Let $\rho:\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow \operatorname{GL}\_2(\bar{\mathbb{Z}}\_p)$ be a $p$-adic Galois representation associated to a $p$-ordinary Hecke eigencuspform, let $\bar{\rho}:\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow \operatorname{GL}\_2(\bar{\mathbb{F}}\_p)$ denote the residual representation. The adjoint representation $\operatorname{Ad}\rho$ is the module of $2\times 2$ trace-zero matrices over $\bar{\mathbb{Q}}\_p$ on which $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ acts by conjugation. There is a relationship between the Selmer group over $\mathbb{Q}$ of the (characteristic zero) adjoint representation $\operatorname{Ad}\rho$ and a certain universal Galois deformation ring associated to $\bar{\rho}$. Namely, the dual of the adjoint Selmer group over $\mathbb{Q}$ is isomorphic to the module of $1$-forms on the deformation ring at the point corresponding to $\rho$. In the case when $\rho$ is unramified outside $p$, a proof of this statement is in Hida's book "Hilbert Modular forms and Iwasawa theory" (cf. Proposition 1.47). The relationship generalizes to totally real fields. The relationship here tells us that since the deformation ring comes from a Hecke algebra, then indeed the Selmer group of the Adjoint representation over $\mathbb{Q}$ is finite (cf. Proposition 1.53), and the same statement holds over a totally real field. This requires the hypothesis that $\rho$ is unramified outside $p$. A similar idea was exploited by Khare and Ramakrishna to show that if one starts with a residual Galois representation which is odd and absolutely irreducible then it may be lifted to a $p$-adic Galois representation for which the adjoint Selmer group is finite, this does not require the hypothesis that $\rho$ is unramified outside $p$.


My question is the following, can one use similar lines of reasoning to say anything about the structure of the adjoint Selmer group over the cyclotomic $\mathbb{Z}\_p$ extension of $\mathbb{Q}$? Has this been explored anywhere?



Adjoint Selmer groups and Deformation rings
As far as I know, it is difficult to extract much information about the adjoint Selmer group over the cyclotomic $\mathbb{Z}\_p$-extension. If the modular form $f$ corresponding to $\rho$ is ordinary at $p$, then one can deduce straightforwardly from the finiteness of the Selmer group over $\mathbb{Q}$ that the cyclotomic Selmer group is co-torsion; but it doesn't seem to be possible to get much more information than this -- in particular, I am not aware of any way to deduce the Iwasawa Main Conjecture for $Ad(\rho)$ by these methods.


The main difficulty is that if you pick an integer $n$ and apply deformation theory to Hilbert modular form given by base-changing $f$ to the $n$-th layer $\mathbb{Q}\_n$ in the cyclotomic $\mathbb{Z}\_p$-extension, it tells you something about the size of $Sel(Ad(f) / \mathbb{Q}\_n)$ as an abelian group, but not about its structure as a $\mathbb{Z}\_p[\mathbb{Z}/p^n]$-module (essentially because the deformation theory doesn't "know" that the Hilbert form is a base-change from $\mathbb{Q}$). So you don't get information about characteristic ideals, etc.


This is actually rather similar to the situation in the Iwasawa theory of the trivial representation (studying class groups in $\mathbb{Z}\_p$-extensions, i.e. the original setting studied by Iwasawa himself). In this setting, the analogue of the results you quote is the [analytic class number formula](https://en.wikipedia.org/wiki/Class_number_formula); and again, this only tells you the cardinality of the Selmer group over $\mathbb{Q}\_n$, but not its Galois module structure. This isn't enough to deduce the Main Conjecture; but it *does* allow you to say that if you have proved "half" of the main conjecture, i.e. one inclusion of ideals, then you must in fact have equality. (So you can prove the Main Conjecture over $\mathbb{Q}\_\infty$ using *any 2 of the following 3 things*: the cyclotomic unit Euler system, Eisenstein congruences a la Ribet/Mazur-Wiles, and the analytic class number formula.)


