# Symbol of the Laplace-Beltrami on $\mathbb{S}^2$

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.

• The symbol of the Laplace operator of a Riemannian metric is the dual of the metric, as a function on the cotangent bundle. – Ben McKay Jul 4 '17 at 20:06
• @BenMcKay if I understand you correctly, then it would be since $ds^2=d\theta^2+ \sin^2(\theta) d\varphi^2$ just $\xi_1^2+\sin^2(x_1)\xi_2^2$ which is weird because it is, structually similar, but eventually different from my result above. – BaoLing Jul 4 '17 at 21:19
• @BenMcKay sorry, would you mind giving a small comment on this? – BaoLing Jul 6 '17 at 15:04
• Your expression above is correct, because the dual of any Riemannian metric is computed using the inverse matrix of the matrix of the Riemannian metric, in any coordinate system. – Ben McKay Jul 6 '17 at 15:16

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 $(\theta,\phi)$ 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 $$\sigma(\xi)=\xi_{\theta}^2 + \frac{\xi_{\phi}^2}{\sin^2 \phi}.$$
The "chart problem" you mentioned is not really a problem, because principal symbols can be defined via local coordinates and glue them together. After all we obtain pseudo-differential operators this way on a manifold anyway. For a slick global definition that employs the "exponential trick", see Rafe Mezzo's answer here. Normalization may be an issue for you as most literature uses $i\xi$ instead of $\xi$ in the powers to make the Fourier transform more slick. It is worth mentioning that the definition exploits exponential trick also carries to the case of manifold with boundary, so it may be something of interest since in this case the definition via local charts is more complicated.