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By definition $M^{2}$ is an Einstein manifold if there exists $\lambda \in C^{\infty}(M)$ such that $$\mbox{Ric}(X,Y)=\lambda\langle X, Y \rangle$$ for all $X,Y \in \Gamma(TM)$.

My question is:

Is it true that $\lambda$ can be a constant function?

I am asking it, because it is very well known that the case $n \geq 3$ is true. And the case $n=2$, is it?

Thank you for your answer.

Rodrigues
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    In dimension $2$ Ric is just the usual Gaussian curvature. When you ask, "is it true that $\lambda$ can be a constant function?" the answer is yes, but it need not be constant. – Ted Shifrin Jun 19 '20 at 21:39
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    In particular, every metric in Einstein in dimension 2 according to your definition. – Michael Albanese Jun 19 '20 at 21:41
  • I'm sorry if i am wrong. If I take $X, Y \in T_{x}M$ such that $\langle X,X \rangle=\langle Y,Y\rangle=1$ and $\langle X,Y\rangle=0$ then $\mbox{Ric}(Y,Y)=\lambda$ and therefore $$\lambda=\mbox{tr}(Z \rightarrow R(Z,Y)Y)=\langle R(X,Y)Y,X)=K(X,Y)$$. Right? Do you have an example that show that $\lambda$ is not constant. https://math.stackexchange.com/users/71348/ted-shifrin – Rodrigues Jun 19 '20 at 21:55
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    Take any surface of non-constant curvature. Have you studied surfaces in $\Bbb R^3$? – Ted Shifrin Jun 19 '20 at 22:57
  • Yes, I have. You are right. Sorry me for my silly question. – Rodrigues Jun 19 '20 at 23:10

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