Skip to main content
added 316 characters in body
Source Link
JS Milne
  • 8.3k
  • 1
  • 43
  • 44

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 ...

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).

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 ...

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 ...

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).

Source Link
JS Milne
  • 8.3k
  • 1
  • 43
  • 44

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 ...