Let $s\_0$ and $s\_1$ be the holomorphic sections of the tautological bundle $O(1)$ over the complex projective line ${\mathbb{CP}}^1$ which correspond to the functions $1$ and $\frac{x\_1}{x\_0}$ in the open set $U\_0= \{x\_0\neq 0\}$. Let $U(z,w)=s\_0(z)\overline{s\_0(w)} + s\_1(z)\overline{s\_1(w)}$ be the bilinear form corresponding to a Fubini-Study metric. Normalizing, we obtain a function $\tilde{U}(z,w)=\frac{u(z,w)}{\sqrt{U(z,z)U(w,w)}}$.


The modulus of $\tilde{U}$ has a geometric interpretation. I asked about this in [this question.](https://mathoverflow.net/questions/64752/is-there-a-complex-analog-of-this-sharpened-cauchy-inequality) By choosing a suitably normalized metric on ${\mathbb{CP}}^1$, we have the identity $|\tilde{U}(z,w)|^2= cos^2 d(z,w)$.


My question is whether the argument of $\tilde{U}$ has an anologous geometric interpretation in terms of "complex-angular" measure between the points $z$ and $w$? 



Geometric interpretation of the argument of the Fubini-Study bilinear form on projective space?
I'm not sure what you mean by complex-angular measure, but there is indeed a geometric interpretation.


Suppose $z$ and $w$ have homogeneous coordinates $[z\_0,z\_1]$ and $[w\_0,w\_1]$, respectively. Then
$$
U(z,w) = 1 + \frac{z\_1}{z\_0} \frac{\overline{w}\_1}{\overline{w}\_0},
$$
and hence
$$
\widetilde{U}(z,w) = \frac{\overline{z}\_0}{|z\_0|} \frac{w\_0}{|w\_0|}
\frac{z\_0 \overline{w}\_0 + z\_1 \overline{w}\_1}{\sqrt{(|z\_0|^2+|z\_1|^2)(|w\_0|^2+|w\_1|^2)}}.
$$
So the question is really asking what the complex phase of $\overline{z}\_0 w\_0 (z\_0 \overline{w}\_0 + z\_1 \overline{w}\_1)$ means.


To put this in a more general setting, let's look at $\mathbb{C}\mathbb{P}^{n-1}$. We'll represent points by unit vectors in $\mathbb{C}^n$, and we'll use the Hermitian form $\langle \cdot, \cdot \rangle$ on $\mathbb{C}^n$. As you pointed out, the absolute value of the inner product $\langle x,y \rangle$ measures the distance between $x$ and $y$ in $\mathbb{C} \mathbb{P}^{n-1}$. The complex phase of $\langle x,y \rangle$ is not well-defined, because we can rotate $x$ and $y$ by unit complex numbers without changing the points in projective space. However, the phase change along a cycle of points is invariant: in the expression
$$
\langle x,y \rangle \langle y,z \rangle \langle z,x \rangle,
$$
phase rotations for $x$, $y$, or $z$ won't change the answer (because they occur once on the linear side of the Hermitian form and once on the conjugate-linear side).


I'm not sure what to call this quantity. I think it's more or less the Pancharatnam phase from physics, but that's a little far afield from what I know. In any case, it gives an invariant for $k$-tuples of points under the action of the unitary group $U(n)$. For two points, you just get $\langle x,y \rangle \langle y,x \rangle = |\langle x,y \rangle|^2$, so it gives the Fubini-Study metric again. For more than three points, you get a perfectly good invariant, but it's not as essential because it can generically be reduced to $2$- and $3$-point invariants. For example,
$$
\langle w,x \rangle \langle x,y \rangle
\langle y,z \rangle \langle z,w \rangle =
\frac{\langle w,x \rangle \langle x,y \rangle
\langle y,w \rangle
 \cdot \langle y,z \rangle \langle z,w \rangle
\langle w,y \rangle
}{\langle y,w \rangle \langle w,y \rangle},
$$
as long as $w$ and $y$ aren't orthogonal. However, when $\langle w,y \rangle = \langle x,z \rangle = 0$, the $4$-point invariant is not determined by lower-order invariants (the $3$-point invariants all vanish, and the $2$-point invariants don't determine the complex phases at all).


The unitary group $U(n)$ maps one $k$-tuple of points in $\mathbb{C} \mathbb{P}^{n-1}$ to another if and only if all the invariants of corresponding sub-tuples agree, and as long as there are no orthogonal pairs of points, it suffices to get agreement for pairs and triples. So these invariants completely classify the orbits of $U(n)$ acting on finite subsets of $\mathbb{C}\mathbb{P}^{n-1}$.


The phase of $\widetilde{U}(z,w)$ in the original example is a special case of this invariant. It's the $3$-point invariant of $[z\_0,z\_1]$, $[w\_0,w\_1]$, and $[1,0]$ (assuming the Hermitian form is conjugate-linear in the second variable). Specifically, that gives
$$
(z\_0 \overline{w}\_0 + z\_1 \overline{w}\_1) w\_0 \overline{z}\_0,
$$
as desired. The appearance of $[1,0]$ here is a little arbitrary, but that's just a consequence of how the original problem was set up.


I've cheated a bit in one respect: the question asked about $\mathbb{C}\mathbb{P}^1$, and on $\mathbb{C} \mathbb{P}^1$ the higher-order invariants are not so exciting, because the orbits of $U(2)$ on arbitrary $k$-tuples of points are simply determined by the pairwise distances between the points. (It's essentially the same as $SO(3)$ acting on $S^2$, in which case this is a familiar fact.) So if we restrict attention to $\mathbb{C}\mathbb{P}^1$, then it's still true that the answer is a $3$-point invariant, but it's not really giving any new information beyond the Fubini-Study metric. However, in higher dimensions it does.


