L5a. The B-field acting on differential forms.

Last time we stopped soon after we got the map.
$$d\kappa: \mathfrak{so}(V\oplus V^*)\subset Cl(V\oplus V^*) \to \mathfrak{so}(V\oplus V^*)$$
Let us see how \( B \) in the left-hand side \(\mathfrak{so}(V\oplus V^*)\) sits inside the right-hand side \( \mathfrak{so}(V\oplus V^*)\), that is, in the Clifford algebra.

Take the simple \(B\)-field \( e^i \wedge e^j  \in \wedge^2 V^*\subset  \mathfrak{so}(V\oplus
V^*)\). This element is completely determined by its action on \( V \): \(e_i\mapsto e^j\), \(e_j\mapsto -e^i\), \(e_k\mapsto 0\) for \(k\neq i,j\). After some observation, we find \( d\kappa^{-1}(e^i\wedge e^j)=e^je^i \), since $$(e^je^i)e_i - e_i(e^je^i)=(e^ie_i+e_ie^i)e^j=e^j,$$ and \( e^j
e^i \) vanishes when acting on \( e_k \) for \( k\neq i,j\).  We then have that a general \( B=\frac{1}{2} \sum B_{ij} e^i\wedge e^j\in \mathfrak{so}(V\oplus V^*) \) corresponds to \( \frac{1}{2}\sum B_{ij} e^je^i \) via \( d\kappa^{-1}\).

So... what? Why do we want to look at \( \mathfrak{so}(V\oplus V^*)\) inside the Clifford algebra? Well, we had an action of the Clifford algebra on \( \wedge^\bullet V^*\). Keeping it simple, without giving details about the \( Cl(V^*)\cdot \textrm{det}(V) \) business, a \( B \in \mathfrak{so}(V\oplus V^*) \)-field  acts on the exterior algebra by $$B\cdot \phi = \frac{1}{2} \sum B_{ij} e^j\wedge (e_i\wedge \phi)= -B\wedge \phi.$$
If we look at \( \exp(B)\in \textrm{Spin}(V\oplus V^*) \) acting on \( \wedge^\bullet V^* \), we have
$$ \exp(B)\cdot \phi=e^{-B}\wedge \phi = (1-B+\frac{1}{2}B^2-\ldots)\wedge \phi.$$

Note that this action matches what we were doing with the maximal isotropic subspaces \( L(E,\epsilon) \). We saw in Problem Sheet #2 that if \( L(E,\epsilon) = \textrm{Ann}(\phi) \), \( e^B L(E,\epsilon) = \textrm{Ann}(e^{-B}\wedge \phi ) \). By applying \(\exp(B)\in \textrm{Spin}(V\oplus V^*)\) to the differential form \( v\cdot \phi \), we have
$$\exp(B)(v\cdot \phi) = \exp(B)(v) \cdot \exp(B)\phi = e^B v \cdot (e^{-B}\wedge \phi ),$$
so if \( v \in \textrm{Ann}(\phi) \), \(e^Bv\in \Ann(e^{-B}\wedge \phi).

Remark: notice the difference between \( \exp(B)\in \textrm{Spin}(V\oplus V^*) \) and \( e^B=\sum_{n=0}^{+\infty} \frac{B^n}{n!} \). The former is the element of the group, the second is the way we write its action explicitely, either on vectors by \( \exp(B)(X+\xi) =  e^B (X+\xi) = X+\xi+i_XB \) or on differential forms by \( \exp(B)(\phi) = e^B \wedge \phi \).