# Which Kahler Manifolds Are Spin?

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.

## 1 Answer

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.

• Can one tell then which Grassmannians Gr(N,K) are spin simply from the values N and K? – Janos Erdmann Feb 14 '17 at 21:19
• if I remember right this is described by M. Cahen, S. Gutt, Spin structures on compact simply connected Rieamannian symmetric spaces, Simon Stevin 62 (1988), 291–330. – 314159. Feb 14 '17 at 21:20
• I can't seem to find a copy of the paper, but I guess it should be equivalent to whether or not (N-K)K is an even of odd integer . . . . – Janos Erdmann Feb 14 '17 at 21:56
• The canonical bundle of the Grassmannian $\mathbb{G}:=\mathrm{Gr}(N,K)$ ($N$-dimensional subspaces of $\mathbb{C}^K$) is $\ (-K)$ times the positive generator of $\mathrm{Pic}(\mathbb{G})$, so $\mathbb{G}$ is spin if and only if $K$ is even. – abx Feb 15 '17 at 5:57
• But that would imply that for complex projective space, the case where K=1, is never spin, which is not true. – Janos Erdmann Feb 15 '17 at 13:51