Principal symbol of an operator on a manifold

Remarks

The right side does not depend on the choices of $f$ and $u$. Take a chart and trivialisations in which $P = \sum_{\abs{\alpha} \le m} A_\alpha(x) \partial^\alpha$. Then, with $\xi = \sum_j \xi_j \, dx_j$, $$\sigma_m(P)(x, \xi) = \sum_{\abs{\alpha} = m} A_\alpha(x) \, \xi^\alpha.$$ Indeed, at $x$ only the terms in which all $m$ derivatives fall on distinct factors $f$ survive, and they give $m! \, \xi^\alpha$. So $\sigma_m(P)$ agrees with the principal symbol on open subsets of $\R^n$, and it is homogeneous of degree $m$ in $\xi$. Many texts include a factor $i^m$, defining the symbol through $D_j = -i \partial_j$; this does not change where the symbol is invertible. The symbol is multiplicative: $\sigma_{m + l}(P Q) = \sigma_m(P) \, \sigma_l(Q)$.

Example: for the exterior derivative $d$ on $k$-forms, $d(f \omega)(x) = df(x) \wedge \omega(x) + f(x) \, d\omega(x) = \xi \wedge \omega(x)$. So $\sigma_1(d)(x, \xi) \omega = \xi \wedge \omega$. For the Laplace–Beltrami operator on functions of a Riemannian manifold, with the sign convention $\Delta = -\operatorname{div} \operatorname{grad} = -\sum_{j,k} g^{jk} \partial_j \partial_k + (\text{lower order})$, one gets $\sigma_2(\Delta)(x, \xi) = -\abs{\xi}_g^2$, or $\abs{\xi}^2_g$ with the factor $i^2$.

Non-example: the full expression $\sum_{\abs{\alpha} \le m} A_\alpha(x) \xi^\alpha$, with the lower-order terms kept, is not the principal symbol. It depends on the chart: for $P = \partial_x^2$ on $\R$ and the new coordinate $y = e^x$, one has $P = y^2 \partial_y^2 + y \partial_y$, whose first-order term is not there in the old coordinate.