Does there exist a functorial splitting for the weight filtration (of singular cohomology)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:45:41Z http://mathoverflow.net/feeds/question/46751 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46751/does-there-exist-a-functorial-splitting-for-the-weight-filtration-of-singular-co Does there exist a functorial splitting for the weight filtration (of singular cohomology)? Mikhail Bondarko 2010-11-20T17:13:12Z 2013-04-23T01:42:52Z <p>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). </p> <p>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?</p> http://mathoverflow.net/questions/46751/does-there-exist-a-functorial-splitting-for-the-weight-filtration-of-singular-co/46760#46760 Answer by Donu Arapura for Does there exist a functorial splitting for the weight filtration (of singular cohomology)? Donu Arapura 2010-11-20T19:02:41Z 2010-11-20T23:44:13Z <p>Mikhail,</p> <p>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. </p> <p>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 <code>$H_1(E-\{p,q\})$</code>. Then there is an exact sequence <code>$$\mathbb{Q}^2\to H_1(E-\{p,q\})\to H_1(E)\to 0$$</code> where the first map sends $(a,b)$ to $a[\gamma_p]+b[\gamma_q]$. A splitting would send <code>$H_1(E-\{p,q\})$</code> 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.</p> <p><strong>Added</strong> 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. </p> http://mathoverflow.net/questions/46751/does-there-exist-a-functorial-splitting-for-the-weight-filtration-of-singular-co/128422#128422 Answer by David Speyer for Does there exist a functorial splitting for the weight filtration (of singular cohomology)? David Speyer 2013-04-23T01:42:52Z 2013-04-23T01:42:52Z <p>There is the Deligne splitting. I take this from Peters and Steenbrink's book, Section 3.1.</p> <p>For a complex variety $X$, we define $I^{p,q} \subseteq H^{\ast}(X, \mathbb{C})$ by <code>$$I^{p,q} := F^p \cap W_{p+q} \cap \left( \overline{F}^q \cap W_{p+q} + \sum_{j \geq 2} \overline{F^{q-j+1}} \cap W_{p+q-j} \right).$$</code></p> <p>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 $F^p = \bigoplus_{r \geq p} \bigoplus_s I^{r,s}$.</p> <p>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.</p> <p>The Deligne splitting only exists with $\mathbb{C}$ coefficients. I think Donu Arapura's answer <a href="http://mathoverflow.net/questions/117432/" rel="nofollow">here</a> is fairly convincing that there is no splitting with $\mathbb{Q}$ or $\mathbb{R}$ coefficients. </p> <p>Disclaimer: I just learned about this today, so I might be missing something.</p>