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Suppose we have a Riemannian manifold ${(M,g)}$, where ${g}$ is the metric of ${M}$. If ${f}$ ${\in}$ ${D(M)}$ (i.e. smooth function on ${M}$), and ${f}$ is positive. So, we can define a new metric ${g'}$ = ${fg}$ (i.e. for given any vector field ${X}$ and ${Y}$ on ${M}$, ${g'(X,Y)} = { f g(X,Y)}$.

I already checked that the new connection $\nabla'$ associated with ${g'}$, we have:

${\nabla'}_X Y$ = ${\nabla}_X Y + S(X,Y)$

where S(X,Y) = $(1/f)$ $[$ ${X(f)Y+(Yf)X-g(X,Y)\operatorname{grad}(f)}$ $]$

and $\operatorname{grad}(f)$ satisfies $X(f) = g(X,\operatorname{grad}(f))$

(Actually, it is an exercise in do Carmo "Riemannian Geometry" , so I just follow the Hint)

Now, I am trying to comput new curvature tensor, sectional curvature, Ricci curvature and scalar curvature.

For curvature tensor $R'$,

My result is $$ \begin{eqnarray} R'(X,Y)Z &=& R(X,Y)Z + {\nabla}_Y S(X,Y) - {\nabla}_X S(Y,Z) \\ &+& S(Y,S(X,Z)) - S(X,S(Y,Z)) \\ &+& S(X,{\nabla}_X Z) - S(X,{\nabla}_Y Z) + S([X,Y],Z) \end{eqnarray} $$

And I don't know how can I do in next step.

How about sectional curvature, Ricci curvature and scalar curvature? Everybody who can give me some hint? Thanks !

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You are nearly there. The last three terms in your expression for the transformed curvature should vanish since the connections are implied to be the Levi-Civita connections of the corresponding metrics, that is torsion free. Thus your formula is correct, and you only need to apply the definitions of the remaining quantities. –  Yuri Vyatkin Jan 11 '12 at 19:53
1  
It would make sense to rename the question as "Conformal transformation of the curvature and related quantities" –  Yuri Vyatkin Jan 11 '12 at 20:03
    
Please notice that $S(X,Y) = (1/2f)[X(f)Y+(Yf)X-g(X,Y)\operatorname{grad}(f)]$ as follows from the Koszul formula –  Yuri Vyatkin Jan 13 '12 at 0:17
    
What is $\nabla_Y S(X,Y)$ in your last formula? I think that you mean $\nabla_Y (S(X,Y))$. And please correct the typos there: it must be $\nabla_Y (S(X,Z))$. –  Yuri Vyatkin Jan 13 '12 at 0:20
    
Did you look at this: en.wikipedia.org/wiki/…? –  Paul Jan 14 '12 at 5:28

1 Answer 1

up vote 6 down vote accepted

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.

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Thank you very much! I will follow your suggestion! :) –  Peter Hu Feb 1 '12 at 14:40

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