This question is about how the principal part (or symbol) is defined on a manifold?-I assume that the answer is: As in $\mathbb{R}^n$ using local coordinates, i.e.


A differential operator $P=\sum\_{|\alpha|\le m} a\_{\alpha}(x)D^{\alpha}$ has by definition the symbol $P\_m(x,\xi)=\sum\_{|\alpha|=m} a\_{\alpha}(x)\xi^{\alpha}.$


Now, I would like to understand this in the case of $\Delta\_{\mathbb{S}^2}$ on the two-sphere. Recall that 
$$\Delta\_{\mathbb{S}^2}=\left(\frac{\partial^{2}}{\partial\vartheta^{2}}+\frac{\cos\vartheta}{\sin\vartheta}\frac{\partial}{\partial\vartheta}+\frac{1}{\sin^{2}\vartheta}\frac{\partial^{2}}{\partial\varphi^{2}}\right)$$


so the first obvious thing would be to consider


$$\left(\xi\_1^2+\frac{1}{\sin^{2}x\_1}\xi\_2^2\right)?$$
The problem seems to be that spherical coordinates do not form a chart, so troubles are at the poles. 
Could anybody here elaborate on that, please?


Edit: Since somebody mentioned spectral theory in the comments before, the symbol I am talking about here is the one used to define the characteristic set on the cotangent bundle.



Symbol of the Laplace-Beltrami on $\mathbb{S}^2$
Your expression above is correct. To see this, note that every Riemannian metric $g=g\_{ij} \, dx^i \, dx^j$ in any coordinate system on any manifold $M$ has "dual metric" a function on the cotangent bundle, given as $g^\*(\xi) = g^{ij} \xi\_i \xi\_j$ for any $\xi=\xi\_i dx^i \in T^\* M$, where $(g^{ij})$ is the inverse metric of $(g\_{ij})$. The symbol of the Laplace operator on functions clearly must be a quadratic function on the cotangent bundle. As a heuristic test, imagine that as we "zoom in", the metric becomes more and more Euclidean, and the symbol becomes more and more like the Euclidean symbol, and so in the limit of rescaling, we see that the symbol is exactly the dual of the metric. (This can be made rigorous.) For the sphere, in [spherical coordinates](https://en.wikipedia.org/wiki/Metric_tensor) $(\theta,\phi)$ the metric matrix is $$(g\_{ij})=\begin{pmatrix}1 & 0 \\ 0 & \sin^2 \theta\end{pmatrix}$$
so the dual is $$(g^{ij})=\begin{pmatrix}1 & 0 \\ 0 & \frac{1}{\sin^2 \theta}\end{pmatrix}$$ so the symbol is $$\sigma(\xi)=\xi\_{\theta}^2 + \frac{\xi\_{\phi}^2}{\sin^2 \phi}.$$


