Maybe this is well known, but i could not find a pointer to some literature:


Let us assume $E$ is a rank n vector bundle on the Riemann sphere $\mathbb{C}\mathbb{P}^1$. We know that $E=\bigoplus\limits\_{i=1}^n L\_i$ for some line bundles $L\_i$.


Now if $E$ is also equipped with a Hermitian metric $h$, does this metric also split?
That is do we have $h=\oplus h\_i$ for Hermitian metrics on the line bundles $L\_i$, so that we have an orthogonal decomposition of $E$? Maybe this only true up to isometry or something?


Or are there non split metrics? Is there an easy example in the case $n=2$? Some references are also welcome.



Do Hermitian metrics also split on the Riemann sphere?
No, such an hermitian metric does not necessarily split, even if $n=2$.
As a matter of fact, vector bundles are a quite rigid sort of objects, while metrics are pretty smooth and can be tweaked locally to have any prescribed look
and a given small enough open subset.


By partitions of unity, it is easy to define metrics that have a prescribed
form on some open sets: if $(U\_\alpha)$ is an open covering, and $h\_\alpha$ is a metric on $U\_\alpha$, and $(\lambda\_\alpha)$ is a partition of unity relative to this open covering, then $\sum h\_\alpha\lambda\_\alpha$ is a metric on $E$ which coincides with $h\_\alpha$
on the set $U\_\alpha$ deprived from the $U\_\beta$, for $\beta\neq\alpha$.
So you can construct a metric on $\bigoplus L\_i$ which is not obviously
diagonal if the metric $h\_\alpha$ is not diagonal (and the $U\_\beta$, for $\beta\neq\alpha$, do not cover $\Sigma$).


Example: Cover the sphere $\Sigma$ by two relatively compact open sets $U\_\alpha$ and $U\_\beta$
on which all the $L\_j$ are trivial, and take non-diagonal metrics $h\_\alpha$
and $h\_\beta$ on $E$. Then the glued metric on $E$ is non-diagonal on $U\_\alpha\setminus U\_\beta$.


However, it might happen that the obtained metric on $E$ is diagonal in some
other presentation of $E$. This is the case if all $L\_j$ are trivial, say,
and if you take a constant non-diagonal metric. Just change the frame and your metric is diagonal.


However, in the important case where the $L\_j$ are pairwise non-isomorphic, and ordered by decreasing degree, the automorphisms of $E$ preserve the filtration 
$0\subset L\_1\subset L\_1\oplus L\_2\subset\dots$, hence are "upper-triangular".
Since a matrix which is both triangular and orthogonal is diagonal,
any metric on $E$ which is not obviously split is non-split. (Thanks to WillSavin's for correcting a mistake in the first version.)


At a higher level, given an exact sequence of vector bundles with hermitian metrics, the theory of Bott-Chern forms allows to measure the deviation of this metric to a split one. See the initial paper of Bott-Chern (*Acta Math.*, 1965) or the paper of Gillet-Soulé (*Annals of maths*, 1990).


