Why does the index of the Dirac operator on a manifold with boundary live inside the Pfaffian line of the boundary Dirac operator? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:04:39Z http://mathoverflow.net/feeds/question/92467 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92467/why-does-the-index-of-the-dirac-operator-on-a-manifold-with-boundary-live-inside Why does the index of the Dirac operator on a manifold with boundary live inside the Pfaffian line of the boundary Dirac operator? Sam Gunningham 2012-03-28T16:42:31Z 2012-03-28T18:36:28Z <p>I am trying to understand a statement from page 72 of <a href="http://math.berkeley.edu/~teichner/Papers/Oxford.pdf" rel="nofollow">What is an elliptic object?</a> by Stolz and Teichner. </p> <p>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 <em>relative index</em> of the Dirac operator $D_\Sigma$ can be interpreted as giving a unit length element in $Pf(D_Y)$.</p> <p>I am having some trouble understanding why this is. Does anyone have a nice explanation (informal is fine) or reference?</p> <p>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.</p>