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?


2 Answers 2



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.

  • $\begingroup$ 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. $\endgroup$ Nov 20, 2010 at 23:00
  • $\begingroup$ 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. $\endgroup$ Nov 20, 2010 at 23:37
  • $\begingroup$ 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. $\endgroup$ Nov 21, 2010 at 0:19
  • $\begingroup$ An observation on elliptic curves over finite fields (that fails over infinite ones): all points are torsion all curves have complex multiplication. $\endgroup$ 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.

Disclaimer: I just learned about this today, so I might be missing something.

  • $\begingroup$ If I recall correctly, there is also a "twist" of the Deligne splitting which is defined over $\mathbb{R}$. This is the content of Proposition 2.20 of Cattani--Kaplan--Schmid's Degeneration of Hodge Structures, if my old notes are to be trusted. This splitting is also functorial (and compatible with tensor products). $\endgroup$ Nov 30, 2022 at 18:14

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