User ruadhai dervan - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-21T22:39:31Zhttp://mathoverflow.net/feeds/user/22294http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/127829/when-constant-scalar-curvature-implies-einstein/127854#127854Answer by Ruadhai Dervan for when constant scalar curvature implies Einstein?Ruadhai Dervan2013-04-17T14:50:57Z2013-04-17T21:24:21Z<p>As an example where this does hold, for $\omega$ a Kähler metric of constant scalar curvature with $\pi c_1(M) = \lambda [\omega]$, then $\omega$ is Kähler-Einstein. This is Proposition 2.12 in Tian's "Canonical metrics in Kähler Geometry".</p>
http://mathoverflow.net/questions/121461/reference-for-notation-h0c-mk/121465#121465Answer by Ruadhai Dervan for Reference for notation $H^0(C, mK)$Ruadhai Dervan2013-02-11T11:20:58Z2013-02-11T13:57:27Z<p>I can't access the draft you've linked but (almost always) this means the vector space of global sections of the $m^{th}$ power of the canonical bundle. Note for a Riemann surface the canonical bundle is just the cotangent bundle, the dual of the tangent bundle. For the general theory of algebraic curves, see e.g. Geometry of Algebraic Curves I by Arabello et al.</p>
http://mathoverflow.net/questions/95503/flow-of-a-hamiltonian-vector-field/95508#95508Answer by Ruadhai Dervan for Flow of a Hamiltonian vector fieldRuadhai Dervan2012-04-29T13:35:23Z2012-04-29T13:35:23Z<p>To answer the question of 'What does the flow of a Hamiltonian vector field correspond to', it's useful to think physically. Exactness means we have some function $H$ which corresponds to the vector field $X_H$. If you take a Darboux coordinate system, then the flow of the Hamiltonian vector field is exactly the solution to Hamilton's equations. Hamilton's principle basically states that the real trajectory of a physical system follows Hamilton's equations, where we interpret $H$ as something like an energy function (equivalent to minimising an energy functional). I hope this helps, it's probably more useful from a physical point of view than a pure mathematical one.</p>
http://mathoverflow.net/questions/92423/general-topology-terminology-questions/92425#92425Answer by Ruadhai Dervan for General topology terminology questionsRuadhai Dervan2012-03-28T00:22:57Z2012-03-28T00:22:57Z<p>In Geometric Invariant Theory, the study of quotients in algebraic geometry, some points are ignored in the quotient (by its construction) that would make the quotient non-Hausdorff. These points are called 'unstable'. Sometimes the set of all unstable points is called the 'unstable locus'. This is of course just a special case of your question, in a slightly different area, but perhaps the terminology is used elsewhere. A good reference for this if you're interested is <a href="http://arxiv.org/abs/math/0512411" rel="nofollow">these notes</a> by Richard Thomas.</p>
http://mathoverflow.net/questions/91942/algebraic-geometry-definition-of-a-morphism/91948#91948Answer by Ruadhai Dervan for Algebraic Geometry - Definition of a MorphismRuadhai Dervan2012-03-22T20:59:07Z2012-03-22T20:59:07Z<p>A regular map $\phi: X \to Y$ of quasi-projective varieties is a continuous map with respect to the Zariski topology such that for $V \subset Y$ an open set and $f$ a regular function on $V$, we have $f\circ \phi$ is regular on $\phi^{-1}V$. This seems to me to be to be exactly what you would want and quite intuitive and understandable. </p>
http://mathoverflow.net/questions/127829/when-constant-scalar-curvature-implies-einstein/127854#127854Comment by Ruadhai DervanRuadhai Dervan2013-04-17T14:52:22Z2013-04-17T14:52:22ZP.S. If anyone would like to remove the e from "Kaehler" and insert an umlaut, they're more than welcome to, as I was unable to.http://mathoverflow.net/questions/125200/coboundaries-and-gluing-in-cech-cohomology-intuitionComment by Ruadhai DervanRuadhai Dervan2013-03-21T20:32:06Z2013-03-21T20:32:06ZWhy is this tagged mathematical physics?http://mathoverflow.net/questions/120062/intersection-of-two-projective-submanifolds-in-pn-treatment-in-shafarevich-boo/120073#120073Comment by Ruadhai DervanRuadhai Dervan2013-01-28T10:06:12Z2013-01-28T10:06:12ZThis is how Shafarevich proves in my version of his book.http://mathoverflow.net/questions/120062/intersection-of-two-projective-submanifolds-in-pn-treatment-in-shafarevich-booComment by Ruadhai DervanRuadhai Dervan2013-01-27T23:04:59Z2013-01-27T23:04:59ZIn my copy of the book 'Basic Algebraic Geometry I: Varieties in Projective Space' by Shafarevich, Theorem 6 is basically what your statement is - this is on page 76. This is the 'Second, Revised and Expanded Edition'. I can type it out for you, if you'd like.http://mathoverflow.net/questions/115220/how-do-fibers-of-the-functor-algebraic-varieties-to-complex-analytic-spaces-loComment by Ruadhai DervanRuadhai Dervan2012-12-03T00:09:19Z2012-12-03T00:09:19ZCould you link to the question you mention about non-injectivity please? Thanks.http://mathoverflow.net/questions/91942/algebraic-geometry-definition-of-a-morphism/91948#91948Comment by Ruadhai DervanRuadhai Dervan2012-03-22T23:50:47Z2012-03-22T23:50:47ZThis thread should be helpful, it's essentially the same question.
<a href="http://mathoverflow.net/questions/1397/morphisms-of-quasi-projective-varieties" rel="nofollow" title="morphisms of quasi projective varieties">mathoverflow.net/questions/1397/…</a>http://mathoverflow.net/questions/91942/algebraic-geometry-definition-of-a-morphism/91948#91948Comment by Ruadhai DervanRuadhai Dervan2012-03-22T23:43:17Z2012-03-22T23:43:17ZYes, that is true. It would also be a nice definition.http://mathoverflow.net/questions/91942/algebraic-geometry-definition-of-a-morphism/91948#91948Comment by Ruadhai DervanRuadhai Dervan2012-03-22T21:12:52Z2012-03-22T21:12:52ZAh, it would be circular if you didn't know what a regular function was. A function $f: X \to \mathbb{A}^1$ on a quasi-projective variety $X$ is called regular if $\forall x\in X$ there is a neighbourhood $U$ of $x$ such that $f|_U$ is the quotient of two homogeneous polynomials $f|_U = F/G$ of the same degree such that $G$ has no zeroes on $U$.
A good reference for this level of algebraic geometry is I.R. Shafarevich - Basic algebraic geometry I. – Ruadhai 0 secs ago