I was wondering if anyone could give me some intuition as to why, for a smooth projective variety $X$ over $\mathbf{C}$ of complex dimension $d$, the Tate twist on $H^n(X(\mathbf{C}),\mathbf{Z})$ to be incorporated to have a pairing with $H\_{d-n}(X(\mathbf{C}),\mathbf{Z})$ into $\mathbf{Z}(-d)$, is related to (tensor powers of) the cohomology group $H^2(\mathbf{P}^1,\mathbf{Z})\simeq H^1(\mathbf{G}\_m,\mathbf{Z})$.


In other words, why does one often define $\mathbf{Z}(-1) := H^2(\mathbf{P}^1,\mathbf{Z})$? Is there some Kunneth formula lurking behind the scene? Should I think about $X(\mathbf{C})$ as a family $X(\mathbf{C})\times \mathbf{P}^1\to \mathbf{P}^1$? Is there a map $H^2((\mathbf{P}^1)^d,\mathbf{Z})\to H^d(X(\mathbf{C}),\mathbf{Z})$? Where does it come from?


I can understand the notation comes from $H\_0(\mathbf{P}^1,\mathbf{Z})\simeq H\_1(\mathbf{G}\_m,\mathbf{Z})\simeq\pi\_1^{\rm ab}(\mathbf{G}\_m) =\pi\_1(\mathbf{G}\_m) := \mathbf{Z}(1)$, the reason I'm asking the question: even if there is $H\_0(X(\mathbf{C}),\mathbf{Z})\to H\_0(\mathbf{P}^1,\mathbf{Z})$, I see no map $X(\mathbf{C})\to\mathbf{P}^1$ in general. Unless somehow $X(\mathbf{C})$ can always be realized as the fiber over some point of $\mathbf{P}^1$ of some map $\mathcal{X}\to\mathbf{P}^1$, with $X$ homotopy equivalent to $\mathcal{X}$. Can one just choose $\mathcal{X} = X(\mathbf{C})\times\mathbf{P}^1$? (can it possibly be so simple? In algebraic geometry, this would amount to taking the trivial deformation of $X\to \*$ to $\mathcal{X} \to\mathbf{P}^1$ along a fixed point $\*\to\mathbf{P}^1$, and trivial deformations strike me as usually not so interesting). I'll appreciate any insight on the matter a lot


Thanks



Tate twists and cohomology of $\mathbf{P}^1$
The Tate twist is what we need to express Poincaré duality without making any choice. Such a choice appears in the choice of an orientation of the affine line minus the origin, and have shadows in the description of the Thom isomorphism, hence, in the description/construction of Gysin maps, trace maps, and so forth. A natural way to make this transparent goes through the theory of Chern classes. In what follows, when I write the symbol $=$, I mean that there is a canonical isomorphism, the construction of which does not involve any choice: it always comes by functoriality.


If we define $\mathbf Z(1)$ as the first homology group of the affine line minus the origin, then it is a free abelian group of rank one, and one defines $\mathbf Z(n)$ as a tensor product of $n$ copies of $\mathbf Z(1)$ for $n\geq 0$, and as the dual of $\mathbf Z(-n)$ for $n<0$. Given an abelian group $A$, one defined $A(n)=A\otimes\mathbf Z(n)$. Since, by definition, cohomology is the dual of homology, and since we are dealing with free groups, we have a canonical identification
$$H^1(\mathbf A^1-\{0\},\mathbf Z)=\mathbf Z(-1)\, .$$
By an elementary Mayer-Vietoris argument, one can deduce that $H^2(\mathbf P^1,\mathbf Z(1))=\mathbf Z$, but that is just a computation, not an explanation.


What precedes means that the classifying space of the topological abelian group $\mathbf C^\times=\mathbf G\_m(\mathbf C)$ is not really a $K(\mathbf Z,2)$, but rather a $K(\mathbf Z(1),2)$. This is why we have Chern classes of line bundles
$$c\_1:Pic(X)=[X,B\mathbf G\_m]\to[X(\mathbf C),B\mathbf C^\times]=H^2(X,\mathbf Z(1))$$
(where $[A,B]$ means the homotopy classes of maps from $A$ to $B$ in the appropriate sense). This is used to prove the projective bundle formula. For a vector bundle $E$ on $X$ of rank $r$, with associated projective bundle $\mathbf P(E)$, if $t$ denotes the first Chern class of the tautological line bundle on $\mathbf P(E)$, then the map $(x\_0,\ldots,x\_{r-1})\mapsto \sum^{r-1}\_{i=0}t^ix\_i$ is an isomorphism:
$$\bigoplus\_{i=0}^{r-1}H^{n-2i}(X,\mathbf Z(-i))=H^n(\mathbf P(E),\mathbf Z)$$
Now, given a closed immersion $i:Z\to X$ between smooth schemes over $\mathbf C$, one defines $H^n(X,Z)=H^n(Z,i^!(\mathbf Z))$. In other words, this is the $n$th cohomology group of the homotopy fibre of the restriction map $R\Gamma(X,\mathbf Z)\to R\Gamma(X-Z,\mathbf Z)$. The Thom isomorphism is a canonical identification:
$$H^n(X,Z)=H^{n-2c}(Z,\mathbf Z(-c))$$
where $c$ is the codimension of $Z$ in $X$. (When $i=s$ is the zero section of a vector bundle $E$ the Thom isomorphism is obtained from the projective bundle formula applied to the direct sum of $E$ and of the trivial line bundle. The general case follows from a deformation to the normal cone argument.)


For $X=\mathbf P^1$ and $Z=\{\infty\}$, this means that
$$H^2(\mathbf P^1,\{\infty\},\mathbf Z)=H^0(\{\infty\},\mathbf Z(-1))=\mathbf Z(-1)\, .$$
The meaning of all this is that $\mathbf Z(-1)$ is not the second cohomology group of the projective line. It is rather the second cohomology group of the projective line pointed at infinity. Of course, by homotopy invariance, there is a canonical isomorphism $H^2(\mathbf P^1,\{\infty\},\mathbf Z)=H^2(\mathbf P^1,\mathbf Z)$, but, in some sense, that is misleading: the canonical identification of $H^2(\mathbf P^1,\{\infty\},\mathbf Z)$ with the dual of $H\_1(\mathbf A^1-\{0\},\mathbf Z)$ is through a (baby version of the) Thom isomorphism, which is yet another expression of the theory of Chern classes.
When we write the motive of $\mathbf P^1$ as a direct sum of the constant motive with a Tate motive, by definition, the Tate motive is the motive of the 
projective line pointed at infinity. Therefore, its Betti realisation goes to $H^2(\mathbf P^1,\{\infty\},\mathbf Z)$.


