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.



Tate twist and comparison between Betti and de Rham cohomology
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.


