Positive frequency solutions in a stationary spacetime I'm studying the basics of QFT in curved spacetime, and for the simplest case of real scalar fields we can define the Klein-Gordon Hermitian form by
$$ (f,g) = i \int_\Sigma \mathrm{d}^3x \sqrt{h} n^\mu ( \overline{f} \partial_\mu g - g \partial_\mu \overline{f}) $$
With $\Sigma$ a spatial hypersurface, $n$ its unit normal and $h$ the determinant of the spatial metric. In a stationary spacetime, we say that a solution $f$ of the Klein-Gordon equation is positive frequency if, for Killing field $k$, we have that
$$ \mathcal{L}_k f = - i \omega f \qquad \omega > 0$$
Question: how can we prove that positive frequency solutions have positive Klein-Gordon 'norm' – that is to say, $(f,f) > 0$?
It's clear to me that if the spacetime is static, or equivalently the metric has no cross-terms between space and time, then the unit normal $n^\mu$ is proportional to the Killing field $k^\mu$ and the Klein-Gordon norm is positive since we we have $n^\mu \partial_\mu \propto \mathcal{L}_k$. However, if this is not the case, then there is no easy way to relate $n$ to $k$, and I cannot see how to proceed.
If staticity is in fact a requirement for this statement to hold, is there an obvious counterexample? That is to say, is there a solution $f$ to the Klein-Gordon equation in a stationary (but not static) spacetime with both $k^\mu \partial_\mu f = -i\omega f$ ($\omega > 0$) and $(f,f)<0$? Thanks for the help.
 A: For a stationary spacetime we can rewrite the metric as
\begin{equation}
g=-\alpha^{2}dt^{2}+h_{ij}( dx^{i}+\beta^{i}dt)( dx^{j}+\beta^{j}dt)
\end{equation}
Now the one-form normal to $\Sigma$ will read
$n^{\flat}=-\alpha dt$ and the corresponding vector field $n^{\sharp}=\frac{1}{\alpha}(k-\beta)$, where $k$ is the stationary Killing vector field.
The KG norm of a positive freq solution $f$ is
\begin{equation}
(f,f)=2\omega \int_{\Sigma} d^{3}x \frac{\sqrt{\det h}}{\alpha} |f|^{2} + I
\end{equation}
where
\begin{equation}
I=-i \int_{\Sigma} d^{3}x \frac{\sqrt{\det h}}{\alpha} (\bar{f}\langle \beta,df\rangle - f \langle \beta, d\bar{f}\rangle ).
\end{equation}
We now need to estimate I. Note that since $f$ is a solution to the KG equation we have
\begin{equation}
\label{intkg}\tag{1}
0=\int_{\Sigma} d^{3}x \sqrt{\det h}\;[\bar{f}(\square - m^{2})f]\alpha.
\end{equation}
Rewriting
\begin{align}
\square f &  = \frac{1}{\sqrt{-\det g}}\partial_{\mu}(\sqrt{-\det g} g^{\mu\nu}\partial_{\nu} f ) \\
& = - \frac{1}{\alpha^{2}}\partial^{2}_{t} f + \frac{1}{\alpha^{2}}\beta^{i}\partial_{i}\partial_{t}f+\frac{1}{\alpha\sqrt{\det h}}\partial_{i}\left(\beta^{i}\frac{\sqrt{\det h}}{\alpha} \partial_{t}f\right)+\frac{1}{\alpha\sqrt{\det h}}\partial_{i}\left(\alpha\sqrt{\det h}g^{ij}\partial_{j}f\right)
\end{align}
and integrating \eqref{intkg} by parts we get
\begin{align}
0=\int d^{3}x \sqrt{\det h} \alpha \left[ \frac{\omega^{2}}{\alpha^2}|f|^{2}-\frac{i\omega}{\alpha^{2}}(\bar{f}\langle \beta,df\rangle - f \langle \beta, d\bar{f}\rangle )-g^{ij}\partial_{i}f\partial_{j}\bar{f}-m^{2}|f|^{2}\right]
\end{align}
and thus
\begin{align}
I=-\omega \int \frac{\det h}{\alpha}|f|^{2}+ \frac{1}{\omega}\int \sqrt{\det h}\alpha \left(m^{2}|f|^2 + g^{ij}\partial_{i}f \partial_{j} \bar{f}\right).
\end{align}
We can substitute this in our original expression for $(f,f)$
\begin{align}
(f,f)=\omega \int d^{3}x \frac{\sqrt{\det h}}{\alpha} |f|^{2}+\frac{1}{\omega}\int d^{3}x \sqrt{\det h} \alpha \left(m^{2} |f|^{2} + g^{ij} \partial_{i}f\partial_{j}\bar{f} \right) \ge 0
\end{align}
