The determinant-line projector is smooth and rank one

In the standard orthonormal wedge basis from [F4], the matrix entry of ${\bigwedge }^{m}T$ in row $I$, column $J$ is $\det T_{I,J}$ by [F3]. Consequently $[({\bigwedge }^{m}T{)}^{\ast }{]}_{I,J}=\overline{\det T_{J,I}}=\det (T^{\ast }{)}_{I,J}$, so $({\bigwedge }^{m}T{)}^{\ast }={\bigwedge }^m(T^{\ast })$. Applying this to $P=P^{\ast }$ shows $Q=Q^{\ast }$.

Functoriality [F2] gives $Q^2={\bigwedge }^m(P^2)={\bigwedge }^{m}P=Q$. Hence $Q$ is a Hermitian idempotent by step 1.1; for $x=Qy$ and $Qz=0$, $⟨x,z⟩=⟨Qy,z⟩=⟨y,Qz⟩=0$, so its range and kernel are orthogonal and it is the orthogonal projection onto its range.

Every decomposable wedge in the image of $Q$ is a wedge of $m$ vectors in $\mathrm{im}P$, so the full image lies in ${\bigwedge }^m\mathrm{im}P$. Conversely, for a basis $v_1,\dots ,v_m$ of the $m$-dimensional range of $P$, $Q(v_1\wedge \cdots \wedge v_m)=v_1\wedge \cdots \wedge v_m\mathrm{\ne }0$. The increasing-wedge basis theorem [F4], applied to this basis of the range, says ${\bigwedge }^m\mathrm{im}P$ is one-dimensional; thus it equals $\mathrm{im}Q$ and $Q$ has rank one.

Each matrix entry of $Q(x)$ is a polynomial minor in the entries of $P(x)$ by [F3], hence is $C^2$. Together with step 2.1 this proves that $Q$ is a $C^2$ rank-one orthogonal projector with the stated range, and its ranges form the determinant line subbundle.