Donaldson Freedman exotic R4

Let $K$ denote the simply connected closed 4- manifold given by

$$K=\{x:y:z:w\in ℂP^3|x^4+y^4+z^4+w^4=0\}$$

Let $E_8$ denote the unique rank 8 unimodular symmetric bilinear form over $\mathbb{Z}$, which is positive definite and with respect to which, the norm of any vector is even. Let $B$ denote the rank 2 bilinear form over $\mathbb{Z}$ which may be represented by the matrix

Then we may regard $H_2(K;\mathbb{Z})$ as a direct sum $M\oplus N$, where the cup product induces the form $E_8\oplus E_8$ on $M$ and $B\oplus B\oplus B$ on $N$ and we have $M$ orthogonal to $N$. (This does not contradict Donaldson’s theorem as $B$ has 1 and -1 as eigenvalues.)

We may choose a (topological) open ball, $U$, in ${\mathrm{\#}}_3{S}^2\times S^2$ which contains a (topological) closed ball, $V$, such that we have a smooth embedding, $f:{\mathrm{\#}}_3{S}^2\times S^2-V\to K$ satisfying the following property:

The map $f$ induces an isomorphism from $H_2({\mathrm{\#}}_3{S}^2\times S^2-U;\mathbb{Z})$ into the summand $N$ .

If we could smoothly embed $S^3$ into $U-V$, enclosing $V$, then by replacing the outside of the embedded $S^3$ with a copy of $B^4$, and regarding $U-V$ aslying in $K$, we obtain a smooth simply connected closed 4- manifold, withbilinear form $E_8\oplus E_8$ induced by the cup product. This contradicts Donaldson’s theorem.

Therefore, $U$ has the property of containing a compact set which is not enclosed by any smoothly embedded $S^3$. Hence $U$ is an exotic ${ℝ}^4$.

By considering the three copies of $B$ one at a time, we could have obtained our exotic ${ℝ}^4$ as an open subset of $S^2\times S^2$.