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.

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.

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.

  • 2
    $\begingroup$ Usually "choosing a canonical representative" need not be possible without tearing things apart. Example: $Gr(1,3)=\mathbb{RP}^2$ can be thought of as "choosing" $3\times 1$ matrix but of course we know $\mathbb{RP}^2$ does not embed into $\mathbb{R}^3$. Embedding a quotient typically needs some room to wrap your original space around so you can make identifications. $\endgroup$ – user10354138 May 14 at 3:40
  • $\begingroup$ Oh, and of course we can project (in the same way as Whitney, or the AG equivalent) to get to the dimension of embedding theorems. $\endgroup$ – user10354138 May 14 at 4:05
  • $\begingroup$ @user10354138, that's a very good point. I am actually more interested in "weighted, signed subspaces," to be honest. This would have dimension $r(k-1)+1$. The Plucker embedding is one way to represent those, but it's very space-inefficient. We could use full-column-rank orthogonal matrices, but then the representation is only unique up to rotation... $\endgroup$ – Mike Battaglia May 14 at 22:22
  • $\begingroup$ I am wondering a simple way to phrase my question. Rather than asking about embedding a "weighted, signed Grassmannian" into Euclidean space, would it be equivalent and simpler to ask about embedding the Grassmannian into projective space? $\endgroup$ – Mike Battaglia May 14 at 22:23

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