Let $A$ and $B$ are two factor von neumann algebras that act on two infinite dimensional Hilbert spaces H and K respectively. Let $\Phi:A\longrightarrow B$ is an additive bijective map with some other conditions that it is not necessary to state them. my question is that if we know $\mathbb{R}I\subseteq \Phi(\mathbb{R}I)$, can we conclude by the bijectivity of $\Phi$ that $\mathbb{R}I=\Phi(\mathbb{R}I)$? (here I is identity operator)
If yes or no, what is your reason?
so thanks for your help.
Let $A=B=\mathbb C$. Fix a Hamel basis $X$ of $\mathbb R$ as a vector space over $\mathbb Q$. Then $X\cup Xi\ $ is a Hamel basis of $\mathbb C$ over $\mathbb Q$.
Since $X$ and $X\cup\{i\}$ have the same cardinality, there exists a bijection $\gamma:X\to X\cup\{i\}$ with $\gamma(1)=i$. Let $\eta:iX\to iX\setminus\{i\}$ be a bijection. These bijections induce a bijection $X\cup Xi\to X\cup Xi$ and thus $\mathbb Q$-linear bijective map (in particular, additive) $\Phi:A\to B$ by $$ \Phi(\sum_jq_jx_j+\sum_kp_ky_ki)=\sum_jq_j\gamma(x_j)+\sum_kp_k\eta(y_k)i $$ for any $q_1,\ldots,q_n,p_1,\ldots,p_m\in\mathbb Q$, $x_1,\ldots,x_n\in X$, $y_1,\ldots,y_m\in X$.
Thus $\Phi$ is an additive bijection $A\to B$ with $\Phi(\mathbb R)=\mathbb R+\mathbb Qi$.
As you can see, additive but not linear is very unnatural.
