Metric on the space $\lbrace (u,v) \in \mathbb{R}^3\times\mathbb{R}^3 , \langle u,v\rangle=0 \rbrace$

Question

Consider the following set : $$\Omega=\lbrace (u,v) \in \mathbb{R}^3\times\mathbb{R}^3 , \langle u,v\rangle=0 \rbrace$$

where $\langle ,\rangle$ is the canonical scalar product of $\mathbb{R}^3$.

• Except that it is the set of all pairs of orthogonal vectors in $\mathbb{R}^3$, what can be said about this specific space?
• Does exist any metric to define a distance $d(x,y)$ between two elements $x,y\in\Omega$?
• Is their any interesting geometrical or topological properties related to such mathematical object?

Answer 1

It is the set of all points $x=(x_1,...,x_6)\in\Bbb R^6$ with

$$x_1x_4+x_2x_5+x_3x_6=0.$$

You can write this as

$$x^T Mx=0,\qquad\text{with } M=\left(\begin{array}{ccc|ccc}&&&1\\&&&&1\\&&&&&1\\\hline 1\\&1\\&&1\end{array}\right).$$

This matrix provides an orthonormal basis of six eigenvectors to non-zero eigenvalues $\{+1,+1,+1,-1,-1,-1\}$. Instead of the standard basis of $\Bbb R^6$, express your vectors in this new basis. So if $x=(\hat x_1,...,\hat x_6)\in\Bbb R^6$ (the coordinates are w.r.t this new basis), then we have

$$x^TMx=0\quad\Longleftrightarrow\quad \hat x_1^2+\hat x_2^2+\hat x_3^2=\hat x_4^2+\hat x_5^2+\hat x_6^2.$$

So you also can say $\Omega=\{(u,v)\in\Bbb R^3\times\Bbb R^3\mid \|u\|^2=\|v\|^2\}$. This shows that $\Omega$ is topologically equivalent to $\Bbb R^3\times \Bbb S^2$ after associating all $\{0\}\times\Bbb S^2$ with a single point. So it is kind of a topological cone but not exactly.

Answer 2

• Question. Except that it is the set of all pairs of orthogonal vectors in $\mathbb{R}^3$, what can be said about this specific space?

• Answer. $\Omega$ is the kernel of bilinear form $K:\mathbb{R}^3\times \mathbb{R}^3\to \mathbb{R}$ such that $K(u,v)=\langle u,v\rangle $.

• Question. Does exist any metric to define a distance $d(x,y)$ between two elements $x,y\in\Omega$?

• Answer. Let $x=(u,v)$ and $y=(w,z)$. Then $$ d(x,y)=\sqrt{\|u-w\|^2_{\mathbb{R}^3}+\|v-z\|^2_{\mathbb{R}^3}} $$
• Question. Is their any interesting geometrical or topological properties related to such mathematical object?
• Answer. As pointed out by Giuseppe Negro, the set $\Omega$ is a quadratic surface in $\mathbb{R}^6$.

Answer 3

The set $\Omega_1=\{(u,v) \in \mathbb{R}^3: \langle u, v \rangle=0, \|u\|=1\}$ is the tangent bundle of $S^2$, so one could use the metric on $TS^2$ to define a metric on $\Omega_1$.

The tangent bundle of $S^2$ is diffeomorphic to the set $z_1^2+z_2^2+z_3^2=1$ in $\mathbb{C}^3$, and so another metric can be defined by using its embedding in $\mathbb{R}^6$.