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Let $X = \mathbb{A}^1_k$ with $k$ infinite and $U = X - \{P,Q\}$ and $\mathbb{Z}_U= i_{!}(\mathbb{Z}|_U)$, $\Bbb{Z}$ the constant sheaf. I want to say that $H^1(X,\mathbb{Z}_U) \neq 0$. If it is zero we see exact sequence

$$0 \to H^0(X,\mathbb{Z}_U) \to H^0(X,\mathbb{Z}) \to H^0(X,\mathbb{Z}_{\{P,Q\}} ) \to0$$

where $\mathbb{Z}_Y$ is $j_\ast(\mathbb{Z}|_{\{P,Q\}})$. The middle term is $\mathbb{Z}$ but can I say right term is $\mathbb{Z} \oplus \mathbb{Z}$? It seems like it's actually $\mathbb{Z}$. I am confused.

Dylan B.
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1 Answers1

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Since your sheaf $\mathbb{Z}_{\{P,Q\}}$ is equal to $j_\ast(\mathbb{Z}|_{\{P,Q\}})$, we have : $$H^0(X,\mathbb{Z}_{\{P,Q\}} ) =H^0(X, j_\ast(\mathbb{Z}|_{\{P,Q\}}))\stackrel {definition}{=}H^0(\{P,Q\}, \mathbb{Z}_X|_{\{P,Q\}})\stackrel {!}{=} H^0(\{P,Q\}, \mathbb{Z})=\mathbb{Z}\oplus \mathbb{Z}$$ This indeed shows that your displayed sequence is not exact (since $H^0(X,\mathbb{Z})=\mathbb Z$ and $\mathbb Z$ cannot be surjected onto $\mathbb{Z}\oplus \mathbb{Z}$) and thus that necessarily $H^1(X,\mathbb{Z}_U) \neq 0$.

Edit
It is actually fairly easy to see that $ H^1(X,\mathbb{Z})=0 $ (because constant sheaves are flasque and thus acyclic on irreducible spaces), and that $H^1(X,\mathbb{Z}_U) \cong \mathbb Z$ since it is the cokernel of the restriction morphism $ H^0(X,\mathbb{Z})= \mathbb{Z} \to H^0(\{P,Q\},\mathbb{Z} )=\mathbb{Z} \oplus \mathbb{Z}: n\mapsto (n,n)$

New edit
As discussed in the comments I have used that the restriction of a constant sheaf $A_X$ on a space $X$ to any subspace $Y$ is the constant sheaf $A_Y$ to deduce $H^0(\{P,Q\}, \mathbb{Z_X}|_{\{P,Q\}})\stackrel {!}{=} H^0(\{P,Q\}, \mathbb{Z})$

Georges Elencwajg
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  • Thank you for answer. I know $H^0(X,j_\ast(\Bbb{Z}|{{P,Q}})) = H^0({P,Q}, \Bbb{Z}|{{P,Q}})$ but why is this equal to $\Bbb{Z} \oplus \Bbb{Z}$? By definition $H^0({P,Q}, \Bbb{Z}|{{P,Q}}) = \varinjlim{U \supseteq {P,Q}} \Bbb{Z}(U)$, every term in the limit is just $\Bbb{Z}$ so how you get $\Bbb{Z} \oplus \Bbb{Z}$? – Dylan B. Nov 28 '13 at 21:39
  • Dear Dylan, $H^0({P,Q}, \mathbb{Z})=\mathbb Z\oplus \mathbb Z=\mathbb Z^2$ because ${P,Q}$ is a discrete space and for a discrete topological space $D$ and a a constant sheaf $C$ on $D$, we have $H^0(D, C)=C^D$ – Georges Elencwajg Nov 28 '13 at 21:46
  • However there is small subtlety because what we have is $H^0({P,Q},\Bbb{Z}|{{P,Q}})$ and not $H^0({P,Q},\Bbb{Z})$. Is it true that $\Bbb{Z}|{{P,Q}}$ is just constant sheaf on ${P,Q}$? – Dylan B. Nov 28 '13 at 21:48
  • Yes there is a subtlety and yes the restriction of a constant sheaf to any subspace is a constant sheaf. My favourite way for seeing this is by interpreting sheaves as étalé spaces: a very old-fashioned point of view which I claim should not be dismissed as an irrelevant historical oddity. – Georges Elencwajg Nov 28 '13 at 21:52
  • Can you please tell me what is wrong with my reasoning: $H^0({P,Q},\Bbb{Z}|{{P,Q}}) = \varinjlim{U \supseteq {P,Q}} \Bbb{Z}(U)$ by definition. For every open set $U$ containing ${P,Q}$, $\Bbb{Z}(U) = \Bbb{Z}$. Why is the limit (and hence global sections) not just $\Bbb{Z}$? This is what is confusing me. – Dylan B. Nov 28 '13 at 21:56
  • The formula in the reasoning in your last comment is not correct: the sections of a sheaf restricted to a subspace $D$ are not given (in general) by the direct limit of the sections on open neighbourhoods of $D$. You have to first compute the sheaf restricted to $D$. Here the restricted sheaf is the constant sheaf $\mathbb Z$ on $D={P,Q}$. – Georges Elencwajg Nov 28 '13 at 22:07
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    I see where I am wrong, what I have defined there is just the inverse image presheaf and not the sheaf itself. The inverse image presheaf in this case is the constant presheaf, and the sheafification of this is indeed the constant sheaf on ${P,Q}$. I got it now, thanks for your patience Mr. Georges. – Dylan B. Nov 28 '13 at 22:13
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    Dear Dylan, I'm happy that I could help you and that we now see eye to eye. And, yes, the difficulty with the now fashionable definition of restriction of a sheaf is (as you quite correctly described) that you first have to define a presheaf through a process involving a direct limit and then sheafify that presheaf. That's why I like the old étalé space approach. – Georges Elencwajg Nov 28 '13 at 22:19