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I am having problems calculating the cohomology with compact support of the open Möbius strip (without the bounding edge).

I am using the Mayer Vietoris sequence: U and V are two open subsets diffeomorphic to $\mathbb{R}^2$ and $U\cap V$ is diffeomorphic to two copies of $\mathbb{R}^2$.

$H^0_C(M)=0$ and that's ok, but then I get the exact sequence

$ 0 \rightarrow H^1_C(M) \rightarrow H^2_C(U\cap V) \rightarrow H^2_C(U)\oplus H^2_C(V) \rightarrow H^2_C(M)\rightarrow 0$

where both $H^2_C(U\cap V)$ and $H^2_C(U)\oplus H^2_C(V)$ have dimension 2.

I would say that the function $\delta:H^2_C(U\cap V) \rightarrow H^2_C(U)\oplus H^2_C(V)$ in the exact sequence sends $(\phi,\psi)$ to $(-j_U(\phi + \psi),j_V(\phi + \psi))$ where $\phi$ and $\psi$ are generators of $H^2_C(U\cap V)$.

So I would say that $Im\ \delta$ is one dimensional and spanned by $(-j_U(\phi + \psi),j_V(\phi + \psi))$ and $\dim H^1_C(M)=\dim H^2_C(M)=1$. But, I have read that all compact cohomology classes of the Möbius strip are zero. So I must be wrong somewhere.

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I think you are probably computing for an open annulus. Once you take orientation data into account, you ought to get something like $\phi\mapsto \phi+\psi$ and $\psi\mapsto \phi-\psi$, which will be two dimensional. –  Grumpy Parsnip Jan 26 '12 at 20:02
    
it makes sense to me, but you mean $\delta(\phi)=(-\phi,\phi)$, $\delta(\psi)=(-\psi,-\psi)$? –  tigu Jan 26 '12 at 20:53
    
Yes that's right. –  Grumpy Parsnip Jan 27 '12 at 13:04
    
Can anyone answer this question decisively? I do not find the answer below persuasive. –  Potato Jun 18 '12 at 3:24

2 Answers 2

Jim's comment is right, the glueing information is hidden in the map $\mathbb R^2 \to \mathbb R^2$ corresponding to the map $\phi : H^2_c(U\cap V) \to H^2_c(U) \oplus H^2_c(V)$.

Recall that if $U$ and $V$ are opens of $\mathbb R^2$, if we have a map $\iota : \mathbb U \to \mathbb V$, it induces a map $\iota^* : \Omega^2_c(\mathbb U) \to \Omega^2_c(\mathbb V)$, such that $\iota^* (f dxdy) = g dxdy$ where $g$ satisfies $f = (g \circ \iota) * J$ where $J$ is the jacobian of $\iota$, and $g=0$ outside the image of $\iota$. Consequently, depending on the (non-changing) sign of $J$, we have $\int_U \omega = \pm \int_V (\iota^*(\omega))$ forall $\omega \in \Omega^2_c(U)$ (the change of variable formula is exactly what we have, but with an absolute value on the jacobian). Therefore you have to keep track wether all your inclusion maps preserve orientation or not.

Write $U \cap V = W = W_1 \cup W_2$, so that $H^2_c(U\cap V) = H^2_c(W_1) \oplus H^2_c(W_2)$. We know that $U,V,W_1,W_2$ are diffeomorphic to $\mathbb R^2$, so the isomorphisms are induced by $\alpha : \omega \in \Omega^2_c(U) \mapsto \int_U \omega$

$\phi$ is given by $\phi(\omega_1 \oplus \omega_2) = (\iota_{W_1 \to U}^*(\omega_1) - \iota_{W_2 \to U}^*(\omega_2), \iota_{W_1 \to V}^*(\omega_1) - \iota_{W_2 \to V}^*(\omega_2))$. So $\alpha \circ \phi (\omega_1 \oplus \omega_2) = (\int_U \iota_{W_1 \to U}^*(\omega_1) - \int_U \iota_{W_2 \to U}^*(\omega_2), \int_V \iota_{W_1 \to V}^*(\omega_1) - \int_V \iota_{W_2 \to V}^*(\omega_2))$.
In the Möbius case, we usually pick maps such that $\iota_{W_1 \to U},\iota_{W_2 \to U},\iota_{W_1 \to V}$ are orientation preserving, and $\iota_{W_2 \to V}$ is orientation-reversing, so we obtain : $\alpha \circ \phi (\omega_1 \oplus \omega_2) = (\int_{W_1} \omega_1 + \int_{W_2} \omega_2, \int_{W_1} \omega_1 - \int_{W_2} \omega_2)$ thus the corresponding map $\tilde{\phi} : \mathbb R^2 \to \mathbb R^2$ is $(x,y) \mapsto (x+y,x-y)$, which is an isomorphism. Therefore, its kernel and cokernel, $H^1_c(M)$ and $H^2_c(M)$, are both zero.

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You have something like $0 \to H^1_C(M) \to \mathbb{R} \oplus \mathbb{R} \to H^2_C(M) \to 0$. Let us consider $\delta \colon \mathbb{R} \oplus \mathbb{R} \to H^2_C(M)$ and $\varphi\colon H^1_C(M) \to \mathbb{R} \oplus \mathbb{R}$. We have that $\ker \delta = \operatorname{Im} \varphi \subseteq \mathbb{R} \oplus \mathbb{R}$ and that $\operatorname{Im} \delta = \ker 0 = H^2_C(M)$, then for the rank-nullity theorem we get $\dim \operatorname{Im} + \dim \ker = \dim V$, which means $\dim \operatorname{Im} \delta + \dim \ker \delta \leq 2$, thus $\dim H^2_C(M) \leq 2$. We also have that $\dim H^1_C(M) = \dim \ker \varphi + \dim \operatorname{Im} \varphi = \dim \operatorname{Im} \varphi = \ker \delta$, so $H_C^1(M)$ and $H^2_C(M)$ must have the same dimension.

If it's $0$, we're done.

Suppose it is $2$, then $H^1_C(M) \cong H^2_C(M) \cong \mathbb{R} \oplus \mathbb{R}$, which means that $0 \to H_C^1(M) \to 0$ is exact; this implies that $H_C^1(M) =0 = H_C^2(M)$, absurd.

Suppose it is $1$, then we have an that if $\alpha$ generates $H^1_C(M)$, then $(\varphi(\alpha), \beta)$ generates $\mathbb{R} \oplus \mathbb{R}$ for some $\beta$ independent of $\alpha$; but then $\delta(\varphi(\alpha),\beta) = \varphi(\alpha) + \beta $ is an isomorphism. Which means that $\delta \varphi$ factors through zero, and is thus the null-map, which is an iso if and only if $H^1_C(M) = H^2_C(M) =0$, which is a contradiction.

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I think your initial exact sequence is wrong. It appears to be missing a term. –  Potato Jun 18 '12 at 3:09
    
@Potato, there is one missing, but it's an isomorphism and "shortening" the sequence doesn't change its exactness, though it changes how the maps behave. –  Andy Jun 18 '12 at 14:18
    
I've never seen that before. It would be nice if you proved that (also I can't remove the downvote unless you edit the answer...) –  Potato Jun 18 '12 at 22:00
    
@Potato, I don't seem to be able to prove it in general right now. Even though I think it's true (I did prove it sometime back). I don't necessarily want you to remove the downvote; maybe I'm wrong and my answer isn't a real answer. –  Andy Jun 19 '12 at 9:25

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