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Let $(M,\theta)$ be a $(2n+1)$-dimensional contact manifold, $\mathcal{C}=\ker\theta$ the contact distribution, and $X\in\mathcal{C}$ a vector field belonging to $\mathcal{C}$. In a couple of minor papers I have found the following notion: $$ \mathrm{type}(X):=\textrm{rank}\{ \theta,X(\theta),\ldots,X^{2n-1}(\theta) \}\, , $$ i.e., an integer $\geq 1$ called "type", which is attached to $X$. (EDIT: by $X^k(\theta)$ I mean the $k^\textrm{th}$ Lie derivative of $\theta$ along $X$.) I found it quite interesting but at the same time doubtful.

The authors of the aforementioned papers claim that it is a contact invariant of $X$ (for instance, $X$ is a contact symmetry iff its type is 1) and use it to obtain certain classification results, but I had a hard time following them.

QUESTION 1: is $\textrm{type}(X)$ indeed a contact-invariant?

(EDIT: the "type" is indeed a function of $p\in M$, and it is globally defined only if such a function is constant.)

In the affirmative case, does anybody know a good reference in contact geometry explaining such a concept with all the due details, including who introduced it in the first place?

QUESTION 2: can the notion of "type" be generalised to more general objects, like $(M,\mathcal{C})$, where $M$ is an arbitrary manifold and $\mathcal{C}$ is a distribution on it?

Roughly speaking, $\mathrm{type}(X)$ should measure how far is $X\in \mathcal{C}$ from being a symmetry of $\mathcal{C}$.

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    $\begingroup$ What does denote $X^k(\theta)$? the $k$-th Lie derivative of $\theta$? $\endgroup$ Mar 23, 2015 at 12:04
  • $\begingroup$ also, this rank depends on the point $p \in M$ $\endgroup$ Mar 23, 2015 at 12:06
  • $\begingroup$ @DanieleZuddas: yes, $X^k(\theta)$ is the $k^\textrm{th}$ derivative, and yes the type is in fact a function of the point $p\in M$. $\endgroup$ Mar 23, 2015 at 13:55

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Using notation $X\theta$ for Lie derivative of a 1-form $\theta$ along a vector field $X$, just check that multiplying $\theta$ by a function and then taking these derivatives pops out lower order terms, which are linear combinations of previous entries in your sequence: $X(f\theta)=(Xf)\theta + f(X\theta)$. This proves that type depends only on the contact structure, since the contact forms of the contact structure are precisely the nonzero multiples of $\theta$. Generalize to any subsheaf $I$ of the cotangent sheaf $\Omega^1_M$ by letting $XI$ be the subsheaf of the cotangent sheaf generated by 1-forms $X^k \theta$ for $\theta$ any local section of $I$ and $k$ any nonnegative integer. For example if $I$ is a vector subbundle of $T^*M$, say with $\theta^i$ a basis of local sections, then any other basis of local sections is $g^i_j \theta^j$ for some invertible matrix of smooth functions $g^i_j$. Therefore $X^k (g^i_j \theta^j)=X^{k-1} ( (X g^i_j) \, \theta^j + g^i_j \, X\theta^j)$ is expressed as a linear combination of the $X^l \theta^i_j$ by induction. I don´t really know how to use this notion of type, since I never ran across it before.

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  • $\begingroup$ I never doubted that type depended only on the conformal class of $\theta$. I wasn't sure it is invariant under the group of contact transformations. However, it is nice that you have such an "universal" sourcebook, and I'm going to have a look at it (I never read it from the beginning to the end, but I know it really contains a lot, so maybe I should do it!) $\endgroup$ Mar 23, 2015 at 15:28
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    $\begingroup$ The contact structure is the same information as the conformal class of $\theta$, since a 1-form vanishes on the contact planes just when it is a multiple of $\theta$. $\endgroup$
    – Ben McKay
    Mar 23, 2015 at 16:45
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    $\begingroup$ The conformal classes of the differential forms $\theta\wedge X\theta,\theta\wedge X\theta\wedge X^2\theta,\ldots$ should give you finer invariants of $X$ than the type... $\endgroup$ Mar 25, 2015 at 12:35

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