# The relationship between spin and spinor curvature

The identity,

$$-\gamma^b{\mathcal{R}}_{ab} = {\mathcal{R}}_{ab}\gamma^b = \frac{1}{2}\gamma^b R_{ab}$$ is presented in the answer to the question Dirac Equation in General Relativity.

How does one prove the identity?

The curvature two-form is defined by $$\newcommand{\Rcal}{\mathcal{R}} \Rcal_{ab} \Psi = [D_a, D_b] \Psi.$$ Here $D_a$ is the covariant derivative without gauge terms. $\Rcal_{ab}$ takes values in the Dirac representation of the Lorentz Lie algebra. Thus, really the relation is $$\mathcal R_{ab \alpha\beta} \Psi_\beta = [D_a, D_b] \Psi_\alpha$$ where $\alpha, \beta$ are Dirac spinor indices. Compare this with the more familiar Riemann tensor $$R_{ab}{}^\mu{}_\nu x^\nu = [D_a, D_b] x^\mu.$$ Since the Riemann tensor is antisymmetric in $\mu,\nu$ we can consider it too to be a 2-form taking values in a representation of the Lorentz Lie algebra.. Of course, this representation is the 4-vector representation.

The Dirac representation of the Lie algebra is realized by $$\epsilon_{\mu\nu} \mapsto \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu$$ that is under an infinitesimal Lorentz transformation by $\epsilon_{\mu\nu}$, $$\Psi \mapsto \Psi + \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu \Psi.$$ This means that $$\Rcal_{ab} = \frac{1}{4}R_{abst}\gamma^s \gamma^t.$$

Now let us define the Ricci tensor by $$R_{ab} = R_a{}^\mu{}_{\mu b} = R_{astb} g^{st}.$$ Then from the fundamental anticommutation relation of the gamma matrices, we can write \begin{align} R_{ab} \gamma^b & = \frac{1}{2} R_{astb} (\gamma^s \gamma^t + \gamma^t \gamma^s) \gamma^b \\ & = \frac{1}{2} R_{astb}\gamma^s\gamma^t \gamma^b - \frac{1}{2} (R_{abst} + R_{atbs}) \gamma^t\gamma^s \gamma^b \tag{1}.\end{align} Here I have used the symmetry of the Riemann tensor, $R_{astb} + R_{abst} + R_{atbs} = 0$. Note that the first term is precisely $2\Rcal_{as}\gamma^s$. Now $R_{atbs} = -R_{atsb}$ and so a relabeling of contracted indices in the last term shows that this term also contributes $2\Rcal_{as}\gamma^s$. For the middle term, use the anticommutation relation to find $$\gamma^t \gamma^s \gamma^b = \gamma^b \gamma^t \gamma^s - 2g^{tb}\gamma^s + 2g^{sb}\gamma^t \tag{2}.$$ Hence, \begin{align}R_{abst}\gamma^t\gamma^s \gamma^b & = -R_{abts} \gamma^b \gamma^t \gamma^s - 2R_a{}^\mu{}_{s \mu} \gamma^s + 2R_a{}^\mu{}_{\mu t}\gamma^t \\ & = -\Rcal_{as}\gamma^s + 4R_{as}\gamma^s. \end{align}

We now have that (1) is $$R_{ab}\gamma^b = 6 \Rcal_{ab}\gamma^b - 2R_{ab}\gamma^b$$ so clearly $$\frac{1}{2}R_{ab}\gamma^b = \Rcal_{ab}\gamma^b \tag{3}$$ which is one of the identities in your question. The other should follow from using the anticommutation relations and (3).

• Why so $\Rcal_{ab} = \frac{1}{4}R_{abst}\gamma^s \gamma^t$ from $\Psi \mapsto \Psi + \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu \Psi$ ? – user48875 Jun 19 '14 at 0:08
• And is there this calculations in the literature? What literature is? – user48875 Jun 19 '14 at 0:11
• Because in the Dirac representation, the element $\epsilon_{\mu\nu}$ of the Lorentz Lie algebra is represented by $\frac{1}{4}\epsilon_{\mu\nu}\gamma^\mu\gamma^\nu$. This is a fairly standard result, if nothing else it should be in Penrose and Rindler. I don't know if the other calculations are anywhere in the literature. – Robin Ekman Jun 19 '14 at 0:16
• My first question may reformulate so: "How does the curvature from element of the Lorentz Lie algebra arise?" I don't know this fairly standart result, unfortunately. Whence does it follow? – user48875 Jun 21 '14 at 9:47
• The Dirac representation is a direct sum of a left-handed and a right-handed spin $1/2$ fermion. Write the $\gamma^\mu$ matrices in the Weyl representation and check that for example $\gamma^1\gamma^2$ generates rotations around the $3$-axis, and $\gamma^0 \gamma^1$ boosts along the $1$-axis (by comparing to how they are represented in for right- and left-handed particles). – Robin Ekman Jun 21 '14 at 10:14