Background: For a smooth proper variety $X$ over an algebraically closed field $k$, we have the etale cohomology groups $H^i(X,\mathbb{Q}_{\ell})$ for $\ell \not= p$. We can use the Kummer exact sequence to show that $H^1(X,\mathbb{Q}_\ell(1))$ is isomorphic to $V_\ell \mathrm{Pic}^0(X)$, where $\mathrm{Pic}^0(X)$ is the Picard variety of $X$, and $V_\ell(-)$ indicates taking the inverse limit of $\ell^n$-torsion points as $n \rightarrow \infty$ (i.e., the Tate module), and tensoring with $\mathbb{Q}$. Moreover, the Kummer exact sequence also shows that for a morphism of smooth proper varieties $f: X \rightarrow Y$, the pullback map $f^*: H^1(Y,\mathbb{Q}_\ell(1)) \rightarrow H^1(X,\mathbb{Q}_\ell(1))$ agrees with the map induced by pullback of line bundles $(L \mapsto f^*L): V_\ell\mathrm{Pic}^0(Y) \longrightarrow V_\ell \mathrm{Pic}^0(X)$.

Suppose moreover that $f: X \rightarrow Y$ is a map of smooth proper varieties of the same dimension $d$. Then Poincare duality shows that $H^1(X,\mathbb{Q}_{\ell}(1))$ is dual to $H^{2d-1}(X,\mathbb{Q}_{\ell}(d-1))$, and we can take the dual to the pullback map $H^{2d-1}(Y,\mathbb{Q}_\ell(d-1)) \rightarrow H^{2d-1}(X,\mathbb{Q}_{\ell}(d-1))$ to get a map $$f_*: H^1(X,\mathbb{Q}_\ell(1)) \rightarrow H^1(Y,\mathbb{Q}_\ell(1)).$$

Here's my question: How do we describe this map in terms of line bundles? Does it agree with the map $V_\ell \mathrm{Pic}^0(X) \rightarrow V_\ell \mathrm{Pic}^0(Y)$ induced by proper pushforward of divisors (i.e., represent each line bundle by a divisor and take its proper pushforward)? This seems extremely likely to me, but I'm not sure how to prove it in general. Notice that this qustion is not the same as the compatibility of cycle class maps with pushforward, since the cycle class of a divisor lives in $H^2$, not $H^1$.

  • $\begingroup$ Have you worked out what happens in the case of, say, elliptic curves, when all the cohomology can be made explicit? Looks to me like it might give you the dual isogeny. $\endgroup$ – Martin Bright Mar 19 '13 at 9:39
  • 2
    $\begingroup$ Clearly the proper pushforward of divisors is the same as the map on $H^1(\mathcal O_X^\times)$ induced by the norm map of $\mathcal O_X^{\times}$, because the divisor of the norm is the proper pushforward of the divisor, and every divisor is locally the divisor of some function. So by the Kummer exact sequence, the proper pushforward on $H^1(X,\mathbb Q_l(1))$ comes from the norm map on $\mu_{l^n}$, which sends a function on $X$ which is an $l^n$th root of unity to its norm, which is also an $l^n$th root of unity. I'm pretty sure this is what you want, but can't prove it. $\endgroup$ – Will Sawin Mar 19 '13 at 17:27
  • $\begingroup$ Will: Yes this is what I had in mind. After asking around my department around my department some more, I think I understand it better: step 1 is to consider the case when $f$ is finite and flat, but $X$ and $Y$ are not necessarily proper: in that case, the proper pushfoward on cohomology is induced by the trace map $f_* f^*F \rightarrow F$, and I think I can show that when $F = \mu_n$, this is given by the norm function as you say. When $f$ has fibers of dimension $\geq 1$, the map must be 0. The last case is when $f$ is generically finite, which one does by shrinking on $X$ and $Y$. $\endgroup$ – Peter Mar 19 '13 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.