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Here are a few more details: Suppose that $F=\mathbb{Q}_X$. In this case, $$ R^qj_*F = \begin{cases} \mathbb{Q}_X & \text{if $q=0$}\\ \mathbb{Q}_Z & \text{if $q= 2 c-1$, $c=codim(Z)$}\\ 0 & \text{otherwise} \end{cases} $$ To see this, first note that the computation is local, so we may replace the pair $(X,Z)$ by $(\mathbb{C}^n, \mathbb{C}^{n-c})$ using the implicit function theorem. After this, it should be easy. With this calculation the $E_2$ term of Leray is concentrated along two lines, and it is not hard to see that it reduces to the above "localization" sequence given in excision in algebraic de Rham cohomology

Notes: Sorry, I had a typo in the formula yesterday. If $Z$ is smooth but not pure dimensional, then $R^qj_*F|_{Z_i}= \mathbb{Q}_{Z_i}$ where $q=2c_i-1$ for each component. The final result will be a bit more cumbersome to state. Some related calculations occur in Deligne's Theorie de Hodge I, II.

Here are a few more details: Suppose that $F=\mathbb{Q}_X$. In this case, $$ R^qj_*F = \begin{cases} \mathbb{Q}_X & \text{if $q=0$}\\ \mathbb{Q}_Z & \text{if $q= 2 c-1$, $c=codim(Z)$}\\ 0 & \text{otherwise} \end{cases} $$ To see this, first note that the computation is local, so we may replace the pair $(X,Z)$ by $(\mathbb{C}^n, \mathbb{C}^{n-c})$ using the implicit function theorem. After this, it should be easy. With this calculation the $E_2$ term of Leray is concentrated along two lines, and it is not hard to see that it reduces to the above "localization" sequence given in excision in algebraic de Rham cohomology

Notes: Sorry, I had a typo in the formula yesterday. If $Z$ is smooth but not pure dimensional, then $R^qj_*F|_{Z_i}= \mathbb{Q}_{Z_i}$ where $q=2c_i-1$ for each component. The final result will be a bit more cumbersome to state. Some related calculations occur in Deligne's Theorie de Hodge I, II.

Martin, my comment was admittedly cryptic, since I generally won't make much effort for anonymous users. But to elaborate slightly, Here are a few more details: Suppose that $F=\mathbb{Q}_X$. In this case, $$ R^qj_*F = \begin{cases} \mathbb{Q}_X & \text{if $q=0$}\\ \mathbb{Q}_Z & \text{if $q= 2 c$, $c=codim(Z)-1$}\\ 0 & \text{otherwise} \end{cases} $$ To see this, first note that the computation is local, so we may replace the pair $(X,Z)$ by $(\mathbb{C}^n, \mathbb{C}^{n-c})$ using the implicit function theorem. After this, it should be easy. With this calculation the $E_2$ term of Leray is concentrated along two lines, and it is not hard to see that it reduces to the above "localization" sequence given in excision in algebraic de Rham cohomology

Here are a few more details: Suppose that $F=\mathbb{Q}_X$. In this case, $$ R^qj_*F = \begin{cases} \mathbb{Q}_X & \text{if $q=0$}\\ \mathbb{Q}_Z & \text{if $q= 2 c-1$, $c=codim(Z)$}\\ 0 & \text{otherwise} \end{cases} $$ To see this, first note that the computation is local, so we may replace the pair $(X,Z)$ by $(\mathbb{C}^n, \mathbb{C}^{n-c})$ using the implicit function theorem. After this, it should be easy. With this calculation the $E_2$ term of Leray is concentrated along two lines, and it is not hard to see that it reduces to the above "localization" sequence given in excision in algebraic de Rham cohomology

Notes: Sorry, I had a typo in the formula yesterday. If $Z$ is smooth but not pure dimensional, then $R^qj_*F|_{Z_i}= \mathbb{Q}_{Z_i}$ where $q=2c_i-1$ for each component. The final result will be a bit more cumbersome to state. Some related calculations occur in Deligne's Theorie de Hodge I, II.

Martin, my comment was admittedly cryptic, since I generally won't make much effort for anonymous users. But to elaborate slightly, I was referring to the case $F=\mathbb{Q}$. In this case, with the help of a tubular neighbourhood, we can see that $j_*F = \mathbb{Q}$ if $q=0$, $R^qj_*F= \mathbb{Q}_Z$ when $q=2 codim(Z)-1$, and the other direct images are zero. Then it is not hard to see that Leray reduces to the above "localization" sequence given in excision in algebraic de Rham cohomology

Martin, my comment was admittedly cryptic, since I generally won't make much effort for anonymous users. But to elaborate slightly, Here are a few more details: Suppose that $F=\mathbb{Q}_X$. In this case, $$ R^qj_*F = \begin{cases} \mathbb{Q}_X & \text{if $q=0$}\\ \mathbb{Q}_Z & \text{if $q= 2 c$, $c=codim(Z)-1$}\\ 0 & \text{otherwise} \end{cases} $$ To see this, first note that the computation is local, so we may replace the pair $(X,Z)$ by $(\mathbb{C}^n, \mathbb{C}^{n-c})$ using the implicit function theorem. After this, it should be easy. With this calculation the $E_2$ term of Leray is concentrated along two lines, and it is not hard to see that it reduces to the above "localization" sequence given in excision in algebraic de Rham cohomology