Let $E→M$ be a plane bundle endowed with an almost complex structure $J.$
$J$ induces a natural positive definite inner product in the associated bundle $End(E)→M$,denoted by $<,>$. More precisely if $A,B \in End(E\_p)$
then chose a frame $e\_1,e\_2$ in $E\_p$ such that 
$$ J(e\_1)=−e\_2, J(e\_2)=e\_1.$$
 This can always be done for an almost complex structure( I believe the name for such a frame is complex frame?). 
With respect to this frame the matrix of the endomrphism $A$ is $a\_{ij}$


and the matrix of $B$ is $b{ij}.$


Define 
$$<A,B>=\sum\_{ij}a\_{ij}b\_{ij}.$$


Now consider the set 
$$P=\{(p,A)| p∈M, A∈End(Ep), <A,J>=0, <A,id>=0,<A,A>=1\}$$
which is a circle fibration above $M.$
Are there any conditions on $J$ that will guarantee that $P$
is trivial?



Triviality of a circle fibration induced by an almost complex structure
There is a decomposition $\mathrm{End}(E)\cong P\oplus\underline{\mathbb R^2}$, where the second summand is trivial and spanned by $\mathrm{id}$ and $J$. The bundle $P$ carries a natural orientation: if $0\ne A\in P\_p$, then $J\circ A\in P\_p$, let $A$, $J\circ A$ form an oriented frame. An oriented real vector bundle is the same as a complex line bundle, and one can identify $P$ with $E\otimes\_{\mathbb C}E$. As such, it is classified by its first Chern class (which is the Euler class in this case). So $P$ is trivial if and only if the first Chern class (or Euler class) of $E$ is a 2-torsion class.


