Talk:Euler characteristic as surgery obstruction (Ex)

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If n is odd, then \chi(X)=\chi(M’)=\chi(M)=0 and there is nothing to check. (This does not matter in the ensuing proof, but might be worth mentioning.)

We will check the effect surgery has on Euler characteristic.

Let S^k\subset M^n be an embedding of a k-sphere into n-manifold M, where the embedded sphere has trivial normal bundle. Say N is obtained from M by surgery along S^k. That is, N\cong M\setminus (S^k\times D^{n-k})\cup (D^{k+1}\times S^{n-k-1}), so \chi(N)=\chi(M)-\chi(S^k)+\chi(S^{n-k-1})\equiv\chi(M)\pmod{2}. Then inductively, since M’ is obtained from M by a finite sequence of surgeries, \chi(X)=\chi(M’)\equiv\chi(M)\pmod{2}.


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