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Let $(M, \xi)$ be a closed contact manifold with co-oriented contact structure $\xi = \ker \alpha$. Let $\mathrm{Cont}(M, \alpha)$ be the group of diffeomorphisms that preserve the contact form $\alpha$, and let $\mathrm{Cont}^+(M, \xi)$ be the diffeomorphisms that preserve $\xi$ with its co-orientation (i.e. that pull $\alpha$ back to a positive multiple of $\alpha$).

Does the inclusion $\mathrm{Cont}(M, \alpha) \hookrightarrow \mathrm{Cont}^+(M, \xi)$ induce a weak homotopy equivalence?

My guess is that this is too much to hope for, but I don't have any candidate for a counter-example.

Motivation: This question came up in understanding the question of when a contact fibre bundle admits a global contact form with diffeomorphic Reeb dynamics on every fibre.

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$Cont(M,\alpha)$ can be very small (finite), and there is not really a Gray stability in this case, so it seems to be no. –  Chris Gerig Apr 12 '12 at 21:22
    
@Chris: Isn't the Reeb flow a 1-parameter subgroup of $Cont(M,\alpha)$? (Hit Cartan's formula for the Lie derivative of $\alpha$ along $R$ with the defining properties of $R$ and you get zero) –  Jonny Evans Apr 12 '12 at 22:36
    
Oh right, $M$ is closed. –  Chris Gerig Apr 13 '12 at 2:46
    
I believe it is true however that generically "most" of $\text{Cont}(M,\alpha)$, or at least its identity component, should be generated by the Reeb flow. (I'm being intentionally imprecise.) –  Chris Wendl Apr 16 '12 at 17:28

2 Answers 2

up vote 8 down vote accepted

One of the problems with this question is that $\text{Cont}(M,\alpha)$ depends heavily on the choice of contact form $\alpha$, so while the previous answer shows that there is a choice of contact form for which $\text{Cont}(M,\alpha)$ and $\text{Cont}^+(M,\xi)$ can't be homotopy equivalent, it's not immediately clear if this is true for all choices of $\alpha$.

That said, here's a connected example for which the inclusion also fails to be surjective on $\pi_0$. (Attribution: this emerged out of discussions with Hansjoerg Geiges.)

Let $(M,\xi)$ be $T^3$ with its standard contact structure $\xi_0$, and using coordinates $(x,y,\theta)$ on $T^3$, write the standard contact form as

$\alpha_0 = \cos(2\pi\theta) dx + \sin(2\pi\theta) dy$.

Now for any $A \in \text{SL}(2,\mathbb{Z})$, the direct sum of $A$ with the identity defines a linear map on $\mathbb{R}^3$ which descends to $T^3$ as a diffeomorphism $f : T^3 \to T^3$. It is easy to show that for any such map, $f^*\alpha_0$ can be deformed through contact forms to $\alpha_0$, hence by Gray's stability theorem, $f$ is isotopic to a contactomorphism $f_0$.

However, if $f_0$ is a strict contactomorphism with respect to $\alpha_0$, then it must preserve the corresponding Reeb vector field, and this is a very strong restriction. It means for instance that the Morse-Bott torus of Reeb orbits at $\{\theta=0\}$ is mapped to another Morse-Bott torus of orbits with the same period, and the only such orbits that exist for $\alpha_0$ point in either the same, opposite or an orthogonal direction. With arguments like this one can show that $f_0$ cannot be a strict contactomorphism unless the matrix $A$ is orthogonal... in fact, I believe it must be a fourth root of the identity.

With a little more work one can say something similar for a much larger set of contact forms, using the fact that a strict contactomorphism must always map Reeb orbits to Reeb orbits of the same period. (I'm fairly sure that for a generic choice of contact form, not only are all Reeb orbits nondegenerate but no two of them have the same period.) Unfortunately, I still don't know how to turn this into any statement for all contact forms, but the evidence is certainly against $\text{Cont}(M,\alpha) \hookrightarrow \text{Cont}^+(M,\xi)$ ever being bijective on $\pi_0$.

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The map isn't surjective on $\pi_{0}$. Take $M$ to be a disjoint union of two 3-spheres, put any contact form $\beta$ on the first sphere, and put $\beta/2$ on the second sphere. There is an element of $\mathrm{Cont}^{+}(M)$ that simply interchanges the two pieces. But this map is not isotopic to a map that preserves the contact form. (Just think of the 3-form $\beta \wedge d\beta$ and the volumes of the two 3-spheres.)

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