@Ted Okay, here we go. First I will set up the general story. The symbol $\Sigma$ is a surface and $M$ is a manifold. The boundary of $M$ is always written $Y$.
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Atiyah-Patodi-Singer is a story about a first order elliptic differential operator acting on sections of vector bundles $\Gamma(E) \to \Gamma(F)$ over your manifold with boundary; for us it's a surface, $\Sigma$. The usual game is to let the function spaces involved be the Sobolev spaces $L^2_k$ for the codomain and $L^2_{k-1}$ for the codomain to get a Fredholm problem. But that's definitely not going to work without some boundary conditions, as we learn from Dirichlet!
To describe the boundary values, let me first make an assumption. First, the metric is of product type near the boundary; let $t$ be the (interior-pointing) normal coordinate. Now, as a general fact, the symbol of $D$ provides a fixed isomorphism $\sigma: E \to F$ so that, for a unique $t$-dependent family of self-adjoint elliptic operators $A_t$ on $E$, our operator $D$ takes the form $\sigma(d/dt + A_t)$. (More precisely, since the symbol is a map $T^*M \otimes E \to F$, we apply it to $dt$.)
Just like the metric, we should demand that $D$ is of product type near the boundary. This means, to us, that $A_t$ is independent of $t$. This will be true in many cases of geometric interest, as the metric was already of product type. We call it $A$. (I called it $L$ earlier, but now I'm trying to follow APS notation instead of the notation I'm used to - I needed to check some of these details.)
The boundary values appropriate to this setup are rather unusual. They are neither Dirichlet nor Neumann, and in fact non-local. Because $A$ is self-adjoint elliptic on a closed manifold, it has discrete real spectrum.
Let $P$ be the subspace of $L^2_{k-1/2}(Y, E)$ spanned by the eigenfunctions of $A$ with non-negative eigenvalue, and let $\Pi_P$ be the $L^2$ projection onto $P$. Our space of functions with boundary condition on $M$, written $L^2_k(M, E; P)$, is the space of $L^2_k$ functions $\phi$ so that $\Pi_P \phi\big|_{Y} = 0$; that is, the boundary value is spanned by the negative eigenvalues. One may similarly define the spaces of smooth functions with desired boundary values.
The Atiyah-Patodi-Singer theorem, finally, is that the map $C^\infty(M, E; P) \to C^\infty(M,F)$ is Fredholm, with index $(\int_M \alpha(D) ) - \frac{h + \eta(0)}{2}$. The first term is precisely the same integral that appears in the usual Atiyah-Singer theorem. The integer $h$ is defined as $\dim \ker A$. And $\eta(0)$ is that complicated spectral invariant of $(Y,A)$. I will describe it again if there is interest.
Let's understand this in the very special case of the Gauss-Bonnet theorem on $\Sigma$, a surface with boundary. I'll assume there's one boundary circle; the general story is not really different. The boundary being of product type means it is isometric to (some interval in) $\Bbb R \times S^1(r)$ for some radius $r$; I will set $r = 1$ for convenience, it's inconsequential. I want to point out right now that the assumption of product-type boundary implies that the geodesic curvature vanishes.
Now for us $E = (\Lambda^0 \oplus \Lambda^2)(T^*\Sigma)$, while $F = \Lambda^1(T^*\Sigma)$. The operator is $(d, d^*)$. In the coordinates $(t, \theta)$ given by $\Bbb R \times S^1$, and the trivialization $\Lambda^1 \cong \langle dt, d\theta \rangle$ and $\Lambda^2 \cong \langle dt \wedge d\theta\rangle$, this operator takes the form $$\begin{pmatrix}d/dt & d/d\theta \\ d/d\theta & -d/dt \end{pmatrix}.$$ Trust me on this - it's just a calculation of the Hodge star.
What's the symbol of this operator? Of course, written as a polynomial, it's $$\begin{pmatrix}t & \theta \\ \theta & -t \end{pmatrix}.$$ Because our basis is an orthonormal basis, plugging in $dt$ we get $$\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}.$$
After applying (the inverse of this) to $D$, we get $d/dt + \begin{pmatrix}0 & d/d\theta \\ -d/d\theta & 0 \end{pmatrix}.$ (This is self-adjoint, even though that minus sign might freak you out!) So that's our operator $A$. We may identify $A$ acting on sections of $\Bbb R^2$ as $-i d/d\theta$ acting on sections of $\Bbb C$. In particular, we may identify an eigenbasis over $\Bbb C$ with $e^{i n x}$ and the eigenvalues as $n$.
It's now easy to calculate that a basis of eigenvalues is given by $(\sin(nx), \cos(nx))$ and $(\cos(nx),-\sin(nx))$, both with eigenvalue $n$ (as well as the constants in each factor). All eigenspaces are 2-dimensional. Because the spectrum is symmetric across $0$, the $\eta$ function vanishes identically.
Let's now calculate the index by hand. Instead of doing the boundary value problem, the best way to calculate this this is to use another theorem in the Atiyah-Patodi-Singer paper, which explicitly identifies the kernel with the kernel of $D$ on $L^2$ sections over the "extended manifold" $\hat \Sigma$, which is $\Sigma$ with infinite cylindrical ends attached.
In fact this may as well be computed as the kernel of $D^*D$; something similar is true for the adjoint, but there's some trickiness in what we mean by "the adjoint", so let me say the final result.
The index of $D$ is the same as $\dim \ker_{\hat \Sigma} D^* D - \dim \ker_{\hat \Sigma} D D^* - h_\infty$. Here $h_\infty = \dim \ker A - h_\infty(E)$, where $h_\infty(E)$ is the dimension of the subspace of $\ker A$ which consists of limiting values of "extended $L^2$ sections of $E$" in the kernel of $D$", where an extended $L^2$ section is the sum of an $L^2$ section and a section which is constant on the ends at an element of $\ker A$.
Both of these operators are the Laplacian, just on different domains ($\Omega^0 \oplus \Omega^2$ vs $\Omega^1$). In particular, we may identify the kernel with harmonic forms. On a noncompact manifold with cylindrical ends, one finds that the $L^2$ harmonic forms are (this is another gorgeous theorem in the APS paper!) isomorphic to the image of $H^*_c$ in $H^*$. APS call this absolute cohomology.
It is of course easy to see that this is zero for our surface in degrees $0$ and $2$. Finally, $h_\infty = 0$, as constants are of course extended $L^2$ sections in the kernel of $D$, and their limiting values span $\ker A$.
Now our surface is $\Sigma_{g,1}$ for some $g$. There is an isomorphism $H^1_c \cong H^1(\Sigma_g) \cong \Bbb R^{2g}$, and $H^1(\Sigma_{g,1}) \cong \Bbb R^{2g}$; in fact, the inclusion map is an isomorphism. Whew! So $\dim\ker_{\hat \Sigma} D D^* = 2g$. So the index is, finally, $I(D) = -2g$. And the Euler characteristic is $\chi(\Sigma_{g,1}) = 1-2g$. Bguh?? What's going on here?
Well, what does the theorem actually say? It says that $I(D) = \int_\Sigma K_g dA - \frac{h + \eta(0)}{2}$ - remember that I said the integral is just the same as in the usual Atiyah-Singer theorem. We saw earlier that $\eta(0) = 0$ in this case. But that term $h$ is not zero! It was the same as $\dim \ker A$, which was $2$. So we get, finally: $\chi(\Sigma) - 1 = I(D) = \int_{\Sigma} K_g dA - 1,$ or rather, $$\chi(\Sigma) = \int_\Sigma K_g dA.$$
The only thing that changes with more boundary circles is that $-1$ becomes $-n$.
In this proof How they show that cartesian product map has an image in direct sum? I know that since u is unit in Q generates Witt group but then I don't know how argument runs? Somebody help me out? Also tell how $\delta_p$ $\epsilon$ = 0 for finitely many p implies it has image in direct sum?
