Are linearizations of involutive PDEs locally solvable? A possibly soft question for you guys and gals. Say a system of analytic PDEs has been completed to involution (in the sense that it's geometric symbol has a Pommaret basis, or has vanishing Spencer/Koszul (co)homology). Is its linearization locally solvable?
Just to include why I'm asking: Lie-Pseudo-groups are determined by an involutive system and their Lie-algebroids by its linearization. If the answer to the question is positive, then since the linearized equation has the same symbol module (at the identity) it also is locally solvable. I have not been able to find this (possible) fact in the literature.
edit: I changed the title of the question as it was pretty bad.
 A: If I understand correctly, you're asking if a nonlinear system of PDE's is involutive, then is the corresponding linearized system also involutive? The answer is yes, but I'm not sure that it's obvious. It is true, as you say, that the symbol, which is the same for both systems, is involutive. However, involutivity also requires conditions on the lower order terms, and these need to be checked. I believe (but am not sure) that this is demonstrated in my monograph "Involutive Hyperbolic Differential Systems" for quasilinear systems.
ADDED: Here's an argument that the homogeneous linearized equation is involutive: Let
$$
\Phi(x, u, \partial u) = 0
$$
be a first order involutive system of PDE's. (If the PDE's are higher order, you can always rewrite the system as a first order one)
Given an involutive symbol, you can count the dimension of the space of formal second order solutions there are. In terms of Cartan characters $s_1, \dots, s_n$ (where $n$ is the number of independent variables), it's given by $s_1 + 2s_2 + \cdots ns_n$.
Since the homogeneous linearized equation has the same symbol, it is involutive, if it has the same dimension of formal second order solutions (which is now a linear subspace of the space of all second order Taylor series for the unknown function).
Given any formal $1$-parameter family $u_t$ of 2nd order solutions to the nonlinear system, you have
$$
\Phi(x, u_t, \partial u_t) = 0.
$$
Differentiating this with respect to $t$, you get
$$
\Phi'(x, u_0, \partial u_0)\dot{u} = 0.
$$
This implies that the dimension of the space of 2nd order solutions $\dot{u}$ to the linearized system is equal to the dimension of the space of 2nd order solutions to the nonlinear system. By Cartan's test for involutivity (I'm sure there's an equivalent condition in the other definitions of involutivity), the homogeneous linearized system is involutive.
A: Take a look at the book Exterior Differential Systems page 71 theorem 2.2 for a precise statement of the Cartan--Kaehler theorem, which says that involutive real analytic torsion-free exterior differential systems have local real analytic integral manifolds. I don't know what a Pommaret basis is, so I can't be sure that involutivity means the same in your sense and in Cartan's sense.
A: With the crucial hypothesis of analyticity, the answer is Yes. I believe that, when the PDEs are presented as an exterior differential system, the result is known as the Cartan-Kähler theorem.
Without recasting the PDEs in terms of exterior differential systems, that is, in the language that you are using, you can find the linear version of this result in Section 2.1 of the review by Spencer (1969) on overdetermined PDE systems. The non-linear version can be found in this paper by Goldschmidt (1967).
Without analyticity, there are further obstructions than lack of involutivity to having smooth solutions, as demonstrated by the famous example by Lewy.
