# Why to introduce spinor fields we need this map in the definition of a spin structure?

Let me start with what I currently understand. Let $${\rm SO}(1,3)$$ be the proper ortochronous Lorentz group. Its universal cover is $${\rm SL}(2,\mathbb{C})$$. The representations of its universal cover are labelled by pairs of integers or half-integers $$(A,B)$$ which label $${\frak su}_\mathbb{C}(2)$$ representations. The representations with $$A+B$$ integer descend to true representations of $${\rm SO}(1,3)$$ but the ones with $$A+B$$ half integer do not and these are spinor representations.

In particular we have for example $$(\frac{1}{2},0)$$ and $$(0,\frac{1}{2})$$ Weyl spinors. The spinor themselves are elements of $$\mathbb{C}^2$$ and given $$\Lambda \in {\rm SL}(2,\mathbb{C})$$ we know how it acts on them through the representations $$D^{(\frac{1}{2},0)}(\Lambda)$$ and $$D^{(0,\frac{1}{2})}(\Lambda)$$.

Now given this setup we would like to talk about spinor fields in some general spacetime $$(M,g)$$. Since spinors are being introduced as elements of a representation space of the universal cover of $${\rm SO}(1,3)$$ it comes as no surprise that the associated fields should come as sections of an associated bundle to a principal $${\rm SL}(2,\mathbb{C})$$-bundle. What happens, though, is that it is often said that one needs a spin structure, which can be defined as follows:

Definition: Let $$(M,g)$$ be a semi-Riemannian manifold of signature $$(t,s)$$ and let $$F(M)$$ be the associated principal $${\rm SO}(t,s)$$-bundle of orthonormal frames. A spin structure on $$(M,g)$$ is a principal $${\rm Spin}(t,s)$$-bundle $$\pi_S:S(M)\to M$$ together with a principal bundle map $$\Phi : S(M)\to F(M)$$ such that $$\Phi(s\cdot \Lambda)=\Phi(s)\cdot \rho(\Lambda),$$ where $$\rho: {\rm Spin}(t,s)\to {\rm SO}(t,s)$$ is the covering map.

What I don't understand is how this structure is used in practice. Why do we need this map $$\Phi : S(M)\to F(M)$$? Why do we need to connect the two bundles together to be able to talk about spinor fields?

Because if we just have one $${\rm Spin}(t,s)$$-bundle - or in signature $$(1,3)$$ one $${\rm SL}(2,\mathbb{C})$$-bundle - it seems we can already take the spinor representations like the Weyl representations and perform the associated bundle construction to build spinor fields. Why apart from that connecting to the frame bundle through this map $$\Phi$$ is necessary?

• You don't want an arbitrary spin bundle; you want one that covers your tangent bundle. – AccidentalFourierTransform Jan 12 at 21:20
• Thinking a little bit more on the subject, is it because we want that the ${\rm SL}(2,\mathbb{C})$ transformations on spinors always be associated to local Lorentz transformations on the local orthonormal frame? Because that is what I think the map $\Phi$ ensures: without it we could act with $\Lambda$ on spinors independently of any local "change of frame" taking place which I think would lend the transformation on the spinors rather weird to interpret I think. – Gold Jan 12 at 21:44
• Yes. Generic Spin(N) bundles are fine, but these define "internal symmetries" instead of "external" ones. For example, this is how you define a gauge theory with gauge group Spin(N). Here the bundle does not follow the tangent one, it is independent. – AccidentalFourierTransform Jan 12 at 21:53

It's dictated by the Lorentz invariance requirement of the Dirac Lagrangian $$L = i\bar{\psi}\not{\partial}\psi - m\bar{\psi}\psi$$
with spinor $$\psi$$ on $$S(M)$$ and the space-time derivative $$\not{\partial}$$ on $$F(M)$$. Lorentz invariance forces you to properly map $$S(M) \to F(M)$$.
• This is backwards because $L$ is not defined until you properly define your spinor bundle. So you cannot use $L$ to argue anything about the spinor bundle, as the latter comes first. – AccidentalFourierTransform Jan 12 at 21:52