@Danu hi, I was just looking at Kaumudi's question about algebraic functions. I wasn't going to risk actually talking about maths - I know my place :-)
$\tau (G)$ is the size of the largest set of independent edges on $G$. $\nu (G)$ is the size of the smallest set of vertices in $G$ such that every edge in $G$ is incident to a vertex in said set.
@BalarkaSen Let $L$ be a line bundle on a compact complex manifold. $L$ is trivial iff $L$ and $L^*$ both admit global sections. One way is really easy because any non-zero constant map to $\Bbb C$ gives a section of the trivial bundle. The converse is slightly more work
Let $s,s'$ be sections of $L,L^*$, and $s_i,s'_i$ the image through a trivialization.
Then I defined $S_i=s_i\cdot s'_i$, and showed that it doesn't transform under transition functions, so the $\{S_i\}$ patch together to a global holomorphic function. The last bit is showing that it's not constant...
For that, I argued as follows: $s$ must hit zero at some point, else $L$ is trivial. Same for $s'$. But they're not the zero section, by assumption, so $S$ is not constant, which yields a contradiction.
Nowhere. Ted asked me, and I can prove it for complex dimension 1 (it's just C-R equations, really). Don't think higher dimension is any harder than that.
Intersection number means and always will mean topological intersection number. The claim I had said that for any two complex submanifolds of a complex manifold (of dual dimension), intersection number is $> 0$. That just means inside eg $\Bbb P^2$ blown up at a point, the exceptional divisor cannot intersect with any other complex submanifold homologous to itself which intersects it with intersection number $-1$.
This in particular means the normal bundle of the exceptional divisor (aka $O(-1)$) can't have a global holomorphic section.
"for any two *transverse complex submanifolds", I meant.
To clarify, there are global topological sections, which means there are topological submanifolds of $\Bbb P^2$ blown up at a point which intersects the exceptional divisor like that. But none of those are complex submanifolds.
@BalarkaSen Let $\gamma_e$ be the $U(1)$-bundle over $S^2$ with Euler class $e$. If $e$ is negative, the total space of $\gamma_e$ is diffeomorphic to $S^2 \times S^1$. But of course, cup product shows that the total space of the associated disc bundles differ.
I really need some help with my assignment. Im terrible at Logic and the ZFC - set theorem I hope you can help check my work so far and help me solve this hard one All the best Ajax
In this exercice we need to prove that:
(1*)Is T a not empty subset of the natural numbers N, that means their ex...
you said you need to prove (1) then your (a) question is something about (2), your (b) question is ... idk there's a (3) and i'm really not sure what's (3), and your (c) question asks for a proof of (3) so even if that made sense idk how it would be different from (b)
Well a) wants the logical proof that (1*) = (2*) Question b) Here is the question to show that (3*) implies (2*) and c) is about proofing the formulae (3*) by induction
Complex structures on Riemann surfaces are the same as orientations + conformal classes of a Riemannian metric. Then the uniformization theorem can be restated as "Every metric on a Riemann surface is conformal to a complete metric of constant sectional curvature."
By the Killing-Hopf theorem, such a simply connected surface is isometric to either the plane, hyperbolic plane, or sphere.
I'm trying to prove the following (simplified). I have a matrix of cells, each containing one (distinct) element and originally they are distributed uniformly at random. Now I do the following: In each "round", I pick two elements at random and swap them. What I want to prove is that the distribution is uniformly random.
So one needs to prove that. You can write down the formula for the change in scalar curvature in dimension 2: $s(e^{2f}g) = e^{-2f}(s(g) + 2\Delta f)$. So one wants to solve the equation $s(g) = \text{sgn}(g)e^{2f} - 2\Delta f$, where $\text{sgn}(g)$ is $0, 1,$ or $-1$.
On a compact manifold this can be solved by calculus of variations techniques. On noncompact manifolds it's a little harder.
I have the following already: the probability of accessing the same column if I pick two different elements is $\frac{1}{N} + \frac{N-1}{N} \cdot \frac{\sqrt{N}-1}{N-1} = \frac{1}{\sqrt{N}}$, where $N$ is the total number of cells in the matrix
cannot use symmetry to prove transitivity as they did in the proof. To prove transitivity they should have introduced 'c' and show that a ~ b and b ~ c and proved that a ~ c.
I also have that if I pick the same element the probability that it will be in the same column is $\frac{1}{\sqrt{N}}$, but I somehow don't see how to generalize these two things
I tried to show that prob. of access i will access column $j_i$ given the condition that acccess i-1 was to some $j_{i-1}$, but somehow I fail at expanding this properly I think.
Been reading the chapter it's in for the analytic side of things I have been studying. Also admittedly feel the theorem is quite curious. (as a minor reason I should be fair about, I'll meet the analyst I'm studying with tomorrow and don't want to go there empty-handed)
Given $X=[0,1]$. We knew that $X \subset X$. Why is X an open subset wrt itself, is it because we cannot go anymore left from e.g. the point 0 thus a ball at 0 cannot extend to points to the left of 0 since there are no such points?
> Let X be a metric space. A ball B of radius r around a point x ∈ X is B = {y ∈ X|d(x, y) < r}.
> Lastly, open sets in spaces X have the following properties: 1. The empty set is open 2. The whole space X is open 3. The union of any collection of open sets is open 4. The intersection of any finite number of open sets is open.
My question concern about point number 2, since if I place a ball on 0, then I can always go infintesimally to the left of 0 and end up nowhere that is not in X
If I have X = [0,1] and I place a ball on 0, I cannot go any left to that because there is nothing there, and not because the points there I will fall outside of X?
@Secret You might as well worry about accidently moving perpendicular to the line segment you are on (note that this does not seem to be a problem for you)
I also just had a paper appear in Documenta (so far just online), and my advances paper should be out on paper soon (whenever the October issue is printed)
Some time later, I might investigate further how people overthink and failed to overthink things, cause I suspect the phenomenon itself might be able to be described by some mathematical space with soem constraints applied to it
Why in the definition of a topology is the requirement that a topology $\tau$, need to be closed under finite union and intersection? What does having union or intersection of sets in $\tau$ that fall outside of the topology $\tau$ mean in terms of the intuitive idea of how two things are connected together?
Hmm, let's see, I will start with the interval [0,1]. So I can e.g. break it into the union of [0,0.5) and [0.5,1] and $\emptyset$. I then call this set S. The intersection of any of them gives $\emptyset$ which is $\in S$ The union of them give my original interval [0,1] which is not in S because of how I slice the interval... I don't felt like there's a gap or something like that here...?
@Secret: In fact, the collection of open sets (the topology) needs to be closed under arbitrary unions and, yes, finite intersections. You can if you want define a topology on $[0,1]$ by having the only open sets be the whole set (you cannot leave that out), $[0,.5)$ and $[.5,1]$. Can you give me continuous functions from that topological space to $\Bbb R$?
Hi @Danu: You awake this time? ... I just spent an hour sorting out some complex geometry for someone (sigh, yet another "small" mistake in Griffiths/Harris).
By the way, I skipped the Veronese & Segre maps because Huybrechts has a huge concentration of typos there so I couldn't make any sense out of what he was typing.
Yeah. Figure out why the $\otimes \mathscr O(-1)$ needs to be there to get a derivative map from $\Bbb C^{n+1}- \{0\}$ to $T\Bbb P^n$.
Probably because Euler thought about the vector field $\sum x^i \frac{\partial}{\partial x^i}$, which is now called the Euler vector field. It is germane here.
Actually the exercise in Huybrechts is a bit different---show that $f\otimes \operatorname{id}_G:E\otimes G\to F\otimes G$ is still injective if $f:E\to F$ is injective
I absolutely, absolutely, absolute detest one of the classic Advanced Calculus texts because he writes out multivariable analysis in terms of $(f,g,h)$ instead of working with vector-valued functions (in arbitrary dimension). Did I say I hate this book?
