Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a complex algebraic variety. The rational cohomology of $X(\mathbb{C})$ carries a canonical filtration called the weight filtration. It also carries a canonical equivalence class of $A_\infty$ structures (extending the cup product ring structure).

How do these two structures interact? Is the weight of a higher multiplication $m_n(x_1, \cdots, x_n)$ determined by the weights of the $x_i$? If the answer is no in general, then can one choose a particular nice $A_\infty$ structure within the equivalence class for which the answer is yes?

share|improve this question
    
A relevant question is: mathoverflow.net/questions/22064/… –  Dan Petersen Apr 24 '13 at 14:07
    
Can you say what you mean by equivalence class of $A_{\infty}$-structures? Do you mean on cochains? You seem to have already passed to cohomology for the weight filtration, in which case you just have $m_{1}$, right? Or do you consider some kind of filtration on a model for cochains? –  dhagbert Apr 24 '13 at 14:53
    
Jan, even without knowing that smooth projective implies formal, smooth projective implies immediately that weight equals degree. –  Jeffrey Giansiracusa Apr 24 '13 at 17:36
    
Dan, thanks for reminding me about that question - I had forgotten about it. Unfortunately nobody ever gave an answer to that one. –  Jeffrey Giansiracusa Apr 24 '13 at 17:37
    
Yes of course, you are right Jeffrey. –  Jan Weidner Apr 24 '13 at 18:18

2 Answers 2

This was meant to be a comment before it got too long.

One of the ways to define a minimal $A_\infty$ model of a cdga (more generally, an $A_\infty$-algebra) $A$ is Merkulov's recipe (see Merkulov http://arxiv.org/pdf/math/9809172.pdf or Chuang and Lazarev http://arxiv.org/pdf/0802.3507.pdf for a summary). All we need is а decomposition $A=W\oplus K$ into subcomplexes with the differential of $A$ vanishing on $W$ (so $W$ can be identified with $H^*(A)$), $K$ acyclic, and a contracting homotopy $h:K\to K$ satisfying $h^2=0$. This decomposition is called the Hodge decomposition since it becomes one when $A=\mathcal{E}^*(M)$, the algebra of complex-valued differential forms on a K\"ahler manifold $M$, in which case $W$ is the subspace of harmonic forms. Note that $W$ needn't be a subalgebra. Note also to find such a decomposition it suffices to find a splitting $A$ into cocycles and some complement and then splitting off the cohomology inside the cocycles, see Chuang, Lazarev, ibid., p. 5.

So if $A$ is some kind of cochain algebra associated to a complex algebraic variety and we take any mixed Hodge structure on $K$ compatible with $h$ and the differential and such that the multpplication $A\otimes A\to A$ is a map of mixed Hodge structures (the latter condition is probably too strong and may be relaxed in some way). Applying Merkulov's formula we get higher multiplications $m_n$ on $H^*(A)$ which are maps of mixed Hodge structures of degree $0$. These maps will be linear combinations of compositions of maps constructed from maps of the form $(0,h)$, multiplications in $A$ and projections to $W$ along $K$. In particular the weight of $m_n(x_1,\ldots,x_n)$ will be the sum of the weights of $x_1,\ldots, x_n$. Of course there is no reason this $A_\infty$ structure should be compatible with morphisms of varieties. I would guess it is possible to choose the splittings so that it is but this would require some more work.

share|improve this answer
    
The last paragraph doesn't sound right to me - shouldn't the $m_n$ be compatible with weights on the nose? –  Dan Petersen Apr 24 '13 at 21:18
    
Dan -- you are right: multiplication is a map of MHS's of degree 0! –  algori Apr 24 '13 at 21:52

This is more a comment than an answer. There are also details which use machinery I am uncomfortable with. I will list them later, maybe some expert can comment?

Let my try to answer the analogous question in the case of of etale cohomology $H^\bullet(X,\overline{ \mathbb Q_l})$, where $X_0$ is a variety over $\mathbb F_q$ and $X$ its basechange to the algebraic closure. Using comparison results, one can maybe deduce the Hodge case, but I doubt that there is a strong enough result in literature. Or maybe one can proceed in a similar way using mixed Hodge modules, I am not sure either.

First of all we may compute $H^\bullet(X,\overline{ \mathbb Q_l})$ as the cohomology of the push forward $f_* \mathbb Q_l$ of the constant sheaf on $X$.

Now I it is true that the derived category $D^b(pt_0,\overline{ \mathbb Q_l})$ is just the derived category of finite dimensional $\overline{ \mathbb Q_l}$ vectorspaces equipped with a (quasi-unipotent) automorphism. In other words it is (a block of) the category of finite dimensional $ \overline{ \mathbb Q_l}[x,x^{-1}]$ modules.

The latter category has cohomological dimension one, since $\overline{ \mathbb Q_l}[x,x^{-1}]$ is a principal ideal domain. In particular any complex is formal.

This means that we find a quasi-isomorphism of complexes, $H^\bullet(X,\overline{ \mathbb Q_l}) \cong f_* \overline{ \mathbb Q_l}$ which is compatible with the Frobenius action!

Now transfering the $A_\infty$-structure along such a quasi-isomorphism, yields $m_n$ which are compatible with the Frobenius action. In particular they have the expected effect on weights i.e. just add them all up!

By the way observe that this is different to the effect of $m_n$ on cohomological degree (add up, add 2-n)! So if $H^\bullet(X,\overline{ \mathbb Q_l})$ is pure (e.g. $X$ smooth, proper) , it must be formal!

Ok, here are the fishy details I am not sure about:

  • First of all we need a dg-algebra structure on $f_* \overline{ \mathbb Q_l}$. For this one needs a dg-enhancement of the usual six functors or an infinity category version.

  • We want to transfer the dg-structure along $H^\bullet(X,\overline{ \mathbb Q_l}) \cong f_* \overline{ \mathbb Q_l}$. I only know how to transfer a dga structure along a homotopy retraction, I never learned how one does it with an arbitrary quasi-isomorphism or under what conditions it is possible.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.