# Holomorphic family of Riemann surfaces

Let $2g+m\ge 3$. A holomorphic family of (non-singular, compact) Riemann surfaces of type $(g,m)$ is a triple $(X,Y,\pi,s_1,\ldots,s_m)$, where $X,Y$ are complex manifolds of (complex) dimension $n+1,n$ respectively, and $\pi:X\to Y$ is a proper holomorphic map with no critical points, the $s_i$ are disjoint holomorphic sections of $\pi$ and the fibres are all of genus $g$.

My question is the following: Suppose $(X,Y,\pi,s_1,\ldots,s_m)$ and $(X',Y,\pi',s_1',\ldots,s_m')$ are two holomorphic families of Riemann surfaces of type $(g,m)$ over the same base $Y$, and suppose $\varphi:X\to X'$ is a ($C^\infty$) diffeomorphism such that $\pi'\circ\varphi = \pi$, and $\varphi$ is a fibrewise biholomorphism (preserving the marked points). Does it follow that the map $\varphi$ is holomorphic?

• Have you thought about the case where $Y$ is the projective line and where $X=X'=Y \times Y$, with map $(z,w) \mapsto (z,W(z,w))$? Then $W(z,w)$ has to be holomorphic in $w$ but not in $z$, if I am reading correctly. Feb 9, 2018 at 12:31
• I guess my question did not list all the necessary assumptions. I think we should assume the stability of the fibres (by adding some marked points if necessary) to rule out your example. I'll edit the question to add the stability assumption. Feb 9, 2018 at 13:43

The result is true under the stability hypothesis. Here's a sketch of the proof. Let us consider the derivative $d\varphi$ as a section of the vector bundle $T^*_X\otimes_{\mathbb R}\varphi^*T_{X'}$ on $X$. Using the complex structures on $X,X'$, we can consider its antilinear part $\bar\partial\varphi\in\Gamma(X,\Omega^{0,1}_X\otimes_{\mathbb C}\varphi^*T_{X'})$. We now make the key observation that $\bar\partial\varphi$ is in fact a section of the subbundle $\pi^*\Omega^{0,1}_Y\otimes_{\mathbb C} \varphi^*T_{X'/Y}$, where we use the notation $T_{X'/Y}$ to denote the bundle of vertical vectors in $X'$ over $Y$.
This is easiest to see using a local coordinate representation of $\varphi$ of the form $(z_1,\ldots,z_n,w)\mapsto(Z_1,\ldots,Z_n,W)= (z_1,\ldots,z_n,g(z_1,\ldots,z_n,w))$, and using the fact that $g$ is holomorphic in $w$. In our local coordinate system, $\bar\partial\varphi = \sum_{j=1}^n\frac{\partial g}{\partial \bar z_j}d\bar z_j\otimes\frac{\partial}{\partial W}$.
It now follows that if we restrict $\bar\partial\varphi$ to any fibre $X_y$ (for some $y\in Y$), we get a section of the vector bundle $\Omega^{0,1}_{Y,y}\otimes_{\mathbb C}\varphi^*{T_{X'_y}}\to X_y$ with the following properties: it is holomorphic and vanishes on the marked points $s_1(z),\ldots,s_m(z)$. By the stability assumption, it now follows that $\bar\partial\varphi=0$, which completes the proof that $\varphi$ is holomorphic.
As pointed out in the comments, if we relax stability, we need not have the result. For example take $X = \mathbb C\times\mathbb C/\Gamma = X'$ where $\Gamma\subset\mathbb C$ is a lattice, and let $\varphi(z,w) = (z,w+|z|^2)$. This is fibrewise holomorphic, but not holomorphic overall. We can see that $\bar\partial\varphi = z\,d\bar z\otimes\frac{\partial}{\partial w}$, and this gives a holomorphic vector field on each fibre (when we drop $d\bar z$).