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Let $g(0)=g_0$ and $Ric(g_0)=\lambda g_0,\lambda\in\mathbb{R}$, the $Ric(g)$ is the Ricci curvature,$g$ is Riemannian metric. How to show that :

The $g(t)=(1-2\lambda t)g_0$ is a solution of $$ \frac{\partial g}{\partial t}=-2Ric(g) $$

Thanks for any detail answer or hint.

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When you multiply the metric by a factor, the Ricci doesn't scale. One way to see this is to look at its expression in the normal coordinates given here.

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    $\begingroup$ As an alternate, one can look at the usual expression for the Levi-Civita connection coefficients, from which it's clear that the connection is scale independent. The (3,1)-curvature tensor is formed just from the connection, so is also scale independent, and the Ricci is just a metric-independent trace of this, which shows Ricci is scale independent. $\endgroup$ – Brian Klatt Sep 30 '15 at 19:12

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