# Can gamma matrices be real in 6 dimensions?

I'm trying to find the really real representation of 6D gamma matrices. The problem is that "do they really exist?"

If yes, then how am I supposed to construct them? Thank you!

• Euclidean or Minkowski signature? – Qmechanic Apr 26 at 10:21
• And also (+,-,-,-,-,-) Minkowski metric or (-,+,+,+,+,+) Minkowski metric? You can only have Majorana reps in Minkowski d=2,3,4 (mod 8), but in all-plus Euclidean there is a purely imaginary gamma rep, and in all-minus Euclidean there is a purely real rep. – mike stone Apr 26 at 10:58
• How about the majorana representation in 6D? Are they purely imaginary? – Super stuff Apr 27 at 1:57
• @ChiralAnomaly Majorana representations are a minefield of inconsistent conventions and careless statements, see physics.stackexchange.com/q/309890/50583 for how different prima facie contradictory claims about the existence of real representations fit together. – ACuriousMind Apr 27 at 12:33
• @Chiral Anomaly. I'd claim Majorana in 2, 3,4 mod 8, but pseudo-Majorana in 0,1,2 mod 8, with p-majorana being necessarily massless as the pseudo "charge conjugation" flips the sign of $\bar\psi \psi$ while the usual one leaves $\bar\psi \psi$ invariant. Usual Majorana gammas are real in East-coast mostly + metric and pure imaginary in the West-coast mostly minus; the conjugaton operation in both cases being an antilinear involution and so defining a "real structure" on the representation space. Is this in agreement with your definitions? – mike stone Apr 27 at 16:58

(Note: This answer only addresses the question about gamma-matrices. The existence and/or properties of the associated spinor spaces and conjugations are not addressed.)

The existence of a real representation depends on the signature. Let $$(p,m)$$ denote the signature in which $$p$$ of the gamma matrices square to $$+I$$ and $$m$$ of them square to $$-I$$, where $$I$$ is the identity matrix. If $$p+m=2n$$ or $$p+m=2n+1$$, then a representation using matrices over $$\mathbb{R}$$ of size $$2^n\times 2^n$$ exists if and only if $$p-m\in\{0,1,2\}$$ (mod 8).

When specialized to the case $$p+m=6$$ that was specified by the OP, this implies that real representations using gamma-matrices of the minimal size $$8\times 8$$ exist only for the signatures $$(4,2)$$, $$(3,3)$$, and $$(0,6)$$. Real representations of this minimal size don't exist for the Minkowski signatures $$(5,1)$$ or $$(1,5)$$.

Here are explicit constructions of real representations with the signatures for which such representation are possible. First define the $$2\times 2$$ matrices $$\begin{gather} X=\left[\begin{matrix} 0&1\cr 1&0\end{matrix}\right] \hskip1cm Y=\left[\begin{matrix} 0&1\cr -1&0\end{matrix}\right] \\ Z=\left[\begin{matrix} 1&0\cr 0&-1\end{matrix}\right] \hskip1cm I=\left[\begin{matrix} 1&0\cr 0&1\end{matrix}\right]. \end{gather}$$ For signature $$(4,2)$$, we have the real representation $$\begin{gather} \gamma_1 = X\otimes X\otimes X \hskip1cm \gamma_2 = X\otimes X\otimes Z \hskip1cm \gamma_3 = X\otimes Z\otimes I \\ \gamma_4 = Z\otimes I\otimes I \hskip1cm \gamma_5 = X\otimes X\otimes Y \hskip1cm \gamma_6 = X\otimes Y\otimes I. \end{gather}$$ This representation uses matrices of size $$8\times 8$$, which is the smallest possible size. To get a real representation with signature $$(3,3)$$, just change $$Z$$ to $$Y$$ in $$\gamma_4$$. To get a real representation with signature $$(0,6)$$, start with the $$(4,2)$$ case and use \begin{align} \hat\gamma_1 &:= \gamma_2\gamma_3\gamma_4 \\ \hat\gamma_2 &:= \gamma_1\gamma_3\gamma_4 \\ \hat\gamma_3 &:= \gamma_1\gamma_2\gamma_4 \\ \hat\gamma_4 &:= \gamma_1\gamma_2\gamma_3\\ \hat\gamma_5 &:= \gamma_5\\ \hat\gamma_6 &:= \gamma_6. \end{align}

• What about $8\times 8$ real Weyl gamma matrices for $d=5+1$? Remember that in $d=5+1$ chirality commutes with complex conjugation so it is possible to take the $4\times 4$ Weyl gamma matrices and replace the imaginary number by a real antimsymmetric $2\times 2$ matrix with determinant equal 1. – Nogueira May 5 at 20:03
• @Nogueira That's a good point. I was thinking of representations of the whole Clifford algebra, not just the even part; but you're right: why not consider chiral spinors, too? My own username should have reminded me of that... :) – Chiral Anomaly May 5 at 21:14

You can always obtain a real representation of a spinor by enlarging the dimension. I presume that this is not what you want. I presume that what you want are $$8\times 8$$ real gamma matrices for euclidean signature or $$4\times 4$$ real gamma matrices for Minkowski signature ($$4\times 4$$ since complex conjugation in $$d=1+5$$ does not flip chirality).

For euclidean signature, organizing the gamma matrices as fermionic creation-annihilation operators

$$b_{i}^{\pm}= \Gamma_{i}\pm i\Gamma_{7-i},\,\,\,\,i=1,2,3$$

a Dirac spinor $$\psi$$ will have indices $$(+++,+--,---,++-)$$ plus permutations of the signs, so we get 8 components, and the gamma matrices will be $$8\times 8$$ complex matrices. Doing a similarity transformation:

$$\psi\rightarrow U \psi \qquad \Gamma^{m}\rightarrow U\Gamma^{m}U^{-1}$$

We want to see if is possible to make the gamma matrices all real. It is convenient to introduce the following $$B_{\pm}$$ matrices:

$$B_{\pm}\Gamma^{m}B^{-1}_{\pm}=\pm (\Gamma^{m})^{*}\qquad B_{+}=\Gamma_{1}\Gamma_2\Gamma_3\qquad B_{-}=\Gamma_{4}\Gamma_5\Gamma_6$$

Under the similarity transformation the $$B_{\pm}$$ matrices should transform as

$$B_{\pm}\rightarrow (U^{-1*}B_{\pm}U)$$

in order to preserve the equation that defines them (they are representation dependent objects). So the similarity transformation we are seeking for is the one that implies

$$(U^{-1*}_{\pm}B_{\pm}U_{\pm}) = 1\qquad B_{\pm}U_{\pm}=U^{*}_{\pm}$$

If we find a $$U$$ that satisfies this equation for $$B_{+}$$ we have the similarity transformation that gives the real gamma matrices and for $$B_{-}$$ pure imaginary matrices.

This equation does not have solution for $$B_{+}$$, only for $$B_{-}$$. In other to see this just compute the condition for integrability of this linear equations:

$$B_{\pm}U_{\pm}=U^{*}_{\pm} \leftrightarrow B_{\pm}B_{\pm}^{*}= 1$$

But $$B_{+}B_{+}^{*}=-1$$ since $$B_{+}$$ is real and $$B_{+}^{2}=-1$$ so there is no similarity transformation $$U_{+}$$ that gives real gamma matrices. Nevertheless $$B_{-}$$ is imaginary so $$B_{-}B_{-}^{*}=+1$$ and the solution is easy to obtain by inspection:

$$U_{-}=B_{+}(1+B_{-})$$

Note that if you hit this equation by $$B_{-}$$ you obtain the complex conjugate of $$U_{-}$$:

$$U_{-}^{*}=B_{+}(1-B_{-})$$

If you want to obtain real gamma matrices for $$SO(6)$$ you may multiply each pure imaginary gamma matrix by $$i$$. This will change the Clifford algebra by an overall minus sign:

$$\{\Gamma^{m}\Gamma^{n}\}=2\eta^{mn}\rightarrow \{\Gamma^{m}\Gamma^{n}\}=-2\eta^{mn}$$

But this will not change the Lorentz transformation and can be absorbed by changing some conventions as the sign that comes in front of the Lorentz generator:

$$\exp\left(\frac{i}{4}[\Gamma_{m},\Gamma_{n}]\theta^{mn}\right)$$

For Minkwoski signature the organization of gamma matrices as fermionic creation-annihilation operators $$b_{i}$$ should be modified in order to preserve the usual creation-annihilation algebra:

$$\{b_{i}^{\pm},b_{j}^{\pm}\}=0\qquad\{b_{i}^{+},b_{j}^{-}\}=\delta_{ij}$$

Let us say that now the $$6$$-th direction is the new $$0$$-th time direction. You can simply do $$\Gamma_{6}\rightarrow i\Gamma_{0}$$ to fix the algebra. This modify the $$b_3^{\pm}$$ which implies that now both $$\Gamma_{0}$$ and $$\Gamma_{3}$$ are real. This will imply that

$$B_{+}=\Gamma_{4}\Gamma_{5}\qquad B_{-}=\Gamma_{0}\Gamma_{1}\Gamma_{2}\Gamma_{3}$$

Now the integrability condition $$B_{\pm}B_{\pm}^{*}=1$$ will not be satisfied for both. We get

$$B_{\pm}(B_{\pm})^{*}=-1$$

Note that $$\Gamma_{0}\Gamma_0=-1$$ since it is equal to $$(i\Gamma_{6})^2=-1$$. This imply that it is impossible to do a similarity transformation that gives real or imaginary gamma matrices in d=5+1 Minkowski space-time starting from $$8\times 8$$ Dirac gamma matrices.

However, if you diagonalize the chirality matrix $$\Gamma=\Gamma_{012345}$$ and look into a eigenspace you obtain $$4\times 4$$ gamma matrices that describes Weyl spinors. Then, if you replace $$i\rightarrow J$$ where

$$J = \begin{bmatrix} 0 & -1\\1 & 0 \end{bmatrix}$$

then you obtain $$8\times 8$$ real chiral gamma matrices.