@TedShifrin I have, like 4 months before a Riemannian geometry conference I signed up for. So I need to learn geometry at some point before that anyway!
Zach: The circumcenter of the isosceles triangle is the center of the original rectangle. And the distance from that to the last vertex (end of fold segment) is clearly different.
@TedShifrin Have you ever thought about/looked at these theorems uniquely characterizing $\Bbb CP^n$ given as few as possible conditions on cohomology? Kind of nice...
It doesn't mean anything very exciting. Take it as meaning that you can almost always find geometric constructions for anything you do homotopically. Cohomology as maps from smooth manifolds, pullback of classes as transverse pullback. Cup product as intersection (literally). I'm sure there's a good way of doing Steenrod squares.
No, Zach, although I'm not sure how I saw it before (other than experimentally), I now have a careful proof that BD' is parallel to the fold, i.e., perpendicular to AC, so D' folds onto B.
@Danu: The Calabi conjecture would say that $c_1$ is represented by a Kähler-Einstein metric, but when is it the induced metric? That only happens if $c_1$ is the hyperplane class.
Yes, we fold along AC for the second fold, Zach, so B meets D' (I now have nailed that down by looking at angles — maybe there was a more obvious way which I saw an hour ago).
Because I end up with vertices A=C', B=D', and E only, Zach.
Neverthless, @MikeM, every piece of published research I've done is regarding submanifolds or subvarieties in some sense or other. Perhaps families thereof.
Assume $X^n$ has an ample line bundle $L$ and $c_1(M)\geq (n+1)c_1(L)$. Then Kobayashi & Ochiai proved that $X$ is $\Bbb CP^n$. Now, I'm constantly seeing the claim that if $c_1(X)>0$ has divisibility $n+1$, it is $\Bbb CP^n$. So that seems to imply you can always finda line bundle realizing the generator.
I'm not going to stop to work this out right now (I want to be watching Federer play), but presumably we're going to invoke Kodaira vanishing somewhere.
I'll fiddle with this for ten minutes then I gotta get back to work.
@Danu A complex structure on a line bundle $E$ is given by a complex connection $\overline{\partial}: \Omega^0(E) \to \Omega^{0,1}(E)$ satisfying $\overline \partial^2 = 0$. Complex connections are affine over $\Omega^{0,1}(\text{End}(E)) = \Omega^{0,1}(M)$. There is a natural correspondence (connection on line bundle E) -> (connection on line bundle $E^{\otimes n}$), which if you pick a base connection $A_0$ on $E$ should be $(A_0 + a) \mapsto A_0^{\otimes n} + na$.
Assuming $E^{\otimes n}$ is holomorphic means I can solve the equation $F_{A_0^{\otimes n} + na} = 0$ in $a$. This equation is literally just $F_{A_0^{\otimes n}} + nd_{A_0^{\otimes n}} a = F_{A_0^{\otimes n}} + nda$. That you can find a way to set this to zero means that $F_{A_0^{\otimes n}} = - nda$ for some choice of $a$.
I think it should be true that $F_{A_0^{\otimes n}}$ should also be $nF_{A_0}$, so that by taking the exact same $a$ you solve the equivalent equation on $E$, obtaining a holomorphic structure on it
In particular this means that if $E$ is a holomorphic line bundle, so are any positive roots of it, yeah?
Actually everything there works fine. I was scared about being able to do this for $n = -1$, but no, every holomorphic line bundle's inverse is still holomorphi
The formulas I wrote there are more or less irrelevant. The point is that the map $\mathcal A_E \to \mathcal A_{E^{\otimes n}}$ is an isomorphism, so you can just take the nth root of a connection at will
The formulas are just there to show that the nth root of a holomorphic thing is still holomorphic
Does anyone know where I could get my hands on a paper where they prove that the two different definitions of the Mandelbrot set: namely the orbit of 0 being bounded with respect to the polynomial P_c and that the corresponding Julia set of P_c being connected are equal?
Any paper would suffice; I tried looking at the references on the Wikipedia articles on the Mandelbrot set and the Julia set as well as trying to search for such a paper online but all sources seemed to make the claim of the sets being equal without any actual proof.
It seems a bit specialized topic, alas. You'll probably have to keep pushing with google-fu.
I imagine it's what the Wikipedia page references here: "As a consequence of the definition of the Mandelbrot set, there is a close correspondence between the geometry of the Mandelbrot set at a given point and the structure of the corresponding Julia set. For instance, a point is in the Mandelbrot set exactly when the corresponding Julia set is connected."
I don't get your question. If you do that to a square, you get the original 4 vertices and the middle point of the square. What is a quadrilateral here?
