# Geometric interpretation of the argument of the Fubini-Study bilinear form on projective space?

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. 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 $w$?

• Hello guys Can anyone help me to clarify, what this bilinear form means. How can we see it geometrically? It looks to me quite similar to standard scalar product, however we do not have the matrix in standard scalar product. Can you please help me if you understand my question? thank you for your answer John – user32661 Mar 30 '13 at 14:20

## 1 Answer

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.