6
$\begingroup$

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?

$\endgroup$
8
  • 3
    $\begingroup$ Fix a symplectic form $\omega $ on $S$. For each metric $g$, there is a unique complex structure $J$ on $S$ such that $\omega (X,JY)=g(X,Y)$; at each point $x\in S$ $J_x$ depends analytically on $g_x$. So the $J$ associated to $tg_1+(1-t)g_0$ depends analytically on $t$, and $\gamma$ is analytic. $\endgroup$ – abx May 1 at 14:57
  • 2
    $\begingroup$ I am a bit confused. $\omega$ should be the area form of $g$ for equality $\omega(X,JY)=g(X,Y)$ to hold. So I don't think you need to fix $\omega$ in advance (the area forms of $g_t$ are all different). However, indeed $J_x$ will depend analytically on $t$. So this observation makes the analyticity of $\gamma$ look more plausible indeed... I wonder how to complete this argument to make it a full proof (I guess this is would require something from PDEs?). $\endgroup$ – aglearner May 1 at 15:12
  • 3
    $\begingroup$ I haven't done the calculation, but the Riemann surface structure associated to a metric is obtained by finding isothermal coordinates, and thus by solving the Beltrami equation. So what you're asking should be equivalent to asking how the solution to the Beltrami equation depends on the initial conditions. This is definitely known (and should e.g. be in Ahlfor's "Lectures on quasiconformal maps). $\endgroup$ – Andy Putman May 1 at 20:46
  • 1
    $\begingroup$ Grisha, "$J$ corresponding to $g$ is just rotation by 90 degrees" that's very true. Now, take $g_t=g_1t+g_0(1-t)$ and take the corresponding path of Riemann surfaces $(S,J_t)$, $t\in (0,1)$. This is a path in the moduli space of Riemann surfaces. For what is the Moduli space of Riemann surfaces you can look here: www-personal.umich.edu/~alexmw/CourseNotes.pdf This space has a natural analytic structure $\endgroup$ – aglearner May 1 at 22:23
  • 1
    $\begingroup$ @GrishaPapayanov You can phrase it in the way you stated. I had in mind the fact that the moduli space of Riemann surfaces has a natural complex holomorphic structure, and any complex holomorphic manifold has a natural real analytic structure. But if you prefer Fenchel-Nielsen coordinates, you can use them. They give the same real analytic structure on the moduli space $\endgroup$ – aglearner May 1 at 22:56
3
+50
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ I think some further explanation is warranted to justify the assertion that $V\to M$ is holomorphic. $\endgroup$ – John Pardon May 2 at 13:49
  • $\begingroup$ @John Pardon: I have added some explanation. I agree that details should be checked, but this is an MO post, not a proof from a textbook. $\endgroup$ – abx May 2 at 16:35
  • $\begingroup$ Dear abx, I am thinking about your answer. For the linear algebra part, I have the space of all Riemannian metrics on $\mathbb R^2$. This can be identified with the positive cone $z^2>x^2+y^2$. The projectivisation of the cone is the hyperbolic plane $H^2$ in the Klein model. A path $tg_1+(1-t)g_0$ in the cone gives a geodesic in the Klein model. Clearly, the real analytic map $[0,1]\to H^2$ that sends $[0,1]$ to a geodesic has a natural complexification (for the natural complex structure on $H^2$). Is this equivalent to the complexification you had in mind? $\endgroup$ – aglearner May 3 at 8:57
  • $\begingroup$ This seems likely, but I am sorry I don't have time right now to do the computation. $\endgroup$ – abx May 3 at 19:08
  • 1
    $\begingroup$ Fine, no problem. As for how I came up with my answer, I think it is quite common to prove real-analyticity of a map by proving that it is the restriction of a holomorphic map. Complex analyticity behaves much better that the real one. $\endgroup$ – abx May 8 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.