+ 39 \lambda^2) + 4 (41 - 83 \lambda + 40 \lambda^2)\bigr)\bigg]\;.
\end{align}

\subsection{Processes with three and four scalar gravitons}

For scattering with three scalar gravitons --- one in the beginning
and two in the end --- the final and initial momenta are related by
\be
\label{alpha3}
k'=\varkappa k~,~~~~~~\varkappa=\left(\frac{1+u_s}{2u_s}\right)^{1/3}\;.
\ee
The angular dependence of the amplitude reads,
\begin{equation}
f_{\a s,ss} = \left(\frac{2 (1- \lambda)}{1 - 3
\lambda}\right)^{3/2}  \frac{(1-x^2)
P_{\a s,ss}(x)}{g_{4}(x)}\;,~~~~~~\a=+,-\;,
\end{equation}
\begin{align}
\label{denom4}
g_{4}(x) &= \bigl((1 + 2 x \varkappa +\varkappa^2)^3 -  (1 -  u_{s}
\varkappa^3)^2\bigr) \bigl(u_{s}^2 (1 + 2 x \varkappa + \varkappa^2)^3 -
(1 -  u_{s} \varkappa^3)^2\bigr) \nonumber \\
& \times \bigl((1 - 2 x \varkappa + \varkappa^2)^3 -  (u_{s} -  u_{s}
\varkappa^3)^2\bigr) \bigl(u_{s}^2 (1 - 2 x \varkappa + \varkappa^2)^3 -
(u_{s} -  u_{s} \varkappa^3)^2\bigr)\;,
\end{align}
and $P_s,ss(x)$ is a 12th order polynomial. For $u_s≠1$ the
denominator has zeros at non-zero angles, and the amplitude vanishes
in the forward and backward limits.

For $u_s = 1$ the amplitude simplifies (though it still remains
quite lengthy):
\begin{align}
f_{+s,ss}=& f_{-s,ss}=f_{ss,+s}=f_{ss,-s}=\notag\\
x^6 (1\! -\! \lambda) \Bigl(119 \!-\! 16 v_{2}^2 (1\! -\!
\lambda)^3 \!-\! 582 \lambda \!+\! 825 \lambda^2 \!-\! 422 \lambda^3 \nonumber
\\
&+ 9 v_{3}^2 (1 \!-\! \lambda)^2 (1\! -\! 6 \lambda)\! -\! 6 v_{3} (30\! -\!
147 \lambda \!+\! 193 \lambda^2 \!-\! 74 \lambda^3) \!-\! 4 v_{2} (1 \!-\!
\lambda) \bigl(35 \!-\! 139 \lambda \!+\! 92 \lambda^2\notag\\
& - v_{3} (3 + 3 \lambda - 6 \lambda^2)\bigr)\Bigr)
+x^4 \Bigl(-275 + 102 v_{3} + 90 v_{3}^2 - 192 v_{1}(1 - \lambda)^3 \bigl(15 - 4
v_{2} (1 - \lambda) \notag\\
&- 6 v_{3} (1 - \lambda) - 29 \lambda\bigr) + 1717 \lambda - 966 v_{3}
\lambda - 234 v_{3}^2 \lambda - 3811 \lambda^2 + 2286 v_{3} \lambda^2
+ 162 v_{3}^2 \lambda^2 \notag\\
&+ 3799 \lambda^3 \!-\! 2058 v_{3} \lambda^3 \!+\! 18 v_{3}^2 \lambda^3 \!-\!
1422 \lambda^4 \!+\! 636 v_{3} \lambda^4
\!-\! 36 v_{3}^2 \lambda^4 \!+\! 16 v_{2}^2 (1 \!-\! \lambda)^3 (17 \!-\! 15
\lambda) \notag\\
&- 4 v_{2} (1 - \lambda) \bigl(161 - 554 \lambda + 653
\lambda^2 - 264 \lambda^3 - 6 v_{3} (1 - \lambda)^2 (19 + 13
\lambda)\bigr)\Bigr) \nonumber \\
&- x^2 \Bigl(167 - 480 v_{3} + 387 v_{3}^2 - 384 v_{1}(1 -
\lambda)^3 \bigl(15 -
8 v_{2} (1 - \lambda) - 9 v_{3} (1 - \lambda) - 31 \lambda\bigr)\notag\\
& - 1381 \lambda + 2214 v_{3} \lambda - 1539 v_{3}^2 \lambda + 3615 \lambda^2 -
3540 v_{3} \lambda^2 + 2295 v_{3}^2 \lambda^2 - 3319 \lambda^3
\nonumber \\
&+ 2322 v_{3} \lambda^3 - 1521 v_{3}^2 \lambda^3 + 902 \lambda^4 -
516 v_{3} \lambda^4 + 378 v_{3}^2 \lambda^4 + 16 v_{2}^2 (1 -
\lambda)^3 (67 - 71 \lambda) \nonumber \\
&- 4 v_{2} (1 - \lambda) \bigl(451 - 1790 \lambda + 2183
\lambda^2 - 836 \lambda^3 - 3 v_{3} (1 - \lambda)^2 (129 - 134
\lambda)\bigr)\Bigr)\notag\\
&+323 - 546 v_{3} + 288 v_{3}^2 - 576 v_{1}(1 - \lambda)^3 \bigl(5 - 4 v_{2} (1 -
\lambda) - 4 v_{3} (1 - \lambda) - 11 \lambda\bigr) \notag\\
&- 2493 \lambda \!+\! 3126 v_{3} \lambda\! -\! 1224 v_{3}^2 \lambda \!+\! 6595
\lambda^2\! -\! 6234 v_{3} \lambda^2 \!+\! 1944 v_{3}^2 \lambda^2 \!-\! 6927
\lambda^3
\!+\! 5226 v_{3} \lambda^3 \notag\\
&- 1368 v_{3}^2 \lambda^3 + 2478
\lambda^4 - 1572 v_{3} \lambda^4 + 360 v_{3}^2 \lambda^4 + 48 v_{2}^2
(1 - \lambda)^3 (17 - 19 \lambda) \notag\\
&- 12 v_{2} (1 - \lambda)
\bigl(101 - 450 \lambda + 609 \lambda^2 - 256 \lambda^3 - 2 v_{3} (1
- \lambda)^2 (46 - 53 \lambda)\bigr)
\bigg]\;.
\end{align}

Finally, we consider the amplitude with four scalar gravitons. The
kinematics in this case is simple, $k'=k$. Still, the amplitude is
rather lengthy since it involves all vertices in an intricate way. In
general it has the form,
\begin{equation}
f_{ss,ss} = \left(\frac{2 (1- \lambda)}{1 - 3
\lambda}\right)^{2}  \frac{P_{ss,ss}(x)}{(1-x^2)^3}\;,
\end{equation}
where $P_ss,ss(x)$ is an even polynomial of degree $8$. The
amplitude has only forward singularities.
For $u_s=1$
we have:
\begin{align}
f_{ss,ss} =& \frac{1}{64(1-\lambda)^2(1-3\lambda)^3(1-x^2)^3}
\bigg[
x^8 (1 - \lambda)^2 \Bigl(185 - 32 v_{2}^2 (1 -
\lambda)^3 - 992 \lambda + 1525 \lambda^2 \notag \\
&- 838 \lambda^3 - 9 v_{3}^2 (1 - \lambda)^2 (1 + 6 \lambda)
- 8 v_{2} (1 - \lambda) \bigl(15 + 6 v_{3} (1 - \lambda) - 68
\lambda + 41 \lambda^2\bigr)\notag\\
& + 12 v_{3} (-15 + 80 \lambda - 103
\lambda^2 + 36 \lambda^3)\Bigr)
+x^6 \Bigl(-500 \!+\! 408 v_{3} \!+\! 171 v_{3}^2 \!+\! 3592 \lambda \!-\! 3624
v_{3} \lambda \notag\\
&- 954 v_{3}^2 \lambda - 9152 \lambda^2 + 11280 v_{3}
\lambda^2 + 2106 v_{3}^2 \lambda^2 + 10584 \lambda^3 - 16344 v_{3}
\lambda^3 - 2304 v_{3}^2 \lambda^3 \notag\\
&- 5340 \lambda^4 \!+\! 11304 v_{3}
\lambda^4 \!+\! 1251 v_{3}^2 \lambda^4 \!+\! 768 \lambda^5
\!-\! 3024 v_{3} \lambda^5 \!-\! 270 v_{3}^2 \lambda^5 \!+\! 192 v_{2}^2 (1\!-\!
\lambda)^4 (1 \!-\! 2 \lambda)\notag\\
& - 384 v_{1} (1 - \lambda)^3 (1 + 2
\lambda - 15 \lambda^2) - 48 v_{2} (1 - \lambda)^2 \bigl(-2 + 34
\lambda - 98 \lambda^2
+ 70 \lambda^3 \notag\\
& - v_{3} (1 - \lambda)^2 (8 - 15
\lambda)\bigr)\Bigr)
+x^4 \Bigl(458 + 585 v_{3}^2 + 73728 v_{1}^2 (1 - \lambda)^5 -
3844 \lambda + 792 v_{3} \lambda \notag\\
& - 2574 v_{3}^2 \lambda + 12580
\lambda^2 - 3840 v_{3} \lambda^2
+ 4446 v_{3}^2 \lambda^2 - 20984 \lambda^3 + 7128 v_{3} \lambda^3
- 3744 v_{3}^2 \lambda^3 \notag\\
&+ 17554 \lambda^4 - 5904 v_{3} \lambda^4
+ 1521 v_{3}^2 \lambda^4 - 5684 \lambda^5
+ 1824 v_{3} \lambda^5 + 64 v_{2}^2 (1 -
\lambda)^4 (124 - 119 \lambda) \notag\\
&- 234 v_{3}^2 \lambda^5 + 384 v_{1} (1 - \lambda)^3
\bigl(-3 + 128 v_{2} (1 - \lambda)^2 + 42 v_{3} (1 - \lambda)^2
+ 30 \lambda - 55 \lambda^2\bigr) \notag\\
&- 16 v_{2} (1 - \lambda)^2
\bigl(3 - 151 \lambda + 469 \lambda^2 - 341 \lambda^3 + 15 v_{3}
(1 - \lambda)^2 (-20 + 17 \lambda)\bigr)\Bigr) \nonumber \\
&+x^2 \Bigl(-300 + 936 v_{3} - 1647 v_{3}^2 - 147456 v_{1}^2 (1 -
\lambda)^5 + 2744 \lambda - 6936 v_{3} \lambda + 8082 v_{3}^2
\lambda \notag\\
& - 9120 + 18288 v_{3} \lambda^2 - 15858 v_{3}^2 \lambda^2 + 13992
\lambda^3 - 22728 v_{3} \lambda^3 + 15552 v_{3}^2 \lambda^3 -
10532 \lambda^4 \nonumber \\
&+ 13656 v_{3} \lambda^4 \!-\! 7623 v_{3}^2 \lambda^4 \!+\! 3200 \lambda^5
\!-\! 3216 v_{3} \lambda^5 \!+\! 1494 v_{3}^2 \lambda^5
\!-\! 64 v_{2}^2 (1 \!-\! \lambda)^4 (255 \!-\! 254 \lambda) \notag\\
&- 384 v_{1}
(1 - \lambda)^3 \bigl(-33 + 256 v_{2} (1 - \lambda)^2 + 84 v_{3}
(1 - \lambda)^2 + 150 \lambda - 137 \lambda^2\bigr) \nonumber \\
&+ 16 v_{2} (1 - \lambda)^2 \bigl(234 - 1290 \lambda + 1978
\lambda^2 - 926 \lambda^3 - 15 v_{3} (1 - \lambda)^2 (44 - 43
\lambda)\bigr)\Bigr) \nonumber \\
&+733 - 1164 v_{3} + 900 v_{3}^2 + 73728 v_{1}^2 (1 - \lambda)^5
- 6890 \lambda + 8448 v_{3} \lambda - 4536 v_{3}^2 \lambda + 23886
\lambda^2 \notag\\
& - 22392 v_{3} \lambda^2 + 9144 v_{3}^2 \lambda^2 - 37880 \lambda^3 + 28080 v_{3}
\lambda^3 - 9216 v_{3}^2 \lambda^3 + 27949 \lambda^4 - 16956 v_{3}
\lambda^4 \notag\\
&+ 4644 v_{3}^2 \lambda^4 - 7814 \lambda^5
+ 3984 v_{3} \lambda^5 - 936 v_{3}^2 \lambda^5 + 32 v_{2}^2 (1 -
\lambda)^4 (257 - 259 \lambda) \notag\\
&+ 384 v_{1} (1 - \lambda)^3
\bigl(-29 + 128 v_{2} (1 - \lambda)^2
+ 42 v_{3} (1 - \lambda)^2 + 122 \lambda - 97 \lambda^2\bigr) \notag\\
&- 8 v_{2} (1 - \lambda)^2 \bigl(459 - 2399 \lambda + 3497 \lambda^2
- 1549 \lambda^3 - 6 v_{3} (1 - \lambda)^2 (113 - 115
\lambda)\bigr) \bigg]\;.
\end{align}

\section{Modes and propagators
with auxiliary field}
\label{app:chi}

The tensor and vector parts of the quadratic Lagrangian following from
the action (\ref{Lchi1}) with the gauge fixing (\ref{eq:chi-GFF}) are
the same as in the original HG action, see Eqs.~(\ref{L2t}),
(\ref{L2v}). The difference, however, occurs in the scalar
sector. Using the same decomposition as in Eqs.~(\ref{eq:mode_decomp})
we obtain,
\be
\label{L2sprim}
\begin{split}
{\tilde{\cal L}_g^{(2s)}}=\frac{1}{2G}\bigg\{&\frac{\dot\psi^2}{2}
+\frac{\dot E^2}{4}-2\chi\dot\psi-\chi\dot E
+\frac{1+\xi}{4\sigma}E\D^3 E\\
&+\dot B\frac{\sigma}{(1+\xi)\D}\dot B+B\D^2 B+2\chi\D B
-2\dot{\bar C}\D\dot C-\frac{2(1+\xi)}{\sigma}\bar C\D^4 C\bigg\}.
\end{split}
\ee
The ghost part, of course, decouples and leads to a simple propagator
which, combined with the vector contribution, gives
\be
\label{chi-ccprop}
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(c_i\)]  -- [color=white] b -- [color=white] c
[particle=\(\bar c_j\)],
a -- [fermion, edge label={}]c,
=G\Big[\delta_{ij} {\cal P}_1+\hat k_i\hat k_j (\tilde\P_0-\P_1)\Big]\;,
\ee
where $_1$ is given in (\ref{proppoles}) and
\be
\label{newproppole}
\tilde\P_0=\frac{i}{\omega^2-\tilde\nu_0 k^6+i\epsilon}~,~~~~~~
\tilde\nu_0\equiv\frac{1+\xi}{\sigma}\;.
\ee
The other components $ψ$, $E$, $B$, $χ$ all mix with each
other. To find their propagators, we switch to the Fourier space and
invert the mixing matrix. Combining with the propagators of tensor and
vector components we arrive at,
\bseq
\label{newprops}
\begin{align}
\label{chichiprop}
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(\chi\)]  -- [color=white] b -- [color=white] c
a -- [scalar, edge label={}]c,
&= G\bigg\{-\frac{i}{3} + \frac{2\nu_{s}}{3} k^6
\mathcal{P}_{s}\bigg\}, \\
\label{chiNprop}
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(\chi\)]  -- [scalar] b -- [double] c
&= iG\frac{\tilde\nu_0k^5 \hat k_i}{3 (\nu_s-\tilde\nu_0)}
\bigg\{2\nu_{s} \mathcal{P}_s - (3\nu_s-\tilde\nu_0) \tilde{\mathcal{P}}_0\bigg\},
\\
\label{chihprop}
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(\chi\)]  -- [scalar] b -- [photon] c
&= iG \frac{2\omega}{3} \bigg\{-\delta_{ij}\mathcal{P}_{s}
+\hat{k}_{i} \hat{k}_{j} \frac{3\nu_s-\tilde\nu_0}{\nu_s-\tilde\nu_0}
\big[\P_s-\tilde\P_0\big]\bigg\},
\\
\label{newNNprop}
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(N_i\)]  -- [color=white] b -- [color=white] c
a -- [double, edge label={}]c,
& = -G\bigg\{ \!\frac{k^4}{ \sigma} (\delta_{ij}\! -\!
\hat{k}_i \hat{k}_j) \mathcal{P}_{1}
\!-\! \frac{\tilde\nu_0 k^4\hat{k}_i \hat{k}_j}{3(\nu_s\!\!-\!\tilde\nu_0)}
\bigg[\frac{2\tilde\nu_0\nu_s}{\nu_s\!\!-\!\tilde\nu_0}\P_s
\!-\!\frac{2\tilde\nu_0^2}{\nu_s\!\!-\!\tilde\nu_0}\tilde\P_0
\!+\!i(3\nu_s\!\!-\!\tilde\nu_0)\omega^2\tilde\P_0^2\bigg]\!\bigg\},
\\
\label{Nhprop}
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(N_i\)]  -- [double] b -- [photon] c
&=  \frac{2G\omega\tilde\nu_0}{3(\nu_s\!-\!\tilde\nu_0)k}\bigg\{
\!\!\!-\!\hat k_i\delta_{jk}\big[\P_s\!-\!\tilde\P_0\big]
\!+\!\hat k_i\hat k_j\hat k_k
\bigg[\frac{\P_s\!-\!\tilde\P_0}{\nu_s\!-\!\tilde\nu_0}
\!+\!i k^6\tilde\P_0^2\bigg]\!\bigg\},
\\
\label{newhhprop}
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(h_{ij}\)]  -- [color=white] b -- [color=white] c
&= 2G\bigg\{(\delta_{i k}\delta_{j l} +
\delta_{i l} \delta_{j k}) \mathcal{P}_{tt} - \delta_{i
j}\delta_{kl} \left[ \mathcal{P}_{tt} -\frac{\P_s}{3}\right]
\nonumber \\
- (\delta_{i k}\hat{k}_{j}\hat{k}_{l}
+ \delta_{i l}\hat{k}_{j}\hat{k}_{k} + \delta_{j
k}\hat{k}_{i}\hat{k}_{l} + \delta_{j l}\hat{k}_{i}\hat{k}_{k})
\big[\mathcal{P}_{tt}-\mathcal{P}_{1} \big] \nonumber \\
+ (\delta_{i j} \hat{k}_{k} \hat{k}_{l} + \delta_{k l}\hat{k}_{i}\hat{k}_{j})
\bigg[\mathcal{P}_{tt}-\frac{3\nu_s-\tilde\nu_0}{3(\nu_s-\tilde\nu_0)}
\mathcal{P}_{s} + \frac{2\tilde\nu_0}{3(\nu_s-\tilde\nu_0)}
\tilde{\mathcal{P}}_0 \bigg]
\nonumber \\
\mathcal{P}_{tt}
+ \frac{(3\nu_s-\tilde\nu_0)^2}{3(\nu_s-\tilde\nu_0)^2} \mathcal{P}_{s}
- 4 \mathcal{P}_{1}\notag\\
- \frac{4\tilde\nu_0(3\nu_s-2\tilde\nu_0)}{3(\nu_s-\tilde\nu_0)^2}
\tilde{\mathcal{P}}_0
+ i \frac{2\tilde\nu_0(3 \nu_s-\tilde\nu_0)k^6}{3(\nu_s-\tilde\nu_0)}
\tilde{\mathcal{P}}_0^2  \bigg]\bigg\}.
\end{align}
\eseq
Here $_tt$ and $_s$ are the same as in (\ref{proppoles}), and
$ν_s$ is given by the $λ→∞$ limit of
Eq.~(\ref{oms}), $ν_s=(8/3)ν_4+ν_5$.

We observe that all propagators (\ref{chi-ccprop}), (\ref{newprops})
are regular in the sense defined in Sec.~\ref{sec:chi}. This follows
from three properties. First, the pole factors $_tt$, $_s$,
$_1$, $_0$ are regular. Second, the propagators scale
homogeneously under the Lifshitz transformations, in the way
compatible with the scaling dimensions of the corresponding
fields. And third, the inverse powers of the spatial momentum $k$
contained in the unit vector $k̂_i$ cancel when we bring the
combinations in the square brackets to the common denominator. As
shown in Ref.~\cite{Barvinsky:2015kil}, the regularity of the
propagators is sufficient for the renormalizability of the theory.

Let us also note the presence of double poles $_0^2$. They
signal presence of a linearly growing gauge mode, similarly as it
happens in the Maxwell theory in general covariant gauges (see
e.g. Sec.~18 of \cite{Strathdee1994}).

Mixing between different components in the Lagrangian (\ref{L2sprim})
implies that the scalar graviton state has overlap not only with the
metric $h_ij$, but also the shift $N_i$ and the field $χ$. To see
this, let us write the eigenmode equations following from
\bseq
\begin{align}
&\omega^2 E-2i\omega\chi-\tilde\nu_0k^6E=0\;,\\
&2i\omega\psi+i\omega E-2k^2 B=0\;.
\end{align}
\eseq
Substituting here the dispersion relation of the scalar graviton,
$ω=√(ν_s)  k^3$, and expressing all fields in terms of
$ψ$ we obtain,
\be
\chi=i\sqrt{\nu_s}\,k^3\psi\;.
\ee
So indeed, for the scalar graviton eigenmode all fields are in general

The situation simplifies considerably if we choose the gauge $ξ=-1$,
entailing $ν̃_0=0$. Then the admixture of the scalar graviton
to the shift vanishes, which also eliminates the mixed propagators
(\ref{chiNprop}), (\ref{Nhprop}). The normalization of the scalar
graviton mode is deduced by imposing the canonical commutations
relations on $ψ$ and its conjugate momentum
\[
\pi_\psi=\frac{\dot\psi-2\chi}{2G}\;.
\]
Collecting everything together, we
find the scalar graviton contribution to the metric and the field
$χ$ in the $ξ=-1$ gauge,
\bseq
\begin{align}
\label{hdecompnew}
&h_{ij}(\x,t)\ni \sqrt{G}\int\frac{d^3k}{(2\pi)^3 2\omega_s}
\,\ve^{(0')}_{ij}\,h_{\k 0'}\,\e^{-i\omega_s
\ve_{ij}^{(0')}=\sqrt{\frac{2}{3}}\big(\delta_{ij}-3\hat k_i\hat
&\chi(\x,t)=\sqrt{G}\int\frac{d^3k}{(2\pi)^3 2\omega_s}\,
i\omega_s\sqrt{\frac{2}{3}}\,h_{\k 0'}\,\e^{-i\omega_s
\end{align}
\eseq
where $h_0̨'$ is the scalar graviton annihilation operator
\be
[h_{\k 0'},h^+_{\k' 0'}]=2\omega_s\, (2\pi)^3\delta(\k-\k')\;.
\ee
This provides us with the expressions for the external lines of the
scalar diagrams for the scalar graviton scattering. The form of the
$h$-line is unchanged, see Eq.~(\ref{extlegsh}), with the polarization
tensor from (\ref{hdecompnew}). Whereas the $χ$-line reads,
\be
\feynmandiagram[baseline=(a.base), horizontal= a to c]{
a [particle =\(\chi\)] -- [color=white] b -- [color=white] c,
a -- [scalar, edge label={}] c,
};\quad= i\sqrt{\frac{2G}{3}}\,\omega\;.
\ee

\bibliographystyle{JHEP}
\bibliography{library.bib}
\end{document}