Lifting a diffeomorphism into a spinor bundle automorphism

Question

I know several papers that treat this, but it seems that most of these papers do things very differently with quite different conclusions, so I am confused.

Basically, when one tries to do classical field theory (as in, the branch of physics) in a mathematically precise manner, one considers a field $\psi$ to be a section of some fiber bundle $\pi_E:E\rightarrow M$ over the spacetime $M$, and writes up a lagrangian $\hat L\in \Omega^n_H(J^1(E))$, which is a horizontal $n$-form on $J^1(E)$ (sometimes $J^2(E)$) such that the action is $$ S[\psi]=\int_M (j^1\psi)^\ast\hat L. $$

If one wishes to involve gravity, one seeks to consider field theories that are diffeomorphism invariant.

If $\pi_E:E\rightarrow M$ is a fiber bundle whose sections are some kind of matter fields, then to have a well-defined concept of diffeomorphism-invariance, for any diffeomorphism $\phi:M\rightarrow M$ one must be able to lift this diffeomorphism into a fiber bundle automorphism $\phi_E\in\text{Aut}(E)$ in a consistent manner.

For the tensor bundles (or indeed any natural bundle), there is a functorial lift given by the tangent map (at least in the case of tensor bundles). For example, if $X\in\Gamma(TM)$ is a smooth vector field, and $\phi:M\rightarrow M$ is a diffeo, then $$ \phi_E=T\phi:TM\rightarrow TM $$ is this vector bundle automorphism.

On the other hand, in physics, very important fields are spinor fields, sections of the spinor bundle $\pi_S:S\rightarrow M$. This, however is not a natural bundle (to my knowledge), and I do not know if there is any "canonical" way to lift diffeomorphisms into $\text{Aut}(S)$.

Since general relativity heavily involves diffeomorphism-freedom, it is extremely important to be able to do that. In particular, I have no idea how to define the stress-energy tensor of a spinor field without representing diffeomorphisms on $S$ somehow.

This situation is further confusing me, since so far I have been an ardent defender of the viewpoint that in the usual local tensor calculus-based formalism, there is no essential difference between "active diffeos" (point transformations) and "passive diffeos" (coordinate transformations).

However the behaviour of a "traditional" spinor field under a coordinate transformation is simple and clear, a spinor field transforms as a scalar under coordinate transformations, and "as a spinor" under changes of orthonormal frames.

However in the modern, invariant viewpoint, one cannot afford this approach. For example, if at $x\in M$, one is given a spinor $\psi\in S_x$ and a vector $v\in T_xM$, and one considers the tensor product $\psi\otimes v\in S_x\otimes T_xM$, then a diffeo will move $v$ to $T\phi(v)\in T_{\phi(x)}M$, but it will not move $\psi$ at all, so this tensor product under a diffeo would become $\psi\otimes T\phi(v)\in S_x\otimes T_{\phi(x)}M$, a product of fibers taken over different base points - clearly undesirable.

Is there any agreed-upon method of dealing with the action of spinors under diffeos? If so, is there a simple way of writing it down/stating it?

Answer 1

There is an intrinsic ambiguity in lifting diffeomorphisms to spinors, since as you say spinor bundles are not natural bundles. But the ambiguity is not large and has more to do with the global properties of spin structures.

First, recall that the oriented orthogonal frame bundle $\mathcal{P}_g M$ of a pseudo-Riemannian manifold $(M,g)$ is an $\mathrm{SO}(p,q)$-principal bundle, where $(p,q)$ represents the signature of the metric $g$. Then recall that a spin structure is a $\mathrm{Spin}(p,q)$-principal bundle $\mathcal{S}M$ together with a principal bundle morphism $\mathcal{S}M \to \mathcal{P}_gM$, which fiberwise restricts to the usual projection homomorphism $\mathrm{Spin}(p,q) \to \mathrm{SO}(p,q)$. Now, given a linear representation of of $\mathrm{Spin}(p,q)$ on a vector space $\Sigma$, in particular a spinor representation, the associated vector bundle $\Sigma_{\mathcal{S}} M = (\mathcal{S} M \times \Sigma)/\mathrm{Spin}(p,q)$ is the corresponding spinor bundle over $M$.

Now, a diffeomorphism $\phi\colon M \to M'$ (at least one that preserves metrics and orientations of oriented pseudo-Riemannian manifolds $(M,g)$ and $(M',g')$, which is all that we will consider for now) induces a morphism of oriented frame bundles $\phi^*\colon \mathcal{P}_g M \to \mathcal{P}_{g'}M$. This is a principal bundle morphism, meaning that it is $\mathrm{SO}(p,q)$-equivariant. Suppose that there exists also a $\mathrm{Spin}(p,q)$-principal bundle morphism $\tilde{\phi}^*\colon \mathcal{S}M \to \mathcal{S}'M'$ of spin structures that covers $\phi^*$. The existence of $\tilde{\phi}^*$ is not automatic, possibly it doesn't exist at all. But if it does exist, then it is not unique, since the $\mathbb{Z}_2$ center of $\mathrm{Spin}(p,q)$ acts non-trivially on $\tilde{\phi}^*$. But that's the only freedom you have; if $\tilde{\phi}^*$ exists, then there exist exactly two possibilities. Basically, it is sufficient to fix the lift of $\phi^*$ at a single point $x\in M$ (there are only two possibilities) and the rest is determined by continuity.

Finally, all that remains to observe is that a morphism of principal bundles induces a morphism of associated bundles. Hence, a spin-lift $\tilde{\phi}^*$ of our diffeomorphism $\phi$ induces a vector bundle morphism $\tilde{\phi}^*_\Sigma \colon \Sigma_{\mathcal{S}} M \to \Sigma_{\mathcal{S}'} M'$. This morphism can be constructed as follows. First, consider the map $$ (\tilde{\phi}^*, \mathrm{id}) \colon \mathcal{S}M \times \Sigma \to \mathcal{S}'M' \times \Sigma . $$ Here, the OP's original intuition that spinors should be treated as "scalar fields" whose pointwise values are unchanged by push-forward along diffeomorphisms is implemented by the fact that the above map acts as the identity on the $\Sigma$ factor. Next, note that $(\tilde{\phi}^*, \mathrm{id})$ is $\mathrm{Spin}(p,q)$-equivariant, and hence induces a map of the associated bundles, which is precisely our desired $\tilde{\phi}^*_\Sigma$ push-forward of spinor fields.

It remains only to recall the OP's final piece of intuition, that spinor fields transform as a "spinor" under changes of orthogonal frames and see how it is implemented in the above construction. Locally, a spin structure is often specified by a local choice of a specific frame field. Say that we have chosen specific frame fields $e^i$ on $(M,g)$ and $e'^i$ on $(M',g')$. Obviously, upon pushing forward $\phi^* e^i = \Phi^i_j e'^j$, where $\Phi$ is an $\mathrm{SO}(p,q)$ valued function, which need not be the identity. So, if $\sigma^\alpha(x)$ is a spinor field, that is, a section of $\Sigma_\mathcal{S} M$, then the push-forward $\sigma' = \tilde{\phi}^*_\Sigma \sigma$ spinor field is defined by the formula $$ \sigma'^\alpha(\phi(x)) = \Phi^\alpha_\beta \sigma^\beta(x) , $$ where $\Phi^\alpha_\beta$ is the matrix representing the pointwise action of $\Phi$ on $\Sigma$. The rigid frame rotation (or "Lorentz transformation") $\Phi$ takes into account the difference between the orthogonal frame $e'^i$ on $(M',g')$ and the push-forward $\phi^* e^i$ from $(M,g)$.

I have one more small remark to make about the meaning of the components $\sigma^\alpha(x)$ on a manifold. On flat space, they would be components with respect to some basis in the vector space $\Sigma$. But, on a manifold, these components obviously need to be interpreted with respect to some (spin-)frame in the spinor bundle $\Sigma_{\mathcal{S}} M$, but a global frame need not always exist, and even locally there is huge freedom in choosing in choosing a spin-frame as a function on $M$. What is important for the above push-forward of spinor fields to make sense, the spin-frames on $\Sigma_{\mathcal{S}} M$ and $\Sigma_{\mathcal{S}'} M'$ need to be chosen such that the $\gamma$-matrices which implement the Clifford multiplication and the action of $TM$ on $\Sigma_{\mathcal{S}} M$ must be fixed. That is, there must be some constant matrix $(\Gamma^i)^\alpha_\beta$ such that $(\gamma^i)^\alpha_\beta(x) = (\Gamma^i)^\alpha_\beta$ with respect to the chosen spin-frame on $M$, while at the same time also $(\gamma'^i)^\alpha_\beta(x') = (\Gamma^i)^\alpha_\beta$ with respect to the chosen spin-frame on $M'$. So the above push-forward rule only works when considering spin-frames of this kind.

Answer 2

As spinors transform with a minus sign under a full rotation, there is no (non-trivial) lift of the action of the group of diffeomorphisms to the spinor bundle (i.e. the spinor bundle is not a natural bundle). This is closely related to the fact that a spinor transforms according to a projective representation of the Lorentz group.

Nonetheless, you can lift vector fields to the spinor bundle (using a connection). Hence, the spinor bundle is an infinitesimally natural bundle and this suffices to construct the stress-energy-momentum tensor. For more details about infinitesimally natural bundles, see Bundles with a lift of infinitesimal diffeomorphisms or Infinitesimally natural principal bundles.