We say that two metrics are affinely equivalent if their Levi-Civita connections coincide. Is it possible that an Einstein (=Ricci tensor is proporional to the metric) is affinely equivalent to a metric which is not Einstein?

Of course, since affinely equivalent metrics have the same Ricci tensor, the question is equivalent to the question whether there exists a metric $g$ such that

(1) its Ricci tensor $Ric$ is nondegenerate (as a bilinear form),

(2) is parallel, $\nabla^g Ric= 0$,

(3) and is not proportional to the metric.

The answer is negative for metrics of Riemannian and Lorentzian signature. In both cases the result follows from the description of affinely equivalent metrics in these signatures and it is hard to generalize it for general signature.

The motivation comes from projective geometry. It is known (Mikes MR0603226 or Kiosak-M arXiv:0806.3169) that if two metrics are projectively, but not affinely equivalent, and one of them is Einstein, then the second is Einstein as well. It is interesting and important to understand whether the assumption that the metrics are not affinely equivalent is necessary.

\Edited after the comment of Robert Bryant: Let us assume in addition that the metric is not the direct product of Einstein metrics

We say that two metrics are affinely equivalent if their Levi-Civita connections coincide. Is it possible that an Einstein (=Ricci tensor is proporional to the metric) is affinely equivalent to a metric which is not Einstein?

Of course, since affinely equivalent metrics have the same Ricci tensor, the question is equivalent to the question whether there exists a metric $g$ such that

(1) its Ricci tensor $Ric$ is nondegenerate (as a bilinear form),

(2) is parallel, $\nabla^g Ric= 0$,

(3) and is not proportional to the metric.

The answer is negative for metrics of Riemannian and Lorentzian signature. In both cases the result follows from the description of affinely equivalent metrics in these signatures and it is hard to generalize it for general signature.

The motivation comes from projective geometry. It is known (Mikes MR0603226 or Kiosak-M arXiv:0806.3169) that if two metrics are projectively, but not affinely equivalent, and one of them is Einstein, then the second is Einstein as well. It is interesting and important to understand whether the assumption that the metrics are not affinely equivalent is necessary.

Edited after the comment of Robert Bryant: Let us assume in addition that the metric is not the direct product of Einstein metrics

We say that two metrics are affinely equivalent if their Levi-Civita connections coincide. Is it possible that an Einstein (=Ricci tensor is proporional to the metric) is affinely equivalent to a metric which is not Einstein?

Of course, since affinely equivalent metrics have the same Ricci tensor, the question is equivalent to the question whether there exists a metric $g$ such that

(1) its Ricci tensor $Ric$ is nondegenerate (as a bilinear form),

(2) is parallel, $\nabla^g Ric= 0$,

(3) and is not proportional to the metric.

The answer is negative for metrics of Riemannian and Lorentzian signature. In both cases the result follows from the description of affinely equivalent metrics in these signatures and it is hard to generalize it for general signature.

The motivation comes from projective geometry. It is known (Mikes MR0603226 or Kiosak-M arXiv:0806.3169) that if two metrics are projectively, but not affinely equivalent, and one of them is Einstein, then the second is Einstein as well. It is interesting and important to understand whether the assumption that the metrics are not affinely equivalent is necessary.

We say that two metrics are affinely equivalent if their Levi-Civita connections coincide. Is it possible that an Einstein (=Ricci tensor is proporional to the metric) is affinely equivalent to a metric which is not Einstein?

Of course, since affinely equivalent metrics have the same Ricci tensor, the question is equivalent to the question whether there exists a metric $g$ such that

(1) its Ricci tensor $Ric$ is nondegenerate (as a bilinear form),

(2) is parallel, $\nabla^g Ric= 0$,

(3) and is not proportional to the metric.

The answer is negative for metrics of Riemannian and Lorentzian signature. In both cases the result follows from the description of affinely equivalent metrics in these signatures and it is hard to generalize it for general signature.

The motivation comes from projective geometry. It is known (Mikes MR0603226 or Kiosak-M arXiv:0806.3169) that if two metrics are projectively, but not affinely equivalent, and one of them is Einstein, then the second is Einstein as well. It is interesting and important to understand whether the assumption that the metrics are not affinely equivalent is necessary.

\Edited after the comment of Robert Bryant: Let us assume in addition that the metric is not the direct product of Einstein metrics