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So I've been trying to understand a proof of the fact that $H^p(M, \mathcal F)=0$ whenever $p\geq 1$ from page 42 of the book "Principles of Algebraic Geometry" by Griffiths and Harris.

The proof is carried out for the sheaf ${\mathcal a}^{r,s}$ of $C^{\infty}$ forms of type $(r,s)$ on $U$ and there is a remark after the proof of how one can prove the result for an arbitrary fine sheaf.

According to them, a fine sheaf is a sheaf for which we have maps $\eta_\alpha:\mathcal F (U_{\alpha})\rightarrow \mathcal F (U)$ for any $U=\cup U_{\alpha}$, such that the support of $(\eta_{\alpha}\sigma)$ is contained in $U_\alpha$ and $\sum {\eta_{\alpha}} (\sigma| _{U_\alpha})=\sigma$ for $\sigma\in \mathcal F (U)$.

So is $(\eta_{\alpha}\sigma)$ a map from the manifold $M$ to $\mathbb R$? I don't understand how $(\eta_{\alpha}\sigma)$ has $M$ as its domain.

Secondly how can they write $(\eta_{\alpha}\sigma)$ when $\sigma\in \mathcal F (U)$? Isn't the domain of $\eta_\alpha$ the group ${\mathcal F }(U_\alpha)$?

Thirdly how does this formulation apply to the case when $\mathcal F$ is the sheaf ${\mathcal a}^{(r,s)}$? We would need maps $\eta_{\alpha}: {\mathcal a}^{(r,s)}(U_{\alpha})\rightarrow {\mathcal a}^{(r,s)}(U)$. But indstead the proof just uses the partion of unity $C^{\infty} $ maps $\rho_{\alpha}$ subordinate to some open cover $\{U_\alpha\}$. So it seems that they haven't used the fact that ${\mathcal a}^{r,s}$ is a fine sheaf.

David Roberts
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Quistis Trepe
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  • I think in order to imitate the partition of unity, G&H abuses the language here. Probably you need a reference for a more detailed version about these. I learned these stuff from Voisin's hodge theory and complex algebraic geometry. – Honglu May 27 '11 at 06:27
  • Yes the problem with learning from Grifiths and Harris is that they constantly abuse notation, making a lot of what they say somewhat elusive. – Quistis Trepe May 27 '11 at 07:33

1 Answers1

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About your third question. Using partition of unity $(\rho_\alpha)_\alpha$ subordinate to $(U_\alpha)_\alpha$ for $\mathcal F$ you can define maps $\eta_\alpha:\mathcal F(U_\alpha)\to\mathcal F(U)$ by taking $s\in\mathcal F(U_\alpha)$ to $\rho_\alpha s$.

About the first two points it seems to me that $(\eta_\alpha\sigma):=\eta_\alpha(\sigma_{|U_\alpha})$.

EDIT: to summarize my answer a fine sheaf is, "by definition", a sheaf having partition of unity.

DamienC
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  • Thanks. Though I still don't understand what it means to say that $support(\eta_\alpha \sigma)\subset U_\alpha$. I can understand what it means in the case where $\mathcal F$ is the sheaf of $C^\infty$ functions, but when it comes to an arbitrary element $\sigma$ of some section $\mathcal F (U)$, I cannot fathom what it means for $\sigma$ to only be non zero on $U_\alpha$, where $U_\alpha$ is some element of a open cover of $M$. – Quistis Trepe May 27 '11 at 13:28
  • For a section $\tau\in \mathcal F(U)$ (e.g. $\tau=(\eta_\alpha\sigma)$), the complement $C$ of its support is defined as follows: a point $x\in U$ lies in $C$ iff there exists an open neighborhood $V$ of $x$ in $U$ such that $tau_{|V}=0$. – DamienC May 27 '11 at 14:56