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(I asked this question on math.stackexchange, but got no reaction in several weeks. So, my conclusion is, that it is harder to answer than I thought, and maybe admissible for the attribute 'research level')

My setting is the following: Given two collections of differential forms $\omega^i$ (with $i=1,...,m$) and $\mathrm d F^j$ (with $j=1,..., n-m$), define $$\eta^i := \sum_{j=1}^m a^i_j\omega^j + \sum_{j=1}^{n-m} b^i_j\mathrm dF^j.$$

The symbols ($a_j^i$ and $b_j^i$) represent functions of $x = (x^1, ..., x^n)$. The matrix $(a_j^i)_{ji}$ is assumed to be regular everywhere.

Roughly speaking, I am interested in the integrability of the Ideal $I_{[0]}:=\{\eta^1, ... \eta^m\}$.

'Normally', this is equivalent to the question: "Does the Frobenius condition $$ \mathrm d \eta^i\wedge \eta^1 \wedge ... \eta ^m =0 $$ hold for $i=1, ..., m$?"

However, in my case time derivatives are admitted: The coefficients $\omega^i = \sum_{j=1}^n \omega_j^i \mathrm d x^j$ depend on the jet coordinates (without $t$): $\omega^i_j = \omega^i_j(x, \dot x, ...)$, where $x$ is itself an $n$-tuple. Additionally, $\mathrm d F^i$ has the form

$$\mathrm d F^i =\mathrm d F^i(x, \dot x) = \sum_{j=1}^n (f^i_j \mathrm d x^j + g_j^i \mathrm d \dot x^j),$$ where the rows $g^i$ of the matrix $(g_j^i)$ are linearly independent.

So, we extend the definition:

$$\eta^i_{[k]} := \sum_{j=1}^m a^i_{[k],j}\,L^k_{D_\chi}\omega^j + \sum_{j=1}^{n-m} b^i_{[k],j}\mathrm L^k_{D_\chi} dF^j.$$

Here, $L_{D_\chi}$ denotes the Lie-derivative with respect to the so called trivial Cartan vectorfiled: $$ D_\chi:= \frac{\partial}{\partial t} + \sum_{j=1}^n \dot{x}^j \frac{\partial}{\partial x^j} + ... \enspace . $$

Applying $L^k_{D_\chi}$ is thus equivalent to the total time derivative of order $k$, $\frac{d^k}{dt^k}$. E.g. for $k=1$ we have: $L_{D_\chi} (c \, \mathrm d x^1) = \dot c \mathrm d x^1 + c \mathrm d \dot x^1$.

Finally, we set $$ \eta^i_{<N>}:= \sum_{k = 0}^N \eta^i_{[k]}.$$

[Edit1] The coefficient functions $b_{[k],j}^i$ are arbitrary, while the functions $a_{[k],j}^i$ must fulfill the following condition: The matrix $$ A:= \sum_{k = 0}^N (a_{[k],j}^i)_{ji} \frac{d^k}{dt^k} \tag{*} $$ which is a polynomial matrix of the time derivative operator $\frac{d}{dt}$ (=$L_{D_\chi}$) must be "unimodular", i.e. it must have an inverse which is also a polynomial matrix of $\frac{d}{dt}$. Unfortunately, unlike the case of constant coefficients, there is no determinant-criterion for unimodularity applicable here.[/Edit1]

Now, I am interested in the (non)-integrability of the algebraic ideal $I = \{\eta^1_{<\infty>}, ..., \eta^n_{<\infty>}\}$.

[Edit3] In other words: Are there functions $h^i = h^i(x, \dot x , ...)$ such that $\mathrm d h^i = \eta^i_{<N>}$ for some fixed $N\in \mathbb N$ and $i = 1, ..., m$. ($a$ and $b$ are arbitrary up to condition (*)) If such functions $h^i$ do not exist: How can this be shown? [/Edit3]

More specifically, given some additional information on $\omega$ and $F$, can one find an upper bound on $N$ above which additional time derivatives "do not help anymore" (to make the wedge product of the Frobenius condition vanish)? Most importantly: is it possible to state the integrability conditions of $I$ in terms of the given $\omega^i, \mathrm d F^i$ and its time derivatives?

Are there known results to problems like this? What would be the right terminology to search for?

[Edit2] The background of this question lays in control theory, in particular in the theory of differential flatness. $\mathrm d F^j$ comes from the system under consideration and $\omega^i$ is the result of an special algorithm for determining a dual basis of the 'tangent system'. If $\eta$ is integrable, then, roughly speaking, $h$ with $\mathrm d h = \eta$ is the desired flat output mapping. [/Edit2]

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    $\begingroup$ Some things are not clear to me: for example the equation $\omega^i=\sum \omega^i_j=dx^j$. Is there an $=$ to much? Also what are the coefficients $a^i_{[k],j}$ and $b^i_{[k],j}$, are they arbitrary? Could you also explain why you make that infinite sum $\eta^i_{<\infty>}$? $\endgroup$ Commented Sep 10, 2014 at 21:13
  • $\begingroup$ @MichaelBächtold Thanks for pointing out the "=" typo and the lack of specification for the $a^i_{[k],j}$. The infinite sum is necessary, because a priori it is not clear, up to which order one must go to get an integrable ideal. Also it is "not very likely" that for $N=100$ the Frobenius condition suddenly holds (for suitable chosen coefficieant functions), I am not aware of any condition to limit $N$. So that is the core of the question. However, some literature hints would also be welcome. $\endgroup$
    – cknoll
    Commented Sep 11, 2014 at 9:23
  • $\begingroup$ Thanks for the attempt to clarify. The construction of the ideal $I$ still seems somewhat unmotivated and out of context to me. Maybe if you add some background you will receive more answers. Also it seems at first sight that an answer to your question might not only depend on the given $\omega$ and $dF$ but also on the $a$ and $b$, does it not? $\endgroup$ Commented Sep 12, 2014 at 8:06
  • $\begingroup$ I added some background information but I fear to provide a substantial motivation I would have to write a lot more. Therefore I tried to state the problem purely mathematical, without referring on motivation. BTW: apart from the unimodularity condition the $a_{[k]_j^i}$ are arbitrary ($b_{[k]_j^i}$ with out restriction). The question is: Do they exist, such that $\eta_{<N>}$ is integrable. Or: how to show that they do not exist. $\endgroup$
    – cknoll
    Commented Sep 13, 2014 at 18:22
  • $\begingroup$ I think it's still a bit unclear what is given and what is variable in your question. Does it boil down to the following? "Do there exist functions $h^i=h^i(x,\dot{x},\ldots)$ (everything else is a function on jet space as well), such that $dh^i = \eta^i$ when evaluated on those $x(t)$ that also satisfy $\omega^i = const$ and $F^j = const$?" So, are you looking for $\eta^i$ such that the answer is positive or are the $\eta^i$ also somehow fixed? $\endgroup$ Commented Sep 13, 2014 at 19:40

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