I'm trying to prove that principal $U(N)$-bundles are in bijective correspondence to complex vector bundles.

I understand the proof that principal $\text{GL}(N, \mathbb{R})$-bundles are equivalent to real vector bundles. Given a principal $\text{GL}(N, \mathbb{R})$-bundle $P$ on a base $X$, use the defining representation of $\text{GL}(N, \mathbb{R})$ on $\mathbb{R}^{N}$ to define an associated real vector bundle $E$ on $X$. Conversely, the frame bundle recovers the original principal bundle.

Obviously, a similar argument holds in the complex case. Now, let $P$ be a principal $U(N)$-bundle on $X$. I can use the defining representation of $U(N)$ on $\mathbb{C}^{N}$ to get an associated complex vector bundle with structure group $U(N)$. Conversely, let $E$ be a complex vector bundle on $X$. I think the key is that every complex vector bundle admits a Hermitian metric which reduces the structure group from $\text{GL}(N, \mathbb{C})$ to $U(N)$. One can then take the frame bundle which will recover the original principal $U(N)$-bundle.

What I'm confused about is that a complex vector bundle may admit many Hermitian metrics, I believe. Do these all give rise to the same principal unitary frame bundle?

A second related question concerns connections and curvature two-forms. Let $P$ be a principal $U(N)$-bundle and $A$ a connection on $P$ with curvature two-form $F_{A}$. The connection is a Lie algebra-valued one-form and $F_{A} \in \Omega(X, adP)$ is a two-form valued in the real vector space $adP$.

Let $E$ be a complex vector bundle equivalent to $P$ under the correspondence described above. Let $\nabla$ be a connection on $E$ with curvature two-form $F_{\nabla}$. We have $F_{\nabla} \in \Omega(X, \text{End}E)$. By the excellent answer to Principal Bundles, Chern Classes, and Abelian Instantons we should have $F_{A} = F_{\nabla}$. In particular, they determine the same Chern classes.

But I'm confused about the relation between the real vector bundle $adP$ and the complex vector bundle $\text{End}E$. Do we have $\text{End}E \cong adP \otimes \mathbb{C}$? Where of course, $E$ and $P$ are equivalent via the above correspondence. To reiterate, the source of my confusion is $F_{\nabla} \in \Omega(X, \text{End}E)$ and $F_{A} \in \Omega(X, adP)$, with $F_{A} = F_{\nabla}$.

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    $\begingroup$ Recall that every real vector bundle has a metric. However in general there are many different inequivalent metrics. As for the question in the last paragraph: ad(P) is an the associated bundle to the adjoint rep of $U(n)$ on $\mathfrak{u}_n$ and $End(E)$ is the associated bundle to the adjoint rep of $GL(n,\mathbb{C})$ on $\mathfrak{gl}_n(\mathbb{C})$. Indeed $U(n)$ is the compact real form of $GL_n$ and the complexification of the adjoint representation is the adjoint representation. In other words $adP$ is a real form for the complex bundle $EndE$ because it is true at the level of reps. $\endgroup$ – Saal Hardali Aug 29 '17 at 20:11
  • $\begingroup$ Thanks! Does $adP$ being a real form for the complex bundle $\text{End}E$ indeed mean that $adP \otimes \mathbb{C} = \text{End}E$? I believe this is exactly what you said, but just to make sure. As for my first question, that is exactly what I was confused about. Given a complex vector bundle (I think) there exists many inequivalent Hermitian metrics which reduce the structure group from $GL(N, \mathbb{C})$ to $U(N)$. But upon passing to the frame bundle, do all these different metrics lead to the same principal bundle of unitary frames? $\endgroup$ – Benighted Aug 29 '17 at 21:52
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    $\begingroup$ This depends what you mean by $adP \otimes \mathbb{C} = End E$. This is true (as I explained) when you regard them as simple vector bundles (no structure attached). If you put an hermitian metric on $E$ then the statement is still true if we regard both sides as unitary bundles. However if we take a different metric on either side they might not be the same. $\endgroup$ – Saal Hardali Aug 29 '17 at 21:56
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    $\begingroup$ Also note that a choice of hermitian metric on $E$ is equivalent to a choice of sub-principal bundle of unitary frames. So different hermitian metrics will lead to different principle bundles of unitary frames. $\endgroup$ – Saal Hardali Aug 29 '17 at 21:56
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    $\begingroup$ Indeed complex vector bundles are not equivalent to unitary principle bundles. The correct statement is that every complex vector bundle has a (possibly non-unique) reduction of structure group to $U(N)$, meaning a hermitian metric. This can be proved by a partition of unity argument or by just observing that $U(n)$ is maximal compact in $GL(n)$ and so the inclusion induces a homotopy equivalence $BU(n) \to BGL(n)$ $\endgroup$ – Saal Hardali Aug 29 '17 at 22:19

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