From what I understand, the Ricci flow
\begin{equation} \frac{\partial g}{\partial t} = -2Ric(g) \end{equation}
always defines a Riemannian metric as long as it exists. I know that the Riemannian condition is an open condition in the space of symmetric two-tensors, but is that all there is to this statement? I know that certain curvature-related quantities must blow up at the first singular time for the Ricci flow, but may this also be understood in terms of when the two-tensor solution for the Ricci flow looses its definiteness?