ScalingStacks

Lemma 3.14 . [027H]

Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.

Complete original source context · Original author HTML

Lemma 3.14.

Let L¯=(L,h)\overline{L}=(L,h) and L¯′=(L′,h′)\overline{L}^{\prime}=(L^{\prime},h^{\prime}) be elements of Pic^C0+​(X)\widehat{\operatorname{Pic}}^{+}_{C^{0}}(X). Then we have the following:

  1. (1)

    μL¯≥0\mu_{\overline{L}}\geq 0 on XanX^{\operatorname{an}}.

  2. (2)

    μL¯​(x)=limn→∞n∈ℕ⁡(L)σL¯⊗n​(x)n{\displaystyle\mu_{\overline{L}}(x)=\lim\limits_{\begin{subarray}{c}n\to\infty\\ n\in{\mathbb{N}}(L)\end{subarray}}\frac{\sigma_{\overline{L}^{\otimes n}}(x)}{n}} for x∈Xanx\in X^{\operatorname{an}}.

  3. (3)

    μL¯⊗L¯′​(x)≤μL¯​(x)+μL¯′​(x)\mu_{\overline{L}\otimes\overline{L}^{\prime}}(x)\leq\mu_{\overline{L}}(x)+\mu_{\overline{L}^{\prime}}(x) for x∈Xanx\in X^{\operatorname{an}}.

  4. (4)

    If L¯≃L¯′\overline{L}\simeq\overline{L}^{\prime}, then μL¯=μL¯′\mu_{\overline{L}}=\mu_{\overline{L}^{\prime}} on XanX^{\operatorname{an}}.

  5. (5)

    For n≥0n\geq 0, μL¯⊗n=n​μL¯\mu_{\overline{L}^{\otimes n}}=n\mu_{\overline{L}} on XanX^{\operatorname{an}}.

Original mathematics by the credited authors. Source-backed reader collection; mathematical self-containment is not assessed.