Let $\pi:X\to \mathbb C^\*$ be a family of projective varieties and $\tau$ be a relative real $(1,1)$-form on $X$, then [we have](http://download.springer.com/static/pdf/300/art%253A10.1007%252Fs00209-015-1484-x.pdf?originUrl=http%3A%2F%2Flink.springer.com%2Farticle%2F10.1007%2Fs00209-015-1484-x&token2=exp=1478124239~acl=%2Fstatic%2Fpdf%2F300%2Fart%25253A10.1007%25252Fs00209-015-1484-x.pdf%3ForiginUrl%3Dhttp%253A%252F%252Flink.springer.com%252Farticle%252F10.1007%252Fs00209-015-1484-x*~hmac=8775a6d7822bb9a1244f0afae2222d08ef326a2c1d464ec2f8ad60adf0bdc622) $$\tau^{n+1}=c(\tau)\tau^n\sqrt{-1}ds\wedge d\bar s$$ 
where $c(\tau)$ is a geodesic curvature. I am looking for a reference to proof it. 


My question is if we change the base to Kahler manifold $Y$, then $$\tau^{n+m}=c(\tau)\tau^n\wedge\omega\_Y^m$$



relative (1,1) form and geodesic curvature
I would recommend you the paper of Schumacher


[Positivity of relative canonical bundles and applications](https://arxiv.org/pdf/1201.2930v1.pdf)


where you can find the statement on page 16 (Lemma 6). In his notation, the formula reads
$$
\omega\_{\mathcal{X}}^{n+1}=\varphi \cdot gdV\sqrt{-1}ds\wedge d\overline{s}
$$
The proof is essentially
the formula for determinants of block matrices, when the lower right entry is invertible. The global (1,1) - form $\omega\_{\mathcal{X}}$ on the total space $\omega\_{\mathcal{X}}$ (your $\tau$) is a genuine form (and not a relative one). The point is that on the right hand side, you find the relative form
$\omega\_{\mathcal{X}/S}^n=gdV$, which is a form of top degree along the fibers. The formula, as stated there, is written down for a one dimensional base $S$. Then you can think of $\varphi$ (your $c(\tau)$) as the geodesic curvature of the curve in some space of hermitian metrics. (In this interpretation, one assumes that the time dependence is only on the real part of the one dimensional complex base direction. Furthermore, one usually takes an analytically trivial fibration for this.) So in my mind, the expression "geodesic curvature" only makes sense for a one dimensional base. For a higher dimensional $S$, you would have something like
$$
\omega\_{\mathcal{X}}^{n+1}= \sum\_{i,\overline{\jmath}}{\varphi\_{i\overline{\jmath}}}\omega\_{\mathcal{X}/S}^n\sqrt{-1}ds^i\wedge ds^{\overline{\jmath}}
$$  

where $\varphi\_{i\overline{\jmath}}=<v\_i,v\_j>\_{\omega\_{\mathcal{X}}}$ is the inner product of the horizontal lifts (with respect to $\omega\_{\mathcal{X}}$) of the coordinate tangent vectors. This is the second interpretation of $\varphi$, which makes sense in any dimension. In this expression, I have omitted all terms which are not of top degree in fiber direction. Note that we have
$$
\omega^{WP}=\int\_{\mathcal{X}/S}{\omega\_{\mathcal{X}}^{n+1}}
$$
for the Weil-Peterson metric on the base. You can also write down a formula for $\omega\_{\mathcal{X}}^{n+m}$, but I don't know if this has any meaning. I hope that will help you. Best wishes!


