I am trying to follow the discussion of Clifford multiplication on page 384 of The Wild World of 4-Manifolds, by Alexandru Scorpan (link, although I hope this will be totally self-contained), and I'm having some trouble understanding how he gets the defining property of the multiplication.
The setup, which I believe I understand, is that a spin structure on a 4-manifold splits because of the isomorphism $$Spin(4)= SU(2)\times SU(2)$$ so we have two $SU(2)$ spinor bundles, $\mathcal{S}^+$ and $\mathcal{S}^-$, with "complex plane fibers" (I think that means $\mathbb{C}^2$-fibers).
Now the part I don't get, is he says there is a "Clifford multiplication" \begin{equation}\tag{1} T_M\times \mathcal{S}^+\to \mathcal{S}^-\qquad \phi\to v*\phi, \end{equation} which, when you combine it with its adjoint $\mathcal{S}^-\to \mathcal{S}^+$, it's clear you must always have \begin{equation}\tag{2} v*(v*\phi)=-|v|^2\phi. \end{equation}
So I certainly recognize the defining property of Clifford algebras in (2), but I don't understand how you can get it in this context. Specifically:
- Does this Clifford multiplication promote the spinors in $\mathcal{S}$ to some kind of module?
- (Related) If $\phi\in\mathcal{S}^+$, how am I to understand $v*\phi\in\mathcal{S}^-$? Is this a particular aspect of the grading of the Clifford algebra that I am missing?
- How exactly is the adjoint map defined? To me, the adjoint of a map $f:X\to Y$ is a map $f^\dagger:Y^*\to X^*$, $f^\dagger(\lambda)\to \lambda\circ f$. So if my multiplication is $f(v,\phi)=v*\phi$, the adjoint would seemingly be $f_v^\dagger(\lambda)=\lambda\circ(v*\phi)$ for $\lambda\in \mathcal{S}^{-*}$. So it would appear he is implying something like $\lambda=-<v,\cdot>$ for a particular tangent vector $v$. But that seems like quite an arbitrary choice here - I must not be understanding what is meant here by "adjoint".