I'm asking about transformation properties of vector and spinor fields. I'm trying to better understand statements like "vectors/spinors transform in a certain way under way under [symmetry operations]" where the symmetry operation might be a rotation or Lorentz transformation.

I want to work in the mathematical context of differential manifolds.

We have a manifold and on that manifold we have (local) coordinates. Within the context of differential geometry I can define the tangent space at each point on the manifold. The tangent space is related to derivatives along coordinate direction on the manifold. So in this way, the tangent space is closely related to, and derived from, coordinates on the manifold. We can undergo a procedure where for each point on the manifold we select a single vector from its tangent bundle. In this way we define a vector field on the manifold. But, importantly, because of the relationship between elements of the tangent space and coordinates I described above, it is possible to derive how vectors transform when we apply coordinate transformations to the manifold.

My question is: Can we do something similar for spinors? That is, can we define a "spinor space" at each point on the manifold whose transformation properties can be derived from some relationship between the spinor space and the coordinates on the manifold? And then can we select a spinor from the spinor space at each point on the manifold to build up a spinor space? OR INSTEAD, do we simply define an abstract spinor space at each point on the manifold and somehow just declare how elements of that space transform when we transform the coordinates on the manifold? That is, can we define spinors in such a way that their transformation properties under coordinate transforms are derived from their definition in relation to a manifold?


1 Answer 1


Briefly, there is an important difference. Let us consider an arbitrary curved $n$-dimensional differential manifold. The structure group is a priori the general linear group $GL(n,\mathbb{R})$.

  1. On one hand, a vector field $X\in\Gamma(TM)$ can be defined directly on $M$.

  2. On the other hand, a Lorentz transformation and a spinor field require additional geometric data and conditions so that (among other things) the structure group of the manifold $M$ is reduced to the Lorentz group $O(n\!-\!1,1;\mathbb{R})$. For more details, see e.g. spinor bundle on Wikipedia.

    An important example is when $M$ is endowed with a choice of vielbein/solder form $e$. Recall that for each spacetime point $p\in M$ a vielbein/solder form $$(T_pM,g_p)\stackrel{e_p}{\longrightarrow}(V,\eta)$$ is essentially a choice of a frame for each tangent space $T_pM$. Here $\eta$ denotes the $n$-dimensional Minkowski metric. We can then use the corresponding real Clifford algebra $Cl(V,\eta)$ and Dirac/gamma matrices to define spinor representations.

  • $\begingroup$ I’m curious if you or someone else could provide more info about what a vielbein or solder structure is or references where I can learn more. Perhaps that’s a better topic for another question? $\endgroup$
    – Jagerber48
    Jul 23, 2023 at 13:44
  • $\begingroup$ to be clear, and address the question: Supposing we do have a differentiable manifold equipped with a vielbein. Is it possible to define a spinor field without reference to its transformation properties, but to then derive those transformation properties from the relationship between the spinor field and the vielbein and manifold? $\endgroup$
    – Jagerber48
    Jul 27, 2023 at 5:07
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jul 27, 2023 at 7:46
  • $\begingroup$ Comments on point 2: the definition of a spinor field and its local Lorentz transformation do NOT require a choice of vielbein/solder form $e$. All you need for a covariant derivative of a spinor field is a spin connection $w$, rather than vielbein/solder form $e$. On the other hand, the Lagrangian/Action of a spinor field does require a choice of vielbein/solder form $e$. The requirement of definition of a spinor field and the requirement of Lagrangian of a spinor field are two different requirements. $\endgroup$
    – MadMax
    Jul 27, 2023 at 17:09
  • $\begingroup$ One may argue that the spin connection $w$ is after all dependent on the vielbein/solder form $e$. However, this dependency is an add-on assumption for a given manifold which is not required in general. $\endgroup$
    – MadMax
    Jul 27, 2023 at 17:10

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