An explicit round-S4 spin twist has charge one and chiral index one

Write a quaternion as $q=x^{4}I+x^a{Q}_a$ with $Q_a=-i{\sigma }_a$, where the Pauli matrices satisfy ${\sigma }_a{\sigma }_b={\delta }_{ab}I+i{ϵ}_{abc}{\sigma }_c$. Thus $Q_a{Q}_b=-{\delta }_{ab}I+{ϵ}_{abc}{Q}_c$, $\mathrm{tr}(Q_a{Q}_b)=-2{\delta }_{ab}$. On ${\mathbb{C}}^4=\mathbb{H}\oplus \mathbb{H}$ take $c(q)=\left(\begin{array}{cc}0& q\\ -\bar{q}& 0\end{array}\right)$. Direct multiplication gives the Clifford relation and $\mathrm{\Gamma }=\mathrm{diag}(-I,I)$, so $S^{+}$ is the second quaternionic factor. The spin action $\mathrm{diag}(a,b)$ conjugates $c(q)$ to $c(aq{b}^{-1})$. Its kernel on vectors is $(1,1),(-1,-1)$; surjectivity onto $SO(4)$ follows since its differential spans the six independent plane rotations, and a compact subgroup with full Lie algebra is open and closed in the connected group $SO(4)$. For an explicit five-dimensional lift extend the matrices by $c_5=i\mathrm{\Gamma }$ and use the antiunitary $K_5=\mathrm{\Gamma }\mathrm{diag}(i{\sigma }_2,i{\sigma }_2)\circ \text{complex conjugation}$; direct multiplication gives $K_5^2=-1$ and $K_5{c}_j=-c_j{K}_5$ for all five Clifford generators. Even products commute with $K_5$. The standard sphere lift exists concretely as the normal-frame reduction of the trivial five-dimensional spin frame: products of two unit Clifford vectors act as plane rotations, generate $SO(5)$, and their kernel is $\{1,-1\}$ (an element commuting with all Clifford vectors is scalar, and in the quaternionic-unitary even algebra such a scalar is real). Restricting the last vector to the outward normal gives exactly this $\mathrm{Spin}(4)$ lift. The invariant alternating form $\epsilon (v,w)=v^T(i{\sigma }_2)w$ on $S^{+}$ identifies $S^{+}$ with its complex dual, so $S^{+}\otimes S^{+}\cong \mathrm{End}(S^{+})$ and $S^{-}\otimes S^{+}\cong \mathrm{Hom}(S^{+},S^{-})$, preserving metrics and connections.

Extend $c$ from vectors to a one-form $\alpha$ by $c(\alpha )=c({\alpha }^{\mathrm{♯}})$, where ${\alpha }^{\mathrm{♯}}$ is its metric-dual vector, and to a two-form $\omega$ by $c(\omega )=\frac{1}{2}{\sum }_{i,j}\omega (e_i,e_j)c(e_i)c(e_j)$ in an orthonormal frame. For an oriented orthonormal coframe let ${\mathrm{\Omega }}_1=e^{14}+e^{23}$, ${\mathrm{\Omega }}_2=e^{24}+e^{31}$, ${\mathrm{\Omega }}_3=e^{34}+e^{12}$. They form an orthogonal self-dual basis, each of squared norm $2$. Matrix multiplication gives $c({\mathrm{\Omega }}_a){\mathrm{\mid }}_{S^{+}}=2Q_a$ and $\mathrm{tr}(Q_a^{\ast }{Q}_b)=2{\delta }_{ab}$. Consequently $F(f,\omega )=fI\mathrm{/}\sqrt{2}+c(\omega ){\mathrm{\mid }}_{S^{+}}\mathrm{/}2$ and $G(\alpha )=c(\alpha ){\mathrm{\mid }}_{S^{+}}\mathrm{/}\sqrt{2}$ are unitary fiber maps. They are independent of spin frame because Clifford multiplication and $\epsilon$ are equivariant. They are parallel because the lifted Levi-Civita connection preserves both operations: its local coefficient $-\frac{1}{4}{\omega }_{ij}c(e_i)c(e_j)$ satisfies the Clifford derivative rule by the commutator identity, and $SU(2)$ preserves $\epsilon$. At a point choose an oriented orthonormal frame with $\mathrm{\nabla }{e}_i=0$ there; such a frame is obtained by parallel transport in geodesic normal coordinates. Directly on the displayed three basis forms, $c(e_i)c(\omega ){\mathrm{\mid }}_{S^{+}}=-2c({\iota }_{e_i}\omega ){\mathrm{\mid }}_{S^{+}}$; linearity proves it for every self-dual two-form. Applying the Dirac operator to $F(f,\omega )$ at that point gives $c(df)\mathrm{/}\sqrt{2}-{\sum }_{i}c({\iota }_{e_i}{\mathrm{\nabla }}_i\omega )=G(df+\sqrt{2}{d}^{\ast }\omega )$. Since this identity is tensorial, it holds globally.

Put $T:H^1(S^4;{\mathrm{\Lambda }}_{\mathbb{C}}^0\oplus {\mathrm{\Lambda }}_{+,\mathbb{C}}^2)\to L^2(S^4;{\mathrm{\Lambda }}_{\mathbb{C}}^1)$, $T(f,\omega )=df+\sqrt{2}{d}^{\ast }\omega$; the first-order estimate makes it bounded. In this proof $T^{\ast }$ denotes the formal $L^2$ adjoint differential expression, rather than the Hilbert adjoint of this bounded map between differently normed spaces. Integration by parts gives $T^{\ast }\alpha =(d^{\ast }\alpha ,\sqrt{2}{P}_{+}d\alpha )$, where $P_{+}=(1+\ast )\mathrm{/}2$. Since $d^2=(d^{\ast }{)}^2=0$, $T^{\ast }T={\mathrm{\Delta }}_0\oplus {\mathrm{\Delta }}_{2,+}$: on a self-dual two-form, $d^{\ast }d\omega =\ast d{d}^{\ast }\omega$, hence $\mathrm{\Delta }\omega =2P_{+}d{d}^{\ast }\omega$. Also $T{T}^{\ast }={\mathrm{\Delta }}_1$, since $2d^{\ast }{P}_{+}d\alpha =d^{\ast }d\alpha +d^{\ast }\ast d\alpha =d^{\ast }d\alpha$ and $d^{\ast }\ast d\alpha =-\ast d(d\alpha )=0$. These exact identities initially hold on smooth sections and extend distributionally on Sobolev sections. Hodge star is parallel and commutes with $\mathrm{\Delta }$, so its self-dual restriction has the same scalar positive principal symbol.

Use the stereographic embedding $x\mapsto (2x\mathrm{/}(1+r^2),(1-r^2)\mathrm{/}(1+r^2))$. At $x=0$, the outward normal is $e_5$ and the tangent vectors are $2e_1,\dots ,2e_4$; moving $e_5$ past four vectors has positive sign, so this chart is compatible with the boundary orientation everywhere by continuity. Its orthonormal coframe is $e^a=2(1+r^2{)}^{-1}d{x}^a$. Cartan's equation $d{e}^a+{\omega }_{ab}\wedge e^b=0$ gives ${\omega }_{ab}=-2(x_{b}d{x}^a-x_{a}d{x}^b)\mathrm{/}(1+r^2)$. The positive spin connection from step 2.1 is therefore $A_{+}={\eta }_{a\mu \nu }{x}^{\nu }{Q}_{a}d{x}^{\mu }\mathrm{/}(1+r^2)=2{\eta }_{a\mu \nu }{x}^{\nu }{T}_{a}d{x}^{\mu }\mathrm{/}(1+r^2)$ with $T_a=Q_a\mathrm{/}2=-i{\sigma }_a\mathrm{/}2$. The two conformal oriented frames are related by the product of reflections $h_{x\mathrm{/}r}{h}_1$ on quaternions, where $h_v(w)=-v\bar{w}v$; their product is $w\mapsto uwu$, with $u=(x^{4}I-i{x}^a{\sigma }_a)\mathrm{/}r$. Its spin lift is $(u,u^{-1})$, so the positive-spin frame has transition $u^{-1}$, exactly the bundle gluing in [F1]. Hence this is the scale-one global BPST connection, not merely a local curvature match. [F1] gives $q=1$ and $⟨c_2(E),[S^4]⟩=-1$ in the declared convention. Equivalently its orthonormal-frame curvature is $-\frac{1}{2}{\sum }_a{Q}_a{\mathrm{\Omega }}_a$, giving $\mathrm{tr}(F\wedge F)=-3d{\mathrm{v}\mathrm{o}\mathrm{l}}_{S^4}$, consistent with that charge.

If $T(f,\omega )=0$ with $(f,\omega )\in H^1$, step 3.1 and weak regularity [F3] make both entries smooth and harmonic. By [F2], $f$ is constant and the harmonic two-form is zero, giving a one-dimensional complex kernel. The same vanishing has an elementary check: $d^{\ast }\omega =0$ and $\ast \omega =\omega$ imply $d\omega =0$; since $H^2=0$, write $\omega =d\beta$, and [F4] gives $\mathrm{\parallel }\omega {\mathrm{\parallel }}_2^2=\int \omega \wedge \bar{\omega }=\int d(\beta \wedge \bar{\omega })=0$. Since $H^1(S^4;\mathbb{C})=0$, [F2] implies $\ker {\mathrm{\Delta }}_1=0$. Its Green inverse $G_1:L^2\to H^2$ from [F2]–[F3] therefore satisfies ${\mathrm{\Delta }}_1{G}_1=I$. By step 3.1, $T^{\ast }{G}_1:L^2\to H^1$ is a bounded right inverse of $T$. Thus $T$ is surjective, its range is closed and its cokernel is zero. It is Fredholm by [F5], with index $1-0=1$. The unitary maps in step 2.1 transfer this conclusion to $D_{A_{+}}^{+}$.

Under the unitary identifications, the full Dirac differential expression is $\mathcal{D}=\left(\begin{array}{cc}0& T^{\ast }\\ T& 0\end{array}\right)$ on smooth sections. Its square expression is the direct sum of the Hodge Laplacians in step 3.1; let $P:H^2\to L^2$ be this second-order operator with domain $H^2$. It is formally symmetric, has scalar positive principal symbol, and is nonnegative because $(Pu,u{)}_2=\mathrm{\parallel }\mathcal{D}u{\mathrm{\parallel }}_2^2$ on smooth sections; [F3] therefore makes $P$ selfadjoint on $H^2$. For $u\in H^2$, Cauchy–Schwarz and nonnegativity give $\mathrm{\parallel }(P+I)u{\mathrm{\parallel }}_2\mathrm{\parallel }u{\mathrm{\parallel }}_2\ge ((P+I)u,u{)}_2\ge \mathrm{\parallel }u{\mathrm{\parallel }}_2^2$, so $P+I$ has closed range and zero kernel. Since $P+I$ is selfadjoint, its range has orthogonal complement $\ker (P+I)=0$ and is therefore dense; hence it is onto. Write $B=(P+I{)}^{-1}:L^2\to H^2$; the elliptic estimate in [F3] makes this inverse bounded. Then $(\mathcal{D}\mp i)B$ maps $L^2$ to $H^1$ and is a right inverse of $\mathcal{D}\pm i$. The block expression is symmetric on $H^1$ by smooth density. If $v\in \mathrm{dom}{\mathcal{D}}^{\ast }$, choose $u\in H^1$ with $(\mathcal{D}+i)u=({\mathcal{D}}^{\ast }+i)v$; then $v-u\in \ker ({\mathcal{D}}^{\ast }+i)=\mathrm{ran}(\mathcal{D}-i{)}^{\perp }=0$. Thus $\mathcal{D}$ is selfadjoint with domain $H^1$. Finally, if $u\in H^1$ and $\mathcal{D}u\in H^1$, then $Pu={\mathcal{D}}^{2}u\in L^2$ distributionally, so elliptic regularity [F3] gives $u\in H^2$; conversely $H^2$ lies in the square domain. Hence the square domain is exactly $H^2$.

The exact isometries and operator identities prove the chiral kernel, cokernel and Fredholm index directly, while step 3.2 supplies the actual global charge-one bundle and connection and step 4.2 the analytic full-Dirac realization. Full AC is used only through the cited analytic and characteristic-form interfaces. No Atiyah–Singer input or physical anomaly formula was assumed.