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).
  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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