# Donaldson Freedman exotic R4

 $K=\{x:y:z:w\in\mathbb{C}P^{3}|x^{4}+y^{4}+z^{4}+w^{4}=0\}$
 $\left(\begin{array}[]{cc}0&1\\ 1&0\end{array}\right)$

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 $\#_{3}S^{2}\times S^{2}$ which contains a (topological) closed ball, $V$, such that we have a smooth embedding, $f:\#_{3}S^{2}\times S^{2}\,\,-\,\,V\to K$ satisfying the following property:

The map $f$ induces an isomorphism from $H_{2}(\#_{3}S^{2}\times S^{2}\,\,-\,\,U;\mathbb{Z})$ 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 $U-V$ as lying in $K$, we obtain a smooth simply connected closed 4- manifold, with bilinear 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 $\mathbb{R}^{4}$.

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

Title Donaldson Freedman exotic R4 DonaldsonFreedmanExoticR4 2013-03-22 15:37:36 2013-03-22 15:37:36 whm22 (2009) whm22 (2009) 13 whm22 (2009) Application msc 57R12 msc 14J80 Donaldsonstheorem exoticR4s ExoticR4s DonaldsonsTheorem