10
$\begingroup$

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.

$\endgroup$

2 Answers 2

12
$\begingroup$

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

$\endgroup$
5
  • 2
    $\begingroup$ 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
    Jun 28, 2017 at 12:31
  • 2
    $\begingroup$ In your general example it should probably read $Q=C^\infty(M)/\mu_x^2$. $\endgroup$ 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
    Jun 29, 2017 at 7:28
  • $\begingroup$ O right, so thats not standard notation I believe. $\endgroup$ 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
    Jun 29, 2017 at 17:35
4
$\begingroup$

What about $\Gamma(M,\mathcal{C}_{M}^{\infty}/\mathcal{I}_{x}^{2})$ where $\mathcal{I}_x$ is the ideal sheaf of a point $x\in M$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.