# rescaled metric quantities on rescaling metrics

I have the following basic, surely stupid, questions. Assume we have a Riemannian metric $g$ on a manifold $M$. let $a\in\mathbb{R}$ a constant and consider the metric $g_1=ag$. Which are the transformation rules for the scalar and sectional curvature and for the Ricci tensor? For the sectional I guess are the same $K_g(\pi)=K_{g_1}(\pi)$, but what about sectional and Ricci tensor. Moreover how does change the metric on tensors: is it true that $g_1(v,w)=a^{l-m}g(v,w)$ if $v,w$ are $(l,m)$-tensors? Thank you

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Warning: If $a \leq 0$, the tensor field $g_1 = a g$ will not be a metric and the notion of 'curvature' is not meaningful for $g_1$. However, for $a > 0$ the question is valid, even if $a$ is a smooth positive function on $M$. This is called a 'conformal change' and you will find many transformation laws under this keyword, for instance http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Under_a_conformal_change
If you look here: http://en.wikipedia.org/wiki/Levi-Civita_connection#Christoffel_symbols you'll see that the Levi-Civita connection for both metrics are the same. Therefore the curvature tensor is the same. Since everything else you mentioned is in terms of the curvature tensor and metric, its easy to see what happens. For example, $K_{g_1} = \frac{1}{a} K_g$.
You also have to be a little careful since the curvature tensors are the same as $(3,1)$ tensors but if you raise or lower indices you will get different tensors (since you're using different metrics to raise/lower).
@unk2: $S_{g_1} = \frac{1}{a} S_g$. –  Eric O. Korman May 5 '11 at 15:59