most recent 30 from http://mathoverflow.net 2013-05-21T20:40:25Z http://mathoverflow.net/feeds/user/3918 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48098/restriction-of-scalars-of-simple-algebraic-groups/48109#48109 Lucio Guerberoff 2010-12-02T22:24:23Z 2010-12-02T22:24:23Z

<p>For the question in the comments, Kevin is right: every connected reductive group over k is an inner form of a unique quasi-split group, up to isomorphism. This follows from an argument using Galois cohomology and the splitting</p> <p>$$ 1\to\operatorname{Int}(G)\to\operatorname{Aut}(G)\to\operatorname{Aut}(\Psi_0(G))\to 1,$$ where $\Psi_0(G)$ is the based root datum of G. More details can be found on the first article of Corvallis, and surely in some other place (maybe Borel-Tits?).</p>

http://mathoverflow.net/questions/47351/how-to-think-of-monodromy-transformations/47365#47365 Lucio Guerberoff 2010-11-25T19:19:17Z 2010-11-25T19:19:17Z

<p>The classical monodromy action is the action of the fundamental group of a (nice) space on the fibers of a covering map. Suppose that $X$ is a (nice) topological space, and let $\pi_1(X,x_0)$ denote the fundamental group of $X$ at the point $x_0$. Let $p:Y\to X$ be a covering map. Then $\pi_1(X,x_0)$ acts on the fibre $Y_{x_0}=p^{-1}(x_0)$ as follows: take a point $y\in Y_{x_0}$ and an element of the fundamental group represented by a loop $\alpha$ at $x_0$; then a basic property of covering maps implies that $\alpha$ lifts to a path in $Y$ starting at $y_0$; the endpoint of this map also lies in the fibre, and it is the result of the action of $\alpha$ on $y$, by definition.</p> <p>This generalizes to a variety of situations. A nice overview is perhaps the book <em>Galois groups and fundamental groups</em> by T. Szamuely.</p>

http://mathoverflow.net/questions/46658/representations-of-reductive-groups-over-arbitrary-fields/46743#46743 Lucio Guerberoff 2010-11-20T16:04:59Z 2010-11-20T16:04:59Z

<p>I cannot make a comment, so I answer Brian here: the Galois action, as Kevin says, is different from the natural action on the Torus. I think it is described in Corvallis articles, and also in Borel-Tits. For this action, if G is a form of H, then it is an inner form if and only if the Galois actions on the root datum are the same.</p>

http://mathoverflow.net/questions/70331/book-on-linear-algebraic-groups-in-scheme-language Lucio Guerberoff 2011-07-14T16:45:43Z 2011-07-14T16:45:43Z

Agreed, Demazure-Gabriel is hard to beat. Whatever you need after it, you can read any other book and translate from that language in your mind to scheme language. Milne's notes also have a promising future.

http://mathoverflow.net/questions/67417/when-does-a-bijective-morphism-of-schemes-induce-an-isomorphism-of-hodge-structur/67423#67423 Lucio Guerberoff 2011-06-10T20:15:59Z 2011-06-10T20:15:59Z

Just to add: at least in the smooth case, the map on the complex analytic manifolds will be a holomorphic bijection, and hence the inverse will also be holomorphic, so the second paragraph applies.

http://mathoverflow.net/questions/65892/differential-forms-on-an-almost-complex-manifold Lucio Guerberoff 2011-05-24T21:06:57Z 2011-05-24T21:06:57Z

Just use induction on the total degree. It's obviously true for total degree 1.