# Cyclic identity of $\gamma$-matrices in 3D and 4D

Reading the book on Supergravity from Freedman & van Proeyen I was very bewildered by the so called cyclic identity of $$\gamma$$-matrices of the Clifford-algebra(eq. 3.67) important in string theory:

$$(\gamma_\mu)_{\alpha(\beta} (\gamma^\mu)_{\gamma\delta)} = 0$$

(the round brackets indicate a symmetrisation operation on the matrix res. bispinor indices) which is according to the book for $$D=3$$ and $$D=4$$ a collorary of the Fierz identity:

$$(\gamma^\mu)_\alpha^{\;\beta} (\gamma_\mu)_\gamma^{\;\delta }=\frac{1}{2^m}\sum_A (-1)^{r_A} (D-2r_A) (\Gamma_A)_\alpha^{\;\delta} (\Gamma^A)_\gamma^{\;\beta}$$

where the $$\Gamma_A$$ the basis elements of the Clifford-algebra, $$r_A$$ their rank (in terms of $$\gamma$$-matrices, i.e. $$r_A=0$$ for $$\Gamma_A=\mathbf{1},\, r_A=1$$ for $$\Gamma^A=\gamma^\mu$$ and $$r_A=2$$ for $$\Gamma^A = \gamma^{[\mu}\gamma^{\nu]}$$ etc.) and $$D$$ the dimension of the considered space. $$D$$ is related to the integer $$m$$ either by $$D=2m$$ or $$D=2m+1$$.

I fear my capacities of proving that are rather limited, but at least I wanted to make a crosscheck. I know that in 3D (where the cyclic identity is supposed to hold) the $$\gamma$$-matrices can be represented by the Pauli-matrices $$\gamma_0 = i\sigma_1$$, $$\gamma_1 = \sigma_2$$ and $$\gamma_2 = \sigma_3$$. I put an imaginary $$i$$ as coefficient of $$\sigma_1$$ in order to make the 3D-$$\gamma$$ matrices Minkowski. Furthermore I assume that covariant and contravariant matrices are related with each other by $$\gamma^{\mu} = \eta^{\mu\nu}\gamma_\nu$$, i.e. by only a sign $$\pm 1$$. Then I write:

$$0=(\gamma_\mu)_{\alpha(\beta} (\gamma^\mu)_{\gamma\delta)}=-(\sigma_1)_{\alpha(\beta}(\sigma^1)_{\gamma\delta)}+(\sigma_2)_{\alpha(\beta}(\sigma^2)_{\gamma\delta)} + (\sigma_3)_{\alpha(\beta}(\sigma^3)_{\gamma\delta)}$$

Then I chose the simplest set of indices $$\alpha=\beta=\gamma=\delta=1$$ and consider that any symmetrisation on equal indices stays without effect:

$$-(\sigma_1)_{11}(\sigma^1)_{11}+(\sigma_2)_{11}(\sigma^2)_{11} + (\sigma_3)_{11}(\sigma^3)_{11}\neq 0$$

since $$(\sigma^3)_{11} = 1$$ whereas $$(\sigma^1)_{11} = (\sigma^2)_{11} = 0$$.

If this cyclic identity is right, then, I guess, I haven't understood something rather fundamental about it. I would very glad if somebody could demonstrate that I am wrong with the found counter-example and even more glad for a proof of it.

As the cyclic identity plays an important role in string theory (string and brane actions according to Freedman & van Proeyen (F&P)) I consider worth to post a "proof" of the cyclic identity, albeit in its certainly rather inelegant way. However, I am unaware of how/if it would work for $$D>4$$.

I had overlooked a detail in the way of lowering indices of $$\gamma$$-matrices that allowed me finally to proof the Fierz rearrangement cyclic identity at least in 3D. As F&P explain in their book for lowering indices of $$\gamma$$-matrices the charge conjugation matrix $$C$$ is applied that is also used for a scalar product of bispinors $$\lambda$$ and $$\chi$$: $$\overline{\lambda} \chi:=\lambda^T C \chi$$. So for $$\gamma$$-matrices one defines (I will use Latin indices for 4-vector indices and Greek letters for bispinor indices.):

$$\gamma^k_{\alpha\beta} = (\gamma_k)_\alpha^\gamma C_{\gamma\beta}\quad\quad (3.62)$$

Taking this into account the cyclic identity requires the following (index $$\nu$$ is not included in the symmetrisation operation):

$$0=(\gamma^k)_{\alpha(\beta} (\gamma_k)_{\gamma\delta)} = (\gamma^k)_\alpha^\mu C_{\mu(\beta}(\gamma_k)_\gamma^\nu C_{\nu\delta)}$$

According to F&P this is true for $$D=3$$ and $$D=4$$, so it is rather easy to check that for $$D=3$$ as the $$\gamma$$-matrices are only 2D. Actually I went through case by case using a 2D-representation of the $$\gamma$$-matrices from N.Lambert "Introduction to Supersymmetry" (Balkan Lectures 2011):

$$\gamma^0 = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$$

$$\gamma^1 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$$

$$\gamma^2 = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$

The use of this representation implies a Minkowski-metric $$\eta_{ik} = diag(-1,+1,+1)$$. Finally, we assume that $$C=\gamma^0$$ given above, this $$C$$ indeed fulfills its definition equation: $$(\gamma^k)^T = - C\gamma^k C^{-1}$$.

My proof is rather primitive, as I checked it for the possible index permutations of 1 and 2: $$(\beta,\gamma,\delta)=(111)$$, $$(112)$$, $$(221)$$ and $$(222)$$. Furthermore there are 2 possibilities for $$\alpha =1, 2$$. The other permutations are due to the symmetrisation operation equivalent to one of the 4 cited cases. For instance for $$(\beta,\gamma,\delta)=(111)$$ we have for $$\alpha=1$$:

$$(\gamma^k)_{1(1} (\gamma_k)_{11)} = \eta_{ik}(\gamma^k)_1^2 C_{2(1}(\gamma^i)_1^2 C_{2 1)} = \eta_{ik}(\gamma^k)_1^2 C_{21}(\gamma^i)_1^2 C_{2 1}=-((\gamma^0)_1^2 )^2 + ((\gamma^1)_1^2 )^2 + ((\gamma^2)_1^2 )^2 = -1 + 1 + 0 =0$$

The case $$(\beta,\gamma,\delta)=(111)$$ seems to be particularly simple as all the indices are same, in particular the symmetrisation operation brackets can be omitted here as they have no effect. For $$\alpha=2$$ the evaluation is rather similar. Evidently I won't demonstrate it for all the cases, but I checked indeed for all the cases.

For $$D=4$$ the same style of proof would be really tedious, but actually more interesting than the pure $$\gamma$$-matrix relation is its application on Grassmann-bispinors $$\lambda_\mu, (\mu=1,2,3)$$(3.68):

$$\gamma_k \lambda_{[1}\overline{\lambda}_2 \gamma^k \lambda_{3]}=0.$$

This can actually be proven with the Fierz-relation (3.70). According to an addendum to chapter 3 of the F&P-book one starts off with $$\gamma_k \lambda_{[1}\overline{\lambda}_{2]}\gamma^k \lambda_3$$:

$$\gamma_k\lambda_{[1}\overline{\lambda}_{2]}\gamma^k \lambda_3 = -\frac{1}{4} \sum_A \gamma_k \Gamma^A\gamma^k \lambda_{3}\overline{\lambda}_{[2} \Gamma_A \lambda_{1]}$$

The sum runs over the basis of the Clifford-algebra. Due $$\gamma$$-matrix properties in D=4 in the term $$\overline{\lambda}_{[2} \Gamma_A \lambda_{1]}$$ only the cases $$\Gamma_A = \gamma^k$$ and $$\gamma^{ik}$$ need to be considered. Due to $$\gamma_j \gamma^{ik}\gamma^j = (D-4)\gamma^{ik} =0$$ we are only left with $$\Gamma_A = \gamma^k$$. Then we get using $$\gamma_k \gamma^i\gamma^k =-2 \gamma^i$$:

$$\gamma_k \lambda_{[1}\overline{\lambda}_{2]}\gamma^k \lambda_3 = -\frac{1}{4} (-2)\gamma^i\lambda_3\overline{\lambda}_{[2}\gamma_i \lambda_{1]}= -\frac{1}{2} \gamma^i\lambda_3\overline{\lambda}_{[1}\gamma_i \lambda_{2]}$$

Finally we get from this:

$$\gamma_k \lambda_{[1}\overline{\lambda}_{2]}\gamma^k \lambda_3 + \gamma_k \lambda_{[1}\overline{\lambda}_{2]}\gamma^k \lambda_3 + \gamma^k\lambda_3\overline{\lambda}_{[1}\gamma_k \lambda_{2]} =0$$

which can be cast in:

$$0=\gamma_k \lambda_{[1}\overline{\lambda}_{2]}\gamma^k \lambda_3 + \gamma_k \lambda_{[2}\overline{\lambda}_{3]}\gamma^k \lambda_1 + \gamma_k \lambda_{[3}\overline{\lambda}_{1]}\gamma^k \lambda_2 =3\gamma_k \lambda_{[1}\overline{\lambda}_2 \gamma^k \lambda_{3]}$$

QED.