I have an issue with an exercise that concerns the behaviour of Riemannian metrics under rescaling.
My issue lies with b). The curvature tensor can be represented in local coordinates as
$$
\tilde{R}^s_{ijk}=\partial_i (\tilde{\Gamma}^s_{jk})
+\sum_l \tilde{\Gamma}^l_{jk} \tilde{\Gamma}^s_{il}
-\partial_j (\tilde{\Gamma}^s_{ik})
-\sum_l \tilde{\Gamma}^l_{ik} \tilde{\Gamma}^s_{jl}.
$$
For the Christoffel Symbols I have found that $\tilde{\Gamma}^k_{ij}=\lambda \Gamma^k_{ij}$. So an application of the product rule yields
$$
\tilde{R}^s_{ijk}
=\lambda
(\partial_i (\Gamma^s_{jk})
+\lambda \sum_l \Gamma^l_{jk} \Gamma^s_{il}
-\partial_j (\Gamma^s_{ik})
– \lambda \sum_l \tilde{\Gamma}^l_{ik} \Gamma^s_{jl})
$$
I am dissatisfied with this result as I would have expected that there would be some sort of relation between $\tilde{R}^s_{ijk}$ and $R^s_{ijk}$. So I am wondering whether there is an error in this.
\Edit: The Christoffel symbols can be computed with the Christoffel formula
$$
\tilde{\Gamma}^k_{ij}
=\frac{1}{2} (\partial_i \tilde{g}(\partial_j, \partial_k))
+\partial_j \tilde{g}(\partial_k, \partial_i)
-\partial_k \tilde{g}(\partial_i,\partial_j))
$$
Since $\lambda$ is a constant factor, we have
$$
\partial_i \tilde{g}(\partial_j, \partial_k)
=\lambda \partial_i g(\partial_j, \partial_k)
$$
and analogously for the other summands. So $\tilde{\Gamma}^k_{ij}=\lambda \Gamma^k_{ij}$.
Best Answer
Let $\tilde \nabla$ be the Levi-Civita connection for $(M,\tilde g)$. Using the definition of the LC connection on a Riemannian manifold as the unique affine connection that is compatible with the metric and symmetric, this suggests we look at these relations for $(\tilde \nabla,\tilde g)$ more closely. By the definition of metric compatibility, we have \begin{align*} \tilde\nabla \tilde g(X,Y) &= \tilde g(\tilde\nabla X,Y) + \tilde g(X,\tilde\nabla Y),\\ \text{hence} \quad \lambda \big[\tilde\nabla g(X,Y)\big] &= \lambda \big[ g(\tilde\nabla X,Y) + g(X,\tilde\nabla Y)\big],\\ \text{therefore,} \quad \tilde\nabla g(X,Y) &= g(\tilde\nabla X,Y) + g(X,\tilde\nabla Y). \end{align*} Thus we see $\tilde\nabla$ is compatible with $g$. Since $\tilde \nabla$ is symmetric by definition, we see that $\tilde\nabla$ satisfies the defining properties of the LC connection $\nabla$ for $(M,g)$, and hence $$\tilde \nabla = \nabla.$$