In which space are we solving the Kähler Ricci flow?

Question

The Kähler Ricci flow on a compact Kähler manifold are formulated as $\frac{\partial}{\partial t}w(t) = -Ric(w)$, $w(0) = w_0$, where $w(t)$ is a family of Kähler metrics and $w_0$ is the initial Kähler metric. But it is not obvious to me in which space we differentiate $w(t)$. My guess is the vector space of the closed real (1,1) forms, but in order to differentiate I need it to be a normed vector space. What is the norm? I know that the first $\bar \partial$-cohomology group must be finite-dimensional, thus it makes sense to differentiate there, but on chain level it is not clear to me how this is formulated.

Answer 1

The object $\omega(t)$ is not a cohomology class, it is a $(1,1)$-form. The time derivative of $\omega(t)$ is taken in each vector space $\Lambda^{1,1} T^*_m M$, pointwise. If $\omega(t)$ is $C^k$, then local coordinate calculation shows that $\operatorname{Ric}_{\omega(t)}$ is $C^{k-2}$. If you vary the choice of a $(1,1)$-form $\eta$ in a Dolbeault cohomology class, you will vary $\operatorname{Ric}_{\eta}$. Indeed, if $\eta$ is a smooth $(1,1)$-form, and you fix a small enough neighborhood of some point, you can make $\eta+i\partial\bar\partial u$ equal, inside that open set, any smooth positive $(1,1)$-form you like, so Ricci flow cannot be defined on cohomology classes. You don't need a norm in a topological vector space to define differentation of maps from an interval of the real line to that vector space. You can, however, easily define a norm on the square integrable sections of any vector bundle with inner product on any compact manifold which has a volume form. So clearly the square integrable $(1,1)$-forms form a Hilbert space. But this might be inconvenient for working with noncompact manifolds, so it is best to think of the Frechet space of $(1,1)$-forms as a topological vector space.