I have a few questions regarding the existence of a spin structure on Kaehlerian and hyperKaehlerian manifolds. I cannot seem to provide a reference for proofs or counterexamples, so references are more than welcome.

Q1: Does every hyperKaehler manifold admit a spin structure?

Q2: Does every Kaehler manifold admit a spin structure?

Q3: Are there any dimension constraints on the existence of a spin structure on such manifolds?

Remark: According to page 85 of Jost's Riemannian Geometry and Geometric Analysis (6th edition), every orientable Riemannian manifold in dimension 4 carries a spin$^c$ structure. Since complex manifolds are orientable, this offers a partial to answer to the above questions. Jost, however, provides no proof or reference.

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    $\begingroup$ Sadly, my library has been depleted, so I no longer have Lawson/Michelsohn's beautiful book Spin Geometry, but that's the standard reference for all such questions. $\endgroup$ Jul 31, 2019 at 17:12
  • $\begingroup$ @TedShifrin Thank you very much for the reference :) $\endgroup$
    – AmorFati
    Aug 2, 2019 at 9:47
  • $\begingroup$ I don't know about hyperKaehler. For Kaehler note that $\mathbb{C}P(n)$ admits a Kaehler structure but for even $n$ is not spin. Its non-spinability can be proven by showing its second Stiefel-Whitney classes are non-zero (see Milnor and Stasheff). For $n$ odd that class is zero and so those complex projective spaces are spin. $\endgroup$
    – Todd N
    Aug 8, 2019 at 16:51
  • $\begingroup$ Apparently all hyperKaehler manifolds are Calabi-Yau's and all CYs are spin. I do not understand the proof of either of those facts yet, I just pieced that together from the hyperKaehler and spin Wikipedia pages. I will try to understand those facts and turn this into an answer at some point. $\endgroup$
    – Todd N
    Aug 9, 2019 at 9:15
  • $\begingroup$ If you can confirm my reasoning in (this question)[math.stackexchange.com/questions/3318145/… for me then I can give an explanation of why hyperKaehlers are spin. $\endgroup$
    – Todd N
    Aug 9, 2019 at 9:56

1 Answer 1


A1: Yes, a hyperKähler manifold is spin. If we take our definition of a hyperKähler manifold to be a $4k$-manifold with holonomy group contained in the symplectic group $Sp(k)$ (see here) and we take our definition of a spin-manifold to be an $n$-manifold $M$ with a reduction of the frame bundle $Fr(TM)$ to $Spin(n)$.

We have the containments $Sp(k) \subset SU(2k) \subset Spin(4k)$ (see here and this question ). Therefore the reduction of the holonomy group of a $4k$-manifold to $Sp(k)$ is more than sufficient to reduce $Fr(TM)$ to $Spin(4k)$. Hence these are spin.

A2: No, a Kähler manifold need not be spin. Complex projective space is Kähler, but $\mathbb{C}P(2n)$ is not spin, as can be proven by computing that its second Stiefel-Whitney class is non-zero (see Milnor and Stasheff). I do not know if there are any nice results like "Kähler + ______ $\implies$ spin." It is worth pointing out that the $\mathbb{C}P(2n+1)$'s are spin (same reference).

I don't know anything about Q3.


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