User:YohanN7/Dirac algebra

Representation spaces of the Lorentz group in the Dirac algebra
The space $U ≈ C^{4}$ is together with the transformations $S(Λ)$ given by


 * $$S(\Lambda) = e^{\frac{i}{2}\omega_{\mu\nu}\sigma^{\mu\nu}}, \Lambda = e^{\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}} , M \in so(3,1)$$,

or, if $Λ$ is space inversion $P$,


 * $$S(P) = \beta.$$

a space of bispinors as detailed in the linked article. Below it will be demonstrated that the Clifford algebra Cℓ4(C) decomposes as a vector space according to


 * $$C\ell(4,\mathbf{C}) = 1 \oplus V_\gamma \oplus V_\sigma \oplus \gamma_5V_\gamma \oplus \gamma_51$$

where the elements transform as:
 * 1-dimensional scalars
 * 4-dimensional vectors
 * 6-dimensional tensors
 * 4-dimensional pseudovectors
 * 1-dimensional pseudoscalars

The 4-Vector representation of SO(3;1)+ in Cℓ4(C)
Let $U$ denote $C^{4}$ endowed with the bispinor representation of the Lorentz group. General results in finite dimensional representation theory show that the induced action on $End(U) ≈ Cℓ_{4}(C)$, given explicitly by

is a representation of $SO(3;1)^{+}$. This is a bona fide representation of $SO(3;1)^{+}$, i.e., it is not projective. This is a consequence of the Lorentz group being doubly connected. But the $γ^{μ}$ form part of the basis for $End(U)$. Therefore, the corresponding map for the $γ^{μ}$ is

Claim: The space $V_{γ}$ spanned by the $γ$, endowed with the Lorentz group action defined above is a 4-vector representation of $SO(3;1)^{+}$. This holds if the equality with a question mark in $$ holds. Proof of claim

In $$ it is asserted, as a guess, that the representation on End(U) of $SO(3;1)^{+}$ reduces to the 4-vector representation on the space spanned by the $γ^{μ}$. Now use the relationship Adexp(X) = exp(ad(X)) between ad and Ad and assume that the $ω_{μν}$ are small.


 * $$e^{-i\omega_{\mu\nu}\sigma^{\mu\nu}}\gamma^{\rho}e^{i\omega_{\mu\nu}\sigma^{\mu\nu}} \overset{?}{=} [e^{i\omega_{\mu\nu}M^{\mu\nu}}]^\rho{}_\sigma \gamma^\sigma \overset{Ad/ad}{\Leftrightarrow}e^{ad_{-i\omega_{\mu\nu}\sigma^{\mu\nu}}}(\gamma^\rho) \overset{?}{=} [e^{i\omega_{\mu\nu}M^{\mu\nu}}]^\rho{}_\sigma\gamma^\sigma

\overset{\omega^{\mu\nu} small}{\Leftrightarrow}$$


 * $$ad_{-i\omega_{\mu\nu}\sigma^{\mu\nu}}(\gamma^\rho) \overset{?}{=} [i\omega_{\mu\nu}M^{\mu\nu}]^\rho{}_\sigma\gamma^\sigma \overset{def}{\Leftrightarrow}

[-i\omega_{\mu\nu}\sigma^{\mu\nu},\gamma^\rho] \overset{?}{=} [i\omega_{\mu\nu}M^{\mu\nu}]^\rho{}_\sigma\gamma^\sigma\overset{linearity}{\Leftrightarrow}$$


 * $$ -[\sigma^{\mu\nu}, \gamma^\rho] \overset{?}{=} [M^{\mu\nu}]^\rho{}_\sigma\gamma^\sigma

\overset{(C2)}\Leftrightarrow -i\gamma^\mu\eta^{\nu\rho} + i\gamma^\nu\eta^{\mu\rho} = [M^{\mu\nu}]^\rho{}_\sigma\gamma^\sigma$$

But,
 * $$ (M^{\mu\nu})^\rho{}_\sigma\gamma^\sigma = i(\eta^{\mu\rho}\delta^\nu_\sigma - \eta^{\nu\rho}\delta^\mu_\sigma)\gamma^\sigma = i\eta^{\mu\rho}\gamma^\nu - i\eta^{\nu\rho}\gamma^\mu,$$

and it follows that the assertion is proved for $ω^{μν}$ small. It holds for arbitrary $ω^{μν}$ by irreducibility of the 4-vector representation and the fact that a connected Lie group is generated by any open set containing the identity. (The Inverse Function Theorem ensures that there is such a neighbourhood of representative matrices.)

This means that the subspace $V_{γ} ⊂ Cℓ_{n}(C)$ is mapped into itself, and further that there is no proper subspace of $V_{γ}$ that is mapped into itself under the action of $SO(3;1)$.

The tensor representation of SO(3;1)+ in Cℓ4(C)
Let $Λ = e^{{{sfrac|i|2}}ω_{μν}M^{μν}}|undefined$ be a Lorentz transformation, and let $S(Λ)$ denote the action of $Λ$ on $U$ and consider how $σ^{μν}$ transform under the induced action on $V_{σ} ⊂ End(U)$.

where the known transformation rule of the $γ^{μ}$ given in $$ has been used. Thus the 6-dimensinal space $V_{σ} = span{σ^{μν}}|undefined$ is a representation space of a $(1,0)⊕(0,1)$ tensor representation.

The full Clifford algebra
For the other representation spaces, it is necessary to consider the higher order elements.

For elements of the Dirac algebra, define the antisymmetrization of products of three and four gamma matrices by

respectively. In the latter equation there are $4! = 24$ terms with a plus or minus sign according to the parity of the permutation taking the indices from the order in the left hand sign to the order appearing in the term. For the $σ^{μν}$, one may in this formalism write

In four spacetime dimensions, there are no totally antisymmetric tensors of higher order than four.

A change of basis in the Clifford algebra
Now by observing that $γτ$ and $γη$ anticommute since τ and η are different in $$ (or the terms would cancel), it is found that all terms can be brought into a particular order of choice with respect to the indices. This order is chosen to be $(0,1,2,3)$. The sign of each term depends on the number n of transpositions of the indices required to obtain the order $(0,1,2,3)$. For n odd the sign is − and for n even the sign is +. This is precisely captured by the totally antisymmetric quantity

Using this, and defining

then corresponding to the chosen order, $$ becomes

Using the same technique for the rank 3 objects one obtains

Proof By considering a term in $$ one observes that for $i ∈ {1,2,3}$



\gamma^\rho\gamma^\sigma\gamma^\tau = \gamma^\rho\gamma^\sigma\gamma^\tau(\gamma^{i})^2 = (\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^i)g^{\mu i}\gamma_\mu = (\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^i)\delta^{\mu i}\gamma_\mu = (\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^i)\gamma_i $$

while



\gamma^\rho\gamma^\sigma\gamma^\tau = -\gamma^\rho\gamma^\sigma\gamma^\tau(\gamma^0)^2 = -(\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^0)g^{\mu 0}\gamma_\mu = -(\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^0)(-\delta^\mu_0)\gamma_\mu = (\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^0)\gamma_0 $$

so



\gamma^\rho\gamma^\sigma\gamma^\eta = (\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^\eta)\gamma_\eta $$

holds for all $η ∈ {0,1,2,3}$. But


 * $$\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^\eta = \epsilon^{\rho\sigma\tau\eta}(\gamma^0\gamma^1\gamma^2\gamma^3) = i\epsilon^{\rho\sigma\tau\eta}\gamma_5,$$

so that


 * $$A^{\rho\sigma\tau} = \gamma^\rho\gamma^\sigma\gamma^\tau - \gamma^\rho\gamma^\tau\gamma^\sigma - \gamma^\sigma\gamma^\rho\gamma^\tau + \gamma^\tau\gamma^\rho\gamma^\sigma + \gamma^\sigma\gamma^\tau\gamma^\rho - \gamma^\tau\gamma^\sigma\gamma^\rho = $$

i\epsilon^{\rho\sigma\tau\eta}\gamma_5 \gamma_\eta - i\epsilon^{\rho\tau\sigma\eta}\gamma_5 \gamma_\eta -i\epsilon^{\sigma\rho\tau\eta}\gamma_5 \gamma_\eta$$
 * $$+ i\epsilon^{\tau\rho\sigma\eta}\gamma_5 \gamma_\eta

+i\epsilon^{\sigma\tau\rho\eta}\gamma_5 \gamma_\eta - i\epsilon^{\tau\sigma\rho\eta}\gamma_5 \gamma_\eta, $$

which, by permuting indices in the Levi-Civita symbol (and counting transpositions), finally becomes


 * $$A^{\rho\sigma\tau} = 3!i\epsilon^{\rho\sigma\tau\eta}\gamma_5 \gamma_\eta$$

Space Inversion
Space inversion (or parity) can be included in this formalism by setting

One finds, using $$

These properties of β are just the right ones for it to be a representative of space inversion as is seen by comparison with an ordinary 4-vector $x^{μ}$ that under parity transforms as $x^{0}→x^{0}$, $x^{i}→x^{i}$. In general, the transformation of a product of gamma matrcies is even or odd depending on how many indices are space indices. Details The $σ^{μν}$ transform according to


 * $$\beta^{-1}(\gamma^\mu\gamma^\nu - \gamma^\mu\gamma^\nu)\beta

= \beta^{-1}\gamma^\mu\gamma^\nu\beta - \beta^{-1}\gamma^\nu\gamma^\mu\beta = [\beta^{-1}\gamma^\mu\beta][\beta^{-1}\gamma^\nu\beta] - [\beta^{-1}\gamma^\nu\beta][\beta^{-1}\gamma^\mu\beta] = \pm(\mu)\gamma^\mu \pm(\nu)\gamma^\nu - \pm(\nu)\gamma^\nu \pm(\mu)\gamma^\mu $$

For μ and ν spacelike, this becomes
 * $$(-\gamma^\mu)(-\gamma^\nu) - (-\gamma^\nu)(-\gamma^\mu) = \sigma^{\mu\nu}.$$

For μ = 0 and ν spacelike, this becomes
 * $$(\gamma^\mu)(-\gamma^\nu) - (-\gamma^\nu)(\gamma^\mu) = -\sigma^{\mu\nu}.$$

For μ and ν both zero, one obtains
 * $$(\gamma^\mu)(\gamma^\nu) - (\gamma^\nu)(\gamma^\mu) = \sigma^{\mu\nu}.$$

Space inversion commutes with the generators of rotation, but the boost operators, anticommute

This is the correct behavior, since in the tandard representation with three-dimensional notation,

Now consider the effect of β on the space $V_{γ_{5}}|undefined$:


 * $$\beta^{-1}\gamma_5\gamma^{\nu}\beta = \beta\gamma_5\beta\beta\gamma^{\nu}\beta = \gamma_5\gamma^{\mu\dagger} = -\gamma_5\gamma^0, \mu = 0, +\gamma_5\gamma^i, \mu = i.$$

This behavior warrants the terminology pseudovector for $γ_{5}γ^{μ}$.

The pseudoscalar representation of O(3;1)+ in Cℓn(C)
For $γ_{5}$ one obtains

for the action of the space inversion matrix β. The behavior of $γ_{5}$ under proper orthocronous Lorentz transformations is simple. One has

Proof This can be seen by first writing out explicitly $$\{\gamma^0,\gamma_5\} = \gamma_0i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_0 = 0,$$ $$\{\gamma^1,\gamma_5\} = \gamma_1i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_1 = 0,$$ $$\{\gamma^2,\gamma_5\} = \gamma_2i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_2 = 0,$$ $$\{\gamma^3,\gamma_5\} = \gamma_3i\gamma_0\gamma_1\gamma_2\gamma_3 + i\gamma_0\gamma_1\gamma_2\gamma_3\gamma_3 = 0,$$ and then counting transpositions in each term required to make the two identical factors adjacent and then using the defining property $$. In each case the two terms will cancel. For example, three transpositions are required in the second term in the first line to bring the leftmost $γ^{0}$ to the spot to the immediate left of the rightmost $γ^{0}$. This introduces a minus sign, so the second term cancels the first. The rest of the cases are handled analogously.

For $σ^{μν}$ the commutator with $γ_{5}$ becomes, using $$

Using the exponential expansion of S in powers of $σ^{μν}$ and $$ the $γ_{5}$ transforms according to

and the space $1_{γ_{5}} = span{γ_{5}}|undefined$ is thus a representation space of the 1-dimensional pseudoscalar representation.

The pseudo-vector representation of O(3;1)+ in Cℓn(C)
Every orthocronous Lorentz transformation can be written wither as Λ or PΛ, where P is space inversion and Λ is orthocronous and proper. The Lorentz transformation properties of $γ_{5}γ^{μ}$ are then found to be either

for proper transformations or

for space inversion. These may be put together in a single equation equation in the context of bilinear covariants, see below.

Summary
The space $U≈C^{4}$ is together with the transformations $S(Λ)$ given by


 * $$S(\Lambda) = e^{\frac{i}{2}\omega_{\mu\nu}\sigma^{\mu\nu}} : \Lambda = e^{\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}}: M \in so(3,1)$$,

or, if Λ is space inversion $$,


 * $$S(P) = \beta.$$

a space of bispinors. Moreover, the Clifford algebra Cℓn(C) decomposes as a vector space according to


 * $$C\ell(4,\mathbf{C}) = 1 \oplus V_\gamma \oplus V_\sigma \oplus \gamma_5V_\gamma \oplus \gamma_51$$

where the elements transform as:
 * 1-dimensional scalars
 * 4-dimensional vectors
 * 6-dimensional tensors
 * 4-dimensional pseudovectors
 * 1-dimensional pseudoscalars

For a description of another type of bispinors, please see the Spinor article. The representation space corresponding to that description sits inside the Clifford algebra and is thus a linear space of matrices, much like the space $V_{γ}$, but instead transforming under the ⊕ representation, just like U in this article.