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Edite According to the essential comment of Ian Agol I revise the question as follows

For a smooth manifold $M$, is there a non identity involution $\theta$ on the lie algebra $\chi^{\infty}(M)$ such that $X$ is topological equivalent to $\theta (X)$, for all smooth vec. field $X$ on $M$?

This question is motivated by the fact that, on $\mathbb{R}^{n}$ the linear vector field $\dot X=AX$ is topological equivalent to the linear vector field $\dot X=-A^{tr}X$.

By topological equivalent, we mean existence of an orbit preserving homeomorphism.

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    $\begingroup$ What do you have in mind for the Killing form for the space of vector fields? $\endgroup$
    – Ian Agol
    Commented Jul 20, 2014 at 17:29
  • $\begingroup$ @IanAgol I apologize for my mistake in the question, I revised it. Thank you for your comment $\endgroup$
    – Ali Taghavi
    Commented Jul 20, 2014 at 20:03
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    $\begingroup$ On odd dimensional spheres, the antipodal map works. $\endgroup$
    – Ian Agol
    Commented Jul 20, 2014 at 23:14
  • $\begingroup$ @IanAgol your very essntial comment about the killing form is a motivation to consider the following:The problem is that we do not have a "trace" for defining the killing form. From this obstraction, we extract the following construction. For a compact Riemannian manifold $M$, one can construct a sobolov hilbert space which contains $\chi^{\infty}(M)$ as a dense space(ex:$H^{\infty}$) such that for each smooth vector field $X$, the operator $ad(X)$ is a bounded operator. Now we search for "trace class" operators, those $X$ for which $ad(X)$ is a trace class operator. $\endgroup$
    – Ali Taghavi
    Commented Jul 22, 2014 at 21:24
  • $\begingroup$ @IanAgol This gives us a Lie subalgebra of $\chi^{\infty}(M)$, consists vector fields for which $ad$ is a trace class operator. Now a good question: To what extent this Lie algebra can be represented? What type of vector fields belong to this Lie algebra? Obviously the Killing form can be defined on this Lie algebra.Can you help to modify this idea? $\endgroup$
    – Ali Taghavi
    Commented Jul 22, 2014 at 21:29

2 Answers 2

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The Lie algebra of vector fields of a smooth manifold determines the manifold with its smooth structure. Below I indicate a series of papers where this is proved in various settings. In particular, this implies that any Lie algebra involution of the Lie algebra of vector fields has to be induced by an involution of the manifold.

  • MR0064764 (16,331a) Shanks, M. E.; Pursell, Lyle E. The Lie algebra of a smooth manifold. Proc. Amer. Math. Soc. 5, (1954). 468–472.

  • MR0375400 (51 #11594)
    Amemiya, Ichiro; Masuda, Kazuo; Shiga, Kōji Lie algebras of differential operators. Osaka J. Math. 12 (1975), 139–172.

  • MR0516602 (80g:57036)
    Grabowski, J. Isomorphisms and ideals of the Lie algebras of vector fields. Invent. Math. 50 (1978/79), no. 1, 13–33.

  • MR2785498 (2012b:58028) Grabowski, Janusz(PL-PAN); Kotov, Alexei(LUX-CUL); Poncin, Norbert(LUX-CUL) Geometric structures encoded in the Lie structure of an Atiyah algebroid. (English summary) Transform. Groups 16 (2011), no. 1, 137–160

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Extending Ian's comment, any involution on $M$ induces an involution on the Lie algebra of the vector fields on $M.$ There should be plenty of those.

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