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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 $Q.$

Request. Construct a non-geometric $C^\infty(M)$-module.

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, Graduate Texts in Mathematics. 220. New York, NY: Springer. xiv, 222 p. (2003). ZBL1021.58001.

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.

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2 Answers 2

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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 $\bigcap_p \mu_p Q = \mu_x/\mu^2_x$.)

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    $\begingroup$ It turns out that this example already appears in the book by Jet Nestruev mentioned by the OP, see Example 11.57 B on page 200. $\endgroup$
    – user85913
    Commented Jun 28, 2017 at 12:31
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    $\begingroup$ In your general example it should probably read $Q=C^\infty(M)/\mu_x^2$. $\endgroup$
    – Michael Bächtold
    Commented Jun 29, 2017 at 7:24
  • $\begingroup$ @MichaelBächtold Indeed. The "$1$" in $\mu^1_x$ was for "first-order derivatives", but this is indeed confusing notation since we're talking about the square of the ideal $\mu_x$. I will edit accordingly. $\endgroup$
    – user85913
    Commented Jun 29, 2017 at 7:28
  • $\begingroup$ O right, so thats not standard notation I believe. $\endgroup$
    – Michael Bächtold
    Commented Jun 29, 2017 at 7:30
  • $\begingroup$ Isn't the standard notation $\mathfrak{m}_x$ for the ideal of germs of functions around $x$ vanishing at $x$, in the local ring $\mathcal{O}_x = \mathcal{C}^{\infty}_x$ at $x$? $\endgroup$
    – Qfwfq
    Commented Jun 29, 2017 at 17:35
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What about $\Gamma(M,\mathcal{C}_{M}^{\infty}/\mathcal{I}_{x}^{2})$ where $\mathcal{I}_x$ is the ideal sheaf of a point $x\in M$?

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