3
$\begingroup$

I would like to know more about the geometry of $\mathbb{R}^2$ equipped with the following inner product $(\mathbf{v},\mathbf{u})=\|\mathbf v\|\cdot \|\mathbf u\|\cos(2\alpha)$, where $\alpha$ is the angle between the vectors $\mathbf v$ and $\mathbf u$. This is not a true inner product since $(a\mathbf v,\mathbf u)=|a|(\mathbf v,\mathbf u)$. In a way, this is an inner product on the space of directions in $\mathbb{R}^2$. Has this been studied, and where could I start looking into literature? I am particularly interested in the possibility of representing the space in terms of spinors.

  • 0
    This looks like a norm on $\mathbb R^2$ that is not induced by an inner product (I haven't checked if this satisfies the triangle inequality). If so, then it induces the usual topology on $\mathbb R^2$, because all norms on finite-dimensional spaces are equivalent.2012-08-21

1 Answers 1

1

Assume this really defines an inner product $\langle\ ,\ \rangle$.

Consider $u=(1,0)$ and $v=(0,1)$. Then the angle between $w=u+v$ and $u$ and the angle between $w$ and $v$ are both $\alpha=\pi/4$ and $\cos(2\alpha)=0$ hence $\langle u,w\rangle=\langle v,w\rangle=0$, which implies that $\langle w,w\rangle=\langle u,w\rangle+\langle v,w\rangle=0$.

This is a contradiction because $\|w\|\ne0$ and the angle that $w$ makes with $w$ is $\beta=0$, and $\cos(2\beta)\ne0$, hence $\langle w,w\rangle=\|w\|\cdot\|w\|\cdot\cos(2\beta)=\|w\|^2\ne0$.

  • 0
    Sorry if I offended you, it was not my intention. I know my question is very vague and that makes it a bad question. That's what happens with vaguely defined research projects :-)2012-04-26