Projective Geometry with Clifford Algebra - lost Inner Product Projective geometry may be studied with the tools of Clifford Algebras by adding a new direction (see for example this article). But as far as I understand it, only blades and null spaces are used for this description of projective geometry. So the inner product in the base vector space, whatever crucial for the clifford product, is no longer meaningful when working in projective geometry.
Are these statements right? Is projective geometry completely unrelated to the metric of the base vector space?
Any suggestion is welcome.
 A: Yes, when working with projective geometry using clifford algebra, one needs to take care that one's results do not depend on the metric of the base space.
However, that doesn't mean one has to avoid the metric completely; rather, one just needs to use metrical products in combinations that are non-metrical.
For example, consider a 2d projective geometry, embedded in 3d Euclidean space.  One can find the intersection of two affine lines as follows: model the two lines as bivectors $A, B$ and then find the dual of the commutator product:
$$p = I^{-1}(A \times B)$$
This does not depend on the metric of the base space--or at least, it doesn't depend on any more than the metric of the projected space.  If the third direction, orthogonal to the projected space, is annihilated in the commutator product, multiplication with the pseudoscalar puts it back.
So when doing projective geometry, we do not need to abandon metrical operations completely, but we do need to ensure that our final results have no net dependence on the metric of the base space.
