Is it necessary that the compactified manifold in string theory must be complex?

Question

I have learned that there are some restrictions imposed on the manifolds which are used to compactify the extra-dimensions of string theory. The most important being the "Ricci flatness" condition. This condition is thought to arise from the fact that vacuum solutions of Einstein's field equations must be Ricci flat (please correct me if I am wrong).

My question, I see Kähler manifolds (specifically Kähler three-folds) are predominantly chosen for Calabi-Yau manifolds in string theory. Are there any compact real Ricci-flat manifolds out there, if yes, why are they not chosen for Calabi-Yau manifolds?

Edit: Does insisting complex manifolds have anything to do with the fact that Hilbert spaces are defined over the field $\Bbb C$?

Does insisting complex manifolds have anything to do with the fact that Hilbert spaces are defined over the field $\Bbb C$? — Eden Zane, Jun 09 '22 at 02:58
$\uparrow $ No. Also note that $M^4$ does not have to be complex. — Qmechanic, Jun 09 '22 at 11:05

score 4 · Accepted Answer · answered Jun 09 '22 at 11:08

If we are talking about the traditional way of realizing the 10-dimensional target space of superstring theory as a product $$M^{10}~=~M^4 \times K^6,$$ where $(K^6,g^{(6)})$ is a compact $6$-dimensional Riemannian manifold, then to have unbroken ${\cal N}=1$ supersymmetry in $4$ spacetime dimensions, the holonomy group of $(K^6,g^{(6)})$ must be the $8$-dimensional Lie group $SU(3)$, which is the hallmark of a Calabi-Yau 3-fold, and which in particular is a complex manifold, cf. OP's title question.

For more details, see e.g., Green, Schwarz and Witten, Superstring theory, chap. 15 and this related Phys.SE question.

Is it necessary that the compactified manifold in string theory must be complex?

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