Suppose $(M^n,g)$ and $(N^n,h)$ are two Riemannian manifolds with metric $g,h$. Assume there is a conformal diffeomorphism $F$ between them, say $F:M\rightarrow N$. It means that $\mu g=F^*h$ for positive function $\mu$ on $M$. Now consider a smooth function $u\in C^\infty(N)$, I am wondering if following two quantities are the same: $$\Delta_h u=\Delta_{\mu g}u\ \circ F$$
Can we check it by using local coordinates or local O.N. basis ?
What I tried after hint by Ted Shifrin, I did the calculation in one dimension, $M$ is one manifold with coordinate $t$, and $N$ is one manifold with coordiante $s$ and with metric $h=l(s)ds^2$, the metric $F^*h=l(F(t))F'(t)^2 dt^2$, then by $$\Delta_h u=\frac{1}{\sqrt{l}}\partial_s(\sqrt{l}\frac{1}{\sqrt{l}}\partial_s u(s))$$ and $$\Delta_{F^*h} u(F(t))=\frac{1}{\sqrt{l(F(t))F'(t)^2}}\partial_s(\sqrt{l(F(t))F'(t)^2}\frac{1}{\sqrt{l(F(t))F'(t)^2}}\partial_s u(F(t)))$$ At point $s=F(t)$, they are same. I realize it may just be a problem about the coordinate change. Higher dimensional case is probably the same.