Donaldson diagonalization theorem for positive-definite four-manifolds

An actual bordism rather than an assumed end. If $b=0$ the free intersection group has rank zero and its empty integral basis gives the claim. Suppose $b>0$. A charge-minus-one bundle exists: glue a trivial bundle on a ball to the trivial bundle on its complement by the unit-quaternion clutching function on the common three-sphere, with the sign used by the explicit BPST construction in the Taubes proof. Its smooth model connection has $c_2[X]=-1$, by the displayed curvature integral $8{\pi }^2$ and the fixed sign convention. Apply [F2] on this actual bundle. It gives a compact oriented smooth five-manifold $W$ whose full boundary is $X$ together with $n$ oriented projective-plane links, where $n$ is the actual number of reducible solution orbits. The determinant orientation can be chosen so the $X$ boundary has its given orientation. Closed components of $W$ have empty boundary and do not affect this assertion.

The signed boundary inequality forces enough reducibles. Positive definiteness makes $\mathrm{sign}(X)=b$. Each projective link has signature ${\epsilon }_i\in \{1,-1\}$ by [F4]. Additivity of the intersection form under finite disjoint union and [F3] give $0=b+{\sum }_{i=1}^n{\epsilon }_i$, hence $b\le n$. The actual reducible classification in [F2] gives $n\le b$, since its $n$ distinct unordered norm-one pairs are mutually orthogonal and nonzero. Thus $n=b$ (and the boundary equation also forces every ${\epsilon }_i=-1$ for this chosen orientation). No numerical spectral-flow label or prior common sign was assumed.

The integral basis, not merely a real diagonalization. Choose one representative $x_i$ from each of these $b$ unordered norm-one pairs in cohomology. They satisfy $x_i\cdot x_j={\delta }_{ij}$ by [F2]. Let $L=H^2(X;\mathbb{Z})\mathrm{/}\mathrm{Tor}$. They are a real basis of $L\otimes \mathbb{R}$ because they are $b$ nonzero orthogonal vectors. For every $y\in L$, its real-basis expansion is $y={\sum }_i(y\cdot x_i)x_i$. Every coefficient is an integer by integrality of [F1], so $y$ belongs to their integral span. Thus these vectors are an integral basis of the full lattice, not a proper finite-index sublattice. Integral Poincare duality identifies this lattice with the homological intersection lattice in [F1]. Transporting this basis gives the asserted identity matrix on $H_2(X;\mathbb{Z})\mathrm{/}\mathrm{Tor}$. The rank-zero case was already covered, and this proves the original diagonalization conclusion.