# Tate twists and cohomology of $\mathbf{P}^1$

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

• It encodes $(2\pi i)^{-d} \int_{X(\mathbf{C}),i}$ (using $i$-oriented integration), which is independent of the choice of $i$. That is, such an operation applied to ${\rm{H}}^{2d}(X(\mathbf{C}), \mathbf{Z}(d))$ yields output in $\mathbf{Z}$. In effect, you shouldn't put the negative Tate twist where you do but rather absorb the positive Tate twist inside the cohomology (since $\mathbf{Z}(d)$ is naturally the orientation sheaf on $X(\mathbf{C})$). – nfdc23 Nov 30 '17 at 7:10
• There are various consistency checks for the twists: They should be compatible with weights of Hodge structures, and the weights in etale cohomology. But to see more directly the relation to $\mathbb{P}^1$, you can use $X\to \mathbb{P^n}\cong S^n\mathbb{P}^1\leftarrow (\mathbb{P}^1)^n$, where the first map is given by inclusion, and the second is a standard identification. – Donu Arapura Nov 30 '17 at 7:16
• $\mathbf{P}^1$ and $\pi_1(\mathbf{C}^{\times})$ have little to do with one another, concerning your question, although I did see in the literature (esp. in $p$-adic Hodge theory, where one has no notion of topological fundamental gp for $p$-adic analytic spaces and formal schemes) and yet one has to introduce an appropriate notion of Tate twist, that indeed the "Tate object" (or Lefschetz motive) is defined by $\mathbf{Z}_p(-1) := H^2(\mathbf{P}^1,\mathbf{Z}_p)$. – user87684 Dec 1 '17 at 7:17
• This has actually nothing to do with $\mathbf{P}^1$, but rather it happens that $H^2(\mathbf{P}^1,\mathbf{Z}_p)$ ought to be the $p$-adic completion of the non-existent $\mathbf{Z}$-dual of $\pi_1(\mathbf{G}_m)$, where $\mathbf{G}_m$ is the $p$-adic analytic, respect. formal, multiplicative gp. Therefore, people often use $H^2(\mathbf{P}^1,\mathbf{Z}_p)$ as a replacement for $\mathbf{Z}_p(-1)$, and I agree it is somewhat misleading. – user87684 Dec 1 '17 at 7:18
• There are two parallel considerations to make. One: for $f : X\to *$ a proper map of $\mathbf{C}$-analytic spaces, consider the Verdier essential pullback $f^{!}$ evaluated on the orientation sheaf $\mathbf{Z}$ of the point $*$. If $X$ is $d$-dimensional and smooth, you can prove $f^{!}\mathbf{Z}\simeq\mathbf{Z}(d)[2d]$ (eg. invoking the fact that $f^!\mathbf{Z}$ must be the dualizing sheaf for the $\mathbf{Z}$-cohomology of $X$ and constructing the (unique!) Verdier trace concretely via integration. – user87684 Dec 1 '17 at 7:23

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 $H^2(\mathbf P^1,\{\infty\},\mathbf Z)$.