TITLE: Twist in identification with singular cohomology
QUESTION [7 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{Q}$ and $$V = H^m(X(\mathbb{C}), \mathbb{Q} \cdot (2\pi i)^r)$$ Then I've seen people write the comparison with complex cohomology (an isomorphism of $\mathbb{Q}_\ell$ vector spaces) as $$V \otimes_\mathbb{Q} \mathbb{Q}_\ell \cong H^m_{et}(X_{\bar{\mathbb{Q}}}, \mathbb{Q}_\ell)(r)$$
Why does twisting on the right hand side correspond to taking coefficients in $\mathbb{Q} \cdot (2\pi i)^r$ instead of $\mathbb{Q}$? What does the twist do, what is it used for?

REPLY [7 votes]: Let me supplement Vivek's nice answer with a few additional comments. People coming from differential geometry typically normalize the $n$th Chern class -- in the Chern-Weil formula -- by a factor of $1/(2\pi i)^n$ to get it to be integral. However, if you want compatibility with algebraic definitions, then you don't want to do this, so then $c_n$ would take values in $H^{2n}(X,(2\pi i)^n\mathbb{Z})$. For example, in additional to the etale  first Chern class explained above, you can use algebraic de Rham cohomology. Here $c_1$ is given by the induced map on $H^2$ associated to 
$$ d\log:\mathcal{O}_X^*[-1]\to \Omega_X^\bullet$$
Continued-- To expand slightly, one can see by a diagram chase that, under the Grothendieck isomorphism $H^2(X,\mathbb{C})\cong H^2(X,\Omega_X^\bullet)$, the image of this $c_1$ lands in $H^2(X,2\pi i\mathbb{Z})$. So by the usual tricks involving the splitting principle, one gets the a similar statement for higher Chern classes. In case, you prefer to avoid Chern classes, it follows (with some work), that the image of the cycle map $CH^n(X)_\mathbb{Q}\to H^{2n}(X, \mathbb{C})$ lands in $H^{2n}(X, \mathbb{Q}(n))$. Again this is consistent with what happens on the etale side.