Showing two forms on a manifold are equal Let $\alpha$ and $\beta$ be two forms on a manifold $M$. To show that they are equal, does it suffice to show that for arbitrary $p\in M$ there exists some chart such that $\alpha_p=\beta_p$. I was told that this is not sufficient. For example, for $f\in C^\infty(M)$, in normal coordinates $x^1,\dots,x^n$ centered at $p$, only at $p$ do we have that the laplacian $\Delta f$ is $$\Delta_pf=\sum_{i,j}g^{ij}(p)\frac{\partial^2 f}{\partial x_i\partial x_j}.$$But I was told this equality is only at $p$ and it is not true that as functions on $M$ these two are equal. I don't understand why not, don't they agree at each point meaning they're the same? What confuses me further is that I was told on a complex manifold it is true that as global functions $$\Delta f=2\sum_{j,k}g^{j\overline k}\frac{\partial^2 f}{\partial z_j\partial \overline z_k}.$$Does the above argument work in normal coordinates on a complex manifold for some reason?
 A: The issue is that the normal coordinates $x^i$ depend on $p$ - I'll denote (some choice of) normal coordinates around $p$ by $x_p^i$ for clarity. 
It's certainly true that at each $p$ your formula holds, so you do have an equation of functions; but the function on the RHS has a lot of dependence on coordinates: if you write all the dependence out explicitly, you get the very ugly
$$ \Delta f|_p = \sum_{i,j} g^{-1}|_p(dx_p^i,dx_p^j) \frac{\partial^2 f}{\partial x_p^i \partial x_p^j}\Big|_p.$$
This is a useful formula for computing the Laplacian at a single point, but if you are interested in it as a global function (or even as a function in a small neighbourhood) then it is pretty much useless - by the time you express the $x_p$ in terms of a single coordinate system you would have been much better off just using the generic formula for the Laplacian.
Regarding the second equation, I think this is only true when the metric is Kähler, which is a very strong condition tying the complex structure to the Riemannian structure. Hopefully someone more versed in complex differential geometry can confirm or correct this.
