MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x$ and $y$ be two points on the unit sphere $S^{n-1}$ in Euclidean space ${\mathbb{R}}^n$. Suppose that the angle $\theta$ between the points $x$ and $y$ is acute, so that the dot product $x\cdot y=\cos\theta>0$, where we treat $x$ and $y$ as unit vectors. The angle $\theta$ can be interpreted as the geodesic distance between $x$ and $y$ in the round metric on the sphere $S^{n-1}$.

The Cauchy inequality applied here states that $x\cdot y\le 1$. Equality holds if and only if $x=y$. (The case where $x$ and $y$ are antipodal is ruled out by the condition that the angle $\theta$ is acute.)

I'm curious as how we can sharpen this inequality to account for the situation when the geodesic distance $\theta$ is large. Noting that $\cos \theta < 1-\theta^2/2$ as $\theta$ is acute, this basic estimate geometrically becomes $x\cdot y <1-\theta^2/2$.

I'm wondering if my lower bound is the best possible. In other words, is it true that $1-x\cdot y=O(\theta^2)$ if the angle $\theta$ is acute?

Is there an analogue of this sharpened inequality when we consider points on the unit sphere $S^{2n-1}$ in complex Euclidean space ${\mathbb{C}}^n$ and apply the complex Cauchy inequality instead?

share|cite|improve this question
Since $x\cdot y = \cos \theta$, isn't the answer to your first question trivially yes by looking at the Taylor expansion of $\cos$? – Yemon Choi May 12 '11 at 6:44
(there are also lower-tech ways of seeing that $1 - \cos\theta = 2\sin^2\theta$ is $O(\theta^2)$) – Yemon Choi May 12 '11 at 6:44
I think the answer to the first question is yes; you can by symmetry always reduce to the $n=2$ case by rotating. – J.C. Ottem May 12 '11 at 6:45
Following JC's comment, it seems that the complex case can be reduced to the ${\mathbb{C}}^2$ by unitary transformations too. – user2529 May 12 '11 at 7:00
So the question is settled then? In $\mathbb{C}^2$ this is a simple computation.. – J.C. Ottem May 12 '11 at 8:55

We checked the book "Complex hyperbolic geometry" by Goldman. In there, it is given that for two points $z,w$ in complex projective space ${\mathbb{CP}}^n$, we have $\cos(d(z,w))=\frac{|\langle z,w\rangle|}{|z||w|}$. Here $d(z,w)$ denotes the geodesic distance on ${\mathbb{CP}}^n$ after a normalization.

Since $\cos\theta\le 1-\theta^2/2$, thus $\frac{|\langle z,w\rangle|}{|z||w|}\le 1-d(z,w)^2/2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.