Are there any examples of D-dimensional Ricci-flat Riemannian (spin) manifolds of dimension D= 2,3,4,5 with the dimension of the space of parallel spinors equal to 1? And the same question for the pseudo-Euclidean case with dimensions D= 4,6. Here the manifolds are not assumed to be compact or complete, or simply connected.
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1$\begingroup$ In the Riemannian case, we can assume the manifold is simply connected. The classification given in "Mckenzie Y. Wang, Parallel spinors and parallel forms" shows that the space cannot be irreducible. $\endgroup$– Paul ReynoldsCommented Oct 2, 2013 at 17:58
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1 Answer
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In the split cases of interest to you ($D=4,6$ and of arbitrary signature), you can find a discussion of the local existence and generality in a 2000 paper of mine entitled Pseudo-Riemannian metrics with parallel spinor fields and vanishing Ricci tensor (http://arxiv.org/abs/math/0004073).
Note that, unlike the Riemannian case, a pseudo-Riemannian metric with a parallel spinor field need not be Ricci-flat; this usually imposes extra conditions.
In the discussion, I point out exactly what the possibilities are for holonomy groups, so that tells you when you can have the dimension of the space of parallel spinor fields equal to $1$. For example, with $D=4$, this can only happen with metrics of type $(2,2)$ and with $D=6$, this can only happen with metrics of type $(3,3)$. (In each case, such examples, even Ricci-flat ones, do exist.)
answered Oct 10, 2013 at 0:01
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$\begingroup$ Many thanks for your nice answer and a reference. In the cases $D = 4$, $(2,2)$ and $D = 6$, $(3,3)$ there are also examples of flat spin manifolds $M^2$ and $M^3$, respectively, where $M = \mathbb R^{(1,1)}_{*}/Z_2 $. The orginal problem for me was to find examples of: i) 5-dimensional Riemannian Ricci-flat spin manifold with only one parallel spinor and ii) 6-dimensional pseudo-Riemannian Ricci-flat spin manifold with either only one parallel spinor or two parallel spinors of opposite chiralities. So the chance is small, but it will be analysed using your classification. $\endgroup$– VladimirCommented Oct 10, 2013 at 12:39