L5b: Differential forms as spinors.

We are going to rebrand the differential forms as spinors. In L4b we saw how \( \textrm{Spin}(V\oplus V^*) \) sits inside \( Cl(V\oplus V^*) \). In other words, the Clifford product is giving a representation \( \textrm{Spin}(V\oplus V^*) \to \textrm{End}(Cl(V\oplus V^*)) \). One can prove that this representation decomposes into \( 2^n \) subrepresentations which are all isomorphic to $$ Cl(V^*)\det(V) = \wedge^\bullet V^* \det(V)\subset Cl(V\oplus V^*).$$

Each of these is the spin representation of the spin group, a representation that does not descend to \(\textrm{SO}(V\oplus V^*) \) (recall that \(\textrm{SO}(V\oplus V^*) \) is not simply connected!). The elements of the spin representation are called spinors. Moreover, the spin representation splits into two irreducible half-spin representations: \( \wedge^{ev} V^* \det(V) \), \( \wedge^{od} V^* \det(V) \), of even and odd spinors.

 Also in L4b, we related the action of \( V \oplus V^* \) on \( \wedge^\bullet V^* \) with the Clifford product of \( Cl(V\oplus V^*) \) on \( Cl(V^*)\det(V) \subset Cl(V\oplus V^*) \). Having this in mind, we will call the differential forms (without the factor \( \det(V) \) spinors.

One last but very important thing about spinors is that they come equipped with a pairing. For differential forms this map $$ (,) : \wedge^\bullet V^* \otimes \wedge^\bullet V^* \to det V $$ is given by $$(\phi, \psi) = (\phi^T\wedge \psi)_{top},$$ where \( \phantom{.}^T \) is the anti-automorphism of the Clifford algebra extending the map \( (v_1 \ldots v_n)^T=v_n\ldots v_1\) and \( top \) denotes the top-degree component of the differential form. Moreover, this pairing is \( \textrm{Spin}_0(V\oplus V^*) \)-invariant.