Can gamma matrices be real in 6 dimensions?

Question

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!

Answer 1

(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}

Answer 2

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.