I can't believe I overlooked this! This is huge!
\( \vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ \sigma}(a) + h_e(\sigma)) \)
\( h_e(\sigma)=\left{\begin{array}{c l}
\exp_\eta^{\circ -\sigma}(e) & \sigma \le 1\\
\exp_\eta^{\circ -1}(e) & \sigma \in [1,2]
\end{array}\right. \)
but since \( \exp_\eta(e) = e \), \( h_e(\sigma) = e \)
therefore:
\( \vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ -\sigma}(a) + e) \)
And therefore \( \vartheta \) is potentially analytic, and isn't piecewise, for all a, and b = e, \( \sigma \in (-\infty, 2] \)
This is crazy.
\( \vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ \sigma}(a) + h_e(\sigma)) \)
\( h_e(\sigma)=\left{\begin{array}{c l}
\exp_\eta^{\circ -\sigma}(e) & \sigma \le 1\\
\exp_\eta^{\circ -1}(e) & \sigma \in [1,2]
\end{array}\right. \)
but since \( \exp_\eta(e) = e \), \( h_e(\sigma) = e \)
therefore:
\( \vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ -\sigma}(a) + e) \)
And therefore \( \vartheta \) is potentially analytic, and isn't piecewise, for all a, and b = e, \( \sigma \in (-\infty, 2] \)
This is crazy.