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:

$$\|\text{Rm}(x, t)\| t = \frac{t}{1+t} \|\text{Rm}(x,0) \| \ \forall (x, t) \in \mathcal{M} \times [0, \infty)$$

But I don't understand how we could bound this. Indeed, since [there are examples of Einstein manifolds with unbounded curvature](https://mathoverflow.net/questions/401683/are-there-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.