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Write $p:U(H)\to PU(H)$ for the scalar quotient. Local strongly continuous sections exist: near a fixed projective unitary, fix its phase by requiring one nonzero matrix coefficient, taken relative to a fixed representative, to be positive real. Thus the pullback $E=\{(z,V):p(V)=\rho (z)\}$ is a locally trivial central extension of the circle by the circle. It is compact, since finitely many closed arcs inside such trivializing neighbourhoods cover the compact base and each pullback is a product with the compact fibre.

Any two lifts commute up to a scalar. Their commutator defines a continuous alternating bicharacter $c(z,w)$ on the base: the scalar commutator is independent of phase choices and multiplicative in each argument. If $z$ is an irrational rotation, its powers are dense and $c(z,z^m)=1$ for every integer $m$. Continuity makes $c(z,w)=1$ for every $w$. Such $z$ are themselves dense, so all the lifts commute, and $E$ is abelian.

The tautological representation $R(z,V)=V$ is strongly continuous. Choose a unit vector $\xi$ and average its rank-one orthogonal projection over Haar measure on $E$:The integrand is trace-norm continuous, so $K$ is positive compact with trace one, and commutes with $R(E)$. A positive eigenvalue has a nonzero finite-dimensional eigenspace invariant under $E$. Commuting unitary matrices on that space have a common unit eigenvector $\eta$. Therefore $R(e)\eta =\chi (e)\eta$ for a continuous character $\chi$ of $E$.

On the central fibre, $R(1,\lambda I)=\lambda I$, so $\chi (1,\lambda I)=\lambda$. Consequentlyis independent of that lift. It is a homomorphism and strongly continuous, because the continuous map on $E$ descends through its quotient to the base. It satisfies $p(U_z)=\rho (z)$. This proves circle projective representations lift to unitary representations using an actual central character, rather than assuming a global choice of phases.

Let $v$ be a real unit vector in $d$-dimensional Euclidean space. Then $P=v{v}^{\mathsf{T}}$ is a rank-one orthogonal projection: $Pv=v$, while $Pw=0$ for $w$ orthogonal to $v$. Consequently the symmetric matrix $A=\alpha I+\beta P$ has eigenvalue $\alpha +\beta$ on the line spanned by $v$ and eigenvalue $\alpha$ on its orthogonal complement. Any orthonormal basis of that complement completes $v$ to an orthonormal eigenbasis. ThusThe trace and determinant formulas remain valid when one of these eigenvalues is zero. When $\beta =0$, all nonzero vectors have the same eigenvalue, and the chosen splitting is optional.