14.1 Differential geometry on Riemann surfaces

Riemann surfaces are real two-dimensional manifolds without boundary. In our study of topology and geometry, we referred to them in various places. Here we summarize the basic facts on Riemann surfaces, which will make this chapter self-contained. We also introduce several new aspects of Riemann surfaces, which provide enough background for the study of bosonic string amplitudes.

14.1.1 Metric and complex structure

Let ${\Sigma }_g$ be a Riemann surface of genus $g$ . It was shown in example 7.9 that we may introduce, in any chart $U$ , the isothermal coordinates $({\xi }^1,{\xi }^2)$ in which the metric is conformally flat:

$$g={\mathrm{e}}^{2\sigma (\xi )}\left(\mathrm{d}{\xi }^1\otimes \mathrm{d}{\xi }^1+\mathrm{d}{\xi }^2\otimes \mathrm{d}{\xi }^2\right).\text{\hspace{1em}(14.2)}$$

Introduce the complex coordinates

$$z={\xi }^1+\mathrm{i}{\xi }^2\bar{z}={\xi }^1-\mathrm{i}{\xi }^2.\text{\hspace{1em}(14.3)}$$

Forms and vectors are spanned by

$$\mathrm{d}z=\mathrm{d}{\xi }^1+\mathrm{id}{\xi }^2\mathrm{d}\bar{z}=\mathrm{d}{\xi }^1-\mathrm{id}{\xi }^2\text{\hspace{1em}(14.4a)}$$

$${\partial }_z=\frac{1}{2}\left(\frac{\partial }{\partial {\xi }^1}-\mathrm{i}\frac{\partial }{\partial {\xi }^2}\right){\partial }_{\bar{z}}=\frac{1}{2}\left(\frac{\partial }{\partial {\xi }^1}+\mathrm{i}\frac{\partial }{\partial {\xi }^2}\right).\text{\hspace{1em}(14.4b)}$$

In terms of the complex coordinates, the metric takes the form

$$g=\frac{1}{2}{\mathrm{e}}^{2\sigma (z,\bar{z})}[\mathrm{d}z\otimes \mathrm{d}\bar{z}+\mathrm{d}\bar{z}\otimes \mathrm{d}z].\text{\hspace{1em}(14.5)}$$

The components of $g$ are

$$g_{z\bar{z}}=g_{\bar{z}z}=\frac{1}{2}{\mathrm{e}}^{2\sigma }{g}_{zz}=g_{\bar{z}\bar{z}}=0\text{\hspace{1em}(14.6a)}$$

$$g^{z\bar{z}}=g^{\bar{z}z}=2{\mathrm{e}}^{-2\sigma }{g}^{zz}=g^{\bar{z}\bar{z}}=0.\text{\hspace{1em}(14.6b)}$$

Let $V$ be another chart of ${\Sigma }_g$ such that $U\cap V\ne \varnothing$ . Let $(w,\bar{w})$ be the complex coordinates in $V$ . The metric in $V$ is

$$g={\mathrm{e}}^{2{\sigma }^{\prime }(w,\bar{w})}\mathrm{d}w\otimes \mathrm{d}\bar{w}.\text{\hspace{1em}(14.7)}$$

The two expressions (14.5) and (14.7) should agree on $U\cap V$ ,

$${\mathrm{e}}^{2\sigma (z,\bar{z})}\mathrm{d}z\otimes \mathrm{d}\bar{z}={\mathrm{e}}^{2{\sigma }^{\prime }(w,\bar{w})}\mathrm{d}w\otimes \mathrm{d}\bar{w}.$$

Since

$$\begin{array}{rl}& \mathrm{d}w\otimes \mathrm{d}\bar{w}=[(\partial w/\partial z)\mathrm{d}z+(\partial w/\partial \bar{z})\mathrm{d}\bar{z}]\otimes [(\partial \bar{w}/\partial z)\mathrm{d}z+(\partial \bar{w}/\partial \bar{z})\mathrm{d}\bar{z}]\\ & \propto \mathrm{d}z\otimes \mathrm{d}\bar{z}\end{array}$$

we must have $\partial w/\partial \bar{z}=\partial \bar{w}/\partial z=0$ . [Another possibility, $\partial w/\partial z=\partial \bar{w}/\partial \bar{z}=0$ is ruled out if $(z,\bar{z})$ and $(w,\bar{w})$ define the same orientation.] Thus, it follows that

$$w=w(z)\bar{w}=\bar{w}(\bar{z})\text{\hspace{1em}(14.8)}$$

which verifies that ${\Sigma }_g$ is a complex manifold. We also have

$${\mathrm{e}}^{2\sigma (z,\bar{z})}={\mathrm{e}}^{2{\sigma }^{\prime }(w,\bar{w})}|\partial w/\partial z{|}^2.\text{\hspace{1em}(14.9)}$$

14.1.2 Vectors, forms and tensors

Let $M={\Sigma }_g$ . The components of vector fields $V^z\partial /\partial z\in T{M}^{+}$ and $V^{\bar{z}}\partial /\partial \bar{z}\in T{M}^{-}$ transform as

$$V^w=\left(\partial w/\partial z\right)V^z{V}^{\bar{w}}=\left(\partial \bar{w}/\partial \bar{z}\right)V^{\bar{z}}.\text{\hspace{1em}(14.10)}$$

The components of differential forms $w_z\mathrm{d}z\in {\Omega }^{1,0}(M)$ and $w_{\bar{z}}\mathrm{d}\bar{z}\in {\Omega }^{0,1}(M)$ transform as

$${\omega }_w={\left(\partial w/\partial z\right)}^{-1}{\omega }_z{\omega }_{\bar{w}}={\left(\partial \bar{w}/\partial \bar{z}\right)}^{-1}{\omega }_{\bar{z}}.\text{\hspace{1em}(14.11)}$$

These are identified with sections of the holomorphic (anti-holomorphic) line bundles over $M={\Sigma }_g$ , for which the transition functions are holomorphic (anti-holomorphic). The metric provides a natural isomorphism between $T{M}^{+}$ and ${\Omega }^{0,1}(M)$ through

$${\omega }_{\bar{z}}=g_{\bar{z}z}{V}^{\bar{z}},V^{\bar{z}}=g^{z\bar{z}}{\omega }_{\bar{z}}.\text{\hspace{1em}(14.12)}$$

Similarly, $T{M}^{-}$ is isomorphic to ${\omega }^{1,0}(M)$ :

$${\omega }_z=g_{z\bar{z}}{V}^{\bar{z}},V^{\bar{z}}=g^{\bar{z}z}{\omega }_z.\text{\hspace{1em}(14.13)}$$

In general, given an arbitrary tensor, the metric allows us to trade all the $\bar{z}$ -indices for $z$ -indices. It is easy to see that

$$T\underset{p_1}{\underbrace{\begin{array}{l}{q}_1\\ z\dots z\\ p_1\end{array}|}}\underset{p_2}{\underbrace{\begin{array}{l}{q}_2\\ \bar{z}\dots \bar{z}\\ \bar{z}\dots \bar{z}\end{array}|}}\to T\underset{p_3}{\underbrace{\begin{array}{l}{q}_1+p_2\\ z\dots z\\ p_1+q_2\end{array}|}}=(g_{z\bar{z}}{)}^{q_2}(g^{z\bar{z}}{)}^{p_2}T\underset{p_1}{\underbrace{\begin{array}{l}{q}_1\\ z\dots z\\ z\dots z\\ \bar{z}\dots \bar{z}\end{array}|}}\underbrace{\begin{array}{l}{q}_2\\ \bar{z}\dots \bar{z}\\ \bar{z}\dots \bar{z}\\ p_2\end{array}|}.\text{\hspace{1em}(14.14)}$$

This correspondence is an isomorphism. For example, observe that

$$T_{z\bar{z}}^{\bar{z}}\to g^{z\bar{z}}{g}_{z\bar{z}}{T}_{z\bar{z}}^{\bar{z}}=T_z{}^z{}_z.$$

Thus, it is only necessary to consider tensors with pure $z$ -indices. For these tensors, we assign the helicity. Since $T$ has $z$ -indices only, it transforms under $z\to w$ as

$$T\to {\left(\frac{\partial w}{\partial z}\right)}^{n}T\text{\hspace{1em}(14.15)}$$

where $n\in \mathbb{Z}$ is given by the number of upper $z$ -indices minus the number of lower $z$ -indices. For example,

$$T_z^{zz}\to T_w^{ww}=\left(\frac{\partial w}{\partial z}\right)T_z^{zz}.$$

All that matters is the difference between the number of upper indices and the number of lower indices. The tensor $T_z^z$ is left invariant under $z\to w$ and is regarded as a scalar. The number $n$ is called the helicity. The set of helicity- $n$ tensors is denoted by ${\mathfrak{T}}^n$ :

$${\mathfrak{T}}^n\equiv \left\{T^{\underset{p}{\underbrace{z\dots z}}}|q-p=n\right\}.\text{\hspace{1em}(14.16)}$$

The helicity characterizes the irreducible representation of $\mathrm{U}(1)=\mathrm{SO}(2)$ .

So far we have assumed $n$ is an integer. It can be shown that $n=\frac{1}{2}$ corresponds to the spinor field on ${\Sigma }_g$ . In fact, the existence of spinors on the Riemann surfaces is guaranteed by the triviality of the second Stiefel-Whitney class of ${\Sigma }_g$ . The set ${\mathfrak{T}}^1$ is identified with the holomorphic line bundle $K$ over ${\Sigma }_g$ . Then ${\mathfrak{T}}^{1/2}$ is the square root of $K$ : $S_{+}^2=K={\mathfrak{T}}^1$ where $S_{+}$ is the positive-chirality spin bundle. Similarly, we have ${\mathfrak{T}}^{-1}=\bar{K}=S_{-}^2$ where $S_{-}$ is the negative-chirality spin bundle.

Example 14.1

In real indices, the helicity $\pm 1$ vectors are given by $V^1\pm \mathrm{i}{V}^2$ . This follows since

$$V^1\frac{\partial }{\partial {\xi }^1}+V^2\frac{\partial }{\partial {\xi }^2}=(V^1+\mathrm{i}{V}^2){\partial }_z+(V^1-\mathrm{i}{V}^2){\partial }_{\bar{z}}.$$

We put $V^z=V^1+\mathrm{i}{V}^2$ and $V^{\bar{z}}=V^1-\mathrm{i}{V}^2\simeq V_z$ . The helicity $\pm 2$ tensors are $T^{11}\pm \mathrm{i}{T}^{22}$ , where $T$ is a symmetric traceless tensor of rank two. In fact, we find

$$\begin{array}{rl}& T^{11}\left(\frac{\partial }{\partial {\xi }^1}\otimes \frac{\partial }{\partial {\xi }^1}-\frac{\partial }{\partial {\xi }^2}\otimes \frac{\partial }{\partial {\xi }^2}\right)+T^{12}\left(\frac{\partial }{\partial {\xi }^1}\otimes \frac{\partial }{\partial {\xi }^2}+\frac{\partial }{\partial {\xi }^2}\otimes \frac{\partial }{\partial {\xi }^1}\right)\\ & =2\left(T^{11}+\mathrm{i}{T}^{12}\right){\partial }_z\otimes {\partial }_z+2\left(T^{11}-\mathrm{i}{T}^{12}\right){\partial }_{\bar{z}}\otimes {\partial }_{\bar{z}}.\end{array}$$

Clearly $T^{zz}=2(T^1+\mathrm{i}{T}^{12})$ has helicity $+2$ and $T^{\bar{z}\bar{z}}=2(T^{11}-\mathrm{i}{T}^{12})$ has helicity $-2$ (note that $g_{\bar{z}\bar{z}}{g}_{z\bar{z}}{T}^{\bar{z}\bar{z}}=T_{zz}$ ).

14.1.3 Covariant derivatives

The only non-vanishing Christoffel symbols of ${\Sigma }_g$ are (see (8.69))

$${\Gamma }_{zz}^z=g^{z\bar{z}}{\partial }_z{g}_{z\bar{z}}=2{\partial }_z\sigma {\Gamma }_{\bar{z}\bar{z}}^{\bar{z}}=g^{\bar{z}z}{\partial }_{\bar{z}}{g}_{\bar{z}z}=2{\partial }_{\bar{z}}\sigma .\text{\hspace{1em}(14.17)}$$

For tensors in ${\mathfrak{T}}^n$ , we define two kinds of covariant derivative: ${\nabla }_{(n)}^z:{\mathfrak{T}}^n\to {\mathfrak{T}}^{n+1}$ and ${\nabla }_z^{(n)}:{\mathfrak{T}}^n\to {\mathfrak{T}}^{n-1}$ . Let

$$T^{\underset{p}{\underbrace{z\dots z}}}\in {\mathfrak{I}}^n(q-p=n).$$

We define

$$\begin{array}{rl}{\nabla }_{(n)}^z{T}_{z\dots z}^{z\dots z}& =g^{z\bar{z}}{\nabla }_{\bar{z}}{T}_{z\dots z}^{z\dots z}\\ & =g^{z\bar{z}}[{\partial }_{\bar{z}}+(q-p){\Gamma }_{\bar{z}z}^z]T_{z\dots z}^{z\dots z}\\ & =g^{z\bar{z}}{\partial }_{\bar{z}}{T}_{z\dots z}^{z\dots z}\end{array}\text{\hspace{1em}(14.18a)}$$

$$\begin{array}{rl}{\nabla }_z^{(n)}{T}_{z\dots z}^{z\dots z}& ={\nabla }_z{T}_{z\dots z}^{z\dots z}\\ & =\left[{\partial }_z+(q-p){\Gamma }_{zz}^z\right]T_{z\dots z}^{z\dots z}\\ & =\left({\partial }_z+2n{\partial }_z\sigma \right)T_{z\dots z}^{z\dots z}.\end{array}\text{\hspace{1em}(14.18b)}$$

2 We use $S_{\pm }$ , instead of ${\Delta }_{\pm }$ , to denote the spin bundles. The symbol ${\Delta }^{\pm }$ is reserved for Laplacians.

In (14.18b), $2n{\partial }_z\sigma$ acts like a gauge potential $\mathcal{A}$ . We also define covariant derivatives with respect to $\bar{z}$

$${\nabla }_{(n)}^{\bar{z}}=g^{\bar{z}z}{\nabla }_z^{(n)},{\nabla }_{\bar{z}}^{(n)}=g_{\bar{z}z}{\nabla }_{(n)}^z.\text{\hspace{1em}(14.19)}$$

The curvature two-form of $K$ and the scalar curvature associated with the Christoffel symbols are

$$\begin{array}{rl}\mathfrak{F}& =R_{zz\bar{z}}^{z}dz\wedge d\bar{z}=-{\partial }_{\bar{z}}(2{\partial }_z\sigma )dz\wedge d\bar{z}\\ & =-2{\partial }_z{\partial }_{\bar{z}}\sigma dz\wedge d\bar{z}\end{array}\text{\hspace{1em}(14.20a)}$$

$$\mathcal{R}=g^{\bar{z}\bar{z}}Ri{c}_{\bar{z}z}+g^{z\bar{z}}Ri{c}_{z\bar{z}}=-8e^{-2\sigma }{\partial }_z{\partial }_{\bar{z}}\sigma .\text{\hspace{1em}(14.20b)}$$

Exercise 14.1

Verify that

$${\nabla }_{(n)}^z=2{\mathrm{e}}^{-2\sigma }{\partial }_{\bar{z}}{\nabla }_z^{(n)}={\mathrm{e}}^{-2n\sigma }{\partial }_z{\mathrm{e}}^{2n\sigma }\text{\hspace{1em}(14.21a)}$$

$${\nabla }_{(n)}^{\bar{z}}=2{\mathrm{e}}^{-2(n+1)\sigma }{\partial }_z{\mathrm{e}}^{2n\sigma }{\nabla }_{\bar{z}}^{(n)}={\partial }_{\bar{z}}.\text{\hspace{1em}(14.21b)}$$

${\nabla }_{(n)}^z$ and ${\nabla }_z^{(n)}$ are mutual adjoints with respect to a properly defined inner product. Let $T,U\in {\mathfrak{T}}^n$ . We require that the inner product be invariant under a holomorphic change of the coordinate $z\to w$ . Since

$$g_{z\bar{z}}\to |\mathrm{d}w/\mathrm{d}z{|}^{-2}{g}_{z\bar{z}}{\mathrm{d}}^{2}z\sqrt{g}\to {\mathrm{d}}^{2}w\sqrt{g}$$

$$\bar{T}\to (\overline{\mathrm{d}w/\mathrm{d}z}{)}^n\bar{T}U\to (\mathrm{d}w/\mathrm{d}z{)}^{n}U.$$

We find the combination

$$(T,U)\equiv \int {\mathrm{d}}^{2}z\sqrt{g}{\left(g_{z\bar{z}}\right)}^n\bar{T}U\text{\hspace{1em}(14.22)}$$

is invariant under holomorphic coordinate transformations. Take $T\in {\mathfrak{T}}^n$ and $U\in {\mathfrak{T}}^{n+1}$ . We find that

$$\begin{array}{rl}(U,{\nabla }_{(n)}^{z}T)& =\int {\mathrm{d}}^{2}z{\mathrm{e}}^{2\sigma }{2}^{-n-1}{\mathrm{e}}^{2(n+1)\sigma }\bar{U}2{\mathrm{e}}^{-2\sigma }{\partial }_{\bar{z}}T\\ & =-2^{-n}\int {\mathrm{d}}^{2}zT{\partial }_{\bar{z}}[{\mathrm{e}}^{(2n+1)\sigma }\bar{U}](\text{p a r t i a l i n t e g r a t i o n})\\ & =-2^{-n}\int {\mathrm{d}}^{2}zT{\mathrm{e}}^{(2n+1)\sigma }\overline{[{\partial }_{z}U+(2n+1)({\partial }_z\sigma )U]}\\ & =-\overline{\int {\mathrm{d}}^{2}z\sqrt{g}(g_{z\bar{z}}{)}^n[{\nabla }_z^{(n+1)}U]\bar{T}}=\overline{(-{\nabla }_z^{(n+1)}U,T)}.\end{array}$$

This shows that

$${\left({\nabla }_{(n)}^z\right)}^{†}=-{\nabla }_z^{(n+1)}.\text{\hspace{1em}(14.23a)}$$

Exercise 14.2

Show that

$${\left({\nabla }_z^{(n)}\right)}^{†}=-{\nabla }_{(n-1)}^z.\text{\hspace{1em}(14.23b)}$$

We define two kinds of Laplacian ${\Delta }_{(n)}^{\pm }:{\mathfrak{T}}^n\to {\mathfrak{T}}^{n\pm 1}\to {\mathfrak{T}}^n$ by

$${\Delta }_{(n)}^{+}\equiv -{\nabla }_z^{(n+1)}{\nabla }_{(n)}^z=-2{\mathrm{e}}^{-2\sigma }[{\partial }_z{\partial }_{\bar{z}}+2n({\partial }_z\sigma ){\partial }_{\bar{z}}]\text{\hspace{1em}(14.24a)}$$

$${\Delta }_{(n)}^{-}\equiv -{\nabla }_{(n-1)}^z{\nabla }_z^{(n)}=-2{\mathrm{e}}^{-2\sigma }[{\partial }_z{\partial }_{\bar{z}}+2n({\partial }_z\sigma ){\partial }_{\bar{z}}+2n({\partial }_z{\partial }_{\bar{z}}\sigma )].\text{\hspace{1em}(14.24b)}$$

Then it follows that

$${\Delta }_{(n)}^{+}-{\Delta }_{(n)}^{-}=4n{\mathrm{e}}^{-2\sigma }\left({\partial }_z{\partial }_{\bar{z}}\sigma \right)=-\frac{1}{2}n\mathcal{R}.\text{\hspace{1em}(14.25)}$$

This shows, in particular, that

$${\Delta }_{(0)}^{+}={\Delta }_{(0)}^{-}(\equiv {\Delta }_{(0)}).\text{\hspace{1em}(14.26)}$$

14.1.4 The Riemann-Roch theorem

Here we derive a version of the Riemann-Roch theorem from the Atiyah-Singer index theorem following D’Hoker and Phong (1988).

Theorem 14.1 · Riemann-Roch theorem

Let ${\Sigma }_g$ be a Riemann surface of genus $g$ . Then the index of the operator ${\nabla }_z^{(n)}$ is

$${\dim }_{\mathbb{C}}\ker {\nabla }_z^{(n)}-{\dim }_{\mathbb{C}}\ker {\nabla }_{(n-1)}^z=(2n-1)(g-1).\text{\hspace{1em}(14.27)}$$

Proof

We use the heat kernel to evaluate the index. We first note that $\ker {\nabla }_z^{(n)}=\ker {\Delta }_{(n)}^{-}$ and $\ker {\nabla }_{(n-1)}^z=\ker {\Delta }_{(n-1)}^{+}$ (see (14.24)). The heat kernel ${\mathcal{K}}_n^{+}$ of ${\Delta }_{(n)}^{+}$ satisfies

$$\left(\frac{\partial }{\partial t}+{\Delta }_{(n)}^{+}\right){\mathcal{K}}_n^{+}(z,w;t)=\left(\frac{\partial }{\partial t}+\Delta -V_n\right){\mathcal{K}}_n^{+}(z,w;t)=0$$

where $\Delta \equiv -2{\partial }_z{\partial }_{\overline{z}}$ is the flat-space Laplacian and

$$V_n\equiv \Delta -{\Delta }_{(n)}^{+}=(1-{\mathrm{e}}^{-2\sigma })\Delta +4n{\mathrm{e}}^{-2\sigma }{\partial }_z\sigma {\partial }_{\bar{z}}.$$

The Laplacian $\Delta$ also defines a heat kernel by

$$\left(\frac{\partial }{\partial t}+\Delta \right)K(z,w;t)=0$$

which is easily solved to yield

$$K(z,w;t)=\frac{1}{4\pi t}{\mathrm{e}}^{-|z-w{|}^2/2t}.$$

The perturbative computation and iteration yield

$$\begin{array}{rl}& {\mathcal{K}}_n^{+}(z,z^{\prime };t)=K(z,z^{\prime };t)\\ & +{\int }_0^{t}ds\int dwK(z,w;t-s)V_n(w){\mathcal{K}}_n^{+}(w,z^{\prime };s)\\ & =K(z,z^{\prime };t)+\int \mathrm{d}s\int \mathrm{d}wK(z,w;t-s)V_n(w)K(w,z^{\prime };s)\\ & +\int \mathrm{d}s\int \mathrm{d}{s}^{\prime }\int \mathrm{d}v\int \mathrm{d}wK(z,v;t-s)V_n(v)\\ & \times K(v,w;s-s^{\prime })V_n(w)K\left(w,z^{\prime };s^{\prime }\right)\\ & +\dots .\end{array}$$

We are particularly interested in ${\mathcal{K}}_n^{+}(z,z;t)$ $t$ being small,

$${\mathcal{K}}_n^{+}(z,z;t)=\frac{1}{4\pi t}+{\int }_0^{t}ds\int dwK(z,w;t-s)V_n(w)K(w,z;s)+\mathcal{O}(t).\text{\hspace{1em}(14.28)}$$

If we take a coordinate system in which $\sigma =0$ at $z$ , we have

$$\begin{array}{rl}& \sigma (w)\simeq 0+{\partial }_z\sigma (w-z)+{\partial }_{\bar{z}}\sigma (\bar{w}-\bar{z})\\ & +\frac{1}{2}\left[{\partial }_z^2\sigma (w-z{)}^2+{\partial }_{\bar{z}}^2\sigma (\bar{w}-\bar{z}{)}^2+2{\partial }_z{\partial }_{\bar{z}}\sigma |w-z{|}^2\right]+\dots .\end{array}$$

Due to rotational symmetry in two-dimensional space, only those terms with one $z$ -derivative and one $\bar{z}$ -derivative survive in the integral in (14.28). Terms proportional to ${\partial }_z\sigma {\partial }_{\bar{z}}\sigma$ cancel between the second and third terms in the expansion and we are left with terms proportional to ${\partial }_z{\partial }_{\bar{z}}\sigma$ . Now we have to evaluate

$$\begin{array}{rl}& {\int }_0^t\mathrm{d}s\int {\mathrm{d}}^{2}wK(z,w;t-s)\\ & \times [2{\partial }_z{\partial }_{\bar{z}}\sigma |\bar{w}-\bar{z}{|}^2{\Delta }_w+4n(\bar{w}-\bar{z}){\partial }_z{\partial }_{\bar{z}}\sigma {\partial }_{\bar{w}}]K(w,z;s)\end{array}$$

From the identities

$$\begin{array}{rl}& \int {\mathrm{d}}^{2}wK(z,w;t-s)|w-z{|}^2{\Delta }_{w}K(w,z;s)\\ & =\frac{1}{16{\pi }^2{s}^2(t-s)}\int {\mathrm{d}}^{2}w|w{|}^2\exp \left(-\frac{t}{2s(t-s)}|w{|}^2\right)\\ & -\frac{1}{32{\pi }^2{s}^3(t-s)}\int d^{2}w|w{|}^4\exp \left(-\frac{t}{2s(t-s)}|w{|}^2\right)\\ & =\frac{(t-s)(2s-t)}{2\pi t^3}\end{array}$$

and

$$\begin{array}{rl}& \int {\mathrm{d}}^{2}wK(z,w;t-s)(\bar{z}-\bar{w}){\partial }_{\bar{w}}K(w,z;s)\\ & =\frac{1}{32{\pi }^2{s}^2(t-s)}\int {\mathrm{d}}^{2}w\exp \left(-\frac{t}{2s(t-s)}|w{|}^2\right)=\frac{t-s}{4\pi t^2}\end{array}$$

we find that

$${\mathcal{K}}_n^{+}(z,z;t)=\frac{1}{4\pi t}+\frac{1+3n}{12\pi }\Delta \sigma +\mathcal{O}(t).\text{\hspace{1em}(14.29a)}$$

We also have the diagonal part of the heat kernel ${\mathcal{K}}_n^{-}$ for ${\Delta }_{(n)}^{-}$ ,

$${\mathcal{K}}_n^{-}(z,z;t)=\frac{1}{4\pi t}+\frac{1-3n}{12\pi }\Delta \sigma +\mathcal{O}(t).\text{\hspace{1em}(14.29b)}$$

From (14.29) and (14.20b), we obtain

$$\begin{array}{rl}\mathrm{ind}{\nabla }_z^{(n)}& =\int {\mathrm{d}}^{2}z\left(\frac{1-3n}{12\pi }-\frac{1+3(n-1)}{12\pi }\right)\Delta \sigma =\frac{1-2n}{8\pi }\int {\mathrm{d}}^{2}x\mathcal{R}\\ & =-\frac{2n-1}{2}\chi ({\Sigma }_g)=(2n-1)(g-1)\end{array}$$

where

$$\chi =\frac{1}{4\pi }\int {\mathrm{d}}^{2}x\mathcal{R}=2-2g$$

is the Euler characteristic of ${\Sigma }_g$ .

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