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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?

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  1. No, you really need linear combinations here.
  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.
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  • $\begingroup$ so how can I interpret jets as homomorphism from $\mathcal(D)$ to $C^{\infty}(M)$? $\endgroup$ – Math_tourist Sep 24 '18 at 6:45

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