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 $\exp(B)(\phi) = e^B \wedge \phi$.