I'm reading a text right now (in complex analysis -- hence the 2$n$ dimensions), and am struggling with a detail in a proof. Now in this proof we're doing a local argument near the boundary of a smoothly bounded domain $\Omega$, so we work in the coordinates of the lower half-space $(t_1,\ldots ,t_{2n-1},r)$ in $\mathbb{R}^{2n}\cong\mathbb{C}^{n}$ where $r<0$. Now in this the proof it's stated that, in terms of tangential and normal derivatives in these local coordinates, the Laplace operator can be written as $$ \Delta =\frac{\partial^2}{\partial\nu^2}+\frac{1}{2g}\frac{\partial g}{\partial\nu}\frac{\partial}{\partial\nu}+\frac{1}{\sqrt{g}}\sum_{j,k=1}^{2n-1}\frac{\partial}{\partial t_k}\left(g^{jk}\sqrt{g}\frac{\partial}{\partial t_j}\right), $$ where $\frac{\partial}{\partial \nu}=\frac{\partial}{\partial r}$ is the normal derivative, $(g_{jk})$ gives the Riemannian metric induced on the level sets of $r$ by the standard Euclidean metric, $(g^{jk})=(g_{jk})^{-1}$, and $g=\det(g_{jk})$.
However, I don't immediately see this. I imagine it's a consequence of the representation $$ \Delta= \frac{1}{\sqrt{\tilde{g}}}\sum_{j,k=1}^{2n}\frac{\partial}{\partial t_k}\left({\tilde{g}}^{jk}\sqrt{\tilde{g}}\frac{\partial}{\partial t_j}\right) $$ (where now $\tilde{g}$ is the induced metric on the lower half-space and $t_{2n}:=r$ for simplicity). I've tried to separate all of the terms in the sum involving the normal derivative but there are these extra terms that I can't seem to get rid of; for example, when trying to pull the $\frac{\partial^2}{\partial\nu^2}$ term out of the sum I get $$ \frac{1}{\sqrt{\tilde{g}}}\frac{\partial}{\partial\nu}\left(\tilde{g}^{2n2n}\sqrt{\tilde{g}}\frac{\partial}{\partial\nu}\right)=\tilde{g}^{2n2n}\frac{\partial^2}{\partial\nu^2}+\frac{\partial\tilde{g}^{2n2n}}{\partial\nu}\frac{\partial}{\partial\nu}+\frac{\tilde{g}^{2n2n}}{2\tilde{g}}\frac{\partial\tilde{g}}{\partial\nu}\frac{\partial}{\partial\nu}. $$ So I suppose I believe the assertion if I can justify to myself that $\tilde{g}^{2n2n}\equiv 1$ and $\tilde{g}^{j2n}\equiv 0$, $1\leq j\leq 2n-1$, which sort of makes sense because of the way the coordinates are chosen, but I'm not fully convinced. I'll be happy to provide more context if necessary. Any help is greatly appreciated. Thanks.