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.
 A: 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.
