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
[Math] rescaled metric quantities on rescaling metrics
differential-geometryriemannian-geometry
Related Solutions
In the setting as stated in the question, this is a sort of muscle exercise, and the answer will not look nice at all. The things get a little simpler if we put $$ f=e^{2 \omega} $$
Using the Koszul formula we obtain $$ \nabla' _X Y = \nabla _X Y + (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{1} $$
Any two linear connections $\nabla'$ and $\nabla$ are related to each other by $$ \nabla' _X Y = \nabla _X Y + S(X,Y) \tag{2} $$ where $S(X,Y)$ is a (1,2)-tensor, the difference tensor of the pair of connections. In our case, the connections are torsion free, and thus the difference tensor is symmetric: $$ S(Y,X)=S(X,Y) $$
Moreover, for torsion-free connections $\nabla'$ and $\nabla$ the corresponding curvature endomorphisms are related via their difference tensor $S$ as $$ R'(X,Y)Z = R(X,Y)Z + \nabla_X S(Y,Z) - \nabla_Y S(X,Z)+S(X,S(Y,Z))-S(Y,S(X,Z)) \tag{3} $$ where by definition $$ \nabla_X S(Y,Z):= \nabla_X (S(Y,Z)) - S(\nabla_X Y,Z) - S(Y, \nabla_X Z) \tag{4} $$
Now it is straightforward to substitute $$ S(X,Y) = (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{5} $$ into (3) and obtain the expression for the conformal transformation of the curvature endomorphisms. It can be found in W.Kühnel's "Differential Geometry", p.349, and I reproduce the formula with some modifications: $$ \begin{align} R'(X,Y)Z &= R(X,Y)Z + g(\nabla_X \operatorname{grad} \omega,Z)Y - g(\nabla_Y \operatorname{grad} \omega,Z)X\\ &+ g(X,Z)\nabla_Y \operatorname{grad} \omega - g(Y,Z)\nabla_X \operatorname{grad} \omega\\ &+ (Y \omega)(Z \omega)X - (X \omega)(Z \omega)Y\\ &- g(\operatorname{grad} \omega, \operatorname{grad} \omega)[g(Y,Z)X - g(X,Z)Y]\\ &+ [(X \omega)g(Y,Z)-(Y \omega)g(X,Z)]\operatorname{grad} \omega \end{align} \tag{6} $$
Using this one can now find expressions for the Ricci curvature, the scalar curvature etc.
Formula (6) suggests the outstanding role of $\operatorname{grad} \omega$ in conformal transformations. This actually is an appearance of $d\omega$ which is canonically identified in the Riemannian geometry with $\operatorname{grad} \omega$ by virtue of the musical isomorphisms.
To get more insights one can observe that the difference tensor $S(X,Y)$ has a more symmetric presentation in the "covariant" form: $$ g(S(X,Y),Z) = d\omega (X) g(Y,Z) + d\omega (Y) g(X,Z) - d\omega (Z) g(X,Y) \tag{7} $$
From this point it is already not that far from realizing that the (0,4)-curvature tensor will have a nicer expression, and in fact it is so.
An ultimate understanding of the conformal transformation of the curvature is obtained by analyzing the algebraic properties of the curvature tensor, the direction that is better covered in the language of the representation theory.
A canonical list used in the references is given in A.Besse's "Einstein manifolds" on pp. 58-59.
As for the sectional curvature, I think that it would be a very straightforward calculation using formula (6) and the definition $$ K = \frac{g(R(X,Y)Y,X)}{g(Y,Y)X - g(X,X)Y} \tag{8} $$
I hope that this answer clarifies the things a little.
To give an answer for the sake of it being here.
The Ricci tensor is defined as usual, as the trace of the Riemann curvature tensor.
Some authors use Ricci curvature to denote the function on the unit tangent bundle: $$ UTM \ni v \mapsto \mathrm{Ric}(v,v)\in \mathbb{R} $$
The values of this function is sometimes called the Ricci curvatures.
Constant Ricci curvature at a point $p$ means that for all $v\in UT_pM$ we have $\mathrm{Ric}(v,v) = c$ for some constant $c$. In three dimensions because the Weyl curvature vanishes identically the Riemann curvature tensor is uniquely determined by the Ricci tensor, and the Ricci tensor being symmetric is uniquely determined by the Ricci curvature function. Hence you can in principle write the Riemann curvature tensor in terms of the Ricci curvature (by polarisation) and show that the sectional curvatures are constant.
Or you can use that in $\mathbb{R}^3$ the space of two-dimensional subspaces is itself three dimensional, so one can actually solve a linear system of equations to write sectional curvatures in terms of Ricci curvatures.
Best Answer
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).