The Spectral Hermitian Surface $(\Bbb C^\times, g_{\infty})$ as an Avatar of a Modular Surface

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Consider $\mathcal F=\lbrace \mathcal L_t=(\Bbb C^{\times},g_t(s)) \rbrace_{t\in\Bbb R_{\gt 0}}$ where $g_t(s)=\vert F_t(s) \vert^2ds\otimes d\bar{s}$ and


$$F_t(s)=\int_{(0,1)} e^{t^2/\log x} \cdot x^{s-1}dx= 2\sqrt{\frac{t^2}{s}}K_1(2\sqrt{t^2s}). $$


For $K_1$ the modified bessel function of the second kind.


Looking at Bessel asymptotics we know that as $s\to 0$, $g_t(s)\sim \frac{1}{|s|^2}ds\otimes d\bar{s}$ (conical) and as $s\to\infty$, $g_t(s)\sim e^{-4\sqrt{t^2s}}ds\otimes d\bar{s}$ (collapsing end), and these asymptotics hold for the later defined $g_{\infty}(s).$ 


Define a trace over $F_t$


$$\mathcal M(s)=\sum_{t=1}^\infty F_t(s)$$


then the following identity is satisfied  


$$\frac{1+2\mathcal M(s)}{1+2\mathcal M(1/s)}= s^{\alpha}$$


where $\alpha$ is the weight. The proof that there exists some $\alpha \in \Bbb R$ such that the identity holds for all $s$ uses Poisson summation. 


Define a new metric which encodes the cumulative effect of the metrics on each of the surfaces $\mathcal L_t$ into one


$$g_{\infty}(s)=:\vert \Phi(s) \vert^2ds\otimes d\bar{s}$$


where we have


$$\Phi(s)=1+2\mathcal M(s)$$ then 


$$\Phi(1/s)= s^{\alpha} \Phi(s)$$


and so if we define the Hermitian metric using


$$h(s)=\vert \Phi(s) \vert^2$$


then


$$g_{\infty}(s)=h(s) \cdot ds \otimes d\bar{s}$$


which transforms like


$$g_{\infty}(1/s)= \vert s \vert^{2(\alpha-2)} \cdot g_{\infty}(s)$$


So we have the conformal Hermitian surface $(\Bbb C^\times,g_{\infty})$ whose metric transforms under  $s\mapsto  1/s$ with a modular type scaling factor. 


Due to the asymptotics of the metric, this surface has an infinite cusp as $s\to\infty$ and expands outward as $s\to 0$ but not as a surface of revolution - the expansion is anisotropic. So the surface is like a warped cylinder that pinches in one direction and expands anisotropically in the other.

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