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Let $K=\mathbb{Q}_p$ and $G$ a split reductive group over $K$ with split maximal torus $T$. Furthermore, $X$ is a smooth projective variety over $K$ with a free $G$-action. Let $U \subset X$ be a $T$-invariant affine subspace, i.e. we have $U=\mathrm{Spec}(A)$ with $A=K[T_0,\ldots, T_N]/\mathfrak{a}$ for some $N \in \mathbb{N}$ and radical ideal $ \mathfrak{a}\subset K[T_0,\ldots, T_N].$

Let $m \in \mathbb{N}$ and $c\in K$ with $\lvert c \rvert_K>1$. We define the affinoid $K$-algebra $$A_m:=K\langle c^{-m}T_0,\ldots, c^{-m}T_N \rangle/(\mathfrak{a})$$ and set $U_m:=\operatorname{Max}A_m$. Hence, $A \subset A_m$ and $U^{\text{cl}} \supset U_m$.

Then, $\mathcal{O}_X(U)$ is a $K$-vector space equipped with a $T$-action. Thus, we have a decomposition into weight spaces $$ \mathcal{O}_X(U)= \bigoplus_{\lambda \in X^*(T)} \mathcal{O}_X(U)_\lambda. $$ Additionally, we equip $\mathcal{O}_X(U)$ with a norm $\lvert \, \, \rvert_m$ such that
$$ \lvert f \rvert_m =\sup_{x \in U_m} \{\lvert f(x)\rvert_K \} $$ for $f \in \mathcal{O}_X(U)$. Note that it is the supremum norm from $A_m$ and that $A$ is dense in $A_m$.

Let $f \in \mathcal{O}_X(U)$ and $f = \sum f_\lambda$ with $f_\lambda \in \mathcal{O}_X(U)_\lambda$. By the properties of the absolute value on $K$, we have $$ \lvert f \rvert_m \leq \sup_{\lambda \in X^*(T)} \vert f_\lambda \rvert_m. $$

But is it true that $$ \lvert f \rvert_m= \sup_{\lambda \in X^*(T)} \vert f_\lambda \rvert_m $$ holds?

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  • $\begingroup$ About the definition of the norm: why should the sup exist? $\endgroup$ Commented Oct 24, 2022 at 12:52
  • $\begingroup$ For the same reason as in the affinoid case. I added to consider only closed points. $\endgroup$
    – KKD
    Commented Oct 24, 2022 at 13:38
  • $\begingroup$ I edited again to consider only points of $U_m$. Now it should be fine. $\endgroup$
    – KKD
    Commented Oct 24, 2022 at 15:12
  • $\begingroup$ The letter $T$ in $T_0, \dotsc, T_N$ has nothing to do with the torus $T$, right? $\endgroup$
    – LSpice
    Commented Oct 24, 2022 at 16:31
  • $\begingroup$ Also, what is the significance of $[]$ vs. $\langle\rangle$ in $K[T_0, \dotsc, T_n]$ vs. $K\langle c^{-m}T_0, \dotsc, c^{-m}T_n\rangle$? Usually I'd think $\langle\rangle$ meant the non-commutative polynomial algebra, but that seems not to be the case here. $\endgroup$
    – LSpice
    Commented Oct 24, 2022 at 16:39

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