2. Grassmannians Reference for This Section: [Hat, Section 1.2], [MS74, Section 5] We Now Turn to a Vital Example of a Vector Bu

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2. Grassmannians Reference for this section: [Hat, Section 1.2], [MS74, Section 5] We now turn to a vital example of a vector bundle which will play a significant role in our future discussions. k k The GrassmannianG n(R ) is the manifold ofn-planes inR . As a set it consists of alln- dimensional subspaces ofR k. To describe it in more detail we mustfirst define the Steifel manifold. k k Definition 2.1. The Stiefel manifoldV n(R ) is the set of orthogonaln-frames ofR . Thus the k k points of it aren-tuples of orthonormal vectors inR . We can think ofV n(R ) as a subset of k 1 n (S − ) ; we let it inherit its topology from this space. We also note that it is a closed subspace of k a compact space, soV n(R ) is compact. k k We have a mapV n(R ) Gn(R ) which takes eachn-frame to the subspace it spans. We let k k k Gn(R ) be construted via the quotient topology fromV n(R ). ThusG n(R ) is also compact. k Lemma 2.2.G n(R ) is Hausdorff. Proof. To prove thatG (Rk) is Hausdorff it suffices to show that for any twon-planesω ,ω n 1 2 ∈ G (Rk) there exists a continuous functionf:G (Rk) R such thatf(ω )=f(ω ). For any n n 1 � 2 pointp R k, letf :G (Rk) R setf (ω) to be the Euclidean distance fromp to theω. This is ∈ p n p continuous because for anyn-frame (v , . . . , v ) V (Rk) representingω we have 1 n ∈ n f (ω)= p p (p v )2 (p v )2. p · − · 1 −···− · n k This is clearly a continuous function�V n(R ) R which gives the same value on each preimage k ofω, so it is a continuous function n(R ) R. Now letp be any point inω 1 which is not inω 2. Thenf (ω ) = 0 butf (ω )= 0, andG we are done. p 1 p 2 � � k Lemma 2.3.G n(R ) is a manifold. k Proof. Letν be ann-plane inR , and letν ⊥ be its orthogonal complement (of dimensionk n). k − Then the set ofn-planes inR which do not intersectν ⊥ is homeomorphic to the set of graphs of linear mapsν ν⊥, which is the space ofn (k n) matrices. This gives a homeomorphism of n(k n) × − a neighborhood ofν toR − , as desired. � k SinceG n(R ) is Hausdorff we can try to construct a CW structure on it. This is relatively simple k i once wefigure out the correct cells to look at. Letp i:R R be the projection onto thefirst i coordinates; thusp k is the identity andp 0 is the trivial map to the point. Asi goes fromk to 0 the dimension of the image of ann-planeω G (Rk) drops fromn to 0; letσ be the smallest ∈ n i integer such that dimp i(ω)=i. The sequenceσ=(σ 1,...,σ n) is called a Schubert symbol. If we k lete(σ) be the subset ofG n(R ) havingσ as their Schubert symbold we note that these are spaces whosen-frames, after reducing into Eschelon form, have columnsσ 1,...,σ n as the pivots; all entries which are not pivot columns can have any real number they wish as the entries, so this subspace is n homeomorphic to a Euclidean space (open cell) of dimension i=1(k (σ i 1) (n i + 1)). The maximal value of this isn(k n). − − − − − � Remark 2.4. Note that this does not rely on any properties ofR other than thatR m is homeomorphic to an open cell of dimensionm. Thus we could have done the exact same analysis for complex Grassmannians inC n and we would have obtained a similar CW structure with cells of slightly different dimensions. n n+1 We have inclusionsG (R ) G (R ) G (R ), whereR = ∞ R induced by n ⊆ n ⊆···⊆ n ∞ ∞ i=1 the inclusionsR n R n+1 as those vectors with 0 as their last coordinate. Each of these respects the CW structure⊆ as defined above, so they are CW inclusions; this produces� a CW structure on Gn(R∞). 6 def Definition 2.5. We defineG n =G n(R∞). Since a Grassmannian is a space encoding information about vector subspaces it comes with a natural definition of a vector bundle. k Definition 2.6. The universal bundleγ nk is a bundlep:γ n Gn(R ) which has over every point the plane that the point represents. More concretely, γ = (ω,v) ω G (Rk), v ω R k G (Rk) R k. nk { | ∈ n ∈ ⊆ }⊆ n × k The mapp:γ nk Gn(R ) is projection onto thefirst coordinate. Whenk= we simply writeγ n instead ofγ n . ∞ ∞ k k Example 2.7. Letn = 1. ThenG 1(R ) isRP ,k-dimensional projective space. We prove that γ1k is nontrivial. Note that a trivial line bundle has a section which is everywhere nonzero; we k show that this is not the case forγ 1k. Suppose thats:RP γ1k is any section, and consider the k k compositionS RP γ1k where thefirst map is the usual double cover. This takes a point x to a pair ( x , t(x)x) for some continuoust:S k R. Note thatt( x) = t(x). SinceS k is {± } − − connected we must havet(x ) = 0 for somex S k. 0 0 ∈ Wefinish up this section with an important observation about Grassmannians. Definition 2.8. LetG be a topological group, and letEG be a weakly contractible space with a freeG-action. The spaceBG is defined to be the quotient ofEG byG.

WhenG is discrete,BG hasπ 0BG= ,π 1BG=G andπ nBG = 0 forn> 1. However, for groups with other topologies this is often∗ not the case. In particular, whenG=O(n) it is still an open question what the homotopy groups are. Theorem 2.9. G BO(n). n � n n+1 Proof. As we had for the Grassmannians, we have a sequence of inclusionsV n(R ) V n(R ) def ⊆ ⊆ V n(R∞). As before, we writeV n =V n(R∞). We have the following facts: ···⊆ k O(n) acts freely onV n(R ). • k k G n(R ) is the quotient ofV n(R ) by the action ofO(n). • The action respects the inclusions ofV ’s andG ’s. • n n Therefore if we can show thatV n is contractible we will be done. k k 1 The mapV n(R ) S − given by projecting ann-frame onto its last vector is aﬁber bundle k 1 k 1 withﬁberV n 1(R − ) (where thisR − is the hyperplane orthogonal to the last vector). Thus there exists a− long exact sequence in homotopy groups k 1 k 1 k k 1 πm+1S − πmVn 1(R − ) πmVn(R ) πmS − . ··· − ··· k 1 k 1 k Sinceπ mS − = 0 form

∗This fact is actually more complicated than it looks. [Hov99, Proposition 2.4.2] shows that this fact holds for compact spaces assuming that the sequential limit is given by closed inclusions of Hausdorﬀ spaces. 7 since theﬁrst map is null-homotopic, the entire map must be as well. Thusπ mVn = 0, as desired. �

Remark 2.10. If we look at complex Grassmannians the same proof works to show thatG n(C∞) BU(n). � This is just a surprising fact for now, but it will become very important later in our discussion ofK-theory.