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Apr 8, 2018 at 18:45 comment added Watson Possibly related: math.stackexchange.com/questions/300586
May 25, 2010 at 14:10 answer added KConrad timeline score: 6
May 25, 2010 at 13:05 answer added Charles Matthews timeline score: 6
May 25, 2010 at 12:36 comment added BCnrd [harmless typo: In the above I should have said "paracompact Hausdorff topological space or manifold..." so that one has partitions of unity so as to kill high cohomology of $O_X$.]
May 25, 2010 at 12:34 comment added BCnrd First, the "twistedness" of the Mobius strip $M$ is encoded in how it fibers over $S^1$ as nontrivial topological (or smooth) line bundle. Second, the exponential sequence $0 \rightarrow O_X \rightarrow O_X^{\times} \rightarrow \mathbf{Z}/2\mathbf{Z} \rightarrow 0$ for any topological space or smooth manifold $X$ (with $O_X$ the sheaf of continuous or smooth functions) yields an isomorphism ${\rm{Pic}}(X) = {\rm{H}}^1(X,O_X^{\times}) \rightarrow {\rm{H}}^1(X,\mathbf{Z}/2\mathbf{Z})$. Thus, the line bundle $M$ represents the nontrivial class in ${\rm{H}}^1(S^1,\mathbf{Z}/2\mathbf{Z})$.
May 25, 2010 at 12:05 history asked Akela CC BY-SA 2.5