I am interested in this statement ( taken from Arthur L. Besse - Einstein Manifolds ):
14.39 Theorem(M. Berger [ Ber 7]). A quaternion-Kähler manifold (M,g) is Einstein (in dimension 4n $\ge$ 8 ).
They have proved that the Ricci tensor is proportional to the metric, let's say:
$$ R_{\mu \nu} = \lambda g_{\mu \nu} $$
Now, by definition, quaternion-Kähler should not have zero Ricci tensor. But they haven't proved that $\lambda$ has to be non-zero.
I think that I am just adhering to the terminology too strictly. In the aforementioned statement, the quaternion-Kähler manifold can turn out to be Ricci flat. The statement is not trying to prove that the Ricci tensor is necessarily non-zero.
Can someone please confirm or correct my understanding? ( I am a physics student who is new to the manifold theory ).
According to the definition Besse gives, a $4n$-dimensional Riemannian manifold $(M, g)$ with $n \geq 2$ is called Quaternion-Kähler if $\operatorname{Hol}(g) \subseteq Sp(n)\cdot Sp(1)$. Berger's Theorem states that $g$ must then be an Einstein metric. Note, a Quaternion-Kähler manifold can have Einstein constant $\lambda = 0$. For an extreme example, consider flat Euclidean space $(\mathbb{R}^{4n}, g_{\text{Eucl}})$ which has $\operatorname{Hol}(g_{\text{Eucl}}) = \{\operatorname{id}\} \subset Sp(n)\cdot Sp(1)$.
However, if $\lambda = 0$, then $(M, g)$ is not only Quaternion-Kähler, it is also locally hyperKähler, see Theorem $14.45$ (also due to Berger). So if $\lambda = 0$ and $M$ is simply connected, then $\operatorname{Hol}(g) \subseteq Sp(n)$.