Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am trying to understand the proof of lemma 3 in the paper "Algebraic cycles and the Hodge structure of a Kuga fiber variety" by B. Brent Gordon, available at The main reason I'm interested in understanding the proof is that if I manage to generalize it I will be able to simplify the proof of one of the theorems in a paper I'm writing.

Let me first recall some of the notation used in that paper. The author sets $M$ to be the quotient of the upper half plane by $$\Gamma(N)=\ker (SL_2(\mathbb{Z})\to SL_2(\mathbb{Z}/n\mathbb{Z})),$$ then he sets $A^1$ to be the universal elliptic curve over $M$ and denotes the $k$-th fibered power of $A^1$ over $M$ as $A$. $M$ has a natural smooth compactification denoted as $\tilde M$. It is obtained by adding the cusp points; these correspond to the orbits of the action of $\Gamma(N)$ on $\mathbb{Q}$. The author takes an arbitrary smooth compactification $\tilde A$ of $A$ over $\tilde M$. The main object of study in the paper is the rational cohomology of $A$ and $\tilde A$.

The statement of Lemma 3 on p. 944 is

There is a natural realization of $\tilde H^{\langle 0,*\rangle}(A, \mathbb{Q})$ in $H^*(\tilde A,\mathbb{Q})$ which is generated by the classes of divisors on $\tilde A$ which are the closures in $\tilde A$ of divisors on $A_\eta$.

(Here $\tilde H^{\langle 0,b\rangle}(A, \mathbb{Q})$ denotes the $n^b$-eigenspace of the image of $H^b(\tilde A,\mathbb{Q})$ in $H^b(A,\mathbb{Q})$ under the map $H^b(A,\mathbb{Q})\to H^b(A,\mathbb{Q})$ induced by the fiberwise multiplication by some $n>1$.)

The proof goes as follows. Let $\eta\in M$ be a point that corresponds to an elliptic curve $E$ without complex multiplication and let $A_\eta$ be $p^{-1}(\eta)$ where $p:A\to M$ is the projection. Note that $A_\eta=E^k$. First the author identifies $\tilde H^{\langle 0,*\rangle}(A, \mathbb{Q})$ with $AH^*(A_\eta,\mathbb{Q})$, which is the subalgebra of $H^*(A_\eta,\mathbb{Q})$ generated by the Poincare duals of all the diagonals and of the divisors of the form $$E\times\cdots\times E\times\{pt\}\times E\times\cdots\times E\subset E^k.$$ Moreover, is has been shown previously in the paper that these divisors extend to divisors in $A$.

So far the argument is clear to me. What confuses me is the next step, which I interpret as follows: We take an element $x\in AH^*(A_\eta,\mathbb{Q})$ which we want to extend to $\tilde A$ and represent it as a linear combination of cycles: $x=\sum a_i c_i$ where $a_i\in\mathbb{Q}$ and $c_i$'s are subvarieties obtained by intersecting the above divisors. We then extend every $c_i$ to a cycle in $A$ and take its closure in $\tilde A$, which we denote as $\tilde c_i$. We then extend $x$ as $\sum a_i \tilde c_i$.

The problem is, the closure of a cycle which is homologous to 0 is not necessarily homologous to 0: take e.g. the closure of a line $\subset \mathbb{A}^2$ in $\mathbb{P}^2$. So my first question is

Question 1: Is there a reason that I'm missing why this procedure would take homologically trivial cycles to homologically trivial cycles?

Of course, one does not really need this to get a lifting: one can simply choose a basis in each $AH^b(A_\eta,\mathbb{Q})$ and lift it element-wise. But this would hardly qualify as "natural" as the result may depend on the choice of the basis.

Another way to lift $AH^*(A_\eta,\mathbb{Q})$ to $H^*(A,\mathbb{Q})$ is as follows. $H^b(\tilde A,\mathbb{Q})$ is a polarized pure Hodge structure and, as shown previously in the paper, $AH^b(A_\eta,\mathbb{Q})$ can be identified with the subspace of $H^b(A,\mathbb{Q})$ that consists of the classes that have the "right" weight, which is the image of $H^b(\tilde A,\mathbb{Q})$ in $H^b(A,\mathbb{Q})$. So one can lift $AH^b(A_\eta,\mathbb{Q})$ as the orthogonal complement of $\ker (H^b(\tilde A,\mathbb{Q})\to H^b(A,\mathbb{Q}))$. So my second question is

Question 2: Is there a reason that I'm missing why this procedure coinsides with the one described above?
share|cite|improve this question
Cat -- for some reason the MO Tex interpreter doesn't like asterisks and curly brackets. If you have a formula containing any of those you should put backticks (`) around it. Sounds like black magic, I know, but at least it works. –  algori Nov 28 '12 at 0:22
.. can't answer any of the questions though. –  algori Nov 28 '12 at 0:48
This feels like something I should be able to answer, but I will not have time to look at it until this weekend. However, here is a small remark: your proposed construction in Q2 doesn't seem to work. The short exact sequence $$ 0 \to ker h^\ast \to H^b (\tilde A) \to \tilde H^b (A) \to 0$$ splits, because of the semisimplicity of the category of pure $\mathbf Q$-Hodge structures, but it does not split in a natural way. Taking the orthogonal complement of a subspace is only well defined in the presence of an inner product. So your proposed construction will not give a "natural realization". –  Dan Petersen Nov 28 '12 at 12:52
@ Dan: I think there is no harm in assuming $\tilde A$ projective. If so, then the cohomology of $\tilde A$ becomes polarized, and one can take the orthogonal complement with respect to the polarization. –  A Confused Cat Nov 29 '12 at 0:52
@ algori: thank you! –  A Confused Cat Nov 29 '12 at 0:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.