Let $Grass(r,k)$ be the set of all $r$-dimensional subspaces of $\Bbb R^k$.
It is well known that $Grass(r,k)$ embeds isometrically as a projective variety into the projectivization of the r'th power of the exterior algebra $\Lambda^r(\Bbb R^k)$; this is the Plücker embedding.
Another way to think of this is that it embeds "weighted, signed to Euclidean space, and I will often equivocate between these two representations.
This is nice and simple to work with, because you can just take the wedge product of a set of vectors to get a representation of the subspace that they span.
The problem is that this representation is extremely space-inefficient. We are taking an $r(k-r)$-dimensional manifold and putting it into a space of ${k}\choose{r}$. This is much larger than the minimum embeddings guaranteed by things like the Whitney and Nash embedding theorems. which can be thought of as guaranteeing a minimum dimension for a "Weighted Grassmannian" to be embedded in Euclidean space.
My question:
What is a better isometric embedding for the Grassmannian as a projective variety into projective real space?
I am mostly interested in representing "weighted, signed subspaces" - in the exterior algebra, if you can take the wedge product of a set of vectors, the $\ell_2$ norm of this multivector represents the volume of the parallelotope generated by the vectors, which is useful. But, I think this is equivalent to just finding a better embedding of the projective Grassmannian into projective real space and then making the homogeneous coordinates non-homogeneous.
This is a paper on on Cremona linearizations that says that there should be a birational mapping on the n'th exterior power that maps the embedded Grassmannian to a linear subspace. So for example, if you map $Grass(2,4)$ using this method, the Pfaffian will map to a linear subspace, so that you can quotient by it to get a space with 5 homogeneous coordinates theoretically equal to the dimension of the embedded Grassmannian as a projective variety.
I really don't know much about Cremona transformations but tried to follow the paper, which gives an explicit construction. It seems only to work for those elements of $\Lambda^r(\Bbb R^k)$ whose first element is nonzero, and I don't see how to get it to work otherwise. It would be nice to have something that works on any element of $\Lambda^r(\Bbb R^k)$, expressed in homogeneous coordinates. I'm also not sure if this is isometric.