bio | website | |
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location | ||
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visits | member for | 4 years, 1 month |
seen | 7 hours ago | |
stats | profile views | 1,207 |
Mar 18 |
answered | In what condition is a conformal flat manifold flat? |
Mar 12 |
awarded | Yearling |
Feb 28 |
comment |
quantitative version of the rigidity of the 2-sphere
I am not an expert on the Ricci flow, but can't you use the existence of the flow for positive curvature metrics and corresponding estimates to deduce "how long" you need to flow, i.e., how close you are to the round sphere? |
Feb 27 |
answered | A question on differential forms and integral invariants |
Feb 7 |
awarded | Scholar |
Feb 7 |
accepted | Riemann's theorem on theta |
Jan 27 |
comment |
Normalizing the value of a principal connection at a point
You both are right, I was not clear enough with my phrasing "connection 1-form with respect to this section vanishes at p," which should mean pullback of the connection form by the section. And Robert's remark shows us how nicely differential calculus can be approximated linearly, I just oversaw it. |
Jan 27 |
answered | Normalizing the value of a principal connection at a point |
Dec 13 |
answered | How to compute the normals to Costa's minimal surface? |
Oct 29 |
comment |
Obstruction to this gauge choice of the connection of a vector bundle
Sorry, my mistake. Of course, you are right, as I should have been aware of if I had read the whole question. |
Oct 28 |
comment |
Obstruction to this gauge choice of the connection of a vector bundle
Somehow, a short computation shows me that your equation $\mathcal L_RA=d_A(I_R A)$ is not gauge invariant, and therefore not even well-defined on an arbitrary vector bundle |
Sep 22 |
awarded | dg.differential-geometry |
Aug 1 |
awarded | Nice Answer |
Jul 25 |
comment |
Formula for the curvature of an induced connection
If $a=0$ then $F_B=\lambda\cdot F_A$ where $\lambda\cdot\colon\mathfrak g\to \mathfrak h$ is the induced map on Lie algebras. For $a\neq0$ you would in general need also to know $f_*A.$ |
Jul 22 |
comment |
$E$ is a holomorphic vector bundle if and only if there is a Dolbeault operator $\bar{\partial}_E$
Over Riemann surfaces there is a proof of this fact which does not use Newlander-Nirenberg: Atiyah, Bott: The Yang-Mills equations over Riemann surfaces. |
Jul 22 |
comment |
$E$ is a holomorphic vector bundle if and only if there is a Dolbeault operator $\bar{\partial}_E$
Dear Tim, I think it is the same complexity to show that flat connections are locally trivial and that flat Riemannian manifolds are locally euclidean: Take a ON parallel frame of the Riemannian manifold, because the connection is torsion free these vectorfields are commuting, so you can integrate them up to obtain a local isometry to euclidean space. |
Jul 11 |
asked | Periods of translation surfaces |
Jun 25 |
awarded | Revival |
Mar 12 |
awarded | Yearling |
Mar 5 |
comment |
Branched Regular Cover over 4-times punctured sphere
Sorry, I was to vague about this: there exists holomorphic coordinates $y$ and $z$ on $\Sigma$ and $P^1$ centered at a branch point resp. branch image such that $f$ is given by $z=y^{g+1}.$ Then, the 4 sheets are exchanged by going around $y=0,$ and such a deck transformations $\varphi_{g+1}$ extends to all of $\Sigma$ holomorphicallysuch that $f\circ\varphi_{g+1}=f.$ This topic is covered in most Riemann surface books, for example the recent one by Donaldson. |