Is there a complex analog of this sharpened Cauchy Inequality? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T10:42:56Z http://mathoverflow.net/feeds/question/64752 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64752/is-there-a-complex-analog-of-this-sharpened-cauchy-inequality Is there a complex analog of this sharpened Cauchy Inequality? Colin Tan 2011-05-12T06:32:37Z 2011-05-28T04:22:14Z <p>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 $S^{n-1}$. </p> <p>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.)</p> <p>I'm curious as how we can sharpen this inequality to account for the situation when the geodesic distance $\theta$ is large. Noting that $\cos \theta &lt; 1-\theta^2/2$ as $\theta$ is acute, this basic estimate geometrically becomes $x\cdot y &lt;1-\theta^2/2$.</p> <p>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?</p> <p>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? </p> http://mathoverflow.net/questions/64752/is-there-a-complex-analog-of-this-sharpened-cauchy-inequality/64959#64959 Answer by Colin Tan for Is there a complex analog of this sharpened Cauchy Inequality? Colin Tan 2011-05-14T03:31:20Z 2011-05-14T03:31:20Z <p>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.</p> <p>Since $\cos\theta\le 1-\theta^2/2$, thus $\frac{|\langle z,w\rangle|}{|z||w|}\le 1-d(z,w)^2/2$.</p>