The fiber potential will be built from the heat kernel of $E$. It has large vertical energy at small times and decays at large times, where it can be cut off, and a lower bound for its logarithmic Hessian controls vertical positivity. To control the radial–vertical block when the time varies, we also need an estimate relating the kernel to its logarithmic gradient.
For $t>0$ define
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23.1.10
\begin{equation} k_ t(w)=\frac2t\sum _{\lambda \in \Lambda } \exp \left(-\frac{|w-\lambda |^2}{t}\right). \end{equation}
This is a smooth positive $\Lambda $-periodic function of integral one: unfolding the integral reduces it to the Gaussian integral on $\mathbb C$, since $\mathrm{d}A_ E=\mathrm{d}a\, \mathrm{d}b/(2\pi )$ for $w=a+\mathrm ib$. For $(p,q)\in \mathbb Z^2$ set
\[ \nu _{p,q}=p+\frac{\mathrm iq}{2\pi },\qquad \chi _{p,q}(w)=\mathrm{e}^{2\pi \mathrm ip\operatorname {Re}w+ \mathrm iq\operatorname {Im}w}. \]
These characters form an orthonormal basis of $L^2(E,\mathrm{d}A_ E)$. Computing the Fourier coefficients by the same unfolding gives
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23.1.11
\begin{equation} k_ t=\sum _{p,q}\mathrm{e}^{-\pi ^2|\nu _{p,q}|^2t}\chi _{p,q}, \qquad \partial _ tk_ t=\partial _ w\partial _{\bar w}k_ t. \end{equation}
All differentiated series converge locally uniformly for $t>0$.
Lemma 23.1.12.14E6CDA9AAC4 Let $0<\eta <1/10$ and define
\[ F_\eta (t,w)=-\eta \int _ t^\infty (k_ r(w)-1)\, \mathrm{d}r. \]
Then $F_\eta $ is smooth and real, and
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23.1.13
\begin{align} (F_\eta )_ t=(F_\eta )_{w\bar w} & =\eta (k_ t-1),\\ (\log k_ t)_{w\bar w}& \geq -1/t,\tag{23.1.14}\\ \int _ E|(F_\eta )_ w(t,w)|^2\mathrm{d}A_ E & =2\eta ^2\log (1/t)+O(\eta ^2)\quad (0<t\leq 1). \tag{23.1.15} \end{align}
The error term in (23.1.15) is uniform in $t$ and $\eta $. For fixed nonnegative integers $a,m$ there is also a uniform estimate
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23.1.16
\begin{equation} \| \partial _ t^ aF_\eta (t,\cdot )\| _{C^ m(E)} \leq C_{a,m}\eta \mathrm{e}^{-t/4}\quad (t\geq 1). \end{equation}
Proof.
Integrating the nonconstant Fourier modes gives
\[ F_\eta (t,w)=-\frac{\eta }{\pi ^2} \sum _{(p,q)\ne (0,0)} \frac{\mathrm{e}^{-\pi ^2|\nu _{p,q}|^2t}}{|\nu _{p,q}|^2}\chi _{p,q}(w). \]
This proves smoothness, reality and (23.1.13). The least nonzero value of $\pi ^2|\nu _{p,q}|^2$ is $1/4$, and any fixed number of differentiations introduces coefficients of polynomial growth in $|\nu _{p,q}|$. Factoring out $\mathrm{e}^{-t/4}$ for $t\geq 1$ therefore leaves a convergent Gaussian series, which proves (23.1.16).
To prove (23.1.14), fix a lift of $w$ and give $\Lambda $ the probability weights
\[ P_\lambda = \frac{\mathrm{e}^{-|w-\lambda |^2/t}}{\sum _{\lambda '}\mathrm{e}^{-|w-\lambda '|^2/t}}. \]
Logarithmic differentiation of the Gaussian sum yields
\[ t\partial _{\bar w}\log k_ t=-\sum _\lambda (w-\lambda )P_\lambda , \]
and one more derivative gives
\[ (\log k_ t)_{w\bar w} =-\frac1t+\frac1{t^2} \left(\sum _\lambda |w-\lambda |^2P_\lambda -\left|\sum _\lambda (w-\lambda )P_\lambda \right|^2\right). \]
The expression in parentheses is a variance and is nonnegative.
Finally, Parseval's identity gives, with $\mathcal E_\eta (t)$ denoting the energy in (23.1.15),
\[ \mathcal E_\eta (t)=\frac{\eta ^2}{\pi ^2} \sum _{(p,q)\ne (0,0)} \frac{\mathrm{e}^{-2\pi ^2|\nu _{p,q}|^2t}}{|\nu _{p,q}|^2}, \qquad \mathcal E_\eta '(t)=-2\eta ^2(k_{2t}(0)-1). \]
The Gaussian formula at the origin implies
\[ k_{2t}(0)=\frac1t+ O\left(t^{-1}\mathrm{e}^{-1/(4t)}\right)\quad (0<t\leq 1). \]
To see this, isolate the zero lattice vector and use $\mathrm{e}^{-|\lambda |^2/(2t)}\leq \mathrm{e}^{-1/(4t)}\mathrm{e}^{-|\lambda |^2/4}$ for $\lambda \ne 0$. The error term is integrable near $t=0$, and $\mathcal E_\eta (1)=O(\eta ^2)$. Integrating the derivative between $t$ and $1$ proves (23.1.15).
$\square$
The Schur complement subtracts a square involving the mixed derivative. When the time parameter is allowed to vary, the heat equation produces a compensating positive square, and the following bound relating the heat kernel to its logarithmic gradient controls the region where that square is small.
Lemma 23.1.17.7C9989D90F66 For a constant $C_0\geq 1$ depending only on $E$,
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23.1.18
\begin{equation} tk_ t(w)\leq C_0\exp \left( -\frac{|t\partial _{\bar w}\log k_ t(w)|^2}{C_0t}\right) \quad (0<t\leq 1). \end{equation}
Proof.
Choose a nearest lattice point $\lambda _0$, and put $d=|w-\lambda _0|$ and $s_\lambda =|w-\lambda |^2-d^2\geq 0$. The distance $d$ is bounded by the diameter of a fundamental rectangle. By the triangle inequality, $s_\lambda \geq |\lambda -\lambda _0|^2/4-C$, and outside a fixed finite set of $\lambda $ we have $s_\lambda \geq |\lambda -\lambda _0|^2/8$. Consequently,
\[ \sum _\lambda (1+|w-\lambda |)\mathrm{e}^{-s_\lambda /t}\leq C, \qquad tk_ t(w)\leq C\mathrm{e}^{-d^2/t}. \]
The probability formula in the previous proof gives, with $G=|t\partial _{\bar w}\log k_ t|$,
\[ G\leq d+\sum _{\lambda \ne \lambda _0}|w-\lambda |\mathrm{e}^{-s_\lambda /t}. \]
If $d\geq 1/4$, this is at most $Cd$. If $d<1/4$, the nonzero lattice distances are at least one and
\[ s_\lambda \geq |\lambda -\lambda _0| (|\lambda -\lambda _0|-2d) \geq \tfrac 12|\lambda -\lambda _0|^2. \]
Summing the resulting Gaussian bound gives $G\leq C(d+\mathrm{e}^{-c/t})$ in all cases. When $G>2C\mathrm{e}^{-c/t}$, we have $d\geq G/(2C)$, and the bound $tk_ t\leq C\mathrm{e}^{-d^2/t}$ gives (23.1.18). In the remaining case $G^2/t$ is bounded, and (23.1.18) follows from $tk_ t\leq C$ after increasing $C_0$.
$\square$
Lemma 23.1.19.AB7F46061184 There is a constant $C$ depending only on $E$ with the following property. For $0<\kappa <1/(200C_0)$, $0<\eta <1/10$, $0<t\leq 1$ and $1/2\leq \mu \leq 1$, define $\sigma >0$ and the functions $h,\zeta $ of $w\in E$ by
\[ \sigma ^2=\frac{2\kappa }{\eta }\log \frac1{\eta t},\qquad h=\eta k_ t,\qquad \zeta =\frac{\sigma }{\mu }\, t\partial _{\bar w}\log k_ t. \]
Then
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23.1.20
\begin{equation} \kappa tk_ t- \frac{\mu ^2(1-\eta )h}{1-\eta +h} \left|\zeta -\frac1{1-\eta }\right|^2 \leq C\kappa \exp \left(-\frac1{C\sigma ^2t}\right). \end{equation}
Proof.
If $\sigma ^2t\geq 1/100$, the bound $tk_ t\leq C_0$ gives (23.1.20) after enlarging $C$. Assume therefore that $\sigma ^2t<1/100$. If $|\zeta -(1-\eta )^{-1}|\geq 1/4$, the subtracted expression is at least $h/[100(1+h)]$. Since $\log (1/(\eta t))>2$, the definition of $\sigma $ gives
\[ \kappa t(1+h)\leq \frac{\eta }{400}+C_0\kappa \eta <\frac{\eta }{100}. \]
Multiplication by $k_ t/(1+h)$ shows that the left side of (23.1.20) is nonpositive.
Otherwise $|\zeta |\geq 3/4$, so
\[ |t\partial _{\bar w}\log k_ t|=\frac{\mu |\zeta |}{\sigma } \geq \frac3{8\sigma }. \]
Discarding the nonnegative square and applying lemma 23.1.17 gives (23.1.20) in this case as well.
$\square$
Pome
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