Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The analogy on the front page of this paper by Bloch and Kriz seems like it's going to be lovely, but I don't get it, because I don't know how to view a torsor for $\mathbb{Q}(1)$ as an extension of mixed Hodge structures.

Here's what I understand. I can view the Riemann surface of the logarithm as a principal $2\pi i\mathbb{Z}$-bundle over the punctured plane. In so doing I get an extension of $\pi_1(\mathbb{C}^\times)$-modules $$ 0 \to M \to E \to \mathbb{Z}\to 0 $$ where $M$ is the group of sections of the pullback of this bundle to the universal cover. Since the pullback bundle trivializes, $M$ is canonically isomorphic to $2 \pi i\mathbb{Z}$. (I get this extension because I get a group cocycle $\pi_1(\mathbb{C}^\times) \to M$ by sending the generator $\gamma$ to $2 \pi i$.)

This is an extension of $\mathbb{Z}[\pi_1(\mathbb{C}^\times)]$-modules, but to put a Hodge structure on $M$ and on $E$, I want an action of $\mathbb{C}^\times$ itself, not its fundamental group.

Finally, I know that $2\pi i \mathbb{Q}$ "is" $\mathbb{Q}(1)$, but I don't actually know this. What I know is that on the one hand the $2n$-th cohomology of an $n$-dimensional smooth projective complex variety is a one-dimensional Hodge structure of weight $2n$ that we call $\mathbb{Q}(-n)$ and that one thinks of this as $\frac{1}{(2 \pi i)^n}\mathbb{Q}$. It seems like magic that just because the discrepancy between branches of the logarithm is $2 \pi i$, we get to say that a Hodge structure of weight $-2$ appears.

share|cite|improve this question

Your Answer


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