Using Mayer-Vietoris to calculate the de Rham cohomology of the Möbius band $M$, what is the choice of separation?
i.e. $M=U\cup V$, which $U$ and $V$ well chosen for calculation?
Calculate the de Rham cohomology of the Möbius band
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3What choices have you tried? – 2012-01-10
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7To summon the muses, you might consider computing the cohomology of a cylinder first. – 2012-01-10
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1Can you write the Möbius band as a union of two open disks, with intersection having two connected components? – 2012-01-10
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0It may also help to draw the band as a square with identifications made, i.e. to break up the square in a way that descends to the quotient. That way you don't have to picture a mobius band in your head. – 2012-01-11
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0(Of course, if the exercise doesn't require you to use Mayer-Vietoris, then there is a much easier way to get the answer...) – 2012-01-12
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0When comes to the cohomology of a cylinder boundary with two circle, I can separate the cylinder as two part U and V , both are homotopy equivalent to circle, and U intersecting V is also homotopy equivalent to a circle. am I right? – 2012-01-12
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0Yes. But the cilinder is also homotopy equivalent to the circle. So you already know the cohomology. You can also write it as the union of two contractible sets. The intersection contains two elements. – 2012-01-12
1 Answers
You can take these two open sets (in blue and red) :

The long exact Mayer-Vietoris sequence is $$0\rightarrow H^0(M)\rightarrow H^0(U)\oplus H^0(V)\rightarrow H^0(U\cap V)\rightarrow H^1(M)\rightarrow H^1(U)\oplus H^1(V)\rightarrow$$
But the Möbius strip is connected, so $H^0(M)=\mathbb R$.
The open sets are connected, so $H^0(U)=H^0(V)=\mathbb R$.
The intersection has two connected components so $H^0(U\cap V)=\mathbb R^2$.
The opens sets are contractible so $H^1(U)=H^1(V)=0$.
So you have the exact sequence $$0\rightarrow\mathbb R\rightarrow \mathbb R^2\rightarrow\mathbb R^2\rightarrow H^1(M)\rightarrow0$$
So you have $1-2+2-\dim H^1(M)=0\Rightarrow \dim H^1(M)=1$.
Next, $$H^1(U)\oplus H^1(V)=0\rightarrow H^1(U\cap V)=0\rightarrow H^2(M)\rightarrow H^2(U)\oplus H^2(V)=0$$ gives you $H^2(M)=0$.
Conclusion: $H^k(M)=\left\{\begin{array}{ll}\mathbb R & \text{ if }k=0,1 \\ 0 & \text{ else}\end{array}\right.$
Remark: without the MV sequence, you could notice that the median circle $\mathbb S^1$ is a deformation retract of the Möbius strip and use https://math.stackexchange.com/a/162378/33615.
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0Why is 1-2+2-dim(H^1) = 0 ? And why is it H^1 (U int. V) = 0 ? – 2015-07-01
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0@WernerGermánBusch The first one is dimensional formula for exact sequence. You can refer to http://math.stackexchange.com/questions/255384/dimensions-of-vector-spaces-in-an-exact-sequence. – 2016-05-31