12
$\begingroup$

This is a question about Faltings' $p$-adic Eichler-Shimura isomorphism from his 1987 article "Hodge-Tate structures and Modular Forms".

Let $N\ge5$, $k\ge2$ be integers. Denote by $X(N)$ the proper modular curve of full level $N$ (over $\mathbb Q$, say), $f\colon\overline E(N)\rightarrow X(N)$ the universal generalized elliptic curve over it and by $e\colon X(N)\rightarrow E(N)$ the unit section. Further let $\omega=e^*\Omega^1_{E(N)/X(N)}$ and $\Omega^1=\Omega^1_{X(N)}$. Let $Y(N)$ be the open modular curve and denote the universal elliptic curve over it still by $f$. Put $V=H^1_{\mathrm{p}}(Y(N)\times\overline{\mathbb Q},\operatorname{Sym}^{k-2}R^1f_*\mathbb Z_p)$, $W_0 = H^0(X(N),\omega^{k-2}\otimes\Omega^1)$ and $W_{k-1}=H^1(X(N),\omega^{2-k})$.

Faltings' $p$-adic Eichler-Shimura isomorphism, which is Thm. 6 (iii) in his paper, states that canonically $$ \mathbb C_p\otimes V(1)\cong \mathbb C_p\otimes W_0 \oplus \mathbb C_p(k-1)\otimes W_{k-1}. $$

It seems to be well-known that this Eichler-Shimura isomorphism "is" the $p$-adic comparison isomorphism for the modular motive ${}^N_kW$ introduced by Scholl in his 1990 paper "Motives for Modular Forms". In trying to make this statement precise, I obtained the following.

The motive introduced by Scholl which he calls ${}^N_kW$ has $V$ as its $p$-adic étale realization and $W:=W_0\oplus W_{k-1}$ as its Hodge realization, with the $W_i$ sitting in degree $i$, respectively. The comparison isomorphism from $p$-adic Hodge theory gives $$ B_{\mathrm{HT}}\otimes V \cong B_{\mathrm{HT}}\otimes W. $$ Taking the degree $0$ part of this one obtains an isomorphism $$ \mathbb C_p\otimes V \cong \mathbb C_p\otimes W_0 \oplus \mathbb C_p(1-k)\otimes W_{k-1}, $$ which differs from Faltings' isomorphism above.

What is the precise relation between Faltings' Eichler-Shimura isomorphism and the general comparison isomorphism? Since I am not familiar with the techniques used by Faltings, I was not able to prove anything about this by myself.

$\endgroup$
  • $\begingroup$ I added a top-level tag. These are those with two-letter prefix, corresponding to arXiv categories. Each question should ideally have (at least) one top-level tag. I hope I picked the right one. If not you can change it easily via an edit. $\endgroup$ – user9072 Feb 16 '16 at 19:32
6
$\begingroup$

The issue seems to be about notation. Your $k$ is what Faltings calls $k + 2$. And what Faltings calls $\underline{V}_k$ is what you would call $V(k-1)$. (Possibly you confused Faltings' $\underline{V}_k$ with his $V_k$?) So when I write out Faltings' comparison isomorphism and then twist, I get (using your notation) \begin{equation} \mathbb{C}_p \otimes V \simeq \mathbb{C}_p \otimes W_{k-1} \oplus \mathbb{C}_p(1-k) \otimes W_0, \end{equation} as one wants.

Faltings' modular forms paper seems to be a prelude (a special case even) of his proof of the general comparison isomorphism which you refer to. So it would be strange if the two comparison maps were not literally the same map.

$\endgroup$
  • $\begingroup$ Thanks for your answer! However, I am still confused. I indeed got confused about $\underline V_k$, but if I just copy Faltings' result and write it in my notation, then I get $\mathbb C_p\otimes V(k-1)\cong \mathbb C_p(k-1)\otimes W_{k-1}\oplus\mathbb C_p\otimes W_0$. So if I twist this $1-k$ times, I obtain $\mathbb C_p\otimes V\cong\mathbb C_p\otimes W_{k-1}\oplus\mathbb C_p(1-k)\otimes W_0$, so the $W_i$ are interchanged. $\endgroup$ – Michael Fütterer Feb 18 '16 at 15:56
  • $\begingroup$ Yeah, I think that's because $W_{k-1}$ is actually in degree 0, and $W_0$ is in degree $k-1$. Think about the weight 2 case. Then $W_1 = H^1(X(N), \mathcal{O}) = F^0/F^1$ is in degree 0, not degree 1. $\endgroup$ – Ari Shnidman Feb 18 '16 at 16:18
  • $\begingroup$ I edited my answer to include the $W_i$'s. $\endgroup$ – Ari Shnidman Feb 18 '16 at 16:24
  • $\begingroup$ Of course, you are right about the degrees. Thank you! $\endgroup$ – Michael Fütterer Feb 18 '16 at 16:27

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.