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

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.

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 If you do not impose a boundary condition then the index is ill defined. Think of the $\bar{\partial}$-operator on the unit disk. Its kernel is infinite dimensional. – Liviu Nicolaescu Mar 28 2012 at 18:01 I understand that the kernel of $D^+_\Sigma$ will be infinite dimensional. Maybe I should have said "relative Index" (that is the terminology used in the paper). My question could be phrased as asking what does this mean. I think that the space $Pf(D_Y)$ is in some sense supposed to represent possible boundary conditions (it is what is assigned to $Y$ in a certain TQFT which assigns +/- 1 to a closed manifold according to the ordinary index of the Dirac operator). – Sam Gunningham Mar 28 2012 at 18:30 I think the confusion arises from the lack of a precise definition of the notion of relative index? Is this relative index defined anywhere in the reference you mentioned? – Liviu Nicolaescu Apr 1 2012 at 13:25 I can't find the definition in the reference which is why I am asking here. The definition is precisely what I want to understand. Sorry for the confusion! – Sam Gunningham Apr 2 2012 at 14:54 I guess it would be best to e-mail to one of the authors this questions. They will clarify this point better than anyone else. – Liviu Nicolaescu Apr 2 2012 at 15:43