I know that type II singularities of the Ricci flow can exist on closed 3-manifolds (e.g. on $S^3$), but on the other hand it seems to me that ODE comparison combined with Hamilton's tensor maximum principle implies that in 3D Ricci flow, there can be no type II singularities.
Please explain where the following argument is flawed:
In 3D Ricci flow, the Ricci tensor evolves by $$ \frac{\partial R_{ij}}{\partial t}=\Delta R_{ij}-Q_{ij}, $$ where $Q_{ij}=6R_{ij}^2-3RR_{ij}+(R^2-2g^{kl}R_{kl}^2)g_{ij}$ (see Hamilton's 1982 paper, section 8). Writing the Ricci tensor as a diagonal matrix $R_{ij}=\mathrm{diag}(\lambda,\mu,\nu)$, $Q_{ij}$ is also diagonal and the top left entry is $$ 2\lambda^2-\mu^2-\nu^2-\lambda\mu-\lambda \nu+2\mu\nu, $$ The other entries are obtained by permuting $\lambda, \mu,\nu$.
Claim 1: Now, for any solution of the ODE system $$ \frac{d}{dt}(\lambda,\mu,\nu)=-Q(\lambda,\mu,\nu), $$ there exist constants $C,T>0$ (determined by the initial conditions) such that the solution can blow up no faster than $\frac{C}{T-t}$ (the type I rate). (This is because $Q$ is quadratic in $\lambda,\mu,\nu$.)
Define the closed subset $\mathcal{Z}$ of the product space of $3\times3$ symmetric matrices (of Ricci curvature tensors with eigenvalues $\lambda\geq \mu\geq \nu$) and the time axis by the inequalities $$ \lambda\leq\frac{C}{T-t}, \qquad \qquad \nu\geq -\frac{C}{T-t}. $$
For suitable choices $C,T>0$ (depending only on the initial conditions), $\mathcal{Z}$ is preserved by the ODE, is closed, and is invariant under parallel transport (it is defined by conditions on eigenvalues). Moreover, for any fixed $t$, the time-slice $\mathcal{Z}(t)$ is convex. (Hamilton 1999 constructs a similar $\mathcal{Z}$ in his proof of the Hamilton-Ivey pinching estimate.)
Claim 2: By Hamilton's tensor maximum principle, the Ricci flow preserves the set $\mathcal{Z}$. Therefore the eigenvalues of the Ricci tensor under 3D Ricci flow cannot blow up faster than the type I rate.
