M2 Smooth Curve Protected Dirac Model

Choose nonzero meromorphic sections $s_L,s_T$ by [F5], using the stated full AC [F1], and write $D_L=(s_L)$, $D_T=(s_T)$. If $s_L^{\mathrm{\prime }}$ is another nonzero meromorphic section, $s_L^{\mathrm{\prime }}\mathrm{/}{s}_L$ is a nonzero meromorphic function and $D_L^{\mathrm{\prime }}-D_L=(s_L^{\mathrm{\prime }}\mathrm{/}{s}_L)$; [F2] gives $\mathrm{deg}{D}_L^{\mathrm{\prime }}=\mathrm{deg}{D}_L$. The same argument applies to $T$, so both line-bundle degrees in the statement are well-defined. Local orders add under tensor products, hence $s_L\otimes s_T$ has divisor $D_L+D_T$, and $\mathrm{deg}V=\mathrm{deg}L+\mathrm{deg}T$.

In the even/odd splitting, the unscaled operator is $R=\left(\begin{array}{cc}0& {\bar{D}}^{\ast }\\ \bar{D}& 0\end{array}\right),$ where $\bar{D}$ is the maximal $L^2$ Dolbeault operator on $V$. By [F12] the common graph domain of $\bar{D}$ and ${\bar{D}}^{\ast }$ is $H^1$, dense in the total $L^2$ space. Integration by parts in the definition of ${\bar{D}}^{\ast }$ [F8, F9] makes $R$ symmetric there. Its principal symbol at a nonzero cotangent vector is the off-diagonal matrix formed from the symbol of ${\bar{\mathrm{\partial }}}_V$ and its adjoint. On a complex curve that scalar symbol is nonzero for every nonzero real cotangent vector, so the matrix is invertible. Thus $R$ is elliptic; [F10] gives its Fredholm $H^1\to L^2$ realization and elliptic regularity.

Let $\eta$ be a nonzero meromorphic differential supplied by [F6] and put $D_K=(\eta )$. Apply [F6] first to $D=0$, using $\mathrm{\ell }(0)=1$, to obtain $h^0(K_C)=g$; apply it next to $D=D_K$, using $\mathcal{O}(D_K)\cong K_C$, to obtain $g-1=\mathrm{deg}{D}_K+1-g$. Thus $\mathrm{deg}{K}_C=\mathrm{deg}{D}_K=2g-2$. The isomorphism $T^{\otimes 2}\cong K_C$ sends the tensor square of $s_T$ to a meromorphic section of $K_C$ with divisor $2D_T$. Therefore $2\mathrm{deg}T=\mathrm{deg}{K}_C=2g-2$, so $\mathrm{deg}T=g-1$.

For either sign, $R\pm i$ has zero kernel: if $(R\pm i)u=0$, symmetry gives $0=⟨(R\pm i)u,u⟩$, whose imaginary part is $\pm \mathrm{\parallel }u{\mathrm{\parallel }}^2$. If $v\in L^2$ is orthogonal to the range of $R\pm i$, testing against smooth sections gives the distributional equation $(R\mp i)v=0$. By [F10] this elliptic order-one equation puts $v$ in $H^1$, after which symmetry gives $v=0$. The Fredholm realization $R\pm i:H^1\to L^2$ from [F10] has closed range; its orthogonal complement is zero, so it is onto. The range criterion [F11] proves that $R$ is self-adjoint. The countable-choice hypothesis of [F11] follows from full AC by [F13]. Multiplication by $\sqrt{2\rho }>0$ preserves symmetry, self-adjointness and the domain, so $Q=\sqrt{2\rho }R$ is self-adjoint and odd.

The even-to-odd block of $Q$ is the positive scalar $\sqrt{2\rho }\bar{D}$, so its kernel is $\ker \bar{D}$, and its cokernel is $\ker {\bar{D}}^{\ast }$: the range is closed by the Fredholm realization [F10], and [F9] identifies its orthogonal complement. The kernel/cokernel are finite-dimensional, and [F12] makes every element of $\ker {\bar{D}}^{\ast }$ smooth. Thus every $L^2$ cokernel class has a smooth representative. Conversely, if a smooth $(0,1)$-form is in $\mathrm{ran}\bar{D}$, [F12] gives it a smooth preimage. Inclusion of smooth forms therefore induces an isomorphism from the smooth Dolbeault quotient to the Hilbert quotient. By [F7], the dimensions are $h^0(C,V)$ and $h^1(C,V)$. Apply [F6] to a divisor $D_V$ representing $V$, supplied by [F5]; the divisor line-bundle identification [F4] gives $h^0(C,V)-h^1(C,V)=\mathrm{deg}{D}_V+1-g=\mathrm{deg}V+1-g$. By steps 1.1 and 2.1, this equals $\mathrm{deg}L+(g-1)+1-g=\mathrm{deg}L$, proving the index formula.