1. Why do we need to introduce the spin connection coefficients $\omega_{\mu \space \space b}^{\space \space a} $ in General Relativity? To me, they just look (mathematically) like the Christoffel symbols and so I'm assuming they aren't tensors? If I'm right in saying that, then do these $\omega_{\mu \space \space b}^{\space \space a} $ transform the same way under general coordinate transformations as the Christoffel symbols, and if so, what's the need in them then?

  2. Finally, what's the relation between the Vierbein $e_{\mu}^a$ and the spin connection (non-mathematically)?

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    $\begingroup$ Here's a quick motivation (maybe not an answer). In the tetrad formalism, the metric is replaced by tetrads $e^a_\mu$ satisfying $g^{\mu\nu} e^a_\mu e^b_\nu = \eta^{ab}$. The original 16 d.o.f. of the tetrad are constrained by 10 equations above, thereby giving 6 new independent d.o.f. The metric $g_{\mu\nu}$ starts out with 10 d.o.f. Thus, we must have 4 new d.o.f. to completely describe the theory. These are precisely the spin connection coefficients. $\endgroup$
    – Prahar
    Commented Apr 15, 2014 at 22:48
  • $\begingroup$ @Prahar what is a d.o.f. ? $\endgroup$
    – magma
    Commented Apr 17, 2014 at 16:21
  • $\begingroup$ degree of freedom $\endgroup$
    – Prahar
    Commented Apr 17, 2014 at 16:23

2 Answers 2


There are both physical and formal reasons to introduce the spin connection.

Physically, we know that there are spin 1/2 particles. A spin 1/2 field cannot be described by anything built from 4-vector fields. You can realize this for example by that 4-vector fields (and so anything built from them) returns to their original value after a $2\pi$ rotation whereas a spin 1/2 field does not (it changes sign). In GR, you will want to take covariant derivatives of this spinor field, this is exactly what the spin connection is for. If all you want to do is vacuum GR then you can do without the spin connection, but if you want to put interesting matter in your spacetime, you need it.

Formally it turns out that some calculations are actually simpler using spinors. The canonical example is the Newman-Penrose formalism [1], a somewhat neater formalism is the Geroch-Held-Penrose formalism [2]. Then there is of course two whole volumes by Penrose and Rindler [3]. One can perhaps understand the utility of spinor formalisms from one arithmetic fact: you deal with 12 (in the GHP formalism, 8) complex quantities instead of 24 real ones; this is few enough that you can have separate names for each quantity which helps the notation immensely. (Imagine having to write $\omega^1{}_{03}$ or similar for every single quantity in your calculation.) The GHP formalism notation is almost as compact as you can hope to achieve given the complexity of GR.

Now since the product of two spinors is a vector, and a null vector at that, the spinor formalisms are extremely well suited to problems with radiation, both gravitational and other. In the GHP formalism every quantity has a direct geometrical interpretation and it is easy to make Ansätze that are both geometrically meaningful and useful for simplifying the equations. An example of this is the integration method of Edgar and Ludwig [4].

The Cartan-Karlhede algorithm for classifying spacetimes is also simpler in spinor form. Part of this is because of the compactness of notation, but part is also that a step in the algorithm is to put tensors in a standard form (for example for Petrov type III you can take the Weyl tensor to have components $\Psi_i = \delta_{i3}$). There are algorithms for doing this with computer algebra for spinors; I do not know about algorithms for doing it with world-vectors.

Now for your more practical questions, yes, the spin connection coefficients are exactly like the Christoffel symbols. Since a world-vector is the product of 2 spinors, you can recover the latter from the former. They do not form a proper tensor and have a transformation law like that of the Christoffel symbols.

The precise relation between the tetrad and the spin connection is, I think, impossible to explain non-mathematically because the proper understanding of it requires thinking about fiber bundles and covering groups. However, vaguely, you can say that just like it is locally possible to take a tetrad, it is always locally possible to find two spinor fields that are everywhere orthonormal (in a certain sense), say $o^A$ and $\iota^A$; these are two-spinors, so $ A = 0,1$. This is called a dyad. They have complex conjugates $\overline{o}^\dot{A}$ and $\overline{\iota}^\dot{A}$. The product of a spinor and a conjugate spinor is a world-vector, so with these you can form four world-vectors, $o^A\overline{o}^\dot{A}$ and so on. It is not too hard to realize that those four form a (null) tetrad.

You could take $-o^A$ and $-\iota^A$ instead and obtain the tetrad. So to each tetrad there are exactly two dyads. You could hand-wavingly say that a dyad is the square root of a tetrad, but the proper, more formal statement is that the spin group is a double cover of the Lorentz group.


  1. Newman, E., & Penrose, R. (2004). An approach to gravitational radiation by a method of spin coefficients. Journal of Mathematical Physics, 3(3), 566-578.
  2. Geroch, R., Held, A., & Penrose, R. (2003). A space‐time calculus based on pairs of null directions. Journal of Mathematical Physics, 14(7), 874-881.
  3. Penrose, R. & Rindler, W. Spinors and Space-time. 2 vols (Cambridge University press, 1984).
  4. Edgar, S. B., & Ludwig, G. (1997). Integration in the GHP formalism III: Finding conformally flat radiation metrics as an example of an “optimal situation”. General Relativity and Gravitation, 29(10), 1309-1328.
  • $\begingroup$ Brilliant answer in terms of why we need this "new" connection. So when I see $\omega$ I can just apply the exact same maths as I would if I see $\Gamma$ (the former being spin connection, the latter being affine connection)? As in, the transformations are the same? They even carry the same signs? $\endgroup$
    – Phibert
    Commented Apr 15, 2014 at 20:27
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    $\begingroup$ Another reason why $\omega^a_b$ are so useful is because when combined with the Cartan/vielbien formalism, they offer the fastest way to compute the curvature forms, which are essential for practically any investigation of a spacetime in GR. $\endgroup$
    – JamalS
    Commented Apr 15, 2014 at 20:48
  • $\begingroup$ Also, the tetrads and spin connection come naturally from gauge principles. If you apply a diffeomorphism and local Poincare transform, you can make the Lagrangian invariant by adding the tetrad and spin connection to it. $\endgroup$
    – Slereah
    Commented Apr 15, 2014 at 21:07
  • $\begingroup$ This has been one of the most clear explanations on the spin connection I've seen anywhere. $\endgroup$
    – lurscher
    Commented May 28, 2021 at 21:28

When you describe the gravity in tetrad formalism, you basically decompose affine connection, $\Gamma$, into a translational part and a Lorentz (or linear) part. So, Christoffel symbols are a (semi-direct) combination of tetrad and spin connection: $$\Gamma = e_a de^a+e_a \omega^a_{\;b} e^b$$ where the spacetime indices are omitted.

Spin connection basically encodes the Lorentz covariance of the theory which means it governs rotations and boosts of observers in spacetime. Generally, it can encode more than the Lorentz symmetry if you don't introduce a metric (see purely affine gravities and Einstein-Eddington theory).

If you consider tetrad and spin connection as a potential field (like the 4-potential, $A_\mu$, in electromagnetism), then torsion and curvature respectively become their field strength tensors (similar to electric and magnetic fields, or $F_{\mu\nu}$).

So, both tetrad and spin connection become seperate fields responsible for some symmetry (like $U(1)$ symmetry in electrodynamics). The symmetry that tetrad possesses is the translation symmetry, and is generated by momentum. On the other hand spin connection carries the spacetime rotations generated by Lorentz group.

Actually it is not necessary to have a spin connection to a gravitational theory in the tetrad formalism. For example Einstein's teleparallel gravity has only tetrads (vielbeins) but spin connections vanish, which implies that the Riemann curvature vanishes identically, but torsion does not.

Because the curvature is defined entirely by spin connection ($R=d \omega + \omega \wedge \omega$), however tetrad dynamically defines the torsion ($T=de+\omega \wedge e$) and it has nothing to do with curvature except being a coframe basis. So, if you need the spacetime to be curved you need to introduce the spin connection as well as tetrad.

By the way, if there exists fermions in your theory you need the spin connection to describe the Dirac term in curved spacetime, since the spin tensor contributes to the torsion algebraically. Hence the name. (see Einstein-Cartan gravity).

  • $\begingroup$ If the tetrads define the torsion and this is 0 in GR, why use tetrads at all? Can you achieve GR with only the spin connection? $\endgroup$
    – Anon21
    Commented Jul 6, 2022 at 13:12
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    $\begingroup$ @Anon21 Torsion is defined as $de+\omega \wedge e$, not directly propotional to tetrad. So, if tortion vanishes, it does not mean tetrad is zero or unneccessary. It only means that the total derivative of tetrad is equivelant to spin connection. From there, one can algebraically construct a spin connection in terms of tetrads (i.e. Levi-Civita spin connection). So, GR is a special case where the choice of a spin connection makes Torsion vanish (along side with non-metricity, of course). $\endgroup$ Commented Apr 13, 2023 at 6:29

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