# Signature of trace of Dirac Matrices

I came across this question in my problem set:

Let $\gamma^\mu$, $\mu=0,1,2,3$ be the Dirac matrices, satisfying: \begin{eqnarray} \gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2\eta^{\mu\nu}I, \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \eta^{\mu\nu}=\mathrm{diag}(+1,-1,-1,-1). \end{eqnarray} And $I$ is the identity matrix. Define the matrices: \begin{eqnarray} \Sigma^{\mu\nu}=[\gamma^\mu,\gamma^\nu], \:\:\:\:\:\:\:\:\:\:\:\:\ \Gamma^{\mu\nu\rho}=\frac{1}{3}(\gamma^\mu\gamma^\nu\gamma^\rho+\textrm{cyclic permutations}),\:\:\:\:\:\:\:\:\:\:\:\:\ \gamma^5=\gamma^0\gamma^1\gamma^2\gamma^3. \end{eqnarray} And denote these elements by $\Gamma^i$. Consider the vector space $\Lambda V$ generated by $\Gamma^i$.

a) Show that $\mathrm{Tr}(\Gamma^i)=0$ for every $i$, but $\mathrm{Tr}((\Gamma^i)^2)\neq0$.

b) Show that $\langle\Gamma^i,\Gamma^j\rangle=\mathrm{Tr}(\Gamma^i\Gamma^j)$ defines a non-degenerate scalar product in $\Lambda V$, with signature (number of positive and negative eigenvalues) given by $8$ and $8$, respectively.

Attempt of a solution: Clearly there are $16$ linearly independent matrices:

$I$ gives 1;

$\gamma^\mu$ gives 4;

$\Sigma^{\mu\nu}$ gives 6;

$\Gamma^{\mu\nu\rho}$ gives 4;

$\gamma^5$ gives 1.

And it is easy to see that $\mathrm{Tr}(\Gamma^i)=0$ (except for $I$) and $\mathrm{Tr}((\Gamma^i)^2)\neq0$. Also it is easy to show that this scalar product is non-degenerate. However I don't know how to compute the signature. What is the definition of the signature? For which matrix should I compute the eigenvalues? $\Lambda V$ is a complex vector space, right? Does the signature really give 8,8? As this is a physics class and not a math class I believe that I the proof should't be too complicated.

• I had to add a $i$ factor in the definition $\Gamma^{\mu\nu\rho}$ to get the correct signature. – user191374 Apr 12 '18 at 15:32

2) The matrix for which you need to compute the eigenvalues is the $16\times 16$ matrix, which I'll call $A$, whose components are $$A^{ij}=Tr(\Gamma^i\Gamma^j)\ \ .$$ The task of computing the eigenvalues will turn out to be easier than it sounds, because it looks like $A$ turns out to be diagonal. (I checked that many of the off-diagonal components are zero, although I didn't check them all.) The gamma matrix trace identities listed in Wikipedia will speed up this task.
1) The definition of the signature, as given in the problem, is the pair of numbers consisting of the number of positive eigenvalues of $A$, and the number of negative eigenvalues of $A$.
3) Yes, $\Lambda V$ is a complex vector space. However, $A$ is real, and its eigenvalues are all real.
4) Yes, presumably $A$ does indeed have $8$ positive and $8$ negative eigenvalues. I personally only computed $12$ out of the $16$ eigenvalues before I ran out of steam, but $8$ positive and $8$ negative eigenvalues was within the range of possible outcomes as of when I decided I'd had enough.