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?

  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.
  • $\begingroup$ so how can I interpret jets as homomorphism from $\mathcal(D)$ to $C^{\infty}(M)$? $\endgroup$ – Math_tourist Sep 24 '18 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.