# Differential operators on a smooth mainifold

Let $$M$$ be a smooth manifold. If I'm not wrong, the set of differential operators on $$M$$ is defined as $$\mathcal{D}_M$$ can be defined by using vector fields. I.e. for each $$D \in \mathcal{D}$$ we have $$D = X_1 \circ \dots \circ X_k$$ for some smooth vector fields $$X_1, \dots X_k$$. Is it correct to think about differential operators in this way?

Moreover I was told that jets are the homomrphism from $$\mathcal{D}$$ to $$C^\infty(M)$$. Can you give me an example of jet?

My idea: take $$f \in C^{\infty}(M)$$. Let's define $$J_f$$ as follows: for every $$D \in \mathcal{D}$$ $$J_f(D) := D(f).$$ This should be an homomorphism from $$\mathcal{D}$$ to $$C^\infty(M)$$, right? Are all jets defined in this way?

2. The 1-jet of $$f\colon (-\epsilon,\epsilon)\to\mathbb{R}^n$$ at $$0$$ is just the tangent vector $$f'(0)$$ at $$f(0)$$. Similarly, the $$k$$-jet are the Taylor expansion up to and including order $$k$$. Differential operator of order $$k$$ are then ($$\mathbb{R}$$-linear, if the D.O. is linear) map from the $$k$$-jet to $$\mathbb{R}$$ (or whatever bundle you like). You can transfer that to manifold settings in the usual way, the gluing is done by the chain rule.
• so how can I interpret jets as homomorphism from $\mathcal(D)$ to $C^{\infty}(M)$? – Math_tourist Sep 24 '18 at 6:45