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Is there any necessary and sufficient condition for existence of $SU(3)$-structure on 6-manifolds $M$?

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up vote 18 down vote accepted

Yes, it is well-known that a $6$-manifold has an $\mathrm{SU}(3)$-structure if and only if it is orientable and spinnable (i.e., it has a spin structure).

The necessity of these two conditions is clear, since $\mathrm{SU}(3)$ is both connected and simply-connected.

For the sufficiency, first note that if $M^6$ is orientable and spinnnable (i.e., the first two Stiefel-Whitney classes of its tangent bundle vanish), then, fixing a Riemannian metric and orientation on $M$, one can choose a spin structure and construct the corresponding spinor bundle $\mathbb{S}\to M$. Since $\mathrm{Spin}(6)\simeq\mathrm{SU}(4)$, this bundle is a complex $4$-plane bundle, which is, of course, a real $8$-plane bundle. Since $8>6$, there is a nonvanishing section of $\mathbb{S}$ over $M$, and this will reduce the structure group of this bundle from $\mathrm{SU}(4)$ to $\mathrm{SU}(3)$. This will, correspondingly, reduce the structure group of the tangent bundle to $\mathrm{SU}(3)$ (since this subgroup does not contain $-I_4\in\mathrm{SU}(4)$). Thus, $M$ carries an $\mathrm{SU}(3)$-reduction of the structure group of the tangent bundle.

NB: This is a classical argument, but I don't know who first wrote it down. Maybe, Alfred Gray did it first, but I am not sure, and I will not have any access to reference books for the next 12 days while I am traveling, so I won't be able to check or give you further information.

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