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I am doing this problem:

Let $g$ be a Riemannian metric on manifold $M$ and $\tilde{g}=e^{2f}g$ another metric conformal to $g$, where $f$ is a smooth function on $M$. Give the relation between the Levi-Civita connection $\nabla$ of $g$ and the Levi-Civita connection $\tilde{\nabla}$ of $\tilde{g}$.

I am currently at this step, but don't know how to continue next:

$2\tilde{g}(\tilde{\nabla}_{X}Y,Z)-2e^{2f}g(\nabla_{X}Y,Z)=X(e^{2f}g(Y,Z))+Y(e^{2f}g(Z,X))-Z(e^{2f}g(X,Y))-e^{2f}Xg(Y,Z)-e^{2f}Yg(Z,X)+e^{2f}Zg(X,Y)$

Any help is appreciated!

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  • $\begingroup$ I posted this calculation a while ago here $\endgroup$ – Yuri Vyatkin Sep 14 '18 at 13:53
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A connection of the conformal metric is determined, as is any connection, by its Christoffel symbols. We have the following formula for Levi-Civita connections:

$$\Gamma_{ij}^k = \frac{1}{2}g^{kl}(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij})$$

Another pretty obvious fact is that $\hat g^{ij} = e^{-2f}g^{ij}$. So for the conformal metric $\hat g$ we can compute the Christoffel symbols: $$ \hat \Gamma_{ij}^k = \frac{1}{2}e^{-2f}g^{kl}(\partial_i (e^{2f}g_{jl}) + \partial_j (e^{2f}g_{il}) - \partial_l (e^{2f}g_{ij})) $$ Now it's just a matter of applying the product rule and using $e^{-2f}e^{2f} = 1$, $g_{ab}g^{bc} = \delta_a^c$, and $g^{ab}\partial_b f = \nabla^a f$ to get:

$$ \hat \Gamma_{ij}^k = \Gamma_{ij}^k + (\partial_i f) \delta_j^k + (\partial_j f)\delta_i^k - (\nabla^k f)g_{ij} $$

If we apply $\hat \nabla$ to any tensor $H^{b_1 ...b_k}_{c_1 ... c_l}$ we have:

$$ \hat \nabla_a H^{b_1 ...b_k}_{c_1 ... c_l} = \partial_a H^{b_1 ...b_k}_{c_1 ... c_l} + \sum_{i=1}^{k} \hat \Gamma_{a p}^{b_i} H^{b_1 ... p ...b_k}_{c_1 ... c_l} - \sum_{j=1}^{l} \hat \Gamma_{a c_j}^{q} H^{b_1 ...b_k}_{c_1 ... q ... c_l} $$

Where $p$ and $q$ are in the i-th and j-th positions, respectively. Hopefully that helps.

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  • $\begingroup$ Welcome to MathSX. Thank you for this answer. It may help some users. $\endgroup$ – Yuri Vyatkin Mar 15 at 9:33
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I'll register a shorter coordinate-free proof. Let the Koszul formula be written as $$2g(\nabla_XY,Z) = A(X,Y,Z) + B(X,Y,Z),$$where $A$ is the part of the Koszul formula containing directional derivatives and $B$ is the part containing Lie brackets. We know that any two connections differ by a tensor, so write $\widetilde{\nabla}_XY = \nabla_XY + T_XY$ -- the goal is to find $T$, and we know that $$2\widetilde{g}(\widetilde{\nabla}_XY,Z) = \widetilde{A}(X,Y,Z) + \widetilde{B}(X,Y,Z).$$Clearly $\widetilde{B}(X,Y,Z) = {\rm e}^{2f}B(X,Y,Z)$, while $$ X\widetilde{g}(Y,Z) = X({\rm e}^{2f})g(Y,Z) + {\rm e}^{2f}X(g(Y,Z))$$says that $\widetilde{A}(X,Y,Z) = X({\rm e}^{2f})g(Y,Z) + Y({\rm e}^{2f})g(X,Z) - Z({\rm e}^{2f})g(X,Y) + {\rm e}^{2f}A(X,Y,Z)$. Thus $$2\widetilde{g}(\widetilde{\nabla}_XY,Z) = X({\rm e}^{2f})g(Y,Z) + Y({\rm e}^{2f})g(X,Z) - Z({\rm e}^{2f})g(X,Y) + 2{\rm e}^{2f}g(\nabla_XY,Z).$$Evaluating $X({\rm e}^{2f}) = 2{\rm e}^{2f}\,X(f)$, etc., and simplyfying $2{\rm e}^{2f}$ on everything, we get $$g(\nabla_XY + T_XY,Z) = X(f)g(Y,Z) + Y(f)g(X,Z) - Z(f)g(X,Y) + g(\nabla_XY,Z).$$Eliminate $\nabla_XY$ from the above and use the definition of $g$-gradient to write the right side in the form $g({\rm something}, Z)$, obtaining $$g(T_XY,Z) = g(X(f)Y + Y(f)X - g(X,Y){\rm grad}(f), Z).$$This means that $$T_XY = X(f)Y + Y(f)X - g(X,Y){\rm grad}(f)$$and hence $$\widetilde{\nabla}_XY = \nabla_XY +X(f)Y + Y(f)X - g(X,Y){\rm grad}(f).$$

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