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A projective frame $\mathcal{F}$ of a projective space $P=P(E)$ defines a basis of the vector space $E$, defined up to multiplication by a scalar. The corresponding dual basis of $E^*$ is well defined up to a scalar, thus projectivizes to a unique projective frame of the dual $P^*$ of $P$, which shall be called the dual frame $\mathcal{F}^*$ of $\mathcal{F}$.

I woundered some time ago how one can construct $\mathcal{F}^*$ from the dual hyperplanes of the elements of $\mathcal{F}$. This is not a difficult question, but surprisingly it is neither completely trivial nor could I find a reference on this problem.

To see why it is not trivial, remember that a projective frame of dimension $n$ has $n+2$ elements; it is trivial to find the $n+1$ first elements of $\mathcal{F}^*$, but the last one is a little bit more difficult to construct.

I have a solution that seems nice to me, in part because it relates several different kind of projective polarities. I think it would be surprising that it is not already well-known, though.

Question: have you ever heard about this problem, and do you now a reference?

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A projective frame in n-dimensional projective space consists of n+1 general points and an additional special point. The frame determines an affine (and Euclidean) structure on the space. This affine structure is the one such that the special point becomes the barycenter of the other points. The affine structure picks out a hyperplane to be the hyperplane at infinity and this hyperplane is the one you seek.

The barycenter and the infinity hyperplane are constructable from each other and thus it makes sense to alternatively define a projective frame as n+1 points and a plane. The dual frame then consists of n+1 hyperplanes and a point.

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  • $\begingroup$ You are right; do you know some reference for this? By the way, there exist several other interpretation of this polarity (for example, by considering the faces of the simplex defined by the $n+1$ first points of the frame as an algebraic hypersurface of degree $n+1$). $\endgroup$ Jan 11, 2011 at 14:49

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