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.