5 Clarifying crediting of results to Plymen, who was responsible for both the spin^c and spin results.

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 explicitly differential-geometric 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 a result of Plymen, originally restated in terms of Morita equivalence (via the dictionary given by the Serre–Swan theorem), 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.

4 Troubles with dashes

Just to elaborate a bit on MTS's answer in explicitly differential-geometric terms, 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 involutionanti-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 a result of Plymen, originally restated in terms of Morita equivalence (via the dictionary given by the Serre--Swan Serre–Swan theorem), 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 \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)$-modules as nothing else than the charge conjugation operator on spinors in mild disguise. 3 added 74 characters in body Just to elaborate a bit on MTS's answer in explicitly differential-geometric terms, let$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}$-linear involution, and hence equip the dual bundle$\mathcal{E}^*$of a$\operatorname{\mathbb{C}l}^{(+)}(M)$-module with the structure of a$\operatorname{\mathbb{C}l}^{(+)}(M)$-module. It is then a result of Plymen, originally restated in terms of Morita equivalence (via the dictionary given by the Serre--Swan theorem), that$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)$-modules. \operatorname{\mathbb{C}l}^{(+)}(M)$-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.

2 added 350 characters in body
1