# what is a spinor structure?

There are of course lots of definitions and references for this, but in the same way that, on a manifold $M$,

• a Riemannian metric is a section of positive definite symmetric bilinear forms on $TM$
• or an almost complex structure is a section $J$ of $\textrm{End}(TM)$ which is everywhere an anti-involution (i.e. $J_x^2 = - \mathrm{Id}_{T_x M}$)
• or an orientation is a non-vanishing section of $\Lambda^m TM$

is

a spin structure, a section of quadratic forms $Q$ on $TM$ (of type $(s,t)$) and a vector bundle $S$ so that $\textrm{End}(S) \simeq \textrm{C}\ell(TM,Q)$?

...or something of the like? any references where it may be stated in this fashion?

• I think that Karoubi's book on K-theory has a definition aling theseclines. – Liviu Nicolaescu Feb 23 '13 at 20:58

Chapter 9 of Elements of Noncommutative Geometry, by Gracia-Bondia, Varilly, and Figueroa, has this perspective on spin$^c$ and spin structures.

The way to think about this algebraically is that the module of (continuous, say) sections of a spinor bundle over a (compact, Riemannian) manifold $M$ is Morita equivalence bimodule for the algebras $C(M)$ and $Cl(M)$, where $C(M)$ is the algebra of continuous functions and $Cl(M)$ is the algebra of continuous sections of the Clifford bundle (formed using the Riemannian metric). You can replace "continuous" with "smooth" here with no problems.

• In your proposed definition R^k has no spin respectively spin^c structure unless k is divisible by 8 respectively 2. – Dmitri Pavlov Feb 24 '13 at 13:48
• @Dmitri: You're correct, I was being a bit imprecise here. As Branimir points out, one should use the even sub-bundle of the Clifford bundle in case $M$ is odd-dimensional, and there are other considerations as well. OP asked for a reference, so I was just briefly summarizing. – MTS Feb 24 '13 at 19:17
• @MTS: As I pointed out in my answer, a much easier way is to define spin and spin^c structures as invertible bimodules between Cl(M) and Cl(F^dim(M)), where F is R respectively C. If dim(M) is divisible by 8 respectively 2 then Cl(F^dim(M)) is Morita equivalent to F, and tensoring with the corresponding bimodule reduces the general definition to the one that you proposed. – Dmitri Pavlov Feb 24 '13 at 21:26
• @Dmitri, that's a nice way of looking at it. As I said, I was just giving a brief summary of the basic idea presented in the reference that I listed. – MTS Feb 25 '13 at 1:48

A spin structure on a real vector space V equipped with a real quadratic form μ is an invertible bimodule (i.e., a Morita equivalence) from Cl(V,μ) to Cl(Rdim(V),ν). Here ν is the direct sum of dim(V) copies of the canonical quadratic form on R.

A spinc structure on a complex vector space V equipped with a complex quadratic form μ is an invertible bimodule from Cl(V,μ) to Cl(Cdim(V),ν). Here ν is the direct sum of dim(V) copies of the canonical quadratic form on C.

Note that spin and spinc structures form a category instead of a mere set, just as we would expect.

Of course, these definitions immediately extend to vector bundles, with the obvious requirement that invertible bimodules form bundles themselves.

A spin or spinc structure on a smooth manifold is a spin or spinc structure on its real or complex tangent bundle.

Just to elaborate a bit in explicitly differential-geometric terms on MTS's answer, which refers to certain results of Plymen's originally restated in terms of Morita equivalence (via the dictionary given by the Serre–Swan theorem), let $M$ be a compact orientable Riemannian manifold, and let $\operatorname{\mathbb{C}l}^{(+)}(M)$ be the finite rank Azumaya bundle given by the complexification of the Clifford bundle $\operatorname{Cl}(M)$ if $\dim M$ is even, and by the complexification of the even subbundle of the Clifford bundle if $\dim M$ is odd. Then $M$ is *spin*$^\mathbb{C}$ if and only if there exists an irreducible $\operatorname{\mathbb{C}l}^{(+)}(M)$-module, that is, a Hermitian vector bundle $\mathcal{S} \to M$ (i.e., a spinor bundle) such that $\operatorname{\mathbb{C}l}^{(+)}(M) \cong \operatorname{End}(\mathcal{S})$.

Now, if what you care about are specifically spin manifolds, one can endow $\operatorname{\mathbb{C}l}^{(+)}(M)$ with a canonical $\mathbb{C}$-linear anti-involution, and hence equip the dual bundle $\mathcal{E}^*$ of a $\operatorname{\mathbb{C}l}^{(+)}(M)$-module with the structure of a $\operatorname{\mathbb{C}l}^{(+)}(M)$-module. It is then , that $M$ is actually spin if and only if there exists an irreducible $\operatorname{\mathbb{C}l}^{(+)}(M)$-module $\mathcal{S} \to M$ such that $\mathcal{S} \cong \mathcal{S}^\ast$ not only as Hermitian vector bundles, but also as $\operatorname{\mathbb{C}l}^{(+)}(M)$-modules—this is, then, the spinor bundle for its corresponding spin structure. Indeed, by the anti-unitary isomorphism $\mathcal{S} \cong \mathcal{S}^\ast$ of Hermitian vector bundles defined by the Hermitian metric, together with a little bit of care, one can recognise such a unitary isomorphism of $\operatorname{\mathbb{C}l}^{(+)}(M)$-modules as nothing else than the charge conjugation operator on spinors in mild disguise.