# Does there exist a functorial splitting for the weight filtration (of singular cohomology)?

There are plenty of examples of varieties whose singular cohomology with rational coefficients considered as a mixed Hodge structure does not decompose as the direct sum of its pure (weight) factors. Yet if we consider the cohomology just as filtered vector spaces over rationals, such a decomposition certainly exist (for any variety).

My question is: could there exist a functorial decomposition like this (say, for the singular cohomology as a functor from the category of all smooth complex varieties), or does there exist some obstruction for such a functorial splitting?

Mikhail,

This is an interesting question. But I think that the answer is no, there would be no functorial splitting of the weight filtration as filtered vector spaces.

This needs a bit of work perhaps, but here is my example. Let $E$ be an elliptic curve. Let the $\sigma$ the involution given by $-1$ in the group law. Choose a non $2$-torsion point $p$. Then $q=\sigma(p)\not= p$. By duality, we may work with homology. Choose small loop $\gamma_p$ about $p$ and let $\gamma_q$ be its image under $\sigma$. Note that the class $[\gamma_p+\gamma_q]=0$ in $H_1(E-\{p,q\})$. Then there is an exact sequence $$\mathbb{Q}^2\to H_1(E-\{p,q\})\to H_1(E)\to 0$$ where the first map sends $(a,b)$ to $a[\gamma_p]+b[\gamma_q]$. A splitting would send $H_1(E-\{p,q\})$ to $W_0=span(\gamma_p)$, or to the anti-invariant part of $\mathbb{Q}^2$ under $\sigma$. However, functoriallity should imply that the splitting ought to be invariant.

Added This example is a bit fishy as it stands (see comments) but I think the basic strategy should work. I'll try to fix it in the morning.

• I am sorry; could you explain the second to last statement? This seems a bit strange: similar arguments should work for an elliptic curve over a finite field, yet in this case there exists a canonical splitting for the filtration. Nov 20, 2010 at 23:00
• OK, I see. I realize that I was probably hasty in the way I set up my example, since Hodge theoretic (and presumably Galois theoretic) extension class corresponding to $$0\to {\mathbb Q}(1)\to H_1(E-\{p,q\})\to H_1(E)\to 0$$ vanishes because p and q are linearly equivalent. Let me try again with E any curve with involution such that $E/\sigma$ has positive genus. Although I should probably work it out carefully. By the way, it may be more efficient to communicate by email. It would be nice to get this example sorted out. Nov 20, 2010 at 23:37
• Dear Donu, I will certainly be very glad to get an e-mail from you, and will write you if I will have anything big on the subject. Nov 21, 2010 at 0:19
• An observation on elliptic curves over finite fields (that fails over infinite ones): all points are torsion all curves have complex multiplication. Nov 21, 2010 at 0:24

There is the Deligne splitting. I take this from Peters and Steenbrink's book, Section 3.1.

For a complex variety $X$, we define $I^{p,q} \subseteq H^{\ast}(X, \mathbb{C})$ by $$I^{p,q} := F^p \cap W_{p+q} \cap \left( \overline{F}^q \cap W_{p+q} + \sum_{j \geq 2} \overline{F^{q-j+1}} \cap W_{p+q-j} \right).$$

Then $H^{\ast}(X, \mathbb{C}) = \bigoplus I^{p,q}$ and $W_k \otimes \mathbb{C} = \bigoplus_{p+q \leq k} I^{p,q}$ and $F^p = \bigoplus_{r \geq p} \bigoplus_s I^{r,s}$.

In particular, defining $U_k = \bigoplus_{p+q=k} I^{p,q}$ gives a splitting of the weight filtration tensored with $\mathbb{C}$. If I am not mistaken, it is functorial.

The Deligne splitting only exists with $\mathbb{C}$ coefficients. I think Donu Arapura's answer here is fairly convincing that there is no splitting with $\mathbb{Q}$ or $\mathbb{R}$ coefficients.