner product. 'almost orthogonality' is a big thing in harmonic analysis for example.
the main justifications for least squares models, (1) it's a button in excel, (2) it makes computation easier.
my daughter has stopped complaining about her leg. her wobbly walk is getting a little smoother but she is not pointing her foot in the right direction. orthopedist says it's not a problem unless it lasts for weeks.
my dad didn't like me visiting even those three times. he had a weirdly idealized conception of college as something you go away to and never come back from. he was also in the process of being divorced by my mom which may have affected his disposition.
my dad had this idea that i was some kind of lothario because when i'd visit home for the weekend i'd stay at female friends houses. it was purely to avoid the awkwardness of being around him.
i think my daughter will have a somewhat better experience.
we don't feed the ducks, but they tend to expect it. i took a great video this weekend of my daughter yelling as a gang of four ducks approached her. "agggh! i don't want them to come close to me!" and hiding behind me.
my mom lived in fear of a goose or swan breaking our arms. no idea how she got that into her hear. she was a logical person, but had a few oddities. she thought one would die of cramps if you went swimming in the hour after eating.
i did an experiment and survived, albeit my dinner contributed to the general atlantic flora & fauna.
If I have a domain $\Omega$, and an open cover $\{U_{\alpha} \}$ of the domain (by open sets of $\Omega$), and functions $f_{\alpha \beta} \in C^{\infty}(U_{\alpha} \cap U_{\beta})$ and a partition of unity for $\Omega$ with respect to the open cover, $\{\psi_{\alpha} \}$, why is $g_{\alpha} = \sum_{\beta} f_{\alpha \beta} \psi_{\beta}$ necessarily smooth on $U_{\alpha}$?
here by $\psi_{\beta} $ it means $\psi_{\beta} \restriction_{U_{\alpha} \cap U_{\beta}}$
oh, is it because if $B(z_0,\epsilon_0) \subset U_{\alpha}$ is precompact, then $g_{\alpha} \restriction B(z_0; \epsilon_0) = \sum_{\beta ; supp(\psi_{\beta}) \cap B(z_0 ; \epsilon_0) \neq \emptyset} (f_{\alpha \beta} \psi_{\beta} \restriction_{U_{\alpha} \cap U_{\beta}}) \restriction B(z_0 , \epsilon_0) \cap (U_{\alpha} \cap U_{\beta})$
and the RHS is a finite sum (because $\{supp(\psi_{\beta}) \}$ is locally finite), and each summand is the restriction of a smooth function on an open set ($U_{\alpha} \cap U_{\beta}$) to a smaller open set
hmm, this is suggestive: the values for odd $k$ are the same as the coefficients in the Taylor series of arctanh(x)^2
okay, so that seems to be coming from the following identity: $$\int_0^1 \frac{x}{1-y^2 x^2}\ln\left(\frac{1+x}{1-x}\right)\,dx = \frac{1}{4y^2}\ln\left(\frac{1+y}{1-y}\right)$$ which is valid for $-1<y<1$ (if you eliminate the removable singularity at $y=0$)
now expand both sides in powers of $y$ and match term-by-term
ah, drat, i transcribed that wrong
RHS should've been $\frac{1}{4y^2}\left[\ln\left(\frac{1+y}{1-y}\right)\right]^2=y^{-2}\tanh^{-1}(y)^2$
If we have a list of jobs ordered according to their finish times (nondecreasing sort), then is it true please that we can find all jobs that are not conflicting with current job based on their finishing times in O(nlogn) please
The point is the jobs already sorted according their their finish time, so no clue why we need binary search to search all elements in O(nlogn) time where $n$ is the number of jobs.
Problem: Suppose $c_1,c_2:(-\epsilon,\epsilon)\rightarrow M$ are smooth curves with $c_1(0)=c_2(0)=p$ , where $M$ is a smooth manifold.
Then, $c_1'(0)=c_2'(0)$ if and only if there exists a chart $(U,\phi)\owns p$ such that
$(\phi\circ c_1)'(0)=(\phi\circ c_2)'(0)$
My attempt:
Forward direction: ...
i thought feller had some version of that in his treatise, which was published in english and ought to have infringing copies available. could be wrong.
i had a print set lost by the USPS. i have a long standing grudge against the postal service.
court of federal claims, here i come. no recipe for success like suing the government in a no-right-to-jury forum, where a representative of the government without life tenure decides how much the government owes you.
