Are there any known examples of Einstein manifolds $(M, g)$ such that $$\sup\_{x \in M} \|\text{Rm}(x) \| = \infty$$
I'm looking for these examples because they might provide a counter-example to a problem of mine, but I can't think of any. Obviously such a manifold can't be compact (and therefore it can't be complete with positive scalar curvature) but I haven't been able to think of any concrete example.



Are there examples of Einstein manifolds with unbounded curvature?
If you don't care about completeness, here's a fairly simple way to construct such examples: Start with a compact Einstein manifold $(M^n,g)$ with Einstein constant $1$ (i.e., $\mathrm{Ric}(g) = (n{-}1)\,g$) that is not conformally flat. Now take the sine-cone, i.e., $\bigl(M\times(0,\pi),h\bigr)$ where $h = \mathrm{d}r^2 + (\sin r)^2\,g$. Then one easily computes that this is also an Einstein manifold with Einstein constant $1$, i.e., $\mathrm{Ric}(h) = n\,h$, but the Weyl curvature of $h$ (which is nonzero since $g$ is not conformally flat) blows up as $r$ approaches either $0$ or $\pi$.


(Also, if one just takes the ordinary cone, $\bigl(M\times(0,\infty),h\bigr)$ where $h = \mathrm{d}r^2 + r^2\,g$, then $h$ will be Ricci-flat, but the Weyl curvature of $h$ will blow up as $r\to 0$.)


Finally, as Anton Petrunin pointed out in the comment below, if $(M,g)$ is complete and Ricci-flat (i.e., Einstein with Einstein constant 0), but not conformally flat (equivalently, not flat), the Riemannian manifold $\bigl(M\times\mathbb{R}, h = \mathrm{d}r^2 + \mathrm{e}^{2r}\,g\bigr)$ will be a complete Einstein manifold with Einstein constant $-1$ whose Weyl curvature has unbounded norm as $r\to-\infty$.


