Let $(X,g)$ be a $m$-dimensional complex, hermitian, spin manifold and let us denote by $S_{\mathbb{C}}$ its complex spinor bundle. Then:

$S_{\mathbb{C}}\simeq \Lambda_{\mathbb{C}}(X)$

Let $\nabla$ be any metric connection on $T_{\mathbb{C}}X$ that lifts to a connection $\nabla^{S_{\mathbb{C}}}$ on $S_{\mathbb{C}}$. Suppose that there is a complex form:

$\Omega\in \Lambda_{\mathbb{C}}(X)$

such that $\nabla\Omega = 0$. Under the isomorphism of the first equation, $\Omega$ corresponds to a spinor $\eta\in \Gamma(S_{\mathbb{C}})$. My question is, does it holds then that $\nabla^{S_{\mathbb{C}}}\eta = 0$? I guess this should be in fact and "if and only if".


  • $\begingroup$ I do not completely understand the assumptions or the logic of your question. Do you mean that $X$ carries a Riemannian metric and $\nabla$ preserves that metric? If this is not assumed, then what do you mean by a spin-structure in this setting and what does it mean that $\nabla$ lifts to a connection on the spin bundle? $\endgroup$ Jun 6, 2015 at 8:32
  • $\begingroup$ @AndreasCap Hi Andreas. Yes, that was implicitly assumed. I will write it more explicitly. $\endgroup$
    – Bilateral
    Jun 7, 2015 at 10:26

1 Answer 1


If you also assume that $\nabla$ preserves the complex structure, it is true that $\nabla^{S_{\mathbb C}}\eta=0$ if and only if $\nabla\Omega=0$. The assumptions mean that the unitary frame bundle of $X$ admits an extension with structure group the appropriate spin group, and that the principal connection inducing $\nabla$ comes from a principal connection on that spin frame bundle. But this means that you can view both complex forms and spinors as associated bundles to the spin frame bundle and the principal connection on the spin frame bundle induces both $\nabla$ and $\nabla^{S_{\mathbb C}}$, and then the statement on parallel sections is clear.


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