# Vector bundles isomorphism

Let $M$ be an orientable manifold of dimension $2m$. Let $(\pi,E,M)$ be a complex line bundle whose fibers are denoted $E_{p}:=\pi^{-1}(\{p\})$ for all $p\in M$. We consider the endomorphism real vector bundle $\text{End}(E)$ whose fibers are the $\Bbb C$-linear endomorphisms of $E_{p}$. Then, given an hermitian product $\langle\cdot,\cdot\rangle$ on $(\pi,E,M)$, we can define the real subbundle $\text{End}(E,\langle\cdot,\cdot\rangle)$ whose fibers are:

$$\text{End}(E,\langle\cdot,\cdot\rangle)_{p}:=\left\{L\in \text{End}(E)_{p}\mid \langle Lv,w\rangle + \langle v,Lw\rangle = 0,\,\forall\,(v,w)\in E_{p}\times E_{p}\right\}$$

It is not difficult to see it is indeed a subbundle. Note that we consider the endomorphism vector bundle as a real vector bundle because the fibers of the subbundle are real vector spaces but not complex ones.

As our initial vector bundle $(\pi,E,M)$ is of rank $1$, we see that $\text{End}(E)_{p}\simeq \Bbb C$ where $\Bbb C$ is seen as a real vector space (endomorphism of $\Bbb C$ are multiplication by a complex). We can easily see that antihermitian maps are multiplication by $z=-\bar{z}$, i.e. by $z\in i\Bbb R$.

We want to show that $\text{End}(E,\langle\cdot,\cdot\rangle) \simeq M\times i\Bbb R$ (isomorphism of real vector bundles over the identity).

From our considerations, we see that the fibers are trivially isomorphic. However, how to show that these isomorphisms of fibers are induced by a smooth map $F:\text{End}(E,\langle\cdot,\cdot\rangle)\to M\times i\Bbb R$? Is it a consequence of the definition of a vector bundle, which implies the existence of local diffeomorphisms mapping fibers isomorphically? I think this kind of argument is flawed as we don't consider the total space by doing so and this argument would apply to show that the Möbius band is isomorphic to the annulus, which is not the case.

My question originated from this document (p. 5, below Definition 3.5).

Any help is appreciated.

• If you have a morphism of bundle which induces an isomorphism on each fiber, this is an isomorphism of vector bundles. – user171326 Apr 12 '17 at 11:54
• @N.H. Thank you four your answer. The thing is I have a collection of isomorphismes on each fiber but how do I know it is induced by a morphism ? – MoebiusCorzer Apr 12 '17 at 11:56
• What is wrong with the map $f \in End(E) \mapsto (p, f(p))_{p \in M}$ ? – user171326 Apr 12 '17 at 12:00
• @N.H. What do you mean by $f(p)$? I mean, I don't know much about $\text{End}(E)$ as a whole. I only have a precise idea of how are its fibers. So what does $f(p)$ means for an element $f\in\text{End}(p)$? – MoebiusCorzer Apr 12 '17 at 12:08
• I did some identifications. But a simpler way is noticing that you want to prove that your vector bundle $End(E)$ is trivial, i.e find a non-zero global section, and such section is given by $s : M \to End(E), p \to \lambda_p : E_p \to E_p$ where $\lambda_p$ is multiplication by $i$. – user171326 Apr 12 '17 at 12:19

The section $\gamma : M \to End(E, \langle \cdot , \cdot \rangle )$ defined by $p \mapsto \lambda_p$ where $\lambda_p : E_p \to E_p, v \mapsto iv$ is non-vanishing and so define an isomorphism between $End(E, \langle \cdot , \cdot \rangle )$ and the trivial bundle, i.e $End(E, \langle \cdot , \cdot \rangle ) \cong M \times i\mathbb R$.