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Let $\overline L= (L, h)$ be a hermitian $C^ \infty$ line bundle on an arithmetic variety $X\to\operatorname{Spec }\mathbb Z$ (I am reasoning in terms of higher Arakelov geometry, like in Gillet & Soule' papers).

$\overline L$ is said to be arithmetically ample if:

  1. $L$ is relatively ample on $X$
  2. $L_{\mathbb C}$ is positive on $X_{\mathbb C}$
  3. A power of $L$ is generated by small sections.

Now assume that $L$ is relatively ample. It is not difficult to show that there exist a scaling $\alpha h$, with $\alpha\in\mathbb R_{>0}$, such that $\overline L_{\alpha}=(L,\alpha h)$ is arithmetically ample.

Then I ask the following:

Are there some $X$(maybe different from $\mathbb P^n_{\mathbb Z}$) and $L$ such that $\overline L_{\alpha}$ is arithmetically ample for any $\alpha>0$.

In other words, is it possible to find an arithmetically ample line bundle such that with any other scaling of the metric, it remains arithmetically ample?

It seems to be "just" a problem regarding successive minima in the theory of lattices... If I try to picture it in my mind, it seems impossible to find a linde bundle with such properties; there should be a lower bound $\alpha_0>0$ such that below that value it is impossible to find generating small sections for the powers of $\overline L_{\alpha}$.

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Let $\overline{L}$ be any arithmetically ample line bundle. In the way you have written it down, $\overline{L}_{\alpha}$ is not arithmetically ample for $\alpha$ sufficiently large, more precisely for $\alpha\ge\alpha_0=\exp\left(\frac{2\overline{L}^{\dim X}}{L_{\mathbb{C}}^{\dim X-1}}\right)$:

Let $s\in H^0(X,L^{\otimes n})$ be any global section of any $n$-th tensor power of $L$. Then we have the following equality of arithmetic intersection numbers $$n\overline{L}^{\dim X}=\overline{L}\cdot\ldots\cdot\overline{L}\cdot\overline{L}^{\otimes n}=\left(\overline{L}|_{\mathrm{div}(s)}\right)^{\dim X-1}-\int_{X(\mathbb{C})}\log |s|c_1\left(\overline{L}\right)^{\dim X-1}.$$ By the arithmetic ampleness $\left(\overline{L}|_{\mathrm{div}(s)}\right)^{\dim X-1}\ge 0$. Thus $$\int_{X(\mathbb{C})}\log |s|c_1\left(\overline{L}\right)^{\dim X-1}\ge -n\overline{L}^{\dim X}.$$ Let us write $|s|_{\alpha}$ for the norm of $s$ with respect to $\overline{L}_{\alpha}^{\otimes n}$. As $|s|=\sqrt{h^{\otimes n}(s,s)}$, this means $|s|_{\alpha}=\sqrt{\alpha^n h^{\otimes n}(s,s)}=\alpha^{n/2}|s|$. Thus \begin{align*} \int_{X(\mathbb{C})}\log |s|_{\alpha}c_1\left(\overline{L}\right)^{\dim X-1}&=\frac{n}{2}\log \alpha\int_{X(\mathbb{C})}c_1\left(\overline{L}\right)^{\dim X-1}+\int_{X(\mathbb{C})}\log |s|c_1\left(\overline{L}\right)^{\dim X-1}\\ &\ge \frac{n}{2}(\log \alpha) L_{\mathbb{C}}^{\dim X-1}-n\overline{L}^{\dim X}\\ &=\frac{n}{2}\left((\log\alpha)L_{\mathbb{C}}^{\dim X-1}-2\overline{L}^{\dim X}\right) \end{align*} If $\alpha\ge\alpha_0=\exp\left(\frac{2\overline{L}^{\dim X}}{L_{\mathbb{C}}^{\dim X-1}}\right)$, then $$\log\sup|s|_{\alpha}\ge\frac{\int_{X(\mathbb{C})}\log |s|_{\alpha}c_1\left(\overline{L}\right)^{\dim X-1}}{\int_{X(\mathbb{C})}c_1\left(\overline{L}\right)^{\dim X-1}}\ge0.$$ Thus, $\sup|s|_{\alpha}\ge 1$ such that $s$ is not strictly small with respect to $\overline{L}_{\alpha}^{\otimes n}$. As $s$ and $n$ were arbitrary, $\overline{L}_{\alpha}^{\otimes n}$ does not have any strictly small global sections for any $n$. In particular, $\overline{L}_{\alpha}$ is not arithmetically ample for every $\alpha\ge\alpha_0$. Note, that by the ampleness of $L_{\mathbb{C}}$ we always have $L_{\mathbb{C}}^{\dim X-1}>0$ such that $\alpha_0$ is always well-defined.

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  • $\begingroup$ Do you think this can explained more plainly, for example by suggesting when $\alpha\rightarrow \infty$, this makes the number of sections with $\sup |s|<1$ smaller? While the question asked by OP is intuitively false by this naive heuristic, it seems the finite place contribution and ampleness is still important here. But can this be done via Minkowski theory, etc? Maybe a silly question... $\endgroup$ Commented Sep 29, 2022 at 4:37
  • $\begingroup$ Right, for increasing $\alpha$ the norm of any section increases (by multiplication by $\alpha^{n/2}$). Thus, for some $\alpha$ and a fixed $n$ no section has sup-norm smaller 1. That's the intuition. My calculation just shows, that this happens uniformly for all $n$ if $\alpha\to\infty$. $\endgroup$ Commented Sep 30, 2022 at 6:24

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