We all know that Tate-Shafarevich group is defined as $Ш(E/K)=\mathrm{Ker}(H^1(K,E)\mapsto \prod_{v}H^1(K_v,E))$ for an abelian variety $A$ defined over a number field $K$, the non-trivial elements of the Tate-Shafarevich group can be thought of as the homogeneous spaces of A that have $K_v$-rational points for every place $v$ of $K$, but no $K$-rational point. And we also have the $p$-part or $n$-part of the group which occurs in the method of Descent, suppose if \phi\colon E/K\to E'/K is an isogeny of elliptic curves, then you have an exact sequence, 0\to E'(K)/\phi(E(K)) \to S^{(\phi)}(E/K) \to \text{Ш}(E/K)[\phi]\to 0 So one can define the $\phi$-part of Sha-group as $Ш_2(E/\mathbb{Q})=[\rm{Sel_{2}}(E/\mathbb{Q})/(E(\mathbb{Q})/\rm{2}E(\mathbb{Q})]$. We know already that $Ш_2(E/\mathbb{Q})$ can be easily proven to be finite as both $\rm{Sel_2}(E/\mathbb{Q})$ and $E(\mathbb{Q})/\rm{2}E(\mathbb{Q})$ are finite.
So my main query is :
"Is there any way to profitably write $Ш(E/\mathbb{Q})$ in terms of $Ш_m(E/\mathbb{Q})$ when we perform an infinite descent. If so that will account for decomposition on Sha."
To add some flavor to it, suppose think that $2$-descent works on elliptic curve $E$, then we obtain the $Ш_2(E/\mathbb{Q})$ and if we proceed for the "Infinite-Descent" we obtain for each $E(\mathbb{Q})/\rm{m} E(\mathbb{Q})$ ($m\ge2$) a corresponding Sha-part i.e $Ш_m(E/\mathbb{Q})$ . So can we write the group as $Ш(E/\mathbb{Q})=\bigcup_m Ш_m(E/\mathbb{Q})$ (that might not represent the union but some way of unifying the local-sha(not to be confused with Hasse-Weil word "local", here local seems to the one at 'm'), that may be either a Direct sum or something like that.
To be sharp
"Are there any references or previously done work for decomposing Tate-Shafarevich group ?". (Please give me the references)
Please correct me if I am in a wrong perspective of these groups and Descent method.
Thank you.