# Examples of odd-dimensional manifolds that do not admit contact structure

I'm having an hard time trying to figuring out a concrete example of an odd-dimensional closed manifold that do not admit any contact structure. Can someone provide me with some examples?

• Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) \times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= non-overtwisted) contact structure. – KSackel Mar 14 at 0:16

According to a well known result of Martinet, every compact orientable $$3$$-dimensional manifold has a contact structure , see also  for various proofs. On the other hand we have

Theorem. For $$n\geq 2$$ there is a closed oriented connected manifold of dimension $$2n+1$$ without a contact structure.

For $$n=2$$, $$SU(3)/SO(3)$$ has no contact structure and for $$n>2$$, $$SU(3)/SO(3)\times\mathbb{S}^{2n-4}$$ has no contact structure, see Proposition 2.4 in .

 H. Geiges, An introduction to contact topology, Cambridge studies in advanced mathematics 109.

 J. Martinet, Formes de contact sur les variétés de dimension 3. Proceedings of Liverpool Singularities Symposium, II (1969/1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.

 R. E. Stong, Contact manifolds. J. Differential Geometry 9 (1974), 219–238.

Although every closed oriented 3 manifold admits infinitely many contact structures, there isa n obstruction to the existence of contact structure on odd dim manifolds of dim$$\geq5$$.

Suppose that $$n>0$$ and $$\xi= ker \alpha$$ is a co-oriented contact structure on $$Y^{2n+1}$$. Then the tangent bundle of $$Y$$ has a splitting $$TY= \xi \oplus \mathbb R$$. Then $$d\alpha$$ is symplectic on $$\xi$$ and thus $$\xi$$ admits a compatible complex vector bundle structure[such a splitting called almost complex structure]. This is equivalent to a regudction group to $$U(n)\times 1 \subset SO(2n+1,\mathbb R)$$. Now lets observe the Chern class of $$\xi$$ viewed as a complex vector bundle over $$Y$$. Moreover $$c_1(\xi)$$ reduced to Stiefel-Whitney $$w_2(Y)\in H^2(Y,\mathbb Z_2)$$. So we can say that if $$Y$$ admits an almost complex structure then $$w_2$$ admits an integral lift which is equivalent of third Stiefel-Whitney class $$W_3(Y)\in H^3(Y,\mathbb Z)$$ vanishes. [Since $$W_3$$ is the image of $$w_2$$ under Bockstein homomorphism $$\beta$$ $$\to H^2(Y,\mathbb Z)\to H^2(Y,\mathbb Z_2)\to_\beta H^3(Y,\mathbb Z)\to$$.] Thus $$W_3=0$$ is a necessary condition.

Now if $$Y= SU(3)/SO(3)$$. Then we have a fibration $$SO(3)\to SU(3)\to Y$$. By using Steenrod squares, $$W_3(Y)$$ is the generator of $$H^3(Y,\mathbb Z)=\mathbb Z_2$$. So this manifold doesnot admit almost contact structure.

Theorem[Borman, Eliashberg,Murphy]- There exists a contact structure in every homotopy class of almost contact structure.