# Lie algebra/group/basis of the four gamma matrices along with the identity?

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

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

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

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.