5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can one tell then which Grassmannians Gr(N,K) are spin simply from the values N and K? $\endgroup$ – Janos Erdmann Feb 14 '17 at 21:19
  • $\begingroup$ 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. $\endgroup$ – 314159. Feb 14 '17 at 21:20
  • $\begingroup$ 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 . . . . $\endgroup$ – Janos Erdmann Feb 14 '17 at 21:56
  • $\begingroup$ 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. $\endgroup$ – abx Feb 15 '17 at 5:57
  • $\begingroup$ But that would imply that for complex projective space, the case where K=1, is never spin, which is not true. $\endgroup$ – Janos Erdmann Feb 15 '17 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.