# Tate twist and comparison between Betti and de Rham cohomology

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.

• 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. – abx Jun 16 '15 at 12:36
• I think that makes sense. Does this mean that the "=" map on the bottom of the diagram is in some way unnatural? – Martin Orr Jun 16 '15 at 13:28
• No, it is an equality. For what I said, you can replace the square by a triangle. – 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.