I am trying to prove two formulations of the Laplace-Beltrami operator $\Delta$ are equivelant: $$\Delta f = \frac{1}{\sqrt{|g|}}\frac{\partial}{\partial x^i}\left(\sqrt{|g|} g^{ij} \frac{\partial f}{\partial x^j}\right) = g^{ij} \left(\frac{\partial^2 f}{\partial x^i \partial x^j} - \Gamma^k_{ij}\frac{\partial f}{\partial x^k}\right) $$ Where I use the Einstein summation convention and $|g|$ refers to the determinant of the metric. I've attempted to follow steps given in the top answer to this post.
However, I can't figure out how to get rid of the $\frac{\partial g_{pq}}{\partial x^i} g^{pq}$ given in the last step.
Any help in proving the equivelance, or how to get rid of the $\frac{\partial g_{pq}}{\partial x^i} g^{pq}$ would be much appreciated!
Thanks
Let me finish the calculations in the linked answer (but the last line there doesn't make sense to me). To save some typing, I will write $\partial_k = \frac{\partial}{\partial x^k}$.
We continue with the calculation from the second last line \begin{align} &\quad \frac{1}{\sqrt{\det g}}\partial_i \big(g^{ij} \sqrt{\det g}\partial_j f\big)\\ &= g^{ij}\partial_{ij} f + (\partial_i g^{ij})\partial_j f + \frac{1}{2}g^{ij}(\partial_i g_{pq})g^{pq}\partial_j f\\ &=g^{ij}\partial_{ij} f + (\partial_i g^{ik})\partial_k f + \frac{1}{2}g^{ik}(\partial_i g_{pq})g^{pq}\partial_k f\quad \text{relabel by }k\\ &=g^{ij}\partial_{ij} f - g^{ip}(\partial_i g_{pq})g^{qk}\partial_k f + \frac{1}{2}g^{ik}(\partial_i g_{pq})g^{pq}\partial_k f \quad\text{differentiate the inverse matrix}\\ &=g^{ij}\partial_{ij} f - g^{qp}(\partial_q g_{pi})g^{ik}\partial_k f + \frac{1}{2}g^{ik}(\partial_i g_{pq})g^{pq}\partial_k f \quad\text{in the second term, }i\leftrightarrow q\\ &=g^{ij}\partial_{ij} f - g^{pq} \frac{1}{2}g^{ik}(\partial_q g_{pi} + \partial_p g_{qi} -\partial_i g_{pq})\partial_k f \qquad p \text{ and }q\text{ are symmetric by }g^{pq}\\ &=g^{ij}\partial_{ij} f - g^{pq}\Gamma_{pq}^k\partial_k f \qquad\qquad\qquad\qquad\quad\text{formula for Christoffel symbols}\\ &=g^{ij}\big(\partial_{ij} f-\Gamma_{ij}^k\partial_k f\big). \end{align} The most tricky part may be the equation $$ g^{pq}(\partial_q g_{pi})=\frac{1}{2}g^{pq}(\partial_q g_{pi}+\partial_p g_{qi}), $$ by the symmetry of $g^{pq}$. (If you are not convinced, write out the the right two terms and do $p\leftrightarrow q$ in the second.)