According to the book "Supergravity" written by Freedman & van Proeyen in 5D for the existence of Majorana spinors it is necessary to introduce so called sympletic ones which requires the introduction a second spinor $\chi^2 $ which is related to the first one $\chi^1$ by (Eq. (3.86)):

$$\chi^i = \epsilon^{ij} B^{-1} (\chi^j)* \tag{1} $$

where $\epsilon^{ij}$ is the antisymmetrical symplectic matrix and $B=it_0 C\gamma^0$ is a matrix which in general $D$ fulfills $B^{\ast} B =-t_1$.

In parenthesis: "normally" one would have for a Majorana spinor: $\psi^\ast = B\psi$ (Eq.3.80 of Supergravity).

$C$ is charge conjugation matrix and $\gamma^0$ the "first" generator of the Clifford algebra in Minkowski-space.

Furthermore there are coefficients $t_r=\pm 1$ which are defined by the symmetry properties of charge conjugation matrix $C$ when it (anti)-commutes with different elements of the Clifford algebra. They are defined by the following relation for any arbitrary dimension at least $D<12$: $$(C\Gamma^{(r))})^T = -t_r C \Gamma^{(r)}$$

where $\Gamma^{(r)}$ is a matrix of the set $\{\Gamma^A =\mathbb{1}, \gamma^\mu, \gamma^{\mu_1,\mu_2},\ldots, \gamma^{\mu_1,\mu_2,\ldots, \mu_D\}}$ of rank $r$.

Actually these coefficients $t_r$ take on different values depending on the dimension of space. In the book "Supergravity" these coefficients are documented in table 3.1 of which I will only extract the values for $D=5$: $t_0=1\quad t_1=1$. So in 5 dimensions it is $B^{\ast} B =-1$

Then I guess, a transformation of the type (1) would look like this:

$$\left(\begin{array}{c} \chi^1 \\ \chi^2\end{array}\right) \rightarrow \left( \begin{array}{cc} 0 & B^{-1}\\ -B^{-1} & 0 \end{array}\right) \left(\begin{array}{c} (\chi^1)^\ast \\ (\chi^2)^\ast \end{array}\right)$$

whereas a concatentation of two such transformations

$$\left(\begin{array}{c} \chi^1 \\ \chi^2\end{array}\right) \rightarrow \left( \begin{array}{cc} 0 & B^{-1}\\ -B^{-1} & 0 \end{array}\right) \left( \begin{array}{cc} 0 & (B^{-1})^\ast \\ -(B^{-1})^\ast & 0 \end{array}\right) \left(\begin{array}{c} (\chi^1)^{\ast\ast} \\ (\chi^2)^{\ast\ast} \end{array}\right) = \left(\begin{array}{c} \chi^1 \\ \chi^2 \end{array}\right)$$

would restore the spinor we started from since in 5D necessarily $B$ fulfills $B^\ast B =-1$.

The concept of symplectic Majorana spinor is confusing for me since by the introduction of a second spinor in order to achieve the Majorana compatibility condition: $B^\ast B =-t_1$ the full spinor $(\chi^1, \chi^2)^T$ (this is apparently what is called symplectic Majorana spinor) now contains 8 components whereas the Clifford Algebra in 5D still operates with $4\times 4$ matrices $\gamma^0, \gamma^1, \gamma^2, \gamma^3,\gamma^5$.

How does this fit together ? Is the form I used for the symplectic Majorana condition correct ?

What is also disturbing is the proliferation of components only 1 dimension more than 4D.

  • $\begingroup$ If the Clifford algebra involves $n \times n$ matrices, a collection of $m$ Dirac spinor fields will have $mn$ complex components. If we like, we can rewrite the theory in terms of $2m$ (symplectic or not depending on dimension / signature) Majorana spinor fields, arriving at $2mn$ real components. I don't see what the problem is. $\endgroup$ Commented Aug 11, 2021 at 15:04

1 Answer 1


I think of the symplectic majorana as two separate Dirac fields that get interchanged under charge conjugation. In other words the index $i$ in your first equation is a "flavour index'' independent of the spinor indices.

You need this doublet in 5,6,7 (mod 8) dimensions as neither the usual Majorana which work in 2, 3, 4 (mod 8) dimensions, nor the (necessarily massless) pseudo-Majorana which work in 0,1,2 (mod 8) dimensions, are available.

The issue is that if we define a $C$ matrix by $C\gamma^\mu C^{-1}= -(\gamma^\mu)^T$ and a $T$ matrix by $T\gamma^\mu T^{-1}= +(\gamma^\mu)^T$ then for the the anticommuting $\psi$-field action density $$ \psi^T C (\gamma^\mu\partial_\mu+m)\psi $$ to be non-zero one needs $C$ to be antisymmetric, which it is in d=2,3,4 (mod 8). $C$ is symmetric in 6,7,8 and non existent in d=1,5.

Similarly, to be non-zero the massless action
$$ \psi^T T(\gamma^\mu\partial_\mu)\psi $$ needs $T$ to be symmetric which it is is in d=0,1,2(mod 8). $T$ is antisymmetric in d= 4,5,6 and non existent in d=3,7. This action cannot have a mass as $\psi^T T\psi$ is zero when $T$ is symmetric.

If you add a flavour index $i=1,2$ and insert an $\epsilon^{ij}$ before the $\gamma^\mu\partial_\mu$, you can get an non-zero action density by using the symmetric $C$ in d=7, the antisymmetric $T$ in 5 dimensions, or either in 6 dimensions. In this last case one can further reduce to get a pair of "symplectic Majorana-Weyl" fermions

  • $\begingroup$ Thank you very much for the answer. I still try to understand above all its last paragraph: Looking at $\psi^T T(\gamma^\mu\partial_\mu)\psi$ what apparently counts is the symmetry of $T\gamma^\mu$ which according to Freedman is antisymmetric in 5D. Then $\psi^T T(\gamma^\mu\partial_\mu)\psi$ should not disappear in the action integral. A mass term $\psi^T T\psi$ doesn't vanish neither, because $T$ is antisymmetric. So why do I need $\epsilon^{ij}$ ? $\endgroup$ Commented Aug 12, 2021 at 9:02
  • $\begingroup$ The derivation showing that $C$ needs to be antisymmetric is at the top of page 49 in my online notes people.physics.illinois.edu/stone/lichnerowicz.pdf. I did not include the derivation requiring $T$ to be symmetric (bottom of page 52) becuase I thought it clear that the sign difference in the def of $T$ and $C$ needs to be compensated by a chage antysmmetric to symmetric . $\endgroup$
    – mike stone
    Commented Aug 12, 2021 at 11:53

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