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I need to do some explicit computation with algebraic de Rham cohomology on some projective varieties, also some open subsets of them.

I don't know the theory well, but I just search some notes to have an idea how to do the computation.

On p.15 (p.18 on PDF file) of Periods and algebraic deRham cohomology by Benjamin Friedrich, there is a long sequence

$ \cdots \rightarrow H_{dR}^{p-1} (D/k) \rightarrow H_{dR}^{p}(X, D/k) \rightarrow H_{dR}^{p} (X/k) \rightarrow H_{dR}^{p} (D/k)\rightarrow H_{dR}^{p+1}(X, D/k) \rightarrow \cdots$

, here $H_{dR}^{p+1}(X, D/k)$ is the relative algebraic de Rham cohomology.

But on pp.10-11 of Note On Algebraic de Rham Cohomology, there is

$ \ \ \ \ \ \ \ \ \ H_{dR}^{i-1} (U) \rightarrow H_{dR}^{i-2}(Y) \rightarrow H_{dR}^{i} (X/k) \rightarrow H_{dR}^{i} (U)$

which is obtained from logarithmic de Rham complex.

It seems to me that these 2 exact sequences are very different. Is the relative algebraic de Rham cohomology $H_{dR}^{p}(X, D/k)$ the same or related to $H_{dR}^{p}(U, D/k)$ if $U = X \ \backslash \ D$ ? And when one wants to do the computation for the open subvariety, which exact sequence is used?

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    $\begingroup$ The second sequence is correct only if Y is smooth in which case the Gysin isomorphism will identify its cohomology with the relative cohomology of (X,U). $\endgroup$
    – naf
    Commented Jun 4, 2013 at 10:04
  • $\begingroup$ @ulrich: Could you give a reference for Gysin isomorphism and (if necessary) how it identifies these things. Also, I think what I need is an identification of the cohomology of $U$ to something, rather than an identification of the cohomology of a closed subscheme to something. $\endgroup$
    – user565739
    Commented Jun 4, 2013 at 16:02

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See Theorem 5.9 of the following article: https://impa.br/wp-content/uploads/2022/01/33CBM17-eBook.pdf. Bott-Tu relative algebraic de Rham cohomology is what Friedlich has studied. I didn' check Godbillon relative algebraic de Rham cohomology is equivalent to one in the note you sited.

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    $\begingroup$ It would be helpful to state the theorem in the answer in case the link fails to work in the future. For the same reason, please include the name of the reference you link to. $\endgroup$ Commented Nov 29, 2023 at 9:32

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