Context:
A solution $(M^n, g(t))$ of the Ricci flow is said to encounter a Type III Singularity if $g(t)$ is defined for all $t \geq 0$ and: $$ \sup _{\mathcal{M}^{n} \times[0, \infty)} \|\operatorname{Rm}(\cdot, t) \| t<\infty $$ Similarly, if $g(t)$ is defined for all $t \geq 0$ but: $$ \sup _{\mathcal{M}^{n} \times[0, \infty)} \|\operatorname{Rm}(\cdot, t) \| t= \infty $$ the solution is said to encounter a Type IIb singularity.
This was taken from the book "The Ricci Flow: An Introduction", by Bennett Chow and Dan Knopf. There, they claim that if $(\mathcal{M}, g_0)$ is an Einstein manifold with $\text{Ric}_{g_0} = \lambda_0 g_0$ and with $g(t) = (1- 2\lambda_0 t)g_0$ as a solution to its Ricci flow, then a Type III singularity happens when $\lambda_0 < 0$ and a Type IIb singularity happens when $\lambda_0 = 0$. Now, in the case $\lambda_0 = 0$, since we have $$\|\text{Rm}(x, t)\|t = t \|\text{Rm}(x,0) \| \ \forall (x, t) \in \mathcal{M} \times [0, \infty)$$ and given that for each $x_0 \in M$, it's clear that $\{\|\text{Rm}(x_0, t)\|t \ \vert \ t \geq 0 \}$ is unbounded, I can see how this case is indeed a Type IIb singularity.
But in the case $\lambda_0 < 0$, we have (assuming without loss of generality that $\lambda_0 = -\frac{1}{2}$):
$$\|\text{Rm}(x, t)\| t = \frac{t}{1+t} \|\text{Rm}(x,0) \| \ \forall (x, t) \in \mathcal{M} \times [0, \infty)$$
But I don't understand how we could bound this. Indeed, since there are examples of Einstein manifolds with unbounded curvature, it seems the authors made a mistake. And actually I can't find any other source where Einstein manifolds with negative scalar curvature are classified as a Type III singularity. So I have two questions:
Am I right and have the authors indeed made a mistake in classifying Einstein solutions with negative scalar curvatue as Type III singularities? Or is there something else going on here I'm not seeing?
What's even the point of classifying a constant (in time) solution as a singularity, like they do for solutions with $\lambda_0 = 0$? This doesn't make sense to me. The solution is eternal and it falls short of what I intuitively think a singularity should be.
