Timeline for The algebraicity of Hodge structure map
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2010 at 19:14 | vote | accept | TJCM | ||
| Mar 17, 2010 at 0:59 | comment | added | TJCM | I see, thank you. Is it in general true for a homomorphism from a reductive group to a linear group, if it is defined over a bigger algebraic closed field, then we can always descent it to a smaller algebraic closed field? (for unipotent groups one may have a lot of non algebraic homomorphisms, e.g. $x\mapsto \pi x$) | |
| Mar 16, 2010 at 19:14 | comment | added | JS Milne | The maps $\mathbb{S}\to G_{\mathbb{R}}$ are defined over $\mathbb{R}$,not $\mathbb{Q}$. However, one of the axioms (2.1.1.1 in Deligne's Corvallis articles; SV1 in my articles) for Shimura varieties implies that the weight homomorphism $w$ of each $h$ factors through the centre of $G$ (hence through connected centre, which is a torus). As the $h$'s are all conjugate, this implies that the $w$'s are all equal, and that the $w$ is defined over the algebraic closure of $\mathbb{Q}$ (it is a homomorphism between tori defined over $\mathbb{Q}$). | |
| Mar 16, 2010 at 14:15 | comment | added | TJCM | Thanks! So when we define a Shimura datum $(G,X)$, the map $\mathbb S\rightarrow G_{\mathbb R}$ is actually defined over $\mathbb Q$, not only over $\mathbb R$? If it is only defined over $\mathbb R$, I still can't assure myself that $\mathbb G_m\rightarrow \mathbb S\rightarrow G_{\mathbb R}$ factors through a $\mathbb Q$ subtorus. | |
| Mar 16, 2010 at 9:27 | history | edited | JS Milne | CC BY-SA 2.5 |
added 316 characters in body
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| Mar 16, 2010 at 8:22 | history | answered | JS Milne | CC BY-SA 2.5 |