Do the four gamma matrices along with the identity element constitute a lie algebra?

With real coefficients we have

$$ \mathbf{v}_{\mathbb{R}}=aI+t\gamma_0+x\gamma_1+y\gamma_2+z\gamma_3 \tag{real coefficients} $$

or using complex coefficients as $$ \mathbf{v}_\mathbb{C}=z_a I+ z_0 \gamma_0+z_1 \gamma_1+z_2\gamma_2+z_2\gamma_3. \tag{complex coefficients} $$

What Lie algebra is associated with $\{1, \gamma_0, \gamma_1,\gamma_2,\gamma_3 \}$?

I am already familiar with this question Do gamma matrices form a basis?, stating that the 16 basis of the Clifford algebra forms a basis of $M(4,\mathbf{C})$, but what about the 5 elements of $\{1, \gamma_0, \gamma_1,\gamma_2,\gamma_3 \}$?

Based on the comments here is the commutator of $\mathbf{v}_{\mathbb{R}}$.

$$ [\mathbf{v}_{1},\mathbf{v}_{2}]=\mathbf{v}_{1}\mathbf{v}_{2}-\mathbf{v}_{2}\mathbf{v}_{1} $$

Using 1+1 to simplify, we have

$$ \begin{eqnarray} [\mathbf{v}_{1},\mathbf{v}_{2}] &&= (a+b\gamma_0)(c+d\gamma_0)-(c+d\gamma_0)(a+b\gamma_0)\\ &&=(ac+ad\gamma_0+bc\gamma_0+bd\gamma_0^2)-(ca+cb\gamma_0+da\gamma_0+db\gamma_0^2)\\ &&=(ac-ac)+(ad-ad)\gamma_0+(bc-bc)\gamma_0+(bd-bd)\gamma_0^2\\ &&=0 \end{eqnarray} $$

  • 2
    $\begingroup$ OP's vector spaces endowed with the commutator do not form a Lie algebra. However, they generate (in the algebra sense) the Lie algebra $gl(4,\mathbb{F})$. $\endgroup$
    – Qmechanic
    Commented Jan 25, 2020 at 16:48
  • $\begingroup$ @Qmechanic can you tell me which property of the lie bracket fails for this vector space? $\endgroup$
    – Anon21
    Commented Jan 25, 2020 at 16:49
  • 4
    $\begingroup$ $[\gamma,\gamma]\neq\sum z_i\gamma_i$ $\endgroup$ Commented Jan 25, 2020 at 16:53
  • $\begingroup$ $\gamma_0$ commutes with itself and with the identity, of course. $\endgroup$ Commented Jan 25, 2020 at 20:44
  • $\begingroup$ @CosmasZachos So then, is the basis $\{1, \gamma_0,\gamma_1,\gamma_2,\gamma_3\}$ a lie algebra? And if so, of which group. $\endgroup$
    – Anon21
    Commented Jan 25, 2020 at 23:28

1 Answer 1


First of all, your set is not closed. For example, $$ [\gamma_0,\gamma_1]=\gamma_0\gamma_1 - \gamma_1\gamma_0 = \gamma_0\gamma_1 + \gamma_0\gamma_1= 2\gamma_0\gamma_1 $$ lies outside the said set. (BTW, the identity element $1$ or $I$ belongs to the Lie group, not the Lie algebra.)

If you set out to find a closed Lie algebra, the above suggests that you have to include $$ \gamma_0\gamma_1 $$ into the mix. And if you goof around further, you would stumble upon the 10-element closed set $$ \{\gamma_0, \gamma_1,\gamma_2,\gamma_3, \gamma_0\gamma_1, \gamma_0\gamma_2, \gamma_0\gamma_3, \gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1 \}, $$ which turns out to be a bona fide Lie algebra.

What could this 10-element Lie algebra be? It's no other than the de Sitter algebra $$ so(1, 4) $$ which corresponds to the 5-dimensional rotation group.

If you are an able college student, you would recognize that the 6-element subset $$ \{\gamma_0\gamma_1, \gamma_0\gamma_2, \gamma_0\gamma_3, \gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1 \} $$ constitutes the Lorentz algebra $so(1,3)$, which is tied to 4-dimensional space-time rotation.

If you are the curious bunch, you might also wonder what can the subset $$ \{\gamma_0, \gamma_1,\gamma_2,\gamma_3\} $$ be?

The straight forward interpretation is that they are the 4 rotations alone the planes spanned by the 5th dimension and each 4 space-time dimension (did we mention that de Sitter is 5-dimensional rotation?). In math jargon, they form the coset $$ so(1, 4)/so(1,3). $$

That said, we can look at them from a different angle: if we re-scale the identity we pondered on earlier $$ [\gamma_0,\gamma_1]=\epsilon\gamma_0\gamma_1 \rightarrow 0 (\epsilon \rightarrow 0) $$ which means your New Year wish is granted, i.e. the gamma matrices commute with each other, we can thus identify $\{\gamma_0, \gamma_1,\gamma_2,\gamma_3\}$ with the space-time translation symmetry (recalling that the Dirac derivative $\not \partial = \gamma^\mu \partial_{\mu}$ couples the space-time translations $\partial_{\mu}$ with the gamma matrices $\gamma^\mu$). Then the whole 10-element de Sitter algebra transmutes into the semi-simple Poincare algebra.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.