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

Question

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

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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 $\{1, \gamma_0, \gamma_1,\gamma_2,\gamma_3 \}$?

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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

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

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.