Let $\pi :C\rightarrow \mathbb{P}^{1}$ be a cyclic cover of degree $m$ of $\mathbb{P}^{1}$. So $C$ has an action of $\mathbb{Z}/m\mathbb{Z}$. Let $\xi$ be a primitive $m$th root of unity. Consider the cohomology group $V_{\mathbb{Q}}=H^{1}(C,\mathbb{Q})$. In the book "Cylic covers, CalabiYau manifolds and complex multiplication" (by Jan Christian Rohde) page 80, we read: "In general there is no $\mathbb{Q}(\xi)$structure on $H^{1}(C,\mathbb{Q})$ which turnes $H^{1}(C,\mathbb{Q})$ into a $\mathbb{Q}(\xi)$vector space " and then he gives a direct sum decomposition with different $\mathbb{Q}(\xi^{r})$structures. But in many books and articles on this subject it is just claimed that $V_{\mathbb{Q}}$ has an induced action of $\mathbb{Q}(\xi)$(Like: de Jong and Noot : "Jacobians with complex multiplication"). Which one is true? However, my second question is: even if there is an action of $\mathbb{Q}(\xi)$, I think we then must have two different $\mathbb{Q}$vector space structures on $V_{\mathbb{Q}}$. The first one is the usual one and one coming from $\mathbb{Q}\hookrightarrow \mathbb{Q}(\xi)$. Since if $dim_{\mathbb{Q}}V_{\mathbb{Q}}=g$, then by the induced action from $\mathbb{Q}\hookrightarrow \mathbb{Q}(\xi)$, it's dimension should be $dim_{\mathbb{Q}}V_{\mathbb{Q}}=m dimV_{\mathbb{Q}(\xi)}$ ($V_{\mathbb{Q}(\xi)}$ means $V_{\mathbb{Q}}$ as a $\mathbb{Q}(\xi)$vector space ). But this two in general do not agree ($2g=g (C)$ and in general is not even divisible by $m$). So the $\mathbb{Q}$structures should be different. Is this true? Can one give a good explanation of this?
I think you will find that the space is a $\mathbb Q[x]/(x^m1)$, which is a product of fields, including $\mathbb Q(\xi)$, where $\xi$ is a primitive $m$th root of unit. Thus it decomposes as a sum of vector spaces over different fields. There is no reason that $\mathbb Q(\xi)$ should be the only field, and thus no reason why $g$ should be a multiple of $\phi(m)$  and indeed, we can easily construct examples where it is not. I hope the examples you found used square brackets instead of round ones!

$\begingroup$ Thanks but I did not get your meaning by the last sentence! As far as I know: $\mathbb{Q}(\xi)=\mathbb{Q}[\xi]$. And also, $\mathbb{Q}[x]/{x^{m}}$ is just an $m$times product of $\mathbb{Q}$. $\endgroup$ – Darius Math Sep 6 '13 at 16:14

$\begingroup$ $Q(\xi)$ is the field extenction generated by a primitive $m$th root of unity. $\mathbb Q[x]/(x^m1)$ is not a field, since it has zero divisors: $(x1)(\sum_{k=1}^{m1} x^k)=0$. The first is a $\phi(m)$dimensional vector space over $\mathbb Q$, and the second is $m$dimensional. $\endgroup$ – Will Sawin Sep 6 '13 at 16:21

$\begingroup$ Of course I did not claim $\mathbb{Q}[x]/(x^{m}1)$ is a field, but it is not equal to $\mathbb{Q}[\xi]$, my meaning was that $\mathbb{Q}[\xi]=\mathbb{Q}[x]/ \Phi_{m1}(x)=\mathbb{Q}(\xi)$ is a field. And it is normally said that there is an action of this on the cohomology. But now this should give (at least) two $\mathbb{Q}$vector space structures. $\endgroup$ – Darius Math Sep 6 '13 at 16:27

$\begingroup$ There is an action of that on the cohomology when $m$ is prime, or in certain other special cases. I think the references you see are either in those cases, or using notation where $\mathbb Q[\xi]\neq \mathbb Q(\xi)$. But the second vector space structure does not exist in general. $\endgroup$ – Will Sawin Sep 6 '13 at 17:20

$\begingroup$ When you write $\mathbb{Q}[x]/x^m$, do you really want $x$ to be nilpotent? $\endgroup$ – S. Carnahan♦ Sep 6 '13 at 18:41
I think I just found my answer. The Jacobians in the family are equipped with an action of the group ring $\mathbb{Z}[\xi_{m}]$. So the action on $V_{\mathbb{Q}}$ is that of the group ring $\mathbb{Q}[\xi_{m}]=\prod_{dm} K_{d}$. So it is a module over this group ring and not just a vector space.