Vector bundles isomorphism

881 Views Asked by At

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.

1

There are 1 best solutions below

0
On BEST ANSWER

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 $End(E, \langle \cdot , \cdot \rangle ) \cong M \times i\mathbb R$.