*2.2. Grassmann and Flag Manifold*

We will show that the map *φ* introduced in Theorem 1 has an interpretation in the projective setting. We still assume *n* ≥ *r* ≥ *s*. The Grassmann manifold Grass(*<sup>n</sup>*,*r*, C) is the set of *r*-dimensional subspace of C*<sup>n</sup>*. This is identified with

$$M(n, r, \mathbb{C})\_{\text{rk} \sim r} / GL(r, \mathbb{C}) \cong \text{Grass}(n, r, \mathbb{C}).$$

Every *r*-dimensional subspace of C*n* is spanned by *r* linear independent column vectors in C*<sup>n</sup>*.

The flag manifold Flag(*n*; *k*1, ... , *km*, C) is the set of flags of type (*k*1, *k*2, ... , *km*), which is defined to be a sequence of subspaces *V*1 ⊂ *V*2 ⊂···⊂ *Vm* of C*<sup>n</sup>*, where 1 ≤ *k*1 < *k*2 < ··· < *km* < *n*, with dim *Vi* = *ki* (*i* = 1, ... , *m*). Grassmann manifold is a special case of flag manifolds with *m* = 1. On the other hand, a flag variety is regarded as the incidence variety of the product of Grassmann manifolds. For example, Flag(*n*; *k*1, *k*2, C) = {(*<sup>V</sup>*1, *<sup>V</sup>*2) ∈ Grass(*<sup>N</sup>*, *k*1, C) × Grass(*<sup>N</sup>*, *k*2, C) | *V*1 ⊂ *<sup>V</sup>*2}. We have an isomorphism

$$(M(\mathfrak{n},r,\mathbb{C})\_{\mathfrak{rk}=r} \times M(r,\mathfrak{s},\mathbb{C})\_{\mathfrak{rk}=s})/(GL(r,\mathbb{C}) \times GL(\mathfrak{s},\mathbb{C})) \cong \text{Flag}(n;s,r,\mathbb{C}).\qed$$

In the following commutative diagram, each space in the upper line, which arises in Theorem 1, is a locally closed subset of an affine space, while each space in the lower line is a projective variety.

$$\begin{array}{ccccc} M(\mathsf{n},r,\mathsf{C})\_{\mathsf{rk}=r} & \longleftarrow & M(\mathsf{n},r,\mathsf{C})\_{\mathsf{rk}=r} \times M(r,\mathsf{s},\mathsf{C})\_{\mathsf{rk}=s} & \xrightarrow{\phi} & M(\mathsf{n},\mathsf{s},\mathsf{C})\_{\mathsf{rk}=s} \\ & \Big\downarrow & & \Big\downarrow & & \Big\downarrow \\ \mathrm{Grass}(\mathsf{n},r,\mathsf{C}) & \longleftarrow & \mathrm{FAlg}(\mathsf{n};\mathsf{s},r,\mathsf{C}) & \longrightarrow & \mathrm{Grass}(\mathsf{n},\mathsf{s},\mathsf{C}). \end{array} \tag{6}$$

The maps in the lower line are given by *V*2 ← (*<sup>V</sup>*1, *<sup>V</sup>*2) → *V*1. This double fibration is often used in Radon transform and Heck correspondence [9].

In the case *r* = 2, the map

$$\mathbb{S}: M(\mathfrak{n}, \mathbf{2}, \mathbb{C})\_{\mathrm{rk} = 2} \longrightarrow \mathrm{Alt}(\mathfrak{n}, \mathbb{C})\_{\mathrm{rk} = 2}$$

induces the Plücker embedding

$$\text{Grass}(n, 2, \mathbb{C}) \longrightarrow \text{Alt}(n, \mathbb{C})\_{\text{rk}=2}/\mathbb{C}^\times \subset \mathbb{P}^{n(n-1)/2 - 1}(\mathbb{C})\text{.}$$

### **3. The Model over Real Numbers**

We now consider the special case *n* = 4,*r* = 2,*s* = 1, and replace C by R. Let *J* = 0 1 −1 0 be the standard non-degenerate skew-symmetric matrix. Note that *JT* = −*J* and det *J* = 1. Let *g* be the diagonal matrix with diagonal entries (1, −1, −1, <sup>−</sup><sup>1</sup>). Finally, we put *g* = 1.

Most of the story in the previous section does hold over the real number field R as well. However, the disconnectedness makes things complicated. For example, although the map V : *M*(2, 1, C) −→ Sym(2, <sup>C</sup>)rk≤1 given by V(*X*) = *XX<sup>T</sup>* is surjective, the map V : *M*(2, 1, R) −→ Sym(2, <sup>R</sup>)rk≤1 is not surjective, because 0 0 0 −1 is not in the image. Inordertoimprovethisdefect,weintroduceanon-zeroscalarmultiplicationsothatwe

 modify the map V by V2 given below (7).
