The Dirac gamma matrices are a set defined by the 16 following matrices: $$\Gamma^{(a)}=\{I_{4x4},\gamma^\mu,\sigma^{\mu\nu},\gamma_5\gamma^\mu,\gamma_5\}.\tag{2.122}$$ Now, I wish to determine the inverse set of gamma matrices, $\Gamma_a$.

According to Ashok Das' Lectures on QFT page 58 equation 2.124, the inverse should be defined as: $$\Gamma_{(a)}=\frac{\Gamma^{(a)}}{Tr(\Gamma^{(a)}\Gamma^{(a)})}\qquad a \text{ not summed}.\tag{2.124}$$ But I don't understand where this comes from, or why it makes sense. If I pick any gamma matrix, say $\gamma_5=\begin{pmatrix} 0 & I_{2x2} \\ I_{2x2} & 0 \end{pmatrix}.$

I can calculate $$\Gamma_a=\frac{\begin{pmatrix} 0 & I_{2x2} \\ I_{2x2} & 0 \end{pmatrix}}{Tr(\begin{pmatrix} 0 & I_{2x2} \\ I_{2x2} & 0 \end{pmatrix}\begin{pmatrix} 0 & I_{2x2} \\ I_{2x2} & 0 \end{pmatrix})}=\frac{\begin{pmatrix} 0 & I_{2x2} \\ I_{2x2} & 0 \end{pmatrix}}{Tr(I_{4x4})}=\frac{1}{4}\begin{pmatrix} 0 & I_{2x2} \\ I_{2x2} & 0 \end{pmatrix}$$ But here, clearly $$\Gamma_a\Gamma^a=\frac{I_{4x4}}{4},$$ which is not what I expect. So how is this properly used? How does one define the inverse Dirac Gamma Matrices?

  • 1
    $\begingroup$ Just a note that I think you'd have the same problem with choosing $I_{4\times 4}$ instead of $\gamma_5$. That might be an easier example to debug. $\endgroup$
    – Andrew
    Commented Jan 4, 2023 at 19:12

1 Answer 1


Yes, Ashok Das should strictly speaking not call (2.124) the "inverse set of matrices"; they are only proportional$^1$ to the inverse. Rather (2.124) is (2.122) where the upper collective index $(a)$ of the 16 matrices (2.122) has been lowered by a metric $g_{(a)(b)}$. The (inverse) metric is here defined as $$g^{(a)(b)}~:=~ {\rm Tr}(\Gamma^{(a)}\Gamma^{(b)}), \qquad a,b~\in~\{1,\ldots,16\},\tag{2.123}$$ which is diagonal. The explicit list of $\Gamma_{(a)}$ is given in (2.126).


$^1$ It is straightforward to check this explicitly by going through the list.

  • $\begingroup$ Is there a reason why for the Gamma matrix lowering (or raising) the index gives us the inverse of that matrix? I mean, is it a coincidence or is that always the case? $\endgroup$ Commented Jan 4, 2023 at 19:26
  • $\begingroup$ @Nick Heumann It is implicit in the Dirac Clifford algebra with one index down and one up, in which case the r.h.s. is a Kronecker δ... $\endgroup$ Commented Jan 4, 2023 at 21:14

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