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Let $M$ be a smooth manifold and $\pi:E \rightarrow M$ a real vector bundle and note $E_x:=\pi^{-1}(x), \forall x\in M$. We set a bundle $\text{End}(E)=E\otimes E^*$. Now suppose there exists a smooth section $J:M\rightarrow\text{End}(E)$ such that for all $x\in M$, $J_x^2=-\text{Id}_{E_x}$.

This gives imidiately that $E$ is of even rang, say $2m$. Now I want to show that the group structure can be reduced to $\text{GL}(m,\mathbb{C})$.

Obviously, we can make $E_x$ a $\mathbb{C}$-vector space by defining $i\cdot v=J_xv$, $\forall v\in E_x$.

Via a trivialisation $\chi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^{2m}$ with $x\in U$ we can make $\{x\}\times\mathbb{R}^{2m}$ a $\mathbb{C}$-vector space by setting $i\cdot (x,v)=(x,\chi\circ J_x\circ\chi^{-1}(x,v))$ and then the trivialisation is actually $\mathbb{C}$-linear on $E_x$

However, this doesn't give a desired reduction of the structure group (I guess) and the problem is because the complex structure of $\mathbb{R}^{2m}$ depends on $x$. If we identify $\mathbb{R}^{2m}$ with $\mathbb{C}^{m}$ via some fixed map $J_0$, then the trivialisation is no longer $\mathbb{C}$-linear.

Another approach I tried is to give a smooth family of complex bases for $U\times\mathbb{C}^{m}$ through the trivialisation (it is immediate to find such family of real bases). No luck so far.

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  • $\begingroup$ Are you familiar with how a metric on a vector bundle reduces the structure group to the orthogonal group? $\endgroup$ – Eric O. Korman Nov 17 '14 at 23:42
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Another approach I tried is to give a smooth family of complex bases for $U×ℂ^m$ through the trivialisation (it is immediate to find such family of real bases).

This is a good idea. Indeed, if you can show that every point has a neighborhood with a (real) frame of the form $\{e_1, Je_1, \ldots, e_n, Je_n\}$ then you are done (then $\{e_1,\ldots,e_n\}$ will be a complex frame).

Start with a trivialization on some subset $U$ of $x$ and let $\{\sigma_1,\ldots,\sigma_{2n}\}$ be a frame. Let $e_1 = \sigma_1$ and check that $e_1(y)$ and $Je_1(y)$ are linearly independent (over the reals) for each $y$. Now at $x$ we have the two linearly independent vectors $e_1(x), Je_1(x)$ and so there must be some $j$ such that $\{e_1(x), Je_1(x), \sigma_j(x)\}$ is linearly independent (if not then $\{e_1(x), Je_1(x)\}$ forms a basis and we are done). Let $e_2 = \sigma_j$ and then you can check that $\{e_1(x), Je_1(x), e_2(x), Je_2(x)\}$ is linearly independent. Continuing in this way by adding in certain $\sigma_k$'s we will have a set of sections $\{e_1, Je_1, \ldots, e_n, Je_n\}$ that is linearly independent at $x$. But by choosing $U$ to be smaller if necessary, we can ensure that the sections are actually linearly independent at every point in $U$. This is because linear independence is specified by the function $\det(e_1, Je_1,\ldots, e_n, Je_n)$ being non-zero, so if this function is non-zero at $x$ it must be non-zero in some neighborhood of $x$.

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  • $\begingroup$ What is $J$ in your construction? Because the section $J_x$ acts on $E_x$ and not on $\mathbb{R^{2m}}$. $\endgroup$ – Lukas Nov 18 '14 at 15:04
  • $\begingroup$ $J$ is the complex structure. The $e_i$ and $\sigma_i$ are local sections of $E$ and so are acted on by $J$. Having a trivialization (i.e. iso from $E$ restricted to $U$ to $U\times \mathbb R^{2n}$) is equivalent to having $2n$ pointwise independent sections of $E$ restricted to $U$. $\endgroup$ – Eric O. Korman Nov 18 '14 at 16:52
  • $\begingroup$ Thanks. This actually worked. $\endgroup$ – Lukas Nov 18 '14 at 21:43

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