Yesterday Paul Lai asked the following

**Question**: Suppose there are distinct points in , can we find a Cartesian coordinates such that for all and distinct ?

As expected, the answer is positive. After discussing with Paul Lai, we arrived at the following

Theorem 1 (Paul Lai’s theorem, 2011)For distinct points in , there are infinitely many Cartesian coordinates systems such that for all and distinct .

(In this note, for a line we mean a one-dimensional subspace (of some ), and by a hyperplane we mean a codimension one subspace (e.g. a plane inside a 3-dimensional space). In particular they all contain . )

*Proof:* First of all note that for distinct and , if and only if and both lies on a coordinate hyperplane , and this is equivalent to the coordinate vector lies on the hyperplane .

Now, since there are only finitely many , the problem is equivalent to finding orthonormal bases such that no lies on a given collection of finitely many hyperplanes. (In our situation, they are and there are at most of them. ) We will use induction on to prove this claim, more precisely we will show: given finitely many hyperplanes in , there are infinitely many orthnormal bases such that no lies on any of these hyperplanes.

For , the hyperplanes are just lines, the collection of all these lines and their orthogonal complements contains only finitely many lines. For any line , , and clearly there are infinitely many such . We can then find a unit vector which generates and which generates . is then our desired basis.

Assume the claim is true up to dimension . Now, let be a line in which does not lie on any of the given hyperplanes and also is not perpendicular to any of them (there are only one line perpendicular to each hyperplane), then is not any of the . In particular, each is a hyperplane in . So by induction assumption we can find infinitely many basis for which no lies on any . We then find a generator for and is the desired basis.

Actually we can do more. Intuitively, for “almost all” choices of the Cartesian coordinates system (i.e. except some very special choices), the conclusion in Theorem 1 holds, i.e. the coordinates of all points are sufficiently distinct. This can be formulated more rigorously as follows.

As remarked in the previous proof, the problem is equivalent to finding (ordered) orthonormal bases which avoids lying on some hyperplanes. The set of all ordered orthonormal basis is exactly the space of orthogonal matrices, as is seen by looking at each column vector of the matrices. It is not hard to see that is a smooth closed (i.e. comapct without boundary) submanifold of dimension (the exact dimension is not important) of the space of all matrices, identified with . Therefore inherits a dimensional volume element (or Hausdorff measure) on it. (I don’t know the relation between this measure and its bi-invariant Haar measure, it seems to me that is not necessarily left or right invariant. ) With this notations, we have

Theorem 2 (Paul Lai’s theorem, 2011)Given finitely (or countably) many hyperplanes in , for almost every , no column of lies in any of these hyperplanes.

Actually we will see in the proof that the exceptional set is nearly a codimension one submanifold, indeed it is a union of finitely (or countably) many codimension one submanifolds.

*Proof:* Clearly it suffices to assume that there is only one given hyperplane . The conclusion follows by the countable subadditivity of the Hausdorff measure. We can further assume that .

For , the first column of lies in if and only if . We will show that is a codimension one smooth submanifold of . Indeed by regular value theorem(a special form of implicit function theorem), it suffices to show that is a regular value of . To see this, suppose such that , i.e. . Clearly there is one , such that (why), we can w.l.o.g. assume . Then

is a curve in through , and

From this we see that is a regular value for , and thus is a smooth codimension one submanifold of . It follows that has dimensional measure zero. With the same argument we conclude that has measure .