Action of $\operatorname{GL}(V)$ on $\mathcal J(V)$ is transitive

Question

Let $V$ be vector space with a complex structure $J$. Define a map $\Psi\colon\operatorname{GL}(V)\longrightarrow\mathcal J(V)$ where $\Psi(\Phi)=\Phi^{-1}J\Phi$ where $\mathcal J(V)$ is the group of complex structures on $V$. I am supposed to prove this map is surjective.

Given a $K\in \mathcal J(V)$ how can construct a $\Phi\in \operatorname{GL}(V)$ such that $\Phi^{-1}J\Phi=K$?

Thanks for any help and suggestions.

Answer 1

Given a complex structure $J$ on the real vector space $V$, you obtain a complex vector space $V_J$ on the same set with scalar multiplication $$(a+bi)v = av+bJ(v).$$ Let $(v_i)_{i\in I}$ be a (complex) basis of $V_J$, then $(v_i, \hat v_i)_{i\in I}$ with $\hat v_i=J(v_i)$ is a (real) basis of $V$. On this basis, $J$ acts like $J(v_i)=\hat v_i$ and $J(\hat v_i)=-v_i$. Doing the same for $V_K$, you get a second basis $(w_i, \hat w_i)_{i\in I}$ of $V$ such that $K(w_i)=\hat w_i$, $K(\hat w_i)=-w_i$. Define $\Phi$ by $w_i\mapsto v_i$ and $\hat w_i\mapsto \hat v_i$.

Answer 2

Hint: Basically, this is the statement that $(V,K)$ admits a complex basis. The map $\Phi$ is obtained taking complex bases $\{v_i\}$ for $(V,J)$ and $\{w_\ell\}$ for $(V,K)$ and then define a real linear map $\Phi$ by sending each $v_i$ to $w_i$ and each $J(v_i)$ to $K(w_i)$.