[proofplan] A Hermitian metric selects, in each fibre, exactly the frames identifying that fibre with $\mathbb{C}^n$ isometrically; these frames form a principal $U(n)$-subbundle because the unitary group is precisely the stabiliser of the standard Hermitian form. Smoothness is checked locally by applying smooth Hermitian Gram-Schmidt to any smooth frame. Conversely, a $U(n)$-reduction lets us declare every reduced frame to be unitary, and this gives a well-defined fibrewise Hermitian [inner product](/page/Inner%20Product) because transition functions lie in $U(n)$. The two constructions are inverse to each other by construction. [/proofplan]

[step:Construct the unitary frame subbundle from a Hermitian metric]Let $h$ be a Hermitian metric on $E$. Let $\operatorname{Fr}_{\mathbb{C}}(E)$ denote the complex frame bundle of $E$, whose fibre over $x\in M$ consists of all complex-linear isomorphisms $u:\mathbb{C}^n\to E_x$. Let $U(n)\subset GL(n,\mathbb{C})$ denote the unitary group preserving the standard Hermitian inner product $\langle\cdot,\cdot\rangle_0$ on $\mathbb{C}^n$, linear in the first argument. Define \begin{align*} P_h:=\{u:\mathbb{C}^n\to E_x \mid x\in M,\ u \text{ is a complex-linear isomorphism, and } h_x(u z,u w)=\langle z,w\rangle_0 \text{ for all } z,w\in\mathbb{C}^n\}. \end{align*} Thus $P_h\subset \operatorname{Fr}_{\mathbb{C}}(E)$ is the set of $h$-unitary complex frames. The right action of $U(n)$ preserves $P_h$. Indeed, if $u\in P_h$ lies over $x\in M$ and $A\in U(n)$, then $uA:\mathbb{C}^n\to E_x$ satisfies \begin{align*} h_x(uA z,uA w)=\langle A z,A w\rangle_0=\langle z,w\rangle_0 \end{align*} for all $z,w\in\mathbb{C}^n$, so $uA\in P_h$. The action is free because the right action of $GL(n,\mathbb{C})$ on frames is free, and it is transitive on each fibre of $P_h$ because two $h$-unitary frames $u,v:\mathbb{C}^n\to E_x$ differ by \begin{align*} A:=u^{-1}\circ v:\mathbb{C}^n\to\mathbb{C}^n. \end{align*} For all $z,w\in\mathbb{C}^n$, \begin{align*} \langle A z,A w\rangle_0=h_x(vz,vw)=\langle z,w\rangle_0, \end{align*} so $A\in U(n)$ and $v=uA$. It remains to record smoothness. Let $V\subset M$ be an [open set](/page/Open%20Set) over which $E$ admits a smooth complex frame \begin{align*} s:V\times\mathbb{C}^n&\to E|_V. \end{align*} Applying the smooth [Hermitian Gram-Schmidt algorithm](/theorems/435) to the ordered smooth frame $s_1,\dots,s_n$ gives a smooth ordered frame $e_1,\dots,e_n$ on $V$ such that \begin{align*} h_x(e_i(x),e_j(x))=\delta_{ij} \end{align*} for every $x\in V$ and every $1\le i,j\le n$. Equivalently, define \begin{align*} e:V\times\mathbb{C}^n&\to E|_V \end{align*} by sending the standard basis vector of $\mathbb{C}^n$ to $e_i(x)$. Then every element of $P_h|_V$ is uniquely of the form $e(x)A$ with $x\in V$ and $A\in U(n)$. Hence the map \begin{align*} \Phi_V:V\times U(n)&\to P_h|_V \end{align*} defined by $\Phi_V(x,A)=e(x)A$ is a smooth local trivialisation. Therefore $P_h\to M$ is a smooth principal $U(n)$-bundle, and its inclusion into $\operatorname{Fr}_{\mathbb{C}}(E)$ is a $U(n)$-equivariant subbundle inclusion.[/step]

[guided]The construction starts from the most direct idea: a Hermitian metric tells us which bases are orthonormal. A complex frame of $E_x$ is a complex-linear isomorphism $u:\mathbb{C}^n\to E_x$. We call it $h$-unitary precisely when it transports the standard Hermitian inner product on $\mathbb{C}^n$ to the Hermitian form $h_x$ on $E_x$: \begin{align*} h_x(u z,u w)=\langle z,w\rangle_0 \end{align*} for every $z,w\in\mathbb{C}^n$. Thus we define \begin{align*} P_h:=\{u:\mathbb{C}^n\to E_x \mid x\in M,\ u \text{ is a complex-linear isomorphism, and } h_x(u z,u w)=\langle z,w\rangle_0 \text{ for all } z,w\in\mathbb{C}^n\}. \end{align*} Why is the structure group $U(n)$ rather than all of $GL(n,\mathbb{C})$? If $A\in U(n)$ and $u\in P_h$, then $A$ preserves $\langle\cdot,\cdot\rangle_0$, so \begin{align*} h_x(uA z,uA w)=\langle A z,A w\rangle_0=\langle z,w\rangle_0. \end{align*} Hence $uA$ is again $h$-unitary. Conversely, if $u$ and $v$ are two $h$-unitary frames over the same point $x$, then the unique change-of-frame map \begin{align*} A:=u^{-1}\circ v:\mathbb{C}^n\to\mathbb{C}^n \end{align*} satisfies \begin{align*} \langle A z,A w\rangle_0=h_x(vz,vw)=\langle z,w\rangle_0 \end{align*} for all $z,w\in\mathbb{C}^n$. This is exactly the defining condition $A\in U(n)$. Therefore each fibre of $P_h$ is a free and transitive right $U(n)$-space. The only remaining point is that $P_h$ is smooth. Work over an open set $V\subset M$ carrying a smooth complex frame $s_1,\dots,s_n$ of $E|_V$. The metric coefficients \begin{align*} x\mapsto h_x(s_i(x),s_j(x)) \end{align*} are smooth functions on $V$ because $h$ is a smooth Hermitian metric. The [Hermitian Gram-Schmidt process](/theorems/435) expresses each orthonormal vector $e_i(x)$ by addition, scalar multiplication, division by positive smooth norms, and previously constructed vectors. Since the original frame is pointwise linearly independent, the norms divided by in Gram-Schmidt are positive at every point of $V$. Hence the resulting vectors $e_1,\dots,e_n$ form a smooth $h$-unitary frame on $V$. Define \begin{align*} e:V\times\mathbb{C}^n&\to E|_V \end{align*} by mapping the standard basis of $\mathbb{C}^n$ to $e_1(x),\dots,e_n(x)$. Every $h$-unitary frame over $x\in V$ is uniquely $e(x)A$ for some $A\in U(n)$, because the change-of-frame map between two unitary frames is unitary. Therefore the map \begin{align*} \Phi_V:V\times U(n)&\to P_h|_V \end{align*} defined by $\Phi_V(x,A)=e(x)A$ is a smooth local trivialisation. This proves that $P_h$ is a smooth principal $U(n)$-subbundle of the complex frame bundle.[/guided]

[step:Construct a Hermitian metric from a unitary reduction] Conversely, let $P\subset \operatorname{Fr}_{\mathbb{C}}(E)$ be a smooth principal $U(n)$-reduction. For $x\in M$, choose any frame $u\in P_x$. For $v,w\in E_x$, define \begin{align*} (h_P)_x(v,w):=\langle u^{-1}v,u^{-1}w\rangle_0. \end{align*} This is independent of the choice of $u$. Indeed, if $u'\in P_x$, then there exists a unique $A\in U(n)$ such that $u'=uA$. Hence \begin{align*} \langle (u')^{-1}v,(u')^{-1}w\rangle_0=\langle A^{-1}u^{-1}v,A^{-1}u^{-1}w\rangle_0=\langle u^{-1}v,u^{-1}w\rangle_0, \end{align*} because $A^{-1}\in U(n)$ preserves $\langle\cdot,\cdot\rangle_0$. For each $x\in M$, the form $(h_P)_x$ is Hermitian and positive definite because it is the pullback of the standard Hermitian inner product along the complex-linear isomorphism $u^{-1}:E_x\to\mathbb{C}^n$. To prove smoothness, let $V\subset M$ be an open set admitting a smooth local section \begin{align*} e:V&\to P. \end{align*} Equivalently, $e$ is a smooth local frame of $E|_V$. If \begin{align*} \sigma,\tau:V&\to E|_V \end{align*} are smooth sections, write their coordinate functions in the frame $e$ as smooth maps \begin{align*} a,b:V&\to\mathbb{C}^n, \end{align*} so that $\sigma(x)=e(x)a(x)$ and $\tau(x)=e(x)b(x)$. Then \begin{align*} (h_P)_x(\sigma(x),\tau(x))=\langle a(x),b(x)\rangle_0, \end{align*} which is a smooth complex-valued function on $V$. Therefore $h_P$ is a smooth Hermitian metric on $E$. [/step]

[step:Verify that the two constructions are inverse] Starting with a Hermitian metric $h$, form $P_h$, and then construct $h_{P_h}$. For $x\in M$, choose $u\in (P_h)_x$. Since $u$ is $h$-unitary, for all $v,w\in E_x$ we may write $z:=u^{-1}v$ and $q:=u^{-1}w$, and obtain \begin{align*} (h_{P_h})_x(v,w)=\langle z,q\rangle_0=h_x(uz,uq)=h_x(v,w). \end{align*} Thus $h_{P_h}=h$. Starting with a $U(n)$-reduction $P$, form $h_P$, and then form $P_{h_P}$. If $u\in P_x$, then by definition of $h_P$, \begin{align*} (h_P)_x(u z,u w)=\langle z,w\rangle_0 \end{align*} for all $z,w\in\mathbb{C}^n$, so $P\subset P_{h_P}$. Conversely, if $v\in (P_{h_P})_x$, choose $u\in P_x$ and define \begin{align*} A:=u^{-1}\circ v:\mathbb{C}^n\to\mathbb{C}^n. \end{align*} For all $z,w\in\mathbb{C}^n$, \begin{align*} \langle A z,A w\rangle_0=(h_P)_x(vz,vw)=\langle z,w\rangle_0. \end{align*} Hence $A\in U(n)$, and therefore $v=uA\in P_x$ because $P_x$ is closed under the right $U(n)$-action. Thus $P_{h_P}\subset P$, so $P_{h_P}=P$. The two assignments are therefore mutually inverse. This proves the natural bijection between Hermitian metrics on $E$ and smooth principal $U(n)$-reductions of $\operatorname{Fr}_{\mathbb{C}}(E)$. [/step]