Question. Would you show how to derive the following?
Starting from the Weyl-basis definitions used in the line shown,
\[ \gamma^0 = I\otimes \tau_1, \qquad \gamma^i = \sigma^i \otimes (-i\tau_2) \quad (i=1,2,3), \]we want to derive
\[ \gamma_5 \equiv i\gamma^0\gamma^1\gamma^2\gamma^3 = I\otimes \tau_3. \]We proceed step by step.
Begin with
\[ \gamma_5 = i\gamma^0\gamma^1\gamma^2\gamma^3 = i\,(I\otimes \tau_1)\, (\sigma^1\otimes (-i\tau_2))\, (\sigma^2\otimes (-i\tau_2))\, (\sigma^3\otimes (-i\tau_2)). \]Now use the tensor-product multiplication rule
\[ (A\otimes B)(C\otimes D)=(AC)\otimes(BD). \]So the product becomes
\[ \gamma_5 = i\, (I\sigma^1\sigma^2\sigma^3) \otimes \bigl(\tau_1(-i\tau_2)(-i\tau_2)(-i\tau_2)\bigr). \]Since \(I\sigma^1\sigma^2\sigma^3=\sigma^1\sigma^2\sigma^3\), this is
\[ \gamma_5 = i\,(\sigma^1\sigma^2\sigma^3)\otimes \bigl(\tau_1(-i)^3\tau_2^3\bigr). \]We now simplify the two factors separately.
Using the Pauli-matrix relation
\[ \sigma^1\sigma^2 = i\sigma^3, \]we get
\[ \sigma^1\sigma^2\sigma^3 = (i\sigma^3)\sigma^3 = i\,(\sigma^3)^2 = iI. \]Thus
\[ \sigma^1\sigma^2\sigma^3 = iI. \]Since \(\tau_2^2=I\), we have
\[ \tau_2^3=\tau_2. \]Also,
\[ (-i)^3 = i. \]So
\[ \tau_1(-i)^3\tau_2^3 = \tau_1(i)\tau_2 = i\,\tau_1\tau_2. \]Now use the Pauli relation
\[ \tau_1\tau_2 = i\tau_3. \]Therefore
\[ i\,\tau_1\tau_2 = i(i\tau_3) = -\tau_3. \]So the second factor is
\[ \tau_1(-i)^3\tau_2^3 = -\tau_3. \]Substituting both factors back in,
\[ \gamma_5 = i\,(iI)\otimes(-\tau_3). \]Since \(i\cdot i = -1\),
\[ \gamma_5 = (-1)I\otimes(-\tau_3) = I\otimes\tau_3. \]Hence
\[ \boxed{\gamma_5 = I\otimes \tau_3}. \]Now write \(I\otimes \tau_3\) explicitly. Since
\[ \tau_3= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}, \]we have
\[ I\otimes \tau_3 = \begin{pmatrix} I & 0\\ 0 & -I \end{pmatrix}. \]So finally,
\[ \boxed{ \gamma_5 = i\gamma^0\gamma^1\gamma^2\gamma^3 = I\otimes \tau_3 = \begin{pmatrix} I & 0\\ 0 & -I \end{pmatrix}} \]This is additional material to complement page 473 of Group Theory in a Nutshell by Anthony Zee.
Created with assistance from ChatGPT.