Let $p$ be a prime, $\Gamma=1+p\mathbb Z_p$ and $\Lambda=\mathbb Z_p[[\Gamma]]$ the Iwasawa algebra. Let $\kappa\colon\Gamma\rightarrow\mathbb Z_p^\times$ be the inclusion. For a character $\varepsilon\colon\Gamma\rightarrow\mathbb Z_p^\times$ and $k\in \mathbb N$, let $x_{k,\varepsilon}$ be the morphism $\Lambda\rightarrow\mathbb Z_p$ induced by $\varepsilon\kappa^k$ and let $P_{k,\varepsilon}$ be its kernel.

This question is about Kitagawa's $\Lambda$-adic modular symbols. To recall the most important notation from his article, let $N\in\mathbb{N}$ with $(N,p)=1$, $\mathrm{MS}(\Gamma_1(Np^r),L_{k-2}(\mathbb Z_p))$ the $\mathbb Z_p$-module of classical modular symbols of weight $k$ and level $Np^r$ and $\overline{\mathcal{MS}}({k-2},\mathbb Z_p)$ be the $p$-adic completion of $\varinjlim_r\ \mathrm{MS}(\Gamma_1(Np^r),L_{k-2}(\mathbb Z_p))$. The latter space has an action of the Hida-Hecke-Algebra of level $Np^\infty$, which I denote simply by $\mathcal H$.

Define further $UM(\mathbb Z_p)=\operatorname{Hom}_{\mathbb Z_p}(\overline{\mathcal{MS}}(0,\mathbb Z_p),\mathbb Z_p)$ with the dual Hecke action and $MS^{\mathrm{ord}}(\Lambda)=\operatorname{Hom}_\Lambda(UM^{\mathrm{ord}}(\mathbb Z_p),\Lambda)$ again with the dual Hecke action, following Kitagawa. Fix a character $\varepsilon\colon\Gamma\rightarrow\mathbb Z_p^\times$ with kernel $1+p^r\mathbb Z_p$ and $k\ge2$. Let $F\colon \mathcal H^{\mathrm{ord}}\rightarrow\Lambda$ be a $\Lambda$-linear morphism and $f_{k,\varepsilon}$ its reduction mod $P_{k,\varepsilon}$ which corresponds to a classical cusp form which we denote by the same symbol. I read that the following control statement holds: $$ MS^{\mathrm{ord}}(\Lambda)[F]\otimes_\Lambda(\Lambda/P_{k,\varepsilon})\cong \mathrm{MS}^{\mathrm{ord}}(\Gamma_1(Np^r),L_{k-2}(\mathbb Z_p))[f_{k,\varepsilon}]. $$ Here $[F]$ resp. $[f_{k,\varepsilon}]$ denotes the subspace where $\mathcal H^{\mathrm{ord}}$ resp. the Hecke algebra of level $Np^r$ and level $k$ act by the eigenvalues given by $F$ resp. $f_{k,\varepsilon}$. I tried to derive this from the statements proved in Kitagawa's article, but I did not succeed. I sketch below my attempt to prove this and my question is: what am I doing wrong?

Start with the exact sequence $0\rightarrow P_{k,\varepsilon}\rightarrow\Lambda\rightarrow\mathbb Z_p\rightarrow0$, where the right map is $x_{k,\varepsilon}$. Consider $\mathbb Z_p$ as a $\Lambda$-algebra via $x_{k,\varepsilon}$ and apply the functor $\operatorname{Hom}_\Lambda(UM^{\mathrm{ord}}(\mathbb Z_p),-)$, which is exact since $UM^{\mathrm{ord}}(\mathbb Z_p)$ is $\Lambda$-free (Prop. 5.7 in Kitagawa). In this way one easily derives isomorphisms $MS^{\mathrm{ord}}(\Lambda)\otimes_\Lambda(\Lambda/P_{k,\varepsilon})\cong\operatorname{Hom}_\Lambda(UM^{\mathrm{ord}}(\mathbb Z_p),\mathbb Z_p)$ and $MS^{\mathrm{ord}}(\Lambda)[F]\otimes_\Lambda(\Lambda/P_{k,\varepsilon})\cong\operatorname{Hom}_\Lambda(UM^{\mathrm{ord}}(\mathbb Z_p),\mathbb Z_p)[F]$.

Next, let $\Gamma$ act on $\overline{\mathcal{MS}}(0,\mathbb Z_p)$ by diamond operators, as in Kitagawa (note that this is not the same as the action through the Hecke algebra $\mathcal H$, which is twisted by a power of $\kappa$). Write $[\varepsilon]$ for the subspace where $\Gamma$ acts by $\varepsilon$. Using the biduality map $\overline{\mathcal{MS}}(0,\mathbb Z_p)\rightarrow\operatorname{Hom}_{\mathbb Z_p}(UM(\mathbb Z_p),\mathbb Z_p)$ one obtains an isomorphism $$\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[\varepsilon\kappa^{k-2}]\cong\operatorname{Hom}_\Lambda(UM^{\mathrm{ord}}(\mathbb Z_p),\mathbb Z_p)$$ what can be derived from the definitions of the various actions. Further one checks that $\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[f_{k,\varepsilon}]\subseteq\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[\varepsilon\kappa^{k-2}]$ (where $[f_{k,\varepsilon}]$ here means the subspace where $\mathcal H$ acts by $x_{k,\varepsilon}\circ F$) and that the restriction of the above isomorphism induces $$\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[f_{k,\varepsilon}]\cong\operatorname{Hom}_\Lambda(UM^{\mathrm{ord}}(\mathbb Z_p),\mathbb Z_p)[F].$$

On the other hand, starting from finite level, we have $\mathrm{MS}^{\mathrm{ord}}(\Gamma_1(Np^r),L_{k-2}(\mathbb Z_p))[f_{k,\varepsilon}]\subseteq\mathrm{MS}^{\mathrm{ord}}(\Gamma_1(Np^r),L_{k-2}(\mathbb Z_p))[\varepsilon]$ and by Thm. 5.5 in Kitagawa the latter space equals $\overline{\mathcal{MS}}^{\mathrm{ord}}(k-2,\mathbb Z_p)[\varepsilon]$. Now by Thm. 5.3 in Kitagawa, we have an isomorphism $ \overline{\mathcal{MS}}^{\mathrm{ord}}(k-2,\mathbb Z_p)\cong\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)(\kappa^{k-2})$, where $(\kappa^{k-2})$ means that the action of $\Gamma$ has been twisted by $\kappa^{k-2}$. This means $\overline{\mathcal{MS}}^{\mathrm{ord}}(k-2,\mathbb Z_p)[\varepsilon]\cong\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[\varepsilon\kappa^{2-k}]$.

Now $\mathrm{MS}^{\mathrm{ord}}(\Gamma_1(Np^r),L_{k-2}(\mathbb Z_p))[f_{k,\varepsilon}]\subseteq\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[\varepsilon\kappa^{2-k}]$, but $MS^{\mathrm{ord}}(\Lambda)\otimes_\Lambda(\Lambda/P_{k,\varepsilon})\cong\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[\varepsilon\kappa^{k-2}]$ and I cannot conclude.

EDIT: From what I wrote above, it follows that $MS^{\mathrm{ord}}(\Lambda)[F]\otimes_\Lambda(\Lambda/P_{k,\varepsilon})\subseteq\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[\varepsilon\kappa^{k-2}]$, but $\mathrm{MS}^{\mathrm{ord}}(\Gamma_1(Np^r),L_{k-2}(\mathbb Z_p))[f_{k,\varepsilon}]\subseteq\overline{\mathcal{MS}}^{\mathrm{ord}}(0,\mathbb Z_p)[\varepsilon\kappa^{2-k}]$. Note the different exponents of the character $\kappa$ ($k-2$ and $2-k$). These spaces in general intersect trivially (unless $k=2$), so if the statement holds in some situation, I must be mistaken somewhere.


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.

Browse other questions tagged or ask your own question.