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Question 1: Let $X$ be a scheme. Then generally for the complex $C^{\bullet}$ in $D^b(X)$, we define $R\Gamma(C^{\bullet})\colon$ = Complex obtained by applying $\Gamma$ to the injective resolution $I^{\bullet}$ of $C^{\bullet}$.

Why is this notation so useful, and do people like to employ $R\Gamma(C^{\bullet})$?

Hypercohomology $H^{i}(C^{\bullet}) \colon=$ $i$-th homology of $R\Gamma(C^{\bullet})$ is useful and defined via $R\Gamma(C^{\bullet})$, but I often see $R\Gamma(C^{\bullet})$ itself without taking its homology.

In what purpose do we need $R\Gamma(C^{\bullet})$?

Question 2: For the complex $C, D \in C^{\bullet}$ in $D^b(X)$, we can define $C ⊗^{L} D$ by using the flat resolution of $C$ (or $D$). Why do we need to take flat resolution not just the tensor product $C ⊗ D$ of two complexes?

I am sorry, I am very weak about these concepts. Hopefully please teach me the reasons of these.

Pierre MATSUMI

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  • $\begingroup$ Just as a general comment, I think many formalisms of things like spectral sequences (for instance, coming from composition of derived functors) become much much simpler conceptually if you employ $R \Gamma(C^{\bullet})$. For instance, $$ R \Gamma( R\mathcal{H}\mathrm{om}(A, \bullet ) ) = R \mathrm{Hom}(A, \bullet ). $$ $\endgroup$ – Karl Schwede Dec 12 '13 at 19:27
  • $\begingroup$ Q2: For one thing, the naive tensor product (without resolving) is not well defined in the derived category. $\endgroup$ – Donu Arapura Dec 12 '13 at 21:28
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The complex $R\Gamma(X,C^\bullet)$ contains more information than its (co)homology. See for example http://arxiv.org/pdf/math/0001045v2.pdf

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  • $\begingroup$ Great Thanks! Pierre $\endgroup$ – Rinmyaku Jan 6 at 16:46

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