3
$\begingroup$

Let's say that Calabi-Yau manifolds are compact Kähler manifolds with a trivial canonical bundle. We know that $H^2(X,\mathcal T_X) = H^1(X,\Omega^2_X)$ is the tangent space to the deformation functor of a Calabi-Yau threefold $X$, and deformations are unobstructed, so the number of moduli for the complex (or algebraic) structure is $h^{1,2}$.

In Yau's popular book on Calabi-Yau manifolds, there was mention both of shape and of size moduli. I assume that the complex structures are parametrized by the shape moduli. I assume that the size moduli are the different ($\mathcal J$-invarant?) Riemannian metrics on it.

Since the number of Kähler classes is $h^{1,1}$, I am wondering if $H^{1}(X,\Omega^1_X)$ could be considered the tangent space to the moduli space (if it exists) of metric structures. Is something like that the case? If so, is the combined moduli space the Cartesian product of the two (or at least the tangent space the product of the two cohomology groups)?

$\endgroup$
1
$\begingroup$

Something like that is somewhat true.

Riffing off of the idea of complex moduli, we can say that the Kähler cone $C(X)$ of a compact complex manifold $X$ should be a "Kähler moduli space" for the manifold. Every Kähler class $\omega$ defines an inner product on the space of real $(1,1)$-classes on $X$. If we identify those classes with the tangent space of $C(X)$ at $\omega$, then we get a Riemannian metric on $C(X)$. That metric can also be obtained by looking at the potential $\omega \mapsto - \log \operatorname{Vol} (X,\omega)$. One can then compute the curvature tensor of this metric, either in cohomology or by picking smooth representatives in each Kähler class. We can also do this over the "complexified" Kähler cone $C(X) \oplus i H^{1,1}(X,\mathbb R)$ and get a Kähler metric.

If we have a family of Kähler manifolds $\mathcal X \to S$ over a smooth base $S$, we can do this construction on the relative Kähler cone $\mathcal C \to S$ whose fiber over $s$ is the Kähler cone of $X_s$. Now, on each point $(\omega, s)$ of $\mathcal C$ we have a Kähler class on $X_s$, and thus a volume form, and thus a metric on $H^0(X_s, K_{X_s})$. We thus get a canonical Hermitian metric on the pullback of $\mathcal R^0 \pi_* K_{\mathcal X/s}$ to $\mathcal C$, which is a kind of relative Weil-Peterson metric.

Doing this over the relative complexified cone, and combining the relative Weil-Peterson metric with the (fiberwise) Kähler metric from before, yields a natural Hermitian metric that somehow combines Kähler and complex moduli. If I recall correctly, it's a Kähler metric when the underlying manifolds are Calabi-Yau (maybe even when they only have zero first Chern class). Waving your hands and shouting "mirror symmetry" should now imply that if such a thing exists, it'd be nice if it had diffeo-geometric interpretations with regards to that metric.

I worked on these ideas in my PhD thesis. I was able to construct the things I described here, and have some unpublished notes that do all of that in cohomology instead of via smooth metrics, but I wasn't able to really harvest anything from this line of investigation before I quit research. YMMV.

$\endgroup$
  • $\begingroup$ Wow, thanks a lot! $\endgroup$ – doetoe Sep 2 '18 at 11:12
  • $\begingroup$ You're welcome. If you pass me your email or something I can point out some references about this stuff, and give you a couple of those ~80% finished notes I mentioned. Maybe you can mine something out of this. $\endgroup$ – Gunnar Þór Magnússon Sep 2 '18 at 13:19
  • $\begingroup$ Great! That is just my username @gmail.com. I'm not sure thought that I'll be able to do much more than glance through them for the time being $\endgroup$ – doetoe Sep 2 '18 at 17:40

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.