Why does the index of the Dirac operator on a manifold with boundary live inside the Pfaffian line of the boundary Dirac operator?

2012-03-28T16:42:31Z

I am trying to understand a statement from page 72 of What is an elliptic object? by Stolz and Teichner.

They have a spin Riemannian 2-manifold $\Sigma$, with boundary 1-manifold $Y$. The Dirac operator $D_Y$ on the (graded, Clifford linear) spinor bundle has a Pfaffian line $Pf(D_Y) = \wedge ^{top} (\ker D^+ _Y)$ (where $D^+_Y$ is $D_Y$ restricted to the even part). It is stated that the relative index of the Dirac operator $D_\Sigma$ can be interpreted as giving a unit length element in $Pf(D_Y)$.

I am having some trouble understanding why this is. Does anyone have a nice explanation (informal is fine) or reference?

EDIT: Note that when $\Sigma$ is closed, the kernel of $D^+_\Sigma$ is finite dimensional, and the index gives an element of $\mathbb Z/2\mathbb Z = KO^{-2}(pt)$. I think this should be thought of as $\pm 1$ inside $\mathbb R = \wedge ^{top}(0)$. The above notion of relative index should generalize this.

Why does the index of the Dirac operator on a manifold with boundary live inside the Pfaffian line of the boundary Dirac operator? - MathOverflow

Why does the index of the Dirac operator on a manifold with boundary live inside the Pfaffian line of the boundary Dirac operator?