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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.

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up vote 2 down vote accepted

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).

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Are those really the only automorphisms? Nontrivial homomorphism from $A$ to $B$ give unipotent automorphisms of $A+B$. If the bundles are nonisomorphic, there must be lots of Homs from lower-degree to higher-degree ones. – Will Sawin Mar 7 '14 at 17:24
Thanks for the hint about Bott-Chern forms. These seem to be exactly what i am looking for. – DonD Mar 7 '14 at 17:26
@WillSawin: You're perfectly right. I'll correct that. On the other hand, non-diagonal unipotent automorphisms are rarely isometric! – ACL Mar 7 '14 at 17:43

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