First I will introduce some notation and definitions.

Fix a level $N$ (take $N=1$ if it makes things easier) and a prime $p$. Let $k$ be a finite field of characteristic $p$ and let $\mathcal{C}$ be the category of complete noetherian local rings $(R, \mathfrak{m}_R)$ together with an isomorphism $R/\mathfrak{m}_R \cong k$. Let $\Lambda$ be an artinian object of $\mathcal{C}$ (I'm interested in cases where $\Lambda = \mathcal{O}_K / \pi^n$ where $K$ is a finite extension of $\mathbb{Q}_p$ with uniformiser $\pi$). Let $S_k(\mathbb{Z}_p)$ denote the set of $q$-expansions of cuspforms of level $N$ and coefficients in $\mathbb{Z}_p$, and for a $\mathbb{Z}_p$-algebra $R$ let $S_k(R) := S_k(\mathbb{Z}_p)\otimes_{\mathbb{Z}_p}R$. Let $S(R) = \sum_k S_k(R)$, and let $\mathbb{T}(R)$ denote the $R$-subalgebra of $\mbox{End}(S(R))$ generated by the Hecke operators $T_n$ for all $n$ such that $(n,p)=1$.

Now if I understand things correctly, it follows from deformation theory of Galois (pseudo-)representations that $\mathbb{T}(\Lambda)$ is a complete noetherian semilocal ring. The maximal ideals correspond to mod $p$ modular representations. Let $\mathfrak{m}$ be a maximal ideal.

My question is the following: can we under some conditions assert that $\mathbb{T}(\Lambda)_\mathfrak{m}$ is smooth or $\mathfrak{m}$-smooth as a $\Lambda$-algebra for all $\Lambda$, or at least for all the rings $\Lambda$ I'm interested in (described above) ?

What I know:
Please feel free to correct me if I'm mixing things up in the following.

Let $\mathbb{T} = \mathbb{T}(W(k))$. Let $\mathcal{R}$ be the universal deformation ring attached to the residual representation corresponding to $\mathfrak{m}$. If the deformation problem is unobstructed, then I know that $\mathcal{R}$ will be a power series ring and therefore smooth. Moreover the restriction of the deformation functor to the category of $\Lambda$-algebras is represented by $R_\Lambda = \mathcal{R}\mathbin{\hat\otimes} \Lambda$ and is therefore smooth again. Generally we have a surjective homomorphism $\mathcal{R}\twoheadrightarrow\mathbb{T}$ but under some conditions we have $\mathcal{R} \cong\mathbb{T}$.

However it seems to me that even if we have $\mathcal{R} \cong\mathbb{T}$, we might not have that $\mathcal{R}_\Lambda \cong \mathbb{T}(\Lambda)$, where $\mathbb{T}_\Lambda$ is as defined above. For an example there is the following paper by Bellaiche and Khare http://people.brandeis.edu/~jbellaic/preprint/Heckealgebra4.pdf

In the cases they consider, $\mathbb{T}(k)_\mathfrak{m}$ (which they call $A_{\overline{\rho}}$) is not the reduction of $\mathbb{T}$ modulo $p$, but a quotient of that modulo some element in the maximal ideal. Actually, under some conditions (e.g. the deformation problem is unobstructed) they identity $\mathbb{T}(k)_\mathfrak{m}$ with the universal ring of pseudo-deformations to $k$-algebras with constant determinant. In these cases, we still can deduced that $\mathbb{T}(k)_\mathfrak{m}$ is smooth.

I would appreciate any hints or references on this question. Thanks.


1 Answer 1


It follows from deformation theory of Galois (pseudo-)representations that $\mathbb T(\Lambda)$ is a complete noetherian semilocal ring. The maximal ideals correspond to mod $p$ modular representations.

The fact that $\mathbb T(\Lambda)$ is a semi-local ring follows from the existence of the natural surjective map $$\mathbb T(\mathcal O_K)/\pi^{n}\longrightarrow \mathbb T(\Lambda)$$ and the fact that system of Hecke eigenvalues (otherwise known as eigenforms) are in finite numbers (not a priori from arguments with pseudo-representations).

Moreover, for stupid reasons of compatibility of coefficients, the maximal ideals of $\mathbb T(\Lambda)$ cannot correspond to mod $p$ modular representations. Now maybe you meant to write "correspond to modular representations with coefficients in $\Lambda$" however, as in your setting the coefficient ring is artinian but not a field, it is not obviously true that such a statement hold: maybe the pseudo-representation $t:G_{\mathbb Q}\longrightarrow \mathbb T(\Lambda)_{\mathfrak m}$ sending $\operatorname{Fr}(\ell)$ to $T(\ell)$ does not lift to an actual representation $\rho_{\mathfrak m}$ with coefficients in $\Lambda_{\mathfrak m}$. Conversely, a single residual representation $\bar{\rho}$ can correspond to several congruent pseudo-representations with coefficients in $\mathbb T(\Lambda)$.

Regarding your actual question, I think the only known line of attack of such problems at present is the one you describe, that is to say the one of Nicolas-Serre-Bellaïche-Khare: either find an explicit description (totally hopeless in your case) or assume that $R(\bar{\rho})=\mathbb T_{\bar{\rho}}$ is smooth and try to relate $\mathbb T_{\bar{\rho}}\otimes_{\mathcal O}\mathcal O/\pi^n$ to $\mathbb T(\Lambda)_{\mathfrak m}$. As I indicated above, this seems to require much clarification as $\mathfrak m$ and $\bar{\rho}$ are not in obvious correspondence anymore (and in particular, I would certainly not bet that unobstructed implies smooth in your setting; I don't see how to rule out the possibility of many congruences between pseudo-representations destroying the smoothness when one passe from $\mathbb T(\mathcal O/\pi)$ to $\mathbb T(\mathcal O/\pi^n)$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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