Talk:Representing homology classes by embedded 2-spheres (Ex)

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1. Let M=8\mathbb{CP}^2\#\overline{\mathbb{CP}}^2. In the previous exercise, we showed that the intersection form 8\langle 1\rangle\oplus\langle-1\rangle Of M/var/www/vhost/map.mpim-bonn.mpg.de/tmp/AppWikiTex/tex_wfON3K is equivalent to E_8\oplus\langle-1\rangle (using e.g. the classification of nondefinite even symmetric forms by rank and signature). Pick a basis \langle y_1,\ldots, y_8, x\rangle Of H_2(M;\mathbb{Z}) so that the intersection form of the y_i is E_8, x\cdot y_i=0 for each i, and x\cdot x=-1. Then x cannot be represented by a smoothly embedded 2-sphere, or else blowing down the 2-sphere would yield a smooth structure on a 4-manifold with intersection form E_8 (violating Donaldson’s theorem).

2. Let M=9\mathbb{CP}^2\#\overline{\mathbb{CP}}^2. Similarly, the intersection form 9\langle 1\rangle\oplus\langle -1\rangle is equivalent to E_8\oplus\langle 1\rangle\oplus\langle -1\rangle. As in Part 1, let x\in H_2(M;\mathbb{Z}) with x\cdot x=-1 represent the \langle -1\rangle of this decomposition. Then x cannot be represented by a smoothly embedded 2-sphere, or else blowing down the 2-sphere would yield a smooth structure of a 4-manifold with intersection form E_8\oplus\langle 1\rangle (again violating Donaldson’s theorem; note that E_8\oplus\langle 1\rangle is still positive-definite and non-diagonalizeable).

3. Let M=8\mathbb{CP}^2\#\overline{\mathbb{CP}}^2\#(S^2\times S^2). The intersection form of M is 8\langle 1\rangle\oplus\langle-1\rangle\oplus\left(\begin{array}{cc}0&1\\1&0\end{array}\right), which similarly is equivalent to E_8\oplus\langle-1\rangle\oplus\left(\begin{array}{cc}0&1\\1&0\end{array}\right). As in parts 1 and 2, let x\in H_2(M;\mathbb{Z}) with x\cdot x=-1 represent the \langle-1\rangle of this decomposition. Then x cannot be represented by a smoothly embedded 2-sphere, or else blowing down this sphere would yield a smooth manifold with intersection form E_8\oplus \left(\begin{array}{cc}0&1\\1&0\end{array}\right) (violating Rokhlin’s theorem, as the signature of this form is 8\not\equiv 0\pmod{16}).


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