In Deligne's paper "Hodge cycles on abelian varieties" (see page 11 of http://jmilne.org/math/Documents/Deligne82.pdf) he says that the following diagram fails to commute by a factor of $(2 \pi i)^m$, where $X$ is a smooth projective variety over $\mathbb{C}$:

$$ \require{AMScd} \begin{CD} H^n_B(X) \otimes \mathbb{C} @>{1 \mapsto (2\pi i)^m}>> H^n_B(X)(m) \otimes \mathbb{C} \\ @VVV @VVV \\ H^n_{dR}(X) @>{=}>> H^n_{dR}(X)(m) \end{CD} $$

Why? How is the vertical map on the right defined? To me the obvious definition is: the map which makes the above diagram commute.

| cite | improve this question | | | | |
  • 1
    $\begingroup$ This is tautological. Put $H:=H^n_B(X)$, and identify $H^n_{dR}(X)$ with $H\otimes _{\mathbb{Q}}\mathbb{C}$ by the left-hand side vertical isomorphism. Then there is one and only one natural isomorphism $H\otimes _{\mathbb{Q}}(2\pi i)^n\mathbb{Q}\otimes _{\mathbb{Q}}\mathbb{C}\rightarrow H\otimes _{\mathbb{Q}}\mathbb{C}$, namely $1\otimes \mu $, where $\mu $ is the multiplication map $(2\pi i)^n\mathbb{Q}\otimes _{\mathbb{Q}}\mathbb{C}\rightarrow \mathbb{C}$. With this definition the assertion is obvious. $\endgroup$ – abx Jun 16 '15 at 12:36
  • $\begingroup$ I think that makes sense. Does this mean that the "=" map on the bottom of the diagram is in some way unnatural? $\endgroup$ – Martin Orr Jun 16 '15 at 13:28
  • $\begingroup$ No, it is an equality. For what I said, you can replace the square by a triangle. $\endgroup$ – abx Jun 16 '15 at 13:58

The sentence before this diagram says: "Although the Tate twist for de Rham cohomology is trivial, it should not be ignored. For example, when $k=\mathbb C$,..."

Let me try to spell out how I read what Deligne is saying. Earlier on the page he writes that $H^n_{dR}(X) = H^n_{dR}(X)(m)$ for all $m$. This is not even an isomorphism but a literal equality. Given this, a reader may ask: why introduce the Tate twist $H^n_{dR}(X)(m)$ in the first place, then? Why not just write $H^n_{dR}(X)$ for all of the Tate twists, given that they are exactly identical to each other? But the reason it makes sense to distinguish them is that $H^n_{dR}(X) \otimes \mathbb C$ comes equipped with a canonical comparison isomorphism with $H^n(X(\mathbb C),\mathbb C)$ and that $H^n_{dR}(X)(m) \otimes \mathbb C$ comes equipped with a canonical isomorphism with $H^n(X(\mathbb C),\mathbb C(m))$; these are not compatible with each other.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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