$L(G)' \cap B \rtimes G = G^B$ if G ICC. If G is not ICC then $L(G)' \cap B \rtimes G = B^G \overline{\otimes} L(FC_G)$

Let G be an ICC (infinite conjugacy class) group and let $\sigma: G \curvearrowright(B, \tau)$ be a trace preserving action. Let L(G) be the group Von Neumann algebra, that is, $L(G)=\overline{span\{\lambda_g\}_{g \in G}}^{WOT}$ where $\lambda : G \to \mathcal{B}(\ell^2(G))$ is left regular representation. Identify $L(G) \hookrightarrow B \rtimes_{\sigma} G=\overline{B[G]}^{WOT}$ by taking $\lambda_g \to u_g=1 \otimes \lambda_g$, in the Crossed Product.

I want to show 2 things:

1) $L(G)' \cap B \rtimes_{\sigma} G = B^{G}$ where $B^G$ is fixed sub algebra of B, by G. (I have answered this part).

2) If the ICC condition is dropped for G is there are more general statement such as $L(G)' \cap B \rtimes_{\sigma} G = B^G \overline{\otimes} L(FC_G)$. Where $L(FC_G)$ is the group Von Neumann algebra of the finite conjugacy classes of G. Or possibly $L(G)' \cap B \rtimes_{\sigma} G = B^{G} \rtimes_{\sigma|_{FC_{G}}} FC_{G}$.

Guess I will answer the first part of my question:

It is clear that $B^G \hookrightarrow B \rtimes_{\sigma} G$ in the natural way, and $B^G \subset L(G)' \cap B \rtimes G$.

Let $x \in L(G)' \cap B \rtimes G$ this has a Fourier Decomposition $x=\sum\limits_{g \in G}{b_g u_g}$, where $b_g \in B$.

Since $x \in L(G)'$ for any canonical unitary $u_h$ we have that $x u_h = u_h x$ thus $(\sum\limits_{g \in G}{b_g u_g})u_h = u_h (\sum\limits_{g \in G}{b_g u_g})$ which gives us that $\sum\limits_{g \in G}{b_g u_{gh}}=\sum\limits_{g \in G}{u_hb_g u_g}= \sum\limits_{g \in G}{\sigma_{h}(b_g) u_{hg}}$. Hitting by $u_{h^{-1}}$ on the right gives us $\sum\limits_{g \in G}{\sigma_h(b_g u_{hgh^{-1}}}=\sum\limits_{g \in G}{b_g u_g}=x$. Making a substitution on the LHS $\lambda=hgh^{-1}$ and noting that $\lambda$ will still run over all elements in G, we end up getting $\sum\limits_{g \in G}{\sigma(b_{hgh^{-1}}) u_g}=\sum\limits_{g \in G}{b_g u_g}$. Since our 2-norm on $B \rtimes_{\sigma} G$ arises from the trace and the action is trace preserving we note that $|| \sigma(b_{hgh^{-1}})||_{2}^{2}=||b_{hgh^{-1}}||_{2}^{2}$.

Thus, we end up with $\sum\limits_{g \in G}{||b_{hgh^{-1}}||_2^2}=\sum\limits_{g \in G}{||b_g||_2^2}$. Since our Fourier Series converge and we have just shown that our coefficients are constant on Conjugacy classes, and the fact that we have ICC condition implies that $\forall b_g \in B$ such that $g \neq e$, $b_g =0$. Thus, we conclude that $b=b_e$ and $b_e \in B$. Moreover, $\sigma(b_e)=b$ thus $b \in B^G$.

Hence, we have that $B^G = L(G)' \cap B \rtimes_{\sigma} G$.

It is clear for the second part of my question that you will still have the same behavior on Infinite Conjugacy classes, but you can have nonzero unitary coefficients in the finite conjugacy classes and still have convergence of the Fourier series in $L^2(B, \tau) \overline{\otimes} \ell^2(G)$.