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Example 2.31 in these notes of Milne is about real Hodge structures and the fact that they can be seen as representations of the Deligne torus $\mathbb{S}$. The pace is a bit too fast for me. I have the following two questions:

  1. Why does $z \in \mathbb{C}^\times$ have to act on $V^{p,q}$ as multiplication by $z^{-p}\overline{z}^{-q}$ and cannot just be multiplication by $z^{p}\overline{z}^{q}$? Would the corresponding map $\mathbb{G}_m \to \mathbb{S}$ be the usual inclusion of $\mathbb{R}$ into $\mathbb{C}$ in that case, rather than being $t \mapsto t^{-1}$?
  2. How is the action on $V$ defined exactly? So far I have only an action on $V \otimes \mathbb{C}$, without knowledge of what happens to pure tensors of the form $v \otimes 1$. If I want to know what $z \cdot v$ is, should I take the real part of $z \cdot (v \otimes 1)$, or the imaginary part, or split $v \otimes 1$ into components of pure weight and do something with them? Or should I just take the $V_{p,p}$ components of $v \otimes 1$ and take their image, since $z$ acts on $V_{p,p}$ as multiplication by $|z|^{-2p} \in \mathbb{R}^\times$? None of these options seems to give the right action (or an action at all).
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  1. Yes, you can leave out the minus signs, but at some point Deligne decided it was better with them in, and most people follow him. Deligne's reasons are a bit obscure, but there's an explanation in Milne's notes Introduction to Shimura Varieties.
  2. The action of $h(z)$ on $V\otimes \mathbb{C}$ commutes with the action of complex conjugation (because of the condition that $V^{p,q}$ is the complex conjugate of $V^{q,p}$), and so preserves $V$.
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