Correction to: A Polyakov Formula for Sectors

1 Correction to: J Geom Anal (2018) 28:1773–1839 https://doi.org/10.1007/s12220-017-9888-y

Let $S_\alpha$ denote a finite circular sector of opening angle $\alpha \in (0,\pi )$ and radius one, and let $e^{-t \Delta _\alpha }$ denote the heat operator associated to the Dirichlet extension of the Laplacian. Based on recent joint work [2] and [3], we discovered an extra contribution to the variational Polyakov formula in [1] coming from the curved boundary component of the sector. Theorems 3 and 4 of [1] should have an added term $+\frac{1}{4\pi }$. This calculation will appear in [2]. The corrected statements of these theorems are given below.

Theorem 1

(Theorem 3 [1]) Let $S_{\pi /2}\subset {\mathbb {R}}^2$ be a circular sector of opening angle $\pi /2$ and radius one. Then the variational Polyakov formula is

$$\begin{aligned} \left. \frac{\partial }{\partial \gamma } \big (-\log (\det (\Delta _{S_\gamma }))\big )\right| _{\gamma =\pi /2} = \frac{-\gamma _e}{4\pi } + \frac{2}{3 \pi }, \end{aligned}$$

where $\gamma _e$ is the Euler-Mascheroni constant.

Theorem 2

(Theorem 4 [1]) Let $0<\alpha < \pi$, and let

$$\begin{aligned} k_{min} = \left\lceil { \frac{-\pi }{2\alpha } }\right\rceil , \text { and } k_{max} = \left\lfloor {\frac{\pi }{2\alpha }}\right\rfloor \text { if } \frac{\pi }{2\alpha } \not \in {\mathbb {Z}}, \text { otherwise } k_{max} = \frac{\pi }{2\alpha } - 1, \end{aligned}$$

and $W_{\alpha } =\left\{ k \in \left( {\mathbb {Z}}\bigcap \left[ k_{min}, k_{max}\right] \right) {\setminus } \left\{ \frac{\ell \pi }{\alpha } \right\} _{\ell \in {\mathbb {Z}}} \right\} .$ Then

$$\begin{aligned} {{\mathcal {S}}}(\alpha )&:=\frac{\partial }{\partial \gamma } \left. \big (-\log (\det (\Delta _{\gamma }))\big )\right| _{\gamma =\alpha } = \frac{1}{3\pi } + \frac{\pi }{12\alpha ^2} \\&\quad + \sum _{k\in W_{\alpha }} \frac{-2\gamma _e + \log (2) - \log \left( {1-\cos (2k\alpha )}\right) }{4 \pi (1-\cos (2k\alpha ))} \\&\quad - (1-\delta _{\alpha , \frac{\pi }{n}})\ \frac{2}{\alpha } \sin (\pi ^2/\alpha ) \int _{-\infty } ^\infty \frac{\gamma _e + \log (2) - \log (1+\cosh (s))}{16\pi (1+\cosh (s))(\cosh (\pi s /\alpha ) - \cos (\pi ^2/\alpha ))} ds, \end{aligned}$$

where $n\in {\mathbb {N}}$ is arbitrary and $\delta _{\alpha , \frac{\pi }{n}}$ denotes the Kronecker delta.

It therefore follows that the list of examples given following Theorem 4 in [1] should be revised accordingly:

• (1) $\alpha =\frac{\pi }{4}$, $W_{\frac{\pi }{4}}= \{-2,\pm 1,\}$, ${{\mathcal {S}}}(\frac{\pi }{4})= \frac{-5\gamma _e}{4\pi } + \frac{\log (2)}{2\pi }+ \frac{5}{3\pi } \sim 0.411167$

• (2) $\alpha =\frac{\pi }{3}$, $W_{\frac{\pi }{3}}= \{-1,1\}$, ${{\mathcal {S}}}(\frac{\pi }{3})= \frac{13}{12\pi } - \frac{2\gamma _e}{3\pi } + \frac{\log (4/3)}{3\pi } \sim 0.252871$

• (3) $\alpha =\frac{\pi }{2}$, $W_{\frac{\pi }{2}}=\{-1\}$, ${{\mathcal {S}}}(\frac{\pi }{2})= \frac{-\gamma _e}{4\pi } + \frac{2}{3\pi } \sim 0.166273$.

• (4) For $\alpha \in ]\frac{\pi }{2},\pi [$, $W_{\alpha }=\emptyset$, but $\sin (\pi ^2/\alpha )\ne 0$. If $\alpha =\frac{2\pi }{3}$, the integral converges rapidly, and a numerical computation gives an approximate value of 0.0075015. Hence ${{\mathcal {S}}}(\frac{2\pi }{3}) \sim \frac{1}{3\pi } + \frac{3}{16 \pi } + \frac{3}{\pi } (0.0075015) \sim 0.1729498$.

2 Misprint

There is a two missing in Equation (1.3) of [1]. That equation should be:

$$\begin{aligned} \partial _t \log \det (\Delta _{g_t}) = {-\frac{1}{12\pi }} \int _{M} \sigma '(t) \ \text {Scal}_t \ dA_{g_t} + \partial _t \log \text {Area}(M,g_t). \end{aligned}$$