One-loop Correction to the AdS/BCFT Partition Function in the Three Dimensional Pure Gravity

Let us consider an Euclidean rotating BTZ black hole in three dimension (Banados:1992gq, ). The metric is given by

	$\displaystyle ds^{2}$	$\displaystyle=$	$\displaystyle\frac{(r^{2}-R_{+}^{2})(r^{2}+R_{-}^{2})}{r^{2}}dt^{2}+\frac{r^{2}}{(r^{2}-R_{+}^{2})(r^{2}+R_{-}^{2})}dr^{2}$		(4)
		$\displaystyle+$	$\displaystyle r^{2}(d\phi-\frac{R_{+}R_{-}}{r^{2}}dt)^{2}.$

Here $R_{-}$ is a real valued parameter and $R_{+}$ is the horizon radius. The absence of a conical singularity constrains the periodicity of time and rotational angle:

$$\left(t,\phi\right)\sim\left(t+\beta,\phi+\theta\right),$$

(5)

where $\beta=\frac{2\pi R_{+}}{R_{+}^{2}+R_{-}^{2}}$ and $\theta=\frac{2\pi R_{-}}{R_{+}^{2}+R_{-}^{2}}$. If we define new coordinate as $\phi^{\prime}\equiv\phi-\frac{\theta}{\beta}t$ , the periodicity can be recast as $\left(t,\phi^{\prime}\right)\sim\left(t+\beta,\phi^{\prime}\right)$ (Carlip:1994gc, ). We also rewrite the metric in terms of this coordinate as

$$g_{ab}=\left(\begin{array}[]{ccc}A&0&B\\ 0&C&0\\ D&0&E\end{array}\right),$$

(6)

where

	$\displaystyle A$	$\displaystyle=$	$\displaystyle\frac{(r^{2}-R_{+}^{2})(r^{2}+R_{-}^{2})}{r^{2}}+r^{2}\left(\frac{R_{-}}{R_{+}}-\frac{R_{+}R_{-}}{r^{2}}\right)^{2},$
	$\displaystyle B$	$\displaystyle=$	$\displaystyle\frac{R_{-}}{R_{+}}r^{2}-R_{+}R_{-},$
	$\displaystyle C$	$\displaystyle=$	$\displaystyle\frac{r^{2}}{(r^{2}-R_{+}^{2})(r^{2}+R_{-}^{2})},$
	$\displaystyle D$	$\displaystyle=$	$\displaystyle\frac{R_{-}}{R_{+}}r^{2}-R_{+}R_{-},$
	$\displaystyle E$	$\displaystyle=$	$\displaystyle r^{2}.$		(7)

Note that det$(g_{ij})=r^{2}.$ To identify the location of an ETW brane we use the map to the Poincare coordinate:

$$\begin{array}[]{cc}\eta=\left(\frac{r^{2}-R_{+}^{2}}{r^{2}+R_{-}^{2}}\right)^{\frac{1}{2}}\cos\left(\frac{R_{+}^{2}+R_{-}^{2}}{R_{+}}t+R_{-}\phi^{\prime}\right)\exp\left(R_{+}\phi^{\prime}\right),\\ x=\left(\frac{r^{2}-R_{+}^{2}}{r^{2}+R_{-}^{2}}\right)^{\frac{1}{2}}\sin\left(\frac{R_{+}^{2}+R_{-}^{2}}{R_{+}}t+R_{-}\phi^{\prime}\right)\exp\left(R_{+}\phi^{\prime}\right),\\ z=\left(\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}}\right)^{\frac{1}{2}}\exp\left(R_{+}\phi^{\prime}\right),\end{array}$$

(8)

which leads to the familiar metric

$$ds^{2}=\frac{dz^{2}+dx^{2}+d\eta^{2}}{z^{2}}.$$

(9)

In this coordinate the identification $(t,\phi^{\prime})\sim(t+\beta,\phi^{\prime})$ is trivial, but the identification $(t,\phi^{\prime})\sim(t,\phi^{\prime}+2\pi)$ is non-trivial. If we define the following complexified coordinate $w=\eta+ix$, then the identification is translated into $(w,z)\sim(we{}^{2\pi(R_{+}+iR_{-})},ze^{2\pi R_{+}})$. From this perspective, we can take the fundamental region as $1\leq\left|w\right|^{2}+z^{2}\leq e^{2\pi R_{+}}$ and the horizon $r=R_{+}$ is mapped to $w=0$ and $1\leq z\leq e{}^{2\pi R_{+}}$.

Let us insert the ETW branes in this setup. The position of ETW branes with a tension $T$ is determined by the Neumann boundary condition in the AdS/BCFT:

$$K_{ab}-Kh_{ab}=-Th_{ab}.$$

(10)

In the Poincare coordinate, the solution to the boundary condition (10) is solved as follows (Akal:2020wfl, ):

$$(z-\alpha)^{2}+(x-p)^{2}+(\eta-q)^{2}=\beta^{2},$$

(11)

where the tension is given by $T=\frac{\alpha}{\beta}$. Here ETW branes, where $\phi^{\prime}$ is constant, are just a sphere of radius 1 for $\phi^{\prime}=0$ and $e^{\pi R_{+}}$ for $\phi^{\prime}=\pi$. Next, we calculate the action

	$\displaystyle S$	$\displaystyle=$	$\displaystyle-\frac{1}{16\pi G}\int_{N}\sqrt{g}(R+2)d^{3}x-\frac{1}{8\pi G}\int_{Q}\sqrt{h}(K-T)d^{2}x$		(12)
		$\displaystyle-$	$\displaystyle\frac{1}{8\pi G}\int_{M}\sqrt{h}Kd^{2}x+\frac{k}{8\pi G}S_{ct},$

where $N$ is the three-dimensional bulk region, $Q$ is the ETW brane and $M$ is the conformal boundary placed at $r=R$, which we later take $R\rightarrow\infty$. $S_{ct}=\int_{M}\sqrt{h}$ is a counter-term constructed only from induced geometric quantities (Balasubramanian:1999re, ).

Consider computing the induced metric and the extrinsic curvature at $r=R$ surface. After some algebras, we get

$$h_{ab}=\left(\begin{array}[]{ccc}A^{\prime}&0&B^{\prime}\\ 0&0&0\\ B&0&C^{\prime}\end{array}\right),$$

(13)

where

	$\displaystyle A^{\prime}$	$\displaystyle=$	$\displaystyle=\frac{(r^{2}-R_{+}^{2})(r^{2}+R_{-}^{2})}{R^{2}}+R^{2}\left(\frac{R_{-}}{R_{+}}-\frac{R_{+}R_{-}}{R^{2}}\right)^{2},$
	$\displaystyle B^{\prime}$	$\displaystyle=$	$\displaystyle\frac{R_{-}}{R_{+}}R^{2}-R_{+}R_{-},$
	$\displaystyle C^{\prime}$	$\displaystyle=$	$\displaystyle R^{2},$		(14)

and the extrinsic curvature reads

$$K_{ab}=\left(\begin{array}[]{cc}\frac{R_{+}^{2}+R_{-}^{2}}{R_{+}^{2}}&\frac{R_{-}}{R_{+}}\\ \frac{R_{-}}{R_{+}}&1\end{array}\right)\sqrt{\left(R^{2}-R_{+}^{2}\right)\left(r^{2}+R_{-}^{2}\right)}.$$

(15)

From this we find

$$\begin{array}[]{cc}\det(h_{ab})=(R^{2}-R_{+}^{2})(R^{2}+R_{-}^{2}),\\ K=\frac{2R^{2}-R_{+}^{2}+R_{-}^{2}}{\sqrt{(R^{2}-R_{+}^{2})(R^{2}+R_{-}^{2})}}.\end{array}$$

(16)

Now, the Einstein-Hilbert action is just the volume integral:

$$-\frac{1}{16\pi G}\int_{R_{+}}^{R}dr\int_{0}^{\beta}dt\int_{0}^{\pi}d\phi^{\prime}(-4r)=\frac{4\pi\beta}{16\pi G}\left(\frac{R^{2}-R_{+}^{2}}{2}\right).$$

(17)

We can also evaluate the GHY term at $r=R$:

	$\displaystyle-\frac{1}{8\pi G}\int_{0}^{\beta}\!dt\int_{\frac{2\pi}{\beta}\left(1-\frac{t}{\beta}\right)}^{\frac{2\pi}{\beta}\left(1-\frac{t}{\beta}\right)+\pi}\!d\phi^{\prime}\left[\sqrt{(R^{2}-R_{+}^{2})(R^{2}+R_{-}^{2})}\right.$
	$\displaystyle\left.\left(\frac{2R^{2}-R_{+}^{2}+R_{-}^{2}}{(R^{2}-R_{+}^{2})(R^{2}+R_{-}^{2})}\!\right)\!\right]\!=\!-\frac{\beta}{8G}(2R^{2}-R_{+}^{2}+R_{-}^{2}).$

Then, we determine the counterterm. That should be constructed from a geometric quantity of the boundary surface. Here as the most simple one ($k=1)$ we take

$$S_{ct}=\int_{M}d^{2}x\sqrt{h}=\beta\pi\sqrt{(R^{2}-R_{+}^{2})(R^{2}+R_{-}^{2})}.$$

(19)

We combine the above results altogether and we get:

$$S=-\frac{\beta}{8G}\left(R^{2}+R_{-}^{2}-\sqrt{(R^{2}-R_{+}^{2})(R^{2}+R_{-}^{2})}\right).$$

(20)

If we take $R\rightarrow\infty$ limit this becomes

$$S\simeq-\frac{\beta\left(R_{+}^{2}+R_{-}^{2}\right)}{16G}=-\frac{\pi R_{+}}{8G}=-\frac{\pi^{2}\beta}{4G(\beta^{2}+\theta^{2})}.$$

(21)

This is what we expected because we chose the position of the ETW branes so that the volume of the bulk space becomes a half of the original volume. In the full AdS case, this is done in (Maldacena:1998bw, ), which is just twice of the action in our calculation.

Next, we consider the general case with non-vanishing $T$. It is natural to assume the rotational symmetry i.e. setting $p=q=0$ in (11). Therefore, we will consider the following equation

$$\left(\!\left(\!\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}}\right)^{\frac{1}{2}}e^{R_{+}\phi^{\prime}}-\alpha\right)^{2}\!+\!\left(\!\frac{r^{2}-R_{+}^{2}}{r^{2}+R_{-}^{2}}\!\right)e^{2R_{+}\phi^{\prime}}=\beta^{2}.$$

(22)

Since the brane is anchored at $\phi^{\prime}=0$ and $\pi$, if we take $r\rightarrow\infty$, then the brane equation becomes

	$\displaystyle\alpha^{2}+1=\beta^{2}$	$\displaystyle\ \ (\phi^{\prime}=0),$
	$\displaystyle\alpha^{2}+\exp(2\pi R_{+})=\beta^{2}$	$\displaystyle\ \ (\phi^{\prime}=\pi).$			(23)

We note that $T$ has range $-1<T<1$ from the above constraint. Using this we can determine the brane configuration: in the case where the brane is anchored at $\phi^{\prime}=0$, the equation becomes

	$\displaystyle\phi^{\prime}$	$\displaystyle=$	$\displaystyle\frac{1}{R_{+}}\log\left(\frac{T}{\sqrt{1-T^{2}}}\sqrt{\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}}}\right.$		(24)
		$\displaystyle+$	$\displaystyle\left.\sqrt{\frac{T^{2}}{1-T^{2}}(\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}})+1}\right)$

and in the case where the brane is anchored at $\phi^{\prime}=\pi$, we have

	$\displaystyle\phi^{\prime}$	$\displaystyle=$	$\displaystyle\frac{1}{R_{+}}\log\left\{e^{\pi R_{+}}\left(\frac{T}{\sqrt{1-T^{2}}}\sqrt{\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}}}\right.\right.$		(25)
		$\displaystyle+$	$\displaystyle\left.\left.\sqrt{\frac{T^{2}}{1-T^{2}}(\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}})+1}\right)\right\}.$

Then the Einstein-Hilbert action can be calculated directly:

	$\displaystyle S_{EH}$	$\displaystyle=$	$\displaystyle-\frac{1}{16\pi G}\int_{R_{+}}^{R_{reg}}dr\int_{0}^{\beta}dt\int d\phi^{\prime}(-4r)$		(26)
		$\displaystyle=$	$\displaystyle\frac{\beta}{8G}(R_{reg}^{2}-R_{+}^{2}).$

This is the same result as $T=0$ case.

Next, we evaluate the brane action. In three dimensions, the extrinsic curvature satisfies $K=2T$. For the brane anchored at $\phi^{\prime}=0$, the induced metric reads:

	$\displaystyle h_{rr}$	$\displaystyle=$	$\displaystyle\frac{r^{4}T^{2}(R_{+}^{2}+R_{-}^{2})}{R_{+}^{2}(r^{2}+R_{-}^{2})^{2}(T^{2}R_{+}^{2}+R_{-}^{2}-r^{2}(T^{2}-1))}$		(27)
		$\displaystyle+$	$\displaystyle\frac{r^{2}}{\left(r^{2}-R_{+}^{2}\right)\left(r^{2}+R_{-}^{2}\right)},$
	$\displaystyle h_{tr}\!$	$\displaystyle=$	$\displaystyle\!\!\frac{R_{-}Tr(R_{+}^{2}-r^{2})\sqrt{R_{+}^{2}+R_{-}^{2}}}{\!R_{+}^{2}\!(r^{2}+R_{-}^{2})\!\sqrt{\!r^{2}(1-T^{2})\!+R_{+}^{2}T^{2}+R_{-}^{2}}},$		(28)
	$\displaystyle h_{tt}$	$\displaystyle=$	$\displaystyle\!\frac{\left(r^{2}-R_{+}^{2}\right)\left(r^{2}+R_{-}^{2}\right)}{r^{2}}\!+\!\left(\frac{R_{-}}{R_{+}}-\frac{R_{+}R_{-}}{r^{2}}\right)^{2}\!r^{2}.$

From this we find

$$\det(h_{ab})=\frac{r^{2}(R_{+}^{2}+R_{-}^{2})}{R_{+}^{2}(r^{2}(1-T^{2})+R_{+}^{2}T^{2}+R_{-}^{2})}.$$

(30)

On the other hand, in the case where the brane is anchored at $\phi^{\prime}=\pi$ , we can repeat the same thing and the result is the same as above. The brane action can be written as

			$\displaystyle S_{brane}$
		$\displaystyle=$	$\displaystyle\frac{T}{8\pi G}\int_{R_{+}}^{R_{reg}}dr\int_{0}^{\beta}dt\sqrt{\frac{r^{2}(R_{+}^{2}+R_{-}^{2})}{R_{+}^{2}(r^{2}(1-T^{2})+R_{+}^{2}T^{2}+R_{-}^{2})}}.$

However, the brane contribution for $\phi^{\prime}=0$ and $\phi^{\prime}=\pi$ is canceled as we note in the calculation of the Einstein-Hilbert action; the brane is curved in the opposite direction with respect to the bulk space. For the other terms on the conformal boundary, we have just the same term as in $T=0$ case because in $r\rightarrow\infty$ limit $\phi^{\prime}$ goes to $0$ and $\pi$.

Combining them, we obtain

	$\displaystyle S$	$\displaystyle=$	$\displaystyle\frac{\beta}{8G}(R_{reg}^{2}-R_{+}^{2})$		(32)
		$\displaystyle-$	$\displaystyle\frac{\beta}{8G}(2R_{reg}^{2}-R_{+}^{2}+R_{-}^{2})$
		$\displaystyle+$	$\displaystyle\frac{k\beta}{8G}\sqrt{(R_{reg}^{2}-R_{+}^{2})(R_{reg}^{2}+R_{-}^{2})}$
		$\displaystyle\rightarrow$	$\displaystyle-\frac{\beta\left(R_{+}^{2}+R_{-}^{2}\right)}{16G},$

where we take $R_{reg}\rightarrow\infty$.

If we assume the tensions of the two branes take different values: $T_{0}$ and $T_{\pi}$, where the branes are anchored at $\phi^{\prime}=0,\pi$. For the Einstein-Hilbert action part the integral becomes

	$\displaystyle S_{EH}$	$\displaystyle=$	$\displaystyle-\frac{1}{16G}\int_{R_{+}}^{R_{reg}}dr\int_{0}^{\beta}dt(-4r)$
			$\displaystyle\cdot\log\!\left[\!\frac{\!\left(\!\frac{T_{\pi}}{\sqrt{1-T_{\pi}^{2}}}\!\sqrt{\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}}}\!+\!\sqrt{\frac{T_{\pi}^{2}}{1-T_{\pi}^{2}}(\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}})\!+\!1}\right)}{\left(\!\frac{T_{0}}{\sqrt{1-T_{0}^{2}}}\!\sqrt{\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}}}\!+\!\sqrt{\frac{T_{0}^{2}}{1-T_{0}^{2}}(\frac{R_{+}^{2}+R_{-}^{2}}{r^{2}+R_{-}^{2}})\!+\!1}\right)\!}\!\right].$

After performing the integral, we finally obtain

	$\displaystyle S$	$\displaystyle=$	$\displaystyle-\frac{\beta\left(R_{+}^{2}+R_{-}^{2}\right)}{16G}$
		$\displaystyle-$	$\displaystyle\frac{\beta\left(R_{+}^{2}+R_{-}^{2}\right)}{16\pi GR_{+}}\!\left(\log\left(\frac{1+T_{\pi}}{1-T_{\pi}}\right)\!-\!\log\left(\frac{1+T_{0}}{1-T_{0}}\!\right)\!\right),$

where we omit the divergent term, which is proportional to $R_{reg}$. For the non-rotating BTZ, this result matches with that in (Fujita:2011fp, ).

The heat kernel method is a convenient way of calculating a one-loop partition function. We follow the analysis in (David:2009xg, ; Giombi:2008vd, ; Gopakumar:2011qs, ; Mann:1996ze, ; Vassilevich:2003xt, ). Let us consider calculating the partition function of a free scalar $\phi$:

$$Z=\int D\phi\,e^{-S(\phi)}.$$

(35)

The action of the scalar field can be rewritten as

$$S(\phi)=\int_{M}d^{3}x\,\sqrt{g}\phi\Delta\phi,$$

(36)

where we omit indices of tensorial structure. Since we are considering a compact space, $\Delta$ has a discrete set of eigenvalues $\lambda_{n}$. The one-loop partition function is expressed as follows:

$$S^{(1)}=-\frac{1}{2}\log\det(\Delta)=-\frac{1}{2}\underset{n}{\sum}\log\lambda_{n}.$$

(37)

If we consider a non-compact space, the spectrum becomes continuous and the one-loop partition function is divergent, which is proportional to the volume. This divergence can be absorbed by the renormalization of the Newton constant.

The heat kernel is defined as

$$K(t,x,y)=\underset{n}{\sum}e^{-\lambda_{n}t}\psi_{n}(x)\psi_{n}(y),$$

(38)

which we usually call a propagator. We can normalize the eigenfunctions as

$$\begin{array}[]{cc}\underset{n}{\sum}\psi_{n}(x)\psi_{n}(y)=\delta^{3}(x-y),\\ \int_{M}d^{3}x\underset{n}{\sum}\sqrt{g}\psi_{n}(x)\psi_{m}(x)=\delta_{nm}.\end{array}$$

(39)

The trace of the heat kernel is given by

$$\int_{M}d^{3}x\underset{n}{\sum}\sqrt{g}K(t,x,x)=\underset{n}{\sum}e^{-\lambda_{n}t}.$$

(40)

Using this we can compute the 1-loop partition function as an integral over $t$:

$$S^{(1)}=-\frac{1}{2}\underset{n}{\sum}\log\lambda_{n}=\frac{1}{2}\int_{+0}^{\infty}\frac{dt}{t}\int_{M}d^{3}x\underset{n}{\sum}\sqrt{g}K(t,x,x).$$

(41)

We can show the above equation by differentiating with respect to $\lambda_{n}$. Note that it is an identity up to an infinite constant. The point is that $K$ satisfies the heat conduction equation

$$\left(\partial_{t}+\Delta_{x}\right)K(t,x,y)=0,$$

(42)

with a boundary condition at $t=0$

$$K(0,x,y)=\delta(x,y).$$

(43)

In this section, we apply the calculation in (Giombi:2008vd, ) to our ETW brane setup. Now consider Poincare $AdS_{3}$ whose metric is given by

$$ds^{2}=\frac{dy^{2}+dzd\overline{z}}{y^{2}}.$$

(44)

Here an ETW brane is placed at $\Re(z)=0$ and the bulk region is defined as $\Re(z)>0$. We note that the ETW brane is connected in the bulk. Since the AdS space is maximally isometric, the geodesic distance $r(x,x^{\prime})$ depends only on the chordal distance $u(x,x^{\prime})$:

$$\begin{array}[]{cc}r(x,x^{\prime})=\mbox{arccosh}\left(1+u(x,x^{\prime})\right),\\ \\ \end{array}$$

(45)

where

$$u(x,x^{\prime})=\frac{\left(y-y^{\prime}\right)^{2}+\left|z-z^{\prime}\right|^{2}}{2yy^{\prime}}.$$

(46)

A thermal AdS can be obtained from an AdS space using the following identification

$$\left(y,z\right)\sim\left(\left|q\right|^{-1}y,q^{-1}z\right),$$

(47)

where $q=e^{2\pi i\tau}$ and $\tau=\tau_{1}+i\tau_{2}$. In the non-zero $\tau_{1}$ case, the boundary of the BCFT wraps around the torus for many times and in this case the region on which BCFT lives, is not clear because the region is not surrounded by the boundaries. To cure this problem later we will restrict to the special case $q=\overline{q}$. Now we consider applying the method of images to the heat kernel method. The tensionless ETW brane is inserted at $\Re(z)=0$. Then, if we consider a mirror position, $z$ and $\overline{z}$ are mapped to $-\overline{z}$ and $-z$, respectively. Then, one-loop partition function becomes

	$\displaystyle S^{(1)}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\int_{+0}^{\infty}\frac{dt}{t}\int_{thermal\,AdS}d^{3}x\underset{n}{\sum}\sqrt{g}\left(K^{\mathbb{\mathbb{H}}/\mathbb{Z}}(t,x,x)\right.$		(48)
		$\displaystyle+$	$\displaystyle\left.K^{\mathbb{\mathbb{H}}/\mathbb{Z}}(t,x^{mirror},x)\right).$

In the tensionless case the boundary condition for the metric is given by $K_{ab}=0$. We note that we treat this condition as an off-shell boundary condition. This means that we must impose the following boundary condition for the perturbation of the metric:

$$\partial_{x}h_{ij}=0.$$

(49)

Since the heat kernel on a thermal AdS can be obtained using the method of images from that of an AdS space, we get

	$\displaystyle S^{(1)}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\int_{+0}^{\infty}\frac{dt}{t}\int_{thermal\,AdS}d^{3}x\underset{n}{\sum}\sqrt{g}\left(K^{\mathbb{\mathbb{H}}}(t,x,\gamma^{n}x)\right.$		(50)
		$\displaystyle+$	$\displaystyle\left.K^{\mathbb{\mathbb{H}}}(t,x^{mirror},\gamma^{n}x)\right).$

For a later convenience we will use a different coordinate:

$$y=\rho\sin\theta,\ \ \ z=\rho\cos\theta e^{i\phi},$$

(51)

where $1\leq\rho\leq e^{2\pi\tau_{2}}$ , $0\leq\theta\leq\frac{\pi}{2}$ and $-\frac{\pi}{2}\leq\phi\leq\frac{\pi}{2}$ . In terms of this coordinate the geodesic distances are given by

$$\begin{array}[]{cc}r(x,\gamma^{n}x)=\mbox{arccosh}\left(\frac{\cosh\,\beta}{\sin^{2}\theta}-\frac{\cos\,\alpha}{\tan^{2}\theta}\right),\\ r(x^{mirror},\gamma^{n}x)=\mbox{arccosh}\left(\frac{\cosh\,\beta}{\sin^{2}\theta}+\frac{\cos(2\phi-\alpha)}{\tan^{2}\theta}\right),\end{array}$$

(52)

where we defined $\alpha=2\pi n\tau_{1}$ and $\beta=2\pi n\tau_{2}$.

The heat kernel of a scalar field on an $AdS_{3}$ space is given by

$$K^{H_{3}}(t,r(x,x^{\prime}))=\frac{e^{-(m^{2}+1)t-\frac{r^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{r}{\sinh(r)}.$$

(53)

Let us consider calculating the ordinary (not the mirror) part. A one-loop determinant can be recast as an integral

$$\begin{array}[]{cc}-\log\det\Delta=Vol(H_{3}/\mathbb{Z})\int_{0}^{\infty}\frac{dt}{t}\frac{e^{-(m^{2}+1)t}}{(4\pi t)^{\frac{3}{2}}}\\ +\underset{n\neq 0}{\sum}\int_{0}^{\infty}\frac{dt}{t}\int\frac{d\rho d\theta d\phi\cos\theta}{\rho\sin^{3}\theta}\frac{e^{-(m^{2}+1)t-\frac{r^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{r}{\sinh(r)},\end{array}$$

(54)

where we split into the zero mode part and non-zero mode $n\neq 0$. The first term can be easily regularized:

$$\int_{0}^{\infty}\frac{dt}{t}\frac{e^{-(m^{2}+1)t}}{(4\pi t)^{\frac{3}{2}}}=\frac{\left(m^{2}+1\right)^{\frac{3}{2}}}{8\pi^{\frac{3}{2}}}\int_{0}^{\infty}dkk^{-\frac{5}{2}}e^{-k}=\frac{\left(m^{2}+1\right)^{\frac{3}{2}}}{6\pi}.$$

(55)

The second term can be calculated directly. Firstly, we change the variable from $\theta$ to $r$ and we get

$$\begin{array}[]{cc}\underset{n\neq 0}{\sum}\int_{0}^{\infty}\frac{dt}{t}\int\frac{d\rho d\theta d\phi\cos\theta}{\rho\sin^{3}\theta}\frac{e^{-(m^{2}+1)t-\frac{r^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{r}{\sinh(r)}\\ =\underset{n\neq 0}{\sum}\int_{0}^{\infty}\frac{dt}{t}\int\frac{d\rho drd\phi}{\rho}\frac{e^{-(m^{2}+1)t-\frac{r^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{r}{2\left(\cosh\beta-\cos\alpha\right)}.\end{array}$$

(56)

Integrating over $r$, $\phi$ and $t$ in order, we can reach the final answer:

$$\begin{array}[]{cc}\underset{n\neq 0}{\sum}\int_{0}^{\infty}\frac{dt}{t}\int\frac{d\rho drd\phi}{\rho}\frac{e^{-(m^{2}+1)t-\frac{r^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{r}{2\left(\cosh\beta-\cos\alpha\right)}\\ =\underset{n\neq 0}{\sum}\frac{\sqrt{\pi}\tau_{2}}{4\left(\cosh\beta-\cos\alpha\right)}\int_{0}^{\infty}\frac{dt}{t}\frac{e^{-(m^{2}+1)t-\frac{\left(2\pi n\tau_{2}\right)^{2}}{4t}}}{t^{\frac{3}{2}}}\\ =\underset{n\neq 0}{\sum}\frac{e^{-2\pi n\tau_{2}\sqrt{m^{2}+1}}}{4n\left(\cosh\beta-\cos\alpha\right)}\\ =\sum_{n=1}^{\infty}\frac{\left|q\right|^{n\left(1+\sqrt{m^{2}+1}\right)}}{n\left|1-q^{n}\right|^{2}}.\end{array}$$

(57)

Now let us consider the mirror part. Firstly, we consider the $n=0$ case. The strategy is that we firstly fix $\phi$ and integrate $\theta$ or $r$ and then integrate over other variables. This leads to the divergent result even though we use a regularization by a gamma function:

$$\begin{array}[]{cc}\int_{0}^{\infty}\frac{dt}{t}\int\frac{d\rho d\theta d\phi\cos\theta}{\rho\sin^{3}\theta}\frac{e^{-(m^{2}+1)t-\frac{r^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{r}{\sinh(r)}\\ =-\frac{\tau_{2}\sqrt{m^{2}+1}}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{d\phi}{1+\cos(2\phi)}.\end{array}$$

(58)

However, this divergence can be eliminated by the renormalization of Newton constant, so we will ignore this term.

Next consider the $n\neq 0$ case. The trick is almost the same, but the situation is slightly changed. The integral over $\phi$ yields

$$\begin{array}[]{cc}\underset{n\neq 0}{\sum}\int_{0}^{\infty}dt2\pi\tau_{2}\frac{e^{-(m^{2}+1)t-\frac{\left(2\pi n\tau_{2}\right)^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{d\phi}{\left|\cos\left(2\phi-\alpha\right)+\cosh\left(\beta\right)\right|}\\ =\underset{n\neq 0}{\sum}\int_{0}^{\infty}dt\,2\pi\tau_{2}\frac{e^{-(m^{2}+1)t-\frac{\left(2\pi n\tau_{2}\right)^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{\pi}{\sinh(\beta)}\\ =\sum_{n=1}^{\infty}\frac{\left|q\right|^{n\left(1+\sqrt{m^{2}+1}\right)}}{n\left(1-\left|q\right|^{2n}\right)}.\end{array}$$

(59)

In this section we consider the vector field contribution to the partition function. For a later convenience we define a set of new real coordinates:

$$\begin{array}[]{cc}x=\Re(z)=\rho\cos\theta\cos\phi,\\ \eta=\Im(z)=\rho\cos\theta\sin\phi.\end{array}$$

(60)

The propagator has two indices and is invariant under an exchange of those indices. Thus we can expand the kernel by the basis of (1,1) symmetric bi-tensor (DHoker:1999bve, ):

$$K_{\mu\nu^{\prime}}(t,x,x^{\prime})=F(t,u)\partial_{\mu}\partial_{\nu^{\prime}}u+\partial_{\mu}\partial_{\nu^{\prime}}S(t,u),$$

(61)

where

$$\begin{array}[]{cc}F(t,r)=-\frac{e^{-\frac{r^{2}}{4t}}}{(4\pi t)^{\frac{3}{2}}}\frac{r}{\sinh r},\\ S(t,r)=\frac{4}{(4\pi)^{\frac{3}{2}}}\frac{e^{-\frac{r^{2}}{4t}}}{\sinh r}\sqrt{t}\int_{0}^{1}d\xi e^{-t(1-\xi)^{2}}\sinh(r\xi).\end{array}$$

(62)

The next step is to calculate the derivatives of $u$. In the Poincare coordinate it can be explicitly given by

			$\displaystyle\partial_{\mu}\partial_{\nu^{\prime}}u$
		$\displaystyle=$	$\displaystyle-\frac{1}{z_{0}w_{0}}\left\{\delta_{\mu\nu^{\prime}}\!+\!\frac{(z-w)_{\mu}}{w_{0}}\delta_{\nu^{\prime}0}+\frac{(w-z)_{\nu^{\prime}}}{z_{0}}\delta_{\mu 0}\!-\!u\delta_{\mu 0}\delta_{\nu^{\prime}0}\right\}.$

Here we note that $z=x$ or $z=x^{mirror}$, and $w=\gamma^{n}x$ . We will present a detailed calculation of these derivatives in the appendix A. Let us consider the ordinary (not the mirror image) contribution. Using those results, we can compute the kernel:

			$\displaystyle\int_{0}^{\infty}\frac{dt}{t}\int d^{3}x\sqrt{g}\sum_{n}g^{\mu\rho}\frac{\partial(\gamma^{n}x)^{\nu^{\prime}}}{\partial x^{\rho}}K_{\mu\nu^{\prime}}^{\mathbb{\mathbb{H}}}(t,r(x,\gamma^{n}x))$
		$\displaystyle=$	$\displaystyle\int_{0}^{\infty}\frac{dt}{t}\int d^{3}x\sqrt{g}\underset{n}{\sum}\left\{\frac{d^{2}}{S}{du^{2}}\left\{(e^{\beta}-\cosh r)(e^{-\beta}\cosh r)\right.\right.$
		$\displaystyle-$	$\displaystyle\left.\left.2\cos\alpha(\cosh r-\cosh\beta)\right\}\right.$
		$\displaystyle+$	$\displaystyle\left.(F+(\frac{dS}{du}))\left(\cosh r-2\cosh\beta-2\cos\alpha\right)\right\}$
		$\displaystyle=$	$\displaystyle Vol(\mathbb{H}/\mathbb{Z})\int_{0}^{\infty}\frac{dt}{t}\frac{(e^{-t}+2+4t)}{(4\pi t)^{\frac{3}{2}}}$
		$\displaystyle+$	$\displaystyle\sum_{n=0}^{\infty}\int_{0}^{\infty}\frac{dt}{t}\frac{2\pi^{2}\tau_{2}}{(\cosh\beta-\cos\alpha)}\frac{e^{-\frac{\beta^{2}}{4t}}}{4\pi^{\frac{3}{2}}\sqrt{t}}(2\cos\alpha+e^{-t}).$

The first term can be regularized by considering a massless limit of a massive field:

			$\displaystyle Vol(\mathbb{H}/\mathbb{Z})\int_{0}^{\infty}\frac{dt}{t}\frac{(e^{-t}+2+4t)}{(4\pi t)^{\frac{3}{2}}}$
		$\displaystyle=$	$\displaystyle\underset{m\rightarrow 0}{lim}Vol(\mathbb{H}/\mathbb{Z})\int_{0}^{\infty}\frac{dt}{t}\frac{e^{-m^{2}t}(e^{-t}+2+4t)}{(4\pi t)^{\frac{3}{2}}}$
		$\displaystyle=$	$\displaystyle Vol(\mathbb{H}/\mathbb{Z})\frac{1}{(4\pi)^{\frac{3}{2}}}(\frac{4\sqrt{\pi}}{3}-8\sqrt{\pi})=-\frac{5}{6\pi}Vol(\mathbb{H}/\mathbb{Z}).$

The second term can be calculated straightforwardly:

$$\begin{array}[]{cc}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{dt}{t}\frac{2\pi^{2}\tau_{2}}{(\cosh\beta-\cos\alpha)}\frac{e^{-\frac{\beta^{2}}{4t}}}{4\pi^{\frac{3}{2}}\sqrt{t}}(2\cos\alpha+e^{-t})\\ =\sum_{n=1}^{\infty}\frac{2\cos\alpha+e^{-\beta}}{2n(\cosh\beta-\cos\alpha)}\\ =\sum_{n=1}^{\infty}\frac{q^{n}+\overline{q}^{n}+\left|q\right|^{2n}}{n\left|1-q^{n}\right|^{2}}.\end{array}$$

(66)

If we omit the term, which is proportional to the volume, the answer is just a half of the original result as we expected.