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I am just studying the Dirac algebra in 4D spacetime composed of four $n\times n$ gamma matrices $\{\gamma^\mu\}=\{\gamma^0,\gamma^1,\gamma^2,\gamma^3\}$ satisfying the following anticommutation relation: $$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf{1}_{n\times n} \ ,$$ where $\eta^{\mu\nu}$ is a Minkowsky metric tensor.

I have learnt from Wikipedia and Weinberg's book to Quantum Field Theory that a 4D Dirac algebra can be spanned by a set of 16 linearly independent, totally antisymmetric tensors: $$ \mathbf{1} \ , \\ \gamma^\mu \ , \\ \gamma^{[\mu}\gamma^{\nu]} \ , \\ \gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]} \ , \\ \gamma^{[\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma]} \ , $$ where the square brackets indicate that we sum all permutations of the indices within the brackets, with a plus or minus sign for even or odd permutations, respectively.

My questions are:

  1. Why is the basis constructed of the totally antisymmetric gamma matrices and not the totally symmetric ones?
  2. How to check that they are linearly independent?
  3. How to check that they form a complete basis, so that we may use these matrices to construct any product of gamma matrices?
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    $\begingroup$ (1) They anticommute, so their symmetrization is zero and their antisymmetrization is trivial. Your expressions just expand to $$\gamma_{i_1}\dotsb\gamma_{i_k}$$ for any $1\leq i_1<\dotsb<i_k\leq4$ and $k=0,1,2,3,4$. (2) Here I give two proofs. I don't know of an "elementary" proof of this fact. My first proof can probably be put in more "concrete" terms. $\endgroup$ Commented Feb 15 at 17:32
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    $\begingroup$ (3) This is easy: $\gamma_i^2$ is a scalar, and the gamma matrices anticommute, so if there are repeated factors you can rearrange the expression to remove them. For example $$\gamma_1\gamma_2\gamma_1=-\gamma_1^2\gamma_2.$$ $\endgroup$ Commented Feb 15 at 17:33
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    $\begingroup$ (1) You're right about the symmetrization. The important thing is that the antisymmetrization gets rid of those terms. (2) Actually, I think this can work. Use the trace form $(X,Y) \mapsto \mathrm{tr}(XY)$. For our purposes we can replace $\gamma^2$ with $\frac1i\gamma^2$ by pulling out a factor of $i$. Then we're working with real matrices, and over all real matrices the trace form is positive-definite, so over any subspace it is nondegenerate. You will find that all the basis elements are orthogonal under this form, and hence they are linearly independent. $\endgroup$ Commented Feb 17 at 15:35
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    $\begingroup$ Should have said $(X, Y) \to \mathrm{tr}(X^TY)$, but otherwise the argument still stands. $\endgroup$ Commented Feb 18 at 6:21
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    $\begingroup$ Maybe. That would give you an inner product, but I am not as familar with inner products as with symmetric bilinear forms. The issue is that, given an inner product space $V$ (in this case the set of 4x4 complex matrices), for any subspace $S$ fix $s\in S$; is it true that $\langle s,s'\rangle = 0$ for all $s'\in S$ iff $s=0$? If this is not true, then orthogonality does not imply linear independence in $S$. We don't know a priori whether or not the Dirac algebra spans all matrices, so we need to assume they only span a subspace. $\endgroup$ Commented Feb 18 at 17:01

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Under fair use for pedagogic purposes, the following is taken (often, but not always, verbatim) from Bethe and Jackiw, Intermediate Quantum Mechanics.

We take the metric $g$ as $$ g(\mu,\nu) = \begin{cases} +1&\mu=\nu = 0, \\ -1&\mu=\nu \in \left\{1,2,3\right\}, \\ 0&\text{otherwise}. \end{cases} $$

We now obtain several theorems about the Dirac matrices. The fundamental relationship for the Dirac matrices is \begin{equation*} \label{eq:sartrywere} \gamma^\mu\gamma^\nu + \gamma^\mu\gamma^\nu = 2\,g^{\mu\nu} \quad \text{and} \quad \gamma_\mu\gamma_\nu + \gamma_\mu\gamma_\nu = 2\,g_{\mu\nu} \tag{1} \end{equation*} Equation (1) is the condition of a Clifford algebra.

We now form $2^4$ matrices from the four matrices $\gamma^\mu$ with the following products: \begin{equation} \label{eq:sfwergerger} k_{\left(d_0,d_1,d_2,d_3\right)}\, {\gamma^{\sigma(0)}}^{d_0} {\gamma^{\sigma(1)}}^{d_1} {\gamma^{\sigma(2)}}^{d_2} {\gamma^{\sigma(3)}}^{d_3} , \end{equation} where $\left(d_0,d_1,d_2,d_3\right)\in\left\{\left(0,0,0,0\right),\left(0,0,0,1\right),\ldots, \left(1,1,1,1\right)\right\}$, $k_{\left(d_0,d_1,d_2,d_3\right)}\in \mathbb{C}\!\setminus \!\left\{0\right\}$, and $\sigma$ is any permutation of $\left(0,1,2,3\right)$. Since the square of each $\gamma^\mu$ equals to $\pm I_4$, we consider only these $4^2$ matrices. Any other product formed with additional multiples of these matrix will be equivalent to one of these $4^2$ matrices up to a multiplicative constant. The order of the factors is immaterial since different matrices either commute or anticommute. Though not necessary, we will enumerate the matrices in the following way. \begin{align*} \Gamma_1 &= I_4, \\ \Gamma_2 &= \gamma^0, \\ \Gamma_3 &= i\gamma^1, \\ \Gamma_4 &= i\gamma^2, \\ \Gamma_5 &= i\gamma^3, \\ \Gamma_6 &= \gamma^0\gamma^1, \\ \Gamma_7 &= \gamma^0\gamma^2, \\ \Gamma_8 &= \gamma^0\gamma^3, \\ \Gamma_9 &= i\gamma^2\gamma^3, \\ \Gamma_{10} &= i\gamma^3\gamma^1, \\ \Gamma_{11} &= i\gamma^1\gamma^2, \\ \Gamma_{12} &= i\gamma^0\gamma^2\gamma^3, \\ \Gamma_{13} &= i\gamma^0\gamma^3\gamma^1, \\ \Gamma_{14} &= i\gamma^0\gamma^1\gamma^2, \\ \Gamma_{15} &= \gamma^1\gamma^2\gamma^3, \\ \Gamma_{16} &= i \gamma^0\gamma^1\gamma^2\gamma^3. \end{align*}

Theorem 1

  • For all pairs $(l,m)$ there exists an $a_{lm}$, where $a_{lm}$ equals $+1$, $+i$, $-1$, or $-i$, such that $\Gamma_l\,\Gamma_m = a_{lm}\,\Gamma_n$.
  • $\Gamma_l\,\Gamma_m = I$ if, and only if, $l=m$.
  • $\Gamma_l\,\Gamma_m = \pm \Gamma_m \,\Gamma_l $.
  • If $\Gamma_l \neq I$, then there always exists a $\Gamma_k$ such that $$\Gamma_k\,\Gamma_l\,\Gamma_k = -\Gamma_l.$$

Proof: These items will be left for the reader to verify.

Theorem 2

For all $l\in\left\{2,\ldots, 16\right\}$ $$ \operatorname{Tr}\left(\Gamma_l\right) = 0. $$

Proof:

Choose a $k$ such that by applying the second item above and the fourth items above we have \begin{align} \operatorname{Tr}\left(\Gamma_l\right) &= \operatorname{Tr}\left(\Gamma_l\,I\right) \\ &= \operatorname{Tr}\left(\Gamma_l\,\Gamma_k\,\Gamma_k\right) &&\text{2nd Item} \\ &= \operatorname{Tr}\left(\Gamma_k\,\Gamma_l\,\Gamma_k\right) &&\text{3rd item} \\ &= \operatorname{Tr}\left(-\Gamma_l\right) &&\text{4th item} \\ &= - \operatorname{Tr}\left(\Gamma_l\right) \end{align} By the transitive property of the equality $- \operatorname{Tr}\left(\Gamma_l\right) = \operatorname{Tr}\left(\Gamma_l\right) $. Thus $ \operatorname{Tr}\left(\Gamma_l\right) =0$ for all the $\Gamma$'s except $\Gamma_1$, which is equal to $I$.

Theorem 3 $$ \sum_{k=1}^{16} x_k\,\Gamma_k = 0 $$ if, and only if, $x_1 = x_2 = \cdots = x_{16}$.

Proof

Choose an arbitrary $m$ and multiply as follows $$ \Gamma_m\sum_{k=1}^{16} x_k\,\Gamma_k = x_m \Gamma_m\Gamma_m + \Gamma_m\sum_{\substack{k=1\\k\neq m}}^{16} x_k\,\Gamma_m \Gamma_k = \Gamma_m\,0. $$ Using the first and second item we have $$ x_m\,I + \Gamma_m\sum_{\substack{k=1\\k\neq m}}^{16} x_k\,a_{mk}\,\Gamma_n = 0. $$ Taking the trace we have that $$ x_m \,\operatorname{Tr}(I) = 0. $$ We must have that $x_m=0$. Since $m$ was arbitrary, symmetrical arguments can be made for each of the sixteen $\Gamma$'s. Since by definition the sequence of vectors $ \mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{n}$ is said to be linearly independent if the equation $$ a_{1}\mathbf {v} _{1}+a_{2}\mathbf {v} _{2}+\cdots +a_{n}\mathbf {v} _{n}=\mathbf {0} , $$ can only be satisfied by a $ a_{i}=0$ for $i=1,\dots ,n$, therefore the $\Gamma$'s are linearly indepdendent.

Corollary 1 The $\Gamma$'s cannot be represented by matrices whose dimension is less than 4-x-4.

Proof: It is impossible to construct 16 linearly independent matrices from matrices with whose dimension is less that $4^2$.

Corollary 2 There exists precisely 16 linear independent 4-x-4 which we can use to represent the 16 $\Gamma$'s.

Proof: It is impossible to construct 16 linearly independent matrices from matrices with whose dimension is less that $4^2$.

p.s. Regarding the question "Why is the basis constructed of the totally antisymmetric gamma matrices and not the totally symmetric ones?", I frankly do not understand what you mean by this. All I will say is that there is no unique representation of the $\Gamma$ matrices.

References

Bethe and Jackiw, Intermediate Quantum Mechanics, pp. 354-6.

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