The algebraicity of Hodge structure map - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T13:45:21Z http://mathoverflow.net/feeds/question/18341 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18341/the-algebraicity-of-hodge-structure-map The algebraicity of Hodge structure map Taisong Jing 2010-03-16T07:03:30Z 2010-03-16T09:27:50Z <p>Let $\mathbb S$ be $\mathbb C^{\times}$'s restriction of scalar to $\mathbb R$. To give a real Hodge structure on an $\mathbb Q$ vector space $V$ is to give a real representation of $\mathbb S$ on $V_{\mathbb R}$. Let $\mathbb G_m\rightarrow \mathbb S\rightarrow GL(V_{\mathbb R})$ be the weight homomorphism. If it is actually defined over $\mathbb Q$ we say the hodge structure is rational. But can we say that the weight homomorphism is always algebraic, that is, defined over $\overline{\mathbb Q}$? Every resource claims this, but I can't see why...</p> http://mathoverflow.net/questions/18341/the-algebraicity-of-hodge-structure-map/18342#18342 Answer by anton for The algebraicity of Hodge structure map anton 2010-03-16T08:10:30Z 2010-03-16T08:10:30Z <p>Who is every source? If you fix a weight $k$, then to give a real Hodge structure is the same as giving a homomorphism <code>$h:{\mathbb G}_m\to GL(V)$</code> which is defined over $\mathbb R$. That does not mean it is defined over $\overline{\mathbb Q}$. Indeed, $h$ doesn't know anything about the $\overline{\mathbb Q}$-structure, so why should it respect it?</p> <p>As an example, let's take $V={\mathbb Q}^2$ and $h$ defined by <code>$$h(a)=A(^a{}_1)A^{-1},\ \ \ A=(^1_0{}^{\pi}_1).$$</code> Then $h$ is not defined over $\overline{\mathbb Q}$. <strong>But</strong> it is conjugate to a homomorphism which is defined over $\mathbb Q$. Maybe that's what you need?</p> http://mathoverflow.net/questions/18341/the-algebraicity-of-hodge-structure-map/18343#18343 Answer by JS Milne for The algebraicity of Hodge structure map JS Milne 2010-03-16T08:22:40Z 2010-03-16T09:27:50Z <p>It looks false to me. Let $V=\mathbb{Q}^{2}$, and let $V(\mathbb{R})=V^{0}\oplus V^{2}$ where $V^{0}$ is the line defined by $y=ex$ and $V^{2}$ is the line defined by $y=\pi x$. Give $V^{0}$ the unique Hodge structure of type $(0,0)$ and $V^{2}$ the unique Hodge structure of type $(1,1)$. To say that $w$ is defined over the subfield $\mathbb{Q}^{\mathrm{al}}$ of $\mathbb{C}$ means that the gradation $V(\mathbb{R})=V^{0}\oplus V^{2}$ arises from a gradation of $V(\mathbb{Q}{}^{\mathrm{al}})$ by tensoring up, but this isn't true. Perhaps the all the "resources" have additional conditions, or perhaps they are all ...</p> <p>Added: When you are defining a Shimura variety, the weight homomorphism w factors through a Q-subtorus of GL(V), and then it is true that w is defined over the algebraic closure of Q (because, for tori T,T', the group Hom(T,T') doesn't change when you pass from one algebraically closed field to a larger field).</p>