2(J_0 J^\sft - JJ_0).
\endmatrix$$

Now, writing any non-zero $v\in V$ as $v= J_0w$, we obtain $$\matrix\format \l &\,\c\,\;& \l \vbox to 12.5pt{} \\
g(v, (J_0 J^\sft - JJ_0)v) &=&  g(J_0w, (J_0 J^\sft - JJ_0)J_0w)= 
\cr
g(J_0w, J_0J^\sft J_0w) + g(J_0w, Jw) &=& 
g(w, J^\sft J_0w) + g(J_0w, Jw) = 
\cr
g(Jw, J_0w) + g(J_0w, Jw) &=& 2\omega(w, Jw)>0.
\endmatrix$$

Thus $\sfL$ maps $\jj_\omega$ into the set $$\calw \deff 
\{\, W \in \endo(V)\;:\; W J_0 =- J_0W,\; \bfone -W^\sft W \gg0 \,\}.
\eqno\rom(1.2.5)$$

It is easy to see that $\calw$ is contractible. Thus it is sufficient to show that $\sfL: \jj_\omega \to \calw$ is a diffeomorphism. Explicit calculations show that the inverse map should be given by $$J= \sfK(W) \deff J_0 (\bfone + W) (\bfone - W)\inv= 
J_0 (\bfone - W)\inv (\bfone + W) = 
J_0 {\bfone + W \over \bfone - W}
\eqno\rom(1.2.6)$$

[*Step 5. Let $W\in \calw$. Then $J= \sfK(W) = J_0 {\bfone + W \over 
\bfone - W}$ is well-defined and $J^2 = -\bfone$*]{}. Really, the condition $\bfone -W^\sft W \gg0$ implies that $\bfone \pm W$ is invertible, and then $$\matrix\format \c &\,\c\,&\l  \vbox to 11.5pt{} \\
J^2 &=& 
-J_0 (\bfone + W) (\bfone - W)\inv J_0\inv 