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I am currently thinking about the Dirac equation in curved (1+1)-dimensional spacetime. First I have tried to understand how vectors can be defined in curved space and how the covariant derivative comes off. In the case of the 2D Dirac equation the partial derivative indeed acts on a spinor; that’s why we need the spin connection here.

I always thought about spacetime and spinor space as two different spaces. When we discussed the spin of the electron, we learned that it is described by what are called spinors, which live in a 2-dimensional complex vector space.

My question is the following:

If the spacetime the electron "lives" in is curved, why does the space the spinor "lives" in also get curved? What is the connection between this spaces? Do both of them get curved in some sense in the same way?

Qmechanic
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Aralian
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    Do you understand more generally what a connection is or is that what the question is about? There isn't really anything special about spinors here as far as I can see - if you replace "spinor" by "vector (field)" everywhere, the mechanism and idea is the same. – ACuriousMind Sep 07 '22 at 09:28
  • Have a look at this answer I posted last year : https://physics.stackexchange.com/q/636894/ – KP99 Sep 07 '22 at 09:29
  • My question is more about this: When you have a curved space time M. I then can define vectors and the covariant derivative of a vector by assigning a tangent space to each point and connecting them via the christoffel symbols. But what does it mean to have spinors in this curved space? Can i also imagine to attach complex 2d tangent spaces to the manifold where the spinors live in and then define a connection? Its just like in my head there are two different spaces. A curved space time on one side and the spinor space on the other. What do they have to do with eachother? – Aralian Sep 07 '22 at 10:00
  • “Can I also imagine to attach complex 2D tangent spaces to the manifold where the spinors live and then define a connection?” Yes. So there aren’t only “two different spaces”. There is a separate spinor space at every point in spacetime, just like for vectors. – Ghoster Sep 07 '22 at 20:40
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    What you need is a spin structure. One needs to construct a principal ${\rm Spin}(1,d)$-bundle ${\cal S}M$ over $M$ whose elements are known as spin frames. It must be such that there is a projection $\Phi:{\cal S}M\to{\cal F}M$ onto the ${\rm SO}(1,d)$-bundle of orthonormal frames such that if $\rho:{\rm Spin}(1,d)\to{\rm SO}(1,d)$ is the covering map one has $$\Phi(s\cdot \Lambda)=\Phi(s)\cdot \rho(\Lambda)$$ where $s\cdot \Lambda$ is the right action of ${\rm Spin}(1,d)$ on spin frames and $\Phi(s)\cdot{\rho(\Lambda)}$ the corresponding action of ${\rm SO}(1,d)$ on orthonormal frames. – Gold Sep 07 '22 at 20:59
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    With this structure in place one may perform the associated bundle construction to build the bundle of spinors. Then at each point of spacetime you'll have one associated vector space, the fibre at the point, containing the spinors at that point. At this point it is the same as you have for vectors living in the tangent space at each point, which is the fibre of the tangent bundle $TM$ at that point. Only for spinors the construction is a bit more involved and may not be possible depending on the topology of $M$. For details you may check Wald's General Relativity Chapter 13. – Gold Sep 07 '22 at 20:59

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