Quantum holographic surface anomalies

Following [23, 24], we look at a bulk geometry asymptotic to $AdS_7\times S^4$. We choose y as the coordinate normal to the boundary, such that the asymptotic form of the metric is [23, 24]

$$\begin{align} G = \frac{R^2}{y^2}\left(\mathrm{d} y^2+\bar G+y^2 \bar{G}^{\left(1\right)}\right) +\frac{R^2}{4}G_{S^4}+{\mathcal{O}}\left(y^2\right)\,. \end{align} \tag{ 2.1 }$$

$\bar G$ is the metric on the boundary of space and $\bar G^{(1)}$ is fixed by the (super)gravity equations to be the Schouten tensor of $\bar G$

$$\begin{align} \bar G_{MN}^{\left(1\right)} = -\bar P_{MN}\left(\bar G\right)\,. \end{align} \tag{ 2.2 }$$

Here we use MN for coordinates on all of (asymptotically) AdS₇.

We study the embedding of a 3d M2-brane with world-volume Σ into this geometry, where the brane ends along a 2d surface $\bar\Sigma$ in the 6d boundary of asymptotically locally AdS₇. We use coordinates τ^a with $a = 1,2$ on the 2d surface and σ^µ with $\mu = 1,2,3$ and $\sigma^3 = y$ on the M2-brane world-volume. The asymptotically locally AdS₇ space is parametrised by coordinates x^M . We take three to be $x^\mu = \sigma^\mu$, and we require that the remaining coordinates $x^{\mu^{^{\prime}}}(\sigma)$ are orthogonal to $\bar{\Sigma}$ at the boundary. Finally we parametrise the S⁴ by zⁱ with $i = 1,\ldots,4$ and we restrict to classical solutions localised at a point on $S^4~z^i(\sigma) = 0$, representing the north pole of S⁴.

The bosonic part of the M2-brane action [49] is the volume form of the induced metric and the pullback of the three-form A₃

$$\begin{align} S_\text{M2} = T_\text{M2} \int_\Sigma \left( \mathrm{vol}_\Sigma + i A_3 \right)\,. \end{align} \tag{ 2.3 }$$

Here A₃ is the potential for the flux

$$\begin{align} F_4 = \textrm{d}A_3 = \frac{3}{8} R^3 \mathrm{vol}_{S^4} +\,{\mathcal{O}}\left(y^2\right)\,, \end{align} \tag{ 2.4 }$$

where $\mathrm{vol}_{S^4}$ is the volume of the unit sphere, as in (2.1). As mentioned above, we assume the surface is localised at a point in S⁴, so the pullback of A₃ in (2.3) vanishes.

The equations of motion for the brane are those of a minimal surface, which can be elegently expressed as the vanishing of the mean curvature vector H^M . Recalling some definitions, the second fundamental form is

$$\begin{align} \mathrm{I\!I}^{M}_{\mu\nu} = \left(\partial_\mu\partial_\nu x^{P} +\partial_\mu x^{Q}\partial_\nu x^{R}\Gamma^{P}{}_{QR}\right) \left(\delta^{M}_{P}-\partial^\rho x^{M}\partial_\rho x^{N}G_{NP}\right)\,, \end{align} \tag{ 2.5 }$$

and using the inverse of the induced metric $g_{\mu\nu}$ we get the mean curvature vector as

$$\begin{align} H^{M} = g^{\mu\nu}\mathrm{I\!I}^{M}_{\mu\nu}\,. \end{align} \tag{ 2.6 }$$

In fact, if the coordinates $x^{\mu^{^{\prime}}}$ are orthogonal to the brane not just at the boundary but everywhere, then the second fundamental form simplifies and the only nonzero components are

$$\begin{align} \mathrm{I\!I}^{\mu^{^{\prime}}}_{\mu \nu} = \left.-\frac{1}{2}G^{\mu^{^{\prime}}\nu^{^{\prime}}}\partial_{\nu^{^{\prime}}}G_{\mu \nu} \right|_\Sigma\,. \end{align} \tag{ 2.7 }$$

We can express $H^{\mu^{^{\prime}}}$ in a Fefferman–Graham expansion as power series in y in (2.1). If $x^{\mu^{^{\prime}}}$ were constant, then it would be the same as the mean curvature on the boundary surface $\bar\Sigma$, so $H^{\mu^{^{\prime}}} = \bar H^{\mu^{^{\prime}}}$ (or strictly speaking, the pullback of $\bar H^{\mu^{^{\prime}}}$, since these two objects are in different bundles). If $x^{\mu^{^{\prime}}}$ is not a constant, we find to lowest nontrivial order in y [23]

$$\begin{align} H^{\mu^{^{\prime}}} = \bar H^{\mu^{^{\prime}}}+y^3\partial_y\left(y^{-3}\partial_yx^{\mu^{^{\prime}}}\right) +{\mathcal{O}}\left(y^2\right)\,. \end{align} \tag{ 2.8 }$$

Imposing the equations of motion sets this to zero and fixes $x^{\mu^{^{\prime}}} = \bar H^{\mu^{^{\prime}}} y^2/4$, as found in [24].

Before imposing the equations of motion and keeping only terms of order ${\mathcal{O}}(y^0)$, the induced metric on the world-volume is

$$\begin{align} \begin{aligned} g_{yy}& = \frac{R^2}{y^2}\left(1 +\partial_y x^{\mu^{^{\prime}}}\partial_y x^{\nu^{^{\prime}}} \bar{g}_{\mu^{^{\prime}}\nu^{^{\prime}}} \right). \\ g_{ab}& = \frac{R^2}{y^2}\left(\bar g_{ab} -y^2\bar P_{ab} -2\bar{\mathrm{I\!I}}^{\mu^{^{\prime}}}_{ab}x^{\nu^{^{\prime}}} \bar{g}_{\mu^{^{\prime}}\nu^{^{\prime}}} \right),\\ g_{ay}& = 0\,, \end{aligned} \end{align} \tag{ 2.9 }$$

Here $\bar g_{ab}$ is the metric on $\bar\Sigma$ and $\bar P_{ab}$ is the pullback of the bulk Schounten tensor to the brane, and likewise for the second fundamental form. We also use $g_{\mu^{^{\prime}}\nu^{^{\prime}}} = G_{\mu^{^{\prime}}\nu^{^{\prime}}}|_{\Sigma}$ for the metric evaluated on the brane, and $\bar{g}_{\mu^{^{\prime}}\nu^{^{\prime}}} = G_{\mu^{^{\prime}}\nu^{^{\prime}}}|_{\bar{\Sigma}}$ for its value at the boundary of AdS.

Plugging in the solution to the asymptotic equations, $x^{\mu^{^{\prime}}} = \bar H^{\mu^{^{\prime}}} y^2/4$, gives the induced metric

$$\begin{align} \begin{aligned} g_{yy}& = \frac{R^2}{y^2}\left(1+\frac{y^2}{4}\bar H^2\right), \\ g_{ab}& = \frac{R^2}{y^2}\left(\bar g_{ab} -\bar P_{ab}y^2 -\frac{y^2}{2}\bar{\mathrm{I\!I}}^{\mu^{^{\prime}}}_{ab}\bar H^{\nu^{^{\prime}}} \bar{g}_{\mu^{^{\prime}}\nu^{^{\prime}}}\right),\\ g_{ay}& = 0\,. \end{aligned} \end{align} \tag{ 2.10 }$$

The classical action (expanding the integrand to order y⁻¹) is now

$$\begin{align}\begin{aligned}{} S_\text{classical} & = T_\text{M2}\int_\Sigma\mathrm{d}^3\sigma\,\frac{R^3}{y^3} \sqrt{\bar g\left(1+\frac{y^2}{4}\bar H^2\right) \left(1-y^2\mathrm{tr}\,\bar P-\frac{y^2}{2}\bar H^2\right)} \\ & = T_\text{M2}R^3\int_0^\infty \frac{\mathrm{d} y}{y^3} \int_{\bar\Sigma}\mathrm{d}^2\tau\,\sqrt{\bar g} \left(1-\frac{y^2}{8}\bar H^2-\frac{y^2}{2}\mathrm{tr}\,\bar P\right)\,. \end{aligned} \end{align} \tag{ 2.11 }$$

Using $T_\text{M2}R^3 = 2N/\pi$, we get a quadratic and logarithmic divergences

$$\begin{align} S_\text{classical} = \frac{N}{\pi\epsilon^2}\mathrm{vol}\left(\bar\Sigma\right) +\frac{N}{4\pi}\int_{\bar\Sigma}\mathrm{d}^2\tau \sqrt{\bar g}\left(\bar H^2+4\mathrm{tr}\,\bar P\right)\log\epsilon\,. \end{align} \tag{ 2.12 }$$

The quadratic divergence can be cancelled by an appropriate Legendre transform [50, 51]. Evaluating $\exp[-S_\text{classical}]$ then gives a power of $1/\epsilon$ which should be minus the anomaly (1.3), so we can identify the leading result at large N for the anomaly coefficients, namely $a_1 = b = 0$ and $a_2 = -N$ as in (1.6) [24].

The induced metric (2.10) is perfectly fine in order to plug into the action and evaluate the classical anomalies. The coordinates and metric have some issues that make them less than ideal for the quantum calculation in section 3. First, the induced metric is not in Fefferman–Graham form, so the asymptotic AdS₃ structure needed in there is not manifest. Second, the quadratic fluctuation action derived in appendix A assumes coordinates tangent and normal to the brane. The fact that $x^{\mu^{^{\prime}}}$ depends on y means that they are not orthogonal.

Both of those issues can be resolved with a change of coordinates to z and $u^{\mu^{^{\prime}}}$ defined via

$$\begin{align} \begin{aligned} x^{\mu^{^{\prime}}} & = u^{\mu^{^{\prime}}} + \frac{y^2}{4} \bar{H}^{\mu^{^{\prime}}}\,, \\ y & = z \exp{\left( -\frac{1}{2} x^{\mu^{^{\prime}}} \bar{H}_{\mu^{^{\prime}}} - \frac{z^2}{16} \bar{H}^2 \right)}\,. \end{aligned} \end{align} \tag{ 2.13 }$$

The coordinates τ^a remain untouched for now. The choice of $u^{\mu^{^{\prime}}}$ is such that it vanishes on the classical solution (2.8) and then the definition of z makes the metric block diagonal and simplifies the induced metric. Plugging this into (2.1), we find to order z⁰

$$\begin{align} \begin{aligned} \frac{G}{R^2} & = \frac{\mathrm{d} z^2}{z^2} + \frac{\textrm{e}^{u^{\rho^{^{\prime}}} \bar{H}_{\rho^{^{\prime}}}}}{z^2} \left[ \left( \bar{G}_{ab} \mathrm{d} \tau^a \mathrm{d} \tau^b + \bar{G}_{\mu^{^{\prime}}\nu^{^{\prime}}} \mathrm{d} u^{\mu^{^{\prime}}} \mathrm{d} u^{\nu^{^{\prime}}} \right) \left( 1 + \frac{z^2}{8} \bar{H}^2 \right) + 2 \bar{G}_{a \mu^{^{\prime}}} \mathrm{d} \tau^a \mathrm{d} u^{\mu^{^{\prime}}} \right]\\ & \quad{} - \frac{1}{2} \bar{H}_{\mu^{^{\prime}}} \bar{H}_{\nu^{^{\prime}}} \mathrm{d} u^{\mu^{^{\prime}}} \mathrm{d} u^{\nu^{^{\prime}}} + \bar{G}_{a \mu^{^{\prime}}} \left( \frac{\mathrm{d} z \mathrm{d} \tau^{a}}{z} \bar{H}^{\mu^{^{\prime}}} - \frac{\mathrm{d} \tau^a \mathrm{d} u^{\nu^{^{\prime}}}}{2} \bar{H}^{\mu^{^{\prime}}} \bar{H}_{\nu^{^{\prime}}} \right) -\bar P + \frac{G_{S^4}}{4}\,. \end{aligned} \end{align} \tag{ 2.14 }$$

Like (2.1), this metric is also in Fefferman–Graham form. Indeed, the Fefferman–Graham expansion is unique only given a boundary metric $\bar G$. The terms in the square bracket in the first line of (2.14) (excluding $z^2\bar H^2/8$) are exactly $\bar G$, so we see that we conformally transformed the boundary metric by $\exp[u^{\rho^{^{\prime}}}\bar H_{\rho^{^{\prime}}}]$. Thus (2.14) is the Fefferman–Graham expansion for a different choice of boundary metric in the same conformal class as $\bar G$.

All the terms of order z⁰ (with the exception of the S⁴ part) are minus the Schouten tensor for the conformally transformed boundary metric. Of course, given that now $u^{\mu^{^{\prime}}} = 0$ on the classical solution, the mean curvature for the surface $\bar\Sigma$ in this conformally transformed metric vanishes.

We can find the induced metric by setting $u^{\mu^{^{\prime}}} = 0$, where also $\bar g_{a\mu^{^{\prime}}} = 0$, resulting in

$$\begin{align} \frac{g}{R^2} = \frac{\mathrm{d} z^2}{z^2} + \frac{\mathrm{d} \tau^a \mathrm{d} \tau^b}{z^2} \left( \bar{g}_{ab} - z^2 \bar\Pi_{ab} \right). \end{align} \tag{ 2.15 }$$

Where we defined

$$\begin{align} \bar\Pi_{ab} = - \frac{1}{8} \bar{H}^2 \bar{g}_{ab} + \frac{1}{2} \bar{\mathrm{I\kern-1ptI}}^{\nu^{^{\prime}}}_{ab} \bar{H}^{\mu^{^{\prime}}} \bar{g}_{\mu^{^{\prime}}\nu^{^{\prime}}} +\bar{P}_{ab}\,, \qquad \mathrm{tr}\,\bar\Pi = \frac{1}{4}H^2+\mathrm{tr}\,\bar P\,. \end{align} \tag{ 2.16 }$$

The induced metric (2.15) is now also in Fefferman–Graham form for an asymptotically locally AdS₃ with boundary metric $\bar g$ on $\bar\Sigma$. Note that the correction $z^2\bar\Pi$ is not the Schouten tensor of $\bar g$, as in the bulk case (2.2), since we do not impose an Einstein equation on the world-volume, but rather the minimal surface equations.

In fact, we could have started the analysis from this statement

Given a surface $\bar\Sigma\subset\bar M_6$ and a conformal class on $\bar M_6$, one can choose a metric in that class such that $\bar\Sigma$ is minimal, so $\bar{H}^{\mu^{^{\prime}}} = 0$. The Graham–Witten solution then is $u^{\mu^{^{\prime}}} = 0$ and the induced metric (2.10) is automatically in Fefferman–Graham form ⁴ .

Note that the conformal transformation used in (2.13) and (2.14) is defined only locally, but that should not matter for our calculation of a local anomaly density.

In practical term, this other approach would set $\bar H^{\mu^{^{\prime}}} = 0$ in all calculations, so $\bar\Pi_{ab} = \bar P_{ab}$. We could also work in this setting and replace $\mathrm{tr}\,\bar P\to \bar H^2/4+\mathrm{tr}\,\bar P$ in the final expressions, as this is the combination that appears in the anomaly (1.5). We did perform the calculation keeping nonzero $\bar H^{\mu^{^{\prime}}}$ just to make sure that we do not make any mistakes.

In practice, solving the heat kernel equation for an arbitrary kinetic operator L on an arbitrary manifold is impossible. Fortunately, for the purpose of extracting anomaly coefficients, we only need the behavior of the differential operators near the boundary of AdS.

We start with a massive scalar laplacian $L = -\Delta + M^2$ on the asymptotically locally AdS₃ world-volume given in (2.15) and further simplify the calculation by choosing Riemann normal coordinates for the metric $\bar g$ about a particular point (corresponding to $\tau_0 = (0,0)$ and at fixed z₀), such that

$$\begin{align} \begin{aligned} g_{zz} & = \frac{R^2}{z^2}\,, \\ g_{az} & = 0 \,, \\ g_{ab} & = \frac{R^2}{z^2}\delta_{ab} -R^2\left(\frac{1}{3z^2}\bar {\mathcal{R}}_{acbd}\tau^c\tau^d +\bar \Pi_{ab} \right). \end{aligned} \end{align} \tag{ 3.3 }$$

For simplicity we set the AdS radius R = 1 in the following. We proceed by considering the parenthesis on the last line as a perturbation about the AdS₃ geometry. More precisely, we treat $(\tau^a,z)$ as homogeneous coordinates and expand in a power series in them, so $g_{\mu\nu} = g_{\mu\nu}^{(-2)}+g_{\mu\nu}^{(0)}+\cdots$ with the index indicating the degree and

$$\begin{align} \begin{aligned} g_{\mu\nu}^{\left(-2\right)}\mathrm{d}\sigma^\mu\mathrm{d}\sigma^\nu & = \frac{1}{z^2}\left(\mathrm{d} z^2+\delta_{ab}\mathrm{d} \tau^a \mathrm{d}\tau^b\right)\,, \\ g_{\mu\nu}^{\left(0\right)}\mathrm{d}\sigma^\mu\mathrm{d}\sigma^\nu & = -\left(\frac{1}{3}\bar {\mathcal{R}}_{acbd} \frac{\tau^c\tau^d}{z^2} +\bar \Pi_{ab} \right)\mathrm{d}\tau^a\mathrm{d}\tau^b\,. \end{aligned} \end{align} \tag{ 3.4 }$$

Likewise, the determinant of the metric is expanded as

$$\begin{align} \begin{aligned} \sqrt{g}^{\left(-3\right)}& = \frac{1}{z^3}\,, \\ \sqrt{g}^{\left(-1\right)} & = -\frac{1}{z}\left(\frac{1}{6}\bar{\mathcal{R}}_{cd}\frac{\tau^c\tau^d}{z^2} +\frac{1}{2} \mathrm{tr}\,\bar\Pi\right). \end{aligned} \end{align} \tag{ 3.5 }$$

We now expand the kinetic operator $L = -\Delta+M^2$ in the same way as

$$\begin{align} L = L^{\left(0\right)}+L^{\left(2\right)}+\cdots \end{align} \tag{ 3.6 }$$

For a massive field, we assume an expansion of the mass term $M^2 = {\mathcal{M}}^{(0)} + {\mathcal{M}}^{(2)} + \cdots$, and the kinetic operator takes the form

$$\begin{align} \begin{aligned} L^{\left(0\right)} & = -\frac{1}{\sqrt{g}^{\left(-3\right)}} \partial_\mu \sqrt{g}^{\left(-3\right)} g^{\left(2\right)\,\mu\nu}\partial_\nu + {\mathcal{M}}^{\left(0\right)}\,, \\ L^{\left(2\right)} & = \frac{\sqrt{g}^{\left(-1\right)}}{\left(\sqrt{g}^{\left(-3\right)} \right)^2} \partial_\mu \sqrt{g}^{\left(-3\right)} g^{\left(2\right)\,\mu\nu}\partial_\nu\\ & \quad -\frac{1}{\sqrt{g}^{\left(-3\right)}} \partial_\mu\left( \sqrt{g}^{\left(-1\right)} g^{\left(2\right)\,\mu\nu} +\sqrt{g}^{\left(-3\right)} g^{\left(4\right)\,\mu\nu}\right)\partial_\nu + {\mathcal{M}}^{\left(2\right)}\,, \end{aligned} \end{align} \tag{ 3.7 }$$

where of course $g^{(4)\,\mu\nu} = -g^{(2)\,\mu \rho} g_{\rho \sigma}^{(0)} g^{(2)\,\sigma \nu}$.

Plugging in the metric (3.4) yields

$$\begin{align} \begin{aligned} L^{\left(0\right)} & = -z^3 \partial_{z} \left( \frac{1}{z} \partial_{z}\right) - z^2\delta^{ab} \partial_a \partial_b + {\mathcal{M}}^{\left(0\right)} \,, \\ L^{\left(2\right)} & = \mathrm{tr}\,\bar{\Pi}\, z^3 \partial_z + \frac{2 z^2}{3} \bar{\mathcal{R}}_{ab} \tau^a \partial_b -\left( \frac{z^2}{3} \bar{\mathcal{R}}_{acbd} \tau^c \tau^d + z^4 \bar\Pi_{ab} \right) \partial_a \partial_b + {\mathcal{M}}^{\left(2\right)}\,. \end{aligned} \end{align} \tag{ 3.8 }$$

We emphasize that the kinetic operator L acts on the point σ, and is evaluated at that point. The quantities (${\mathcal{M}}, \bar\Pi, \cdots$) appearing in $L^{(2)}$ are all evaluated at σ₀.

The heat kernel itself can also be expanded as

$$\begin{align} K\left(t;\sigma,\sigma_0\right) = K^{\left(0\right)}\left(t;\sigma,\sigma_0\right) +K^{\left(2\right)}\left(t;\sigma,\sigma_0\right) + \cdots \end{align} \tag{ 3.9 }$$

To calculate the anomaly we only need those first two terms. $K^{(0)}$ is of homogeneous degree zero, so the volume integral in (3.2) has a quadratic divergence. It may also contribute logarithmic divergences from the subleading terms in the volume form $\sqrt{g}^{(-1)}$ (3.5). $K^{(2)}$ is quadratic, so like the y² terms in the classical action (2.11), gives logarithmic divergences contributing to the conformal anomaly.

The heat kernel equation (3.1) then reduces to the recursive relations

$$\begin{align} \left(L^{\left(0\right)} + \partial_t\right) K^{\left(0\right)} = 0\,, \qquad \left(L^{\left(0\right)} + \partial_t\right) K^{\left(2\right)} = -L^{\left(2\right)} K^{\left(0\right)}\,. \end{align} \tag{ 3.10 }$$

The first equation is the AdS₃ heat kernel equation, its solution is given by [44] (also [45, 46])

$$\begin{align} K^{\left(0\right)}\left(t;\sigma,\sigma_0\right) = \frac{1}{\left(4\pi t\right)^{3/2}} \frac{\rho}{\sinh \rho} \exp\left( -\left(1+{\mathcal{M}}^{\left(0\right)}\right)t - \frac{\rho^2}{4t} \right) \,, \end{align} \tag{ 3.11 }$$

where ρ is the geodesic distance in AdS₃ between σ and σ₀

$$\begin{align} \rho = \mathrm{arccosh}\left( \frac{z^2 + z_0^2 + |\tau - \tau_0|^2}{2 z z_0} \right)\,. \end{align} \tag{ 3.12 }$$

The equation for $K^{(2)}$ can be solved by observing that $K^{(0)}(t^{^{\prime}} - t;\sigma,\sigma_0)\Theta(t^{^{\prime}}- t)$ (with Θ the Heaviside step function) is a Green's function for $L^{(0)}+\partial_t$, i.e.

$$\begin{align} \left(L^{\left(0\right)} + \partial_t\right) K^{\left(0\right)}\left(t^{^{\prime}}-t;\sigma,\sigma_0\right) \Theta\left(t^{^{\prime}}-t\right) = \frac{1}{\sqrt{g}^{\left(-3\right)}}\delta^{\left(3\right)}\left(\sigma-\sigma_0\right)\delta\left(t^{^{\prime}} - t\right)\,. \end{align} \tag{ 3.13 }$$

This allows us to express $K^{(2)}$ as the integral

$$\begin{align} \begin{aligned} K^{\left(2\right)}\left(t;\sigma,\sigma_0\right) & = -\int_0^t \mathrm{d} t^{^{\prime}} \int_\Sigma \mathrm{d}^3 \sigma^{^{\prime}}\sqrt{g}^{\left(-3\right)} K^{\left(0\right)}\left(t - t^{^{\prime}};\sigma,\sigma^{^{\prime}}\right) L^{\left(2\right)}_{\sigma^{^{\prime}}} K^{\left(0\right)}\left(t^{^{\prime}};\sigma^{^{\prime}},\sigma_0\right)\\ &\quad -\frac{\sqrt{g}^{\left(-1\right)}}{\sqrt{g}^{\left(-3\right)}}K^{\left(0\right)}\left(t;\sigma,\sigma_0\right) \,, \end{aligned} \end{align} \tag{ 3.14 }$$

where the second line ensures that

$$\begin{align} \lim_{t \to 0} K^{\left(2\right)}\left(t;\sigma,\sigma_0\right) = -\frac{\sqrt{g}^{\left(-1\right)}}{\left(\sqrt{g}^{\left(-3\right)}\right)^2}\delta^{\left(3\right)}\left(\sigma-\sigma_0\right)\,, \end{align} \tag{ 3.15 }$$

so that the boundary conditions at t = 0 (3.1) are compatible with the subleading terms in the measure $\sqrt{g}$ (3.5).

To evaluate the determinant we need (3.2) the coincident point limit of the heat kernel $K(t;\sigma_0,\sigma_0)$, also known as the trace of the heat kernel. For $K^{(0)}$, we have the explicit expression (3.11) from which we get

$$\begin{align} \lim_{\sigma \to \sigma_0} K^{\left(0\right)}\left(t;\sigma,\sigma_0\right) = \frac{1}{\left(4\pi t\right)^{3/2}} \exp\left( -\left(1+{\mathcal{M}}^{\left(0\right)}\right)t\right)\,. \end{align} \tag{ 3.16 }$$

For $K^{(2)}$ we can use the integral expression (3.14) to evaluate the coincident limit. $K^{(0)}$ depends on the coordinates only through the geodesic distance $\rho(\sigma^{^{\prime}},\sigma_0)$, so it is natural to change coordinates for AdS₃ to

$$\begin{align} \mathrm{d} s^2 = \left(\mathrm{d} \rho^2 + \sinh^2{\rho}\left(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2\right)\right)\,. \end{align} \tag{ 3.17 }$$

The change of variables is given by

$$\begin{align} z = \frac{z_0}{\cosh{\rho} - \sinh \rho \cos \theta}\,, \qquad \tau_a = \frac{z_0 \sinh \rho \sin{\theta}\, e_a}{\cosh{\rho} - \sinh{\rho} \cos{\theta}}\,, \end{align} \tag{ 3.18 }$$

where $e_a = (\cos\phi,\sin\phi)_a$.

We now can express $L^{(2)}$ (3.8) in these coordinates, but since we act with it on $K^{(0)}$, we only need to keep derivatives with respect to ρ. Those are

$$\begin{align} \begin{aligned} L^{\left(2\right)}f\left(\rho\right) & = -z_0^2 \left[ \frac{\bar{\mathcal{R}} \sinh{\rho} \sin^2{\theta}} {6 \left(\cosh{\rho} - \sinh{\rho} \cos{\theta}\right)^3} \partial_\rho + \mathrm{tr}\,\bar\Pi \frac{\coth{\rho}}{\left( \cosh{\rho} - \sinh{\rho} \cos{\theta} \right)^2} \partial_\rho \right.\\ & \quad + \left. \bar{\Pi}_{ab} \frac{\sin^2{\theta}\, e_a e_b}{\left(\cosh{\rho} - \sinh{\rho} \cos\theta\right)^4} \left( \partial_\rho^2 - \coth{\rho} \partial_\rho \right) \right]f\left(\rho\right) +{\mathcal{M}}^{\left(2\right)}f\left(\rho\right) \,. \end{aligned} \end{align} \tag{ 3.19 }$$

The integral in (3.14) is

$$\begin{align} - \int_0^t \mathrm{d} t^{^{\prime}} \int_\Sigma \mathrm{d} \rho \,\mathrm{d} \theta \,\mathrm{d} \phi \sinh^2{\rho} \sin{\theta} K^{\left(0\right)}\left(t-t^{^{\prime}};\rho\right) L^{\left(2\right)} K^{\left(0\right)}\left(t^{^{\prime}};\rho\right)\,. \end{align} \tag{ 3.20 }$$

The φ integral is easy to do using

$$\begin{align} \int_0^{2\pi} \mathrm{d} \phi\, e_a e_b = \pi \delta_{ab}\,. \end{align} \tag{ 3.21 }$$

The θ integral can be done as well to get

$$\begin{align} \begin{aligned} -4\pi & \int_0^t \mathrm{d} t^{^{\prime}} \int_0^\infty \mathrm{d} \rho \sinh^2{\rho}\, K^{\left(0\right)}\left(t-t^{^{\prime}};\rho\right) \\ & \times\left[ z_0^2 \left(-\frac{1}{3}\mathrm{tr}\,\bar\Pi \left( \partial_\rho^2 + 2 \coth{\rho} \partial_\rho \right) + \frac{\bar{\mathcal{R}}}{6} \frac{\cosh{\rho} \sinh{\rho} - \rho}{\sinh^2{\rho}} \partial_\rho \right) + \left\langle {\mathcal{M}}^{\left(2\right)} \right\rangle \right] K^{\left(0\right)}\left(t^{^{\prime}};\rho\right)\,. \end{aligned} \end{align} \tag{ 3.22 }$$

$\left\langle {\mathcal{M}}^{(2)} \right\rangle$ denotes the average of ${\mathcal{M}}^{(2)}$ over the spherical coordinates. We can simplify further the coefficient of $\mathrm{tr}\,\bar\Pi$ by using the heat kernel equation (3.10), which in these coordinates is

$$\begin{align} \left( -\partial_\rho^2 - 2 \coth{\rho}\, \partial_\rho + {\mathcal{M}}^{\left(0\right)} + \partial_t^{^{\prime}} \right) K^{\left(0\right)}\left(t^{^{\prime}};\rho\right) = 0\,, \end{align} \tag{ 3.23 }$$

which removes both first and second order ρ derivatives and we obtain

$$\begin{align} \begin{aligned} & 4\pi \int_0^t \mathrm{d} t^{^{\prime}} \int_0^\infty \mathrm{d} \rho \sinh^2{\rho}\, K^{\left(0\right)}\left(t-t^{^{\prime}};\rho\right)\\ & \quad \times \left[ \frac{z_0^2}{3} \mathrm{tr}\,\bar\Pi\, \left(\partial_{t^{^{\prime}}} + {\mathcal{M}}^{\left(0\right)} \right) - \frac{z_0^2}{6}\bar{\mathcal{R}} \frac{\cosh{\rho} \sinh{\rho} - \rho}{\sinh^2{\rho}} \partial_\rho - \left\langle {\mathcal{M}}^{\left(2\right)} \right\rangle \right] K^{\left(0\right)}\left(t^{^{\prime}};\rho\right)\,. \end{aligned} \end{align} \tag{ 3.24 }$$

The remaining integrals can be evaluated without much difficulty after plugging in the expression for $K^{(0)}$ in (3.11). A more elegant way to evaluate them is to use partial integration to reduce them to a convolution of heat kernels. Consider first the term proportional to $\mathrm{tr}\,{\bar{\Pi}}$. We can take out the tʹ derivative as

$$\begin{align} \begin{aligned} & \frac{4\pi z_0^2}{3} \mathrm{tr}\,\bar\Pi\, \int_0^t \mathrm{d} t^{^{\prime}} \int_0^\infty \mathrm{d} \rho \sinh^2{\rho} K^{\left(0\right)}\left(t - t^{^{\prime}};\rho\right) \left(\partial_{t^{^{\prime}}} + {\mathcal{M}}^{\left(0\right)} \right) K^{\left(0\right)}\left(t^{^{\prime}};\rho\right) \\ & \quad = \frac{4 \pi z_0^2}{3}\mathrm{tr}\,\bar\Pi \int_0^t \mathrm{d} t^{^{\prime}} \left(\partial_{t^{^{\prime}}}+\partial_t + {\mathcal{M}}^{\left(0\right)} \right) \int_0^\infty \mathrm{d} \rho \sinh^2{\rho} K^{\left(0\right)}\left(t - t^{^{\prime}};\rho\right) K^{\left(0\right)}\left(t^{^{\prime}};\rho\right)\,. \end{aligned} \end{align} \tag{ 3.25 }$$

Now the ρ integral is the convolution of two heat kernels (up to the factor 4π from the spherical integral), giving simply $K^{(0)}(t;\sigma_0,\sigma_0)$. The tʹ dependence completely drops out and the integral over it gives t, leaving

$$\begin{align} \frac{z_0^2}{3} \mathrm{tr}\,\bar\Pi\, t \left(\partial_t + {\mathcal{M}}^{\left(0\right)}\right) K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right)\,. \end{align} \tag{ 3.26 }$$

Second, for the term proportional to $\bar{\mathcal{R}}$, with an exchange of $t^{^{\prime}}\to t-t^{^{\prime}}$ the derivative can be moved to the other $K^{(0)}$, so we can write it as

$$\begin{align} \begin{aligned} & -\frac{\pi z_0^2}{3} \bar{\mathcal{R}} \int_0^{t} \mathrm{d} t^{^{\prime}} \int_0^\infty \mathrm{d} \rho \left( \cosh{\rho} \sinh{\rho} - \rho \right) \partial_\rho\left(K^{\left(0\right)}\left(t - t^{^{\prime}};\rho\right) K^{\left(0\right)}\left(t^{^{\prime}};\rho\right)\right)\\ & \quad = \frac{2\pi z_0^2}{3} \bar{\mathcal{R}} \int_0^{t} \mathrm{d} t^{^{\prime}} \int_0^\infty \mathrm{d} \rho \sinh^2{\rho}\, K^{\left(0\right)}\left(t - t^{^{\prime}};\rho\right) K^{\left(0\right)}\left(t^{^{\prime}};\rho\right) = \frac{z_0^2}{6} \bar{\mathcal{R}} t K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right)\,. \end{aligned} \end{align} \tag{ 3.27 }$$

The last term in (3.24) is that proportional to ${\mathcal{M}}^{(2)}$, and assuming $\left\langle {\mathcal{M}}^{(2)} \right\rangle$ does not depend on ρ (as we show in the case of interest below), it is directly the convolution of the two heat kernels.

Adding (3.26) and (3.27) and the contribution from ${\mathcal{M}}^{(2)}$ and plugging the value of the heat kernel at coincident points (3.16) gives

$$\begin{align} \left[ -\frac{z_0^2}{3} \mathrm{tr}\,\bar\Pi\, \left(\frac{3}{2t}+1\right) + \frac{z_0^2}{6} \bar{\mathcal{R}} + \left\langle {\mathcal{M}}^{\left(2\right)} \right\rangle \right] \frac{\exp\left( -\left(1+{\mathcal{M}}^{\left(0\right)}\right)t \right)} {\left(4 \pi\right)^{3/2} t^{1/2}}\,. \end{align} \tag{ 3.28 }$$

Finally, adding the term which corrects the boundary conditions from the second line of (3.14), it exactly cancels the $3/2t$ term and we get

$$\begin{align} K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) = \left[ \frac{z_0^2}{6}\left(\bar{\mathcal{R}} - 2 \mathrm{tr}\,\bar\Pi\right) - \left\langle {\mathcal{M}}^{\left(2\right)} \right\rangle \right] \frac{\exp\left( -\left(1+{\mathcal{M}}^{\left(0\right)}\right)t \right)} {\left(4 \pi\right)^{3/2} t^{1/2}}\,. \end{align} \tag{ 3.29 }$$

Note the appearance of the combination $\bar{\mathcal{R}} - 2 \mathrm{tr}\,{\bar{\Pi}}$. The correction to the AdS₃ heat kernel coming from geometric terms is expected to be proportional to this combination (up to perhaps also a contribution from terms involving the Weyl tensor), because of the following consistency check. Take the surface to be a sphere $\bar{\Sigma} = S^2$, with corresponding classical geometry AdS₃ [26]. In that case the heat kernel is exactly $K^{(0)}$, so we should find $K^{(2)} = 0$. Indeed, although both $\bar{\mathcal{R}} = 2$ and $\mathrm{tr}\,{\bar{\Pi}} = 1$ are nonzero for the sphere, the combination $\bar{\mathcal{R}} - 2 \mathrm{tr}\,{\bar{\Pi}}$ vanishes.

For the evaluation of the determinant (3.2), we need the t integral of the trace of the heat kernel, (3.16) and (3.29), which are simple gamma function integrals. Reinstating powers of R we get

$$\begin{align} \int_0^\infty \frac{\mathrm{d} t}{t}\, K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right) & = \frac{\left(1+{\mathcal{M}}^{\left(0\right)} R^2\right)^{\frac{3}{2}}}{6\pi R^3}\,, \end{align} \tag{ 3.30 }$$

$$\begin{align} \int_0^\infty \frac{\mathrm{d} t}{t}\, K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) & = \frac{\sqrt{1 + {\mathcal{M}}^{\left(0\right)} R^2}}{4\pi R} \left[ \frac{z_0^2}{6R^2} \left(2\mathrm{tr}\,\bar\Pi - \bar{\mathcal{R}}\right) + \left\langle {\mathcal{M}}^{\left(2\right)} \right\rangle \right]\,. \end{align} \tag{ 3.31 }$$

We can use the results above to evaluate the determinant of bosonic fluctuations of M2-branes describing surface operators. The quadratic action is derived in appendix A and consists of eight bosonic and eight fermionic modes. The bosonic modes are in turn split into the four modes related to the S⁴ part of the geometry and four directions transverse to the M2-brane world-volume within AdS₇.

The first four bosonic modes are massless scalars, and their kinetic operator is simply $L = -\Delta$, see (A.6). The path integral over these fluctations contributes $-\frac{4}{2} \log \det(-\Delta)$ to the log of the partition function, which can be evaluated using (3.2). The integrand contains both a cubic and a simple pole in z. Performing the z integral, the cubic pole leads to a quadratic divergence in the cutoff, which we drop (it can be removed by a local counterterm). The simple pole leads to $\log\epsilon$ divergence, and leaves an integral over $\bar{\Sigma}$ as in the expression for the anomaly (1.4). The contribution of these four modes to the anomaly density is therefore

$$\begin{align} {\mathcal{A}}^{S^4} & = -\frac{4}{2} \substack{\mathrm{res}\\{z_0 \to 0}}\left[ \int_0^\infty \frac{\mathrm{d} t}{t} \frac{\sqrt{g}^{\left(-3\right)} + \sqrt{g}^{\left(-1\right)} + \cdots}{\sqrt{\bar{g}}} \left( K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right) + K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) + \cdots \right) \right]\nonumber\\ & = -2 \substack{\mathrm{res}\\{z_0 \to 0}} \left[ \left( -\frac{1}{2z_0} \mathrm{tr}\,{\bar{\Pi}} \right) \int_0^\infty \frac{\mathrm{d} t}{t} K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right) + \frac{1}{z_0^3} \int_0^\infty \frac{\mathrm{d} t}{t} K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) + \cdots \right]\nonumber\\ & = \frac{1}{4\pi} \frac{1}{3} \bar{\mathcal{R}}\,. \end{align} \tag{ 3.32 }$$

The second line has the metric in Riemann normal coordinates (3.5), and in going to the last line we use the expressions for the integrated heat kernel (3.30) and (3.31).

The second set of four bosonic modes parametrise the normal bundle $N\Sigma$ to the world-volume of the M2-brane. The kinetic operator derived in (A.6) acts on the fiber bundle (with fiber $V = \mathbb{R}^4$), and the path integral over the fluctuations contributes a factor $-\frac{1}{2} \log\det_V(-\Delta + M^2)$ to the log of the partition function. Evaluating this determinant requires a slight generalisation of the formalism for the scalars.

The heat kernel is still defined by (3.1), with the initial condition multiplied on the right hand side by the identity matrix $\unicode{x1D7D9}_{V}$, and the determinant of L has the aditional trace

$$\begin{align} \log\det\nolimits_V\left(L\right) = - \int_\Sigma \mathrm{d}^3 \sigma \sqrt{g} \int_0^\infty \frac{\mathrm{d} t}{t} \mathrm{tr}\,_V{K\left(t;\sigma,\sigma\right)}\,. \end{align} \tag{ 3.33 }$$

The class of differential operators we need to consider are laplacians on the normal bundle along with a mass term. They take the form

$$\begin{align} L = -\frac{1}{\sqrt{g}} D_{\mu} \left( \sqrt{g} g^{\mu\nu} D_\nu \right) + M^2\,, \end{align} \tag{ 3.34 }$$

with D_µ the covariant derivative acting on fluctuations $\zeta^{m^{^{\prime}}}$

$$\begin{align} D_\mu \zeta^{m^{^{\prime}}} \equiv \partial_\mu \zeta^{m^{^{\prime}}} + \omega^{m^{^{\prime}}n^{^{\prime}}}_{\mu} \zeta^{n^{^{\prime}}}\,, \end{align} \tag{ 3.35 }$$

where $\omega_\mu^{m^{^{\prime}}n^{^{\prime}}}$ is the spin connection on the normal bundle. To evaluate it, we pick a vielbein for the metric G (2.14). We then expand it at u = 0 in $(z,\tau^a)$ as

$$\begin{align} E = E^{\left(-1\right)} + E^{\left(1\right)} + \cdots\,, \end{align} \tag{ 3.36 }$$

where $E^{(-1)}$ is a vielbein for the AdS₇ metric, for instance

$$\begin{align} E^{\left(-1\right)\,m} = \frac{1}{z} \delta_\mu^m \mathrm{d} \sigma^\mu\,, \qquad E^{\left(-1\right)\,m^{^{\prime}}} = \frac{1}{z} \bar{E}_{\mu^{^{\prime}}}^{m^{^{\prime}}} \mathrm{d} u^{\mu^{^{\prime}}}\,. \end{align} \tag{ 3.37 }$$

The relevant spin connection is defined by

$$\begin{align} \omega_\mu^{m^{^{\prime}}n^{^{\prime}}} = E_{\mu^{^{\prime}}}^{m^{^{\prime}}} \left( \partial_\mu E^{\mu^{^{\prime}} n^{^{\prime}}} + \Gamma^{\mu^{^{\prime}}}_{\mu \rho^{^{\prime}}} E^{\rho^{^{\prime}} n^{^{\prime}}} \right). \end{align} \tag{ 3.38 }$$

Importantly, this vanishes using the leading order vielbein $E^{(-1)}$, so $\omega^{(-1)} = 0$. If we expand L (3.34), to leading order we recover four copies of the scalar laplacian on AdS₃ (3.8). The mass matrix ${\mathcal{M}}^{(0)}_{m^{^{\prime}}n^{^{\prime}}}$ is diagonal with eigenvalues 3 [42]. Likewise $K^{(0)}$ is a diagonal matrix with entries as in (3.11). Had $\omega^{(-1)}$ not vanished, we would have had to solve for the vector heat kernel, in analogy with the spinors in the next section. Instead, the nontrivial bundle only affects $L^{(2)}$ and $K^{(2)}$.

The first subleading correction is $L^{(2)}$ as in (3.7) with the additional contribution from the spin connection

$$\begin{align} -\frac{1}{\sqrt{g}^{\left(-3\right)}} \partial_{\mu} \left( \sqrt{g}^{\left(-3\right)} g^{\left(2\right)\,\mu\nu} \right) \omega_\mu^{\left(1\right)\,m^{^{\prime}}n^{^{\prime}}} - 2 g^{\left(2\right)\,\mu\nu} \omega^{\left(1\right)\,m^{^{\prime}}n^{^{\prime}}}_\mu \partial_\nu\,. \end{align} \tag{ 3.39 }$$

Here $\omega^{(1)}$ is the subleading correction to the spin connection. $K^{(2)}$ is then given by (3.14), and we are interested in the trace of the heat kernel $\mathrm{tr}\,_V{K^{(2)}}$. Using that ${\mathcal{M}}^{(0)}$ is a diagonal matrix (and so is $K^{(0)}$), while $\omega^{(1)}$ is antisymmetric, we see that these additional contributions simply drop out of the trace

$$\begin{align} \mathrm{tr}\,_V{K^{\left(0\right)} \omega^{\left(1\right)} K^{\left(0\right)}} = 0\,. \end{align} \tag{ 3.40 }$$

We therefore do not even need to write down explicit expressions for the corrections $E^{(1)}$ and $\omega_\mu^{(1)\,m^{^{\prime}}n^{^{\prime}}}$. Instead, $K^{(2)}$ is a straightforward generalisation of the scalar case, where the only nontrivial matrix structure is in ${\mathcal{M}}^{(2)}$. The result is the same as (3.29)

$$\begin{align} \mathrm{tr}\,_V{K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right)} = \mathrm{tr}\,_V\left[ \left( \frac{z_0^2}{6}\left(\bar{\mathcal{R}} - 2 \mathrm{tr}\,\bar\Pi\right) - \left\langle {\mathcal{M}}^{\left(2\right)} \right\rangle \right) \frac{\exp\left( -\left(1+{\mathcal{M}}^{\left(0\right)}\right)t \right)} {\left(4 \pi\right)^{3/2} t^{1/2}} \right]\,. \end{align} \tag{ 3.41 }$$

For the M2-brane, the mass matrix M² is a function over the world-volume Σ derived in (A.6). Its trace over the four transverse AdS₇ modes is

$$\begin{align} M^2_{m^{^{\prime}}m^{^{\prime}}} = - g^{\mu\nu} G^{\mu^{^{\prime}}\nu^{^{\prime}}} R_{\mu^{^{\prime}} \mu \nu^{^{\prime}} \nu} - g^{\mu \nu} g^{\rho \sigma} G_{\mu^{^{\prime}}\nu^{^{\prime}}} \tilde{\mathrm{I\kern-1ptI}}_{\mu\rho}^{\mu^{^{\prime}}} \tilde{\mathrm{I\kern-1ptI}}_{\sigma \nu}^{\nu^{^{\prime}}}\,, \end{align} \tag{ 3.42 }$$

where here $R_{\mu^{^{\prime}}\mu\nu^{^{\prime}}\nu}$ is the bulk Riemann tensor (as opposed to the world-volume Riemann tensor ${\mathcal{R}}$). In order to read ${\mathcal{M}}^{(0)}$ and ${\mathcal{M}}^{(2)}$ we rewrite M² as follows. Recall that µ, ν are the M2-brane directions and $\mu^{^{\prime}}$, $\nu^{^{\prime}}$ the transverse ones, so we can write

$$\begin{align} G^{\mu^{^{\prime}}\nu^{^{\prime}}} R_{\mu^{^{\prime}} \mu \nu^{^{\prime}} \nu} = R_{\mu\nu} - g^{\rho \sigma} R_{\mu \rho \nu \sigma}\,. \end{align} \tag{ 3.43 }$$

The Ricci tensor is fixed by the bulk equations of motion

$$\begin{align} R_{\mu\nu} = -6 G_{\mu\nu} = - 6 g_{\mu\nu}\,, \end{align} \tag{ 3.44 }$$

and for the components of the Riemann tensor, we relate them to the world-volume curvature ${\mathcal{R}}$ via the Gauss-Codazzi equation

$$\begin{align} R_{\mu \rho \nu \sigma} = {\mathcal{R}}_{\mu \rho \nu \sigma} + 2 \mathrm{I\kern-1ptI}^{\mu^{^{\prime}}}_{\mu [\sigma} \mathrm{I\kern-1ptI}^{\nu^{^{\prime}}}_{\nu] \rho} G_{\mu^{^{\prime}}\nu^{^{\prime}}}\,. \end{align} \tag{ 3.45 }$$

Together we then find

$$\begin{align} M_{m^{^{\prime}}m^{^{\prime}}}^2 = 18 + \mathcal{R}\,, \end{align} \tag{ 3.46 }$$

which is now expressed solely in terms of quantities depending on the induced metric g (3.3). It is easy to show that the world-volume Ricci scalar has expansion $\mathcal{R} = -6 + \mathcal{R}^{(2)} + \cdots$ with

$$\begin{align} \mathcal{R}^{\left(2\right)} = \frac{z_0^2}{\left( \cosh \rho - \cos \theta \sinh \rho \right)^2} \left(\bar{\mathcal{R}} - 2 \mathrm{tr}\, \bar{\Pi}\right)\,. \end{align} \tag{ 3.47 }$$

The leading trace of the mass matrix is $\mathrm{tr}\,{{\mathcal{M}}^{(0)}} = 12$, as expected for four modes of mass-squared 3. The subleading correction is

$$\begin{align} \mathrm{tr}\,{{\mathcal{M}}^{\left(2\right)}} = \mathcal{R}^{\left(2\right)} = \frac{z_0^2}{\left(\cosh{\rho} - \cos{\theta} \sinh{\rho}\right)^2} \left(\bar{\mathcal{R}} - 2 \mathrm{tr}\,{\bar{\Pi}}\right)\,. \end{align} \tag{ 3.48 }$$

Integrating over S²,

$$\begin{align} \left\langle \mathrm{tr}\,{{\mathcal{M}}^{\left(2\right)}} \right\rangle & = \frac{1}{4\pi} \int \mathrm{d} \theta\, \mathrm{d} \phi \sin{\theta} \mathrm{tr}\,{{\mathcal{M}}^{\left(2\right)}} = z_0^2 \left( \bar{\mathcal{R}} - 2 \mathrm{tr}\,{\bar{\Pi}} \right)\,. \end{align} \tag{ 3.49 }$$

As we anticipated in (3.27), this does not depend on ρ.

To get the contribution to the anomaly, we again take the t integral of the trace of the heat kernel (3.30) and (3.31), and look for the residue at $z_0\to0$ as in (3.32). We get

$$\begin{align} \begin{aligned} {\mathcal{A}}^{N\Sigma} & = -\frac{1}{2} \substack{\mathrm{res}\\{z_0 \to 0}} \left[ \left( -\frac{1}{2z_0} \mathrm{tr}\,{\bar{\Pi}} \right) \int_0^\infty \frac{\mathrm{d} t}{t} \mathrm{tr}\,_V K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right)\right.\\ & \quad + \left. \frac{1}{z_0^3} \int_0^\infty \frac{\mathrm{d} t}{t} \mathrm{tr}\,_V K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) + \cdots \right]\\ & = \frac{1}{4\pi}\left( -\frac{1}{3} \bar{\mathcal{R}} + 6 \mathrm{tr}\,{\bar{\Pi}} \right). \end{aligned} \end{align} \tag{ 3.50 }$$

A similar analysis can be performed for spinors. The Dirac operator in euclidean signature is $\cancel{D} - M$ and it acts on the spinor bundle S with fiber $\mathbb{C}^8$ corresponding to the (chiral) spinor representations of SO(3), $SO(4)_N$ for normal directions and $SO(4)_S$ for S⁴ directions. We are interested in the determinant of its square

$$\begin{align} \det\left( -\cancel{D}^{{\,\,}2} + M^2 \right) \equiv \det\left( L_F \right)\,. \end{align} \tag{ 3.51 }$$

The square of the Dirac operator is related to the spinor laplacian via a generalisation of the Lichnerowicz formula (see e.g. [52]). Using γ_m and $\rho_{m^{^{\prime}}}$ for the gamma matrices of SO(3) and $SO(4)_N$ respectively (so $\left\{\gamma_m, \gamma_n \right\} = 2\delta_{mn}$ and similarly for ρ) and $\gamma_{mn} = \frac{1}{2} \left[ \gamma_m, \gamma_n \right]$ for the antisymmetrised product, it reads

$$\begin{align} \cancel{D}^{{\,\,}2} = \Delta_{1/2} - \frac{\mathcal{R}}{4} + \frac{1}{8} \gamma_{m n} \rho_{m^{^{\prime}}n^{^{\prime}}} R_{m n m^{^{\prime}}n^{^{\prime}}}\,. \end{align} \tag{ 3.52 }$$

$\Delta_{1/2}$ is spinor laplacian given in terms of the spinor covariant derivative as

$$\begin{align} \Delta_{1/2} = \frac{1}{\sqrt{g}} D_\mu \left( \sqrt{g}g^{\mu\nu} D_\nu \right)\,, \qquad D_\mu \psi = \left( \partial_\mu - \frac{1}{4} \gamma_{mn} \omega_{\mu}^{mn} - \frac{1}{4} \rho_{m^{^{\prime}}n^{^{\prime}}} \omega_{\mu}^{m^{^{\prime}}n^{^{\prime}}} \right) \psi\,, \end{align} \tag{ 3.53 }$$

and R is the curvature 2-form of the normal bundle

$$\begin{align} R_{\mu \nu}^{m^{^{\prime}}n^{^{\prime}}} = 2 \partial_{[\mu} \omega_{\nu]}^{m^{^{\prime}}n^{^{\prime}}} + 2 \omega_{[\mu}^{m^{^{\prime}}p^{^{\prime}}} \omega_{\nu]}^{p^{^{\prime}}n^{^{\prime}}}\,. \end{align} \tag{ 3.54 }$$

Note that we assume here again that the embedding in S⁴ is trivial, so we do not see its spin connection.

To proceed, we again look for a near-AdS₃ expansion of the operator $L_F = -\cancel{D}^{{\,\,}2} + M^2$ and the corresponding heat kernel. We also restrict to constant $M^2 = {\mathcal{M}}^{(0)}$. We then require the generalisation of section 3.1 to the case of spinor bundles on AdS₃.

To leading order, the metric (3.4) is AdS₃ and the differential operator $L^{(0)}$ is the AdS₃ spinor laplacian. The spinor heat kernel derived in [53] (see also [54, 55]) transforms nontrivially as a bispinor at points σ and σ₀. It takes the factorised form

$$\begin{align} K^{\left(0\right)}\left(t;\sigma,\sigma_0\right) & = U\left(\sigma, \sigma_0\right) k_F\left(t;\rho\right)\,, \end{align} \tag{ 3.55 }$$

where U encodes the bispinor transformations and k_F is a function of t and the geodesic distance ρ between σ and σ₀ only. On a curved background, U can be obtained by parallel transporting a spinor along the geodesic connecting σ and σ₀

$$\begin{align} U\left(\sigma, \sigma_0\right) = \mathcal{P} \exp{\int_{\sigma_0}^{\sigma} \mathrm{d} x^\mu\, \frac{1}{2} \omega_\mu^{mn} \gamma_{mn}}\,. \end{align} \tag{ 3.56 }$$

To write explicit expressions, we use the coordinates in (3.17) with σ₀ the origin of AdS₃ and the vielbein

$$\begin{align} \begin{aligned} \textrm{e}^1 & = \sin\theta \cos \phi\,\mathrm{d} \rho + \sinh{\rho}\left( \cos{\theta} \cos{\phi}\,\mathrm{d} \theta - \sin{\theta} \sin{\phi}\,\mathrm{d} \phi \right)\,,\\ \textrm{e}^2 & = \sin{\theta} \sin{\phi}\,\mathrm{d} \rho + \sinh{\rho} \left( \cos{\theta} \sin{\phi}\,\mathrm{d} \theta + \sin{\theta} \cos{\phi}\,\mathrm{d} \phi \right)\,,\\ \textrm{e}^3 & = \cos{\theta}\,\mathrm{d} \rho - \sinh{\rho} \sin{\theta}\,\mathrm{d} \theta\,. \end{aligned} \end{align} \tag{ 3.57 }$$

This frame is diagonal in projective coordinates and the corresponding spin connection is

$$\begin{align} \begin{aligned} \omega^{12} & = -2 \sinh^2{\frac{\rho}{2}} \sin^2{\theta}\,\mathrm{d} \phi\,,\\ \omega^{13} & = 2 \sinh^2{\frac{\rho}{2}} \left( \cos{\phi}\,\mathrm{d}\theta - \cos{\theta} \sin{\theta} \sin{\phi}\,\mathrm{d} \phi \right),\\ \omega^{23} & = 2 \sinh^2{\frac{\rho}{2}} \left( \sin{\phi}\,\mathrm{d}\theta + \cos{\theta} \sin{\theta} \cos{\phi}\,\mathrm{d}\phi \right). \end{aligned} \end{align} \tag{ 3.58 }$$

Now the geodesic connecting the origin to a point $\sigma = (\rho, \theta, \phi)$ is along the vector $\partial_\rho$. Since the spin connection along that direction is trivial $\omega_\rho^{mn} = 0$, this frame is known as a parallel frame and the matrix U (3.56) reduces to the identity matrix ⁵

$$\begin{align} U\left(\sigma, \sigma_0\right) = \unicode{x1D7D9}_8\,. \end{align} \tag{ 3.59 }$$

Using this frame, the heat kernel equation reduces to

$$\begin{align} \begin{aligned} &\left(L_F^{\left(0\right)}+\partial_t\right) K^{\left(0\right)}\left(t;\sigma,\sigma_0\right) = \left[-\partial_\rho^2 - 2 \coth{\rho} \,\partial_\rho + \frac{1}{2} \tanh^2\frac{\rho}{2} - \frac{3}{2} + {\mathcal{M}}^{\left(0\right)} + \partial_t \right] k_F\left(t;\rho\right) = 0\,, \end{aligned} \end{align} \tag{ 3.60 }$$

where $\frac{1}{2} \tanh^2({\rho}/{2})$ is the leading order contribution of $g^{\mu\nu} \omega_\mu^{mn} \omega_{\nu}^{mn}/8$ and $-{3}/{2} = {\mathcal{R}^{(0)}}/{4}$ is the scalar curvature of AdS₃. If we define

$$\begin{align} k_F\left(t;\rho\right) = -\frac{1}{4\pi\sinh\left(\rho/2\right)}\partial_\rho\left(\frac{h\left(t;\rho\right)}{\cosh\left(\rho/2\right)}\right), \end{align} \tag{ 3.61 }$$

then h satisfies the one-dimensional heat kernel equation

$$\begin{align} \left[-\partial_\rho^2 + {\mathcal{M}}^{\left(0\right)} + \partial_t \right] h\left(t;\rho\right) = 0\,. \end{align} \tag{ 3.62 }$$

A solution to this equation is

$$\begin{align} h\left(t;\rho\right) = \frac{\exp{\left( -{\mathcal{M}}^{\left(0\right)} t - {\rho^2}/{4t} \right)}} {\sqrt{4\pi t}}\,, \end{align} \tag{ 3.63 }$$

and its derivative

$$\begin{align} k_F\left(t;\rho\right) & = \frac{\rho + t \tanh\left({\rho/2}\right)}{\sinh{\rho}} \frac{\exp{\left( -{\mathcal{M}}^{\left(0\right)} t - {\rho^2}/{4t} \right)}}{\left(4\pi t\right)^{3/2}}\,, \end{align} \tag{ 3.64 }$$

satisfies the initial condition (3.1) [53].

To find the subleading correction to the heat kernel $K^{(2)}$, we can again use (3.14), which extends to the spinor case. One obvious difference is that the differential operator (3.52) contains extra terms compared to the scalar laplacian

$$\begin{align} \begin{aligned} L_F^{\left(2\right)} = \left[ L^{\left(2\right)} + \frac{1}{8} g^{\left(4\right)\,\mu\nu} \omega_\mu^{\left(-1\right)\,mn} \omega_{\nu}^{\left(-1\right)\,mn} + \frac{1}{4} g^{\left(2\right)\,\mu\nu}\omega_\mu^{\left(1\right)\,mn} \omega_{\nu}^{\left(-1\right)\,mn} + \frac{\mathcal{R}^{\left(2\right)}}{4} \right] \unicode{x1D7D9}_8 + \cdots \end{aligned} \end{align} \tag{ 3.65 }$$

where the ellipses involve terms with single gamma matrices which vanish upon taking the trace over the spinor bundle. One should also worry about the matrix U. As stated, in our frame $U(\sigma^{^{\prime}},\sigma_0) = \unicode{x1D7D9}$, but the same is not generally true for $U(\sigma,\sigma^{^{\prime}})$. As we only require $K^{(2)}(\sigma_0,\sigma_0)$, the second heat kernel comes with $U(\sigma_0,\sigma^{^{\prime}})$, which is also the identity.

The correction to the metric in (3.65) is given in (3.4) and that of the scalar curvature is (3.47). To obtain that of the spin connection, use that the vielbein for the asymptotically AdS₃ metric (3.4) is

$$\begin{align} \textrm{e}^{m}_\mu = \textrm{e}^{\left(-1\right)\,m}_{\mu} + \frac{1}{2} g_{\mu\nu}^{\left(0\right)} \textrm{e}^{\left(1\right)\,\nu}_m + \cdots \end{align} \tag{ 3.66 }$$

with $\textrm{e}^{(-1)\,m}_\mu$ the leading vielbein defined in (3.57) and $\textrm{e}^{(1)\,\mu}_m$ its inverse. Note that $g^{(0)}$ can be written in the $\rho, \theta, \phi$ coordinates using the change of variables (3.18).

The expression for $K^{(2)}$ is then as in (3.14). Explicitly,

$$\begin{align} -\int_0^t \mathrm{d} t^{^{\prime}} \int_\Sigma \mathrm{d} \rho \,\mathrm{d} \theta \,\mathrm{d} \phi \sinh^2{\rho} \sin{\theta}\, k_F\left(t - t^{^{\prime}};\sigma_0,\sigma^{^{\prime}}\right) \mathrm{tr}_S L_F^{\left(2\right)} k_F\left(t^{^{\prime}};\sigma^{^{\prime}},\sigma_0\right)\,. \end{align} \tag{ 3.67 }$$

We then need to do the spherical integral, as in (3.22). For the scalar laplacian, we get the same as above and for the other terms in (3.65) we find

$$\begin{align} \frac{1}{8} \int \!\!\sin{\theta} \,\mathrm{d} \theta\, \mathrm{d} \phi\, g^{\left(4\right)\,\mu\nu} \omega_\mu^{\left(-1\right)\,mn} \omega_{\nu}^{\left(-1\right)\,mn} & = \frac{2 \pi}{3} \bar{\Pi}\tanh^2\frac{\rho}{2} \!+ \!\frac{\pi}{6} \bar{\mathcal{R}}\left(\rho \cosh{\rho}\! -\! \sinh{\rho}\right)\frac{\sinh\left(\rho/2\right)}{\cosh^3\left(\rho/2\right)}\,, \end{align} \tag{ 3.68 }$$

$$\begin{align} \frac{1}{4} \int \sin{\theta} \,\mathrm{d} \theta \,\mathrm{d} \phi\, g^{\left(2\right)\,\mu\nu} \omega_\mu^{\left(1\right)\,mn} \omega_{\nu}^{\left(1\right)\,mn} & = -\frac{\pi}{3}\bar{\mathcal{R}} \rho \tanh{\frac{\rho}{2}} \,, \end{align} \tag{ 3.69 }$$

$$\begin{align} \frac{1}{4} \int \sin{\theta} \,\mathrm{d} \theta\, \mathrm{d} \phi\, \mathcal{R}^{\left(2\right)} & = \pi \left(\bar{\mathcal{R}} - 2 \bar{\Pi} \right)\,. \end{align} \tag{ 3.70 }$$

Assembling the results, we obtain

$$\begin{align} \begin{aligned} -4\pi & \int_0^t \mathrm{d} t^{^{\prime}} \int_0^\infty \mathrm{d} \rho \sinh^2{\rho}\, k_F\left(t-t^{^{\prime}};\rho\right) \left[ \frac{z_0^2}{3}\mathrm{tr}\,\bar\Pi \left( -\partial_\rho^2 - 2 \coth{\rho}\,\partial_\rho + \frac{1}{2} \tanh^2\frac{\rho}{2} - \frac{3}{2} \right) \right.\\ & \quad \left. + z_0^2 \bar{\mathcal{R}} \left( \frac{1}{6} \frac{\cosh{\rho} \sinh{\rho} - \rho}{\sinh^2\rho} \partial_\rho + \frac{1}{4} - \frac{\rho + \sinh{\rho}}{24} \frac{\sinh\left(\rho/2\right)}{\cosh^3\left(\rho/2\right)} \right) \right] k_F\left(t^{^{\prime}};\rho\right) \mathrm{tr}_S\left(\unicode{x1D7D9}\right) \,. \end{aligned} \end{align} \tag{ 3.71 }$$

This integral is evaluated explicitly in appendix B. Adding (B.3) and (B.7) together with the boundary term appearing in the second line of (3.14), we finally get the trace of the heat kernel

$$\begin{align} \mathrm{tr}_S K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) & = - \frac{z_0^2}{12} \left(\bar{\mathcal{R}} - 2 \mathrm{tr}\,{\bar{\Pi}}\right) \frac{\textrm{e}^{-{\mathcal{M}}^{\left(0\right)} t}}{\left(4\pi\right)^{3/2} t^{1/2}} \mathrm{tr}_S{(\unicode{x1D7D9})}\,. \end{align} \tag{ 3.72 }$$

Its t integral is easy to evaluate and gives (reinstating powers of R)

$$\begin{align} \int_0^\infty \frac{\mathrm{d} t}{t} \mathrm{tr}_S K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right) & = \frac{\sqrt{{\mathcal{M}}^{\left(0\right)}}}{24\pi R^2} \left(4 {\mathcal{M}}^{\left(0\right)} R^2 - 3\right) \mathrm{tr}_S\left(\unicode{x1D7D9}\right) \,, \end{align} \tag{ 3.73 }$$

$$\begin{align} \int_0^\infty \frac{\mathrm{d} t}{t} \mathrm{tr}_S K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) & = \frac{z_0^2}{48\pi R^2} \left(\bar{\mathcal{R}} - 2 \mathrm{tr}\,{\bar{\Pi}}\right) \sqrt{{\mathcal{M}}^{\left(0\right)}} \mathrm{tr}_S{\left(\unicode{x1D7D9}\right)} \,. \end{align} \tag{ 3.74 }$$

Getting from this to the contribution of the fermions to the anomaly of the surface operator is immediate. The Dirac operator governing the quadratic fermionic fluctuations for an M2-brane is derived in appendix A, and from equation (A.17) we can read ${\mathcal{M}}^{(0)} = 9/4$.

Evaluating the path integral yields the pfaffian of the Dirac operator, or $\frac{1}{2} \log \det_S(- \cancel{D}^{{\,\,}2} + 9/4)$. The contribution of the determinant to the anomaly is obtained from the pole of the z integral as in the bosonic case (3.32). Using (3.73) and (3.74), and $\mathrm{tr}_S(\unicode{x1D7D9}) = 8$ we get

$$\begin{align} {\mathcal{A}}^{F} & = \frac{1}{2} \substack{\mathrm{res}\\{z_0 \to 0}} \left[ \left( -\frac{1}{2z_0} \mathrm{tr}\,{\bar{\Pi}} \right) \int_0^\infty \frac{\mathrm{d} t}{t} \mathrm{tr}_S K^{\left(0\right)}\left(t;\sigma_0,\sigma_0\right) + \frac{1}{z_0^3} \int_0^\infty \frac{\mathrm{d} t}{t} \mathrm{tr}_S K^{\left(2\right)}\left(t;\sigma_0,\sigma_0\right) + \cdots \right] \nonumber\\ & = \frac{1}{4\pi} \left( \frac{1}{2} \bar{\mathcal{R}} - 4 \mathrm{tr}\,{\bar{\Pi}} \right). \end{align} \tag{ 3.75 }$$

We consider fluctuations $u^{\mu^{^{\prime}}}, z^i$ around a classical solution described by world-volume coordinates x^µ as in the metric (A.1). The M2-brane action (2.3) contains two terms, the volume form and a term coupling to the pullback of the three-form A₃. Because we restrict to classical solutions localised at a point on S⁴, there are no tangent vectors along S⁴ and the pullback of A₃ is at least cubic in the fluctuations, and we can neglect it. The quadratic terms arising from the volume form is obtained in [73] ⁶ (see also [74])

$$\begin{align} & \frac{T_\text{M2}}{2} \int_\Sigma \mathrm{vol}_\Sigma \left[ \partial_\mu u^{\mu^{^{\prime}}} \partial_\nu u^{\nu^{^{\prime}}} g^{\mu\nu} g_{\mu^{^{\prime}} \nu^{^{\prime}}} + \partial_\mu z^i \partial_\nu z^i g^{\mu\nu}\right.\nonumber\\ & \quad - \left.\left( R_{\mu^{^{\prime}} \mu \nu^{^{\prime}} \nu} + g_{\mu^{^{\prime}} \rho^{^{\prime}}} g_{\nu^{^{\prime}} \sigma^{^{\prime}}} \tilde{\mathrm{I\kern-1ptI}}^{\rho^{^{\prime}}}_{\mu\rho} \tilde{\mathrm{I\kern-1ptI}}^{\sigma^{^{\prime}}}_{\sigma \nu} g^{\rho \sigma} \right) g^{\mu\nu} u^{\mu^{^{\prime}}} u^{\nu^{^{\prime}}} \right], \end{align} \tag{ A.2 }$$

with $\tilde{\mathrm{I\kern-1ptI}}$ the traceless part of the second fundamental form and R the Riemann tensor.

To put the kinetic term in canonical form, we introduce the vielbeine $E^{\mu^{^{\prime}}}_{m^{^{\prime}}}(\sigma)$ and parametrise u as

$$\begin{align} u^{\mu^{^{\prime}}} = E^{\mu^{^{\prime}}}_{m^{^{\prime}}}\zeta^{m^{^{\prime}}}\,. \end{align} \tag{ A.3 }$$

The derivatives are then expressed as

$$\begin{align} \partial_\mu u^{\mu^{^{\prime}}} = E^{\mu^{^{\prime}}}_{m^{^{\prime}}} D_\mu \zeta^{m^{^{\prime}}}\,, \qquad D_\mu \zeta^{m^{^{\prime}}} \equiv \partial_\mu \zeta^{m^{^{\prime}}} + \omega^{m^{^{\prime}}n^{^{\prime}}}_{\mu} \zeta^{n^{^{\prime}}}\,, \end{align} \tag{ A.4 }$$

with ω the spin connection on the normal bundle. With these substitutions the quadratic action becomes

$$\begin{align} \frac{T_\text{M2}}{2} \int_\Sigma \mathrm{vol}_\Sigma \left[ D_\mu \zeta^{m^{^{\prime}}} D_\nu \zeta^{m^{^{\prime}}} g^{\mu\nu} - \left( E^{\mu^{^{\prime}}}_{m^{^{\prime}}} E^{\nu^{^{\prime}}}_{n^{^{\prime}}} R_{\mu^{^{\prime}} \mu \nu^{^{\prime}} \nu} + E_{\mu^{^{\prime}}m^{^{\prime}}} E_{\nu^{^{\prime}}n^{^{\prime}}} \tilde{\mathrm{I\kern-1ptI}}^{\mu^{^{\prime}}}_{\mu\rho} \tilde{\mathrm{I\kern-1ptI}}^{\nu^{^{\prime}}}_{\sigma \nu} g^{\rho \sigma} \right) g^{\mu\nu} \zeta^{m^{^{\prime}}} \zeta^{n^{^{\prime}}} \right]. \end{align} \tag{ A.5 }$$

Finally, to bring this to the form of a differential operator, we integrate the kinetic term by parts. This yields a boundary term, and assuming it vanishes, we get [73]

$$\begin{align} \begin{aligned} S_\text{fluc}^\text{(bos)} & = \frac{T_\text{M2}}{2} \int_\Sigma \mathrm{vol}_\Sigma \zeta^{m^{^{\prime}}} \left[ - \Delta_{m^{^{\prime}}n^{^{\prime}}} + M^2_{m^{^{\prime}}n^{^{\prime}}} \right] \zeta^{n^{^{\prime}}} - z^i \Delta z^i \,,\\ M^2_{m^{^{\prime}}n^{^{\prime}}} & = -g^{\mu\nu} \left( E^{\mu^{^{\prime}}}_{m^{^{\prime}}} E^{\nu^{^{\prime}}}_{n^{^{\prime}}} R_{\mu^{^{\prime}} \mu \nu^{^{\prime}} \nu} + E_{\mu^{^{\prime}}m^{^{\prime}}} E_{\nu^{^{\prime}}n^{^{\prime}}} \tilde{\mathrm{I\kern-1ptI}}^{\mu^{^{\prime}}}_{\mu\rho} \tilde{\mathrm{I\kern-1ptI}}^{\nu^{^{\prime}}}_{\sigma \nu} g^{\rho \sigma}\right), \end{aligned} \end{align} \tag{ A.6 }$$

with Δ the usual laplacian and $\Delta_{m^{^{\prime}}n^{^{\prime}}}$ the vector laplacian acting on $\zeta^{m^{^{\prime}}}$ (the covariant derivative is defined in (A.4))

$$\begin{align} \Delta_{m^{^{\prime}}n^{^{\prime}}}\zeta^{n^{^{\prime}}} = \frac{1}{\sqrt{g}} \left[D_{\mu} \left( \sqrt{g} g^{\mu\nu} D_\nu \right)\zeta\right]_{m^{^{\prime}}}\,. \end{align} \tag{ A.7 }$$

The fermionic part of the M2-brane action is obtained in [49] and is expressed in terms of an 11d spinor $\Psi(\sigma)$ (whose indices we suppress). On $AdS_7 \times S^4~\Psi$ is in the spinor representation of $\mathfrak{so}(1,6)$ and $\mathfrak{so}(4)$. To quadratic order and working in lorentzian signature, it reads [75, 76] (see also [77])

$$\begin{align} T_\text{M2} \int_\Sigma \mathrm{d}^3 \sigma\sqrt{-g} \left[ \bar{\Psi} \left( g^{\rho \mu} \Gamma_\mu - \frac{1}{2} \varepsilon^{\mu\nu\rho} \Gamma_{\mu\nu} \right) \left(D_\rho + \frac{1}{288} \left( {\Gamma_\rho}^{NPQR} - 8 \delta_\rho^{[N} \Gamma^{PQR]} \right) F_{NPQR} \right) \Psi \right] \,. \end{align} \tag{ A.8 }$$

Here we use $M,N,\ldots$ for coordinates on the full 11d space. $\Gamma^M$ are the 11d gamma matrices and we use the shorthand for antisymmetrised indices $\Gamma_{\mu \nu} \equiv \Gamma_{[\mu} \Gamma_{\nu]}$ (with the appropriate 1/2). The spinor covariant derivative is

$$\begin{align} D_\rho \Psi = \partial_\rho \Psi - \frac{1}{4} \omega_\rho^{MN} \Gamma_{MN} \Psi \,, \end{align} \tag{ A.9 }$$

where to avoid introducing extra notations, we use curved space indices on the spin connection and gamma matrices.

As mentioned previously, we restrict to M2-branes located at a point on S⁴, so the term proportional to $\delta_\rho^{[N} \Gamma^{PQR]}$ vanishes and the covariant derivative (A.9) includes only the AdS₇ spin connection. We can also simplify the remaining flux term by noting that $\Gamma_\rho$ commutes with $\Gamma^{NPQR}$ and using the explicit expression for the flux (2.4) we get

$$\begin{align} \frac{1}{288} \Gamma_{\rho} \Gamma^{NPQR} F_{NPQR} = \frac{1}{2R} \Gamma_{\rho} \Gamma_{789 \natural}\,. \end{align} \tag{ A.10 }$$

To proceed, we can pick a frame where $E_\mu^{m^{^{\prime}}} = E_{\mu^{^{\prime}}}^m = 0$. So the gamma matrices used above can now be expressed in terms of those with flat space indices as $\Gamma_\mu = E_\mu^m \Gamma_m$ and $\Gamma_{\mu^{^{\prime}}} = E_{\mu^{^{\prime}}}^{m^{^{\prime}}}\Gamma_{m^{^{\prime}}}$. We can further specify the frame by requiring that the flat space gamma matrices $\Gamma_m$, $\Gamma_{m^{^{\prime}}}$ and $\Gamma_i$ are given by ⁷

$$\begin{align} \Gamma_m = \gamma_m \otimes \rho_* \otimes \tau_*\,, \qquad \Gamma_{m^{^{\prime}}} = \unicode{x1D7D9}_2 \otimes \rho_{m^{^{\prime}}} \otimes \tau_*\,, \qquad \Gamma_{i} = \unicode{x1D7D9}_2 \otimes \unicode{x1D7D9}_4 \otimes \tau_i\,, \end{align} \tag{ A.11 }$$

with γ_m , $\rho_{m^{^{\prime}}}$ and τ_i respectively the gamma matrices for $SO(1,2)$, $SO(4)_N$ and $SO(4)_S$, and $\rho_* \equiv \rho_1 \rho_2 \rho_3 \rho_4$, $\tau_* \equiv \tau_1 \tau_2 \tau_3 \tau_4$. Note that $\rho_*^2 = \tau_*^2 = 1$. With this basis the lagrangian becomes

$$\begin{align} \begin{aligned} & \bar{\Psi} \left( \gamma^p \rho_* \tau_* - \frac{1}{2} \gamma_{m n} \varepsilon^{mnp} \right) E_p^\mu \left(\partial_\mu \psi - \frac{1}{4} \gamma_{qr} \omega_\mu^{qr} - \frac{1}{2} \gamma_q \rho_* \rho_{r^{^{\prime}}} \omega_\mu^{qr^{^{\prime}}} - \frac{1}{4} \rho_{q^{^{\prime}}r^{^{\prime}}} \omega_\mu^{q^{^{\prime}}r^{^{\prime}}} \right) \Psi\\ & \quad + \frac{3}{R} \bar{\Psi} \left( 1 - \gamma_{012} \rho_* \tau_* \right) \tau_* \Psi\,. \end{aligned} \end{align} \tag{ A.12 }$$

$\omega_\mu^{qr}$ and $\omega_\mu^{q^{^{\prime}}r^{^{\prime}}}$ are respectively the spin connections on the world-volume and on the normal bundle. $\omega_{\mu}^{qr^{^{\prime}}}$ can be expressed as

$$\begin{align} \omega_\mu^{q r^{^{\prime}}} = E^{\sigma q} E_{\rho^{^{\prime}}}^{r^{^{\prime}}} \Gamma_{\mu\sigma}^{\rho^{^{\prime}}} = - E^{\sigma q} E_{\rho^{^{\prime}}}^{r^{^{\prime}}} \mathrm{I\kern-1ptI}_{\mu\sigma}^{\rho^{^{\prime}}}\,, \end{align} \tag{ A.13 }$$

where in the first equality we used the definition of the spin connection in terms of a frame (analogous to (3.38)), and in the second we used that the second fundamental form (2.7) is also the Christoffel symbol for the metric G.

With some gamma matrix algebra and using that $\gamma_{012} = 1$, this can be rewritten as

$$\begin{align} \begin{aligned} &-\bar{\Psi} \left(1 - \rho_* \tau_*\right) \gamma^\mu \left(\partial_\mu - \frac{1}{4} \gamma_{mn} \omega_\mu^{mn} - \frac{1}{4} \rho_{m^{^{\prime}}n^{^{\prime}}} \omega_\mu^{m^{^{\prime}}n^{^{\prime}}} \right) \Psi +\frac{3}{2R} \bar{\Psi} \left(1 - \rho_* \tau_*\right) \tau_* \Psi\\ & \quad - \frac{1}{2} \bar{\Psi} \left(1 - \rho_* \tau_*\right) \rho_* \rho_{m^{^{\prime}}} \Psi H^{m^{^{\prime}}}\,, \end{aligned} \end{align} \tag{ A.14 }$$

and recall that the mean curvature vector H vanishes because of the equations of motion. Finally, the M2-brane action has κ-symmetry, and we can fix a gauge by imposing the condition

$$\begin{align} \left(1 + \frac{1}{6} E_{\mu}^m E_{\nu}^n E_\rho^p \Gamma_{mnp} \varepsilon^{\mu\nu\rho} \right) \Psi = \left( 1 + \rho_* \tau_* \right) \Psi = 0\,. \end{align} \tag{ A.15 }$$

This condition is solved by decomposing $\Psi = \frac{1}{\sqrt{2}}(\psi_+ + \psi_-)$ in terms of spinors $\psi_\pm$ with chirality

$$\begin{align} \tau_* \psi_\pm = -\rho _* \psi_{\pm} = \pm \psi_\pm. \end{align} \tag{ A.16 }$$

We then obtain the gauge-fixed quadratic action

$$\begin{align} \begin{aligned} S_\text{fluc}^\text{(fer)} & = -T_\text{M2} \int_\Sigma \mathrm{d}^3 \sigma\sqrt{-g} \left[ \bar{\psi}_+ \left( \cancel{D} + \frac{3}{2R} \right) \psi_+ + \bar{\psi}_- \left( \cancel{D} - \frac{3}{2R} \right) \psi_- \right]\,,\\ \cancel{D} &\equiv \gamma^\mu D_\mu = \gamma^\mu \left(\partial_\mu - \frac{1}{4} \gamma_{mn} \omega_\mu^{mn} - \frac{1}{4} \rho_{m^{^{\prime}}n^{^{\prime}}} \omega_\mu^{m^{^{\prime}}n^{^{\prime}}} \right)\,. \end{aligned} \end{align} \tag{ A.17 }$$

We can read the relevant determinant for the analysis of section 3 from this action. After Wick rotation to euclidean signature, the path integral over quadratic fluctuations gives the pfaffian

$$\begin{align} \mathrm{Pf}\, \left(\cancel{D} + \frac{3}{2R} \right) \mathrm{Pf}\, \left(\cancel{D} - \frac{3}{2R} \right)\,. \end{align} \tag{ A.18 }$$

Using the identity

$$\begin{align} \mathrm{Pf}\, \left(\cancel{D} \pm \frac{3}{2R} \right)^2 = \det \left(\cancel{D} \pm \frac{3}{2R} \right) = \det{}^{1/2} \left( - \cancel{D}^{{\,\,}2} + \frac{9}{4R^2} \right)\,, \end{align} \tag{ A.19 }$$

we can write

$$\begin{align} \mathrm{Pf}\, \left(\cancel{D} + \frac{3}{2R} \right) \mathrm{Pf}\, \left(\cancel{D} - \frac{3}{2R} \right) = \det{}^{1/2}\left( - \cancel{D}^{{\,\,}2} + \frac{9}{4R^2} \right)\,, \end{align} \tag{ A.20 }$$

up to a possible overall sign ambiguity in the path integral.

We note that our gauge-fixed action (A.17) and the determinant (A.20) agree with existing results for fermionic fluctuations derived for specific geometries: they reproduces that of the parallel planes [57] and the Dirac operator for AdS₃ [42].