Paths $tg_1+(1-t)g_0$ in the moduli space of Riemann surfaces Suppose $S$ is a smooth compact oriented surface without boundary. Let $g_0$ and $g_1$ be two smooth Riemannian metrics on $S$. Consider the interpolating path of metrics $g_t=g_1t+g_0(1-t)$. Recall that a Riemannian metric $g$ on an oriented surface defines a unique (integrable) almost complex structure $J$ satisfying $J(g)=g$, $J^2=-1$. So we get a path $\gamma: [0,1]\to M$ in the moduli space of Riemann surfaces.
Question. Is it true that $\gamma$ is a real analytic path in $M$? If so, how can I convince myself in this? (the statement strikes me as counter-intuitive...) (We recall that the moduli space of Riemann surfaces (of fixed genus) has a natural real analytic structure (for example given by Fenchel–Nielsen coordinates))
PS. The comment of abx below suddenly makes this statement much more plausible for me. Indeed if we look at the path of $J_t$, then at each point $x\in S$, ${J_t}_x$ depends analytically on $t$ (by obvious linear algebra). But still, how to go from here to saying that the path in the moduli space is real analytic?
 A: Answering your PS: as you point out,  the complex structure $J_t$ is given in each coordinate chart by a matrix which depends real-analytically on $t$. Now you can find a neighborhood $U$ of $[0,1]$ in $\mathbb{C}$ and extend  $J_t$ as a holomorphic function of $t\in U$; by analytic continuation it will still be a complex structure. Covering $S$ by finitely many coordinate charts and taking the intersection of the corresponding $U$ you get a holomorphic family of complex structures $J_t$ on $S$, for $t$ in a  certain neighborhood $V$ of $[0,1]$ in $\mathbb{C}$. This gives a holomorphic map $V\rightarrow M$ which extends your path $\gamma$, hence $\gamma$ is real-analytic.
Edit: To see that the classifying map $V\rightarrow M$ is holomorphic, define an almost complex structure $\mathscr{J}$ on $V\times S$ as follows. For $(t,x)$ in $V\times S$, $\mathscr{J}_{(t,x)}$ is the endomorphism $(I(t),J_{t}(x))$ of $T_{(t,x)}(V\times S)=T_t(V)\oplus T_x(S)$, where $I$ is the standard complex structure on $\Bbb{C}$. Integrability should not be difficult to check, since the action on $T(V)$ and on $T(S)$ are separated. Thus we have a complex structure on $X:=V\times S$, and the projection $X\rightarrow V$ is holomorphic. So this gives a holomorphic family of Riemann surfaces, hence a holomorphic map to the moduli space.
