Let $M$ be a smooth manifold and let $Q$ be an arbitrary $C^\infty(M)$-module. $Q$ is called geometric if $$\bigcap\_{p\in M}\mu\_pQ=0,$$ where $\mu\_p$ is an ideal in $C^\infty(M)$ of functions vanishing at point $p.$ 


From the definition, $q\in\mu\_pQ$ iff $q=\sum\_{i=1}^Nf\_iq\_i$ for some $f\_i$'s from $\mu\_p$ and $q\_i$'s from $Q.$



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> **Request.** Construct a non-geometric $C^\infty(M)$-module. 
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For any vector bundle $E\to M$ the $C^\infty(M)$-module $\Gamma(E)$ is geometric. This is because, if $\xi\in\mu\_p\Gamma(E),$ then $\xi(p)=0.$ Hence we have to look for non-geometric modules in a different way.


I came across this notion in the following Jet Nestruev's book:


*Nestruev, Jet*, [**Smooth manifolds and observables**](http://dx.doi.org/10.1007/b98871), Graduate Texts in Mathematics. 220. New York, NY: Springer. xiv, 222 p. (2003). [ZBL1021.58001](https://zbmath.org/?q=an:1021.58001).


**Remark about the notion:**
I investigated more or less Vinogradov's bibliography and I guess this notion first appeared in the following book:


*Krasil’shchik, I.S.; Lychagin, V.V.; Vinogradov, A.M.*, Geometry of jet spaces and nonlinear partial differential equations. Transl. from the Russian by A. B. Sosinskij, Advanced Studies in Contemporary Mathematics, 1. New York etc.: Gordon and Breach Science Publishers. xx, 441 p. (1986). [ZBL0722.35001](https://zbmath.org/?q=an:0722.35001).



An example of a non-geometric $C^\infty(M)$-module
Let $M$ be the unit circle in $\mathbb C$, and consider the algebra homomorphism $C^\infty(M)\to M\_2(\mathbb R)$ given by $$ f\mapsto \begin{bmatrix} f(1) & \frac{df}{d\theta}(1) \\ 0 & f(1)\end{bmatrix}$$ (where $\theta$ is the angular coordinate on $M$). This homomorphism makes $\mathbb R^2$ into a $C^\infty(M)$-module with $\bigcap\_p \mu\_p \mathbb R^2 = \left[\begin{smallmatrix} \mathbb R \\ 0\end{smallmatrix}\right]$.


(In general, fix $x\in M$ and let $\mu^2\_x\subset C^\infty(M)$ be the ideal of functions $f$ such that $f$ and all of its first-order derivatives vanish at $x$. Then take $Q=C^\infty(M)/\mu^2\_x$. I believe that then $\bigcap\_p \mu\_p Q = \mu\_x/\mu^2\_x$.)


