I'm trying to learn about spin bundles and find myself very confused by the construction and motivation of the spin connection.

The places where I've seen the spin connection formulated, it seems to be explained by the need to account for the change in the vierbeins - i.e. the rotation of the orthonormal frames from point-to-point. But I don't understand why this is related to spinors. It seems to me that an object which carries a frame index just indicates a vector in the tangent space whose components are considered with reference to the orthonormal basis. So a covariant derivative of a vector component should include a factor of the spin connection whenever you are considering such a component. The frame bundle's associated vector bundle under the fundamental representation is just the tangent space. Why does a frame index indicate a half-integer spin?

As I understand it, the spin frame bundle is a lift of the frame bundle using a double cover map. Where does the connection there come from? Is it a pullback of the one on the frame bundle, or do we construct a new one?

  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/199793/50583, at least my answer there should answer part of what you're asking here $\endgroup$
    – ACuriousMind
    Jan 18, 2020 at 9:32
  • $\begingroup$ @ACuriousMind Right, this seems to affirm the idea of the spin connection as associated with the tetrad basis for vectors. But the spinor aspect is obscure to me. $\endgroup$
    – quixot
    Jan 18, 2020 at 10:59
  • $\begingroup$ given that $\frak{spin}(n)\cong\frak{so}(n)$, shouldn't any connection on your SO(n)-bundle lift to a connection on your Spin(n)-bundle? $\endgroup$
    – Christoph
    Jan 18, 2020 at 13:48
  • 1
    $\begingroup$ So is the 'spin connection' the one on the SO(n) bundle or the one on the Spin(n) bundle? Since it seems we also need it to describe the variation in vector components if they are in the frame basis. $\endgroup$
    – quixot
    Jan 19, 2020 at 10:01
  • $\begingroup$ I vote we call "spin bundles" instead "spindles" from now on, then with our vernacular involving terms like "fibers, bundles, spin and spindle," it literally sounds like we're discussing weaving the fabric of the universe lol!! $\endgroup$
    – R. Rankin
    Dec 19, 2020 at 4:06

2 Answers 2


I explain how to obtain the spin connection from the ordinary Levi-Civita connection in this answer of mine.

As for why we need the "spin connection" to express spinors, note that spinors by definition transform in a particular representation of the Lorentz algebra $\mathfrak{so}(1,3)$ (or its generalizations). But this representation is not a representation of the general linear group $\mathfrak{gl}(4)$ that the Levi-Civita connection is valued in, so the Levi-Civita connection cannot act on these things. But in order to express a covariant derivative, we need a notion of the connection acting on our spinors, and so we reduce the structure group to get a $\mathfrak{so}(1,3)$-valued connection that can act on our spinors.


A spinor bundle is a vector bundle associated to a principal bundle whose structure group is a spin group. There is a constraint for the spacetime manifold to support such a principal bundle. It must be a spinnable manifold. Actually, as a point of physical nomenclature, the usual phrase is: the manifold is spin. But as a point of use of the English language, it would be better described - as I just did - as spinnable. Here, both the semantics and the grammar of the phrasing is correct. Physics is hard enough without having to parse badly phrased language.

A spin connection is then a connection on this principal bundle. It is equivalently a spin algebra valued connection form. And as the algebra of covering groups are isomorphic to the algebra of the base group, that is spin(p,q) =~ so(p,q), then this is equivalently a special orthogonal algebra valued connection form. This suggests we have dispensed with the spin structure. But this is not correct. For the connection form to be a connection on a principal bundle it still has to transform correctly under the spin group. Thus the spin structure is still neccessary.

However, because spin connections are equivalently valued in so(p,q) algebras. Then actual so(p,q) connections which transform under SO(p,q) are also called spin connections. This can be justifiably - in my opinion - be called a sloppy use of language. Its easy to get confused here between spin connections and rotation group connections - as you've noted.


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