2
$\begingroup$

Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) of $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for maps: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps of abelian sheaves on $Y'$: $$ g_1^*, g_2^* : f^{-1}\hat{\Omega^*_X} \to \hat{\Omega^{*}_{X'}} $$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy explicitly if possible as well...

$\endgroup$

1 Answer 1

2
$\begingroup$

I don't think they are homotopic in general. If I remember correctly, the argument of independence is as follows: consider $p_1, p_2 : X\times X\to X$. Then maps $g_1, g_2 : X'\to X$ induce a map $g : X'\to X\times X$. Since $g = (f, f)$ when restricted to $Y'$, $g$ induces a map $g^* : \hat{\Omega^{*}_{X\times X}}\to \hat{\Omega^{*}_{X'}}$, where the first hat means completion with respect to $Y$ embedded diagonally in $X\times X$. Hartshorne results (the one you quoted) says that $p_1^*, p_2^* : \hat{\Omega^{*}_{X}}\to\hat{\Omega^{*}_{X\times X}}$ (again completed with respect to $Y$) are quasi-isomorphisms. Since $g_1 = p_1g$ and $g_2=p_2 g$, we see that on the derived category the maps $g_1^*$ and $g_2^*$ are isomorphic (meaning that there is a commutative square having $g_1^*$, $g_2^*$ on the left and right and having isomorphisms on the top and bottom. Moreover this square is ''functorial'' (i.e., if you have another lift, there is a cocycle condition). I hope the answer is clear; unfortunately it is hard to draw a diagram here...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.