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Tate twists in singular and de Rham cohomology

Posted by Martin Orr on Friday, 19 June 2015 at 19:30

Tate twists in singular cohomology are a device for dealing with factors of 2 \pi i which come up whenever we compare singular and de Rham cohomology of complex projective varieties. In this post I will explain the problem, including calculating the 2 \pi i in the case of \mathbb{P}^1, and define Tate twists to solve it.

In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, we could just write out factors of 2 \pi i everywhere. On the other hand, there is also a notion of Tate twists for \ell-adic cohomology, which cannot be omitted so easily, and which I will discuss in a subsequent post.

The problem

Let k be a field of characteristic zero which can be embedded in \mathbb{C} and let X be a smooth projective variety over k. For each embedding \sigma \colon k \hookrightarrow \mathbb{C}, there is a comparison isomorphism between de Rham and singular cohomology of X^\sigma:  \sigma^* \colon H^n_{dR}(X/k) \otimes_{k,\sigma} \mathbb{C} \to H^n(X^\sigma(\mathbb{C}), \mathbb{Q}) \otimes_{\mathbb{Q}} \mathbb{C}.

As I briefly mentioned previously, for each k-algebraic subvariety Z \subset X of codimension p, we can define a so-called de Rham cycle class cl_{dR}(Z) \in H^{2p}_{dR}(X/k).

If we choose an embedding \sigma \colon k \hookrightarrow \mathbb{C}, we can also define a cycle class for Z^\sigma in singular cohomology (which appears in the Hodge conjecture). But there are two different normalisations for the cycle class map into singular cohomology.

First, there is the topologist's version cl_{top}(Z^\sigma), defined by taking the fundamental class of Z^\sigma in singular homology and then using Poincaré duality. This is an integral element of singular cohomology:  cl_{top}(Z^\sigma) \in H^{2p}(X^\sigma, \mathbb{Z}).

However, it turns out that the topologist's cycle class map is not compatible with the de Rham cycle class map. They differ by a factor of (2\pi i)^p:  \sigma^*(cl_{dR}(Z)) = (2\pi i)^p \, cl_{top}(Z^\sigma). I will prove this below for the class of a point in \mathbb{P}^1.

The solution: Tate twists

We fix this problem by changing the normalisation of the cycle class map for singular cohomology. Why do we change the normalisation for singular cohomology instead of for de Rham cohomology? Because this normalisation is something we will only do over \mathbb{C}, and the definition of the de Rham cycle class map works over all fields.

If we simply multiplied cl_{top} by (2\pi i)^p, then we would solve the compatibility issue, but it would no longer take values in H^{2p}(X^\sigma, \mathbb{Z}) (and hence its values would not be Hodge classes). So instead we introduce a new object called a Tate twist.

The Tate twist for singular cohomology is a \mathbb{Z}-Hodge structure, denoted \mathbb{Z}(1), with underlying \mathbb{Z}-module 2\pi i\mathbb{Z} and with Hodge type (-1, -1).

As pieces of notation, we define  \mathbb{Z}(m) = \mathbb{Z}(1)^{\otimes m} for each positive integer m, and define \mathbb{Z}(-m) to be the dual of \mathbb{Z}(m). We also define \mathbb{Q}(m) to be \mathbb{Q}-Hodge structure induced by \mathbb{Z}(m), and  H^n(X^\sigma, \mathbb{Z})(m) = H^n(X^\sigma, \mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Z}(m).

We define the algebraic geometer's cycle class map for singular cohomology to be  cl_B(Z^\sigma) = cl_{top}(Z^\sigma) \otimes (2 \pi i)^p \in H^{2p}(X^\sigma, \mathbb{Z})(p). (The subscript B is for Betti.)

It is useful to also define Tate twists for de Rham cohomology, even though they are trivial. We thus define H^n_{dR}(X/k)(m) to be equal to H^n_{dR}(X/k) (not just isomorphic). The reason why we want to use the label (m) for a Tate twist in de Rham cohomology is so that we remember which isomorphism to use to compare it with singular cohomology.

I was a bit confused about this until I asked a MathOverflow question a few days ago which helped me clear it up. The comparison isomorphism  H^n(X^\sigma, \mathbb{Z})(m) \otimes_{\mathbb{Z}} \mathbb{C} = H^n(X^\sigma, \mathbb{C}) \otimes_{\mathbb{Z}} \mathbb{Z}(m) \to H^n_{dR}(X^\sigma/\mathbb{C})(m) is defined as the untwisted comparison isomorphism  \sigma^{*-1} \colon H^n(X^\sigma, \mathbb{C}) \to H^n_{dR}(X^\sigma/\mathbb{C}) multiplied by the inclusion (2 \pi i)^m \mathbb{Z} \hookrightarrow \mathbb{C}.

With this comparison isomorphism, we get the desired compatibility between cycle class maps:  cl_B(Z^\sigma) = \sigma^*(cl_{dR}(Z)) \in H^{2p}(X^\sigma, \mathbb{C})(p).

Because the Tate twist \mathbb{Z}(1) is defined to have Hodge type (-1, -1), the Hodge structure H^{2p}(X^\sigma, \mathbb{Z})(p) has weight 0. We define a Hodge class in H^{2p}(X^\sigma, \mathbb{Z})(p) to be an element of H^{2p}(X^\sigma, \mathbb{Z})(p) with Hodge type (0,0). The image of cl_B consists of Hodge classes.

We could have carried out all of our previous two posts on absolute Hodge classes using H^{2p}(X^\sigma, \mathbb{Z})(p) in place of H^{2p}(X^\sigma, \mathbb{Z}); the changes would be essentially just notational.

Calculating the de Rham cycle class map for \mathbb{P}^1

I am going to justify the fact that the de Rham cycle class map and the topologist's cycle class map into H^2(\mathbb{P}^1) differ by a factor of 2 \pi i. One often sees the 2 \pi i justified by some sort of vague reference to the integral round a loop in \mathbb{G}_m(\mathbb{C}), but that doesn't seem to fit our setting of smooth projective varieties. So I am going to write out this calculation carefully (except that I will not be careful about signs - there are several sign conventions along the way and I don't think it is worth keeping track of them). Thanks to Jack Shotton who helped with this.

First, cl_{top}(pt) is a generator of H^2(\mathbb{P}^1, \mathbb{Z}). If we choose a 2-form \zeta_{top} representing the corresponding element of H^2_{dR}(\mathbb{P}^1/\mathbb{C}), the definition of the comparison isomorphism says that  \int_{\mathbb{P}^1} \zeta_{top} = 1.

The hard work is to calculate cl_{dR}(pt). In particular, we want to show that  \int_{\mathbb{P}^1} cl_{dR}(pt) = 2 \pi i.

First, we interpret our point as a divisor of degree 1 on \mathbb{P}^1. Under the standard isomorphism \operatorname{Pic}(\mathbb{P}^1) \cong H^1(\mathbb{P}^1, \mathcal{O}^\times), a divisor of degree 1 corresponds to the Čech cocycle  z \in \mathcal{O}^\times(U_1 \cap U_2) with respect to the open cover U_1 = \mathbb{P}^1 - \{ 0 \}, U_1 = \mathbb{P}^1 - \{ \infty \} (this is one place where I can never remember the sign convention).

We then apply the map d \operatorname{log} \colon \mathcal{O}^\times \to \Omega^1 to get the class in H^1(\mathbb{P}^1, \Omega^1) represented by the Čech cocyle  \frac{dz}{z} \in \Omega^1(U_1 \cap U_2).

The degeneracy of the Hodge-de Rham spectral sequence gives us an isomorphism H^1(\mathbb{P}^1, \Omega^1) \to H^2_{dR}(\mathbb{P}^1). To compute this isomorphism analytically over \mathbb{C}, we compare the Čech and Dolbeault complexes for \Omega^1:

   \Omega^1(\mathbb{P}^1)       \ar[rr]^-{\txt{restrict}}   \ar@{=}[d]
 & \Omega^1(U_1) \times \Omega^1(U_2)
                \ar[rr]^-{(\omega_1, \omega_2) \mapsto \omega_1 - \omega_2}
                \ar[d]^{\substack{(\omega_1, \omega_2)  \\  \mapsto  \\  f_1 \omega_1 + f_2 \omega_2}}
 & \Omega^1(U_1 \cap U_2)
                \ar[d]^{\substack{\omega  \\  \mapsto  \\  (\bar\partial f_1) \omega}}
 & 0                            \ar[d]
\\ \Omega^1(\mathbb{P}^1)       \ar@{^{(}->}[rr]
 & \mathcal{A}^{1,0}(\mathbb{P}^1)  \ar[rr]^{\bar\partial}
 & \mathcal{A}^{1,1}(\mathbb{P}^1)  \ar[r]^{\bar\partial}
 & 0

Here \mathcal{A}^{p,q} denotes the sheaf of C^\infty (p,q)-forms. We have chosen a partition of unity f_1, f_2 \in C^\infty(\mathbb{P}^1(\mathbb{C}), \mathbb{R}) such that f_1 + f_2 = 1, f_1 vanishes on a neighbourhood of zero and f_2 vanishes on a neighbourhood of \infty. Note that \bar\partial (f_1 + f_2) = 0 and this implies that the square in the centre of the diagram commutes.

We conclude that cl_{dR}(pt) is represented by the C^\infty (1,1)-form  \zeta_{dR} = (\bar\partial f_1) \frac{dz}{z} = \frac{\partial f_1}{\partial \bar{z}} \frac{d\bar{z} \, dz}{z}. We want to compute the integral of \zeta_{dR} over \mathbb{P}^1(\mathbb{C}), and show that this is equal to 2 \pi i.

Choose a loop \gamma which generates \pi_1(U_1 \cap U_2). Let D_1 and D_2 be the two regions into which \gamma divides \mathbb{P}^1(\mathbb{C}), labelled such that 0 \in D_1 and \infty \in D_2.

On D_1, the 1-form f_1 dz/z is well-defined and has total derivative equal to \zeta_{dR}. Similarly, on D_2, the 1-form -f_2 dz/z is well-defined and has total derivative equal to \zeta_{dR}.

Hence, using Stokes' theorem, we get  \int_{\mathbb{P}^1} \zeta_{dR}
    = \int_{D_1} d \left( f_1 \frac{dz}{z} \right) + \int_{D_2} d \left( -f_2 \frac{dz}{z} \right)
    = \int_\gamma f_1 \frac{dz}{z} + \int_\gamma f_2 \frac{dz}{z}. (The minus sign in the second integrand cancels with the fact that the boundary of D_2 goes backwards round \gamma.)

Since f_1 + f_2 = 1, we get  \int_{\mathbb{P}^1} \zeta_{dR} = \int_\gamma \frac{dz}{z} = 2 \pi i.

Tags abelian-varieties, alg-geom, hodge, maths


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