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Let $g$ be a Riemannian metric on $M$ and let $\tilde{g}=f^{2}g$ where $f$ is a smooth function that is never zero. let $\nabla$ and $\nabla'$ be the Riemannain connections of $g$ and $\tilde{g}$ on $M$, give the relation between $\nabla$ and $\nabla'$.

So I'm not really sure what to do here but I tryed to express each connection in terms of its Christoffel symbols and locall expression for a connection

What I get when I do that is the following let $\Gamma^{k}_{i,j}$ and $'\Gamma^{k}_{i,j}$ denote the Christoffel of each connections then I get that $$'\Gamma^{k}_{i,j}=\Gamma^{k}_{i,j}+\dfrac{1}{f}g^{k,l}(g_{i,j}\partial_{i}f+g_{i,l}\partial_{j}f-g_{i,j}\partial_{l}f) $$

Now when I plug this into the local expression for the connection of $\nabla'$ I get

$$ \nabla'_{X}Y=\nabla_{X}Y + big ~mess$$

Is there a better way of doing this?

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    $\begingroup$ You can get the Levi-Civita connection from the metric using the Koszul formula, and I actually think that would be a bit cleaner. Have you tried using this? $\endgroup$ Commented Feb 3, 2014 at 2:03
  • $\begingroup$ No I haven't I'm not that familiar with the Koszul formula but thanks for the tip I will try this out. $\endgroup$ Commented Feb 3, 2014 at 2:09
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    $\begingroup$ I assume you're not used to the differential forms/moving frames approach. BTW, such conformal changes of metric are usually written $\tilde g = e^{2\phi}g$ to avoid the logarithmic derivatives that show up this way. $\endgroup$ Commented Feb 3, 2014 at 2:46
  • $\begingroup$ I do know what differential forms and local frame are but I don't know what moving frames are, sorry. But what is this moving frames approach? $\endgroup$ Commented Feb 3, 2014 at 3:43

1 Answer 1

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The formula for the conformal rescaling of the Levi-Civita connection is an essential tool in Riemannian geometry, and its derivation is given in many sources.

As Isaac Solomon and Ted Shifrin have mentioned in the comments, a slick way to derive it is to consider $f = e^{\omega}$ and use the Koszul formula. The result will be in the form: $$ \nabla' _X Y = \nabla _X Y + (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{1} $$

Proof. The Koszul formula (see e.g. here) gives the following expression for the Levi-Civita connection $\nabla$ of the metric $g$: $$ \begin{align} 2 g(\nabla_X Y, Z) & = X \, g(Y,Z) + Y \, g(Z,X) - Z \, g(X,Y) \\ \tag{2} &- g(X,[Y,Z]) + g(Y,[Z,X]) + g(Z,[X,Y]) \end{align} $$

Let $\nabla'$ be the Levi-Civita connection for the metric $g' = e^{2\omega}g$. Substituting these objects into (2) $$ \begin{align} 2 e^{2 \omega} g(\nabla'_X Y, Z) & = X \left( e^{2 \omega} g(Y,Z) \right) + Y \left( e^{2 \omega} g(Z,X) \right) - Z \left( e^{2 \omega} g(X,Y) \right) \\ &- e^{2 \omega} g(X,[Y,Z]) + e^{2 \omega} g(Y,[Z,X]) + e^{2 \omega} g(Z,[X,Y]) \end{align} $$ and computing the derivatives using the product rule, we obtain $$ \begin{align} 2 e^{2 \omega} g(\nabla'_X Y, Z) & = e^{2 \omega} X g(Y,Z) + e^{2 \omega} Y g(Z,X) - e^{2 \omega} Z g(X,Y) \\ & + 2 e^{2 \omega} g(Y,Z) \, X \omega + 2 e^{2 \omega} g(Z,X) \, Y \omega - 2 e^{2 \omega} g(X,Y) \, Z \omega \\ &- e^{2 \omega} g(X,[Y,Z]) + e^{2 \omega} g(Y,[Z,X]) + e^{2 \omega} g(Z,[X,Y]) \end{align} $$

In the last display we can divide both sides of the equation by $e^{2 \omega}$, which is a strictly positive function, to get $$ \begin{align} 2 g(\nabla'_X Y, Z) & = X g(Y,Z) + Y g(Z,X) -Z g(X,Y) \\ & + 2 g(Y,Z) \, X \omega + 2 g(Z,X) \, Y \omega - 2 g(X,Y) \, Z \omega \\ &- g(X,[Y,Z]) + g(Y,[Z,X]) + g(Z,[X,Y]) \end{align} $$

Using the Koszul formula (2) again we rewrite the above expression as $$ \begin{align} 2 g(\nabla'_X Y, Z) & = 2 g(\nabla_X Y, Z) + 2 g(Y,Z) \, X \omega + 2 g(Z,X) \, Y \omega - 2 g(X,Y) \, Z \omega \end{align} $$ which is equivalent to (1) because vector field $Z$ is arbitrary, $g$ is non-degenerate, and $Z \omega = \mathrm{d} \omega (Z)$. Recall also that $\operatorname{grad} \omega = (\mathrm{d} \omega)^{\sharp}$.


The version of this formula in terms of coordinates and the Christoffel symbols is obtained by a similar calculation, the result will be $$ '\Gamma^{k}_{ij}=\Gamma^{k}_{ij} + \delta_{i}^{k} \partial_j \omega + \delta_{j}^{k} \partial_i \omega - g_{i j} g^{k l} \partial_{l} \omega $$

This can be also obtained as a consequence of (1).

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  • $\begingroup$ Thanks for this it was very helpful. $\endgroup$ Commented Feb 4, 2014 at 21:16
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    $\begingroup$ Those who prefer Christoffel symbols may take a look at this answer. $\endgroup$ Commented Mar 15, 2019 at 9:36

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