Of course, this is the short-distance behaviour of the fermionic Green's function $iS(x, y)$ and we recognise the usual leading singularity $\simeq i\nabla\ell^2/\ell^2$ of the Dirac Green's function, as well as a sub-leading logarithmic singularity. Just as for $\zeta_+$, it follows that if we define

$$ \zeta_{-}^{\text{reg}}(s, x, y) = \zeta_{-}(s, x, y) - \zeta_{-}^{\text{sing}}(s, x, y) \quad (4.101) $$

we have subtracted all the potential singularities of this zeta-function, namely the pole at $s = \frac{1}{2}$ that is present for $x = y$ and the short-distance singularity of the Green's function which is present for $s = \frac{1}{2}$. In particular, we see that the following limit exists:

$$ \lim_{s \to 0} \left( \zeta_{-} \left( s + \frac{1}{2}, x, x \right) + s \zeta'_{-} \left( s + \frac{1}{2}, x, x \right) \right) = \zeta_{-}^{\text{reg}} \left( \frac{1}{2}, x, x \right) + \frac{m}{4\pi} (\gamma - \log \mu^2) i\gamma_{*}. \quad (4.102) $$

Using $\zeta_{-}(s, x, x) = im\gamma_*\zeta_{+}(s + \frac{1}{2}, x, x)$, cf (4.33), we conclude that also

$$ \zeta_{-}^{\text{reg}}(\frac{1}{2}, x, x) = im\gamma_{*} \zeta_{+}^{\text{reg}}(1, x, x). \quad (4.103) $$

Note that, contrary to the singular parts, to evaluate the regular parts $\zeta_{\pm}^{\text{reg}}$ the knowledge of the small-$t$ asymptotics of the heat kernels is not enough, but requires some knowledge about all eigenvalues and eigenfunctions. One way this information is coded is in the regularized and so-called renormalized Green's functions which we will discuss next.

## 4.5 Renormalized Green's functions

### 4.5.1 $G_R$, $G_ζ$, $S_R$ and $S_ζ$

For the Green's function $G(x, y) = \zeta_+(1, x, y)$ of $D^2$, we may define a regularized Green's function $G^{\text{reg}}(x, y)$ by subtracting the short-distance singularity, cf (4.90)

$$ G^{\text{reg}}(x, y) = G(x, y) - G^{\text{sing}}(x, y) = G(x, y) + \frac{1}{4\pi} \log \frac{\mu^2 \ell^2(x, y)}{4} \mathbf{1}_{2 \times 2}. \quad (4.104) $$

Note that this is similar to, but different from $\zeta_+^{\text{reg}}(1, x, y) = G(x, y) - \frac{1}{4\pi} E_1(\frac{\mu^2 \ell^2(x, y)}{4}) \mathbf{1}_{2 \times 2}$. But just as $\zeta_+^{\text{reg}}(1, x, y)$, the regularized Green's function $G^{\text{reg}}(x, y)$ has a well-defined limit as $x \to y$. The so-called renormalized Green's function at coinciding points $G_R$ then is simply defined as

$$ G_R(y) = \lim_{x \to y} G^{\text{reg}}(x, y). \quad (4.105) $$

In complete analogy, we define $S^{\text{reg}}(x, y)$ and $S_R(y)$:

$$ S^{\text{reg}}(x, y) = S(x, y) - S^{\text{sing}}(x, y) = S(x, y) + \frac{1}{4\pi} (i\nabla_x + m\gamma_*) \log \frac{\mu^2 \ell^2(x, y)}{4} \mathbf{1}_{2 \times 2}. \quad (4.106) $$

Again, this is similar to, but different from $-i\zeta_-^{\text{reg}}(\frac{1}{2}, x, y) = S(x, y) - \frac{1}{4\pi}(i\nabla_x + m\gamma_*) E_1(\frac{\mu^2 \ell^2(x, y)}{4})$. The renormalized Green's function at coinciding points $S_R$ then is defined as

$$ S_R(y) = \lim_{x \to y} S^{\text{reg}}(x, y). \quad (4.107) $$
Recall from (4.56) that tr $\gamma_*$ $S(x, y) = m \operatorname{tr} G(x, y)$. The same relation obviously also holds between $S^{\text{sing}}$ and $G^{\text{sing}}$. Hence

$$ \operatorname{tr} \gamma_* S^{\text{reg}}(x, y) = m \operatorname{tr} G^{\text{reg}}(x, y). \quad (4.108) $$

If we take the limit $x \to y$ of the last relation we get $\operatorname{tr} \gamma_* S_R(y) = m \operatorname{tr} G_R(y)$, but we will prove the more general relation without the traces below in (4.114).

It is also easy to express $G_R$ and $S_R$ in terms of $\zeta_+^{\text{reg}}(1, x, x)$ and $\zeta_-^{\text{reg}}(\frac{1}{2}, x, x)$. Indeed,

$$ \begin{aligned} G^{\text{reg}}(x, y) &= \zeta_{+}^{\text{reg}}(1, x, y) + \frac{1}{4\pi} \left( E_1 \left( \frac{\mu^2 \ell^2(x, y)}{4} \right) + \log \frac{\mu^2 \ell^2(x, y)}{4} \right) \mathbf{1}_{2 \times 2} \\ S^{\text{reg}}(x, y) &= -i \zeta_{-}^{\text{reg}} \left( \frac{1}{2}, x, y \right) + \frac{1}{4\pi} (i \nabla_x + m \gamma_*) \left( E_1 \left( \frac{\mu^2 \ell^2(x, y)}{4} \right) + \log \frac{\mu^2 \ell^2(x, y)}{4} \right). \end{aligned} \quad (4.109) $$

Taking the limit $x \to y$ and using the asymptotics (4.89) of $E_1$ we get

$$ \begin{aligned} G_R(x) &= \zeta_+^{\text{reg}}(1, x, x) - \frac{\gamma}{4\pi} \mathbf{1}_{2 \times 2} \\ S_R(x) &= -i \zeta_-^{\text{reg}} \left( \frac{1}{2}, x, x \right) - \frac{m\gamma}{4\pi} \gamma_* . \end{aligned} \quad (4.110) $$

For completeness, let us mention that one sometimes defines different “renormalized” Green’s functions $G_\zeta(x)$ and $S_\zeta(x)$ from the $\zeta_\pm(s, x, x)$ by subtracting simply the poles and then letting $s \to 1$ or $s \to \frac{1}{2}$:

$$ \begin{aligned} G_\zeta(x) &= \lim_{s \to 1} \left( (\mu^2)^{s-1} \zeta_+(s, x, x) - \frac{\mathbf{1}_{2 \times 2}}{(4\pi)(s-1)} \right), \\ S_\zeta(x) &= \lim_{s \to \frac{1}{2}} \left( -i (\mu^2)^{s-1/2} \zeta_-(s, x, x) - \frac{m\gamma_*}{4\pi(s-\frac{1}{2})} \right). \end{aligned} \quad (4.111) $$

It follows, using (4.92) and (4.93) that

$$ \begin{aligned} G_{\zeta}(x) &= \zeta_{+}^{\text{reg}}(1, x, x) + \frac{\gamma}{4\pi} \mathbf{1}_{2 \times 2} &&\Rightarrow G_{\zeta}(x) = G_{R}(x) + \frac{\gamma}{2\pi} \mathbf{1}_{2 \times 2}, \\ S_{\zeta}(x) &= -i \zeta_{-}^{\text{reg}}(\frac{1}{2}, x, x) + \frac{\gamma}{4\pi} m \gamma_{*} &&\Rightarrow S_{\zeta}(x) = S_{R}(x) + \frac{\gamma}{2\pi} m \gamma_{*}. \end{aligned} \quad (4.112) $$

We can now re-express the relation (4.94) in terms of $G_R$ or $G_\zeta$ and equivalently the relation (4.102) in terms of $S_R$ or $S_\zeta$ as follows

$$ \begin{aligned} \lim_{s \to 0} \left( \zeta_+(s+1, x, x) + s \zeta'_+(s+1, x, x) \right) &= \zeta_+^{\text{reg}}(1, x, x) + \frac{1}{4\pi} (\gamma - \log \mu^2) \mathbf{1}_{2 \times 2} \\ &= G_R(x) + \frac{1}{4\pi} (2\gamma - \log \mu^2) \mathbf{1}_{2 \times 2} = G_\zeta(x) - \frac{1}{4\pi} \log \mu^2 \mathbf{1}_{2 \times 2}, \\ \lim_{s \to 0} \left( \zeta_-(s+1/2, x, x) + s \zeta'_-(s+1/2, x, x) \right) &= \zeta_-^{\text{reg}}(1/2, x, x) + \frac{m}{4\pi} (\gamma - \log \mu^2) i \gamma_* \\ &= i S_R(x) + \frac{im}{4\pi} (2\gamma - \log \mu^2) \gamma_* = i S_\zeta(x) - \frac{im}{4\pi} \log \mu^2 \gamma_* . \end{aligned} \quad (4.113) $$
Moreover, it follows from (4.33) and (4.111) that $S_ζ(x) = mγ_*G_ζ(x)$ and then by (4.112) also that $S_R(x) = mγ_*G_R(x)$ (which also follows from (4.110) and (4.103)):

$$S_ζ(x) = mγ_*G_ζ(x), \quad S_R(x) = mγ_*G_R(x). \qquad (4.114)$$

Finally note that if the metric has sufficiently many isometries these renormalized Green's function must be constants. This is the case for the round sphere and for the flat torus. Of course, they still depend on the geometric data of these surfaces, like the radius of the round sphere, or the periods of the flat torus, but not on the point $x$. Their explicit form can be quite non-trivial even for such simple geometries. Let us show this for our preferred example of the flat torus, now with periods $2a$ and $2b$.

### 4.5.2 Green's function $\tilde{G}(x, y)$ and $\tilde{G}_R(x)$ on the flat torus for $m = 0$

Let us first show how one can obtain the Green's function of the squared Dirac operator in the massless case on a “rectangular” flat torus with periods $2a$ and $2b$, i.e. with $\tau = i\frac{b}{a}$. Since we look at the massless case, we will determine the Green's function with the zero-mode part excluded, i.e. $\tilde{G}(x, y)$. Being flat, there is no spin-connection and no curvature and we have, as above, that $D^2 = \mathbf{1}_{2\times2}(-\partial_\nu\partial^\nu) = -\mathbf{1}_{2\times2}\Delta_{\text{scalar}}$. Hence, the problem is identical to finding the Green's function of the scalar Laplace operator on the flat torus. The non-trivial part is to reconcile the appropriate short-distance singularity and the periodicity of the torus. But this problem has a well-known solution in terms of theta functions and the corresponding computation can e.g. be found in [15]. Using complex coordinates $z = z^1 + iz^2$, where $z^i = x^i - y^i$ one finds $\tilde{G}(x, y) = g(z, \bar{z})\mathbf{1}_{2\times2}$ with

$$g(z, \bar{z}) = \frac{(\mathrm{Im} \, z)^2}{8ab} - \frac{1}{4\pi} \log \left| \theta_1 \left( \frac{z}{2a} \middle| i \frac{b}{a} \right) \right|^2 \equiv \frac{(\mathrm{Im} \, z)^2}{8ab} - \frac{1}{4\pi} \log \left[ \theta_1 \left( \frac{z}{2a} \middle| i \frac{b}{a} \right) \theta_1 \left( \frac{\bar{z}}{2a} \middle| i \frac{b}{a} \right) \right], \qquad (4.115)$$

where the theta function $\theta_1$ is defined by [33]

$$\theta_1(\nu|\tau) = 2q^{1/4} \sum_{n=0}^{\infty} (-)^n q^{n(n+1)} \sin(2n+1)\pi\nu, \quad q = e^{i\pi\tau}. \qquad (4.116)$$

Indeed, the scalar Laplace operator on flat space in the complex coordinates is simply $\Delta_{\text{scalar}} = 4\partial_z\partial_{\bar{z}}$ and then, for $z, \bar{z} \neq 0$, we have $-4\partial_z\partial_{\bar{z}}g(z, \bar{z}) = -4\partial_z\partial_{\bar{z}}\frac{(\mathrm{Im} \, z)^2}{8ab} = \frac{1}{4ab} = \frac{1}{A}$. On the other hand, $g(z, \bar{z}) \sim_{|z|\to 0} -\frac{1}{4\pi} \log|z|^2$ which is the required short-distance singularity. Hence, one has correctly $-4\partial_z\partial_{\bar{z}}g(z, \bar{z}) = \delta^{(2)}(z) - \frac{1}{A}$.

As familiar by now, to get the renormalized Green's function one must subtract the short-distance singularity $-\frac{1}{4\pi} \log \frac{\mu^2|z|^2}{4} \mathbf{1}_{2\times2}$ and take the limit $z \to 0$ with the result

$$\tilde{G}_{\mathrm{R}}^{m=0}(x) = -\frac{1}{2\pi} \log \left( \frac{2\pi}{a\mu} q^{\frac{1}{4}} \sum_{n=0}^{\infty} (-)^n (2n+1)q^{n(n+1)} \right) \mathbf{1}_{2\times2}, \quad q = e^{-\pi b/a}, \quad (\text{flat torus}). \qquad (4.117)$$

Obviously, this does not depend on the point $x$ on the torus, but it depends quite non-trivially on the shape of the torus (i.e. on $b/a$), as well as on its volume $((2\pi)^2 ab)$.

What about $\tilde{S}_R^{m=0}(x)$? Equation (4.114) states that in the massive theory $S_R = m\gamma_*G_R$. Now from our explicit study of the flat torus we know that the zero-mode parts
of G and S are related as $S^{(0)} = \frac{1}{m}(\psi_0^\dagger\psi_0 - \psi_0^*(\psi_0^*)^\dagger) = \frac{1}{m}\gamma_*(\psi_0^\dagger\psi_0 + \psi_0^*(\psi_0^*)^\dagger) = m\gamma^*G^{(0)}$, a relation we will also prove in more generality below. Subtracting this relation from $S_R = m\gamma_*G_R$ we get $\tilde{S}_R = m\gamma_*\tilde{G}_R$. Now both, $\tilde{S}_R$ and $\tilde{G}_R$ have a smooth limit as $m \to 0$, and we can conclude that $\tilde{S}_R^{m=0}(x) = 0$. Below, in subsection 5.3, we will study the zero-mode parts of the Green's functions more generally and show that $\tilde{S}_R^{m=0}(x) = 0$ always holds.

# 5 Determining the gravitational action

We now want to determine the variation of the effective gravitational action under infinitesimal conformal variations and write this variation in such a form that it can be integrated to obtain the effective action $S_{grav}[g, \hat{g}]$. We will do this in an expansion in powers of $m^2$. For this expansion to make sense, all quantities entering this expansion must have a well-defined limit as $m \to 0$, i.e. the expansion is done in terms of quantities defined in the massless theory. Thus, for genus one and larger, one must carefully subtract the zero-mode parts from all Green's functions and zeta-functions.

We will first rewrite the variation of the gravitational action in terms of the renormalized Green's function $G_R$ and its variation $\delta G_R$. The latter can be re-expressed in an expansion to all orders in $m^2$ involving higher Green's functions and zeta functions, still of the massive theory. Proceeding this way first, provides a more streamlined presentation while being already a useful result for spherical topology where the $m \to 0$ limit exists for all Green's functions and zeta-functions without the need to subtract any zero-mode parts. Then, however, we repeat the analogous computation on arbitrary Riemann surfaces, taking into account the presence of zero-modes that must first be subtracted from these quantities. This uses the properties of the zero-mode projectors worked out in subsection 2.5. The corresponding computations is a bit more cumbersome, but the final result will be only slightly more complicated. Finally, we obtain the gravitational action as an infinite expansion in powers of $m^2$, with a finite radius of convergence, valid on a manifold of arbitrary genus. In particular, at order $m^2$, we obtain a Mabuchi-like term, as well as terms that are multi-local in $\sigma$ and involve the Green's functions for the reference metric $\hat{g}$, as well as a finite number of area-like parameters.

## 5.1 Expressing the variation of $S_{grav}$ in terms of $G_R$

Recall our formula (3.24) for the gravitational action in terms of $\zeta(0)$ and $\zeta'(0)$ which gives

$$ \delta S_{grav} = \frac{1}{4} \delta \zeta'(0) + \frac{1}{4} \log \mu^2 \delta \zeta(0). \qquad (5.1) $$

The variations of the zeta functions under infinitesimal conformal rescalings with $\delta\sigma(x)$ have been worked out above and are given in (4.64) and (4.67) or (4.62). $\delta\zeta(0)$ was shown to be simply $-\frac{m^2}{\pi} \int \sqrt{g} \delta\sigma = -\frac{m^2}{2\pi}\delta A$. Combining (4.62) with (4.40) and (4.113) we get

$$ \delta\zeta'(0) = -\int \sqrt{g} \delta\sigma \left(\frac{\mathcal{R}}{12\pi} + \frac{m^2}{\pi}\right) - 2m^2 \int \sqrt{g} \delta\sigma \left(\text{tr} G_R + \frac{1}{2\pi}(2\gamma - \log \mu^2)\right). \qquad (5.2) $$
Thus

$$ \delta S_{\text{grav}} = -\frac{2\gamma + 1}{8\pi} m^2 \delta A - \frac{1}{48\pi} \int \sqrt{g} \delta\sigma \mathcal{R} - \frac{m^2}{2} \int \sqrt{g} \delta\sigma \operatorname{tr} G_R. \quad (5.3) $$

The term ~ $\delta A$ is usually referred to as the variation of a cosmological constant term, while the term ~ $\int \sqrt{g} \delta\sigma \mathcal{R}$ is the variation of the Liouville action (1.4), see (1.17):

$$ \int \sqrt{g} \delta\sigma \mathcal{R} = \delta S_{\text{Liouville}}. \quad (5.4) $$

Note that the coefficient of $\delta S_{\text{Liouville}}$ in $\delta S_{\text{grav}}$ is exactly $\frac{1}{2}$ the one that occurs in the gravitational action of a bosonic scalar matter field. It is interesting to trace back how this coefficient occurs. The term ~ $\mathcal{R}$ originates from $\zeta_+(0, x, x)$ which was given in terms of the heat kernel coefficient $a_1 \equiv F_1(x, x)$. For the bosonic scalar field and scalar Laplacian $a_1 = \frac{\mathcal{R}}{6}$. At present, $a_1$ is $\frac{\mathcal{R}}{6} - \frac{\mathcal{R}}{4} = -\frac{\mathcal{R}}{12}$. But because we are dealing with fermions, the gravitational action has an overall minus sign, so that in the end one gets a $+\frac{\mathcal{R}}{12}$, i.e. exactly one half the bosonic result.<fnref n="25" /> Of course, this is consistent with the well-known central charges 1 and $\frac{1}{2}$ of the conformal algebra for a single massless scalar and massless Majorana fermion. We conclude that

$$ \delta \left[ S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \frac{m^2}{8\pi} (2\gamma + 1) A \right] = -\frac{m^2}{2} \int \sqrt{g} \delta\sigma \operatorname{tr} G_R. \quad (5.5) $$

It remains to characterise the right-hand side of this equations and express it as a total variation of some appropriate quantity. Obviously, we have (using $\delta\sqrt{g} = 2\sqrt{g}\delta\sigma$)

$$ \int \sqrt{g} \delta\sigma \operatorname{tr} G_R = \frac{1}{2} \delta \int \sqrt{g} \operatorname{tr} G_R - \frac{1}{2} \int \sqrt{g} \delta \operatorname{tr} G_R, \quad (5.6) $$

so that we can rewrite (5.5) as

$$ \delta \left[ S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \frac{m^2}{8\pi} (2\gamma + 1) A + \frac{m^2}{4} \int \sqrt{g} \operatorname{tr} G_R \right] = \frac{m^2}{4} \int \sqrt{g} \delta \operatorname{tr} G_R. \quad (5.7) $$

## 5.2 Conformal variations of the Green's functions

We will need to study the variations of the Green's functions $S(x, y)$ and $G(x, y)$ under conformal transformations. Since $G(x, y) = \int d^2z \sqrt{g(z)} S(x, z)S(z, y)$ it will be enough in the first place to obtain the variation of $S(x, y)$, from which the variation of $G(x, y)$ can then be deduced. In this subsection we assume again that $m \neq 0$ so that there are no zero-modes of the Dirac operator $D = i\nabla^\dagger + m\gamma_\mu$. Generically there are, of course, zero-modes of $i\nabla^\dagger$ as discussed in subsection 2.5. In the next subsection, when we want to do a small mass expansion, we will explicitly separate the contributions of these zero-modes of $i\nabla^\dagger$, that contribute terms of order $\frac{1}{m}$ to the Green's function $S(x, y)$ and terms of order $\frac{1}{m^2}$ to the Green's function $G(x, y)$. But at present, we will not need to do this separation, as long as $m \neq 0$.

<footnote n="25">²⁵Of course, we also had the $\frac{1}{4}$ multiplying the $\delta\zeta'(0)$ and $\delta\zeta(0)$ instead of a $\frac{1}{2}$ in the bosonic case. But this extra $\frac{1}{2}$ is offset by a factor 2 coming from the Dirac traces.</footnote>
### 5.2.1 Conformal variations of $S(x, y)$, $G(x, y)$ and $G_R(y)$

Recall that $S(x, y)$ is the solution of $D_x S(x, y) = (i \nabla_x + m\gamma_*)S(x, y) = \delta^{(2)}(x-y)/\sqrt{g}$, see eq. (4.47). We want to determine the variation of $S$ under a conformal rescaling. Consider two metrics $g$ and $g'$ related by an infinitesimal conformal rescaling $g' = e^{2\delta\sigma}g$. Then, of course $\sqrt{g'} = e^{2\delta\sigma}\sqrt{g}$ and from (2.79), $\nabla' = e^{-3\delta\sigma/2}\nabla e^{\delta\sigma/2}$. Thus the corresponding Green's functions $S$ and $S'$ satisfy

$$ (i\nabla_x + m\gamma_*)S(x, y) = \frac{\delta^{(2)}(x-y)}{\sqrt{g}}, \\ (ie^{-3\delta\sigma(x)/2}\nabla e^{\delta\sigma(x)/2} + m\gamma_*)S'(x, y) = \frac{\delta^{(2)}(x-y)}{\sqrt{g}}e^{-3\delta\sigma(x)/2}e^{-\delta\sigma(y)/2}. \quad (5.8) $$

It looks as if for $m = 0$ one would simply have $S'(x, y) = e^{-\delta\sigma(x)/2}S(x, y)e^{-\delta\sigma(y)/2}$. But for $m = 0$ we must precisely subtract the zero-mode contribution and deal with $\tilde{S}$ and $\tilde{S}'$ instead and then the relation becomes again more complicated due to the conformal transformation of the zero-mode projector. This will be dealt with in the next two subsections. But this remark motivates the following definition of $\hat{\delta}S$:

$$ S'(x, y) \equiv S(x, y) + \delta S(x, y) = e^{-\delta\sigma(x)/2} S(x, y) e^{-\delta\sigma(y)/2} + \hat{\delta}S(x, y) \\ \Leftrightarrow \delta S(x, y) = -\frac{1}{2}(\delta\sigma(x) + \delta\sigma(y))S(x, y) + \hat{\delta}S(x, y). \quad (5.9) $$

We then insert this $S'(x, y)$ into the second equation (5.8) and develop to first order in the variation. This yields

$$ D_x \hat{\delta}S(x, y) = (i\nabla_x + m\gamma_*) \hat{\delta}S(x, y) = -m\delta\sigma(x) \gamma_* S(x, y), \quad (5.10) $$

which is solved by multiplying with the Green's function of $D$ and integrating:

$$ \int d^2z \sqrt{g(z)} S(x, z) D_z \hat{\delta}S(z, y) = -m \int d^2z \sqrt{g(z)} \delta\sigma(z) S(x, z) \gamma_* S(z, y). \quad (5.11) $$

Using the hermiticity of $D_z$ the terms on the left-hand side is $\int d^2z \sqrt{g(z)} (D_z S(z, x))^{\dagger} \hat{\delta}S(z, y) = \hat{\delta}S(x, y)$, so that

$$ \hat{\delta}S(x, y) = -m \int d^2z \sqrt{g} \delta\sigma(z) S(x, z) \gamma_* S(z, y), \quad (5.12) $$

and

$$ \delta S(x, y) = -\frac{1}{2}(\delta\sigma(x) + \delta\sigma(y))S(x, y) - m \int d^2z \sqrt{g} \delta\sigma(z) S(x, z) \gamma_* S(z, y). \quad (5.13) $$

One can now continue from this to obtain the variation of $S_R(x)$ by subtracting the variation of the singular part of $S(x, y)$, and then deduce the variation of $\text{tr} \gamma_* S_R = \text{tr} G_R$. We have tried this avenue, but it turns out not to be the most straightforward one. Instead, we will determine the variation of $G(x, y)$ from the variation of $S(x, y)$ and then deduce the variation of $G_R(x)$ by subtracting the variation of the singular part of $G(x, y)$. Recall that $G(x, y) = \int dz S(x, z)S(z, y)$, where here and in the following, to simplify the notations, we
abbreviate the integration measure $d^2z\sqrt{g(z)}$ by writing simply $dz$. Of course, one should not forget then that $dz$ changes under a conformal rescaling:

$$ \int dz \approx \int d^2z \sqrt{g(z)}, \quad \delta dz = dz 2\delta\sigma(z). \qquad (5.14) $$

The variation of $G(x, y)$ is then given by

$$ \begin{aligned} \delta G(x, y) &= \delta \int dz S(x, z) S(z, y) = \int dz \left[ 2\delta\sigma(z) S(x, z) S(z, y) + \delta S(x, z) S(z, y) + S(x, z) \delta S(z, y) \right] \\ &= \int dz \left[ \left( \delta\sigma(z) - \frac{\delta\sigma(x)}{2} - \frac{\delta\sigma(y)}{2} \right) S(x, z) S(z, y) + \hat{\delta}S(x, z) S(z, y) + S(x, z) \hat{\delta}S(z, y) \right]. \end{aligned} \qquad (5.15) $$

Now the short-distance singularity of $G(x, y)$ is $G^{\text{sing}}(x, y) = -\frac{1}{4\pi} \log \frac{\mu^2 \ell^2(x, y)}{4} \mathbf{1}_{2 \times 2}$ and we need to compute its variation

$$ \delta G^{\text{sing}}(x, y) = -\frac{1}{4\pi} \frac{\delta \ell^2(x, y)}{\ell^2(x, y)} \mathbf{1}_{2 \times 2}. \qquad (5.16) $$

Actually, we only need this in the limit $x \to y$, i.e. when $x$ and $y$ are only separated by an infinitesimal $dx$. But then the (geodesic) distance between the two points is just given by $\ell^2(x, x+dx) = g_{\mu\nu}dx^\mu dx^\nu = e^{2\sigma(x)}\hat{g}_{\mu\nu}dx^\mu dx^\nu$. Obviously then $\delta\ell^2(x, x+dx) = 2\delta\sigma(x)\ell^2(x, x+dx)$. Somewhat more generally, for $x$ close to $y$, one can show that (see e.g. the appendix A1 of [34]):

$$ \frac{\delta \ell^2(x, y)}{\ell^2(x, y)} = \delta\sigma(x) + \delta\sigma(y) + \mathcal{O}(\ell^2). \quad \Rightarrow \quad \delta G^{\text{sing}}(x, y) = -\frac{1}{4\pi}(\delta\sigma(x) + \delta\sigma(y) + \mathcal{O}(\ell^2)) \mathbf{1}_{2 \times 2}. \qquad (5.17) $$

We see that the variation of the singular part is non-singular. Hence it must be that $\delta G(x, y)$ as given in (5.15) also is non-singular. Indeed, the terms involving $\hat{\delta}S$ are easily seen to be non-singular, so only the first term in (5.15) could potentially be singular as $x \to y$. Indeed $S(y, z)S(z, y)$ is divergent as $\sim \frac{1}{\ell^2(z, y)}$ when the integration variable $z$ gets close to $y$. But then the factor $\delta\sigma(z) - \delta\sigma(y)$ vanishes as $z - y \sim \ell(z, y)$ so that the integrand behaves as $\frac{1}{\ell(z, y)}$ which is an integrable singularity in 2 dimensions, leading to a finite integral over $d^2z$. We may thus take the limit $x \to y$:

$$ \lim_{x \to y} \delta G(x, y) = \int dz \left[ (\delta\sigma(z) - \delta\sigma(y))S(y, z)S(z, y) + \hat{\delta}S(y, z)S(z, y) + S(y, z)\hat{\delta}S(z, y) \right], \qquad (5.18) $$

and subtracting (5.17) with $x = y$ we get

$$ \delta G_{\text{R}}(y) = \int dz \left[ (\delta\sigma(z) - \delta\sigma(y))S(y, z)S(z, y) + \hat{\delta}S(y, z)S(z, y) + S(y, z)\hat{\delta}S(z, y) \right] + \frac{\delta\sigma(y)}{2\pi} \mathbf{1}_{2 \times 2}. \qquad (5.19) $$

If we now take the trace and integrate over $y$, the first term in the integrand is odd under the exchange of $y$ and $z$ and thus does not contribute. Using the cyclicity of the trace and exchanging again $y$ and $z$, the second and third terms are easily seen to be identical. We arrive at

$$ \int dy \delta \operatorname{tr} G_{\text{R}}(y) = 2 \int dy dz \operatorname{tr} S(y, z) \hat{\delta}S(z, y) + \frac{\delta A}{2\pi}. \qquad (5.20) $$
We now insert the explicit expression (5.12) for $\hat{\delta}S$ to get

$$ 
\begin{aligned}
\int dy \delta \operatorname{tr} G_R(y) &= \frac{\delta A}{2\pi} - 2m \int dy \int dz \int du \delta\sigma(u) \operatorname{tr} S(y, z) S(z, u) \gamma_* S(u, y) \\
&= \frac{\delta A}{2\pi} - 2m \int dy du \delta\sigma(u) \operatorname{tr} G(y, u) \gamma_* S(u, y).
\end{aligned} 
\quad (5.21) $$

In the previous section, we have shown in (4.56) that $\operatorname{tr} \gamma_* S(x, y) = m \operatorname{tr} G(x, y)$. The argument was based on the relation $S(x, y) = (i\not\nabla_x + m\gamma_*)G(x, y)$ and the fact that $G$ only contains the matrices $\mathbf{1}_{2\times2}$ and $\gamma_*$ and that in the end only $\mathbf{1}_{2\times2}$ has a non-vanishing trace. One can show in completely the same way that

$$ \operatorname{tr} \gamma_* S(x, y) G(u_1, v_1) \dots G(u_n, v_n) = m \operatorname{tr} G(x, y) G(u_1, v_1) \dots G(u_n, v_n). \quad (5.22) $$

This allows us to rewrite (5.21) as

$$ 
\begin{aligned}
\int dy \delta \operatorname{tr} G_R(y) &= \frac{\delta A}{2\pi} - 2m^2 \int dy du \delta\sigma(u) \operatorname{tr} G(u, y) G(y, u) \\
&= \frac{\delta A}{2\pi} - 2m^2 \int du \delta\sigma(u) \operatorname{tr} G_2(u, u),
\end{aligned} 
\quad (5.23) $$

where we used the orthonormality of the eigenfunctions to rewrite the right-hand side in terms of a higher Green's function $G_2$:

$$ \int dy G(u, y) G(y, v) = \sum_n \frac{1}{\lambda_n^4} (\chi_n(u) \chi_n^\dagger(v) + \phi_n(u) \phi_n^\dagger(v)) \equiv G_2(u, v) \equiv \zeta_+(2, u, v). \quad (5.24) $$

Let us note for later reference that one can equally easily show that

$$ \int du dv S(x, u) G(u, v) S(v, y) = G_2(x, y) \equiv \zeta_+(2, x, y). \quad (5.25) $$

Finally, inserting (5.23) into (5.7), we get

$$ \delta \left[ S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \frac{\gamma}{4\pi} m^2 A + \frac{m^2}{4} \int dy \operatorname{tr} G_R(y) \right] = -\frac{m^4}{2} \int du \delta\sigma(u) \operatorname{tr} G_2(u, u), \quad (5.26) $$

If there are no zero-modes of $i\not\nabla$, which is the case for spherical topology, then $S$ and $G$ have finite limits as $m \to 0$ and one can assert that the term on the right-hand side of (5.26) is of order $m^4$. In this case, if we satisfy ourselves with an order $m^2$ calculation, we may just drop the right-hand side, and moreover replace $G_R$ by the corresponding quantity in the massless theory. However, for genus one or larger, we have zero-modes of $i\not\nabla$ and one must separate the zero-mode and the non zero-mode parts to isolate the contributions that are truly of order $m^2$. This separation will be the subject of subsection 5.3.

### 5.2.2 Conformal variation of the higher Green's function $G_n$, $n \ge 2$

But let us go on and determine $\int du \delta \operatorname{tr} G_n(u, u)$ for $n \ge 2$. We have

$$ G_n(u_1, u_1) = \int du_2 \dots du_n G(u_1, u_2) G(u_2, u_3) \dots G(u_n, u_1), \quad (5.27) $$
so that

$$ 
\begin{aligned}
\int du_1 \delta \operatorname{tr} G_n(u_1, u_1)) &= \int du_1 \dots du_n \left[ 2 \sum_{i=2}^{n} \delta\sigma(u_i) \operatorname{tr} G(u_1, u_2) \dots G(u_n, u_1) \right. \\
&\qquad \left. + \sum_{i=1}^{n} \operatorname{tr} G(u_1, u_2) \dots \delta G(u_i, u_{i+1}) \dots G(u_n, u_1) \right] \\
&= \int du_1 \left[ 2(n-1)\delta\sigma(u_1) \operatorname{tr} G_n(u_1, u_1) + n \int du_1 du_2 \operatorname{tr} \delta G(u_1, u_2) G_{n-1}(u_2, u_1) \right], \quad (5.28)
\end{aligned}
$$

where in the last step we used the cyclicity of the trace. The variation $\delta G(z, y)$ was determined in (5.15) and it follows that

$$ 
\begin{aligned}
& \int du_1 du_2 \operatorname{tr} (\delta G(u_1, u_2)) G_{n-1}(u_2, u_1) \\
&= \int dv du_1 du_2 (\delta\sigma(v) - \frac{\delta\sigma(u_1) + \delta\sigma(u_2)}{2}) \operatorname{tr} S(u_1, v) S(v, u_2) G_{n-1}(u_2, u_1) \\
&\quad + \int dv du_1 du_2 \operatorname{tr} (\hat{\delta}S(u_1, v) S(v, u_2) + S(u_1, v) \hat{\delta}S(v, u_2)) G_{n-1}(u_2, u_1). \quad (5.29)
\end{aligned}
$$

The different pieces of the first term on the right-hand side can be rewritten, using the cyclicity of the trace and an identity analogous to (5.25), as $\int dv du_1 du_2 \delta\sigma(v) \operatorname{tr} S(u_1, v) S(v, u_2) G_{n-1}(u_2, u_1) = \int dv \delta\sigma(v) \operatorname{tr} G_n(v, v)$, and similarly for $\int dv du_1 du_2 \delta\sigma(u_1) \operatorname{tr} S(u_1, v) S(v, u_2) G_{n-1}(u_2, u_1) = \int du_1 \delta\sigma(u_1) \operatorname{tr} G_n(u_1, u_1)$, etc, and we see that the terms on the right-hand side of (5.29) involving the explicit $\delta\sigma$ cancel. Hence only the terms involving $\hat{\delta}S$ remain. Thus

$$ 
\begin{aligned}
& \int du_1 du_2 \operatorname{tr} (\delta G(u_1, u_2)) G_{n-1}(u_2, \dots, u_1) \\
&= -m \int dv dw du_1 du_2 \delta\sigma(w) \operatorname{tr} \left( S(u_1, w) \gamma_* S(w, v) S(v, u_2) + S(u_1, v) S(v, w) \gamma_* S(w, u_2) \right) G_{n-1}(u_2, u_1) \\
&= -m \int dw du_1 \delta\sigma(w) \operatorname{tr} \left( \gamma_* G_n(w, u_1) S(u_1, w) + \gamma_* S(w, u_1) G_n(u_1, w) \right) \\
&= -2m^2 \int dw \delta\sigma(w) \operatorname{tr} G_{n+1}(w, w). \quad (5.30)
\end{aligned}
$$

We see that we can rewrite (5.28) as

$$ 
\int du \delta \operatorname{tr} G_n(u, u) - 2(n-1) \int du \delta\sigma(u) \operatorname{tr} G_n(u, u) = -2n m^2 \int du \delta\sigma(u) \operatorname{tr} G_{n+1}(u, u). \quad (5.31)
$$

Now, on the left-hand side we combine $\int du \delta \operatorname{tr} G_n + 2 \int du \delta\sigma(u) \operatorname{tr} G_n = \delta \int du \operatorname{tr} G_n$ so that, after dividing by $2n$:

$$ 
\int du \delta\sigma(u) \operatorname{tr} G_n = \delta \int du \frac{1}{2n} \operatorname{tr} G_n + m^2 \int du \delta\sigma(u) \operatorname{tr} G_{n+1}, \quad (5.32)
$$

where we dropped the arguments, all Green's functions being obviously evaluated at coinciding points $u$. If we multiply this equation by $m^{2n}$ and iterate it $N-n+1$ times we
get

$$m^{2n} \int du \delta\sigma(u) \operatorname{tr} G_n = \delta \int du \sum_{k=n}^{N} \frac{m^{2k}}{2k} \operatorname{tr} G_k + m^{2(N+1)} \int du \delta\sigma(u) \operatorname{tr} G_{N+1}. \quad (5.33)$$

We now use this relation for $n = 2$ to rewrite the right-hand side of (5.26) to get

$$\delta \left[ S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \frac{\gamma}{4\pi} m^2 A + \frac{1}{4} \int d^2 y \sqrt{g} \left( m^2 \operatorname{tr} G_R(y) + \sum_{k=2}^{N} \frac{m^{2k}}{k} \operatorname{tr} G_k(y, y) \right) \right] = -\frac{m^{2(N+1)}}{2} \int d^2 y \sqrt{g} \delta\sigma(y) \operatorname{tr} G_{N+1}(y, y), \quad (5.34)$$

where on the left-hand side one can replace $\int d^2y \sqrt{g} \operatorname{tr} G_k(y, y) = \zeta(k)$.

### 5.2.3 Variation of the gravitational action to all orders in $m^2$ for spherical topology

Contrary to what this expansion (5.34) might suggest, it is important to realise that the quantities $G_R$ and $\zeta(k)$ are the quantities computed in the massive theory. However, in the absence of zero-modes of $i\nabla\nabla$ (i.e. for spherical topology), they all have a finite limit as $m$ goes to zero. Below, in subsubsection 5.3.5, we will give a precise bound on the remainder term on the right-hand side of (5.34). In particular, for $\frac{m^2A}{a_*^2} < 1$, where $a_*$ is the smallest eigenvalue of $D$ for zero mass and unit area $A = 1$, this term goes to zero as $N \to \infty$. Hence

$$\delta \left[ S_{\text{grav}}^{\text{sphere}} + \frac{1}{48\pi} S_{\text{Liouville}} + \frac{\gamma}{4\pi} m^2 A + \frac{m^2}{4} \int \sqrt{g} \operatorname{tr} G_R(y) + \sum_{k=2}^{\infty} \frac{m^{2k}}{4k} \zeta(k) \right] = 0 \quad (5.35)$$

It is relatively straightforward to re-express $G_R$ and $\zeta(k)$ in terms of the same set of quantities in the massless theory. We will do this below for the case of arbitrary genus.

## 5.3 Variation of the gravitational action for arbitrary genus

### 5.3.1 Zero-mode and non-zero-mode parts of the Green's functions

As discussed in subsection 2.5, for genus one or larger, there are zero-modes $\psi_{0,i}$ of $i\nabla\nabla$. They may be chosen as having definite chirality, i.e. be eigenstates of $\gamma_*$.* There are as many positive as negative chirality zero-modes. For the arbitrary torus there is exactly one zero-mode of each chirality and we have obtained them explicitly. In general, since $i\nabla\nabla\psi_{0,i} = 0$ we have $D\psi_{0,i} = m\gamma_*\psi_{0,i} = \pm m\psi_{0,i}$, i.e $\lambda_0 = m$. Recall the spectral decomposition (4.49) of the Green's functions in terms of the eigenfunctions and eigenvalues. It follows that the zero-mode parts of the Green's functions $S$ and $G$ behave as $\frac{1}{m}$ and $\frac{1}{m^2}$ and do not have a limit as $m \to 0$. Since we want to do a small-$m$ expansion, we have to separate the Green's functions into their zero-mode and their non-zero-mode parts. In terms of the projectors $P_0$ and $Q_0$ on the zero-modes defined in (2.102) and (2.106), the zero-mode contribution to the Green's functions $G$ and $S$ simply are $\frac{P_0}{\lambda_0^2} = \frac{P_0}{m^2}$ and $\frac{Q_0}{\lambda_0} = \frac{Q_0}{m}$. Obviously then, one
has the decomposition

$$
\begin{align}
G(x, y) &= G_{(0)}(x, y) + \tilde{G}(x, y), & G_{(0)}(x, y) &= \frac{P_0(x, y)}{m^2}, \nonumber \\
S(x, y) &= S_{(0)}(x, y) + \tilde{S}(x, y), & S_{(0)}(x, y) &= \frac{Q_0(x, y)}{m}, \tag{5.36}
\end{align}
$$

where $\tilde{G}(x, y)$ and $\tilde{S}(x, y)$ are the contributions of the non-zero-modes for which the $\lambda_n$
have finite limits as $m \to 0$. Hence, $\tilde{G}(x, y)$ and $\tilde{S}(x, y)$ have finite limits as $m \to 0$, which
obviously is not the case for $G^{(0)}$ and $S^{(0)}$.

As already noted, the zero-modes are orthogonal to the non-zero modes so that

$$
\begin{align}
\int dz S_{(0)}(x, z) \tilde{S}(z, y) &= 0, & \int dz G_{(0)}(x, z) \tilde{G}(z, y) &= 0, \nonumber \\
\int dz S_{(0)}(x, z) \tilde{G}(z, y) &= 0, & \int dz G_{(0)}(x, z) \tilde{S}(z, y) &= 0. \tag{5.37}
\end{align}
$$

Furthermore (cf (2.107))

$$
\int dz S_{(0)}(x, z)S_{(0)}(z, y) = \frac{1}{m^2} \int dz Q_0(x, z)Q_0(z, y) = \frac{1}{m^2} P_0(x, y) = G_{(0)}(z, y). \quad (5.38)
$$

We also have $G(x, y) = \int dz S(x, z)S(z, y)$. If we substitute $S = \tilde{S} + S_{(0)}$ and use the
orthogonality between the zero and non-zero-modes, eq. (5.37), we get

$$
\tilde{G}(x, y) = \int dz \tilde{S}(x, z) \tilde{S}(z, y). \tag{5.39}
$$

We can similarly separate all higher Green's functions as $G_n(x, y) = G_n^{(0)}(x, y) + \tilde{G}_n(x, y)$
and we have separate relations for the zero-mode and the non-zero-mode parts:

$$
\begin{align}
\tilde{G}_{n+1}(x, y) &= \int dz \tilde{G}(x, z) \tilde{G}_n(z, y) = \int dz du \tilde{S}(x, z) \tilde{G}_n(z, u) \tilde{S}(u, y), \nonumber \\
G_{n+1}^{(0)}(x, y) &= \int dz G_{(0)}(x, z) \tilde{G}_n^{(0)}(z, y) = \int dz du S_{(0)}(x, z) G_n^{(0)}(z, u) S_{(0)}(u, y). \tag{5.40}
\end{align}
$$

Finally note that the zero-mode parts satisfy $i\nabla\tilde{S}_{(0)} = i\nabla\tilde{G}_{(0)} = 0$ and recall from (2.107) that $Q_0 = \gamma_* P_0$ so that $S_{(0)} = m\gamma_* G_{(0)} = (i\nabla + m\gamma_*)G_{(0)}$. Subtracting this identity from $S = (i\nabla + m\gamma_*)G$ results in

$$
\tilde{S}(x, y) = (i\nabla_x + m\gamma_*)\tilde{G}(x, y) \equiv D_x\tilde{G}(x, y), \quad (5.41)
$$

so that the zero-mode parts and the non-zero-mode parts satisfy this same relation sepa-
rately. Of course, all these relations also follow directly from the spectral representations
of the Green's functions.

We now want to rewrite the various relations of the previous subsection in terms of
these $\tilde{S}(x, y)$ and $\tilde{G}(x, y)$ for which the $m \to 0$ limits exist. In the end we will find that we
obtain almost the same result as in (5.34) or (5.35) except that all quantities are replaced
by the corresponding quantities with tildes referring to the non-zero-modes. This result
can be understood as being due to the fact that when varying the gravitational action the
eigenvalue $\lambda_0$ corresponding to the zero-modes does not change and that the zero-mode and non-zero-mode contributions are orthogonal.

But let us proceed. First note that the zero-mode pieces do not contribute to the short-distance singularities. Hence $G(x, y)$ and $\tilde{G}(x, y)$ have the same short-distance singularity $-\frac{1}{4\pi} \log \frac{\mu^2 \ell^2(x, y)}{4} \mathbf{1}_{2 \times 2}$. Similarly $S(x, y)$ and $\tilde{S}(x, y)$ have the same short-distance singularity $-\frac{1}{4\pi} (i \nabla_x + m\gamma_*) \log \frac{\mu^2 \ell^2(x, y)}{4}$. Thus

$$ S^{\text{reg}}(x, y) = S_{(0)}(x, y) + \tilde{S}(x, y) + \frac{1}{4\pi} (i \nabla_x + m\gamma_*) \log \frac{\mu^2 \ell^2(x, y)}{4} = \frac{Q_0(x, y)}{m} + \tilde{S}^{\text{reg}}(x, y), \\ G^{\text{reg}}(x, y) = G_{(0)}(x, y) + \tilde{G}(x, y) + \frac{1}{4\pi} \log \frac{\mu^2 \ell^2(x, y)}{4} \mathbf{1}_{2 \times 2} = \frac{P_0(x, y)}{m^2} + \tilde{G}^{\text{reg}}(x, y). \quad (5.42) $$

By taking the coincidence limit $x \to y$ we then get

$$ S_R(y) = \frac{Q_0(y, y)}{m} + \tilde{S}_R(y), \quad G_R(y) = \frac{P_0(y, y)}{m^2} + \tilde{G}_R(y). \quad (5.43) $$

Also, since $Q_0 = \gamma_* P_0$, the relation (4.114) translates into

$$ \tilde{S}_R = m\gamma_*\tilde{G}_R. \quad (5.44) $$

But both, $\tilde{S}_R$ and $\tilde{G}_R$ have a smooth limit as $m \to 0$, and we find that $\tilde{S}_R$ vanishes in the massless theory:

$$ \tilde{S}_R^{m=0}(x) = 0. \quad (5.45) $$

We will need the variations of the zero-mode parts of the Green's functions $G_{(0)}(x, y)$ and $S_{(0)}(x, y)$ under conformal transformations. As we have just seen, the latter are given in terms of the zero-mode projectors $P_0(x, y)$ and $Q_0(x, y) = \gamma_* P_0(x, y)$. But the conformal variations of $P_0(x, y)$ have been studied in some detail in subsection 2.5.4, see in particular (2.115) and (2.116).

### 5.3.2 Expressing the variation of the gravitational action in terms of $\tilde{G}_R$

The variation of the gravitational action, eq. (5.5) contains the term $-\frac{m^2}{2} \int \sqrt{g} \delta\sigma \operatorname{tr} G_R$ on its right-hand side. We now use (5.43) and (2.116) to rewrite this as

$$ \begin{aligned} -\frac{m^2}{2} \int dy \, \delta\sigma(y) \operatorname{tr} G_R(y) &= -\frac{m^2}{2} \int dy \, \delta\sigma(y) \operatorname{tr} \left( \frac{P_0(y, y)}{m^2} + \tilde{G}_R(y) \right) \\ &= -\delta \left( \log \det \mathcal{P}_0 + \frac{m^2}{4} \int dy \operatorname{tr} \tilde{G}_R(y) \right) + \frac{m^2}{4} \int dy \, \delta \operatorname{tr} \tilde{G}_R(y). \end{aligned} \quad (5.46) $$

We can then rewrite our initial equation (5.7) as

$$ \delta \left( S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \log \det \mathcal{P}_0 + \frac{m^2}{8\pi} (2\gamma+1) A + \frac{m^2}{4} \int \sqrt{g} \operatorname{tr} \tilde{G}_R \right) = \frac{m^2}{4} \int \sqrt{g} \delta \operatorname{tr} \tilde{G}_R. \quad (5.47) $$

This is as (5.7) but with $G_R$ replaced by $\tilde{G}_R$ and the extra term $\delta \log \det \mathcal{P}_0$ on the left-hand-side.
At present, this term $\delta \log \det \mathcal{P}_0$ has emerged from separating the zero-mode piece from $G_R$. But we have seen in section 3 that, since $\delta\lambda_0 = 0$, we may write $\delta S_{\text{grav}}$ from the beginning in terms of quantities that do not involve the zero-modes. Then in eq. (5.5) it is directly $\tilde{G}_R$ that appears. So where does this term $\delta \log \det \mathcal{P}_0$ then come from? Recall that the variation of the gravitational action was $\delta S_{\text{grav}} = \frac{1}{4}\delta\zeta'(0) + \frac{1}{4}\log \mu^2 \delta\zeta(0)$, cf (5.1). First, $\delta\zeta(0)$ only involved the residue of the pole of $\zeta_+(s, x, x)$ which is the same as the residue of the pole of $\tilde{\zeta}_+(s, x, x)$. Second, when expressing $\delta\zeta'(0)$ directly in terms of the quantities without zero-modes as given in the second line of (4.67), namely

$$ \frac{1}{4}\delta\zeta'(0) = \frac{1}{2} \int \delta\sigma \operatorname{tr} \tilde{\zeta}_+(0, x, x) - \frac{m^2}{2} \int \delta\sigma \operatorname{tr} \tilde{\zeta}_+^{\text{reg}}(1, x, x) - 2m^2 C_+ \delta A, \quad (5.48) $$

the last two terms on the right-hand side will be expressed in terms of $\int \delta\sigma \tilde{G}_R$ with no zero-mode contribution subtracted. On the other hand, $\zeta_+(0, x, x)$ was given in terms of the heat kernel coefficient $\frac{1}{4\pi}F_1(x, x)$, cf (4.40), as $\zeta_+(0, x, x) = -\frac{1}{4\pi}(\frac{\mathcal{R}(x)}{12} + m^2) \mathbf{1}_{2 \times 2}$. Now the corresponding $\tilde{\zeta}_+(0, x, x)$ is obtained from the heat kernel $\tilde{K}_+(t, x, x)$ with its zero-mode part subtracted. But the latter is just $e^{-m^2 t} P_0(x, x) = P_0(x, x)(1 + \mathcal{O}(t))$. It follows that $\frac{1}{4\pi}\tilde{F}_1(x, x) = \frac{1}{4\pi}F_1(x, x) - P_0(x, x)$ and, hence,

$$ \tilde{\zeta}_+(0, x, x) = -\frac{1}{4\pi} \left( \frac{\mathcal{R}(x)}{12} + m^2 \right) \mathbf{1}_{2 \times 2} - P_0(x, x), \quad (5.49) $$

so that

$$ \frac{1}{2} \int dx \, \delta\sigma(x) \operatorname{tr} \tilde{\zeta}_+(0, x, x) = -\frac{1}{48\pi} \delta S_{\text{Liouville}} - \frac{m^2}{8\pi} \delta A - \delta \log \det \mathcal{P}_0, \quad (5.50) $$

where we used (2.116). Thus one gets again all the terms of (5.47). This latter computation also shows that this zero-mode related term $\log \det \mathcal{P}_0$ is actually already present in the massless theory and should have appeared together with the Liouville action. But, as mentioned earlier, in the massless theory one may redefine the matter partition function by an appropriate factor to precisely cancel this term.

Starting from (5.47), we can now proceed as in the previous subsection and express $\int \sqrt{g} \delta \operatorname{tr} \tilde{G}_R$ in terms of a total variation and $\int \sqrt{g} \delta \operatorname{tr} \tilde{G}_2(y, y)$, which then is again expressed as a total variation and a term $\int \sqrt{g} \delta \operatorname{tr} \tilde{G}_3(y, y)$, etc. We have to be careful to subtract any zero-mode contributions where they might appear, but the orthogonality (5.37) of the zero and non-zero mode parts of the Green's functions essentially ensures that the variations of the non-zero-mode quantities only involve other non-zero-mode quantities.

To begin with, we have

$$ \delta \tilde{G}_R(y) = \lim_{x \to y} \delta \left( \tilde{G}(x, y) + \frac{1}{4\pi} \log \frac{\mu^2 \ell^2(x, y)}{4} \mathbf{1}_{2 \times 2} \right). \quad (5.51) $$

We already observed above that $\delta(\frac{1}{4\pi} \log \frac{\mu^2 \ell^2(x, y)}{4}) \sim_{x \to y} \frac{\delta\sigma(y)}{2\pi}$ is non-singular and, hence, the same must be true for $\delta \tilde{G}(x, y)$. From (5.39) we know that $\tilde{G}(x, y) = \int dz \tilde{S}(x, z) \tilde{S}(z, y)$, and we can set $x = y$ after taking the variation, so that

$$ \delta \tilde{G}_R(y) = \int dz (2\delta\sigma(z) \tilde{S}(y, z) \tilde{S}(z, y) + \delta \tilde{S}(y, z) \tilde{S}(z, y) + \tilde{S}(y, z) \delta \tilde{S}(z, y)) + \frac{\delta\sigma(y)}{2\pi} \mathbf{1}_{2 \times 2}. \quad (5.52) $$
### 5.3.3 Determining $\delta \tilde{S}$

We then need to determine $\delta \tilde{S}(x, y)$. To do this, we just repeat the steps leading to (5.12) and (5.13), but now subtracting the zero-mode parts. We start with $(i\nabla_x + m\gamma_*)S(x, y) = \delta^{(2)}(x-y)/\sqrt{g}$ and separate the non-zero and zero-mode parts: $S = \tilde{S}+S_{(0)}$. The zero-mode piece $S_{(0)} = \frac{Q_0}{m}$ satisfies $i\nabla_x S_{(0)}(x, y) = 0$ so that (recall $\gamma_* Q_0 = P_0$)

$$(i\nabla_x + m\gamma_*)\tilde{S}(x, y) = \frac{\delta^{(2)}(x-y)}{\sqrt{g}} - m\gamma_* S_{(0)}(x, y) = \frac{\delta^{(2)}(x-y)}{\sqrt{g}} - P_0(x, y). \quad (5.53)$$

Of course, this just expresses that $D_x \tilde{S}(x, y)$ gives the completeness relation without the zero-modes. Consider now a metric $g'$ related to $g$ by an infinitesimal conformal rescaling $g' = e^{2\delta\sigma}g$. Then, as before, $\sqrt{g'} = e^{2\delta\sigma}\sqrt{g}$ and $\nabla' = e^{-3\delta\sigma/2}\nabla e^{\delta\sigma/2}$. We have, of course, $\tilde{S}'(x, y) = S'(x, y) - S'_{(0)}(x, y)$ where $i\nabla' S'_{(0)}(x, y) = 0$. Then (5.53) becomes

$$(ie^{-3\delta\sigma(x)/2} \nabla e^{\delta\sigma(x)/2} + m\gamma_*) \tilde{S}'(x, y) = \frac{\delta^{(2)}(x-y)}{\sqrt{g}} e^{-3\delta\sigma(x)/2} e^{-\delta\sigma(y)/2} - P'_0(x, y). \quad (5.54)$$

Much as in (5.9), we set

$$ \begin{gathered} \tilde{S}'(x, y) = \tilde{S}(x, y) + \delta \tilde{S}(x, y) = e^{-\delta \sigma(x)/2} (\tilde{S}(x, y) + \hat{\delta} \tilde{S}(x, y)) e^{-\delta \sigma(y)/2} \\ \Leftrightarrow \quad \delta \tilde{S}(x, y) = -\frac{1}{2}(\delta \sigma(x) + \delta \sigma(y)) \tilde{S}(x, y) + \hat{\delta} \tilde{S}(x, y), \end{gathered} \quad (5.55) $$

where the equivalence between the two equations holds up to terms of second order in the variations. Developing (5.54) to first order in the variations and comparing with (5.53) we get

$$ \begin{align} D_x \hat{\delta} \tilde{S}(x, y) &\equiv (i \nabla_x + m \gamma_*) \hat{\delta} \tilde{S}(x, y) \tag{5.56} \\ &= -m \delta \sigma(x) \gamma_* \tilde{S}(x, y) - \left( \frac{3}{2} \delta \sigma(x) + \frac{1}{2} \delta \sigma(y) \right) P_0(x, y) - \delta P_0(x, y), \nonumber \end{align} $$

where $\delta P_0 = P'_0 - P_0$, so that

$$ \begin{align} \int dz \tilde{S}(x, z) D_z \hat{\delta} \tilde{S}(z, y) &= -m \int dz \tilde{S}(x, z) \delta \sigma(z) \gamma_* \tilde{S}(z, y) \nonumber \\ &\quad - \int dz \tilde{S}(x, z) \left[ \left( \frac{3}{2} \delta \sigma(z) + \frac{1}{2} \delta \sigma(y) \right) P_0(z, y) + \delta P_0(z, y) \right]. \tag{5.57} \end{align} $$

On the left-hand side, as before, we use the hermiticity of $D_z$ and let it act on $\tilde{S}(x, z)$, according to (5.53):

$$ \int dz \tilde{S}(x, z) D_z \hat{\delta} \tilde{S}(z, y) = \hat{\delta} \tilde{S}(x, y) - \int dz P_0(x, z) \hat{\delta} \tilde{S}(z, y). \quad (5.58) $$

Hence,

$$ \begin{align} \hat{\delta} \tilde{S}(x, y) &= -m \int dz \tilde{S}(x, z) \delta \sigma(z) \gamma_* \tilde{S}(z, y) \nonumber \\ &\quad - \int dz \tilde{S}(x, z) \left[ \left( \frac{3}{2} \delta \sigma(z) + \frac{1}{2} \delta \sigma(y) \right) P_0(z, y) + \delta P_0(z, y) \right] \nonumber \\ &\quad + \int dz P_0(x, z) \hat{\delta} \tilde{S}(z, y). \tag{5.59} \end{align} $$
Recall the orthogonality relation between $\tilde{S}$ and $G_{(0)} \sim P_0$ which we write for metric $g$ and for the metric $g' = e^{2\delta\sigma}g$ (with $dz$ still standing for $d^2z\sqrt{g(z)}$):

$$ \int dz P_0(x, z) \tilde{S}(z, y) = 0, \quad \int dz e^{3\delta\sigma(z)/2} e^{-\delta\sigma(y)/2} P_0'(x, z) (\tilde{S}(z, y) + \hat{\delta}\tilde{S}(z, y)) = 0. \quad (5.60) $$

First, this shows that the term $\sim \delta\sigma(y)$ in (5.59) does not contribute. Next, developing the second equation to first order and using the first one we get

$$ \int dz P_0(x, z) \hat{\delta}\tilde{S}(z, y) = - \int dz \left( \frac{3}{2} \delta\sigma(z) P_0(x, z) \tilde{S}(z, y) + \delta P_0(x, z) \tilde{S}(z, y) \right). \quad (5.61) $$

Hence,

$$ \begin{aligned} \hat{\delta}\tilde{S}(x, y) = & -m \int dz \delta\sigma(z) \tilde{S}(x, z) \gamma_* \tilde{S}(z, y) \\ & - \int dz \left[ \frac{3}{2} \delta\sigma(z) (\tilde{S}(x, z) P_0(z, y) + P_0(x, z) \tilde{S}(z, y)) + \tilde{S}(x, z) \delta P_0(z, y) + \delta P_0(x, z) \tilde{S}(z, y) \right]. \end{aligned} \quad (5.62) $$

Now, $\delta P_0(x, z)$ has been worked out in (2.115). It contains a piece $-\frac{1}{2}(\delta\sigma(x)+\delta\sigma(z))P_0(x, z)$ as well as a piece $\sum_{i,j} a_{ij} \psi_{0,i}(x) \psi_{0,j}^\dagger(z)$ where the $a_{ij}$ are some constants. Thus, when multiplied by $\tilde{S}(z, y)$ and integrated over $dz$ the second piece does not contribute. Similarly for $\int dz \tilde{S}(x, z) \delta P_0(z, y)$. Thus we can rewrite (5.62) as

$$ \hat{\delta}\tilde{S}(x, y) = - \int dz \delta\sigma(z) \left[ m \tilde{S}(x, z) \gamma_* \tilde{S}(z, y) + \tilde{S}(x, z) P_0(z, y) + P_0(x, z) \tilde{S}(z, y) \right]. \quad (5.63) $$

This, together with the second equation (5.55) determines $\delta\tilde{S}$ entirely in terms of $\tilde{S}$, $\delta\sigma$ and $P_0$. Note that we cannot (yet) use the “orthogonality” of $\tilde{S}$ and $P_0$ since the integral over $dz$ also involves $\delta\sigma(z)$. However, we will see in the sequel that the $P_0$ terms drop out from the relevant quantities we will compute here.

### 5.3.4 Determining $\delta\tilde{G}_R$

We have already given $\delta\tilde{G}_R$ in terms of $\delta\tilde{S}$ and $\delta\sigma$ in (5.52). Using now the second equation (5.55) to express $\delta\tilde{S}$ in terms of $\delta\sigma$ and $\hat{\delta}\tilde{S}$, this becomes

$$ \begin{aligned} \delta\tilde{G}_R(y) = & \int dz \left( (\delta\sigma(z) - \delta\sigma(y))\tilde{S}(y, z)\tilde{S}(z, y) + \hat{\delta}\tilde{S}(x, z)\tilde{S}(z, y) + \tilde{S}(x, y)\hat{\delta}\tilde{S}(z, y) \right) \\ & + \frac{\delta\sigma(y)}{2\pi} \mathbf{1}_{2\times2}. \end{aligned} \quad (5.64) $$

Let us remark for further reference that if we do not take the $x \to y$ limit and discard the last term we also get

$$ \delta\tilde{G}(x, y) = \int dz \left( \left( \delta\sigma(z) - \frac{\delta\sigma(x)}{2} - \frac{\delta\sigma(y)}{2} \right) \tilde{S}(x, z) \tilde{S}(z, y) + \hat{\delta}\tilde{S}(x, z) \tilde{S}(z, y) + \tilde{S}(x, z) \hat{\delta}\tilde{S}(z, y) \right). \quad (5.65) $$

If we take the trace of $\delta\tilde{G}_R$ and integrate, as before, the term $\delta\sigma(z) - \delta\sigma(y)$ vanishes by antisymmetry, and we are left with

$$ \int dy \delta \operatorname{tr} \tilde{G}_R(y) = 2 \int dy dz \operatorname{tr} \hat{\delta}\tilde{S}(y, z) \tilde{S}(z, y) + \frac{\delta A}{2\pi}. \quad (5.66) $$
Next, we insert $\hat{\delta}\tilde{S}$ from (5.63):

$$ \int dy \delta \operatorname{tr} \tilde{G}_{\mathrm{R}}(y) = -2 \int dy \, dz \, du \, \delta\sigma(u) \operatorname{tr} \left( m \tilde{S}(y, u) \gamma_\ast \tilde{S}(u, z) \right. \\ \left. \qquad + \tilde{S}(y, u) P_0(u, z) + P_0(y, u) \tilde{S}(u, z) \right) \tilde{S}(z, y) + \frac{\delta A}{2\pi}. \qquad (5.67) $$

We see that the terms involving $P_0$ now vanish: $\int dz P_0(u, z) \tilde{S}(z, y) = 0$ and (using the cyclicity of the trace) $\int dy \tilde{S}(z, y) P_0(y, u) = 0$. Furthermore, we can replace $\int dy \tilde{S}(z, y) \tilde{S}(y, u)$ by $\tilde{G}(z, u)$. We are left with

$$ \begin{aligned} \int dy \delta \operatorname{tr} \tilde{G}_{\mathrm{R}}(y) &= -2m \int dz \, du \, \delta\sigma(u) \operatorname{tr} \gamma_\ast \tilde{S}(u, z) \tilde{G}(z, u) + \frac{\delta A}{2\pi} \\ &= -2m^2 \int dz \, du \, \delta\sigma(u) \operatorname{tr} \tilde{G}(u, z) \tilde{G}(z, u) + \frac{\delta A}{2\pi} = -2m^2 \int du \, \delta\sigma(u) \operatorname{tr} \tilde{G}_2(u, u) + \frac{\delta A}{2\pi}, \end{aligned} \qquad (5.68) $$

where we used, much as for $S$ and $G$, that $\tilde{S}(u, z) = (i\not\nabla_u + m\gamma_\ast)\tilde{G}(u, z)$ and the fact that only the $m\gamma_\ast\tilde{G}$ survives in the trace. Inserting this into (5.47) we get

$$ \begin{aligned} \delta \left( S_{\mathrm{grav}} + \frac{1}{48\pi} S_{\mathrm{Liouville}} + \log \det \mathcal{P}_0 + \frac{\gamma}{4\pi} m^2 A + \frac{m^2}{4} \int du \operatorname{tr} \tilde{G}_{\mathrm{R}}(u) \right) \\ = -\frac{m^4}{2} \int du \, \delta\sigma(u) \operatorname{tr} \tilde{G}_2(u, u). \end{aligned} \qquad (5.69) $$

Of course, this is very similar to (5.26), except that now all Green's functions have their zero-modes excluded and, instead, we have the zero-mode related piece $\log \det \mathcal{P}_0$.

Since $\tilde{G}_2$ has a finite limit as $m \to 0$, the right-hand side is a genuine order $m^4$-term and one can read the complete variation of $S_{\mathrm{grav}}$ up to and including the order $m^2$-terms from the left-hand side. But we also want to obtain the higher-order contributions which requires to study similarly the variation of the higher $\tilde{G}_n$.

### 5.3.5 Conformal variation of the higher $\tilde{G}_n(x, y)$, $n \ge 2$

As before, we now want to obtain the conformal variations of the non-zero-mode parts of the higher Green's functions. It follows immediately from the orthonormality of the zero-mode and non-zero-mode parts of $G$ that

$$ \begin{aligned} \tilde{G}_n(x, y) &= G_n(x, y) - G_n^{(0)}(x, y) \equiv G_n(x, y) - \frac{P_0(x, y)}{m^{2n}} \\ &= \int du_2 \dots du_n \tilde{G}(x, u_2) \tilde{G}(u_2, u_3) \dots \tilde{G}(u_n, y). \end{aligned} \qquad (5.70) $$

Then, as before for $G_n$ in (5.28), we now have for $\tilde{G}_n$

$$ \begin{aligned} \int du_1 \delta \operatorname{tr} \tilde{G}_n(u_1, u_1) ={}& (2(n-1) \int du \, \delta\sigma(u) \operatorname{tr} \tilde{G}_{(n)}(u, u) \\ & + n \int du_1 du_2 \operatorname{tr} \delta \tilde{G}(u_1, u_2) \tilde{G}_{(n-1)}(u_2, u_1). \end{aligned} \qquad (5.71) $$
Let us evaluate the second term on the right-hand side of this equation. $\delta\tilde{G}$ was given in (5.65) so that

$$ 
\begin{aligned}
& \int dx dy \operatorname{tr} \delta\tilde{G}(x, y)\tilde{G}_{(n-1)}(y, x) \\
& = \int dx dy dz \left[ \begin{aligned}
& \left(\delta\sigma(z) - \frac{\delta\sigma(x)}{2} - \frac{\delta\sigma(y)}{2}\right) \operatorname{tr} \tilde{S}(x, z)\tilde{S}(z, y)\tilde{G}_{(n-1)}(y, x) \\
& + \operatorname{tr} (\hat{\delta}\tilde{S}(x, z)\tilde{S}(z, y) + \tilde{S}(x, z)\hat{\delta}\tilde{S}(z, y))\tilde{G}_{(n-1)}(y, x)
\end{aligned} \right].
\end{aligned} 
\quad (5.72) $$

Using the cyclicity of the trace we can show again that the term involving explicitly the $\delta\sigma$ vanishes and only the terms involving $\hat{\delta}\tilde{S}$ remain. Using (5.63) these terms are

$$ 
\begin{aligned}
& \int dx dy dz \operatorname{tr} (\hat{\delta}\tilde{S}(x, z)\tilde{S}(z, y) + \tilde{S}(x, z)\hat{\delta}\tilde{S}(z, y))\tilde{G}_{(n-1)}(y, x) \\
& = - \int dx dy du dz \delta\sigma(z) \operatorname{tr} [m\tilde{S}(x, z)\gamma_*\tilde{S}(z, u) + \tilde{S}(x, z)P_0(z, u) + P_0(x, z)\tilde{S}(z, u)] \times \\
& \qquad \times [\tilde{S}(u, y)\tilde{G}_{(n-1)}(y, x) + \tilde{G}_{(n-1)}(u, y)\tilde{S}(y, x)].
\end{aligned} 
\quad (5.73) $$

Again, by the cyclicity of the trace, we see that the terms involving $P_0$ always appear as $\int du P_0(z, u)\tilde{S}(u, y) = 0$ or $\int dx \tilde{G}_{(n-1)}(y, x)P_0(x, z) = 0$, etc. Hence they give vanishing contributions and (5.73) reduces to

$$ 
\begin{aligned}
(5.73) &= -m \int dx dz \delta\sigma(z) \operatorname{tr} (\tilde{S}(x, z)\gamma_*\tilde{G}_n(z, x) + \gamma_*\tilde{S}(z, x)\tilde{G}_n(x, z)) \\
&= -m^2 \int dx dz \delta\sigma(z) \operatorname{tr} (\tilde{G}(x, z)\tilde{G}_n(z, x) + \tilde{G}(z, x)\tilde{G}_n(x, z)) \\
&= -2m^2 \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_{n+1}(y, y),
\end{aligned} 
\quad (5.74) $$

where in the next to last step we used the by now familiar argument to replace $\tilde{S}\gamma_*$ and $\gamma_*\tilde{S}$ by $m\tilde{G}$ inside the trace. Putting the pieces together, using (5.71), (5.72) and (5.74), we get

$$ 
\int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_n(y, y) = 2(n-1) \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_n(y, y) - 2n m^2 \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_{n+1}(y, y), 
\quad (5.75) $$

exactly as in (5.31) but now for the $\tilde{G}_n$ instead of the $G_n$. One can then do the same manipulations, obtaining the analogue relations of (5.32) (again we do drop the arguments $y$)

$$ 
\int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_n = \frac{1}{2n} \delta \int dy \operatorname{tr} \tilde{G}_n + m^2 \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_{n+1}. 
\quad (5.76) $$

This can be iterated to obtain a relation analogous to (5.33):

$$ m^{2n} \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_n = \delta \int dy \sum_{k=n}^{N} \frac{m^{2k}}{2k} \operatorname{tr} \tilde{G}_k + m^{2(N+1)} \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_{N+1}. 
\quad (5.77) $$

This looks like an expansion in powers of $m^2$, but one should be aware that the Green's functions $\tilde{G}_k$ are the Green's function of the massive theory. However, and this was the
whole point of having subtracted the zero-modes, each $\tilde{G}_k$ now has a finite limit as $m \to 0$ and can be expanded itself in powers of $m^2$, as we will do shortly.

Let us give an upper bound on the term on the right-hand side. We have $\int dy \operatorname{tr} \tilde{G}_{N+1} = \sum_{n \neq 0} \frac{2}{\lambda_n^{2(N+1)}} = \tilde{\zeta}(N+1)$. Furthermore, $\lambda_n^2 = \lambda_n^2|_{m=0} + m^2$ and the eigenvalues of the massless theory have a simple scaling with the area, namely$^{26}$

$$ \lambda_n^2|_{m=0}^A = \frac{1}{A} \lambda_n^2|_{m=0}^{A=1}. \qquad (5.78) $$

If we denote $M_{\delta\sigma} = \max|\delta\sigma| \ge 0$ we have the following bound

$$ \begin{aligned} \left| \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_{N+1} \right| &\le M_{\delta\sigma} \left| \int dy \operatorname{tr} \tilde{G}_{N+1} \right| = M_{\delta\sigma} \tilde{\zeta}(N+1) \le M_{\delta\sigma} \tilde{\zeta}_{m=0}(N+1) \\ &= A^{N+1} M_{\delta\sigma} \tilde{\zeta}_{m=0}^{A=1}(N+1). \end{aligned} \qquad (5.79) $$

Now, just as Riemann's zeta-function $\zeta_R(s)$ has a finite limit as $s \to \infty$, our $\tilde{\zeta}_{m=0}^{A=1}(N+1)$ has a well-defined "limit behaviour" as $N \to \infty$ and this is given in terms of the smallest eigenvalue that contributes. If we denote $a_0 \equiv \lambda_1|_{m=0}^{A=1}$ is the smallest (positive) non-zero eigenvalue for zero mass and unit area $A = 1$, and $d_{a_0}$ its degeneracy, then $\tilde{\zeta}_{m=0}^{A=1}(N+1) \sim_{N \to \infty} 2\frac{d_{a_0}}{a_0^{2(N+1)}}$. Hence

$$ \left| m^{2(N+1)} \int dy \delta\sigma(y) \operatorname{tr} \tilde{G}_{N+1} \right| \le m^{2(N+1)} A^{N+1} M_{\delta\sigma} \tilde{\zeta}_{m=0}^{A=1}(N+1) \sim_{N \to \infty} 2d_{a_0} M_{\delta\sigma} \left(\frac{m^2 A}{a_0^2}\right)^{N+1}. \qquad (5.80) $$

Thus for small enough mass, $m^2 A < a_0^2$ the right-hand side goes to zero as $N \to \infty$ and we may drop the corresponding term in (5.77). We may similarly estimate the radius of convergence of the series

$$ \int dy \sum_{k=n}^{\infty} \frac{m^{2k}}{2k} \operatorname{tr} \tilde{G}_k = \sum_{k=n}^{\infty} \frac{m^{2k}}{2k} \tilde{\zeta}(k) = \sum_{k=n}^{\infty} \frac{(m^2 A)^k}{2k} \tilde{\zeta}^{A=1}(k). \qquad (5.81) $$

If we now denote by $a$ the smallest positive eigenvalue (of the massive theory) appearing in $\tilde{\zeta}^{A=1}$ (such that its zero-mass limit is $a_0 > 0$), then the series converges for $|\frac{m^2 A}{a_0^2}| < 1$. We conclude that for small enough mass we can take the $N \to \infty$ limit in (5.77) and obtain

$$ m^{2n} \int d^2 y \sqrt{g} \delta\sigma(y) \operatorname{tr} \tilde{G}_n = \delta \int d^2 y \sqrt{g} \sum_{k=n}^{\infty} \frac{m^{2k}}{2k} \operatorname{tr} \tilde{G}_k. \qquad (5.82) $$

### 5.3.6 Variation of the gravitational action to all orders in $m^2$ for arbitrary genus

We insert this result (5.82) for $n = 2$ into eq. (5.69) and get

$$ \begin{aligned} & \delta \left[ S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \log \det \mathcal{P}_0 + \frac{\gamma}{4\pi} m^2 A \right. \\ & \left. \qquad + \frac{1}{4} \int d^2 u \sqrt{g} \left( m^2 \operatorname{tr} \tilde{G}_R(u) + \sum_{k=2}^{\infty} \frac{m^{2k}}{k} \operatorname{tr} \tilde{G}_k(u, u) \right) \right] = 0. \end{aligned} \qquad (5.83) $$

<footnote>$^{26}$This follows rather trivially from the fact that a conformal rescaling with constant $\sigma$ changes $A \to e^{2\sigma} A$ and $\bar{\nu} \to e^{-\sigma} \bar{\nu}$ so that $\bar{\nu}^2 \to e^{-2\sigma} \bar{\nu}^2$ and, hence, $\lambda_n^2 \to e^{-2\sigma} \lambda_n^2$.</footnote>
As before, the integrals of the traces of the higher Green's functions $\tilde{G}_k$ at coinciding points are just the $\tilde{\zeta}(k)$ so that equivalently

$$ \delta \left[ S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \log \det \mathcal{P}_0 + \frac{\gamma}{4\pi} m^2 A + \frac{m^2}{4} \int d^2 u \sqrt{g} \operatorname{tr} \tilde{G}_R(u) + \sum_{k=2}^{\infty} \frac{m^{2k}}{4k} \tilde{\zeta}(k) \right] = 0. \qquad (5.84) $$

This looks very similar to the result (5.35) where we did not separate the zero-mode contributions. The only difference (apart from replacing the $G$'s by the $\tilde{G}$'s, or replacing the $\zeta(k)$ by the $\tilde{\zeta}(k)$) is the presence of the zero-mode term $\log \det \mathcal{P}_0$ which only contributes at order $m^0$. Indeed, we have seen in section 4.2 that $\delta\lambda_0 = 0$ so that $\delta\zeta(s) = \delta\tilde{\zeta}(s)$.

As already remarked above, $\tilde{G}_R$ and the $\tilde{\zeta}(k)$ are the quantities defined with non-vanishing mass. Since we subtracted the zero-mode parts, these quantities have a well-defined series expansion in $m^2$ involving quantities defined in the massless theory. We will now work out these expansions. First, we have $\lambda_n^2 = \lambda_{n,(m=0)}^2 + m^2$, cf (2.92), so that

$$ \begin{aligned} \frac{1}{\lambda_n^{2k}} &= \frac{1}{(\lambda_{n,(m=0)}^2 + m^2)^k} = \frac{1}{\lambda_{n,(m=0)}^{2k}} \left(1 + \frac{m^2}{\lambda_{n,(m=0)}^2}\right)^{-k} \\ &= \frac{1}{\lambda_{n,(m=0)}^{2k}} \sum_{r=0}^{\infty} (-)^r \frac{(k+r-1)!}{(k-1)! r!} \left(\frac{m^2}{\lambda_{n,(m=0)}^2}\right)^r. \end{aligned} \qquad (5.85) $$

Then

$$ \begin{aligned} \sum_{k=2}^{\infty} \frac{m^{2k}}{k} \frac{1}{\lambda_n^{2k}} &= \sum_{k=2}^{\infty} \sum_{r=0}^{\infty} (-)^r \frac{(k+r)!}{k! r!} \frac{1}{k+r} \left( \frac{m^2}{\lambda_{n,(m=0)}^2} \right)^{r+k} \\ &= \sum_{l=2}^{\infty} \left( \sum_{r=0}^{l-2} (-)^r \frac{l!}{(l-r)! r!} \right) \frac{1}{l} \left( \frac{m^2}{\lambda_{n,(m=0)}^2} \right)^l \\ &= \sum_{l=2}^{\infty} \left[ \sum_{r=0}^{l} (-)^r \frac{l!}{(l-r)! r!} - (-)^{l-1} l - (-)^l \right] \frac{1}{l} \left( \frac{m^2}{\lambda_{n,(m=0)}^2} \right)^l \\ &= \sum_{l=2}^{\infty} \left[ 0 - (-)^{l-1} l - (-)^l \right] \frac{1}{l} \left( \frac{m^2}{\lambda_{n,(m=0)}^2} \right)^l \\ &= \sum_{l=2}^{\infty} (-)^l \frac{l-1}{l} \left( \frac{m^2}{\lambda_{n,(m=0)}^2} \right)^l. \end{aligned} \qquad (5.86) $$

If we now sum over $n \neq 0$ we get the corresponding relation for the sum of zeta-functions

$$ \sum_{k=2}^{\infty} \frac{m^{2k}}{k} \tilde{\zeta}(k) = \sum_{k=2}^{\infty} (-)^k \frac{k-1}{k} m^{2k} \tilde{\zeta}_{m=0}(k). \qquad (5.87) $$

Note that to lowest order in $m$ this is just $\frac{m^4}{2} \tilde{\zeta}(2) = \frac{2-1}{2} m^4 \tilde{\zeta}_{m=0}(2) + \mathcal{O}(m^6)$, which is indeed obviously correct, and it is similarly easy to directly check also the next order.

We may similarly proceed with $\tilde{G}_R$. First note that the short distance singularity of $\tilde{G}(x, y)$ is the same for vanishing and for non-vanishing mass. Indeed, this singularity
depends on the 2-derivative part of $D^2$ and is not changed by adding the mass or not. On the other hand, as already used repeatedly, the $\tilde{G}_r(x, y)$ have no short-distance singularities for $r \ge 2$ and the $\tilde{G}_r(y, y)$ are finite. Recall the obvious fact that if $\psi_n$ is an eigenfunction of $D^2$ for $m \ne 0$, i.e. $D^2\psi_n = \lambda_n^2\psi_n$ where $D^2 = -\nabla^2 + m^2$, then this same $\psi_n$ is also an eigenfunction of $D^2$ for $m = 0$, i.e. of $-\nabla^2$ with eigenvalue $\lambda_n^2 - m^2 = \lambda_{n,(m=0)}^2$. (Note that this argument is correct for the eigenfunctions of $D^2$, but not for the eigenfunctions of $D$ and the following reasoning for $\tilde{G}$ can not be transposed to $\tilde{S}$.) Hence, using (5.85) for $k=1$

$$ 
\begin{aligned}
\tilde{G}(x, y) &= \sum_{n \ne 0} \frac{1}{\lambda_n^2} (\psi_n(x) \psi_n^\dagger(y) + \psi_n^*(x) (\psi_n^*)^\dagger(y)) \\
&= \sum_{r=0}^{\infty} (-)^r m^{2r} \sum_{n \ne 0} \frac{1}{\lambda_{n,(m=0)}^{2r+2}} (\psi_n(x) \psi_n^\dagger(y) + \psi_n^*(x) (\psi_n^*)^\dagger(y)) \\
&= \sum_{r=0}^{\infty} (-)^r m^{2r} \tilde{G}_{r+1}^{m=0}(x, y) = \tilde{G}^{m=0}(x, y) + \sum_{r=1}^{\infty} (-)^r m^{2r} \tilde{G}_{r+1}^{m=0}(x, y). \quad (5.88)
\end{aligned}
$$

Subtracting the short-distance singularity and letting then $x \to y$ gives

$$ \tilde{G}_R(y) = \tilde{G}_R^{m=0}(y) + \sum_{r=1}^{\infty} (-)^r m^{2r} \tilde{G}_{r+1}^{m=0}(y, y), \quad (5.89) $$

so that

$$ m^2 \int d^2 y \sqrt{g} \tilde{G}_R(y) = m^2 \int d^2 y \sqrt{g} \tilde{G}_R^{m=0}(y) - \sum_{k=2}^{\infty} (-)^k m^{2k} \tilde{\zeta}_{m=0}(k). \quad (5.90) $$

We can then rewrite (5.84) as

$$ \delta \left[ S_{\text{grav}} + \frac{1}{48\pi} S_{\text{Liouville}} + \log \det \mathcal{P}_0 + \frac{\gamma}{4\pi} m^2 A \right. \\ \left. + \frac{m^2}{4} \int d^2 u \sqrt{g} \operatorname{tr} \tilde{G}_R^{m=0}(u) + \sum_{k=2}^{\infty} \frac{(-)^{k+1}}{4k} m^{2k} \tilde{\zeta}_{m=0}(k) \right] = 0. \quad (5.91) $$

This looks exactly like (5.84) - which was written in terms of the massive Green's function and massive zeta-function - except that now the terms in the sum over $k$ have an alternating sign!

The last sum in this equation (5.91) ressembles, of course, the Taylor expansion of $\log(1+x)$ and, indeed, formally we have

$$ 
\begin{aligned}
\log \left( \frac{\operatorname{Det}' D^2}{\operatorname{Det}' D_{m=0}^2} \right) &= 2 \log \prod_{n \ne 0} \frac{\lambda_n^2}{\lambda_{n,m=0}^2} = 2 \sum_{n \ne 0} \log \left( 1 + \frac{m^2}{\lambda_{n,m=0}^2} \right) = 2 \sum_{n \ne 0} \sum_{k=1}^{\infty} \frac{(-)^{k+1}}{k} \frac{m^{2k}}{\lambda_{n,m=0}^{2k}} \\
&= \sum_{k=1}^{\infty} \frac{(-)^{k+1}}{k} m^{2k} \tilde{\zeta}_{m=0}(k). \quad (5.92)
\end{aligned}
$$

It looks a bit as if we have been going in circles and we could have written this expansion right away at the beginning. However, this formal manipulation results in a sum over $k$
including also the singular $k = 1$ term $\tilde{\zeta}_{m=0}(1)$. Our more careful treatment has produced instead, the integral of the Green's function $\tilde{G}_R^{m=0}$, which can be rewritten in terms of $\tilde{G}_\zeta^{m=0}$ which in turn is related to the regular part of $\tilde{\zeta}_{m=0}$, i.e. $\tilde{\zeta}_{m=0}^{\text{reg}}(1)$, as well as the piece $\log \det \mathcal{P}_0$ that originated from the zero-modes.

## 5.4 Integrating the variation of the gravitational action

We have obtained the variation of the gravitational action - to all orders in an expansion in powers of $m^2$ - under an infinitesimal variation of the conformal factor, corresponding to an infinitesimal variation of the metric. This variation was expressed in terms of variations of integrals of the renormalized Green's function $\tilde{G}_R^{m=0}$ of the massless theory and of the higher Green's functions $\tilde{G}_n^{m=0}(x, x)$ ($n \ge 2$), or equivalently of the finite values of zeta function $\tilde{\zeta}_{m=0}(n)$, also of the massless theory. We can then immediately "integrate" these infinitesimal variations to obtain $S_{\text{grav}}[g, \hat{g}]$. Note that $\mathcal{P}_0$ as defined in (2.111) depends on $\hat{g}$ (through the zero-modes $\hat{\psi}_{0,i}$) and $\sigma$. We write $\mathcal{P}_0[g] = \mathcal{P}_0[\hat{g}, \sigma] = \mathcal{P}_0[g, \hat{g}]$. Since the zero-modes $\hat{\psi}_{0,i}$ are orthonormalized for the metric $\hat{g}$ we obviously have $\mathcal{P}_0[\hat{g}] = \mathcal{P}_0[\hat{g}, \hat{g}] = \mathbf{1}$. It follows that $\delta \log \det \mathcal{P}_0$ integrates to $\log \frac{\det \mathcal{P}_0[g, \hat{g}]}{\det \mathcal{P}_0[g, \hat{g}]} = \log \det \mathcal{P}_0[g, \hat{g}]$. We get

$$ S_{\text{grav}}[g, \hat{g}] = -\frac{1}{48\pi} S_{\text{Liouville}}[g, \hat{g}] - \log \det \mathcal{P}_0[g, \hat{g}] - \frac{\gamma}{4\pi} m^2(A - \hat{A}) \\ -\frac{m^2}{4} (\mathcal{G}[g] - \mathcal{G}[\hat{g}]) + \sum_{k=2}^{\infty} \frac{(-)^k}{4k} m^{2k} (\tilde{\zeta}_{m=0}^g(k) - \tilde{\zeta}_{m=0}^{\hat{g}}(k)), \quad (5.93) $$

where we defined

$$ \mathcal{G}[g] = \int d^2u \sqrt{g(u)} \operatorname{tr} \tilde{G}_R^{m=0}[g](u), \quad (5.94) $$

with the obvious notation that $\tilde{G}_R^{m=0}[g]$ is the renormalized Green's function of the massless theory, with the zero-mode subtracted, for the metric $g$. Let us insist that the right-hand side of (5.93) is entirely expressed in terms of quantities defined in the massless theory. We now want to determine, in particular, the dependence of the functional $\mathcal{G}[g]$ on the conformal factor $\sigma$.

Just as the Liouville action only involves the local quantities $\sigma$, $\hat{g}$ and $\hat{R}$, ideally one would like to express the higher-order terms similarly in terms of such local quantities. For the simpler case of the effective gravitational action obtained by integrating out a massive scalar field [13] it turned out that the order $m^2$-term could be expressed in terms of the Mabuchi action which is local in $\sigma$ and the Kähler potential $\Phi$, and some relatively simple term that involved $\tilde{G}_R^{m=0}[\hat{g}]$. (In this case $\tilde{G}_R^{m=0}$ is the renormalized Green's function of the scalar Laplacian). For such a massive scalar field the higher-order (in $m^2$) terms were computed in [14] and were expressed in terms of higher Green's functions which encode much non-local information about the manifold.

At present, $\tilde{G}^{m=0}$ is the Green's function of $D^2 = -\Delta_{\text{sp}} + \frac{R}{4}$ which, in particular, involves the spinoral Laplacian which is different from the scalar Laplacian. In order to express $\mathcal{G}[g]$ in terms of local quantities and possibly $\tilde{G}_R^{m=0}[\hat{g}]$ we have tried to determine the variation of $\tilde{G}_R^{m=0}$ and tried to write it in such a form that one can integrate the infinitesimal variations to obtain its variation for finite $\sigma$. We tried to mimic the computation for the
scalar field where it was useful to express the variation of the conformal factor as the scalar Laplacian of the variation of the Kähler potential and then integrate by parts this scalar Laplacian. However, at present, this turns out to be more complicated since $\tilde{G}$ is not the Greens's function of the scalar Laplacian, and we did not manage to obtain a useful form of the variation of $\tilde{G}_{\text{R}}^{m=0}$ along this line.

Here, we will instead proceed along a different avenue which, nevertheless, will allow us to express $\mathcal{G}[g] - \mathcal{G}[\hat{g}]$ in terms of the (local) conformal factor $\sigma$, as well as several quantities pertaining to the metric $\hat{g}$ only, such as the Green's function $\tilde{S}_{\hat{g}}$, as well as the zero-mode projector $\bar{P}_0$ which involves the finitely many area-like (non-local) parameters $\mathcal{P}_{0,ij}$ and its inverse matrix.

To simplify the notations in this subsection we will introduce further abbreviations for the different integration measures including various powers of $e^{\sigma}$:

$$d\hat{z} = d^2z\sqrt{\hat{g}(z)}, \quad d\hat{z}_{\sigma} = d^2z\sqrt{\hat{g}(z)}e^{\sigma(z)}, \quad dz = d^2z\sqrt{g(z)} \equiv d^2z\sqrt{\hat{g}(z)}e^{2\sigma(z)}. \quad (5.95)$$

### 5.4.1 The variations of $\tilde{S}$ under finite conformal rescalings for $m=0$

We now want to relate $\tilde{S} \equiv \tilde{S}_g$, which is the Green's function for the metric $g$, to $\tilde{S} \equiv \tilde{S}_{\hat{g}}$, which is the Green's function for the metric $\hat{g}$, where as usual $g = e^{2\sigma}\hat{g}$.

We will again exploit $\nabla_x = e^{-3\sigma(x)/2}\nabla_x e^{\sigma(x)/2}$ but now for finite $\sigma(x)$. This relation will be particularly fruitful when relating the Green's function of the massless theory. Throughout the remainder of this subsection all Green's functions $\tilde{S}, \tilde{\tilde{S}}$ and $\tilde{G}$ refer to the massless Green's functions, although we will not indicate it explicitly any more. For $\sigma = 0$, i.e. for the metric $\hat{g}$, we have

$$i\nabla_x \tilde{\tilde{S}}(x, y) = \frac{\delta^{(2)}(x-y)}{\sqrt{\hat{g}}} - \hat{P}_0(x, y), \quad (5.96)$$

where $\hat{P}_0$ is the projector on the zero-modes of $i\nabla_x$, as given in (2.114). Recall also from (2.112) that the projector on the zero-modes of $i\nabla_x$ is $P_0(x, y) = e^{-\sigma(x)/2}\bar{P}_0(x, y)e^{-\sigma(y)/2}$, where $\bar{P}_0$ was defined in (2.113). In particular it involves the “area-like” ortho-normalisation factors $\mathcal{P}_{0,ij} = \int d\hat{z}e^{\sigma(z)}\hat{\psi}_{0,i}^{\dagger}(z)\hat{\psi}_{0,j}(z)$, cf (2.111). The differential equation for $\tilde{S}$ then is

$$\begin{align}
i\nabla_x \tilde{\tilde{S}}(x, y) &= \frac{\delta^{(2)}(x-y)}{\sqrt{\hat{g}}} - P_0(x, y) \nonumber \\
\Leftrightarrow \quad i\nabla_x \tilde{\tilde{S}}(x, y) e^{\sigma(x)/2} \tilde{\tilde{S}}(x, y) e^{\sigma(y)/2} &= \frac{\delta^{(2)}(x-y)}{\sqrt{\hat{g}}} - e^{\sigma(x)} \bar{P}_0(x, y). \tag{5.97}
\end{align}$$

We now set

$$\bar{S}(x, y) = e^{\sigma(x)/2} \tilde{\tilde{S}}(x, y) e^{\sigma(y)/2} \quad \Leftrightarrow \quad \tilde{\tilde{S}}(x, y) = e^{-\sigma(x)/2} \bar{S}(x, y) e^{-\sigma(y)/2}, \quad (5.98)$$

so that (5.97) becomes

$$i\nabla_x \bar{S}(x, y) = \frac{\delta^{(2)}(x-y)}{\sqrt{\hat{g}}} - e^{\sigma(x)} \bar{P}_0(x, y). \quad (5.99)$$
This is very similar to (5.96) except for the term involving the zero-mode “projector” $\overline{P}_0$.
We then set

$$ \overline{S}(x, y) = \tilde{S}(x, y) + T(x, y), \qquad (5.100) $$

and obtain the following equation for $T$:

$$ i\overline{\nabla}_x T(x, y) = \overline{P}_0(x, y) - e^{\sigma(x)} \overline{P}_0(x, y). \qquad (5.101) $$

Changing $x \to z$ and multiplying with $\tilde{\overline{S}}(x, z)$ and integrating we get

$$ \int d\hat{z} \tilde{\overline{S}}(x, z) i\overline{\nabla}_z T(z, y) = - \int d\hat{z} \tilde{\overline{S}}(x, z) e^{\sigma(z)} \overline{P}_0(z, y) = - \int d\hat{z}_\sigma \tilde{\overline{S}}(x, z) \overline{P}_0(z, y), \quad (5.102) $$

where the piece $\sim \overline{P}_0$ dropped out since $\int d\hat{z} \tilde{\overline{S}}(x, z) \overline{P}_0(z, y) = 0$. On the left-hand side we integrate by parts the $i\overline{\nabla}_z$ and use (5.96) to get

$$ T(x, y) - \int d\hat{z} \overline{P}_0(x, z) T(z, y) = - \int d\hat{z} \tilde{\overline{S}}(x, z) e^{\sigma(z)} \overline{P}_0(z, y). \qquad (5.103) $$

This determines $T$ up to a certain zero-mode part which is subtracted on the left-hand side. Just like $\overline{S}$ and $\tilde{\overline{S}}$, $T$ must also satisfy $(T(x, y))^{\dagger} = T(y, x)$, and we can also expand $T(x, y)$ on products of eigenfunctions of $i\overline{\nabla}$. We separate the zero-mode and non-zero-mode parts, and we have schematically

$$ T(x, y) = \sum_{i,j} \hat{\Psi}_{0,i}(x) t_{ij} \hat{\Psi}_{0,j}^{\dagger}(y) + \sum_{i,n} \hat{\Psi}_{0,i}(x) c_{in} \hat{\Psi}_{n}^{\dagger}(y) + \sum_{i,n} \hat{\Psi}_{n}(x) c_{in}^{*} \hat{\Psi}_{0,i}^{\dagger}(y) \\ + \sum_{n,k} \hat{\Psi}_{n}(x) d_{nm} \hat{\Psi}_{m}^{\dagger}(y), \qquad (5.104) $$

where $\hat{\Psi}_{0,i}$ stands generically for the zero-modes $\hat{\psi}_{0,i}$ and $\hat{\psi}_{0,i}^{*}$, and $\hat{\Psi}_{n}$ stands generically for the non zero-modes $\hat{\psi}_{n}$ and $\hat{\psi}_{n}^{*}$. Then, the left-hand side of equation (5.103) only contains the terms with the $c_{in}^{*}$ and $d_{nm}$ while the right-hand side contains terms of the form $\hat{\Psi}_{n}(x) \hat{\Psi}_{0,i}^{\dagger}(y)$ only. This shows that $d_{nm} = 0$, and that all $c_{in}^{*}$ are determined, and hence also all $c_{in}$. Thus, (5.103) determines $T$ uniquely, up to the pure zero-mode piece involving the $t_{ij}$. The latter corresponds to the arbitrary solution to the homogeneous equation $T(x, y) - \int d\hat{z} \overline{P}_0(x, z) T(z, y) = 0$ one may always add. Now, obviously, as for $\tilde{\overline{S}}$, we do not want to include such a pure zero-mode piece in $T$ and we make the choice to set it to zero. Then $T$ only contains the pieces with $c_{in}$ and $c_{in}^{*}$ and, hence

$$ T(x, y) = - \int d\hat{z}_{\sigma} (\tilde{\overline{S}}(x, z) \overline{P}_{0}(z, y) + \overline{P}_{0}(x, z) \tilde{\overline{S}}(z, y)). \qquad (5.105) $$

Obviously, this satisfies $T(x, y)^{\dagger} = T(y, x)$ and one may, of course, check directly that it is a solution$^{27}$ of (5.101). Inserting the solution (5.105) into (5.100), and using (5.98), we get

$$ \tilde{S}(x, y) = e^{-\frac{\sigma(x)}{2}} \left( \tilde{\overline{S}}(x, y) - \int d\hat{z}_{\sigma} \left( \tilde{\overline{S}}(x, z) \overline{P}_{0}(z, y) + \overline{P}_{0}(x, z) \tilde{\overline{S}}(z, y) \right) \right) e^{-\frac{\sigma(y)}{2}}. \qquad (5.106) $$

<footnote>$^{27}$Indeed, since $i\overline{\nabla}_x \overline{P}_0(x, y) = 0$ one has $i\overline{\nabla}_x T(x, y) = -e^{\sigma(x)} \overline{P}(x, y) + \int d\hat{z} e^{\sigma(z)} \overline{P}_0(x, z) \overline{P}_0(z, y)$, and it is not difficult to show that $\int d\hat{z} e^{\sigma(z)} \overline{P}_0(x, z) \overline{P}_0(z, y) = \overline{P}_0(x, y)$.</footnote>
This can be equivalently rewritten in terms of

$$S(x, y) \equiv e^{-\frac{\sigma(x)}{2}} \tilde{S}(x, y) e^{-\frac{\sigma(y)}{2}} \qquad (5.107)$$

as

$$\tilde{S}(x, y) = S(x, y) - \int dz (S(x, z)P_0(z, y) + P_0(x, z)S(z, y)), \qquad (5.108)$$

which shows that in the absence of zero-modes (spherical topology) one simply has $\tilde{S} = S$.

Both (5.106) and (5.108) express $\tilde{S}$ solely in terms of the conformal factor $\sigma$ and quantities computed with the reference metric $\hat{g}$, i.e. $\tilde{S}$ and the zero-modes $\hat{\psi}_{0,i}$. To make the dependencies slightly more explicit we introduce the notations $\tilde{S} \equiv \tilde{S}_g$, $\tilde{S} \equiv \tilde{S}_{\hat{g}}$ and $\tilde{P}_0 \equiv \tilde{P}_{0,\hat{g},\sigma}$, so that (5.106) is rewritten once more as

$$\tilde{S}_g(x, y) = e^{-\frac{\sigma(x)}{2}} \left( \tilde{S}_{\hat{g}}(x, y) - \int d\hat{z}_{\sigma} \left( \tilde{S}_{\hat{g}}(x, z) \tilde{P}_{0,\hat{g},\sigma}(z, y) + \tilde{P}_{0,\hat{g},\sigma}(x, z) \tilde{S}_{\hat{g}}(z, y) \right) \right) e^{-\frac{\sigma(y)}{2}}. \qquad (5.109)$$

Let us insist, that without the presence of the factor $e^{\sigma(z)}$ (hidden in the integration measure $d\hat{z}_{\sigma}$) the integral of $\tilde{S}_{\hat{g}}\tilde{P}$ and of $\tilde{P}_0\tilde{S}_{\hat{g}}$ would of course vanish due to the orthogonality of the zero and non zero-modes.

### 5.4.2 The variations of $\tilde{G}$ and $\tilde{G}_R$ under finite conformal rescalings for $m = 0$

The dependence of the Green's function $\tilde{G}_g(x, y)$ on the conformal factor can then be obtained straightforwardly by inserting the previous result for $\tilde{S}_g$ into $\tilde{G}_g(x, y) = \int d^2z \sqrt{\hat{g}} e^{2\sigma(z)} \tilde{S}_g(x, z) \tilde{S}_g(z, y)$:

$$\tilde{G}_g(x, y) = e^{-\frac{\sigma(x)}{2}} \int d\hat{z}_{\sigma} \left( \tilde{S}_{\hat{g}}(x, z) - \int d\hat{u}_{\sigma} \left( \tilde{S}_{\hat{g}}(x, u) \tilde{P}_{0,\hat{g},\sigma}(u, z) + \tilde{P}_{0,\hat{g},\sigma}(x, u) \tilde{S}_{\hat{g}}(u, z) \right) \right) \\ \times \left( \tilde{S}_{\hat{g}}(z, y) - \int d\hat{v}_{\sigma} \left( \tilde{S}_{\hat{g}}(z, v) \tilde{P}_{0,\hat{g},\sigma}(v, y) + \tilde{P}_{0,\hat{g},\sigma}(z, v) \tilde{S}_{\hat{g}}(v, y) \right) \right) e^{-\frac{\sigma(y)}{2}}. \qquad (5.110)$$

Developing the product of the two brackets we get 9 terms. To further simplify the notation, we will suppress the arguments in the multiple integrals. Then we find

$$\tilde{G}_g(x, y) = e^{-\frac{\sigma(x)}{2}} \left[ \begin{aligned} & \int d\hat{z}_{\sigma} \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} - \int d\hat{z}_{\sigma} d\hat{u}_{\sigma} \left( \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \tilde{P}_0 + \tilde{S}_{\hat{g}} \tilde{P}_0 \tilde{S}_{\hat{g}} + \tilde{S}_{\hat{g}} \tilde{P}_0 \tilde{S}_{\hat{g}} + \tilde{P}_0 \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \right) \\ & + \int d\hat{z}_{\sigma} d\hat{u}_{\sigma} d\hat{v}_{\sigma} \left( \tilde{S}_{\hat{g}} \tilde{P}_0 \tilde{S}_{\hat{g}} \tilde{P}_0 + \tilde{S}_{\hat{g}} \tilde{P}_0 \tilde{P}_0 \tilde{S}_{\hat{g}} + \tilde{P}_0 \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \tilde{P}_0 + \tilde{P}_0 \tilde{S}_{\hat{g}} \tilde{P}_0 \tilde{S}_{\hat{g}} \right) \end{aligned} \right] e^{-\frac{\sigma(y)}{2}}. \qquad (5.111)$$

Again, due to the presence of the $e^{\sigma}$ factors (hidden in $du_{\sigma}$ etc) for each integration, no simplification occurs$^{28}$ in the integrals of products of $\tilde{S}_{\hat{g}}$ and $\tilde{P}_0$.

The short-distance singularity of (5.111) comes from the first term in the bracket and we know that it is $-\frac{1}{4\pi} \log \frac{\mu^2 \ell_g^2(x, y)}{4} \mathbf{1}_{2 \times 2}$. As already discussed, we have for finite conformal

<footnote>$^{28}$The only slight simplification comes from $\int d\hat{u}_{\sigma} \tilde{P}_0(z, u) \tilde{P}_0(u, v) = \tilde{P}_0(z, v)$ which one can show by rewriting $\tilde{P}_0$ in terms of $P_0$ and using that $\int d^2u \sqrt{\hat{g}} P_0(z, u) P_0(u, v) = P_0(z, v)$.</footnote>
transformations

$$ \begin{aligned} \ell_g^2(x, y) \sim_{x \to y} e^{\sigma(x)+\sigma(y)} \ell_g^2(x, y) &\Rightarrow -\frac{1}{4\pi} \log \frac{\mu^2 \ell_g^2(x, y)}{4} \tag{5.112} \\ \sim_{x \to y} -\frac{1}{4\pi} \log \frac{\mu^2 \ell_g^2(x, y)}{4} - \frac{\sigma(x) + \sigma(y)}{4\pi}. \end{aligned} $$

We see again that the singular parts of $\tilde{G}_g$ and $\tilde{G}_{\hat{g}}$ are the same and it follows that $\tilde{G}_g(x, y) - \tilde{G}_{\hat{g}}(x, y)$ has no short-distance singularity. We also see that

$$ \tilde{G}_{\mathrm{R}}[g](y) - \tilde{G}_{\mathrm{R}}[\hat{g}](y) = \lim_{x \to y} \left( \tilde{G}_g(x, y) - \tilde{G}_{\hat{g}}(x, y) + \frac{\sigma(y)}{2\pi} \mathbf{1}_{2 \times 2} \right). \quad (5.113) $$

Combining this with (5.111) we get

$$ \begin{aligned} \tilde{G}_{\mathrm{R}}[g](y) ={}& \tilde{G}_{\mathrm{R}}[\hat{g}](y) + \frac{\sigma(y)}{2\pi} \mathbf{1}_{2 \times 2} + \int d\hat{z} (e^{\sigma(z)-\sigma(y)} - 1) \tilde{S}_{\hat{g}}(y,z) \tilde{S}_{\hat{g}}(z,y) \\ & - e^{-\sigma(y)} \int d\hat{z}_{\sigma} d\hat{u}_{\sigma} \left( \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \bar{P}_{0} + \tilde{S}_{\hat{g}} \bar{P}_{0} \tilde{S}_{\hat{g}} + \tilde{S}_{\hat{g}} \bar{P}_{0} \tilde{S}_{\hat{g}} + \bar{P}_{0} \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \right) \Big|_{x=y} \\ & + e^{-\sigma(y)} \int d\hat{z}_{\sigma} d\hat{u}_{\sigma} d\hat{v}_{\sigma} \left( \tilde{S}_{\hat{g}} \bar{P}_{0} \tilde{S}_{\hat{g}} \bar{P}_{0} + \tilde{S}_{\hat{g}} \bar{P}_{0} \bar{P}_{0} \tilde{S}_{\hat{g}} + \bar{P}_{0} \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \bar{P}_{0} + \bar{P}_{0} \tilde{S}_{\hat{g}} \bar{P}_{0} \tilde{S}_{\hat{g}} \right) \Big|_{x=y}. \end{aligned} \quad (5.114) $$

The functional $\mathcal{G}[g] = \int d^2y \sqrt{g} \operatorname{tr} \tilde{G}_{\mathrm{R}}[g](y) = \int d\hat{y} e^{2\sigma(y)} \operatorname{tr} \tilde{G}_{\mathrm{R}}[g](y) \equiv \int d\hat{y}_{\sigma} e^{\sigma(y)} \operatorname{tr} \tilde{G}_{\mathrm{R}}[g](y)$ then is easily obtained from the previous relation. In particular, one can now use the cyclic-ity of the trace to simplify the many terms involving $\bar{P}_0$. Writing out explicitly all factors of $e^{\sigma}$ we get

$$ \begin{aligned} \mathcal{G}[g] ={}& \mathcal{G}[\hat{g}] + \int d\hat{y} (e^{2\sigma} - 1) \operatorname{tr} \tilde{G}_{\mathrm{R}}[\hat{g}](y) + \frac{1}{\pi} \int d\hat{y} e^{2\sigma} \sigma \\ & + \int d\hat{y} d\hat{z} (e^{\sigma(z)+\sigma(y)} - e^{2\sigma(y)}) \operatorname{tr} \tilde{S}_{\hat{g}}(y,z) \tilde{S}_{\hat{g}}(z,y) \\ & - 4 \int d\hat{y} d\hat{z} d\hat{u} e^{\sigma(y)+\sigma(z)+\sigma(u)} \operatorname{tr} \tilde{S}_{\hat{g}}(y,z) \tilde{S}_{\hat{g}}(z,u) \bar{P}_0(u,y) \\ & + 2 \int d\hat{y} d\hat{z} d\hat{u} d\hat{v} e^{\sigma(y)+\sigma(z)+\sigma(u)+\sigma(v)} \operatorname{tr} (\tilde{S}_{\hat{g}} \bar{P}_0 \tilde{S}_{\hat{g}} \bar{P}_0 + \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \bar{P}_0 \bar{P}_0). \end{aligned} \quad (5.115) $$

Again, in the absence of zero-modes, i.e. for spherical topology, the last two integrals are absent. Let us recall, once more, that all quantities in this equation are those of the massless theory.
### 5.4.3 The gravitational action at order $m^2$

We can then insert this result into (5.93) to get the contributions to the effective gravitational action at order $m^2$:

$$ S_{\text{grav}}[g, \hat{g}] = -\frac{1}{48\pi} S_{\text{Liouville}}[g, \hat{g}] - \log \det \mathcal{P}_0[g, \hat{g}] - \frac{\gamma}{4\pi} m^2(A - \hat{A}) \\ -\frac{m^2}{4} \left[ \begin{aligned} & \int d^2y \sqrt{\hat{g}}(e^{2\sigma} - 1) \operatorname{tr} \tilde{G}_{\text{R}}[\hat{g}](y) + \frac{1}{\pi} \int d^2z \sqrt{\hat{g}} e^{2\sigma} \sigma \\ & + \int d^2y \sqrt{\hat{g}} d^2z \sqrt{\hat{g}} (e^{\sigma(z)+\sigma(y)} - e^{2\sigma(y)}) \operatorname{tr} \tilde{S}_{\hat{g}}(y, z) \tilde{S}_{\hat{g}}(z, y) \\ & -4 \int d^2y \sqrt{\hat{g}} d^2z \sqrt{\hat{g}} d^2u \sqrt{\hat{g}} e^{\sigma(y)+\sigma(z)+\sigma(u)} \operatorname{tr} \tilde{S}_{\hat{g}}(y, z) \tilde{S}_{\hat{g}}(z, u) \bar{P}_0(u, y) \\ & +2 \int d^2y \sqrt{\hat{g}} d^2z \sqrt{\hat{g}} d^2u \sqrt{\hat{g}} d^2v \sqrt{\hat{g}} e^{\sigma(y)+\sigma(z)+\sigma(u)+\sigma(v)} \operatorname{tr} (\tilde{S}_{\hat{g}} \bar{P}_0 \tilde{S}_{\hat{g}} \bar{P}_0 + \tilde{S}_{\hat{g}} \tilde{S}_{\hat{g}} \bar{P}_0 \bar{P}_0) \end{aligned} \right] \\ + \mathcal{O}(m^4). \qquad (5.116) $$

We see that at order $m^0$ we get the Liouville action and a specific zero-mode contribution involving the area-like constants $\mathcal{P}_{0,ij}$. Of course, if present, the zero-modes are zero-modes of the massless Dirac operator $i\nabla$, while the massive Dirac operator never has zero-modes. In the massless theory, one does not do the functional integral over these zero-modes and then there is a certain freedom how exactly we define the matter partition function. In particular, one can then redefine the matter partition by dividing $Z_{\text{mat}}[g]$ by $\det \mathcal{P}_0[g]$ and then the corresponding term would have been absent in the effective gravitational action. However, we are really computing in the massive theory and then the massive Dirac operator has no zero-modes. Hence, there is no reason to change the definition of the matter partition function and the term $\log \det \mathcal{P}_0[g, \hat{g}]$ should be genuinely present.

At order $m^2$ we find the cosmological constant term $\sim m^2(A - \hat{A})$, but also a term $\int \sqrt{\hat{g}} e^{2\sigma} \sigma$ characteristic of the Mabuchi action

$$ S_M[g, \hat{g}] = \frac{4}{A} \int \sqrt{\hat{g}} e^{2\sigma} \sigma + \dots, \qquad (5.117) $$

Obviously, this term is local in $\sigma$. If we trace things back, this term originated from the conformal transformation of the singular part of the Green's function $\tilde{G}$ one had to subtract to get the renormalized Green's function $\tilde{G}_{\text{R}}$ (see e.g. (5.113)). It is interesting to note, that the Mabuchi action appears at present with a coefficient $-\frac{m^2 A}{16\pi}$ while in the case of the massive scalar field one exactly obtained the opposite coefficient $+\frac{m^2 A}{16\pi}$.

The other terms present in (5.116) are multi-local in $\sigma$ and also involve the various Green's function for the reference metric $\hat{g}$, as well as the area-like parameters $\mathcal{P}_{0,ij}$. Such “non-local” terms involving the Green's functions on the manifold also are present in the effective gravitational action for massive scalars [14] at higher orders in $m^2$, starting at $m^4$.

One might wonder why our gravitational action (5.116) at order $m^2$ involves the zero-mode projectors when we have shown that corresponding variation (5.91) could be written solely in terms of $\mathcal{G}[g] = \int \sqrt{\hat{g}} \tilde{G}_{\text{R}}[\hat{g}]$? The reason is that upon integrating the variation from $\sigma = 0$ to $\sigma$ (from $\hat{g}$ to $g$) the zero-mode projector changes in a non-trivial way, and subtracting the zero-modes from $\mathcal{G}[g]$ is not the same as subtracting them from $\mathcal{G}[\hat{g}]$.
# 6 Discussion and outlook

In these notes, we have studied the effective gravitational action for massive fermions in two dimensions, continuing and completing what had been initiated in [1]. The appropriate mass term is a Majorana type mass term $\int \psi^\dagger m \gamma_\mu \psi$, and the spectral analysis we performed was based on the Dirac operator $D = i \not{\nabla} + m\gamma_\mu$ whose eigenfunctions necessarily are complex. What might have looked as a simple generalisation of the massive scalar case, actually turned out to be technically quite involved. We performed a detailed study of the corresponding Green's functions, local zeta-functions and local heat kernels of this Dirac operator $D$ and of its square $D^2$. One of the crucial insights of the present paper was to obtain a simple relation, not only between the Greens's functions $S$ of $D$ and $G$ of $D^2$, but also between the corresponding local zeta-functions $\zeta_-(s, x, y)$ and $\zeta_+(s + 1/2, x, y)$. By the (inverse) Mellin transform this allowed us to obtain much information about the local heat kernel $K_-(t, x, y)$ from the small-$t$ expansion of the heat kernel $K_+(t, x, y)$ which we worked out in the appendix. (This confirms some of the conjectures made in [1] but also infirms some others.) In particular, we could obtain all the necessary information about the short-distance singularities and the poles of the local zeta-functions $\zeta_\pm(s, x, x)$ at coinciding points. Most importantly we studied the variations of these quantities under infinitesimal conformal rescalings of the metric, and then “integrated” these infinitesimal variations to get the finite effective gravitational action $S[g, \hat{g}]$.

The gravitational action was obtained by performing an expansion in powers of the (small) mass and, accordingly, we need to define all our Green's and local zeta functions also for vanishing mass. In this case there are generically (for genus larger or equal to one) zero-modes of the massless Dirac operator that must be excluded from the spectral decomposition of these Green's or local zeta functions and, hence, a detailed understanding of the contributions of these fermionic zero-modes was necessary. Of course the number and properties of the zero-modes crucially depend on the topology of the manifold. Nevertheless, we were able to sufficiently characterise these zero-mode parts of the Green's functions, allowing us to make general statements for arbitrary genus. In particular, the dependence of the zero-modes (and only of the zero-modes) on the conformal factor turned out to be rather simple, which allowed us to completely characterise the different projectors on these zero-modes we needed to introduce. This was one of the important new insights in these notes as compared to what was done in [1].

We could thus obtain an expansion of the gravitational action, for arbitrary genus, to all powers in $m^2$, involving higher and higher Green's functions at coinciding points. At order $m^0$ the effective gravitational action contains the well-known Liouville action with the appropriate coefficient for a Majorana fermion, namely $1/2$ times the one for a single scalar field. There is also a certain contribution involving the normalisation factors of the zero-modes which depend on a set of “area-like” parameters. While in the massless theory this contribution could be removed by a correspondingly different definition of the matter partition function from the outset, there is no reason to do this in the massive theory and then these terms constitute a genuinely new contribution at order $m^0$.

At order $m^2$ we found a local contribution $\int \sqrt{\hat{g}} \sigma e^{2\sigma}$, characteristic of the Mabuchi action. This term appeared with the same coefficient as for a massive scalar field, but with
the opposite sign, as one might perhaps have expected. But we also found, at this order $m^2$, contributions that can be expressed in terms of integrals and multiple integrals involving the conformal factor $\sigma$ at the different integration points, and the Green's function of $D$ for the background metric $\hat{g}$ and the renormalized Green's function of $D^2$ at coinciding points, also in the background metric $\hat{g}$, as well as on the area-like parameters $P_{0,ij}$. To appreciate the degree of non-locality of these additional terms, one must remember that the knowledge of a Green's function encodes much non-local information about the manifold. But the Green's function which appear are only those for metric $\hat{g}$, so in this sense these contributions are “non-local in $\hat{g}$” only, and local or multi-local in the conformal factor $\sigma$. Such multi-local or non-local terms are to be expected and were also present in the scalar case starting at order $m^4$ [14]. Finally we argued that the area-like constants $P_{0,ij}$ that appear in the zero-mode projectors are not more or less non-local than a term $\frac{1}{A}$.

The effective gravitational action is to be used in the next step, if one wants to compute the functional integral over the geometries. This will involve in particular a functional integration over the conformal factor with an appropriate measure $D\sigma$. The form of the effective gravitational action we have obtained (at order $m^2$) displays this $\sigma$-dependence either explicitly, or through the zero-mode projector $\bar{P}_0$ that depends on the $P_{0,ij}$, having it separated from the quantities that only depend on the fixed “background” metric $\hat{g}$. We hope to return to this issue elsewhere.

## Acknowledgments

M.N. is supported by a “Laboratoire d’Excellence” grant of the physics department of the ENS.

## A Small-$t$ expansion for heat kernels of type $K_+(t, x, y)$

In this appendix we will show how the small-$t$ expansion of heat kernels for general (positive) second-order differential operators with a standard initial condition can be obtained by a recursive scheme. We then specialise these results to the squared Dirac operator in two dimensions. This will yield the expansion of $K_+(t, x, y)$. There is an abundant literature about small-$t$ expansions of heat kernels, see [23, 24, 30, 31], which focus much on the “diagonal” elements, i.e. $x = y$. Here we will follow the more pedestrian approach of references [34, 35] from which many of the formulae are borrowed. Our computation will be facilitated by using Riemann normal coordinates which we discuss first.

### A.1 Riemann normal coordinates

It is often useful to go to a special coordinate system for which the metric in the vicinity of a given point is as close to the Euclidean metric as possible. In general relativity, these are the coordinates of a freely falling observer. More generally, they are called Riemann normal coordinates. We will discuss here their definition and properties in an arbitrary dimension $n$.
**Definition and basic relations.** We fix any point as the origin (which would have coordinates $y$ in a general coordinate system). Riemann normal coordinates $x$ around this point are defined by taking $n$ independent geodesics through this point and defining the geodesic distance from the origin along the $k^{\text{th}}$ geodesic as the new coordinate $x^k$. Taking independent geodesics means that their tangent vectors at the origin are $n$ linearly independent vectors. We may furthermore chose these tangent vectors as orthogonal. Then by rescaling the $x^k$ appropriately one can achieve that in these new coordinates the geodesic distance of the point $x$ from the origin is$^{29}$

$$ \ell^2(x, 0) = x^i x^i \equiv x^2. \qquad (A.1) $$

This looks like the standard Euclidean distance, but it does not mean that the space is flat. It only means that the metric at the origin is $\delta_{ij}$ and that its first derivatives, and hence the Christoffel symbols, vanish at the origin. However, this change of coordinates cannot remove the curvature and, indeed, one can show that the Taylor expansion of the metric around $x=0$ is given by (see e.g. [36])

$$ g_{ij}(x) = \delta_{ij} - \frac{1}{3} R_{ikjl} x^k x^l - \frac{1}{6} R_{ikjl;m} x^k x^l x^m \\ + \left[ \frac{2}{45} R_{ikrl} R^r_{mjn} - \frac{1}{20} R_{ikjl;mn} \right] x^k x^l x^m x^n + O(x^5), \qquad (A.2) $$

where $R^i_{jkl}$ is the Riemann curvature tensor, defined as

$$ R^i_{jkl} = \partial_k \Gamma^i_{lj} - \partial_l \Gamma^i_{jk} + \Gamma^i_{kr} \Gamma^r_{lj} - \Gamma^i_{lr} \Gamma^r_{kj}, \qquad (A.3) $$

satisfying $R_{ijkl} = -R_{jkl} = -R_{ijlk} = R_{klij}$. Here $(...);_{mn} = \nabla_n \nabla_m(...)$ denotes covariant derivatives, and

$$ \mathcal{R}_{kl} = R^j_{kjl} \equiv R_{kjl}^j, \quad \mathcal{R} = \mathcal{R}_k^j, \qquad (A.4) $$

denote the Ricci tensor and Ricci curvature scalar. All curvature tensors in (A.2) are evaluated at $x=0$. In particular, since $g_{ij}(0) = \delta_{ij}$ we do not have to distinguish upper and lower indices on these tensors in this formula. The inverse metric is

$$ g^{ij}(x) = \delta_{ij} + \frac{1}{3} R_{ikjl} x^k x^l + \frac{1}{6} R_{ikjl;m} x^k x^l x^m \\ + \left[ \frac{1}{15} R_{ikrl} R^r_{mjn} + \frac{1}{20} R_{ikjl;mn} \right] x^k x^l x^m x^n + O(x^5). \qquad (A.5) $$

Note that in these coordinates one has $g_{ij}(x)x^j = x^i$ and $g^{ij}(x)x^j = x^i$ since all other terms involve symmetric products of the coordinates $x$ contracted with antisymmetric curvature tensors. The square root of the determinant of the metric is

$$ \sqrt{g(x)} = 1 - \frac{1}{6} \mathcal{R}_{kl} x^k x^l - \frac{1}{12} \mathcal{R}_{kl;m} x^k x^l x^m \\ + \left[ \frac{1}{72} \mathcal{R}_{kl} \mathcal{R}_{mn} - \frac{1}{40} \mathcal{R}_{kl;mn} - \frac{1}{180} R^r_{kls} R^s_{mnr} \right] x^k x^l x^m x^n + O(x^5), \qquad (A.6) $$

<footnote>$^{29}$In this appendix we use latin indices $i, j, k, l, ...$ with the present normal coordinates. Of course, they are “coordinate indices”, not to be confused with the “flat” indices $a, b, c$ of the local orthonormal frames used in section 2.</footnote>
Note that the terms $O(x^5)$ are terms $O(R^3, \nabla R^2)$, i.e. involving at least 3 curvature tensors or 2 curvature tensor and a covariant derivative.

**Scalar Laplace operator.** The scalar Laplace operator is given in general by

$$ \Delta = \frac{1}{\sqrt{g}} \partial_i (\sqrt{g} g^{ij} \partial_j) = g^{ij}(x) \partial_i \partial_j + f^j(x) \partial_j, \quad f^j = -g^{kl} \Gamma^j_{kl} = \partial_i g^{ij} + g^{ij} \partial_i \log \sqrt{g}. \quad (A.7) $$

Note that the scalar Laplacian is a negative operator, i.e. its eigenvalues are $\le 0$ with a single zero-mode given by the constant function. In order to have a positive (or rather non-negative) operator the Laplacian is sometimes defined with the opposite sign. We will keep our definition (A.7) so that the non-negative operator is $(-\Delta)$. It is indeed $-\Delta$ that appears in our squared Dirac operator. Using (A.5) and (A.6), a straightforward computation yields for the function $f^j(x)$ in the present normal coordinates

$$ f^j(x) = -\frac{2}{3} \mathcal{R}_{jk} x^k + \left[ \frac{1}{12} \mathcal{R}_{kl;j} - \frac{1}{2} \mathcal{R}_{kj;l} \right] x^k x^l - \left[ \frac{1}{5} \mathcal{R}_{jl;mn} + \frac{1}{40} \mathcal{R}_{mn;lj} - \frac{3}{40} \mathcal{R}_{mn;jl} \right. \\ \left. - \frac{23}{180} \mathcal{R}_{ls} R^s_{mnj} + \frac{4}{45} R_{rjls} R_{smnr} \right] x^l x^m x^n + O(x^4). \quad (A.8) $$

This, together with (A.5), yields the expansion of the scalar Laplace operator in the vicinity of $x = 0$, up to terms $O(x^4)$ or $O(R^3, \nabla R^2)$. In particular, we have $\Delta \ell^2(x, 0) = [g^{ij}(x) \partial_i \partial_j + f^j(x) \partial_j] (x^k x^k) = 2 [g^{jj}(x) + f^j(x) x^j]$, so that (in $n$ dimensions)

$$ \Delta \ell^2(x, 0) - 2n = -\frac{2}{3} \mathcal{R}_{kl} x^k x^l - \frac{1}{2} \mathcal{R}_{kl;m} x^k x^l x^m - \left[ \frac{1}{5} \mathcal{R}_{kl;mn} + \frac{2}{45} R^r_{kls} R^s_{mnr} \right] x^k x^l x^m x^n \\ + O(x^5). \quad (A.9) $$

**Spinorial Laplace operator in 2 dimensions.** We also need to work out similarly the spinorial Laplace operator (2.48) in normal coordinates. We will do this only for the case of interest in this paper, namely in 2 dimensions and only up to the order we need. Recall that we had

$$ \Delta_{\text{sp}} = \mathbf{1}_{2 \times 2} \left( \Delta_{\text{scalar}} - \frac{1}{16} \omega^{\mu} \omega_{\mu} \right) - \frac{i}{4} \gamma_{*} \left( (\nabla_{\mu} \omega^{\mu}) + 2 \omega^{\mu} \partial_{\mu} \right). \quad (A.10) $$

So the first thing we need to figure out is the expansion of the $\omega_{\mu} = 2\omega_{\mu}^{12}$ around the origin in Riemann normal coordinates. Rather then using some general formula, we will determine the zwei-beins $e^a$ and then solve the zero-torsion condition to the order we are interested in. It will turn out that we only need to determine $\omega_{\mu}$ up to first order in $x$. This means that we will need the expansions of the $e^a$ up to and including terms of order
$(x)^2$. From (A.2) we get

$$
\begin{align}
e^a \otimes e^a &= g_{ij} \, dx^i \otimes dx^j = dx^i \otimes dx^i - \frac{1}{3} R_{ikjl} x^k x^l \, dx^i \otimes dx^j + \mathcal{O}(x^3) \nonumber \\
&= dx^i \otimes dx^i - \frac{1}{3} \left( R_{1212} x^2 x^2 dx^1 \otimes dx^1 + R_{1221} x^2 x^1 dx^1 \otimes dx^2 \right. \nonumber \\
&\qquad \left. + R_{2112} x^1 x^2 dx^2 \otimes dx^1 + R_{2121} x^1 x^1 dx^2 \otimes dx^2 \right) + \mathcal{O}(x^3) \nonumber \\
&= dx^i \otimes dx^i - \frac{\mathcal{R}}{6} \left( x^2 x^2 dx^1 \otimes dx^1 - x^1 x^2 (dx^1 \otimes dx^2 + dx^2 \otimes dx^1) \right. \nonumber \\
&\qquad \left. + x^1 x^1 dx^2 \otimes dx^2 \right) + \mathcal{O}(x^3) \tag{A.11}
\end{align}
$$

where we used $R_{1212} = -R_{2112} = -R_{1221} = R_{2121} = \frac{1}{2}\mathcal{R}$. From this we read

$$
\begin{align}
e^1 &= dx^1 - \frac{\mathcal{R}}{12} (x^2 x^2 dx^1 - x^1 x^2 dx^2) + \mathcal{O}(x^3), \nonumber \\
e^2 &= dx^2 - \frac{\mathcal{R}}{12} (x^1 x^1 dx^2 - x^1 x^2 dx^1) + \mathcal{O}(x^3). \tag{A.12}
\end{align}
$$

Next,

$$
de^1 = \frac{\mathcal{R}}{4} x^2 dx^1 \wedge dx^2 + \mathcal{O}(x^2), \quad de^2 = -\frac{\mathcal{R}}{4} x^1 dx^1 \wedge dx^2 + \mathcal{O}(x^2). \tag{A.13}
$$

The zero-torsion conditions $de^1 + \omega^{12} \wedge e^2 = 0 = de^2 - \omega^{12} \wedge e^1$ yield

$$
\begin{align}
\omega^{12} = -\frac{\mathcal{R}}{4}x^2dx^1 + \frac{\mathcal{R}}{4}x^1dx^2 + \mathcal{O}((x)^2) &\Leftrightarrow \omega_1^{12} = -\frac{\mathcal{R}}{4}x^2 + \mathcal{O}((x)^2), \tag{A.14} \\
\omega_2^{12} = \frac{\mathcal{R}}{4}x^1 + \mathcal{O}((x)^2). &
\end{align}
$$

Recall that we defined $\omega_\mu = 2\omega_\mu^{12}$ and, hence, to this order

$$
\omega_1 = \omega^1 = -\frac{\mathcal{R}}{2} x^2 + \mathcal{O}((x)^2), \quad \omega_2 = \omega^2 = \frac{\mathcal{R}}{2} x^1 + \mathcal{O}((x)^2). \tag{A.15}
$$

Since the Christoffel symbols are $\mathcal{O}(x)$ it follows that

$$
\omega^{\mu}\omega_{\mu} = \mathcal{O}((x)^2), \quad \nabla_{\mu}\omega^{\mu} = \partial_{\mu}\omega^{\mu} + \mathcal{O}((x)^2) = \mathcal{O}(x), \quad \omega^{\mu}\partial_{\mu} = \frac{\mathcal{R}}{2}(x^1\partial_2 - x^2\partial_1) + \mathcal{O}((x)^2\partial). \tag{A.16}
$$

A.2 Small-$t$ expansion of the heat kernel

We now turn to the small-$t$ expansion of the heat kernels $K(t, x, y)$ for a general class of second-order differential operators $\mathcal{D}$ on an $n$-dimensional manifold, satisfying the standard initial condition. This yields heat kernels generalising our $K_+(t, x, y)$. We then specialise these results to the squared Dirac operator $\mathcal{D} = D^2$ in 2 dimensions.

We will suppose that the second-order differential operator is (the negative) of a gen-
eralised Laplace type, i.e. of the form

$$
-\mathcal{D} = \mathbf{1} g^{ij}(x) \partial_i \partial_j + A^i(x) \partial_i + B(x), \quad i, j = 1, \dots, n, \tag{A.17}
$$
with $\mathbf{1}$, $A^i(x)$ and $B(x)$ being $p \times p$-matrices and $g^{ij}$ a positive (inverse) metric. Note the minus-sign on the left-hand side needed so that $\mathcal{D}$ can be a positive differential operator for appropriate $A^i$ and $B$.

We want to find the solution $K(t, x, y)$ of the corresponding heat equation

$$ \frac{d}{dt} K(t, x, y) = -\mathcal{D}K(t, x, y), \qquad (A.18) $$

with initial condition

$$ K(t, x, y) \sim \delta^{(d)}(x-y) [g(x)g(y)]^{-1/4} \mathbf{1} \quad \text{as } t \to 0. \qquad (A.19) $$

If $\mathcal{D}$ is hermitian and we denote by $\Lambda_n$ and $\varphi_n$ its eigenvalues and $(p$-component) eigenfunctions, then

$$ K(t, x, y) = \sum_n e^{-\Lambda_n t} \varphi_n(x) \varphi_n^\dagger(y) \qquad (A.20) $$

is the solution to the heat equation with the correct initial condition. Indeed, $\frac{d}{dt} K(t, x, y) = -\sum_n \Lambda_n e^{-\Lambda_n t} \varphi_n(x) \varphi_n^\dagger(y) = -\mathcal{D} \sum_n e^{-\Lambda_n t} \varphi_n(x) \varphi_n^\dagger(y) = -\mathcal{D} K(t, x, y)$ and for $t \to 0$ one gets $K(t, x, y) \to \sum_n \varphi_n(x) \varphi_n^\dagger(y)$ which by the completeness relation equals the right-hand side of (A.19). As is clear from (A.19) and (A.20) this heat kernel generalises indeed what we called $K_+$ in the main text. The other heat kernel $K_-$ will be studied to a certain extent in appendix B.

For $A^i = B = 0$, $g^{ij} = \delta^{ij}$ the differential operator is just $p$ copies of the flat space Laplace operator and the well-known solution is $K(t, x, y) = (4\pi t)^{-n/2} \exp\left(-\frac{(x-y)^2}{4t}\right) \mathbf{1}$. This must also be the leading small $t$, small distance behaviour on a curved manifold, if $(x-y)^2$ is replaced by the square of the geodesic distance $\ell^2(x, y)$ between $x$ and $y$. Corrections to this leading behaviour can then be obtained as a perturbative expansion in $t$ and $\ell(x, y)$.

**Asymptotic expansion and recursion relations.** We will now show that the heat kernel $K(t, x, y)$ admits an asymptotic small $t$ expansion around the “flat-space” solution. We work out the recursion relations between the expansion coefficients in general. Then we use normal coordinates to explicitly solve these recursion relations and obtain the first few coefficients explicitly. This discussion is essentially taken from [34, 35], but generalises it slightly.

We search for a solution of (A.18), (A.19) of the form

$$ K(t, x, y) = (4\pi t)^{-n/2} \exp\left(-\frac{\ell^2(x, y)}{4t}\right) F(t, x, y). \qquad (A.21) $$

where $\ell(x, y)$ is the geodesic distance between $x$ and $y$ and $F$ is expanded in integer non-negative powers of $t$,

$$ F(t, x, y) = \sum_{r=0}^{\infty} F_r(x, y) t^r. \qquad (A.22) $$

$K$ being a $p \times p$-matrix, the same is true for the coefficient functions $F_r(x, y)$
Let us begin by computing $g^{ij}\partial_i\partial_j K$:

$$ g^{ij}(x)\partial_i\partial_j\left(\frac{e^{-\ell^2/(4t)}F}{t^{n/2}}\right) \qquad (A.23) \\ = \frac{e^{-\ell^2/(4t)}}{t^{n/2}}\left[\frac{g^{ij}(x)\partial_i\ell^2\partial_j\ell^2}{16t^2}F - \frac{g^{ij}(x)\partial_i\partial_j\ell^2}{4t}F - \frac{g^{ij}(x)\partial_i\ell^2}{2t}\partial_jF + g^{ij}(x)\partial_i\partial_jF\right], $$

where $\partial_i \equiv \frac{\partial}{\partial x^i}$. It is not too difficult to show that

$$ g^{ij}(x)\partial_i\ell(x, y)\partial_j\ell(x, y) = 1 \quad \Rightarrow \quad g^{ij}(x)\partial_i\ell^2\partial_j\ell^2 = 4\ell^2. \qquad (A.24) $$

(On the other hand, there is no such simple result for $g^{ij}(x)\partial_i\partial_j\ell^2$.) The remaining pieces in $\mathcal{D}$ give

$$ (A^i(x)\partial_i + B(x))\left(\frac{e^{-\ell^2/(4t)}F}{t^{n/2}}\right) = \frac{e^{-\ell^2/(4t)}}{t^{n/2}}\left[-\frac{A^i(x)\partial_i\ell^2}{4t}F + A^i\partial_iF + B(x)F\right], \quad (A.25) $$

Adding the two parts we get

$$ -\mathcal{D}_x \left( \frac{e^{-\ell^2/(4t)}F}{t^{n/2}} \right) = \frac{e^{-\ell^2/(4t)}}{t^{n/2}} \left[ \begin{aligned} & \ell^2 F - \frac{g^{ij}(x)\partial_i\partial_j\ell^2}{4t}F - \frac{g^{ij}(x)\partial_i\ell^2}{2t}\partial_jF - \frac{A^i(x)\partial_i\ell^2}{4t}F \\ & + \underbrace{g^{ij}(x)\partial_i\partial_jF + A^i\partial_iF + B(x)F}_{\mathcal{D}_x F} \end{aligned} \right], \quad (A.26) $$

On the other hand,

$$ \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{e^{-\ell^2/(4t)}F}{t^{n/2}} \right) = \frac{e^{-\ell^2/(4t)}}{t^{n/2}} \left[ \frac{\ell^2}{4t^2} F - \frac{n}{2t} F + \frac{\mathrm{d}F}{\mathrm{d}t} \right], \quad (A.27) $$

Upon inserting (A.26) and (A.27) into the heat equation (A.18) we see that the leading singular terms $\sim \frac{1}{\ell^2}$ in the square bracket cancel, and the heat equation amounts to

$$ \frac{g^{ij}(x)\partial_i\partial_j\ell^2 - 2n}{4t}F + \frac{g^{ij}(x)\partial_i\ell^2}{2t}\partial_jF + \frac{A^i(x)\partial_i\ell^2}{4t}F + \frac{\mathrm{d}F}{\mathrm{d}t} = \mathcal{D}_x F. \quad (A.28) $$

The terms on the left-hand side either have an explicit $\frac{1}{t}$ or a derivative $\frac{\mathrm{d}}{\mathrm{d}t}$ which also decreases the power of $t$ by one unit. Inserting the expansion (A.22) of $F$ yields the desired recursion relations:

$$ \frac{1}{4}(g^{ij}(x)\partial_i\partial_j\ell^2 - 2n + A^i(x)\partial_i\ell^2)F_0 + \frac{1}{2}g^{ij}(x)\partial_i\ell^2\partial_jF_0 = 0, \quad (A.29) $$

$$ \frac{1}{4}(g^{ij}(x)\partial_i\partial_j\ell^2 - 2n + A^i(x)\partial_i\ell^2)F_r + \frac{1}{2}g^{ij}(x)\partial_i\ell^2\partial_jF_r + rF_r = \mathcal{D}_x F_{r-1}, \quad (r \ge 1), \quad (A.30) $$

(where again $\partial_i \equiv \frac{\partial}{\partial x^i}$.) The recursion relation (A.30) can be somewhat simplified by writing $F_r = F_0\tilde{F}_r$. One then gets equivalently

$$ \frac{1}{2}g^{ij}(x)\partial_i\ell^2\partial_j\tilde{F}_r + r\tilde{F}_r = F_0^{-1}\mathcal{D}_x(F_0\tilde{F}_{r-1}). \quad (A.31) $$
Before we turn to solving these recursion relations, let us note the obvious fact that $K$ will be exponentially small unless $\ell^2/t$ is not too large. This means that as an order of magnitude estimate we must have $\ell \le \sqrt{t}$, and the small $t$ expansion must also be a small $\ell$ expansion. Thus if one is interested in expanding $F$ up to order $t^2$ it is consistent to expand $F_0(x, y)$ in powers of $(x - y) \sim \ell$ up to fourth order, expand $F_1$ in powers of $(x - y) \sim \ell$ up to second order, and for $F_2$ only keep $F_2(y, y)$. For the purpose of the main text it will be sufficient to expand to order $t$ only, which means obtaining $F_0$ up to order $\ell^2$ and $F_1$ only to order 0, i.e. $F_1(y, y)$. Since this small $t$ / small $\ell$ expansion only explores the vicinity of the point $y$, it is much useful to introduce Riemann normal coordinates around $y$ which will simplify the calculations quite a bit.

**Solution to the recursion relations of the heat equation in normal coordinates.**

Let us then introduce normal coordinates centered in $y$, i.e $F_r(x, y) \to F_r(x, 0)$. Recall that $\ell^2 = x^i x^i$ so that $\partial_i \ell^2 = 2x^i$, and $g^{ij} \partial_j \ell^2 = 2x^i$ with all other terms cancelling due to the antisymmetry properties of the Riemann tensor, cf (A.5). From this same equation one also gets

$$g^{ij} \partial_i \partial_j \ell^2 = g^{ij} 2 \delta_{ij} = 2g^{ii} = 2\delta^{ii} + \frac{2}{3} R_{ikil} x^k x^l + \mathcal{O}(x^3) = 2n + \frac{2}{3} R_{kl} x^k x^l + \mathcal{O}(x^3). \quad (A.32)$$

Then relation (A.29) for $F_0$ gives

$$x^j \partial_j F_0 = -\left(\frac{1}{6} R_{kl} x^k x^l + \frac{1}{2} A^l x^l + \mathcal{O}(x^3)\right) F_0. \quad (A.33)$$

Let us assume that

$$A^i(x) = a_{ij} x^j + \mathcal{O}(x^2), \quad (A.34)$$

with matrix-valued coefficients $a_{ij}$. Then $A^l x^l = a_{kl} x^k x^l + \mathcal{O}(x^3)$ and the $F_0$-equation is

$$x^j \partial_j F_0 = -\left(\frac{1}{6} R_{kl} x^k x^l + \frac{1}{2} a_{kl} x^k x^l + \mathcal{O}(x^3)\right) F_0, \quad (A.35)$$

with solution

$$F_0 = 1 - \frac{1}{12} R_{kl} x^k x^l - \frac{1}{4} a_{kl} x^k x^l + \mathcal{O}(x^3). \quad (A.36)$$

Next, the recursion relation for $F_1$ reads

$$(\frac{1}{6} R_{kl} x^k x^l + \frac{1}{2} a_{kl} x^k x^l + \mathcal{O}(x^3)) F_1 + x^j \partial_j F_1 + F_1 = \mathcal{D}_x F_0. \quad (A.37)$$

If we only want to get $F_1$ to order $(x - y)^0$ this becomes

$$F_1 + \mathcal{O}(x) = \mathcal{D}_x F_0. \quad (A.38)$$

We have

$$\mathcal{D}_x F_0 = g^{ij} \left(-\frac{1}{6} R_{ij} - \frac{1}{2} a_{ij}\right) + B + \mathcal{O}(x^2) = -\frac{1}{6} R - \frac{1}{2} g^{ij} a_{ij} + B + \mathcal{O}(x^2), \quad (A.39)$$
so that finally

$$F_1 = -\frac{1}{6}\mathcal{R} - \frac{1}{2}g^{ij}a_{ij} + B + \mathcal{O}(x). \qquad (A.40)$$

As a simple check of these formula, let us choose $-\mathcal{D} = \Delta_{\text{scalar}}$, i.e. $p = 1$ (no matrices) and $A^i = f^i = -\frac{2}{3}\mathcal{R}_{ij}x^j + \mathcal{O}((x)^2)$ and $B = 0$, according to (A.7) and (A.8). Then $a_{ij} = -\frac{2}{3}\mathcal{R}_{ij}$ and

$$F_0^\Delta = 1 + \frac{1}{12}\mathcal{R}_{kl}x^kx^l + \mathcal{O}(x^3), \quad F_1^\Delta = +\frac{1}{6}\mathcal{R} + \mathcal{O}(x), \qquad (A.41)$$

which are indeed the correct well-known values for the scalar Laplace operator.

Note that in the formula for the $F_r$, the $\mathcal{R}_{kl}$, $\mathcal{R}$ and $a_{kl}$ are to be taken at $y = 0$ which was the origin of our normal coordinates. More generally, if we use normal coordinates around an arbitrary point $y$ (without shifting the coordinates) the $\mathcal{R}_{kl}$, $\mathcal{R}$ and $a_{kl}$ are to be evaluated at $y$ and the $x^k$ appearing should be $(x - y)^k$. Thus

$$\begin{aligned} F_0(x, y) &= 1 - \left( \frac{1}{12} \mathcal{R}_{kl}(y) + \frac{1}{4} a_{kl}(y) \right) (x - y)^k (x - y)^l + \mathcal{O}((x - y)^3) \\ F_1(x, y) &= -\frac{1}{6} \mathcal{R}(y) - \frac{1}{2} g^{ij} a_{ij}(y) + B(y) + \mathcal{O}(x - y). \end{aligned} \qquad (A.42)$$

In particular, at coinciding points we have

$$F_0(y, y) = 1, \quad F_1(y, y) = -\frac{1}{6}\mathcal{R}(y) - \frac{1}{2}g^{ij}a_{ij}(y) + B(y). \qquad (A.43)$$

This is independent of the normal coordinates used and true in an arbitrary coordinate system.

**The heat kernel expansion for the 2-dimensional Dirac operator $D^2$.** Let us finally specify our formula to 2 dimensions and $\mathcal{D} = D^2$, the squared Dirac operator, given in (2.47) and (2.48) (we now use again $\mu, \nu, \dots$ for the coordinate indices):

$$\begin{aligned} D^2 &= -\Delta_{\text{sp}} + \frac{1}{4}\mathcal{R} + m^2 = -\Delta_{\text{scalar}} + \frac{1}{16}\omega^\mu\omega_\mu + \frac{i}{4}\gamma_*((\nabla_\mu\omega^\mu) + 2\omega^\mu\partial_\mu) + \frac{1}{4}\mathcal{R} + m^2 \\ &= -g^{\mu\nu}\partial_\mu\partial_\nu - (f^\mu - \frac{i}{2}\omega^\mu\gamma_*)\partial_\mu + \frac{1}{16}\omega^\mu\omega_\mu + \frac{i}{4}\gamma_* (\nabla_\mu\omega^\mu) + \frac{1}{4}\mathcal{R} + m^2, \end{aligned} \qquad (A.44)$$

where $f^\mu = \partial_\nu g^{\mu\nu} + g^{\mu\nu}\partial_\nu \log \sqrt{g}$ was given in (A.7). From this we identify

$$A^\mu = f^\mu - \frac{i}{2}\omega^\mu\gamma_*, \quad B = -\frac{1}{16}\omega^\mu\omega_\mu - \frac{i}{4}\gamma_* (\nabla_\mu\omega^\mu) - \frac{1}{4}\mathcal{R} - m^2. \qquad (A.45)$$

We know from (A.8) that in normal coordinates around $y = 0$ we have $f^\mu = -\frac{2}{3}\mathcal{R}_{\mu\nu}x^\nu + \mathcal{O}((x)^2)$ and from (A.15) that $\omega^1 = -\frac{\mathcal{R}}{2}x^2 + \mathcal{O}((x)^2)$ and $\omega^2 = \frac{\mathcal{R}}{2}x^1 + \mathcal{O}((x)^2)$. Recall that $a_{\mu\nu}$ was defined as $A^\mu(x) = a_{\mu\nu}x^\nu + \mathcal{O}((x)^2)$, so that

$$a_{11} = -\frac{2}{3}\mathcal{R}_{11}, \quad a_{12} = -\frac{2}{3}\mathcal{R}_{12} + \frac{i}{4}\mathcal{R}\gamma_*, \quad a_{21} = -\frac{2}{3}\mathcal{R}_{12} - \frac{i}{4}\mathcal{R}\gamma_*, \quad a_{22} = -\frac{2}{3}\mathcal{R}_{22}. \qquad (A.46)$$

It follows that (still in normal coordinates)

$$g^{\mu\nu}a_{\mu\nu} = -\frac{2}{3}\mathcal{R} + \mathcal{O}((x)^2). \qquad (A.47)$$
Also (see (A.16))

$$B = -\frac{1}{4} \mathcal{R} - m^2 + \mathcal{O}(x). \qquad (A.48)$$

It is maybe a bit disappointing, but we see that, at this order, no effect of the spin-connection $\omega^\mu$ and of the $\gamma_*$-matrix structure survives. We finally find for the first two heat kernel coefficients at coinciding points (cf (A.43)):

$$F_0(y, y) = 1, \quad F_1(y, y) = -\frac{1}{12} \mathcal{R}(y) - m^2. \qquad (A.49)$$

Let us emphasise again that these results concern the heat kernel $K_+$.

## B The small-$t$ asymptotics of $K_-(t, x, y)$ for the flat torus

In this appendix, we will similarly try to obtain the small-$t$ asymptotics of $K_-$, Although $K_-(t, x, y)$ satisfies the same differential equation as $K_+(t, x, y)$, it does not have any “useful” initial condition and the method of the previous appendix then does not apply. Instead, we will focus on the case of the flat (square) torus where we explicitly know the eigenvalues and eigenfunctions.

As discussed in section 4, it will be enough to determine the small-$t$ asymptotics of $\mathcal{L}$. However, as we will see, this turns out to be still rather non-trivial. Before discussing the more complicated case of $x \neq y$, let us first study the case of coinciding points $x = y$. Then $K_-(t, x, x)$ is given by (4.75) in terms of $\mathcal{L}(t, 0)$ which is

$$\mathcal{L}(t, 0) = \sum_{n_1, n_2} \frac{e^{-\lambda_{\vec{n}}^2 t}}{\lambda_{\vec{n}}} = \sum_{n_1, n_2} \frac{e^{-(n_1^2+n_2^2+m^2)t}}{\sqrt{n_1^2+n_2^2+m^2}}. \qquad (B.1)$$

The exponential provides an effective cut-off to the sum, restricting it to values of $n_1, n_2$ that are less than a few $1/\sqrt{t}$. For small $t$ these are still very many contributions. One may then try to estimate this sum by replacing it by the corresponding integral. For $m$ not too small this turns out to be quite a good approximation. However, as $m \to 0$, the sum is dominated by the single point $n_1 = n_2 = 0$ contributing $1/m$. Separating this term we write

$$\mathcal{L}(t, 0) = \frac{e^{-m^2 t}}{m} + \sum_{(n_1, n_2) \neq (0, 0)} \frac{e^{-(n_1^2 + n_2^2 + m^2)t}}{\sqrt{n_1^2 + n_2^2 + m^2}} \approx \frac{e^{-m^2 t}}{m} + \int_{(p_1, p_2) \notin D_1} d^2 p \frac{e^{-(p_1^2 + p_2^2 + m^2)t}}{\sqrt{p_1^2 + p_2^2 + m^2}}, \qquad (B.2)$$

where $D_1$ denotes a disc of unit area (radius $r_0 = \pi^{-1/2}$) around the origin, corresponding to the “area” of the missing value $n_1 = n_2 = 0$. Upon going to polar coordinates ($p_1 =$
$p \cos \phi$, $p_2 = p \sin \phi$), this becomes

$$
\begin{align}
\mathcal{L}(t, 0) &\approx \frac{e^{-m^2 t}}{m} + 2\pi \int_{r_0}^{\infty} dp \frac{p e^{-t(p^2+m^2)}}{\sqrt{p^2+m^2}} = \frac{e^{-m^2 t}}{m} + \pi \int_{r_0^2}^{\infty} d\xi \frac{e^{-t(\xi+m^2)}}{\sqrt{\xi+m^2}} \nonumber \\
&= \frac{e^{-m^2 t}}{m} + \pi \int_{r_0^2+m^2}^{\infty} d\xi \frac{e^{-t\xi}}{\sqrt{\xi}} \nonumber \\
&= \frac{e^{-m^2 t}}{m} + 2\pi \int_{\sqrt{r_0^2+m^2}}^{\infty} dp e^{-tp^2} = \frac{e^{-m^2 t}}{m} + 2\pi \left( \frac{1}{2}\sqrt{\frac{\pi}{t}} - \int_{0}^{\sqrt{r_0^2+m^2}} dp e^{-tp^2} \right) \nonumber \\
&= \frac{e^{-m^2 t}}{m} + \frac{\pi^{3/2}}{\sqrt{t}} - 2\pi \sqrt{r_0^2+m^2} \sum_{n=0}^{\infty} \frac{(-)^n (r_0^2+m^2)^n}{n!(2n+1)} t^n. \tag{B.3}
\end{align}
$$

We see that $\mathcal{L}(t, 0)$ contains integer powers of $t$, as well as a term $\sim t^{-1/2}$. The integer powers and in particular the order $t^0$ term (except the $\frac{1}{m}$) are affected by the ambiguity in choosing $r_0$ and this reflects the error made in replacing the sum by an integral. So the only thing we can safely conclude is

$$
\mathcal{L}(t, 0) \approx \frac{1}{m} + \sqrt{\frac{\pi^3}{t}} + \mathcal{O}(t^0, m^0). \qquad (B.4)
$$

By eq. (4.75) this directly gives the corresponding asymptotics of $K_-(t, x, x)$ as given in (4.82).

Instead of approximating the sum by an integral, we could try to use the Poisson sum-
mation formula. This formula gives the sum as another infinite sum of Fourier coefficients
$\hat{f}_k$. These Fourier coefficients are given by integrals and the Fourier coefficient $\hat{f}_0$ actually
is just the integral approximation to the original sum. Thus, adding the (infinite number)
of higher Fourier coefficients could be viewed as providing the correction terms for having
replaced the sum by the integral $\hat{f}_0$. Unfortunately we will not be able to compute exactly
the $\hat{f}_k$ for $k \neq 0$. Let us nevertheless outline this computation. The computation will be
the same for $\mathcal{L}(t, 0)$ and $\mathcal{L}(t, z)$, so we will do the general case allowing non-vanishing $z$.
We will use the notation $\vec{z} = (z^1, z^2)$, $\vec{n} = (n_1, n_2)$, $\vec{p} = (p^1, p^2)$ as well as $\vec{k} = (k_1, k_2)$.
Thus we write

$$
\mathcal{L}(t, z) = \sum_{n_1, n_2} f(\vec{n}; t, \vec{z}), \quad f(\vec{p}; t, \vec{z}) = \frac{e^{-t(\vec{p}^2 + m^2)} e^{i\vec{p} \cdot \vec{z}}}{\sqrt{\vec{p}^2 + m^2}}. \quad (B.5)
$$

By the Poisson summation formula one then has

$$
\mathcal{L}(t, z) = \sum_{k_1, k_2} \hat{f}(\vec{k}; t, \vec{z}), \qquad (B.6)
$$

where

$$
\hat{f}(\vec{k}; t, \vec{z}) = \int d^2 p e^{2\pi i \vec{k} \cdot \vec{p}} f(\vec{p}; t, \vec{z}) = \int d^2 p \frac{e^{-t(\vec{p}^2 + m^2)} e^{i\vec{p} \cdot (\vec{z} + 2\pi\vec{k})}}{\sqrt{\vec{p}^2 + m^2}} = \hat{f}(0; t, \vec{z} + 2\pi\vec{k}). \quad (B.7)
$$

We see that the $\hat{f}$ only depend on $\vec{z}$ and $\vec{k}$ through the combination $\vec{u} \equiv \vec{z} + 2\pi\vec{k}$. This
is why evaluating them at $\vec{z} = 0$ or $\vec{z} \neq 0$ essentially presents the same difficulty since
we need to evaluate the $\hat{f}(0, t, \vec{u})$ at non-vanishing $\vec{u}$. Using again polar coordinates $p^1 = p \cos \phi$, $p^2 = p \sin \phi$, and denoting $p = |\vec{p}|$ and similarly $u = |\vec{u}|$, one has

$$ \hat{f}(0; t, \vec{u}) = \int_{0}^{\infty} dp \int_{0}^{2\pi} d\phi \frac{p e^{-t(p^2+m^2)}}{\sqrt{p^2+m^2}} e^{ipu \cos \phi} \quad (B.8) $$

For $\vec{u} = 0$ this is indeed the integral we computed above. For $\vec{z} \neq 0$ or $\vec{k} \neq 0$, however, we have $\vec{u} \neq 0$ and the angular dependence makes the integral much more difficult to compute. The angular $\phi$-integral can be done exactly as $\int_0^{2\pi} d\phi e^{ipu \cos \phi} = 2\pi J_0(pu)$, where $J_0$ is a Bessel function of first kind, so that

$$ \hat{f}(0; t, \vec{u}) = 2\pi \int_{0}^{\infty} dp \frac{p e^{-t(p^2+m^2)}}{\sqrt{p^2+m^2}} J_0(pu). \quad (B.9) $$

The Bessel function $J_0(\xi)$ is real, with $J_0(0) = 1$, and as $\xi$ increases it decreases and has its first zero around $\xi \approx 2.5$, and it then oscillates with an amplitude that decreases $\sim \xi^{-1/2}$. These oscillations will strongly diminish the contribution to the integral of those values of $p$ that are larger than a few $1/u$. On the other hand, the exponential $e^{-tp^2}$ also restricts the $p$-values to less than a few $1/\sqrt{t}$. Roughly speaking, there is an upper cutoff of the order of $\min(1/\vec{u}, 1/\sqrt{t})$. There does not seem to be any known expression for the integral (B.9) for $\vec{u} \neq 0$. For $\vec{u} = 0$ the exact result for this integral can be read from our above computations in (B.2) and (B.3) (with $r_0 = 0$ and without the $1/m$-term):

$$ \hat{f}(\vec{0}, t, \vec{0}) = \frac{\pi^{3/2}}{\sqrt{t}} - 2\pi \int_{0}^{m} dp e^{-tp^2} = \frac{\pi^{3/2}}{\sqrt{t}} - 2\pi m \sum_{n=0}^{\infty} \frac{(-)^n m^{2n}}{n!(2n+1)} t^n. \quad (B.10) $$

Let us now assume that $\vec{u} \neq 0$. We want to study again the small-$t$ asymptotics, at fixed $\vec{u} \neq 0$. This means in particular that we can assume $1/\sqrt{t} \gg 1/u$ and thus the effective cut-off scale for the integral (B.9) is set by $1/u$. The difficulty in evaluating the integral comes (among others) from the factor $p/\sqrt{p^2+m^2}$, which equals 0 at $p = 0$ and is close to 1 for $p \gg m$. Thus, for small mass $m$ one could expect that replacing this factor by 1 will induce an error of order $m$ only. But this means that the leading term actually computes the result for $m = 0$ and we know that the original sum has a term $1/m^2$ which becomes singular in this limit. One could then compute the integral $2\pi \int_0^\infty dp e^{-tp^2} J_0(pu)$ exactly (ref. [33], section 7.7.3, eq 23) and obtain $\pi^{3/2}/\sqrt{t} e^{-z^2/(8t)} I_0(z^2/8t) + \mathcal{O}(m)$, where $I_0$ is the modified Bessel function of the first kind, and then $\hat{f}(0; t, \vec{u}) \sim_{t \to 0} 2\pi u + \mathcal{O}(m)$. By the Poisson formula (B.6) and (B.7), we need so sum all $\hat{f}(0; t, \vec{z} + 2\pi \vec{k})$ to obtain $\mathcal{L}(t, z)$ and the resulting sum obviously diverges.

Let us then proceed differently: if we are only interested in the leading small-$t$ result, in the integral (B.9) we may simply replace $e^{-tp^2}$ by 1. Indeed, recall that this integral is effectively cut off at $1/u \ll 1/\sqrt{t}$. Then we may express the result in terms of a known integral:

$$ \hat{f}(0; t, \vec{u}) = 2\pi \int_{0}^{\infty} dp \frac{p}{\sqrt{p^2+m^2}} J_0(pu) (1 + \mathcal{O}(t)) = \frac{2\pi}{u} e^{-mu} (1 + \mathcal{O}(t)). \quad (B.11) $$
In the full Poisson sum for $\mathcal{L}(t, z)$ the exponential factor $e^{-mu}$ ensures the convergence:

$$ \mathcal{L}(t, z) \sim_{t \to 0} 2\pi \sum_{(k_1, k_2)} \frac{e^{-m|\vec{z} + 2\pi\vec{k}|}}{|\vec{z} + 2\pi\vec{k}|}, \quad (z \neq 0). \qquad (B.12) $$

Now, on the torus, $z^i = x^i - y^i \in [0, 2\pi)$ and $|\vec{z} + 2\pi\vec{k}| = \sqrt{(x^1 - y^1 + 2\pi k_1)^2 + (x^2 - y^2 + 2\pi k_2)^2}$. The sum then is dominated by the value of $\vec{k}$ such that this expression takes the smallest possible value, which corresponds to the geodesic distance $\ell(x, y)$ on the torus. Hence

$$ \mathcal{L}(t, z) \sim_{t \to 0} 2\pi \frac{e^{-m\ell(x,y)}}{\ell(x,y)} (1 + \mathcal{O}(e^{-ma})), \quad a > 0, \quad x \neq y. \qquad (B.13) $$

Again, for $m \to 0$, the sub-leading terms no longer are sub-leading and we get back the divergence of $\mathcal{L}$.

We will be mainly interested in the regime where both $t$ and $\ell(x, y)$ are small. Then we have seen that $\mathcal{L}(t, 0) \sim_{t \to 0} \frac{1}{m} + \frac{\pi^{3/2}}{\sqrt{t}}$ and, for $\ell(x, y) \neq 0$ but small, $\mathcal{L}(0, z) \sim_{t \to 0} \frac{2\pi}{\ell(x, y)}$. A somewhat ad hoc formula for $\mathcal{L}(t, z)$ that incorporates both behaviours is

$$ \mathcal{L}(t, z) \sim_{\text{small } t, \text{ small } \ell(x,y)} \left(\sqrt{\frac{\pi}{t}} + \frac{1}{\pi m}\right) \int_{-\pi/2}^{\pi/2} d\phi e^{-\frac{\ell^2(x,y)}{4t}} \sin^2 \phi. \qquad (B.14) $$

Indeed, for $x = y$, i.e. $\ell^2(x, y) = 0$, the $\phi$-integral trivially gives $\pi$ and one gets correctly $\frac{\pi^{3/2}}{\sqrt{t}} + \frac{1}{m}$. On the other hand for small but fixed $\ell(x, y) \neq 0$ and $t \to 0$, the integral can be evaluated by a saddle-point approximation and yields $\int_{-\infty}^{\infty} d\phi e^{-\frac{\ell^2(x,y)}{4t}} \phi^2 = \sqrt{\frac{4\pi t}{\ell^2(x,y)}}$, so that $\mathcal{L}(t, z) \sim (\sqrt{\frac{\pi}{t}} + \frac{1}{\pi m}) \sqrt{\frac{4\pi t}{\ell^2(x,y)}} = \frac{2\pi}{\ell(x,y)} + \mathcal{O}(\sqrt{t})$, as it should.

Recall that $K_-(t, x, y)$ was given by $D_x L(t, x, y) = (i\sigma_x \frac{\partial}{\partial x^1} + i\sigma_z \frac{\partial}{\partial x^2} + m\gamma_*)L(t, x, y)$ where

$$ L(t, x, y) = \frac{i}{4\pi^2} \mathcal{L}(t, x, y) \mathbf{1}_{2 \times 2} \sim_{t \to 0} \frac{i e^{-m\ell(x,y)}}{2\pi \ell(x,y)} \mathbf{1}_{2 \times 2} (1 + \mathcal{O}(e^{-ma})), \quad x \neq y, \qquad (B.15) $$

It is straightforward to work out the corresponding asymptotics of $K_-(t, x, y)$ for $x \neq y$, but we will not do it here. Instead, in the main text we obtain the relevant information directly from the relation $\zeta_-(s, x, y) = -D_x \zeta_+(s + \frac{1}{2}, x, y)$.

**Open Access.** This article is distributed under the terms of the Creative Commons Attribution License (CC-BY 4.0), which permits any use, distribution and reproduction in any medium, provided the original author(s) and source are credited.

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