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This is a follow up questions to this answer. In Minkowski spacetime, we require our action to have few active symmetries :

  1. Symmetries under isometries of the metric

  2. Symmetry under global $U(1)$ transforms

Now, for tensor fields, both of these active transforms are easy to generalise to general manifolds, as isometries are just special cases of diffeomorphisms and we know how to transform a tensor field under diffeomorphisms. Also, tensor fields don't change under $U(1)$ transforms.

But for fields that are sections of a $U(1)$ bundle or a Spin-bundle, the generalisation isn't obvious, I think.

Suppose we have a section of a $U(1)$ or a spin bundle (not the Principal bundle but an associated bundle) with local trivialisations $\{U_i, \psi _i\}$, with transition functions $t_{ij}$ between charts overlaps. A section of these bundles can be given by functions $f_i$ on each $U_i$ that are valued in some vector space on which the group is represented.

How should these sections transform under active transforms like isometries of the metric and global $U(1)$ transform? (Also, Isometries are still well defined but I'm not sure what a global $U(1)$ transform would mean in this context).

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    $\begingroup$ The spinor bundle is a complex vector bundle, so you can act with $U(1)\subset \mathbb C$ by scalar multiplication. Invariance under $U(1)$ follows from sesquilinearity of $\gamma: S\otimes \bar S \to \mathbb R^4$ (and the descended map using the associated bundle construction). The same holds for any hermitian vector bundle. $\endgroup$ Commented Dec 2 at 9:03
  • $\begingroup$ The reason is roughly that for hermitian vector bundles, the transition functions take values in $U(n)$, and the subgroup $U(1)$ is central. So you can 'pull' the $U(1)$ action past the transition map. $\endgroup$ Commented Dec 2 at 9:08
  • $\begingroup$ For a truly general manifold, there is no hope if you want to have even free dynamics for spinor or gauge bundles. You probably need a locally pseudo-Riemannian metric. (Even for mild generalizations, like Berwald manifolds, these bundles have not been found and very well may not exist.) You can get around some degeneracy problems and still define spinors though. See, for example, this paper (and references therein): $\endgroup$ Commented Dec 2 at 21:27

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As stated in the comments, $U(1)$ may be taken to act by scalar multiplication. More generally, any isomorphism of $G$-representations $V\to V$ induces an isomorphism of associated vector bundles (it is $G$-invariant, and therefore descends to the fibres of the vector bundle $E = P\times V / G$). These isomorphisms can be deemed 'global symmetries'.

If $V$ is an irreducible complex representation of a compact Lie group $G$, the group of these isomorphisms is one of $\mathbb R^\times, \mathbb C^\times, \mathbb H^\times$. If one additionally requires that the representation is unitary, some inner product needs to be preserved, such that the isomorphism groups are reduced accordingly to their maximal compact subgroups $\mathbb Z_2, U(1), Sp(1) = SU(2)$. For reducible representations, the groups of $G$-linear isomorphisms can be larger.

EDIT

For the isometries, the story is a little more complicated. Let $\Lambda: Spin(M) \to SO(M)$ and $\lambda: Spin(n)\to SO(n)$ be the double covers that define your spin structure ($SO(M)$ is the bundle of orthonormal frames). An isometry is a diffeomorphism $f$ such that the differential $Df$ defines at any point an isometry $(T_x M, g_x) \cong (T_{f(x)} M, g_{f(x)})$. This amounts to a map $\phi\in C^\infty(SO(M),SO(M))^{SO(n)}$ (the superscript indicates equivariancy, that is, $\phi(pg) = \phi(p)g$). If you are lucky, $\phi$ lifts to a spin isometry -- that is, some map $\Phi\in C^\infty (Spin(M), Spin(M))^{Spin(n)} $ such that $$\Lambda^* \phi = \lambda \circ \Phi.$$ In this case, the lift acts on spinors by left multiplication. Generally, there are topological obstructions, parametrised by $H^1(M,\mathbb Z_2)$.

EDIT

This lift acts explicitly as follows: Recall that the fibres have a simply transitive $Spin(n)$ action. Thus, at every point, $\Phi(p) = p.g(p)$ for some $g(p) \in Spin(n)$. Putting this together, we find that $\Phi$ can be rewritten as a map $\tilde\Phi \in C^\infty(Spin(M),Spin(n))$, and equivariance of $\Phi$ translates to equivariance of $\tilde\Phi$ where $Spin(n)$ acts on itself by conjugation (thus $\tilde\Phi(pg) = g^{-1}\tilde\Phi(p) g$). This map takes values in $Spin(n)$, such that it acts on spinors $\psi = [p,s] \in Spin(M)\times\mathbb S/ Spin(n)$ by left multiplication $\tilde\Phi \cdot \psi = [p,\tilde\Phi(p) s]$.

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  • $\begingroup$ Thanks. This handes the global $U(1)$ transform case. Also, what about active transforms under the isometry group of the metric? How do the sections of G-bundles like U(1) and Spin bundles transform under the isometries? $\endgroup$ Commented Dec 2 at 11:37
  • $\begingroup$ @RyderRude I hope my edit answers your question. The $U(1)$ (more generally $G$) part does not 'see' isometries, but the spin part does. $\endgroup$ Commented Dec 2 at 13:18
  • $\begingroup$ Thanks. I don't quite understand what the lift is in case it exists. Can you please add that? $\endgroup$ Commented Dec 2 at 14:15
  • $\begingroup$ In a local trivialisation, this amounts to lifting a group homomorphism $SO(n) \to SO(n)$ to a group homomorphism $Spin(n) \to Spin(n)$ (the group homomorphism is given by conjugation by the isometry). Thus, for some $g\in SO(n)$, the group hom is $h\mapsto ghg^{-1}$. You now need to find some $\tilde g \in Spin(n)$ such that $\lambda(\tilde g ) = g$ (this is unique up to $\pm 1$). This always works locally, but may fail globally as it may not be possible to pick a consistent lift. $\endgroup$ Commented 2 days ago

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