Do the four gamma matrices form a basis for the set of matrices $GL(4,\mathbb{C})$? I was actually trying to evaluate a term like $\gamma^0 M^\dagger \gamma^0$ in a representation independent way, where $M, M^\dagger$ are $4\times 4$ matrices.

  • $\begingroup$ To form a complete basis of 𝑀(4,ℂ), are the coefficients of the 16-elements vector real or complex? $\endgroup$
    – Anon21
    May 3, 2020 at 15:27

3 Answers 3


As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way

  • 1 the identity matrix $\mathbb{1}$
  • 4 matrices $\gamma^\mu$
  • 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$
  • 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$
  • 1 matrix $\sigma^{\mu\nu\rho\delta}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\delta]}=i\epsilon^{\mu\nu\rho\delta}\gamma^5$

these 16 matrices form the basis that we were looking for.

Furthermore they are used to construct the spinor bilinears multiplying by $\bar{\psi}$ on the left and $\psi$ on the right, which transform in the Lorentz indices as follows

  • $\bar{\psi}\psi$ scalar
  • $\bar{\psi}\gamma^\mu\psi$ vector
  • $\bar{\psi}\sigma^{\mu\nu}\psi$ 2nd rank (antisymmetric) tensor
  • $\bar{\psi}\sigma^{\mu\nu\rho}\psi$ pseudovector
  • $\bar{\psi}\gamma^5\psi$ pseudoscalar

the fact that they form a basis of of $M(4,\mathbb{C})$ is very important because these are then the only independent spinor bilinears (i.e $\bar{\psi}M\psi$) that can be constructed, any other can be expressed a linear combination of these. A different issue is if it would make any sense to sum any of these since they are different types of tensors under Lorentz group transformations.

  • $\begingroup$ I can't help but notice that a 2nd rank symmetric tensor is conspicuously absent can you form one from the anticommutator and the spinors as in the above? $\endgroup$
    – R. Rankin
    Dec 19, 2018 at 7:18
  • $\begingroup$ What does the bracket notation $^{[\mu}\gamma^{\nu]}$ in $\gamma^{[\mu}\gamma^{\nu]}$ means? $\endgroup$
    – Anon21
    Aug 1, 2019 at 13:09
  • $\begingroup$ @AlexandreH.Tremblay It's just the antisymmetrization $\gamma^{[\mu} \gamma^{\nu]} = \frac{1}{2} (\gamma^{\mu} \gamma^{\nu} - \gamma^{\nu} \gamma^{\mu})$ $\endgroup$
    – jpm
    Aug 9, 2019 at 8:33

To complement V. Moretti's excellent answer, it's worth emphasizing that the dimension of the four-by-four complex matrices $\mathbb C^{4\times 4}$, when seen as a vector space over $\mathbb C$, is $4\!\times\!4=16$. As such, a set of four matrices (i.e. vectors in $\mathbb C^{4\times 4}$) can never be a basis for it.

It's also worth saying that the general linear group $\text{GL}(4,\mathbb C)\subset\mathbb C^{4\times 4}$, i.e. the four-by-four matrices with nonzero determinant, is not a vector space (for one, it doesn't have a zero), and it is therefore misleading to speak of a basis for it. That said, it is still possible to ask for a minimal set of vectors (i.e. matrices) whose span will contain $\text{GL}(4,\mathbb C)$; this turns out to require a full basis of $\mathbb C^{4\times 4}$ because the matrices you 'skip', $\mathbb C^{4\times 4}\setminus\text{GL}(4,\mathbb C)$, have measure zero, so $\text{GL}(4,\mathbb C)$ is a complex manifold of dimension 16 and requires that many parameters to describe.


No they do not, due to dimensional reasons, but they are generators of the algebra. That is, $I$ and the products of $\gamma^a$ (products of one, two, three and four matrices) form such a basis.

NOTE ADDED. As Emilio Pisanty correctly remarked (also making some further interesting comments) $GL(4, \mathbb C)$ is not a linear space so questions about bases of it are inappropriate. In fact I implicitly interpreted that $GL(4,\mathbb C)$ as $M(4, \mathbb C)$, the complex algebra of $4\times 4$ complex matrices which, by definition, is also a complex vector space.

  • $\begingroup$ What is the difference between the notations $GL(4,\mathbb{C})$ and $M(4,\mathbb{C})$? Is the objection of basis due to the fact that former is a group and not a linear vector space? $\endgroup$
    – SRS
    Nov 8, 2016 at 10:08
  • 1
    $\begingroup$ Yes, $GL(n, K)$ is a group, $M(n, K)$ is an associative unital algebra, but it is not a group when forgetting the linear structure since it also includes non-invertible matrices differently from $GL(n, K)$. $\endgroup$ Nov 8, 2016 at 10:18

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