[proofplan] We first put a smooth fibre metric on $E$, using paracompactness to assemble local Euclidean metrics by a smooth [partition of unity](/page/Partition%20of%20Unity). With this metric in hand, we define $G$ fibrewise as the orthogonal complement $F^\perp$. The main point is to verify that this fibrewise construction is smooth: locally, the metric gives a smooth projection from $E$ onto $F$, and the kernel of this projection is a smooth subbundle. Finally, the positive-definiteness of the metric gives the fibrewise direct-sum decomposition. [/proofplan]

[step:Construct a smooth fibre metric on $E$] Let $r=\operatorname{rank}E$. We first construct a smooth fibre metric \begin{align*} h_p:E_p\times E_p\to \mathbb{R},\qquad p\in M, \end{align*} on $E$. Choose a smooth trivializing cover $\{U_a\}_{a\in A}$ of $E$, with smooth vector bundle trivializations \begin{align*} \Psi_a:E|_{U_a}\to U_a\times \mathbb{R}^r. \end{align*} Since $M$ is paracompact and smooth, choose a locally finite smooth partition of unity $\{\rho_a\}_{a\in A}$ subordinate to $\{U_a\}_{a\in A}$. Thus each $\rho_a:M\to [0,1]$ is smooth, $\operatorname{supp}\rho_a\subset U_a$, the family is locally finite, and \begin{align*} \sum_{a\in A}\rho_a(p)=1 \end{align*} for every $p\in M$. For $p\in U_a$, let \begin{align*} h_{a,p}:E_p\times E_p\to \mathbb{R} \end{align*} be the pullback of the standard Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^r$ through the linear isomorphism on fibres induced by $\Psi_a$. More explicitly, if \begin{align*} \Psi_a(v)=(p,\xi),\qquad \Psi_a(w)=(p,\eta), \end{align*} with $\xi,\eta\in\mathbb{R}^r$, define \begin{align*} h_{a,p}(v,w)=\sum_{\ell=1}^r \xi_\ell\eta_\ell. \end{align*} Now define $h_p:E_p\times E_p\to \mathbb{R}$ by \begin{align*} h_p(v,w)=\sum_{a\in A}\rho_a(p)\,h_{a,p}(v,w). \end{align*} Terms with $\rho_a(p)=0$ are omitted. The sum is finite near each point because the partition of unity is locally finite, so $h$ is smooth in every local trivialization. For each $p\in M$, the form $h_p$ is symmetric and bilinear because it is a finite linear combination of symmetric bilinear forms. If $v\in E_p$ and $v\neq 0$, then \begin{align*} h_p(v,v)=\sum_{a\in A}\rho_a(p)\,h_{a,p}(v,v). \end{align*} Every summand is non-negative. Since $\sum_a\rho_a(p)=1$, there exists $a\in A$ with $\rho_a(p)>0$; for this index, $h_{a,p}(v,v)>0$ because $h_{a,p}$ is positive definite. Hence $h_p(v,v)>0$. Therefore $h$ is a smooth fibre metric on $E$. [/step]

[step:Define the candidate complement as the orthogonal complement of $F$] For each $p\in M$, define \begin{align*} G_p=\{v\in E_p: h_p(v,w)=0 \text{ for every } w\in F_p\}. \end{align*} Let \begin{align*} G=\bigsqcup_{p\in M}G_p\subset E. \end{align*} We will prove that $G$ is a smooth vector subbundle of $E$ and that it is a fibrewise complement to $F$. [/step]

[step:Express $G$ locally as the kernel of a smooth projection onto $F$]Fix $p_0\in M$. Since $F\subset E$ is a smooth vector subbundle, there exists an open neighbourhood $U\subset M$ of $p_0$ and smooth sections \begin{align*} s_1,\dots,s_k:U\to F|_U \end{align*} forming a frame for $F|_U$, where $k=\operatorname{rank}F$. Shrinking $U$ if necessary, extend this frame to a smooth frame \begin{align*} s_1,\dots,s_k,e_{k+1},\dots,e_r:U\to E|_U \end{align*} for $E|_U$. Define the smooth Gram matrix map $A:U\to \mathbb{R}^{k\times k}$ by requiring, for each $p\in U$, that \begin{align*} A(p)_{ij}=h_p(s_i(p),s_j(p)) \end{align*} for $1\le i,j\le k$. For each $p\in U$, the vectors $s_1(p),\dots,s_k(p)$ are linearly independent and $h_p$ is positive definite, so $A(p)$ is a symmetric positive-definite matrix. Hence $A(p)$ is invertible for every $p\in U$, and the inverse matrix map \begin{align*} A^{-1}:U\to \mathbb{R}^{k\times k} \end{align*} is smooth. Define a smooth bundle map over $U$ \begin{align*} P:E|_U&\to F|_U \end{align*} as follows. For $v\in E_p$ with $p\in U$, set \begin{align*} P(v)=\sum_{i=1}^k\sum_{j=1}^k s_i(p)\,(A(p)^{-1})_{ij}\,h_p(v,s_j(p)). \end{align*} The map $P$ is smooth because it is written in local frames using smooth coefficient functions. For every $1\le m\le k$, the definition of $P$ gives \begin{align*} P(s_m(p))=\sum_{i=1}^k\sum_{j=1}^k s_i(p)\,(A(p)^{-1})_{ij}\,h_p(s_m(p),s_j(p)). \end{align*} Using the definition of $A(p)_{mj}$, this becomes \begin{align*} P(s_m(p))=\sum_{i=1}^k\sum_{j=1}^k s_i(p)\,(A(p)^{-1})_{ij}\,A(p)_{mj}. \end{align*} The matrix identity $A(p)^{-1}A(p)=I_k$, using symmetry of $A(p)$ to match the displayed indices, gives \begin{align*} P(s_m(p))=s_m(p). \end{align*} Thus $P$ restricts to the identity on $F|_U$. We now identify its kernel. If $v\in E_p$, then $P(v)=0$ if and only if, for every $1\le m\le k$, \begin{align*} 0=h_p(P(v),s_m(p)). \end{align*} By the definition of $P$, the right-hand side is \begin{align*} \sum_{i=1}^k\sum_{j=1}^k (A(p)^{-1})_{ij}\,h_p(v,s_j(p))\,h_p(s_i(p),s_m(p)). \end{align*} Since $P(v)\in F_p$ and $P$ is the $h_p$-[orthogonal projection](/theorems/437) onto $F_p$, this is equivalent to \begin{align*} h_p(v,s_m(p))=0 \end{align*} for every $1\le m\le k$. Because $s_1(p),\dots,s_k(p)$ form a basis of $F_p$, this is equivalent to $v\in G_p$. Therefore \begin{align*} G|_U=\ker P. \end{align*}[/step]

[guided]The goal is to show that the fibrewise definition of $G_p$ varies smoothly with $p$. A fibrewise orthogonal complement is automatically a vector subspace in each fibre, but smoothness of the collection of these subspaces requires a local frame. Fix $p_0\in M$. Since $F$ is a smooth subbundle, choose an open neighbourhood $U\subset M$ of $p_0$ and a smooth local frame \begin{align*} s_1,\dots,s_k:U\to F|_U \end{align*} for $F|_U$, where $k=\operatorname{rank}F$. After shrinking $U$ within a bundle-trivializing neighbourhood for $E$, extend this to a smooth local frame \begin{align*} s_1,\dots,s_k,e_{k+1},\dots,e_r:U\to E|_U \end{align*} for $E|_U$. The metric lets us project onto the span of $s_1,\dots,s_k$, but because this frame need not be orthonormal, we must correct by the Gram matrix. Define $A:U\to \mathbb{R}^{k\times k}$ by specifying its entries as \begin{align*} A(p)_{ij}=h_p(s_i(p),s_j(p)). \end{align*} For each $p\in U$, the matrix $A(p)$ is positive definite: if $\lambda=(\lambda_1,\dots,\lambda_k)\in\mathbb{R}^k$ is nonzero, then the vector \begin{align*} w=\sum_{i=1}^k\lambda_i s_i(p) \end{align*} is nonzero in $F_p$, and \begin{align*} \sum_{i=1}^k\sum_{j=1}^k \lambda_i A(p)_{ij}\lambda_j =h_p(w,w)>0. \end{align*} Hence $A(p)$ is invertible for every $p\in U$. Since the entries of $A$ are smooth and matrix inversion is smooth on the [open set](/page/Open%20Set) of invertible matrices, $A^{-1}:U\to\mathbb{R}^{k\times k}$ is smooth. Now define \begin{align*} P:E|_U&\to F|_U \end{align*} by \begin{align*} P(v)=\sum_{i=1}^k\sum_{j=1}^k s_i(p)\,(A(p)^{-1})_{ij}\,h_p(v,s_j(p)), \end{align*} for $v\in E_p$ and $p\in U$. This formula is the usual orthogonal projection formula in a non-[orthonormal basis](/page/Orthonormal%20Basis). It is smooth because the sections $s_i$, the metric coefficients $h_p(v,s_j(p))$, and the entries of $A^{-1}$ are smooth in the chosen local frame. We verify that $P$ is the projection onto $F|_U$. For $1\le m\le k$, the definition of $P$ gives \begin{align*} P(s_m(p))=\sum_{i=1}^k\sum_{j=1}^k s_i(p)\,(A(p)^{-1})_{ij}\,h_p(s_m(p),s_j(p)). \end{align*} Using $A(p)_{mj}=h_p(s_m(p),s_j(p))$, we get \begin{align*} P(s_m(p))=\sum_{i=1}^k\sum_{j=1}^k s_i(p)\,(A(p)^{-1})_{ij}\,A(p)_{mj}. \end{align*} By $A(p)^{-1}A(p)=I_k$, with symmetry of $A(p)$ matching the displayed indices, \begin{align*} P(s_m(p))=s_m(p). \end{align*} Thus $P$ is the identity on $F_p$. Finally, we compare $\ker P$ with the orthogonal complement. If $v\in E_p$ satisfies $P(v)=0$, then $P(v)$ has zero inner product with each basis vector $s_m(p)$. Since $P(v)$ is the unique element of $F_p$ with the same inner products against the basis vectors $s_m(p)$ as $v$, it follows that, for each $1\le m\le k$, \begin{align*} h_p(v,s_m(p))=0. \end{align*} Hence $v$ is orthogonal to every vector in $F_p$, so $v\in G_p$. Conversely, if $v\in G_p$, then $h_p(v,s_j(p))=0$ for every $j$, and the defining formula gives \begin{align*} P(v)=0. \end{align*} Therefore $G|_U=\ker P$.[/guided]

[step:Build a smooth local frame for $G$] For each $a\in\{k+1,\dots,r\}$, define a smooth section $g_a:U\to E|_U$ by \begin{align*} g_a(p)=e_a(p)-P(e_a(p)). \end{align*} Since $P(e_a(p))\in F_p$ and $P$ is the identity on $F_p$, we have \begin{align*} P(g_a(p))=P(e_a(p))-P(P(e_a(p)))=0. \end{align*} Thus $g_a(p)\in G_p$ for every $p\in U$. We show that $g_{k+1},\dots,g_r$ form a frame for $G|_U$. First suppose \begin{align*} \sum_{a=k+1}^r c_a g_a(p)=0 \end{align*} for [real numbers](/page/Real%20Numbers) $c_{k+1},\dots,c_r$. Then \begin{align*} \sum_{a=k+1}^r c_a e_a(p) = P\left(\sum_{a=k+1}^r c_a e_a(p)\right)\in F_p. \end{align*} Since \begin{align*} s_1(p),\dots,s_k(p),e_{k+1}(p),\dots,e_r(p) \end{align*} is a basis of $E_p$, the only linear combination of $e_{k+1}(p),\dots,e_r(p)$ lying in $F_p=\operatorname{span}\{s_1(p),\dots,s_k(p)\}$ is the zero vector. Hence $c_a=0$ for every $a$. Now let $v\in G_p$. Write $v$ in the local frame for $E$: \begin{align*} v=\sum_{i=1}^k b_i s_i(p)+\sum_{a=k+1}^r c_a e_a(p). \end{align*} Since $v\in G_p=\ker P$, applying $P$ gives \begin{align*} 0=P(v)=\sum_{i=1}^k b_i s_i(p)+\sum_{a=k+1}^r c_a P(e_a(p)). \end{align*} Therefore \begin{align*} \sum_{i=1}^k b_i s_i(p)=-\sum_{a=k+1}^r c_a P(e_a(p)). \end{align*} Substituting this into the expression for $v$ yields \begin{align*} v=\sum_{a=k+1}^r c_a\bigl(e_a(p)-P(e_a(p))\bigr). \end{align*} By the definition of $g_a$, this is \begin{align*} v=\sum_{a=k+1}^r c_a g_a(p). \end{align*} Thus $g_{k+1},\dots,g_r$ span $G_p$ and are linearly independent for every $p\in U$. Therefore $G$ is a smooth vector subbundle of $E$. [/step]

[step:Verify the fibrewise direct-sum decomposition] Fix $p\in M$. We prove \begin{align*} E_p=F_p\oplus G_p. \end{align*} First, if $v\in F_p\cap G_p$, then $v\in F_p$ and $v$ is orthogonal to every vector in $F_p$. Taking $w=v$ in the definition of $G_p$ gives \begin{align*} h_p(v,v)=0. \end{align*} Since $h_p$ is positive definite, $v=0$. Hence \begin{align*} F_p\cap G_p=\{0\}. \end{align*} Next let $v\in E_p$. Choose a neighbourhood $U$ of $p$ as above and use the local projection $P:E|_U\to F|_U$. Then \begin{align*} v=P(v)+(v-P(v)). \end{align*} The first term lies in $F_p$. The second term lies in $G_p$ because \begin{align*} P(v-P(v))=P(v)-P(P(v))=P(v)-P(v)=0. \end{align*} Therefore every $v\in E_p$ is a sum of an element of $F_p$ and an element of $G_p$. Together with the zero-intersection result, this proves \begin{align*} E_p=F_p\oplus G_p \end{align*} for every $p\in M$. [/step]

[step:Identify the direct sum as a vector bundle isomorphism] Define the fibrewise addition map $\Phi:F\oplus G\to E$ by \begin{align*} \Phi(u,v)=u+v. \end{align*} This map is smooth and fibre-preserving because addition in a smooth vector bundle is smooth. For every $p\in M$, the induced [linear map](/page/Linear%20Map) \begin{align*} \Phi_p:F_p\oplus G_p\to E_p \end{align*} is an isomorphism by the direct-sum decomposition just proved. Hence $\Phi$ is an isomorphism of smooth vector bundles. Therefore $G$ is a smooth complement to $F$ in $E$. [/step]