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A common way to put coordinates on $\mathbb P^k\mathbb R$ is to choose $k+2$ points (such that no one of them lies on the hyperplane generated by any $k$ of the others) and interpret them as the origin $(0, 0, \dots, 0, 1)$, the unit versors $(1, 0, \dots, 0, 1)$, $(0, 1, \dots, 0, 1)$, ..., $(0, 0, \dots, 1, 1)$ and the unity point $(1, 1, \dots, 1, 1)$ of an Euclidean space $\mathbb R^k \subset \mathbb P^k \mathbb R$ (with the immersion $(x_1, \dots, x_k) \mapsto (x_1, \dots, x_k, 1)$).

Another way to specify a projective reference is to take the points corresponding to the origin and the unit versors of $\mathbb R^k$ and specify furthermore the coordinate infinity points $(1, 0, \dots, 0, 0)$, $(0, 1, \dots, 0, 0)$, ..., $(0, 0, \dots, 1, 0)$, with of course the prescription that each of them lies on the line connecting the origin with the appropriate unit versor (and does not coincide with any of them).

Suppose now that you want to pass from one system to the other using only synthetic methods (that is, just by computing spans and intersections of projective subspaces). This means that you want to recover the unity point knowing the coordinate infinity points (and the origin and unit versors) and viceversa.

It is easy to build the unity point: you take the intersection of all the $k$ planes spanned by $k-1$ coordinate infinity points and the unit versor corresponding to the missing direction.

However, I could not devise a synthetic construction to obtain the coordinate infinity points from the unity point. Is anybody able to provide one? (or provide some proof that there cannot be one, although that would appear really strange to me)

Remark. If you have the coordnate infinity points and the unit versor, the origin is redundant (since all the axes can be recovered directly and then intersected). Here I am not bothering with optimality, uniqueness or things like that, I am just curious of how to construct the coordinate infinity points.

Giovanni Mascellani
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1 Answers1

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If I understood your question well, it is enough to exhibit $k$ independent points on the hyperplane at infinity: then their span is the hyperplane at infinity and you just need to intersect this hyperplane with every line joining the origin and one of the unit versors.

To do it, consider the plane spanned by the origin and $k-1$ of the unit versors, say all but the $i$-th unit versor, and call this hyperplane $H_i$. Consider also the line $l_i$ joining the unit point and the $i$-th versor.

This line clearly doesn't lie on $H_i$ since it contains at least two points not on $H_i$. So there's a point of intersection $H_i\cap l_i$. If the $i$-th unit versor is $(0,\dots,1,\dots, 0,1)$ with the first $1$ in the $i$-th position, then $H_i\cap l_i$ is exactly point $(1,1,\dots,1,0,1,\dots,0)$, with the first $0$ in the $i$-th position: this point lies, as you can see, at infinity, and letting $i$ vary you get independent points.

Andrea Bianchi
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