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. 



In which space are we solving the Kähler Ricci flow?
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.


