# What transformation gives a Weyl-like representation by flipping $\gamma^0$ and $\gamma^5$?

The usual Weyl representation of the Dirac matrices is defined like this: $$\tag{1}\gamma_W^a = T_W \, \gamma^a \, T_W^{-1},$$ where \begin{align}\tag{2} T_W &= \frac{1}{\sqrt{2}} (1 + \gamma^5 \, \gamma^0), & T_W^{-1} &= \frac{1}{\sqrt{2}} (1 - \gamma^5 \, \gamma^0) \equiv T_W^{\dagger}. \end{align} We then get a kind of rotation in the Dirac matrices space (notice the sign in $$\gamma_W^5$$): \begin{align}\tag{3} \gamma_W^0 &= \gamma^5, &\gamma_W^i &= \gamma^i, &\gamma_W^5 &= -\, \gamma^0. \end{align} This is the Weyl representation of the Dirac matrices.

Now, I wonder if there's a similar transformation that would perform a flipping of $$\gamma^0$$ and $$\gamma^5$$, instead of a rotation in the $$(\gamma^0, \, \gamma^5)$$ "plane". I'm looking for a matrix $$V$$ (probably unitary) such that \begin{align} \gamma_V^0 &= V \, \gamma^0 \, V^{-1} = \gamma^5, \tag{4} \\[1ex] \gamma_V^i &= V \, \gamma^i \, V^{-1} = \gamma^i, \tag{5} \\[1ex] \gamma_V^5 &= V \, \gamma^5 \, V^{-1} = \gamma^0. \tag{6} \end{align} Is such a transformation possible, using some unitary matrix $$V$$? How can we find it explicitely?

Transformations (4) and (6) imply that both $$\gamma^0$$ and $$\gamma^5$$ commute with the matrix $$V^2 \equiv V \, V$$: \begin{align}\tag{7} V^2 \, \gamma^0 &= \gamma^0 \, V^2, & V^2 \, \gamma^5 &= \gamma^5 \, V^2. \end{align} My intuition tells me that there's no unitary matrix $$V$$ satisfying (4)-(6), but I'm probably wrong. The sign in (3) pisses me off!

Indeed, you are doomed. There is no such V.

Suppose there were an equivalence (4,5,6).

Then consider $$\gamma_5 = i \gamma_0 \gamma_1 \gamma_2 \gamma_3$$, as well as its transform, independently of basis or representation, $$V\gamma_5 V^{-1} = i V\gamma_0 \gamma_1 \gamma_2 \gamma_3 V^{-1} \implies \\ \gamma_0 = i \gamma_5 \gamma_1 \gamma_2 \gamma_3 \\ = i \gamma_5 \gamma_0\gamma_0\gamma_1 \gamma_2 \gamma_3= \gamma_5\gamma_0\gamma_5= - \gamma_0.$$

• Wow! Excellent demonstration! Thank you very much, it's a nice one!
– Cham
Mar 6 '20 at 17:41
• Thanks. The tensor product rep of the Dirac rep all but leads you to it. Mar 6 '20 at 17:43
• I think that this demonstration is well worth a notice in a book. Maybe as an exercice.
– Cham
Mar 6 '20 at 18:04