TITLE: Which Kahler Manifolds Are Spin?
QUESTION [5 upvotes]: As is well-known (see here for a M.O. question) all Kahler manifolds are $spin^c$. I would like to ask which are in fact $spin$. 
Taking my motivation from the case of complex projective space, I make the following (naive) conjecture:

Conjecture: A compact $2n$-dimensional Kahler manifold $M$ is spin, if there exists a line bundle $L$ over $M$ such that 
  $$
L \otimes L \simeq \Omega^{(0,n)},
$$
  where $\Omega^{(a,b)}$ denotes the space of $(a,b)$-forms.

Can someone tell me if this is true or nor, and if not, what is an easy counter-example.

REPLY [5 votes]: By a classical paper of Atiyah (http://www.maths.ed.ac.uk/~aar/papers/atiyahspin.pdf)
the spin structures on a compact complex manifold  $(M^{2n},J)$ are in bijective correspondence with isomorphism classes of holomorphic line bundles $\cal{L}$ such that $\cal{L}\otimes\cal{L}=K$ where ${\cal{K}}=\Lambda^{n}(T^{*}M^{1, 0})$ is the canonical line bundle. In particular, an almost complex manifold  admits a spin structure if and only if $\cal{K}$ admits a square root, i.e. there exists a complex line bundle $L$ such that $L^{\otimes 2}=\cal{K}$.
Of course not any Kahler manifold is spin, for example there are several flag manifolds which are not spin, since their first Chern class is not even. 
added Example: For the coset $M=G/K=SU(n)/S(U(p)\times U(n−p))$ one can show that admits a unique  $SU(n)$-invariant spin structure, if and only if $n$ is even.