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Let $k$ be a field and $X$ be a proper $k$ scheme. Futhermore let $\mathcal{H}$ be a coherent $\mathcal{O}_X$-module. Grothendieck's Finiteness Theorem says that for all $i \ge 0$ the cohomology groups $H^i(X, \mathcal{H})$ are finite vector spaces.

Let $\mathcal{F}, \mathcal{G}$ be two inverible sheaves. Consider $H^1(X, \mathcal{F}^{\vee} \otimes \mathcal{G}) = k^n$. Under some conditions for $X$ we have the identification $H^1(X, \mathcal{F}^{\vee} \otimes \mathcal{G}) = Ext^1 _X(\mathcal{O}_X, \mathcal{F}^{\vee} \otimes \mathcal{G})= Ext^1 _X (\mathcal{F},\mathcal{G})$.

$Ext^1 _X (\mathcal{F},\mathcal{G})$ describes the extension classes of $\mathcal{F}$ by $\mathcal{G}$.

By definition two extensions $\mathcal{L}, \mathcal{K}$ are equivalent if there exist a sheaf isomorphism $d: \mathcal{L} \to \mathcal{L}$ such that following diagram commutes:

$$ \require{AMScd} \begin{CD} 0 @>{} >> \mathcal{F} @>{a} >> \mathcal{L} @>{a} >> \mathcal{G} @>{} >> 0\\ @VV0V @VVidV @VVdV @VVidV @VV0V \\ 0 @>{} >> \mathcal{F} @>{b}>> \mathcal{K} @>{a} >> \mathcal{G} @>{} >> 0; \end{CD} $$

Now the question: Generally, $Ext^1 _X (\mathcal{F},\mathcal{G})$ is only a set consisting of all extension equivalence classes. But here - since it is isomorphic the vector space $k^n$ - I'm curious if this extra vector space structure tells "more" about the shape of the extension classes.

For example: How are the equivalence classes related to each other which belong in $k^n$ to the same line $l \subset k^n$. Is there a special connection between them in the commuting diagram? Or same subspace?

Or: What can we say about $d$ between two aquivalence classes lying in the same line? In $k^n$ they can be transformed to each other but simple scalar multiplication $ \cdot k$. Does this tell something about $d$?

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    $\begingroup$ Yes, $d$ is scalar multiplication by the same constant and thus if non-zero, $L=K$. $\endgroup$ – Mohan Oct 8 '18 at 8:04
  • $\begingroup$ @Mohan: Do you know a source where that is proved rigorouosly? Despite of the intuitive plausibility that it should hold it seems that to show it formally one need to work with concrete construction of derived functors. $\endgroup$ – KarlPeter Oct 8 '18 at 10:30
  • $\begingroup$ Most books on homological algebra will prove the local version and the global version follows. $\endgroup$ – Mohan Oct 8 '18 at 11:28

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