Now, a neat trick (except it isn't). Since \( \mathbb{I} (\mathbb{\overline{I}}) = \mathbb{\overline{I}} \) and \( \mathbb{I}(-\mathbb{I}) = -\mathbb{I} \) then we have an equivalence between the above permutation rep of V, and a 2-dimensional one, as: \( V = \{\mathbb{I}, \, -\mathbb{I},\, \mathbb{\overline{I}},\, -\mathbb{\overline{I}}\} \sim \{ (\mathbb{I}\otimes\mathbb{I}),\, (\mathbb{I}\otimes\mathbb{\overline{I}}),\, (\mathbb{\overline{I}}\otimes \mathbb{I}),\, (\mathbb{\overline{I}} \otimes \mathbb{\overline{I}}) \} \)
It looks like going from the 4-dimensional to the 2-dimensional matrices is a matter of replacing \( \otimes \mathbb{I} \) with 1, and \( \otimes \overline {\mathbb{I}} \) with -1, "on the right". But hang on, you can also replace \( \mathbb{I}\otimes \) with 1, and \( \overline {\mathbb{I}}\otimes \) with -1, "on the left". There's a symmetry (no, really).
