Smooth sliceness and four-ball genus obstructions

Choose an integral basis of $L$ and let $B$ be the integer $n\times n$ matrix whose columns are the images of that basis under $f$. Since $f$ is isometric, the Gram matrix of $L$ is $B^{\mathsf{T}}B$.

Taking determinants gives $1=\mathrm{\mid }\det (B^{\mathsf{T}}B)\mathrm{\mid }=\mathrm{\mid }\det B{\mathrm{\mid }}^2$, hence $\det B=1$ or $-1$.

By [F1] and step 2.1, the image subgroup $B{\mathbb{Z}}^n=f(L)$ has index $\mathrm{\mid }\det B\mathrm{\mid }=1$ in ${\mathbb{Z}}^n$. The quotient therefore has one element, which means $B{\mathbb{Z}}^n={\mathbb{Z}}^n$ and $f(L)={\mathbb{Z}}^n$.

By [F1] and [A1], $H_1(X;\mathbb{Z})=0$, so the Ext term in [F3] is zero and evaluation identifies $H^2(X;\mathbb{Z})$ with $\mathrm{Hom}(H_2(X;\mathbb{Z}),\mathbb{Z})$.

By [F4] and [A1], $H_2(X;\mathbb{Z})$ is finitely generated. Its quotient $L=H_2(X;\mathbb{Z})\mathrm{/}\mathrm{Tor}$ is therefore free of finite rank, and every homomorphism from $H_2$ to $\mathbb{Z}$ factors uniquely through $L$. Thus $\mathrm{Hom}(H_2,\mathbb{Z})=\mathrm{Hom}(L,\mathbb{Z})$.

Under the isomorphism [F2], the map $L\to \mathrm{Hom}(L,\mathbb{Z})$ sending $\bar{x}$ to $(\bar{y}\mapsto Q_X(\bar{x},\bar{y}))$ is the composite of the inverse Poincaré-duality map $H_2\to H^2$ with the evaluation map of [F3]; [F5] identifies that composite with the intersection pairing. Both component maps are isomorphisms by steps 1.1–1.2 and [F2]. Hence the pairing map is an isomorphism.

Construct the required Clifford module explicitly. Put $V_0^{+}={\mathrm{\Lambda }}^1{T}^{\ast }X\otimes \mathbb{C}$ and $V_0^{-}=({\mathrm{\Lambda }}^0\oplus {\mathrm{\Lambda }}^{-})T^{\ast }X\otimes \mathbb{C}$, with their metric Hermitian forms. For a real covector $\xi$ define $B_{\xi }a=(-{\iota }_{{\xi }^{\mathrm{♯}}}a,\sqrt{2}{\pi }_{-}(\xi \wedge a))$ and $c(\xi )(a,b)=(-B_{\xi }^{\ast }b,B_{\xi }a)$ on $V_0^{+}\oplus V_0^{-}$. In an oriented orthonormal frame with $\xi =t{e}^1$, the scalar part of $B_{\xi }a$ has squared norm $t^2\mathrm{\mid }{a}_1{\mathrm{\mid }}^2$, while the three forms $\sqrt{2}{\pi }_{-}(e^1\wedge e^j)$, $j=2,3,4$, are orthonormal. Hence $B_{\xi }^{\ast }{B}_{\xi }=\mathrm{\mid }\xi {\mathrm{\mid }}^{2}I$. Both spaces have dimension four, so for $\xi \mathrm{\ne }0$ also $B_{\xi }{B}_{\xi }^{\ast }=\mathrm{\mid }\xi {\mathrm{\mid }}^{2}I$; the zero case is immediate. Polarization gives the Clifford relation $c(\xi )c(\eta )+c(\eta )c(\xi )=-2⟨\xi ,\eta ⟩I$, and the displayed block matrix is skew-Hermitian and odd. Wedge, contraction and ${\pi }_{-}$ are parallel for the Levi–Civita connection, so the induced connection is Clifford-compatible. Its chiral Dirac operator is exactly ${\tilde{D}}_0=d^{\ast }\oplus \sqrt{2}{d}^{-}$. This is an actual geometric Clifford module, not a postulated Spin factorization.

Let $E=P{\times }_{SU(2)}{\mathbb{C}}^2$ and $G=\mathrm{ad}P\otimes \mathbb{C}$, with their induced unitary connections. The bundle $G$ is the trace-zero endomorphism bundle of $E$. Set $Z=-F_E\mathrm{/}(2\pi i)$, a traceless two-by-two matrix of commuting two-forms. The determinant convention gives $c_2(E)=-\frac{1}{2}\mathrm{tr}(Z^2)$ as a form. For a traceless diagonal matrix with entries $u,-u$, its commutator action on trace-zero endomorphisms has eigenvalues $0,2u,-2u$. Therefore $\mathrm{tr}(\mathrm{ad}Z)=0$ and $\mathrm{tr}((\mathrm{ad}Z{)}^2)=4\mathrm{tr}(Z^2)$. These are polynomial identities for all traceless complex matrices: matrices with distinct eigenvalues are dense, and continuity gives the rest. Polarizing the degree-two identity gives the same identity for matrices of two-forms. It follows, through degree four, that $\mathrm{ch}(G)=3-4c_2(E)$, with no degree-two term. This fixes the factor and sign directly from the stated Chern normalization.

Fix its chirality sign and multiplicity. Write $c_j=c(e^j)$. Applying $c_4{c}_3{c}_2{c}_1$ to $e^1\in V_0^{+}$ gives successively $(-1,0)$, $-e^2$, $(0,(e^{23}-e^{14})\mathrm{/}\sqrt{2})$, and $-e^1$. Reversal of four distinct generators has positive sign, so $c_1{c}_2{c}_3{c}_4{e}^1=-e^1$. The construction is $SO(4)$-equivariant, and oriented rotations carry $e^1$ to every unit one-form; therefore $-c_1{c}_2{c}_3{c}_4=I$ on $V_0^{+}$. This volume operator anticommutes with every $c_j$; since $c_1$ identifies the two summands, it is $-I$ on $V_0^{-}$. Thus the given grading IS the reference chirality $i^2{c}_1{c}_2{c}_3{c}_4$. In the local factorization in [F1], ${\mathrm{\Gamma }}_{V_0}={\mathrm{\Gamma }}_S\otimes {\mathrm{\Gamma }}_W$ consequently forces ${\mathrm{\Gamma }}_W=I$. The local multiplicity has rank $8\mathrm{/}4=2$ and is entirely even. If $\mathrm{\Theta }=\mathrm{ch}(V_0\mathrm{/}S)$, then ${\mathrm{\Theta }}_0=2$. This conclusion patches without choosing a global spinor bundle.

Calibrate the untwisted index by harmonic forms, rather than import an Euler/signature index formula. On a closed manifold, ${\int }_{X}da\wedge da={\int }_{X}d(a\wedge da)=0$. Since $\alpha \wedge \alpha =(\mathrm{\mid }{\alpha }^{+}{\mathrm{\mid }}^2-\mathrm{\mid }{\alpha }^{-}{\mathrm{\mid }}^2)d\mathrm{v}\mathrm{o}\mathrm{l}$ for real two-forms, polarization and complexification give $2\mathrm{\parallel }{d}^{-}a{\mathrm{\parallel }}_2^2=\mathrm{\parallel }da{\mathrm{\parallel }}_2^2$. Hence the smooth kernel of ${\tilde{D}}_0$ is exactly the harmonic one-forms. Its formal adjoint is $(f,\omega )\mapsto df+\sqrt{2}{d}^{\ast }\omega$, with $\omega$ anti-self-dual. Pairing this equation with $df$ gives $\mathrm{\parallel }df{\mathrm{\parallel }}_2^2=0$, because $⟨d^{\ast }\omega ,df⟩=⟨\omega ,d(df)⟩=0$. Thus $f$ is constant and $d^{\ast }\omega =0$. For an anti-self-dual two-form the latter is equivalent to $d\omega =0$, so $\omega$ is harmonic. The kernel and cokernel identification in [F4] identifies the Hilbert kernel and cokernel with these smooth spaces. On harmonic two-forms the Hodge star preserves the space and splits the intersection form into positive and negative definite summands: its value on ${\omega }^{+}+{\omega }^{-}$ is $\mathrm{\parallel }{\omega }^{+}{\mathrm{\parallel }}_2^2-\mathrm{\parallel }{\omega }^{-}{\mathrm{\parallel }}_2^2$. By [F2] their negative dimension is precisely $b_{-}(X)$. Consequently ${\mathrm{ind}}_{\mathbb{C}}{\tilde{D}}_0=b_1(X)-1-b_{-}(X)$.

Twist the actual module of step1.1 by $G$, with the product connection. On its local multiplicity factor the curvature is $F^{V_0\mathrm{/}S}\otimes I+I\otimes F_G$, and the two factors commute. The finite exponential and tensor-trace identities therefore give $\mathrm{ch}((V_0\otimes G)\mathrm{/}S)=\mathrm{\Theta }\mathrm{ch}(G)$. The identities are invariant under the local multiplicity transitions in [F1], so they are global form identities, requiring no Spin structure. Applying the actual analytic index theorem to the twisted and untwisted modules yields ${\mathrm{ind}}_{\mathbb{C}}{\tilde{D}}_A=3{\mathrm{ind}}_{\mathbb{C}}{\tilde{D}}_0-4{\int }_X{\widehat{A}}_0{\mathrm{\Theta }}_0{c}_2(E)$. Here ${\tilde{D}}_A=d_A^{\ast }\oplus \sqrt{2}{d}_A^{-}$, ${\widehat{A}}_0=1$, and ${\mathrm{\Theta }}_0=2$ by step2.1. Using [F3] to evaluate the Chern form gives ${\mathrm{ind}}_{\mathbb{C}}{\tilde{D}}_A=3(b_1-1-b_{-})-8⟨c_2(P),[X]⟩$. No assertion that $d_A^2=0$ was used: the Clifford operator and its index calculation apply to every smooth connection.

Multiplication of the target summand ${\mathrm{\Lambda }}^{-}\otimes \mathrm{ad}P$ by $\sqrt{2}$ is a bounded invertible map on $L^2$, so ${\tilde{D}}_A$ and $D_A$ have the same kernel and isomorphic cokernels on their exact first-Sobolev domains. The complex operator just computed is the complexification of the real one. If a complexified equation holds, its real and imaginary parts satisfy the real equation; hence its kernel is the complexification of the real kernel, and the same holds for the orthogonal cokernel using the adjoint. Finite complex dimensions of these spaces equal the corresponding real dimensions. Closed range transfers to the real operator by restricting the complex estimate, or by taking real parts of convergent-image preimages in the kernel complement. Thus $D_A:H^1\to L^2$ is real Fredholm and has index $8k-3(1-b_1+b_{-})$, with $k=-⟨c_2(P),[X]⟩$. This is not a claim about a KO-index theorem.

If ${\pi }_1(X)=0$, every closed real one-form is exact: define a primitive by path integration from a basepoint; nullhomotopy and local primitives make the integral independent of path. Thus $b_1=0$ by [F2]. Positive definiteness gives $b_{-}=0$, and the prescribed charge is $k=1$. The proved formula gives $8-3=5$. Full AC has entered exactly through [F1]–[F3]'s analytic index, harmonic representatives and characteristic-class comparison; the finite symbol, chirality, trace and sign calculations add no choice.

The collar of $M$ lets every absolute two-cycle be moved into $\mathrm{int}W$. The signed transverse intersection number is integer-valued and bilinear. Since $H_2(M;\mathbb{Q})=0$, [F1] shows that $H_2(W;\mathbb{Q})\to H_2(W,M;\mathbb{Q})$ is an isomorphism.

The pairing map over $\mathbb{Q}$ is the composite $H_2(W;\mathbb{Q})\to H_2(W,M;\mathbb{Q})\stackrel{{\mathrm{PD}}^{-1}}{\to }{H}^2(W;\mathbb{Q})\to {\mathrm{Hom}}_{\mathbb{Q}}(H_2(W;\mathbb{Q}),\mathbb{Q})$. Every map in this composite is an isomorphism by step 1.1, [F2], and [F3], so the form is rationally nondegenerate.

By [F4] and [A1], $H_2(W;\mathbb{Z})$ is finitely generated, so its torsion-free quotient is a finite-rank free abelian group. If $x$ has finite order $m$, bilinearity gives $m{Q}_W(x,y)=Q_W(mx,y)=0$ in $\mathbb{Z}$, hence $Q_W(x,y)=0$; the same argument applies in the other variable. Thus the form descends to the torsion-free quotient. The four-dimensional cup-product sign is $(-1{)}^{2\cdot 2}=+1$, so the form is symmetric.

Let $A_7$ be the $7\times 7$ leading principal block on vertices $1,\dots ,7$. Its leading principal determinants $d_k$ satisfy $d_0=1$, $d_1=2$, and $d_k=2d_{k-1}-d_{k-2}$ for $2\le k\le 7$, by expansion along the last row of the tridiagonal chain matrix. Induction gives $d_k=k+1$, so these seven determinants are positive.

The vector $x=A_7^{-1}{e}_3$ has coordinates $x_i=\min (i,3)(8-\max (i,3))\mathrm{/}8$ for $1\le i\le 7$: substitution into $2x_i-x_{i-1}-x_{i+1}={\delta }_{i3}$ with $x_0=x_8=0$ verifies $A_{7}x=e_3$. Hence $e_3^{\mathsf{T}}{A}_7^{-1}{e}_3=15\mathrm{/}8$. The Schur-complement determinant of $C=\left(\begin{array}{cc}{A}_7& -e_3\\ -e_3^{\mathsf{T}}& 2\end{array}\right)$ is $\det (A_7)(2-15\mathrm{/}8)=8\mathrm{/}8=1$.

By Sylvester's criterion, the positive leading principal determinants $d_1,\dots ,d_7$ from step 1.1 and $\det C=1$ from step 2.1 make $C$ positive definite. For any $z\in {\mathbb{Z}}^8$, $z^{\mathsf{T}}Cz=2{\sum }_i{z}_i^2-2{\sum }_{\{i,j\}\text{ an edge}}{z}_i{z}_j$ is even. Every nonzero vector therefore has positive even integral norm, hence norm at least $2$; in particular there is no norm-one vector.

The standard lattice ${\mathbb{Z}}^8$ contains norm-one basis vectors, so no isometry from $L_{E_8}$ onto ${\mathbb{Z}}^8$ exists by step 3.1. If there were an isometric full-rank embedding, its Gram determinant would be $1$ by step 2.1, and [F2] would force its image to be all of ${\mathbb{Z}}^8$, which would be an isometry, again contradicting step 3.1.

The actual BPST bundle and almost solution. Fix $p\in X$ and oriented normal coordinates, with $g_{ij}={\delta }_{ij}+O(r^2)$. In quaternionic coordinates the regular BPST potential is $I_{\lambda }=\mathrm{Im}(qd\bar{q})\mathrm{/}({\lambda }^2+\mathrm{\mid }q{\mathrm{\mid }}^2)$. Write $q=rv$, $\mathrm{\mid }v\mathrm{\mid }=1$, and $\theta =vd\bar{v}$, so $d\theta =-\theta \wedge \theta$. Differentiation gives $F_{I_{\lambda }}={\lambda }^2({\lambda }^2+r^2{)}^{-2}\mathrm{Im}(dq\wedge d\bar{q})$, which is self-dual in the oriented Euclidean metric. Its Euclidean norm square is $48{\lambda }^4\mathrm{/}({\lambda }^2+r^2{)}^4$, with integral $8{\pi }^2$, and its three curvature components at zero span $\mathfrak{s}\mathfrak{u}(2)$. Gauge transformation by $v$ gives the singular potential ${\lambda }^2({\lambda }^2+r^2{)}^{-1}\bar{v}dv$, bounded by $C{\lambda }^2\mathrm{/}{r}^3$ when $r\ge \lambda$. Fix a small coordinate radius $\rho$, independent of $\lambda$, cut off this singular potential on $\rho <r<2\rho$ to zero, and use the regular frame inside. The fixed transition $v$ on their overlap constructs one smooth bundle $P^{\mathrm{\prime }}$ independent of $\lambda$ and smooth connections $A_{\lambda }$ which equal the product connection outside the chart. It is this singular potential that is cut off; a cutoff of a regular potential is not assumed to commute with a gauge transformation.

Small error for the given curved metric. The Hodge projection differs from its Euclidean value by $O(r^2)$ in the inner ball. Thus $E_{\lambda }=F_{A_{\lambda }}^{-}$ satisfies $\mathrm{\mid }{E}_{\lambda }\mathrm{\mid }\le C{\lambda }^2{r}^2\mathrm{/}({\lambda }^2+r^2{)}^2\le C{\lambda }^2\mathrm{/}({\lambda }^2+r^2)$ there, and $\mathrm{\mid }{E}_{\lambda }\mathrm{\mid }\le C_{\rho }{\lambda }^2$ on the fixed cutoff annulus. Direct radial integration gives $\mathrm{\parallel }{E}_{\lambda }{\mathrm{\parallel }}_2\le C{\lambda }^2\sqrt{1+\mathrm{\mid }\log \lambda \mathrm{\mid }}$ and $\mathrm{\parallel }{F}_{A_{\lambda }}{\mathrm{\parallel }}_2\le \mathrm{\Lambda }$ uniformly. For the Green norm use [F2]'s kernel bound. If the pole has $\mathrm{\mid }x\mathrm{\mid }\le 2\lambda$, split at radius $4\lambda$; the inner contribution is $O({\lambda }^2)$ and the outer shells contribute $C{\lambda }^2{\int }_{\lambda }^{2\rho }dr\mathrm{/}r$. If $s=\mathrm{\mid }x\mathrm{\mid }>2\lambda$, split into $\mathrm{\mid }x-y\mathrm{\mid }<s\mathrm{/}2$, $\mathrm{\mid }y\mathrm{\mid }<2s$ outside that set, and $\mathrm{\mid }y\mathrm{\mid }\ge 2s$. The first integral is at most $C({\lambda }^2\mathrm{/}{s}^2)s^2=C{\lambda }^2$, the second at most $C{s}^{-2}{\lambda }^2{\int }_0^{2s}{r}^3\mathrm{/}({\lambda }^2+r^2)dr\le C{\lambda }^2$, and the third is bounded by the same logarithmic shell integral. A pole outside the coordinate region only improves the bound, and the fixed annulus contributes $O({\lambda }^2)$. The non-singular kernel term is smaller; metric volumes are uniformly comparable. Consequently $\mathrm{\parallel }{E}_{\lambda }{\mathrm{\parallel }}_{\mathrm{♯}}\le C{\lambda }^2(1+\mathrm{\mid }\log \lambda \mathrm{\mid })$ and $\mathrm{\parallel }{E}_{\lambda }{\mathrm{\parallel }}_B\to 0$ in the actual fixed metric.

The charge of the constructed bundle. The bundle $P^{\mathrm{\prime }}$ and its transition map in step 1.1 are fixed as $\lambda$ varies. Its Chern–Weil integral is consequently an integer independent of $\lambda$. On the fixed inner ball, the explicit Euclidean self-dual BPST integral tends to $-1$ in the convention $c_2=(8{\pi }^2{)}^{-1}\mathrm{tr}(F\wedge F)$. On the cutoff annulus its curvature is $O_{\rho }({\lambda }^2)$, so its Chern integral tends to zero. The integral is metric independent. Hence $c_2(P^{\mathrm{\prime }})[X]=-1$ exactly. This computes the charge of the actual fixed bundle used by the contraction, not merely that of an abstract model.

Uniform spectral gap with no hidden inverse assumption. From [F2]'s Weitzenböck formula, scalar Sobolev applied to $\mathrm{\mid }\xi \mathrm{\mid }$ using Kato, and $\mathrm{\parallel }{E}_{\lambda }{\mathrm{\parallel }}_2\to 0$, absorb $C\mathrm{\parallel }{E}_{\lambda }{\mathrm{\parallel }}_2\mathrm{\parallel }\xi {\mathrm{\parallel }}_4^2$ to get $\mathrm{\parallel }{\mathrm{\nabla }}_{A_{\lambda }}\xi {\mathrm{\parallel }}_2^2\le C(⟨P_{A_{\lambda }}\xi ,\xi ⟩+\mathrm{\parallel }\xi {\mathrm{\parallel }}_2^2)$. If no uniform positive gap existed as $\lambda \downarrow 0$, take $\mathrm{\parallel }{\xi }_n{\mathrm{\parallel }}_2=1$ and $\mathrm{\parallel }(d_{A_{{\lambda }_n}}^{-}{)}^{\ast }{\xi }_n{\mathrm{\parallel }}_2\to 0$. The displayed estimate and Kato give a uniform $L^4$ bound, so the $L^2$ mass in every radius-$r$ ball is at most $C{r}^2$. In the global singular/product frame away from $p$, $A_{{\lambda }_n}\to 0$ smoothly by step 1.1. Local $H^1$ bounds and Rellich therefore give a strongly locally $L^2$ convergent subsequence to an ordinary adjoint-valued anti-self-dual two-form $\xi$ with $d^{\ast }\xi =0$. The small-ball bound shows $\mathrm{\parallel }\xi {\mathrm{\parallel }}_2=1$, with no lost mass. Lower semicontinuity on each punctured compact set gives a finite ordinary $H^1$ norm off $p$, uniformly as the deleted radius tends to zero. Multiply by cutoffs vanishing near $p$: their error is bounded in $L^2$ by the $L^4$ norm of $\xi$ on a shrinking annulus times the bounded $L^4$ norm of the cutoff gradient, and tends to zero by absolute continuity. Thus $\xi$ extends in $H^1$ with $d^{\ast }\xi =0$ globally. Anti-self-duality then gives $d\xi =0$, so [F3] makes it harmonic, contrary to $b_{-}=0$ and $\mathrm{\parallel }\xi {\mathrm{\parallel }}_2=1$. Hence a fixed gap $c>0$ exists for all sufficiently small $\lambda$. This was proved before choosing the correction threshold depending on $c$.

Smooth solutions and actual irreducibility. Steps 2.1–3.1 satisfy every hypothesis of [F2] for small $\lambda$. It produces smooth self-dual $A_{\lambda }+a_{\lambda }$ on the same $P^{\mathrm{\prime }}$, with $\mathrm{\parallel }{a}_{\lambda }{\mathrm{\parallel }}_4\to 0$. If arbitrarily small members were reducible, choose unit parallel adjoint sections $u_{\lambda }$. Pull them back to the scale-$\lambda$ unit ball in the regular frame, where $A_{\lambda }$ is exactly $I_1$; four-dimensional $L^4$ norms of one-forms are invariant under this scaling and the fixed metric is uniformly comparable. Thus $\mathrm{\parallel }{\mathrm{\nabla }}_{I_1}{u}_{\lambda }{\mathrm{\parallel }}_4\le C\mathrm{\parallel }{a}_{\lambda }{\mathrm{\parallel }}_4\to 0$. Since $\mathrm{\mid }{u}_{\lambda }\mathrm{\mid }=1$, their ordinary $W^{1,4}$ norms are bounded in that fixed smooth frame. Rellich in $H^1\to L^2$, followed by $\mathrm{\parallel }{u}_i-u_j{\mathrm{\parallel }}_4^4\le C\mathrm{\parallel }{u}_i-u_j{\mathrm{\parallel }}_2^2$, gives a strongly $L^4$ convergent subsequence. Its limit is an $I_1$-parallel unit section; the first-order equation with smooth coefficients bootstraps it to a smooth section. Commuting derivatives at zero makes its value commute with all three curvature values of $I_1$. Those span $\mathfrak{s}\mathfrak{u}(2)$, whose common centralizer in its Lie algebra is zero, a contradiction. Therefore sufficiently small corrected members are irreducible.

Comparison with the specified bundle. Use [F5]'s finite smooth triangulation of $X$. An $SU(2)$ bundle is the unit sphere bundle of its defining quaternionic line, since $SU(2)$ acts freely and transitively on its unit vectors. It is trivial on the three-skeleton: the obstruction groups ${\pi }_0,{\pi }_1,{\pi }_2$ of $SU(2)=S^3$ vanish. On each oriented four-cell a three-skeleton trivialization has a boundary unit-vector map $S^3\to S^3$, with integer degree. These degrees form its four-cochain obstruction; changing a three-cell trivialization changes it by that cellular coboundary, by [F5]. Here is the general Chern identification: in the pulled-back trivial bundle over one characteristic four-disk, extend the boundary unit section as a vector-valued map and make it transverse to zero in the interior, retaining its boundary homotopy class. The boundary degree is the sum of its local zero degrees, by excising disjoint small zero-balls and the boundary homology relation. Those local degrees are precisely the relative pullback of the oriented Thom class. Thus this cellular cocycle represents the Euler class of the underlying complex-oriented real four-plane bundle, and equals $c_2$ by [F4]. This is a general relative-cell computation, not an inference from the BPST calibration. For two bundles, the obstruction to extending a three-skeleton isomorphism over a four-cell is the difference of these two degrees; pointwise multiplication of the two boundary $S^3$ maps adds their degrees, by [F5]. By [F6] and step 2.2, $P$ and $P^{\mathrm{\prime }}$ have the same degree-four class. Adjusting the isomorphism on the three-cells kills the difference cochain by its coboundary, and degree classification then extends it over every four-cell. Smooth it as an everywhere nonzero section of the quaternionic homomorphism bundle, using finite smooth charts and a partition of unity, and normalize its quaternionic norm. Uniform approximation preserves nonvanishing; unit quaternionic maps are determinant-one unitary isomorphisms. Transfer the smooth irreducible self-dual connection of step 4.1 through this actual smooth bundle isomorphism to $P$. This proves the full given-metric, given-bundle conclusion.

Apply [F2] to the integral unimodular form $\omega =L_{E_8}$, which satisfies its exact algebraic hypothesis by [F1]. It gives an oriented closed simply connected topological four-manifold $Y_{E_8}$ with intersection form $L_{E_8}$. Simply connected includes path connectedness, so $Y_{E_8}$ has every property required in the statement.

The source also gives smoothness away from a point, but no smooth structure on the whole closed manifold is asserted or used. This establishes precisely the stated topological realization.

By [F3], the quotient and equation bundle are smooth in the stated $H^3\mathrm{/}{H}^4$ spaces. In a slice centered at $A$ the equation is $F_A^{-}+d_A^{-}a+(a\wedge a{)}^{-}$. The derivative of each allowed perturbation factors through the compact inclusion $H^3\hookrightarrow H^2$, and [F4] proves Fredholmness of the slice derivative with the same index as [F1]. Formula [F2] gives that index as five.

Apply [F4] with the exact finite larger and smaller slice balls in the Statement. Its compact zero-container follows from the uniform $H^3$ bound on nearby perturbation values: in each smaller slice ball the curvature equation and $d_A^{\ast }a=0$ give a uniform $H^4$ bound on the solution $a$, hence compactness in $H^3$. Compact-set derivative control then preserves surjectivity at the remaining zeros. Density follows from finite cokernel-complement perturbations and finite-dimensional Sard. Thus the claimed set is open dense, and its zero set in $V$ is a smooth five-manifold by step 1.1.

By [F5], the unperturbed real determinant line is orientable with every gauge transformation preserving the orientation. Scaling the compact perturbation derivative from zero to its actual value preserves this property. At a transverse zero the kernel of the slice derivative is the moduli tangent space and its cokernel is zero, so the determinant orientation gives a tangent orientation. The extra scalar gauge block is invertible and has its fixed orientation. Therefore the actual unbased moduli manifold is orientable. The based quotient only pulls this construction back and adds the three framing directions described in [F3]; it is not used to change the index in step 1.1. This proves all conclusions.