How can I compute the de Rham cohomology of $\mathbb{R}^3$ minus n lines through the origin? I would like to do it with the Mayer-Vietoris sequence (which is the only thing I know to calculate cohomology besides homotopy invariance). For the case n=1 I have homotopy invariance with $\mathbb{R}^2\setminus{0}$ and this is omotopic to the circle so i can compute the cohomology. Is this homotopy still true in the general case?
De Rham cohomology of the euclidean space without n lines
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algebraic-geometry
differential-geometry
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0I think this has been asked here before... Maybe you can find it answered and all? – 2012-05-30
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0In any case: you can deform the complement of a finite set of lines through the origin to the complement if a finite set of points in the sphere. The latter space is diffeomorphic to the complement of a finite set of points in the plane, and you can compute its cohomology using Mayer-Vietoris and an induction. – 2012-05-30
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0Instead of thinking about lines you can think about cylinders meeting around the origin. For the case n=3 you get something homotopy equivalent to a cube with a disc removed at each side. – 2012-05-30
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0Also, it should be clear how you can go on from here to general $n$. (On a second note this should be algebraic-topology, not differential-geometry.) – 2012-05-30
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0@MarianoSuárez-Alvarez: i searched for the question but i didn't find it. Anyway i already computed the cohomology of the complement of a finite sets of points in the plane. How can I demonstrate the homotopy equivalence? And why isn't it homotopic to the complement of only a point (i imagine to retract all the lines at the origin..) – 2012-05-30
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0Consider the map F(t,x)=(1-t)x+tx/|x|. This is a deformation retract to the sphere with points removed, and hence a homotopy equivalence. – 2012-05-30