Given an Einstein manifold with constant negative Ricci curvature, so $Ric = \lambda g$ with $\lambda <0$, if I change the value of $\lambda$, the solution will only be the one given by its rescaled metric, or it is not always so simple, because there can be more Einstein manifolds corresponding to that value of $\lambda$ and not only the one to which it is sufficient to rescaling the metric?
Einstein manifold varying the $\lambda$ value
curvaturedifferential-geometryriemannian-geometry
Best Answer
First of all, if $g$ is Einstein with Einstein constant $\lambda$, then $\mu\cdot g$ is Einstein with Einstein constant $\frac{\lambda}{\mu}$, then you are right while saying that an Einstein metric gives solution up to rescaling to any Einstein equation.
But there is no need for this metric to be unique. For example, on a compact surface of genus $\gamma \geqslant 2$, there exist many non-conformal hypebolic metrics. They are all Einstein as they are all of constant sectionnal curvature $-1$.
More generally, if there exist two different (say, non-conformal) metric $g_1$ and $g_2$ that are Einstein on $M$ with negative constant, that is $Ric_{g_i} = \lambda_ig_i$and $\lambda_i<0$, the equation $$Ric_g = \Lambda g,~~~\Lambda <0$$ has at least two solution, which are are $\frac{\lambda_i}{\Lambda}g_i$.