Ambitwistor strings are chiral, infinite tension analogues of conventional string theory whose target space is the space of complex null geodesics and whose spectrum consists exclusively of massless states. At genus zero, these strings underpin the Cachazo-He-Yuan formulæ for tree level scattering of gravitons, gluons and scalars. In this paper we extend these formulæ in a number of directions. Firstly, we consider Ramond sector vertex operators and construct simple amplitudes involving space-time fermions. These agree with tree amplitudes in ten dimensional supergravity and super Yang-Mills. We then show that, after the usual GSO projections, the ambitwistor string partition function is modular invariant. We consider the scattering equations at genus one, and calculate one loop scattering amplitudes for NS-NS external states in the Type II ambitwistor string. We conjecture that these give new representations of (the integrand of) one loop supergravity amplitudes and we show that they have the expected behaviour under factorization of the worldsheet in both non-separating and separating degenerations.
Article funded by SCOAP3
= \frac{\tr\left(\mathsf{T}^{a_1}\mathsf{T}^{a_2}\mathsf{T}^{a_3}\right)}{z_{12}z_{23}z_{31}}, \end{equation*} and including the $c$ and $\tilde c$ ghost contributions leaves us with the heterotic amplitude \be\label{h3pt2} \cM^{{\rm het}}(1_{f},2_{f},3_{b}) =\delta^{10}\left(\sum_{i=1}^{3}k_i\right)\;\tr\left(\mathsf{T}^{a_1}\mathsf{T}^{a_2}\mathsf{T}^{a_3}\right)\;\zeta_{1}\cdot \slashed{\epsilon}_{3}\cdot\zeta_{2}\,, \ee the correct amplitude for two gauginos and a gauge boson. Likewise, the three-point amplitude for two gravitini and a graviton in the type II model is given by \be\label{II3pt} \cM^{\mathrm{II}}(1_{f},2_{f},3_{b}) =\delta^{10}\left(\sum_{i=1}^{3}k_i\right)\;\zeta_{1}\cdot \slashed{\epsilon}_{3}\cdot\zeta_{2}\ \tilde{\epsilon}_{1\mu}\tilde{\epsilon}_{2\nu}\tilde{\epsilon}_{3\rho} T^{\mu\nu\rho}, \ee where $T^{\mu\nu\rho}$ is built from the metric and momenta as \be T^{\mu\nu\rho}=\eta^{\mu\nu}(k_1-k_2)^{\rho}+\eta^{\nu\rho}(k_2-k_3)^{\mu}+\eta^{\rho\mu}(k_3-k_1)^{\nu}\,. \ee Thus~\eqref{II3pt} is the correct amplitude for the scattering of two gravitinos and one graviton. Four point amplitudes involving two bosons and two fermions may also be computed by following the usual steps in string theory. The main tool is the Ward identity~\cite{Knizhnik-ml-1984nr}: \be \left\la\Theta_{1\alpha}\Theta_{2\beta}\,\psi^{\mu}_{3}\,\psi^{\nu}_{4}\,\psi^{\rho a}_{4}\right\ra =\sum_{i\neq 4}\frac{M^{\nu\rho}_{i}}{z_{4i}}\left\la\Theta_{1\alpha}\,\Theta_{2\beta}\,\psi^{\mu}_{3}\right\ra\,, \ee where $M_i$ is a rotation matrix acting on the insertion at site $i$. This allows us to compute the leading trace contribution to the four point amplitude in the heterotic model \be\label{h4pt} \begin{aligned} &\cM^{\rm het}(1_{f},2_{f},3_{b},4_{b}) =\left\la c_{1}\tilde{c}_{1}\widehat{U}^{-1/2}_{1}\;c_{2}\tilde{c}_{2}\widehat{U}^{-1/2}_{2}\;c_{3}\tilde{c}_{3}V^{-1}_{3} \int_{\Sigma}\bar\delta(k_4\cdot P)\,V_{4}\right\ra\\ &\ =\delta^{10}\!\left(\sum_{i=1}^{4}k_{i}\right)\;\tr\left(\mathsf{T}^{a_1}\mathsf{T}^{a_2}\mathsf{T}^{a_3}\mathsf{T}^{a_4}\right) \int \bar{\delta}\!\left(\sum_{i=1}^{3}\frac{k_{4}\cdot k_{i}}{z_{4i}}\right)\frac{z_{31}}{z_{34}z_{41}} \times \\ &\hspace{2cm}\left[\zeta_{1}\!\cdot\!\slashed{\epsilon}_{3}\!\cdot\!\zeta_{2}\sum_{j=1}^{3}\frac{\epsilon_{4}\cdot k_i}{z_{4i}}+\left(\frac{\zeta_{1}\!\cdot\!\gamma^{[\nu\sigma]}\!\cdot\!\slashed{\epsilon}_{3}\!\cdot\!\zeta_{2}}{z_{41}}-\frac{\zeta_{2}\!\cdot\!\gamma^{[\nu\sigma]}\!\cdot\!\slashed{\epsilon}_{3}\!\cdot\!\zeta_{1}}{z_{42}}\right)k_{4\;\nu}\epsilon_{4\;\sigma}\right. \\ &\hspace{2.5cm}\left. \ +\ \frac{k_{4}\!\cdot\!\epsilon_{3}\ \zeta_{1}\!\cdot\!\slashed{\epsilon}_{4}\!\cdot\!\zeta_{2}-\epsilon_{4}\!\cdot\!\epsilon_{3} \ \zeta_{1}\!\cdot\! \slashed{k}_{4}\!\cdot\!\zeta_{2}}{z_{43}}\right] + \hbox{permutations} + \hbox{multi-trace}\,. \end{aligned} \ee We have confirmed that upon reducing to four dimensional kinematics (always possible for four particle scattering) this expression produces the correct amplitude for the scattering of two gauginos and two gluons. Similarly, one can compare~\eqref{h4pt} with the single trace contribution to the $\alpha'\rightarrow 0$ limit of 4-point heterotic string amplitudes given in e.g.~\cite{Friedan-ml-1985ge,Polchinski-ml-1998rr}. A similar computation confirms that the type II amplitude \be \begin{aligned}\label{II4pt} &\cM^{\rm II}(1_{f},2_{f},3_{b},4_{b})=\left\la c_{1}\tilde{c}_{1}\widehat{U}^{-1/2,-1}_{1}\;c_{2}\tilde{c}_{2}\widehat{U}^{-1/2,-1}_{2}\;c_{3}\tilde{c}_{3} V^{-1,0}_{3}\;\int_{\Sigma}\bar\delta(k_4\cdot P)\, V_{4}\right\ra \\ &\ =\delta^{10}\left(\sum_{i=1}^{4}k_{i}\right)\;\int \bar{\delta}\left(\sum_{i=1}^{3}\frac{k_{4}\cdot k_{i}}{z_{4i}}\right)z_{23}z_{31}\\ &\hspace{2cm} \times\ \left[\zeta_{1}\cdot\slashed{\epsilon}_{3}\cdot\zeta_{2}\sum_{j=1}^{3}\frac{\epsilon_{4}\cdot k_i}{z_{4i}}+\left(\frac{\zeta_{1}\cdot\gamma^{[\nu\sigma]}\cdot\slashed{\epsilon}_{3}\cdot\zeta_{2}}{z_{41}}-\frac{\zeta_{2}\cdot\gamma^{[\nu\sigma]}\cdot\slashed{\epsilon}_{3}\cdot\zeta_{1}}{z_{42}}\right)k_{4\;\nu}\epsilon_{4\;\sigma}\right. \\ &\hspace{3cm} \left. +\frac{k_{4}\cdot\epsilon_{3} \zeta_{1}\cdot\slashed{\epsilon}_{4}\cdot\zeta_{2}-\epsilon_{4}\cdot\epsilon_{3} \zeta_{1}\cdot\slashed{k}_{4}\cdot\zeta_{2}}{z_{43}}\right] \ \times \mathrm{Pf}\left(\widetilde{M}^{12}_{12}\right) \end{aligned} \ee is the correct one for the scattering of two gravitinos and two gravitons. Here, the $8\times 8$ matrix $\widetilde{M}$ is the same as the one appearing in the formul\ae\ of Cachazo et al.~\cite{Cachazo-ml-2013hca} and accounts for the contribution to the amplitude from the NS $\tilde \Psi$ fields. ]]>
. \end{align} Evaluating this correlator leads again to Pfaffians of $2n\times2n$ matrices. For the $\Psi$ system we obtain the matrix \begin{align}\label{oddM} M=\begin{pmatrix} A & -C^{\rm T}\\ C & B \end{pmatrix}\,, \end{align} where the entries now depend on the $\Psi$ zero modes $\Psi_0$. For $i\neq j$ we have \be\label{oddentries} \begin{aligned} A_{ij}&=k_i\cdot k_j\;S_1(z_{ij};\tau)+k_i\!\cdot\!\Psi_0\ k_j\!\cdot\!\Psi_0\qquad &i,j\neq 1\\ B_{ij}&=\epsilon_i\cdot\epsilon_j\;S_1(z_{ij};\tau)+\epsilon_i\!\cdot\!\Psi_0\ \epsilon_j\!\cdot\!\Psi_0 &\\ C_{ij}&=\epsilon_i\cdot k_j\;S_1(z_{ij};\tau)+\epsilon_i\!\cdot\!\Psi_0\ k_j\!\cdot\!\Psi_0\,, & \end{aligned} \ee whenever $i\neq1$, and diagonal entries \be\label{odddiagC} \begin{aligned} C_{ii}=-\epsilon_i\!\cdot\! P(z_0)\rd z_i-\sum_{j\neq i}^n\epsilon_i\!\cdot\! k_j\ S(z_{ij};\tau)\,, \end{aligned} \ee again for $i\neq1$. When $i=1$, the entries of $A$ and $C$ are modified to \be\label{odd1entries} \begin{aligned} A_{1j}&= P(z_0)\!\cdot\! k_j\ S_1(z_{0j})+P(z_0)\!\cdot\!\Psi_0\ k_j\!\cdot\!\Psi_0\\ C_{11}&=\epsilon_i\!\cdot\! P(z_0)\ S_1(z_{10})+\epsilon_i\!\cdot\!\Psi_0\ P(z_0)\!\cdot\!\Psi_0\, , \end{aligned} \ee as they originate from contractions involving the picture changing operator. In these expressions, $S_1(z_{ij};\tau)$ is the free fermion propagator \be\label{oddprop} S_1(z_{ij};\tau):=\left(\frac{\theta'_1(z_i-z_j;\tau)}{\theta_{1}(z_i-z_j;\tau)}+4\pi\frac{{\rm Im}(z_i-z_j)}{{\rm Im}(\tau)}\right)\,\sqrt{\rd z_i}\sqrt{\rd z_j} \ee orthogonal to the zero mode. Again we have chosen to treat this as a half-form in each of $z_i$ and $z_j$, making it invariant under modular transformations. Note also that the zero mode $\Psi_0^\mu = \Psi_{0z}^\mu\sqrt{\rd z}$, where $\Psi^\mu_{0z}$ are anticommuting constants. After performing all contractions to obtain the Pfaffian of $M$ (and a Pfaffian of a similar matrix $\widetilde M$), we must still perform the path integral over all the fields. Here we find simply \begin{align} \frac{{\det'(\delbar_{\,T_\Sigma})}^2}{{\det'(\delbar_{\,\mathcal{O}})}^{10}} \frac{{\rm Pf}(\delbar_{K_\Sigma^{1/2}})^{10}}{\det(\delbar_{T_\Sigma^{1/2}})} \frac{{\rm Pf}(\delbar_{K_\Sigma^{1/2}})^{10}}{\det(\delbar_{T_\Sigma^{1/2}})}=1\,, \end{align} where we have used the fact that $K_\Sigma^{1/2} = T_\Sigma^{1/2}=\cO$ for the odd spin structure. Finally then, including the integration over zero modes, the contribution of the odd spin structure to $n\geq5$ particle amplitudes is \begin{multline}\label{oddamp2} \cM^{1;\;\rm odd}_n=\delta^{10}\left(\sum k_i\right)\int \rd^{10}p\, \rd^{10}\Psi_{0}\, \rd^{10}\tilde\Psi_{0}\,\rd\tau\,\bar\delta(P^2(z_1))\prod_{i=2}^n\bar\delta(k_i\cdot P(z_i)) \\ \times {\rm Pf}(M)\; {\rm Pf}(\tilde M)\;\frac{\rd z_1}{(\rd z_0)^3}, \end{multline} where $\rd^{10}\Psi_0$ and $\rd^{10}\tilde{\Psi}_{0}$ are the integrals over the $\Psi$ and $\tilde\Psi$ zero modes, while the ratio $\rd z_1/(\rd z_0)^3$ arises from the zero modes of the ghost and antighosts in the picture changing operators. It is easy to see that~\eqref{oddamp2} is invariant under $\tau\rightarrow\tau +1$. Under $\tau\rightarrow -1/\tau$, invariance of $p\rd z$ again implies that $\rd^{10}p\rightarrow\tau^{10}\,\rd^{10}p$. Likewise, invariance of $\Psi_0\sqrt{\rd z}$ implies that the Berezinian integration $\rd^{10}\Psi_{0}\rightarrow\tau^{-5}\,\rd^{10}\Psi_{0}$, and similarly for the $\tilde\Psi$ zero modes. Therefore, under $\tau\to-1/\tau$, \be \rd^{10}p\,\rd^{10}\Psi_0\,\rd^{10}\tilde\Psi_0\,\rd\tau \to \frac{1}{\tau^2}\rd^{10}p\,\rd^{10}\Psi_0\,\rd^{10}\tilde\Psi_0\,\rd\tau\,. \ee Since the Pfaffians and $\delta$-functions are modular invariant, the only remaining factor comes from the ghost zero mode contribution $\rd z_1/(\rd z_0)^3$. This produces the missing $\tau^2$ and renders the result modular invariant. ]]>