Closed-surface real Cauchy–Riemann index and smooth adjoint obstructions

Choose a unitary connection on $E$. On the complex surface of real dimension two, the compatible Levi–Civita connection preserves $j$: in an oriented orthonormal frame its connection matrix commutes with the quarter-turn. Thus $V={\mathrm{\Lambda }}^{0,\ast }{T}^{\ast }\mathrm{\Sigma }\otimes E$, with even part $E$ and odd part $G$, has the unitary product connection and Clifford action $c(\alpha )=\sqrt{2}({\alpha }^{0,1}\wedge -{\iota }_{({\alpha }^{1,0}{)}^{\mathrm{♯}}})$. In a unitary coframe these matrices square to minus the squared covector norm, are skew Hermitian and interchange the two degrees. The connection preserves this action because it preserves $j$, the metric and wedge/contraction. Directly on degree-zero sections its chiral operator is $D_V^{+}=\sqrt{2}{\bar{\mathrm{\partial }}}_{\mathrm{\nabla }}$. These verify all module hypotheses of [F1]; no Spin structure on the surface is required.

We record weak first-order regularity, including the initial regularity needed for an adjoint. In conformal source and complex bundle coordinates, multiply the equation by a smooth invertible scalar factor to put its principal part in the form $D_0={\mathrm{\partial }}_x+i{\mathrm{\partial }}_y$ (for the adjoint, reverse the $x$ coordinate). A smooth real-linear zeroth-order matrix remains. If a compactly supported distribution $w\in H^s({\mathbb{R}}^2)$ satisfies $(D_0+C)w=f$, use a smooth frequency cutoff $\chi$, zero near zero and one outside the radius-two disk. The multiplier $Q(\xi )=\chi (\xi )\sigma (D_0)(\xi {)}^{-1}$ maps $H^s$ boundedly to $H^{s+1}$ for every integer $s$, by its explicit weighted $L^2$ Fourier bound. Its complementary low-frequency multiplier $1-Q\sigma (D_0)$ maps $H^s$ to every higher order. For nested spatial cutoffs the identity $\rho w=Q(D)(\rho f-\rho Cw+[D_0,\rho ]w)+(1-Q(D)D_0)(\rho w)$ gains one derivative whenever $f$ has the current order. Smooth coefficient multiplication is bounded at nonnegative integer orders by Leibniz; the commutator is zeroth order. Starting at $L^2$ this proves smoothness for a homogeneous weak equation, and starting at $H^1$ proves the estimate $\mathrm{\parallel }w{\mathrm{\parallel }}_{H^{q+1}(K)}\le C(\mathrm{\parallel }Pw{\mathrm{\parallel }}_{H^q(U)}+\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(U)})$ on nested patches. Fourier Cauchy–Schwarz and [F2]'s embeddings turn arbitrary high Sobolev order into smoothness. A finite patch cover gives the global estimate. This argument uses explicit multipliers only, not an unproved general first-order parametrix theorem.

We compute its relative curvature rather than assume a vector-bundle Riemann–Roch formula. Write $F^T$ for the curvature of $T^{1,0}\mathrm{\Sigma }$. The line ${\mathrm{\Lambda }}^{0,1}{T}^{\ast }\mathrm{\Sigma }$ has the same curvature, so the curvature on the two summands of $V$ is $F^E$ and $F^E+F^{T}I$. In an oriented orthonormal frame $c(e^1)c(e^2)$ has diagonal entries $-i,i$. Its traceless spin-connection curvature is therefore $\mathrm{diag}(-F^T\mathrm{/}2,F^T\mathrm{/}2)$: this follows also by taking the commutator with the two Clifford matrices, which reproduces their covariant derivatives. Subtraction leaves the same multiplicity curvature $F^E+(F^T\mathrm{/}2)I$ in both summands. Consequently $\widehat{A}$ has degree-zero part1 and no degree-two part, and the degree-two relative character is $c_1(E)+(n\mathrm{/}2)c_1(T\mathrm{\Sigma })$ with [F1]'s normalization. By [F3] its integral is $⟨c_1(E),[\mathrm{\Sigma }]⟩+n(1-g)$. Thus [F1] gives this COMPLEX index for ${\bar{\mathrm{\partial }}}_{\mathrm{\nabla }}$. Its real index is twice that number.

In a local real frame write $2P={\mathrm{\partial }}_x+J{\mathrm{\partial }}_y+A$ and let $G_0$ be the smooth positive pairing-density matrix. An $L^2$ annihilator $\eta$ obeys $-v_x-{\mathrm{\partial }}_y(J^{\mathsf{T}}v)+A^{\mathsf{T}}v=0$ distributionally, where $v=G_0\eta$. Equivalently $v_x+J^{\mathsf{T}}{v}_y+({\mathrm{\partial }}_y{J}^{\mathsf{T}}-A^{\mathsf{T}})v=0$. Since $(J^{\mathsf{T}}{)}^2=-I$, a smooth local frame change reduces its principal part to the constant one in step 1.2; explicitly $\mathrm{\Phi }=(I-J^{\mathsf{T}}i)\mathrm{/}2$ intertwines $i$ and $J^{\mathsf{T}}$ near a point in coordinates making the latter equal to $i$ there. Hence $\eta$ is smooth. If it vanishes on an open set, [F4] applied in these actual local adjoint coordinates makes it vanish on the connected surface. There was no assumption that an arbitrary higher-Sobolev dual functional is initially an $L^2$ function.

The operator $P-{\bar{\mathrm{\partial }}}_{\mathrm{\nabla }}$ has zero principal symbol and so is a smooth real-linear multiplication map. It factors as the compact inclusion $H^1(E)\hookrightarrow L^2(E)$ followed by bounded multiplication. By [F2], $P:H^1\to L^2$ is therefore Fredholm of the real index in the Statement. This includes the zero-rank case, in which both spaces and index are zero.

Thus the higher kernels are the same smooth finite kernel, and $\mathrm{Ran}{P}_q=H^q\cap \mathrm{Ran}{P}_0$: if an $H^q$ target has an $H^1$ preimage, step 1.2 promotes it to $H^{q+1}$. This intersection is closed. Smooth targets are dense in $L^2$ by [F2], so their images span the finite-dimensional quotient $L^2\mathrm{/}\mathrm{Ran}{P}_0$; a dense linear subspace of a finite-dimensional space is the whole space. Inclusion consequently identifies the higher quotient with that same finite cokernel. This proves all asserted Fredholm realizations and indices.

Let $q:L^2\to L^2\mathrm{/}\mathrm{Ran}{P}_0$. Then $P(H^1)+V=q^{-1}(q(V))$. The image $q(V)$ is a subspace of a finite-dimensional space, hence closed, so the augmented range is closed. Hilbert orthogonal projection says it is all $L^2$ exactly when it has no nonzero annihilator. Any such annihilator already annihilates $P$ and is therefore the smooth section in step 2.2. For finite-dimensional smooth $V$, the higher and $L^2$ augmented ranges have the same quotient by step 4.1. Surjectivity consequently agrees at all levels. A surjective bounded operator between these Hilbert spaces has a bounded right inverse on its kernel orthogonal complement, by the bounded inverse theorem. This proves the augmented assertions as well.