A CW Structure on a Grassmannian Define the Grassmannian Gr K(Rn) to Be the Space of K Dimensional Vec- Tor Subspaces of Rn

A CW structure on a Grassmannian

n Deﬁne the Grassmannian Grk(R ) to be the space of k dimensional vec- n n n−1 tor subspaces of R . For example, Gr1(R ) = RP . The topology may n be given by expressing Grk(R ) as a quotient of the Stiefel manifold of orthonormal k frames in Rn,

n n Vk(R ) = {(v1, . . . , vk): vi ∈ R , vi · vj = δi,j} which is topologized as a closed subset of the product of k copies of Sn−1. It n and its quotient space Grk(R ) are compact Hausdorff spaces. The Grassmannian is very symmetric—it has a transitive action by the Lie group SO(n) of rotations in Rn—but to define a CW structure on it we must break this symmetry. This symmetry breaking occurs by picking a complete flag in Rn. Any one will do (and they acted on freely and transitively by SO(n)), so let’s just agree to use the flag determined by the standard ordered basis: so 0 = R0 ⊂ R1 ⊂ R2 ⊂ · · · ⊂ Rn i n where R is the subspace of R spanned by {e1, . . . , ei}. The ith coordinate i i−1 function xi maps R surjectively to R with kernel R . n Let V ∈ Grk(R ). Intersecting V with this flag gives a filtration of V ,

0 = V0 ⊆ V1 ⊆ · · · ⊆ V

i with Vi = V ∩ R . It’s still the case that Vi−1 = ker (xi : Vi → R), but the restriction of xi to Vi is no longer necessarily surjective. When it is zero, Vi−1 = Vi. When it is surjective, dim Vi = 1 + dim Vi−1. The k-plane V determines a weakly increasing sequence of integers

0 ≤ a1 ≤ a2 ≤ · · · ≤ ak ≤ n − k by requiring that i + ai is the smallest index j for which dim Vj = i. This sequence a = (a1, . . . , ak) is the type of V . Write ˚e(a) for the subset of n Grk(R ) consisting of all k-planes of type a. They will be the “open cells” n in a CW decomposition of Grk(R ). n n The flag determines a section of the projection Vk(R ) → Grk(R ) (which, to be sure, is only continuous when restricted to a fixed ˚e(a)), by assigning to a k-plane V the frame v1, . . . , vk, described as follows. The first vector, v1, is the unit basis vector for V1+a1 (which is one-dimensional) with positive

(1 + a1)st entry; the next vector, v2, is the unit vector in V2+a2 which is orthogonal to v1 and has positive (2 + a2)nd entry; etc. For example, the 2-planes in R4 are spanned by unique (orthonormal) frames which form columns in matrices of the following shapes.  p ∗   p ∗   p ∗   ∗ ∗   ∗ ∗   ∗ ∗   p   ∗   ∗   p ∗   p ∗   ∗ ∗    ,   ,   ,   ,   ,      p   ∗   p   ∗   p ∗  p p p where p denotes a positive real, ∗ any real, and empty spaces are 0. The types are (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).

As V runs over k-planes in ˚e(a), the vector v1 has a1 degrees of freedom: the ﬁrst a1 entries can make up any vector of norm less than 1, and the bottom entry is then determined. The vector v2 has 2 + a2 nonzero entries, but is subject to the two conditions that it be orthogonal to v1 and of unit length; so it enjoys a2 degrees of freedom. We thus expect dim˚e(a) = kak, where kak = a1 + ··· + ak Therefore we deﬁne

n [ SkdGrk(R ) = ˚e(a) kak=d

n Theorem. This filtration defines a finite CW structure on Grk(R ). Proof. We will construct a pushout diagram

` d−1 ` d kak=d Sa / kak=d Da

f g Skd−1 / Skd and show that for each a,(e(a), ∂e(a)) ∼= (Dd,Sd−1). Deﬁne e(a) to be the closure of ˚e(a) regarded as a subset of the product on k copies of Sn−1:

n e(a) = {(v1, ··· , vk): vi ∈ R , vi · vj = δi,j, vi · bi ≥ 0}

2 where bi = ei+ai . The subspace ∂e(a) is the subset where some vi · bi = 0. What needs to be proved is that e(a) ∼= Dd. We will prove that it is homeomorphic to a product of the disks

ai i+ai D = {v ∈ R : kvk = 1, bj · v = 0 for j < i, bi · v ≥ 0}

a1 Do this by induction on k. When k = 1, e(a1) = D . So we now need to 0 ak 0 construct a homeomorphism e(a ) × D → e(a), where a = (a1, . . . , ak−1). This homeomorphism will have the form

0 0 0 (v , v) 7→ (v ,Tv0 v) , v = (v1, . . . , vk−1) where Tv0 is an n × n rotation matrix such that

(k−1)+ak−1 k+ak Tv0 bi = vi for i < k , Tv0 x − x ∈ R for x ∈ R

Then vi · Tv0 v = Tv0 bi · Tv0 v = bi · v = 0,

(k−1)+ak−1 and, since bk is orthogonal to R ,

vi · Tv0 v = bk · v ≥ 0 which checks what is needed. The operator T is constructed as the composite

Tv0 = T (bk−1, vk−1) ··· T (b2, v2)T (b1, v1) where, for any two unit vectors b, v ∈ Rn with b + v 6= 0, T (b, v) is the rotation matrix which sends b to v and is the identity on the orthogonal complement of the span of b and v.