= Solution
{c}
The inverse of the $(v,\lambda)$ <metric tensor> block is
$$
(g^{AB})=\begin{pmatrix}0&1/m\\1/m&\lambda^2/m^2\end{pmatrix}.
$$
The normal <covector> $d\lambda$ has squared norm $\lambda^2/m^2$. Therefore
$$
\boxed{\{\lambda=\lambda_0\}\text{ is timelike for }\lambda_0\ne0,
\quad\{\lambda=0\}\text{ is null}.}
$$
More explicitly, the induced <metric tensor> is $-\lambda_0^2dv^2+m^2d\Omega_2^2$; at zero its sole zero direction is $\partial_v$. The coefficients are independent of $v$, so $K=\partial_v$ is a <Killing vector field>. On $\lambda=0$,
$$
K^2=-\lambda^2=0,\qquad
K^\flat=-\lambda^2dv+m\,d\lambda=m\,d\lambda.
$$
Hence $K$ is both tangent and normal to this <null hypersurface>, proving it is a <Killing horizon>. The definition of <surface gravity> is $\nabla_KK=\kappa K$ on the <Killing horizon>, or equivalently $d(K^2)=-2\kappa K^\flat$. Since $d(K^2)=-2\lambda\,d\lambda$ vanishes there while $K^\flat\ne0$,
$$
\boxed{\kappa=0.}
$$
Thus this is a <degenerate Killing horizon>, and $v$ is an <affine parameter> on each of its <null geodesics>.
For a concrete <horizon frequency comparison>, fix the same angular position for a narrow high-<frequency> <light> beam and two freely falling radial observers. Work in <geometric optics>, with the beam's affinely transported <wavevector> tangent to the <Killing horizon>, $k=qK$. Here $q>0$ is constant because $\nabla_KK=0$. Choose the observers as translates of one <timelike geodesic> by the <isometry> generated by $K$, crossing at $v_1$ and $v_2$. Give them the same positive <Killing energy> per unit <rest mass>, $E=-K\cdot W$. At their crossings their normalized <four-velocities> have
$$
W^\lambda=-\frac Em,\qquad W^v=\frac1{2E},\qquad
W^\theta=W^\varphi=0,
$$
which satisfy $g(W,W)=2mW^vW^\lambda=-1$. They can be realized by local <geodesics> with precisely these initial data. The measured <angular frequency> is
$$
\omega=-k\cdot W=qE,\qquad
\boxed{\frac{\omega_2}{\omega_1}=1.}
$$
To connect this measurement to <surface gravity>, if $\nabla_KK=\kappa K$ with constant $\kappa$, affine transport of $k=q(v)K$ would give $dq/dv=-\kappa q$. For observers related by the same <Killing vector field> translation, $-K\cdot W$ agrees at the crossings, giving $\omega_2/\omega_1=e^{-\kappa(v_2-v_1)}$. Here there is \b[no accumulated <redshift> along the <Killing horizon>] for these equally prepared observers. Arbitrary observers with different <four-velocities> can still measure different <frequencies> through their different <Killing energies>.
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