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Bergman kernels on punctured Riemann surfaces

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In this paper we consider a punctured Riemann surface endowed with a Hermitian metric that equals the Poincaré metric near the punctures, and a holomorphic line bundle that polarizes the metric. We show that the Bergman kernel can be localized around the singularities and its local model is the Bergman kernel of the punctured unit disc endowed with the standard Poincaré metric. One of the technical tools is a new weighted elliptic estimate near the punctures, which is uniform with respect to the tensor power. As a consequence, we obtain an optimal uniform estimate of the supremum norm of the Bergman kernel, involving a fractional growth order of the tensor power. This holds in particular for the Bergman kernel of cusp forms of high weight of non-cocompact geometrically finite Fuchsian groups of first kind without elliptic elements.

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Fig. 1
Fig. 2


  1. 1.

    As for the higher-dimensional local reference metric, which has to be taken as a perturbation of the product of the one-dimensional Poincaré cusp metric with some smooth metric on the divisor, its Bergman kernel can surely be properly understood as well, but with other approaches than the explicit description of this paper.

  2. 2.

    Conversely, one might imagine the parameter \(\alpha \) mentioned above (smooth case) go to 0 at speed 1/p, possibly together with the approximation of some Poincaré type metric by carefully chosen smooth metrics, to try and understand how the partial Bergman kernels approximate \(B_p\); for sure, some subtleties will occur along this double limit process, of which the wild transition region observed in Fig. 1 might be an artefact.


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H. A. is thankful to the University of Cologne where this paper was partly written; he would also like to thank Michael Singer for inspiring conversations. G. M. acknowledges support from Université Paris Diderot–Paris 7 (now Université de Paris) where this paper was partly written and warmly thanks the project Analyse Complexe et Géométrie for hospitality over many years.

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Correspondence to Xiaonan Ma.

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Hugues Auvray is partially supported by ANR contract ANR-14-CE25-0010. Xiaonan Ma is partially supported by ANR contract ANR-14-CE25-0012-01, NNSFC No. 11528103, No. 11829102 and funded through the Institutional Strategy of the University of Cologne within the German Excellence Initiative. George Marinescu is partially supported by CRC TRR 191.

Communicated by Ngaiming Mok.

A Proof of Lemma 3.4

A Proof of Lemma 3.4

We prove in this appendix the existence of a constant C such that for all \(\zeta >0\) and all \(p\ge 1\),

$$\begin{aligned} p(1+p(1-\zeta )^2)|\delta _p(\zeta )| \le C, \end{aligned}$$

where we recall the notation from (3.23):

$$\begin{aligned} \delta _p(\zeta ) = e^{p(1-\zeta +\log \zeta )} - e^{-\frac{p}{2}(1-\zeta )^2} \left( 1-\frac{p}{3}(1-\zeta )^3 \right) . \end{aligned}$$

Clearly, (A.1) holds for any fixed p, that is: for any \(p\ge 1\), there exists \(C_p\) such that for all \(\zeta >0\), \(p(1+p(1-\zeta )^2)|\delta _p(\zeta )| \le C_p\). We thus want to show that the \(C_p\) can be chosen independent of p; this we do arguing by contradiction: we hence assume that there exists a positive sequence \((\zeta _p)_{p\ge 1}\) such that, up to passing to a subsequence,

$$\begin{aligned} p(1+p(1-\zeta _p)^2)|\delta _p(\zeta _p)| \xrightarrow {p\rightarrow \infty } \infty . \end{aligned}$$

Up to passing to a subsequence, \((\zeta _p)\) converges in \([0,\infty ]\); we first distinguish the three cases \((\zeta _p)\rightarrow 0\), \((\zeta _p)\rightarrow \infty \), and \((\zeta _p)\rightarrow \ell \in \mathbb {R}_{>0}\).

  1. 1.

    \((\zeta _p)\rightarrow 0\): here, \(p(1+p(1-\zeta _p)^2)\sim p^2\), whereas \(e^{p(1-\zeta _p+\log \zeta _p)}\le e^{-p}\) for p large, hence

    $$\begin{aligned} p(1+p(1-\zeta _p)^2)e^{p(1-\zeta _p+\log \zeta _p)} \longrightarrow 0; \end{aligned}$$

    likewise, \(e^{-\frac{p}{2}(1-\zeta )^2}\le e^{-\frac{p}{4}}\) for p large, and \(\big (1-\frac{p}{3}(1-\zeta )^3\big )\sim -\frac{p}{3}\), hence

    $$\begin{aligned} p(1+p(1-\zeta _p)^2)e^{-\frac{p}{2}(1-\zeta _p)^2} \left( 1-\frac{p}{3}(1-\zeta _p)^3 \right) \longrightarrow 0. \end{aligned}$$

    This way, \(p(1+p(1-\zeta _p)^2)|\delta _p(\zeta _p)|\rightarrow 0\), which contradicts (A.2).

  2. 2.

    \((\zeta _p)\rightarrow \infty \): one has \(p(1+p(1-\zeta _p)^2)\sim p^2\zeta _p^2\) and \(e^{p(1-\zeta _p+\log \zeta _p)}\le e^{-\frac{p\zeta _p}{2}}\) for p large, hence

    $$\begin{aligned} p(1+p(1-\zeta _p)^2)e^{p(1-\zeta _p+\log \zeta _p)} \lesssim p^2\zeta _p^2e^{-\frac{p\zeta _p}{2}} \longrightarrow 0; \end{aligned}$$

    moreover \(e^{-\frac{p}{2}(1-\zeta _{p})^2} \le e^{-\frac{p\zeta _p}{4}}\) for p large, and \(\big (1-\frac{p}{3}(1-\zeta _{p})^3\big )\sim \frac{p}{3}\zeta _p^3\), hence

    $$\begin{aligned}&p(1+p(1-\zeta _p)^2)e^{-\frac{p}{2}(1-\zeta _p)^2} \left( 1-\frac{p}{3}(1-\zeta _p)^3 \right) \\&\quad \lesssim \frac{p}{3}\zeta _p^3e^{-\frac{p\zeta _p}{4}} = \frac{1}{3p^2} (p\zeta _p)^3e^{-\frac{p\zeta _p}{4}} \longrightarrow 0. \end{aligned}$$

    Here again, \(p(1+p(1-\zeta _p)^2)|\delta _p(\zeta _p)|\rightarrow 0\), and (A.2) is contradicted.

  3. 3.

    \((\zeta _p)\rightarrow \ell \): one must deal here with the dichotomy \(\ell \ne 1\Big {/}\ell =1\).

    1. (a)

      \(\ell \ne 1\): \(p(1+p(1-\zeta _p)^2) \sim (1-\ell )^2p^2\), and as \(\zeta \mapsto 1-\zeta +\log \zeta \) is strictly convex and attains 0 at \(\zeta =1\), \(e^{p(1-\zeta _p+\log \zeta _p)}\le e^{-\varepsilon p}\) for p large, with some \(\varepsilon >0\), hence

      $$\begin{aligned} p(1+p(1-\zeta _p)^2)e^{p(1-\zeta _p+\log \zeta _p)} \lesssim (1-\ell )^2p^2e^{-\varepsilon p} \longrightarrow 0; \end{aligned}$$

      furthermore, \(e^{-\frac{p}{2}(1-\zeta _{p})^2}\le e^{-\frac{p(1-\ell )^2}{4}}\) for p large, and \(\big (1-\frac{p}{3}(1-\zeta _{p})^3\big )\sim \frac{p}{3}(1-\ell )^3\), hence

      $$\begin{aligned}&p\left( 1+p(1-\zeta _p)^2 \right) e^{-\frac{p}{2}(1-\zeta _p)^2} \left| 1-\frac{p}{3}(1-\zeta _p)^3 \right| \nonumber \\&\quad \lesssim \frac{p^3}{3}|1-\ell |^5e^{-\frac{p(1-\ell )^2}{4}} \longrightarrow 0. \end{aligned}$$

      Once more, this yields \(p(1+p(1-\zeta _p)^2)|\delta _p(\zeta _p)|\rightarrow 0\), and a contradiction to (A.2).

    2. (b)

      \(\ell = 1\): setting \(z_p=\zeta _p-1\rightarrow 0\), we must one last time distinguish between three different cases: \(|pz_p^3|\rightarrow \infty \), \(pz_p^3\rightarrow \mu =\lambda ^3\in \mathbb {R}^*\) and \(pz_p^3\rightarrow 0\) (up to passing to a subsequence).

      1. (i)

        \(|pz_p^3|\rightarrow \infty \): in particular, \(pz_p^2=|pz_p^3|^{2/3}p^{1/3}\rightarrow \infty \). Here \(p(1+p(1-\zeta _p)^2)\sim p^2z_p^2\) and \(1-\zeta _p+\log \zeta _p = -z_p +\log (1+z_p) \le -\frac{z_p^2}{3}\) for p large, thus

        $$\begin{aligned}&p(1+p(1-\zeta _p)^2)e^{p(1-\zeta _p+\log \zeta _p)} \lesssim p^2z_p^2e^{-\frac{pz_p^2}{3}} = p^2z_p^2e^{-\frac{|pz_p^3|^{2/3}p^{1/3}}{3}} \nonumber \\&\quad \le p^2z_p^2e^{-(|pz_p^3|^{2/3}+p^{1/3})} =p^{4/3}e^{-p^{1/3}} |pz_p^3|^{2/3}e^{-|pz_p^3|^{2/3}} \longrightarrow 0,\nonumber \\ \end{aligned}$$

        where the last inequality holds as soon as \(p^{1/3}\) and \(|pz_p^3|^{2/3}\ge 6\). As for the other summand,

        $$\begin{aligned}&p(1+p(1-\zeta _p)^2) e^{-\frac{pz_p^2}{2}} \left| 1-\frac{p}{3}z_p^3 \right| \nonumber \\&\quad \sim |p^3z_p^5|e^{-\frac{pz_p^2}{2}} = p^{4/3}|pz_p^3|^{5/3}e^{-\frac{|pz_p^3|^{2/3}p^{1/3}}{2}} \nonumber \\&\quad \le p^{4/3}|pz_p^3|^{5/3}e^{-(|pz_p^3|^{2/3}+p^{1/3})} =p^{4/3}e^{-p^{1/3}} |pz_p^3|^{5/3}e^{-|pz_p^3|^{2/3}} \longrightarrow 0, \end{aligned}$$

        (the last inequality holds as soon as \(p^{1/3}\) and \(|pz_p^3|^{2/3}\ge 6\)). In conclusion, \(p(1+p(1-\zeta _p)^2)|\delta _p(\zeta _p)|\rightarrow 0\), in contradiction with (A.2).

      2. (ii)

        \(pz_p^3\rightarrow \mu =\lambda ^3 \ne 0\), that is: \(z_p\sim \lambda p^{-1/3}\). First, \(p(1+p(1-\zeta _p)^2) \sim p^2z_p^2 \sim \lambda ^2p^{4/3}\); moreover \(1-\zeta _p+\log \zeta _p = -z_p +\log (1+z_p) = - \frac{z_p^2}{2} + \frac{z_p^3}{3} + \mathcal {O}(p^{-4/3})\), hence

        $$\begin{aligned} \begin{aligned} p(1+p(1-\zeta _p)^2)|\delta _p(\zeta _p)|&\sim \lambda ^2p^{4/3}e^{-\frac{pz_p^2}{2}} \left| e^{\frac{pz_p^3}{3} + \mathcal {O}(p^{-1/3})}-1 +\frac{pz_p^3}{3} \right| \\&\lesssim \lambda ^2p^{4/3}e^{-\frac{\lambda ^2p^{1/3}}{3}} \left| e^{\mu /3}-1-\frac{\mu }{3} \right| \longrightarrow 0, \end{aligned}\end{aligned}$$

        a contradiction with (A.2).

      3. (iii)

        \(pz_p^3\rightarrow 0\): In this very last case, where again \(pz_p^4=pz_p^3\cdot z_p\rightarrow 0\),

        $$\begin{aligned} \delta _p(\zeta _p)&= e^{-\frac{pz_p^2}{2}} \left( e^{p(\frac{z_p^3}{3}+\mathcal {O}(z_p^4))} - 1-\frac{p}{3}z_p^3 \right) \nonumber \\&= e^{-\frac{pz_p^2}{2}} \left( \left( 1+\frac{pz_p^3}{3} +\mathcal {O}(p^2z_p^6)\right) \left( 1+\mathcal {O}(pz_p^4)\right) - 1-\frac{p}{3}z_p^3 \right) \nonumber \\&= e^{-\frac{pz_p^2}{2}}\left( \mathcal {O}(p^2z_p^6) +\mathcal {O}(pz_p^4)\right) , \end{aligned}$$


        $$\begin{aligned} (p+p^2z_p^2)\delta _p(\zeta _p) =&\mathcal {O}\left( p^3z_p^6e^{-\frac{pz_p^2}{2}}\right) + \mathcal {O}\left( p^2z_p^4e^{-\frac{pz_p^2}{2}}\right) + \mathcal {O}\left( p^4z_p^8e^{-\frac{pz_p^2}{2}}\right) \nonumber \\ =&\mathcal {O}(1) \end{aligned}$$

        independently of the behavior of \(p^2z_p^2\), as \(z\mapsto z^ke^{-\frac{z^2}{2}}\), \(k=2,3,4\), are bounded on \(\mathbb {R}\). In other words, \(p(1+p(1-\zeta _p)^2)\delta _p(\zeta _p) =\mathcal {O}(1)\), and this final contradiction of (A.2) ends the proof of Lemma 3.4. \(\square \)

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Auvray, H., Ma, X. & Marinescu, G. Bergman kernels on punctured Riemann surfaces. Math. Ann. (2020).

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