I am trying to prove that if $M$ is an even dimensional manifold, and the bundle of linear frames of $TM$ admits a reduction of its structure to group to $GL_n(\mathbb{C})$, then $M$ admit's an almost complex structure.
In theory I feel like this should be easy, however I am struggling on the details. I know that if the frame bundle admits a reduction of its structure group to $GL_n(\mathbb{C})$ then we have that $TM$ admits a covering of vector bundle charts $(U,\phi)_i$ such that the transition functions take values in $GL_n(\mathbb{C})$. Let $\tilde{I}$ be a complex structure on $\mathbb{R}^{2n}$, and then $GL_n(\mathbb{C})\subset GL_{2n}(\mathbb{R})$ is the subgroup of linear transformations which commute $\tilde{I}$. I thus want to define $I$ in each bundle chart by: \begin{align*} I_U=\phi^{-1}\circ(\text{Id}_U\times \tilde{I})\circ \phi \end{align*} Checking that this is independent of bundle chart then seems too easy as if $(V,\psi)$ is another bundle chart such that $U\cap V\neq \emptyset$ then: \begin{align*} (\psi^{-1}\circ \phi)\circ I_U\circ (\phi^{-1}\circ \psi)=I_V \end{align*} trivially, so a partition of unity argument implies that this defines a global $(1,1)$ tensor field on $M$. It is also easy to see that in any chart $I_U\circ I_U=-\text{Id}$. However, I have never used the fact the transition functions take values in $GL_n(\mathbb{C})$, so I suspect I am missing something. Do I need to check that $I_V\circ I_U=-\text{Id}$ as well for some reason?