An Elementary Introduction to the Grassmann Manifolds XX Brazilian Topology Meeting UFPR

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An elementary introduction to the Grassmann manifolds

XX Brazilian Topology Meeting UFPR - UTFPR, Curitiba, 2016

Carlos Eduardo Dur´anFern´andez

Universidade Federal do Parana´

Contents

Preface ix Chapter 1. Basic constructions 1 1.1. Linear Algebra 4 1.2. Topology 8 1.3. An Euclidean presentation of Grassmann manifolds 14 Chapter 2. Geometry 17 2.1. An application to ordinary diﬀerential equations 17 2.2. Homogeneous spaces and Cartan-Klein geometry 21 2.3. Congruence of curves in the Grassmann manifold 23 2.4. Jets of curves in Grassmann manifolds 27 2.5. Notable subsets of the Grassmann manifold 29 Chapter 3. Algebraic Topology 35 3.1. Combinatorial structure 35 3.2. Homotopy 42 Bibliography 51

vii

Preface

There is an enormous amount of mathematics that can be treated and exemplified in the Grassmann manifolds. The purpose of these notes is to give an introduction to the Grassmann manifolds that is as elementary as possible, while also showing the way to more advanced constructions. Thus, the present work can be considerd as a “pre- quel” to more specific treatments, for example the books of Milnor and Stasheff [20] for applications to the topology of vector bundles and carachteristic classes, Piccione and Tausk [23] for the applications to Geometry and Calculus of Variations, and [11] for the algebraic geometry viewpoint. We also introduce the reader to the invariant theory of curves in the Grassmannian [2, 7, 8]) and the Grassmannian of C∗-algebra [4, 24] The main leitmotiv is: how to actually get a grasp on a Grassmann manifolds and compute? In the author’s experience, the jump from abstract definition to confident computing, an ability needed for quality research, is a difficult one; here we describe how this jump can be made for Grassmann manifolds. The work is divided in three chapters: in the first one we give the basic definitions and, more importantly, models and parametrization of the Grassmann manifolds. Then in chapter 2 we keep identifying objects in the Grassmannian in terms of quantities living in more familiar spaces; in particular derivatives of curves and give applications to the congruence problem. Finally chapter 3 is intended to give a flavor to the finer topology of the Grassmann manifold: we give the standard CW decomposition in terms of Schubert cells (homology) and describe the isoclinic spheres that have been used as geometric representatives of stable homotopy groups. These notes are the companion of a minicourse given at the XX Brazilian Topology Meeting. I would like to thank Eduardo Hoefel and the organizers for the invitation to teach this minicourse.

Carlos Eduardo Dur´anFern´andez

CHAPTER 1

Basic constructions

n The Grassmann manifold Grk(R ) is the set whose points are the 1 k-dimensional subspaces of Rn. A look at the simplest case gives some flavor of what is to come: There is only a single 1-dimensional subspace of R1; therefore the 1 1 Grassmannian Gr1(R ) is just the singleton {R }; that was too simple. 2 Consider then the next case: the Grassmannian Gr1(R ); the interesting structure of the Grassmann already appears in this level. Each point of the this Grassmann manifold is a straight line through the 2 2 origin in R . We now describe Gr1(R ) in terms we can manage: We can first think of describing a point ` in this Grassmannian by its equation, `(m) = {(x, y) ∈ R : y = mx} , where m is the slope of the line. Thus we are lead to think that the 2 Gr1(R ) is parametrized by the real line R through the slope, ` 7→ m. This is almost right: this parametrization misses the vertical line {x = 0}. We deal with this problem as follows: not that `(m) approximates the vertical line as m → +∞, and also as m → −∞. This suggests the following: extend the real line by a single “point at infinity” ∞, 2 and then Gr1(R ) is completely parametrized by this space; in a more pedantic language, we have Proposition 1.1. The map R 3 m 7→ `, where ` is the unique straight line through the origin with slope m, and ∞ 7→ {x = 0} gives 2 a natural bijection between R ∪ ∞ and the Grassmannian Gr1(R ). Remark 1.2. The slope m of a non-vertical line ` is the unique number characterized by the relation (1, m) ∈ `. Here the unreachable character of vertical line is patent: a vector (1, m) can never belong in the line {x = 0}. A similar phenomenom with more involved combina- torics will be studied in the next section and chapter 3.

1Clearly, this definition can be extended: we can talk about the Grassmann manifolds of quite general vector spaces, e.g., [4], where they consider the Grass- mann manifold of certain infinite dimensional vector spaces. We can also consider vector spaces over fields other than the real numbers R. However in this introduction we prefer to fix ideas by just considering Rn. 1 2 1. BASIC CONSTRUCTIONS

Note that, since large (in absolute value) slope m makes `(m) close to the vertical line, we must think of the point at inﬁnity as being close to large slopes, and thus we have that the Grassmann manifold 2 Gr1(R ) is a circle: the two points of the drawn semi-circle represent the same line {y = 0} and therefore must be identiﬁed. The meaning

2 Figure 1. Gr1(R ) is a circle of “is a circle” above will be made precise in section 1.2. Let us try now a diﬀerent approach that bypasses the problem of the vertical line: consider now a line given by its symmetrical equation,

`(α, β) = {(x, y) ∈ R : αx + βy = 0} . Now, as it should be, the vertical line {x = 0} is on equal footing as with the other. The numbers α and β are arbitrary, with the exception of α = β = 0 (since 0 = 0 describes all of R2 and not just a line). However, we have introduced a new problem: the lines

`(2, 3) = {(x, y) ∈ R : 2x + 3y = 0} and `(4, 6) = {(x, y) ∈ R : 4x + 6y = 0} are the same line. As is the usual praxis in such cases, we deal with this ambiguity by ﬁxing an equivalence relation.

Exercise. Show that `(α, β) = `(γ, δ) if and only if there is 0 6= λ ∈ R such that λα = γ, λβ = δ. (The same λ must work simultaneously for both conditions!) Thus we have 1. BASIC CONSTRUCTIONS 3

Proposition 1.3. The map [(α, β)] 7→ `(α, β) induces a natural bi- 2 2 ~ jection between Gr1(R ) and the set of equivalence classes R −{0}/ ∼, where (α, β) ∼ (γ, δ) if and only if there is 0 6= λ ∈ R such that λα = γ, λβ = δ. 2 How do we recognize Gr1(R ) as a circle in this presentation? In order do accomplish this, we need to give a slightly diﬀerent -more precisely, a dual- interpretation of the bijection above. Two points uniquely determine a straight line; since we already now that the origin (0, 0) is a point through which the line passes, we only need another diﬀerent point (a, b). Two such points (a, b) and (c, d) determine the same line if and only if they satisfy the same kind of relation as in proposition 1.3: there is 0 √6= λ ∈ R such that λa = c, λb = d. Given (a, b) 6= (0, 0), taking λ = ( a2 + b2)−1 we get an element in the same equivalence class that lies in the unit circle S1, and these elements cover all the equivalence classes. Thus proposition 1.3 takes the form Proposition 1.4. The map (a, b) 7→ `(a, b), where (a, b) ∈ S1, 2 induces a natural bijection between Gr1(R ) and the set of equivalence classes S1/ ', where (a, b) ' (c, d) if and only if (a, b) = ±(c, d).

2 1 Figure 2. Gr1(R ) is identified with S /± 2 Proposition 1.4 tells us that Gr1(R ) is in a natural bijective correspondence with the circle S1 with antipodal points identified. This is easily seen to be a circle again: all equivalence classes are contained in the right half circle {(a, b) ∈ S1 : b ≥ 0}, and now each (a, b) represents exactly one equivalence class with the exception of (0, ±1) which represent the same vertical line. The fact that this is a circle is left as a visualization exercise for now; later in section 1.2 we will construct the tools necessary to give precise statements on these identifications of the Grassmann manifolds. 4 1. BASIC CONSTRUCTIONS

Exercise 1.5. Relate all the presentations of the Grassmannian 2 Gr1(R ) above; that is, how is the same line ` represented with the diﬀerent coordinates given by propositions 1.1, 1.3 and 1.4?

n+1 Remark 1.6. The Grassmannians Gr1(R ) are called real projective spaces and are denoted by RP n. In general, this kind of process of “describing an object in manageable terms” is usually called coordinatization or parametrization. We have thus “coordinatized the real projective line RP 1”.

1.1. Linear Algebra We now begin the study proper of the generic Grassmann mani- n folds. The methods used to coordinatize Grk(R ) are similar in spirit to the introduction on RP 1 above, with the difference that the previous section was written geometrically (note that we wrote about “lines” instead of “one-dimensional subspaces”), whereas for general Grassmann manifolds we use a more involved linear algebraic approach. Consider first the construction used in proposition 1.4, but without normalizing. Note that we tagged a line by giving a non-zero vector on the line; that is, we gave a basis for the given one-dimensional subspace. Let us adapt this method for the general case: given a k-dimensional n subspace p of R , we can choose an ordered basis (~v1, . . . ,~vk) of p; this motivates the following definition:

n Definition 1.7. The Stiefel manifold Stk(R ) is the set of k-tuples of linearly independent vectors in Rn. n n ~ Remark 1.8. The Stiefel manifold St1(R ) is just R − {0}. Remark 1.9. In geometrical contexts, an element of a Stiefel manifold is sometimes called a frame.

n n There is a tautological function π : Stk(R ) → Grk(R ) given by

π(~v1, . . . ,~vk) = span(~v1, . . . ,~vk) , that is, π sends A to its column space. We now face the same problem as in the RP 1 case: algebraically, when do two elements (~v1, . . . ,~vk) and (~w1, . . . , ~wk) generate the same n subspace? The answer is easily found if one writes Stk(R ) adequately: n we organize an element (~v1, . . . ,~vk) ∈ Stk(R ) as the (n × k)-matrix A whose columns are the vector vi: A = ~v1| ... |~vk , 1.1. LINEAR ALGEBRA 5 where the vertical bars just denote that we are emphasizing the matrix A written as columns. The condition of the vectors being linearly independent translates to the matrix A having maximal rank k. From now on each time we mention the Stiefel manifold it will be organized in matrix form as described above. Now reflecting on the way matrices are multiplied, we observe that if we transform A with right multiplication, that is, if B = AX, where X is a (k × k) matrix, the columns of B are linear combinations of the columns of A, and thus the column space of B is contained in the column space of A. The matrix X is invertible if and only if the columns spaces of A and B are actually the same. Then it is easy to show that n Proposition 1.10. The map π induces a bijection between Grk(R ) n and the quotient Stk(R )/ ∼, where A ∼ B if and only if there is an invertible (k × k) matrix X such that BX = A. Note that the multiplication by λ on the previous section is on the left because there we wrote the vector as row vectors instead of column vectors. In the presence of the standard Euclidean structure on Rn, the reduction to S1 is extended as follows: n n Definition 1.11. The orthogonal Stiefel manifold Vk(R ) ⊂ Stk(R ) > is the set of (n×k)-matrices A satisfying A A = Ik, the (k×k) identity matrix. n n Remark 1.12. The orthogonal Stiefel manifold V1(R ) ⊂ St1(R ) is just the unit sphere Sn−1 ⊂ Rn − {~0}. We now have the exact generalization of proposition 1.4: n Proposition 1.13. The map π induces a bijection between Grk(R ) n and the quotient Vk(R )/ ', where A'B if and only if there is an orthogonal (k × k) matrix X such that BX = A. Exercise 1.14. Construct the appropriate quotient and prove proposition 1.13 for an arbitrary Euclidean inner product on Rn. Let us now see how the “slope” coordinatization of proposition 1.1 generalizes. Recall that in the introductory RP 1-case, finite slopes did not reach all planes, and the set of reachable planes was identified with R. The following definitions generalize these sets of “reachable” and “unreachable” planes:

Definition 1.15. The top cell UV associated to V is the set n ~ UV = {p ∈ Grk(R ): p ∩ V = {0}} . 6 1. BASIC CONSTRUCTIONS

The train TV of V is the complement of the top cell, that is, n TV = {p ∈ Grk(R ) : dim(p ∩ V ) > 0} . The low dimension of the RP 1 case hides the complexity of the train: the train of the vertical line {x = 0} in R2 is the line itself. For higher dimension, the train is stratified by the increasing dimensions of p ∩ V . n Let us fix these ideas with some concrete planes in Grk(R ): fix the n plane H = span(~e1, . . .~ek) ∈ Grk(R ), where ~ei denotes the canonical basis of Rn; that is, n H = {~v = (x1, . . . , xn) ∈ R : xi = 0 for i > k} . The plane H should be visualized as the “horizontal” line {y = 0} of the previous section on RP 1. In that first instance, the vertical line {x = 0} could not be reached by the slope parametrization. A similar situation occurs here, with the difference that the “problematic” set is combinatorially more involved. n n Consider the complementary plane V ∈ Grn−k(R ), H ⊕ V = R , n V = {~v = (x1, . . . , xn) ∈ R : xi = 0 for i ≤ k} , which is the analog of the vertical plane {x = 0} of the RP 1 case (but note that here the dimensions of the horizontal and vertical planes are possibly different!) We now follow the lead of remark 1.2: there are many “slopes” with respect to H and V which collectively characterize the planes in the top cell UV . Indeed, we have n Lemma 1.16. Let p ∈ Grk(R ) be a plane in the top cell of V . Then for each i = 1, . . . , k, there is a unique vector ~zi = (0,..., 0, z1i, . . . z(n−k)i) ∈ V such that ~ei + ~zi ∈ p. The ordered set

{~e1 + ~z1, . . . ,~ek + ~zk} is a basis of p. Proof. Consider the projection π deﬁned by

π(x1, . . . xk, xk+1, . . . , xn) = (x1, . . . xk, 0,..., 0) , which has image equal to H and kernel equal to V . Since p ∩ V = {0}, ker π|p = {0} and then π|V : p → H is an isomorphism. Let ~vi = −1 (π|p) (~ei). By construction, vi ∈ p is uniquely determined by ~ei and has the form ~ei + ~zi, ~zi ∈ V . Lemma 1.16 furnishes a great oportunity to introduce a recurring theme in computations in Grassmann manifolds: writing conditions in terms of the Stiefel manifold. Let p ∈ Grkn, and A = (~v1, . . .~vk) ∈ 1.1. LINEAR ALGEBRA 7

n Stk(R ) be a matrix representing p, that is, the columns of A span p. How is the condition “p is in the top cell UV ” expressed in terms of A? We distinguish the “top” and “bottom” parts of A, the   v11 v12 ··· v1k  . . .   . . .     vk1 vk2 ··· vkk  A =   v(k+1)1 v(k+1)2 ··· v(k+1)k  . . .   . . .  vkn vkn ··· vkn and we observe that p ∩ V 6= {0} means that a linear combination of the columns of A is in V , that is, it has zeros in the ﬁrst k components. That means that p ∈ UV if and only if the square matrix

v11 v12 ··· v1k . . . X =  . . .  vk1 vk2 ··· vkk is invertible. Note that the condition of the top square part of A being invertible only depends on the span of the columns; that is, it is a n n condition on Stk(R ) that “descends” to a condition on Grk(R ). Using proposition 1.10, we see that the columns B = AX−1 also span p, and this matrix has the form  1 0 ··· 0   0 1 ··· 0   . . .   . . .   . . .    B =  0 0 ··· 1  z z ··· z   (k+1)1 (k+1)2 (k+1)k  . . .   . . .  zkn zkn ··· zkn where the zij are exactly the ones appearing in lemma 1.16. We have, in fact, constructed canonical way of uniquely writing a plane in the n top cell UT as the class of a matrix in Stk(R ); we have then proved n n Theorem 1.17. Let H ∈ Grk(R ), V ∈ Grn−k(R ) be complementary subspaces as above, and let Mn−k,k(R) denote the space of (n−k)×k n matrices. Then the map φH,V : Mn−k,k(R) → Grk(R ) given by 1 φ (Z) = column space of k×k H,V Z is injective, and its image is the top cell UV . 8 1. BASIC CONSTRUCTIONS

Note that Z = 0 corresponds to H. This is almost the generalization of proposition 1.1; there we established that the only missing line, the vertical line, could be safely assigned the slope ∞; whereas in proposition 1.17 the missing planes, which are exactly the members of the train TV , have much combinatorial complexity and will be studied in chapter 3. Note that H corresponds to Z = 0 in this parametrization. n We ﬁnish this section on linear-algebraic parametrizations of Grk(R ) with a change of point of view. Let S : H → V by a linear transfor- n mation. The graph of S is the set ΓS ⊂ H ⊕ V = R given by

ΓS = {x + S(x): x ∈ H} . n The graph is then a map Γ : Lin(H,V ) → Grk(R ); note that Γ0 = H. If we fix the canonical basis of H and V above, then linear transformations correspond to (n − k) × k matrices; then lemma 1.16, can be interpreted as a linear transformation Sp : H → V defined by the linear extension of Sp(~ei) = ~zi and extend by linearity. n n Theorem 1.18. Let H ∈ Grk(R ), V ∈ Grn−k(R ) be complementary subspaces as above. Then the graph map Γ: Lin(H,V ) → n Grk(R ), is injective, and its image is the top cell UV . Exercise 1.19. Prove the results in this section with the words “as n n above” removed, that is, with H ∈ Grk(R ), V ∈ Grn−k(R ) arbitrary complementary subspaces. Exercise 1.20. Repeat exercise 1.5 in the general situation given here. 1.2. Topology Up to now we have refrained on purpose to include any topological concepts; hence the word “bijection” in many of the results in n the previous sections. We now introduce a topology on Grk(R ); this topology should reflect our intuitions of k-planes through the origin being “close”. Let us give a brief review of the quotient topology, which is essential in the study of the topology of the Grassmann manifolds. Let X be a topological space, ∼ an equivalence relation on X and Y = X/ ∼ the quotient, that is, the set of all equivalence classes. The projection to the quotient is the map π : X → Y that sends an element to its class. The quotient topology on Y is defined as follows: a set U is open in Y if and only if π−1(U) is open in X. It is then, by definition, the largest topology that makes the projection to the quotient continuous. The key hint on thinking on quotient spaces is: don’t. Even when working on the quotient, one usually visualizes everything in the space 1.2. TOPOLOGY 9

X. Let us begin with the open sets: it is not true that the open sets of Y are projections of open sets of X; but this can be remedied: an set C in X is saturated if π−1(π(C)) = C, which translates to the property: if C contains a point x then it also contains all points x1 with x1 ∼ x. For the equivalence relation ~0 6= (x, y) ∼ (λx, λy) of the dual of proposition 1.3, the following picture illustrates the saturation concept:

Figure 3. A saturated open set for the quotient R2/ ∼

It is quite useful to characterize topological spaces by the set of its continuous functions. For quotient spaces, we have Proposition 1.21. Let π : X → X/ ∼ be the projection to the quotient. The quotient topology is characterized by: a function f : X/ ∼→ Z is continuous if and only if the composition f ◦ π : X → Z is continuous. Proposition 1.21 in practice means that the best way of defining functions from quotients X/ ∼ is to define them as functions on X that are constant on equivalence classes; that is, f(x) = f(x1) whenever x ∼ x1. We will have plenty of opportunities to use this technique. We now define the canonical topology of Grassmann manifold: note first that the space Mnk(R) is n × k real matrices is isomorphic, as a kn vector space, to R . We endow Mnk(R) with the topology induced by this identification.

n Exercise 1.22. Show that Stk(R ) is an open subset of the space of (n × k) matrices. Hint: the rank of A is the same as the rank of A>A. The topology of the Stiefel manifold is the one induced as an open subset of Mnk(R). 10 1. BASIC CONSTRUCTIONS

n Definition 1.23. The Grassmann Grk(R ) manifold is the topo- n logical quotient of the Stiefel manifold Stk(R ) under the equivalence relation ∼ of proposition 1.10. Let us now establish a few facts about the Grassmann manifolds as topological spaces; this is very good training on the workings of the quotient topology.

n Exercise 1.24. Show that the Stiefel mainfold Stk(R ) is connected if n > 1. Then, the Grassmann manifold, being a continuous image of a connected space, is connected. Consider now Rn endowed with the standard (or any other) Eu- n clidean inner product, so that the presentation of Grk(R ) of proposition 1.13 makes sense. We have

n n Proposition 1.25. The quotients Vk(R )/ ' and Stk(R )/ ∼ are homeomorphic. Proof. True to our motto of working upstairs and not in the quo- n n n tient, consider the maps i : Vk(R ) → Stk(R ) and GS : Stk(R ) → n Vk(R ) be the inclusion and the Gram-Schmidt process, respectively. Exercise 1.26. Show that the Gram-Schmidt process is continuous.

Exercise 1.27. Show that of x ∼ x1 then GS(x) ' GS(x1) (the opposite direction u ' u1 then i(u) ∼ i(u1) is trivially true).

Denote by π∼ and π' the projections to the quotient of the respective equivalence relations ∼ and '. Exercise 1.27 says that π' ◦ GS and π∼ ◦ i are constant on equivalence classes, thus inducing continu- n n n ous functions i0 : Vk(R )/ '→ Stk(R )/ ∼ and GS0 : Stk(R )/ ∼→ n n Vk(R ). Since GS ◦ i is the identity of Vk(R ), GS0 ◦ i0 is the identity n of Vk(R )/ '. In the other direction, η = i ◦ GS is not the identity n n of Stk(R ), but for all x ∈ Stk(R ), x ∼ η(x). Thus i0 ◦ GS0 is the n identity of Stk(R ). n Therefore the quotient Vk(R ) is also the Grassmann manifold, topologically. However geometrically is a very diﬀerent space.

n Corollary 1.28. The Grassmann manifold Grk(R ) is compact. n n Proof. Grk(R ) = π'(Vk(R )) is the continuous image of the com- n pact (prove it!) space Vk(R ). 1.2. TOPOLOGY 11

Exercise 1.29. A function f : Rn → Rm is homogeneous of degree k if f(λ~x) = λkf(~x). Show that a homogeneous function induces a continuous map fˆ : RP n−1 → RP m−1. Exercise 1.30. The Veronese embedding: consider the map V : R3 → R6 given by V (x, y, z) = (x2 − y2, x2 − z2, y2 − z2, 6xy, 6xz, 6yz) induces an injective map RP 2 → RP 5. The following construction is quite important: it establishes the Grassmann manifold as a homogeneous manifold, where we can apply geometrical methods. Let Tˆ : Rn → Rn be an invertible linear transformation, represented in the canonical basis by a (n × n) invertible matrix T . Then the map n n LT : Stk(R ) → Stk(R ), LT A = T A, takes classes to classes and n n therefore induces a continuous map `T : Grk(R ) → Grk(R ). The inverse of this map is `T −1 , and therefore `T is a homeomorphism. We have ˜ n ˜ n Lemma 1.31. Let H, H ∈ Grk(R ), V, V ∈ Grn−k(R ), such that Rn = H⊕V = H˜ ⊕V˜ . Then there is an invertible linear transformation Tˆ : Rn → Rn such that Tˆ(H) = H,˜ Tˆ(V ) = V˜ Proof. Choose ordered basis

•{~v1, . . . ,~vk} of H, •{~vk+1, . . . ,~vn} of V , ˜ •{~w1, . . . , ~wk} of H, ˜ •{~wk+1, . . . , ~wn} of V .

Then the assignment ~vi → ~wi extended by linearity defines an invertible n linear transformation of R that satisfies the conclusion. The lemma means that, for many applications, one can “adjust” n ˆ some situations in Grk(R ) to a simple form via a suitable T . As an example, we finish this section with another relatively simple topological property of the Grassmann manifolds: k Theorem 1.32. Let V ∈ Grn−k(R ). Then the top cell UV is open n and dense in Grk(R ). Proof. By lemma 1.31, we can assume that V is the span of the last (n − k)-vectors of the canonical basis of Rn, and let H be the span n of the first k-vectors in R . The inverse image of the top cell UV is n then the saturated set in Stk(R ) given by ˜ n UV = {A ∈ Stk(R ) : the top (k × k) square of A is invertible } , 12 1. BASIC CONSTRUCTIONS as in the discussion after lemma 1.16; therefore the top cell is open on n n Grk(R ). As for the density, given A ∈ Stk(R ), consider the matrix 1k×k A = A + , 0(n−k)×k where is not an eigenvalue of the top (k × k) square of A. Now we can completely understand the top cells of a Grassmann manifold; again by lemma 1.31 we assume the fixed V = span(~ek+1, . . . ,~en). Theorem 1.33. The slope parametrization of Theorem 1.17 induces kn a homeomorphism between Mn×k(R) ' R and the top cell UV .

Proof. Theorem 1.17 shows that the map φH,V is continuous and injective. Its inverse ψH,V is induced by the map from the inverse image ˜ Xk×k UV of the top cell as follows: if A = , then ψH,V (A) = Y(n−k)×k −1 Y(n−k)×kXk×k. Theorems 1.32 and 1.33 have several important consequences: Definition 1.34. A topological manifold is a paracompact Haus- dorff topological space that is locally homeomorphic to Rn. Remark 1.35. By the theorem of invariance of domain (Rn is homeomorphic to Rm implies m = n), in the connected case the number n is well-defined and is called the dimension of the manifold. We have n Theorem 1.36. The Grassmann manifold Grk(R ) is a topological manifold of dimension nk. Proof. We have already established the compacteness (which imply paracompactness) and conectedness of the Grassmann manifolds. The local homeomorphisms with Rnk are furnished by proposition 1.33, since top cells UV with varying V covers the Grassmann manifolds. The Hausdorff condition is left as an exercise (be careful: continuous images of Hausdorff spaces are not necessarely Hausdorff). Definition 1.37. A compactification of a Hausdorff topological space X is a map h : X → Y such that: • h is a homeomorphism onto its image, • h(X) is dense in Y , • Y is compact. Thus, a compactification adds “infinitely distant” points to X in order to embedd it into a compact manifold. 1.2. TOPOLOGY 13

Exercise 1.38. Show that the closed interval [−1, 1] is a compact- 2 iﬁcation of the real line R, by taking h(x) = π arctan(x). We then have

n Theorem 1.39. The Grassmann manifold Grk(R ) is a compactification of Rnk. Proof. Follows directly from Theorem 1.32. Definition 1.40. The one point compactification of a space X is the space Y = X ∪ {∞}, the map h : X → Y being the tautological inclusion, and the topology of Y given by: a neighborhood of points in X are just the neighborhoods of X, and neighborhoods of ∞ are given by the complments of compact sets in X. Exercise 1.41. Let N = (0, 1) be the “north pole” of the circle S1. Show that the sterographic projection S described below gives a homeomorphism between S1 and the one-point compactifiction of R. Generalize to a homeomorphism between the one point compactification of Rn and the unit sphere Sn.

Figure 4. The stereographic projection

2 The next two exercises show, in different ways, that Gr1(R ) = RP 1 is homeomorphic to the circle S1; thus we have understood and identified our first Grassmann manifold. Exercise 1.42. Show that map φ : RP 1 → one point compactification of R given by ( slope of ` if ` is not vertical, φ(`) = ∞ if ` is vertical. induces a homeomorphism between RP 1 and the one point compactification of R, and by transitivity and exercise 1.41, a homeomorphism 14 1. BASIC CONSTRUCTIONS between RP 1 and the circle S1. Write out all the formulas and inverses, and be very careful with the continuity at the vertical line/north pole/point at infinity. We now identify RP 1 as S1 using the quotient presentation. Define the function f : R2 − {0} → R2 by (x2 − y2, 2xy) f(x, y) = x2 + y2 Exercise 1.43. Show that: • |f(x, y)| = 1 for all x, y, • f is onto, • f(x, y) = f(˜x, y˜) if and only if (x, y) and (˜x, y˜) lie in the same line through the origin. • f induces a homeomorphism between RP 1 and S1. Hint: the formula for f is just the formula for the complex function z 7→ (z/|z|)2. Exercise 1.42 can be generalized: Exercise 1.44. Redo the beggining of this chapter substituting the reals for the complex field C. Show that CP 1 is homemorphic to the sphere S2 ⊂ R3 ; in this context we call S2 the Riemann sphere. Relate to your complex analysis course. Do the same for the Hamilton quaternions H, but remember that they are not commutative. Try to do the same with the Cayley octonions O, which are neither commutative nor associative. See [3]. We finish this section with the following exercise which will be useful in chapter 3: Exercise 1.45. The hereditary properties of Grassmann manifolds: n n+s Show that the inclusion R → R ,(x1, . . . , xn) 7→ (x1, . . . , xn, 0,... 0), n n+s induces a continuous, injective map Grk(R ) → Grk(R ), which is a homeomorphism onto its image. Generalize for arbitrary injective linear maps.

1.3. An Euclidean presentation of Grassmann manifolds We have been careful about using only the vector space structure of Rn, the exception being when we described the Grassmann manifold as n a quotient of the orthogonal Stiefel manifold Vk(R ), in order to prove compactness. If add an Euclidean structure to Rn (that is, a positive deﬁnite inner product), we can give a more confortable alternate de- n2 scription as a subset of the space Mn(R) ' R of n × n matrices. This 1.3. AN EUCLIDEAN PRESENTATION OF GRASSMANN MANIFOLDS 15 is done as follows: consider the space Q of all projections q : Rn → Rn, that is, linear maps satisfying q2 = q. A classical exercise in linear algebra is then

Exercise 1.46. Show that a projection q decomposes Rn = Im(q)⊕ Ker(q), and q|Im(q) acts as the identity. Conversely, given a decomposition Rn = V ⊕ W , then deﬁning q(~x) = ~x for ~x ∈ V and q(~x) = 0 for ~x ∈ W deﬁnes a projection q. Thus the set of projections is in bijective correspondence with the set of decomposition of Rn in two complementary subspaces.

n n There is then a map Q → ∪k=0Grk(R ), given by q 7→ Im(q) (the map q 7→ Ker(q) would also work, but we need to ﬁx one of them). This map is clearly onto, but highly non-injective (given a subspace V , there are many complements W with Rn = V ⊕ W ). Thus Q is “bigger” than the Grassmannian, altough it is an interesting space in itself [25]. Exercise 1.47. Show that if T is an invertible linear transformation and p is a projection then Im(T pT −1) = T (Im(p)).

Suppose now that we introduce an Euclidean inner product on Rn. Then given V , there is a canonical complement, namely its orthogonal V ⊥. Fix an orthonormal basis of Rn so that we translate linear transformations to matrices.

Exercise 1.48. Fix an orthonormal basis of Rn. Show that a (n × n) matrix p corresponds in this basis to a projection with orthogonal image and kernel if and only p2 = p = p>. The previous exercise leads to the set

2 > Π = {p ∈ Mn(R): p = p = p } The set Π is not a ﬁxed Grassmannian but contains all of them: Exercise 1.49. Show that Π has n+1 connected components, and p, p˜ belong in the same connected component if and only if rank(p) = rank(˜p).

n n Consider now the map φ :Π → ∪k=0Grk(R ) given as above sending a matrix to its column space (which is the image of the linear transformation it represents). Exercise 1.50. Show that φ restricted to each connected component is a homeomorphism. Find an explicit formula for its inverse. 16 1. BASIC CONSTRUCTIONS

Exercise 1.51. Show that the action of orthogonal matrices (as linear trasnformations of Rn) on the Grassmannian is translated to conjugation by orthogonal matrices on the space of orthogonal projections (Hint: use exercise 1.47) Note that we can, at least for the topological properties, forget about the images and just use the algebraic properties (square and the transpose) of the matrices p. This leads to a far-reaching generalization of the Grassmann manifolds; in any algebraic structure where we have an square and a “transpose” we can deﬁne a Grassmann manifold; if in addition we have a suitable norm we can also perform analysis and topology. This is the Grassmannian of a C∗-algebra; see e.g. [4]. CHAPTER 2

Geometry

In this section we study some aspects of the Geometry of the Grass- mann manifolds. The main thrust of the first chapter was the recognition and parametrization “up to order zero”, that is, recognizing the points of the Grassmann manifolds. In this chapter we will look at the first and second order information, that is, first and second derivatives, necessary for their (differential) geometric study. In addition, we also describe some important relatives of the Grass- mann manifolds, obtained when we impose some additional structure in Rn (inner product, complex structure, symplectic form, etc.)

2.1. An application to ordinary differential equations We begin this chapter with the application of Grassmann manifolds that has motivated the author through the years. Consider a second order, ordinary differential equation: γ00(t) = F (t, γ(t), γ0(t)) . These equations are ubiquitous in physics (Newtonian mechanicas, F = ma, etc.), and Geometry (equation of geodesics). In general, their study is extremely difficult. One technique that helps in the understanding of these equations is the linearization: a canonical process where one obtains from the equation above a linear differential equation of the form (2.1) y00(t) + p(t)y0(t) + q(t)y(t) = 0 . For simplicity, we assume that the coefficients are defined for all t ∈ R (and then, a theorem for linear ordinary differential equations implies that the solutions are defined for all t ∈ R too). It is a very important problem in this area to find or estimate the conjugate points: given t0 ∈ R, find t1 > t0 (if it exists) such that there is a non-trivial solution of of (2.1) that is zero both at t = t0 and t = t1. Example 2.1. If p(t) = 0 and q(t) = λ2 is a positive constant, then the solutions of (2.1) that vanish at t = t0 have the form y(t) = a sin(λ(t − t0)) , where a ∈ R is non-zero for non-trivial solutions. 17 18 2. GEOMETRY

Therefore the forward conjugate points appear at tk = t0 + kπ/λ, 0 6= k ∈ N. Note that, the bigger λ is, the sooner conjugate points appear. Note also that if p = 0 and q(t) = −λ2 is a negative constant, then the solutions of (2.1) are given by y(t) = a sinh(λ(t − t0)) , which do not have conjugate points. Note that, in the positive case, the bigger the constant λ2, the earlier the first conjugate point appears. Generalizing this, a typical theorem is the following consequence of the Sturm comparison theorem (see for example chapter X of [15]): Theorem 2.2. Let y00 + R(t)y = 0 be a second order ordinary differential equation. If the function R(t) satisfies R(t) ≥ λ2 > 0, then the the first conjugate point of t0 appears not later than t0 + π/λ. Let us study the conjugate point problem in a geometric way using the Grassmann manifold. For a single second order differential equation, we actually use RP 1. Recall that the space of solutions of the equation (2.1) is two dimensional, generated for example by two solutions ah(t), av(t) satisfying the initial conditions 0 0 ah(t0) = 1, ah(t0) = 0 and av(t0) = 0, av(t0) = 1 .

Let us draw the evolution of the solutions ah, av simultaneously a a curve in the plane R2; that is, consider the curve in R2 given by a (t) A(t) = v . ah(t) Note that this curve never touches the origin (because of uniqueness of solutions of ordinary diﬀerential equations).

Figure 1. Simultaneous drawing of the linearly independent solutions 2.1. AN APPLICATION TO ORDINARY DIFFERENTIAL EQUATIONS 19

Now the key insight is that the appearance of a conjugate point t1 only depends on the line through the origin that passes through A(t):

Figure 2. The induced curve of lines Indeed, any non-trivial solution that vanishes at zero is of the form cav for some constant c ∈ R, and therefore we can just use av itself for detecting conjugate points. We have that A(t0) spans the vertical line {x = 0}, and a conjugate point appears when A(t) is vertical again:

Figure 3. A conjugate point

We call the class `(t) of A(t) in RP 1 the Jacobi curve associated to the linear second order differential equation (2.1), and we have seen that conjugate points only depend on the Jacobi curve. Exercise 2.3. Find an analytical condition on a frame A(t) that translates the following picture (figure 4). Show that this condition only depends on the class `(t) of A(t) in RP 1. Use the Wronskian (e.g., [15]) to show that such a shape of the curve A cannot happen if the frame A(t) is obtained with the independent solution of a differential equation as in av, ah. 20 2. GEOMETRY

Figure 4. Jacobi curve coming back

The previous exercises illustrates how the topology of the Grass- mannian appears in the study of conjugate points: in order for a conjugate point to appear, the Jacobi curve must do a “complete turn” around RP 1. For readers knowledgeable about the fundamental group, it means that the Jacobi curve up to the first conjugate point generates 1 π1RP . Let us describe the higher dimensional case: consider now a system of second order homogeneous linear differential equations like equation (2.1), (2.2) ~y00(t) + ~y0(t)P (t) + ~y(t)Q(t) = 0 , where ~y(t) are row vectors in Rn and P and Q are (n × n)-matrix valued functions. Conjugate points are defined now in the same way: t1 is conjugate to t0 if there is a solution ~y(t) of (2.2) vanishing at both t1 and t0. We now generalize the one dimensional projective treatment: consider the frame   a11(t) . . . a1n  . .   . .     an1(t) ann(t)  A(t) =   a(n+1)1(t) a(n+1)n(t)  . .   . .  a(2n)1(t) a(2n)n(t) where • The rows of A(t) are a system of linearly independent solutions of (2.2). • The first n rows are solutions that vanish at t = t0. Focusing on the columns, the frame A(t) is a map into the Stiefel 2n manifold Stn(R ) and its projection to the Grassmann manifold is the 2.2. HOMOGENEOUS SPACES AND CARTAN-KLEIN GEOMETRY 21

2n Jacobi curve `(t) with values in the “half-Grassmannian” Grn(R ) Now in contrast with the one-dimensional case, it is not enough to just consider each individual row solution: a conjugate point will typically appear as a linear combination of the ﬁrst n rows that also vanishes for some other point t1. In order to deal with this complication, we projectivize: The space span(~e(n+1), . . .~e2n} will be called the vertical subspace V ; in our setup then V = `(t0). Then a point t1 is conjugate to t0 if and only if `(t1) ∩ V 6= {0} , that is, if `(t) intersects the train of V . We have then expressed the conjugate point problem in terms of the self-intersection properties of a curve in the Grassmann manifold. This is a far-reaching viewpoint, for example, it leads to the Maslov index in the Calculus of Variations (see, e.g. [23]). This projectivized viewpoint also extends to higher order equations and variational problems, see [33] for a very classical exposition and [8] and [7] for contemporary methods. We are thus moved to study the geometry of curves in Grassmann manifolds. Before that, in the next short section we will see what we mean by Geometry.

2.2. Homogeneous spaces and Cartan-Klein geometry Let X be any set. A group of transformations G on X is a subset of the space of bijections of X that is closed under compositions and inverses (in particular, the identity transformation is in G). Usually, the subset reflects some kind of structure of the space X: homeomorphisms for topological spaces, isometries for metric spaces, etc. Felix Klein defined a geometry as the study of properties of a space that are invariant under a group of transformations. Example 2.4. Plane Euclidean geometry: (in modern terms) the space is the plane R2 and the group of transformations is the Euclidean group of distance-preserving maps of R2 (endowed with its canonical norm); this group is generated by reflections and includes translations and rotations. An example of invariant property is the angle between two lines. Example 2.5. Plane affine geometry: the space is also the plane R2 , but the group is the affine group of transformations that preserve straight lines. This group is generated by invertible linear transformations and translations. The angle between two lines is not invariant under an affine transformation. An example of affine invariant is the 22 2. GEOMETRY

ab ratio ac between the lengths of segmentes defined by three colinear points a, b, c. A central problem is the congruence problem: given two objects in a class in the given geometry, when does there exist a transformation taking one to the other? Example 2.6. Two triangles in Euclidean geometry are congruent if and only if the length of the sides are equal. Two (non-degenerate) triangles in affine geometry are always congruent. In the previous example, the three numbers given by the length of the sides is complete invariant: a set of computable quantities that determine weather or not two objects are congruent. The specific subset of the bijections of a space can be abstracted away; given a group G a left action (resp. right action) of G on a space X is a group homomorphism (resp. antihomomorphism) of G into the bijections of X. Again, typically the action falls into a rather restricted subset of bijections (homeomorphisms, linear maps, etc.) A left action can be transformed into a right action and vice-versa by precomposing with the group inverse map, whih is an antihomomorphism, but we rather not do it in our context: left and right actions come concretely by left and right multiplication of matrices. It is customary to denote a group action by a binary operation: given say a left action which is a homomorphism φ : G → bij(X), we write for example g · x instead of φ(g)(x). The quotient of an action is the set of classes by the equivalence relation x ' y if and only if there is g ∈ G such that g · x = y, that is, the quotient is the space of orbits. Of course, when everything in sight is topological, we put the quotient topology on the quotient. Our main example is the left action of (n × n) invertible matrices, and the right action of (k×k) invertible matrices on the Stiefel manifold n Stk(R ) as in chapter 1. The quotient by the right action gives the n n Grassmann manifold Grk(R ), and the left action defines Grk(R ) as a Klein geometry. Remark 2.7. It is equally valid to consider the Grassmann manifold as a geometry using only orthogonal matrices, but this gives a completely different geometry. This illustrates the principle of the same set with different symmetry groups must be considered a different space. The congruence problem we study in the next section is about curves in the Grassmann manifold, motivated by section 2.1. We want to construct a complete invariant that decides: given two curves n `1, `2 :[a, b] → Grk(R ), is there a matrix T ∈ GL(n, R) such that 2.3. CONGRUENCE OF CURVES IN THE GRASSMANN MANIFOLD 23

T `1 = `2? The context is diﬀerentiable in nature; when the objects and the methods are diﬀerentiable the study of invariants is usually called Cartan-Klein geometry; see e.g. [21, 27].

2.3. Congruence of curves in the Grassmann manifold In this section we illustrate the equivalence problem for parametrized curves in the Grassmannian. The main references we use are • For RP 1, the work of Flanders ([10]), 2n • For the half-Grassmannian Grn(R ), [2], kn • For the “divisible” Grassmannian Grn(R ), [8], and the references contained in these works (in particular in the Appen- dix of [2]). It is worth emphasizing that for the theory of equivalence for general maps into Grassmann manifolds is a wide open area of research. In line with the spirit of these notes, the main objective is to learn how to effectively compute in a Grassmann manifold. The constructions described in this section illustrates a main theme: compute in the Stiefel manifold quantities that only depend on the span of the columns. The first order of business is to do a rough classification of how n “twisted” a given curve is: let A : R → Stk(R ) be a differentiable curve. The linear flag associated to A is the set of inclusions

0 0 00 n {0} ⊂ cs(A(t)) ⊂ cs(A(t)|A (t)) ⊂ cs(A(t)|A (t)|A (t)) ⊂ · · · ⊂ R , where “cs” stands for “column space of” and the vertical bar denotes juxtaposition. The following exercise is then one fundamental instance of the principle “compute in Stiefel, descend to Grassmann”:

Exercise 2.8. Show that, if B(t) = A(t)X(t) where X(t) is a (k × k) invertible diﬀerentiable matrix, then the canonical ﬂags of B(t) and A coincide.

Exercise 2.8 means that the canonical ﬂag depends only on the curve `(t) in the Grassmannian given by the span of the columns of A. The dimensions of these spaces can be put in an ascending sequence of integers 0 < k ≤ k1 ≤ · · · ≤ kr ≤ n which is called the Young diagram of `(t) and measures how free the derivatives ` are ([17]). As is usual in equivalence problems, the nicest invariants occur in a generic set of maximally twisted curves, since each successive derivative furnishes more and more information. 24 2. GEOMETRY

kn Definition 2.9. A curve `(t) in the divisible Grassmannian Grn(R ) is said to be fanning if its Young diagram is the maximal sequence 0 < n < 2n < ··· < (k − 1)n < kn. This means that each derivative contributes as much as possible to the dimension increase of the flag, and therefore a fanning curve is “maximally twisted” Exercise 2.10. Relate exercises 2.3 and 2.8. Show that the analytic condition of exercise 2.3 is that the curve is not fanning. We now turn to the equivalence problem. We take the route that makes the presentation more elementary and compact, by means of doing the RP 1-case. The known higher order cases have similar intuitions but progressively higher combinatorial difficulties due to wors- ening non-commutativity properties. We began this section associating a curve in RP 1 to a second- order, linear homogeneous differential equation. We now proceed in the other direction: let `(t) be a fanning curve in RP 1. Choose a frame A(t) ∈ R2 − {~0} spanning `(t). The fanning hypothesis means that the vectors A(t) and A0(t) are linearly independent and therefore a basis of R2. Any vector ~v ∈ R2 can be expressed in a unique way as a linear combination of A(t) and A0(t), in particular the acceleration A00(t). Then we have structural equation (2.3) A00(t) + 2A0(t)p(t) + A(t)q(t) = 0 for some real-valued functions p(t), q(t) uniquely defined by A(t). The factor of 2 in the middle one is convenient for some computations and is consistent with most of the literature. We write p and q on the right so that they can be considered (1 × 1) matrices which will generalize to higher dimensions, but each (1 × 1) row of A satisfies a the second order differential equation of the form a00(t) + 2p(t)a0(t) + q(t)a(t) = 0 . Note that if T : R2 → R2 is an invertible matrix, then if Aˆ = T A, then by multiplying both sides of (2.3) we see that the coefficients of the structure equation satisfied by A˜ are still the same p(t) and q(t), and they are therefore invariant under the action of GL(2, R). It is tempting to declare then that these coefficients are the invariants we are seeking, and indeed, if our equivalence problem was of fanning curves of frames in R2 endowed with the action of GL(2, R) then this would be right. Exercise 2.11. Prove the assertion of the last phrase of the previous paragraph. 2.3. CONGRUENCE OF CURVES IN THE GRASSMANN MANIFOLD 25

However, the coefficients p and q change when we use two different frames to describe the same curve in RP 1. Let us see what happens if we change the frame A by B(t) = A(t)X(t)−1 where X(t) 6= 0: A(t)q(t) = B(t)X(t)q(t) 2A0(t)p(t) = 2B0(t)X(t)p(t) + 2B(t)X0(t)p(t) A00(t) = B00(t)X(t) + 2B0(t)X0(t) + B(t)X00(t) adding these equations, we have that B satisfies the differential equation with new coefficients X0(t) X0(t) X00(t) B00(t)+2B0(t) + p(t) +B(t) q(t) + 2 p(t) + = 0 . X(t) X(t) X(t) Or, introducing new notation, B00(t) + B0(t)ˆp(t) + B(t)ˆq(t) = 0, where X0(t) pˆ(t) = + p(t)(2.4) X(t) X0(t) X00(t) (2.5) qˆ(t) = q(t) + 2 p(t) + X(t) X(t) We have Theorem 2.12. Two fanning frames A(t) and B(t) represent the same curve in RP 1 if and only if the coefficients of their respective equations (2.3) are related by the equations (2.4) and (2.5) for some non-vanishing function X(t). Proof. The “only if” part is the computation above. For the “if” part, the multiplier X(t) determined by the differential equation 0 X(t)(ˆp(t) − p(t)) = X (t) satisfies A(t) = B(t)X(t). We want to find a quantity that is invariant under the transformations (p, q) 7→ (ˆp, qˆ). A standard way of doing this is to find a normal form for the coefficients p and q: what is the “nicest” form we can find for equation (2.3) by transforming the coefficients by the rules given R by equations (2.4) and (2.5). By setting X(t) = e− p(s)ds we can make the coefficient of the first derivative disappear, and then B satisfies the simpler-looking differential equation (2.6) B00(t) + B(t)Q(t) = 0 A fanning frame B is said to be in normal form if it satisfies a differential equation as (2.6), with no first derivative term.

Lemma 2.13. Let `(t) be a curve in RP 1. Then (1) There is a normal frame spanning `(t). 26 2. GEOMETRY

(2) Two normal frames spanning ` differ by a multiplicative constant. Proof. For (1), let A(t) be an arbitrary frame spanning `(t). Let B(t) = A(t)X−1(t) as above. For (2), let C(t), D(t) be two normal frames spanning `(t). Since they both span the same curve, there is a non-vanishing real function X(t) such that C(t) = D(t)X(t). Since they are both normal the coefficients of the first derivative in their respective 0 structure equations vanish. Then by equation (2.4), X (t) = 0. Exercise 2.14. There is a white lie in the argument above: how does one know that there exists a frame spanning `? Use the methods of chapter 1 in order to find first local, and then global, frames spanning `. Note that, in the reduction to normal form procedure, Q(t) is related to the original coefficients coming from A by Q(t) = q(t) + 2 0 p (t) − p (t). Given a frame A satisfying (2.3), the function SA(t) := q(t) + p2(t) − p0(t) is called the Schwarzian of A. Lemma 2.13 then shows that the Schwarzian depends only on `(t) and not on the specific frame: indeed, the association ` 7→ normal frame spanning ` is almost canonical: by lemma 2.13, any two normal frames differ by a constant; and then their respective structure equations (2.6) are the same and therefore gives the same Schwarzian. Exercise 2.15. Use equations (2.4) and (2.6) to show by direct computation that the Schwarzian only depends on the curve `(t). We are now ready for the solution of the equivalence problem:

1 Theorem 2.16. Two curves `1(t), `2(t):[a, b] → RP are congruent if and only if their Schwarzians coincide. Proof. The “only if” part follows from the discussion after equation (2.3) and from the Schwarzian being well-defined for curves in 1 RP . In order to do the “if” part, let A1(t), A2(t) be normal frames spanning `1(t) and `2(t), respectively. We first find T ∈ GL(2, R) such 0 0 that at the initial point a, T A1(a) = A2(a),T A1(a) = A2(a). Exercise 2.17. Show that such a matrix T can always be found.

We now show that T `1(t) = `2(t) for all t ∈ [a, b]. Indeed, T A1(t) and A2(t) satisfy the same structure equation, since the Schwarzians are equal by hypothesis, and they have the same initial conditions at t = a; by uniqueness of solution of ordinary diﬀerential equations, T A1(t) = A2(t) and a fortiori T `1(t) = `2(t) for all t ∈ [a, b]. 2.4. JETS OF CURVES IN GRASSMANN MANIFOLDS 27

And we have found the complete invariant we wanted: the Schwarzian. The next step in the study of invariants is usually understanding curves with notable invariants. For example, fanning curves in RP 1 with vanishing Schwarzian are the quotients of normal frames satisfying 00 2 A(t) = 0, that is, there are linearly independent vectors ~v0,~v1 ∈ R such that A(t) = ~v0 + t~v1. We finish this section relating our discussion with the classical Schwarzian from complex analysis. Consider a curve `(t):[a, b] → R a (t) such that `(a) is the x-axis. Let A(t) = 1 be a (not necessarily a2(t) normal) frame spanning `(t). We have that a1(a) 6= 0 and by continuity a1(t) 6= 0 in some neighbourhood of t = a. Then the frame 1 1 M(t) = A(t) = , a1(t) m(t) where m(t) = a2(t) is defined around t = a and spans the same line a1(t) `(t). Note that this is the slope parametrization of the very beginning of this book. Taking the necessary derivatives, we see that the structure equation of M is M00(t) + 2M00(t)p(t) = 0 , m00(t) where p(t) = − 2m0(t) . Exercise 2.18. Show that the fanning condition translates to m0(t) 6= 0 in this presentation and therefore dividing by m0(t) is allowed. Therefore our Schwarzian is given by ( ) 1 m00(t)0 m00(t)2 Q(t) = − + , 4 m0(t) m0(t) which is, modulo constants due to our choice of normalizations, the Schwarzian of classical complex analysis. Note that, in the slope parametrization, curves with zero Scwharzian are exactly the Möbiustransforma- tions a + tb t 7→ c + td where (a, b), (c, d) are the components of the vectors ~v0 and ~v1 above.

2.4. Jets of curves in Grassmann manifolds In the previous section we have been taking first and second derivatives of frames, which live in an open subset of RN and therefore we 28 2. GEOMETRY know how to take derivatives. Here we construct some tools that al- low the recognition of these derivatives as derivatives of curves in the Grassmann manifold itself. The presentation is necessarily informal, since a complete treatment needs the theory of differentiable manifolds and spaces of jets. The interested reader can find an unified exposition in the book [16]. Let X and M be differentiable manifolds (objects where it makes sense to talk about derivatives). The space of J r(X; M) of r-jets of maps of X onto M is roughly described as follows: each element of J r(X; M) contains the information of the value of a function f : X → M at a given point x ∈ X plus the information of its first r derivatives.

Example 2.19. The space J k(R; Rn) is simply (Rn)k+1. Given a curve γ : R → Rn, its k-jet at a point, say at t = 0, is the evaluated vector of derivatives (γ(0), γ0(0), . . . γ(k)(0)). This also applies to any open subset of Rn.

n Let now `(t): R → Grk(R ) be a diﬀerentiable curve in the Grass- mann manifold. How to take the derivative of a curve of subspaces is not immediately obvious. We try to follow the method we have n been using: let A(t): R → Stk(R ) be a frame spanning `(t). We are tempted now to deﬁne the derivative of ` as A0. However, there is the problem is that many frames span the same curve; indeed, from chapter 1 we know exactly all other frames spanning `(t): they are of the form B(t) = A(t)X(t) where X is an invertible (k × k)-matrix. We must think of A(t) and B(t) as the same curve in the Grassmann manifold, and then they must have “the same” derivative. Taking an actual derivative, and remembering that Leibniz’s rule applies to matrix multiplication, we have

B0(t) = A0(t)X(t) + A(t)X0(t) , B00(t) = A00X(t) + 2A0(t)X0(t) + A(t)X00(t) , . (2.7) . r X r B(r)(t) = A(r−s)(t)X(s)(t) . s s=0

We now deal with two simple spaces of jets: ﬁrst the space of jets in r n r the Stiefel manifolds J (R; Stk(R ) and J (R; GL(k, R)) of invertible (k ×k) matrices. The reason we called them “simple” is that the target 2 spaces are open subsets of Rkn and Rk , respectively, and then example 2.5. NOTABLE SUBSETS OF THE GRASSMANN MANIFOLD 29

2.19 applies:

r n n J (R; Stk(R ) = {(A0, A1,..., Ar) , A0 ∈ Stk(R )} , r J (R; GL(k, R)) = {(X0,X1,...,Xr) , det(X0) 6= 0} , and the important insight is that J r(R; GL(k, R)) is a group, with multiplication induced by Leibniz’s rule: (2.8) r X r (X ,X ,...,X )(Y ,Y ,...,Y ) = (X Y ,X Y +X Y ,..., X Y ) . 0 1 r 0 1 r 0 0 0 1 1 0 s r−s s s=0

r n Given now (A0, A1,..., Ar) ∈ J (R, Stk(R ) and (X0,X1,...,Xr) ∈ J r(R, GL(k, R), we deﬁne the operation • by

(A0, A1,..., Ar) • (X0,X1,...,Xr) = (B0, A1,..., Br) , Pj r where Bj = s=0 j Aj−sXs, that is, deﬁned by Leibniz rule again (but be careful: formula (2.7) relates functions of t, whereas this last equations relate matrices that are functions and its derivatives evaluated at a point.

Exercise 2.20. Show that the association • deﬁnes a group action r r n of th group J (R; GL(k, R)) on J (R; Stk(R ). Then we can recognize the space of derivatives up to order r of curves in the Grassmann manifold also as a quotient:

r n r n r The space J (R, Grk(R )) can be identiﬁed with the quotient J (R; Stk(R )/J (R; GL(k, R)) under the •-action.

2.5. Notable subsets of the Grassmann manifold n Up to now we have been considering the Grassmann manifold Grk(R ), where there is no other structure in Rn other than the vector space structure. The only exception being the standard inner product for considering orthogonal Stiefel manifolds and orthogonal projections, which was used only to give alternative presentations. In this section we impose additional structure in the Grassmann manifold arising from special vector spaces. It is psychologically convenient now to change the notation slightly and consider Grk(V ), the Grassmannian of a vector space V , with the same deﬁnition: it is the set of k-dimensional subspaces in V ; if V is a real, n-dimensional vector n space, after ﬁxing a base we recover Grk(R ). 30 2. GEOMETRY

2.5.1. Other fields. The first change, which we mention only in passing, is to consider vector spaces over other fields; in geometry and topology we consider mainly the complex field (C) and the Hamilton quaternions H (which do not form a field but a division ring, having non-commutative multiplication; however most of linear algebra can be recovered, with the notable exception of the determinant, []). Then, n n fixing a basis, we have the spaces Grk(C ) and Grk(H ).

Exercise 2.21. Redo this book replacing the reals with the complex numbers everywhere. In particular, exercise 1.41 is transformed to a homeomorphism between CP 1 and the Riemann sphere S2. In particular, the results at the end of section 2.3 translate to classical M¨obius transformations and the “extended complex plane” acquires a meaning: it is extended the “missing line” of a given slope parametrization which is on equal footing with all other points. In the case of quaternions, prove that HP 1 is homeomorphic to the sphere S4.

The projection from Stiefel to Grassmann in these cases are extremely rich in properties. The Hopf fibrations S3 → S2 and S7 → S4 2 1 2 1 can be seen as V1(C ) → CP and V1(H ) → HP , respectively. This last fibration has contribuited many constructions in geometry, and differential topology; we just mention a topic dear to the author, the geometry of exotic spheres [5, 6, 9, 12, 13, 18, 28]. The Hopf fibrations are also the substrate of many construction in Physics; see [31]. Another construction is to try to extend the concepts of this book using the Cayley octonions Ca, an 8-dimensional real division algebra which is neither commutative nor associative. In contrast with the quaternion case, the lack of associativity is a heavy blow: it means that the relation ~v ∼ ~w if and only if there is λ 6= 0 with λ~v = ~w is not an equivalence relation. However, with some care one can recover the Cayley projective line CaP 1 and the Cayley projective plane CaP 2, see [3]. Now this section is called “some geometrically defined subsets of the Grassmann manifold” and all we have done up to now is extend the concept to other fields. However, by applying the forgetful functor we can always consider these Grassmannians as subsets of some real Grassmann manifold: since a complex (resp. quaternionic) vector space of dimension n is (canonically) a real vector space of dimension 2n (resp 4n), we have

n 2n n 4n Grk(C ) ⊂ Gr2k(R ) , Grk(H ) ⊂ Gr4k(R ) . 2.5. NOTABLE SUBSETS OF THE GRASSMANN MANIFOLD 31

We can “half forget” and consider a complex vector space as a real vector space and the multiplication by i is a linear map with square -1. In general, a complex structure on a real vector space V is a map J : V → V such that J 2 = −1. Such a vector space always has even dimension 2n. Then we can deﬁne the following subset: let

Xk = {` ∈ Gr2k(V ): ` = J`} . Alternatively, a given a complex structure on a real vector space we can deﬁne a complex vector space structure by setting (a+ib)~v = a~v +bJ~v, and then the set Xk above is identiﬁed with Grk(V ), where now V is considered a complex vector space. For the quaternions, a quaternionic structure on a real vector space is a pair (I,J) of linear transformations of V satisfying I2 = J 2 = −1, IJ = −JI (then IJ = K as a third element and {1, I, J, K} span an algebra isomorphic to the quaternion algebra). A similar discussion as the complex case applies.

2.5.2. The Lagrangian Grassmannian. The most notable subset of the Grassmann manifold is the Lagrangian Grassmannian. Here we give a brief introduction, limiting ourselves to the study of the slope parametrization, referring the reader to the comprehensive book [23] for an in-depth study. Definition 2.22. A symplectic form on a real vector space W is a bilinear form ω : W × W → R that is: (1) Antisymetric: ω(~v, ~w) = −ω(~w,~v). (2) Non-degenerate: ω(~v, ~w) = 0 for all ~w ∈ V implies ~v = ~0. The pair (W, ω) is called a symplectic vector space.

Example 2.23. Let W = Rn × Rn and ω be given by n X ω0((~x,~y), (~v, ~w)) = xiwi − yivi . i=1 It can be shown that every symplectic vector space is equivalent to the canonical example above (and, in particular, a symplectic vector space is always even-dimensional). Given a symplectic vector space (W, ω), we can deﬁne “orthogonality” just as in Euclidean vector space: if X is a subset of W , then X⊥ = {~v ∈ W : ω(~v, x) = 0 for all x ∈ X} . Exercise 2.24. Show that if X is a subspace of a symplectic vector space, then dim X + dim X⊥ = dim V , and (X⊥)⊥ = X. 32 2. GEOMETRY

However, the antisymmetry of the symplectic form goes against Euclidean intuition: Definition 2.25. Let (W, ω) be a symplectic vector space. A subspace ` ⊂ W is said to be • Isotropic if ` ⊂ `⊥ , • Coisotropic if `⊥ ⊂ `, • Lagrangian if ` = `⊥. Lemma 2.26. A Lagrangian subspace of a symplectic vector space (W, ω) has dimension half of the dimension of W .

Proof. Follows directly from exercise 2.24.

The Lagrangian Grassmannian Λn(W, ω) is the set of all Lagrangian subspaces of a symplectic vector space (W, ω). By lemma 2.26, Λn(W, ω) is contained in the half-Grassmannian Grn(W ) where dim W = 2n. The reason the Lagrangian Grassmannian is so important is that, in the same vein we projectivized differential equations in section 2.1, this projectivization naturally falls inside the Lagrangian Grassman- nian when the differential equation comes from a first order variational principle, as in Classical Mechanics and Differential Geometry (see e.g. [23]). Generalizations to higher order variational principles are found in [7]. Let us study the Lagrangian Grassmanian Λn of the canonical sym- n n plectic vector space (R × R , ω0). Note that each of the subspaces H = Rn × {~0} and V = {~0} × Rn are Lagrangian subspaces. The n canonical basis of R will be split in two: we call ~ei, i = 1, . . . n the ba- ~ ~ ~ sis of H and fi, i = 1, . . . n the basis for V . Together, ~e1, . . .~en, f1,... fn form a symplectic basis: they satisfy ~ ~ • ω0(~ei,~ej) = ω0(fi, fj) = 0 , ~ • ω0(~ei, fj) = δij (Kronecker delta). We use the complementary planes Rn ×{~0} and {~0}×Rn. By theo- 2n rem 1.17, a dense open neighbourhood of H in Grn(R ) is parametrized by n × n matrices, the map being given by 1 φ (Z) = column space of k×k , H,V Z with H corresponding to Z = 0. Now the question is, which of those planes are Lagrangian? We have

Theorem 2.27. The plane φH,V (Z) is Lagrangian if and only if the matrix Z is symmetric. 2.5. NOTABLE SUBSETS OF THE GRASSMANN MANIFOLD 33

th Proof. Let ~ci be the i column of the matrix above. Then ~ei ~ci = P ~ , r zirfr ~ where we have written the vectors ~ei, fr as column vectors. Now computing the symplectic form on a pair ~ci,~cj of such columns, we have X ~ ~ ω0(~ci,~cj) = ω0(~ei + zirfr,~ej + zjsfs) = zji + zij . r,s

Thus if ` = φH,V (Z) is Lagrangian, ω0|ell = 0 and Z must be symmet- ⊥ ric. Conversely, if Z is symmetric then ω0|` = 0 and then ` ⊂ `, but ⊥ then since dim ` = n we have that ` = ` . In accordance with the spirit of this notes, we have stated both the statement and the proof of Theorem 2.27 in a fixed basis of fixed complementary subspaces. In reality, this characterization/parametrization on works for any pair of complementary Lagrangian subspaces on a symplectic vector space (W, ω), and is invariantly defined: an open and dense set of Lagrangian planes is parametrized by the set of symmetric bilinear forms on one of them. Concretely, in the symplectic situation we can use the symplectic form to reparametrize the graph map of theorem 1.18, T : H → V 7→ ΓT , as follows: we send T to the bilinear form βT on H, ~ ~ ~ ~ βT (h1, h2) = ω(h1,T h2) .

We then have that βT is a symmetric bilinear form if and only if the graph ΓT is a Lagrangian plane. This characterization is extremely useful: see for example [22] for the study of conjugate points and [1] for the geometry of curves in the Lagrangian Grassmanian.

CHAPTER 3

Algebraic Topology

Here we describe some algebraic topological features of the Grass- mann manifold. Roughly speaking, elementary algebraic topology has two modes: Homology, with a somewhat combinatorial flavor, and Ho- motopy which is more geometric. At more advanced levels these dis- tinctions get less sharp; see for example [29] for combinatorial ideas in Homotopy Theory. We do not have the space in these notes for a description of Homology and Homotopy and we keep the discussion of these general theories informal; a complete study can be found for example in the textbook [14]. What we will do is to describe two constructions that exemplify the Homology and Homotopy of the Grass- mann manifold: for homology, its CW decomposition, a way of combinatorially constructing the Grassmannian from simpler pieces that allows the computation of its homology. In relation to homotopy, we will describe specific generators of some homotopy groups of the Grass- mann manifolds, which have been crucial in Differential Geometry and curvature. In this section, it is much more convenient to consider an auxiliary Euclidean strcuture of Rn, say the standard one, as opposed to the previous section, where we were careful to consider the Grass- mann manifolds as a geometry over the general linear group.

3.1. Combinatorial structure

3.1.1. CW complexes. Recall that the closed disk Dn is the space n n D = {~x ∈ R : ||~x|| ≤ 1} with boundary the unit sphere n−1 n S = {~x ∈ R : ||~x|| = 1} and interior the ball n n B = {~x ∈ R : ||~x|| < 1} . The author vividly recalls his Algebraic Topology teacher stating: “I am a very humble man. My only ambition in Topology is to understand disks and their boundaries”, only later realizing that this is an 35 36 3. ALGEBRAIC TOPOLOGY extremely arrogant statement: by mixing disks glued at their boundaries, one gets essentially all spaces of interest in Algebraic Topology [19]. This is the idea behind CW complexes. We give some informal discussion of CW complexes ﬁrst and then proceed with the precise deﬁnitions. Roughly speaking, a CW complex is built inductively adding disks of dimension n to an already built CW complex of lesser “dimension”. By convention, o 0-disk is a point. The 0-skeleton of a CW complex is a discrete set of points:

Figure 1. A 0-skeleton Then we will glue some 1-disks (the 1-disk is just the interval [−1, 1]) to the 0-skeleton through the boundary:

Figure 2. A 1-skeleton Note that the 1-disk that is “hanging” form the third point was attached by gluing the two boundary points to the same point. Now we glue some 2-disks to the already constructed 1-skeleton: Where the 2-disk that was attached at the right was glues by sending all the boundary to the fourth point, whereas the one at the left was attached by sending the boundary to the circle formed by previous 1-disks. and so on. We only deal with ﬁnite CW complexes: at each stage we only glue a ﬁnite amount of disks, and we stop at a certain dimension 3.1. COMBINATORIAL STRUCTURE 37

Figure 3. A 2-skeleton n which is called the dimension of the CW complex. For infinite CW complexes, one has to be careful with the topology one puts in it; see for example appendix A of [14]. After definition 3.2 it will be implicit that our CW complexes are always finite. Let us express the 2-sphere S2 as a CW complexes in two different ways: First,

Figure 4. One CW-decomposition of the 2-sphere thus the 0-skeleton is a point, the 1-skeleton is the same as the 0- skeleton, and the 2-skeleton was obtained by gluing just one 2-cell. Another way is the following given in ﬁgure 5, where each successive skeleton is obtained by gluing two cells of the corresponding dimension Exercise 3.1. Extend these examples to the n-sphere. Observe that the ﬁrst way is simpler, however the second way has the advantage of considering each hemisphere separately. We will see that this motivates the CW structure on the projective spaces. Note 38 3. ALGEBRAIC TOPOLOGY

Figure 5. Another CW-decomposition of the 2-sphere that the same space can have diﬀerent decompositions as CW complexes. Let us ﬁnish this informal discussion with a CW decomposition of the 2-torus: where we have homeomorphically “squared” the 2-disk.

Figure 6. CW-decomposition of the torus

We adopt the following definition, adapting proposition A.2 in [14] to the finite case: Definition 3.2. A finite CW complex is a Hausdorff topological n space X and a finite collection of maps φα,n : D → X, called the characteristic maps, satisfying:

(1) Each φα,n restricts to a homeomorphism from the interior of n n (D ) onto its image, called a cell eα ⊂ X , and these cells are all disjoint and their union is X. n (n−1) (2) For each cell eα, φα,n(S ) is contained in the union of cells of dimension less than n. 3.1. COMBINATORIAL STRUCTURE 39

(3) X is covered by the images of the φα,n. Given a CW complex X, the union of all cells of dimension less than or equal to k is called the k-skeleton Xk. The maximum dimension n of a disk appearing in the characteristic maps is called the dimension of the CW complex. The restrictions φα,n|Sn−1 are called the attaching maps. The deﬁnition then also expresses the idea of CW complex can be tough as gluing n-disks at their boundary to the (n − 1) skeleton through the attaching maps. Exercise 3.3. Find explicit formulas for the characteristic and attaching maps in the examples. Exercise 3.4. Show that in a CW complex, the diﬀerence of sets k Xk − Xk−1 is homeomorphic to the disjoint union of open balls B . 3.1.2. Schubert symbols and Schubert cells. We now proceed to construct a standard CW structure on the Grassmann manifolds. As n+1 n usual, it is instructive to begin with Gr1(R ) = RP . Proposition 3.5. The projective spaces RP n admit a CW decomposition with exactly one cell in each dimension 0, 1, . . . n. Proof. The proof is by induction on n. Recall that by exercise 1.45, the inclusions

0 1 n RP ⊂ RP ⊂ · · · ⊂ RP are induced by the chain of inclusions

1 2 n+1 R ⊂ R ⊂ · · · ⊂ R . The space RP 0 is a point, that is, a 0-cell. Suppose now RP n−1 admits a decomposition satisfying the conclusion of the theorem. All we have to n n do now is to construct a characteristic map φn : D → RP satisfying conditions (1) and (2) of definition 3.2, and condition (3) is translated to RP n = RP n−1 ∪ φ(Dn). A natural candidate is the top cell of definition 1.15 (hence the name): note that the top cell in the case of RP n is exactly RP n−RP n−1, and it is homeomorphic to Rn by the slope parametrization. All that is needed then is to find an adequate map n n n φn : D → RP such that φn maps (int(D )) homeomorphically onto the top cell and the boundary. The points in RP n = RP n−1 ∪ φ(Dn) are the projection of the saturated open set Un+1 in the the Stiefel n+1 manifold St1(R ) defined by n+1 Un+1 = {~v = (v1, . . . , vn+1) ∈ St1(R ): vn+1 6= 0} . 40 3. ALGEBRAIC TOPOLOGY

n n+1 Consider the map Φn : D → St1(R ), given by q X 2 Φn(x1, . . . , xn) = (x1, . . . , xn, 1 − xi ) . Note that the image of Φ actually falls inside the orthogonal Stiefel n+1 manifold V1(R ), that is, the unit sphere, and the last coordinate is n in fact positive. Also, for ~x in the interior of D ,Φn(~x) ∈ Un+1. We can now deﬁne φn to be composition of Φn with the projection from Stiefel to Grassmann; this is the analytic content of ﬁgure 1 in chapter 1.

Exercise 3.6. Show that the functions φn inductively defined above satisfy the definition of carachteristic maps of a CW decomposition. k+1 Prove that the attaching maps are given by the projection V1(R ) → RP k, that is, Sk → RP k. We now proceed to the general Grassmann manifolds; our discussion is based on chapter 6 of [20] . The inclusions 0 1 n R ⊂ R ⊂ · · · ⊂ R n still plays a lead role: given a plane ` ∈ Grk(R ), we consider the sequence of integers 1 2 n−1 n 0 ≤ dim(R ∩`) ≤ dim(R ∩`) ≤ · · · ≤ dim(R ∩`) ≤ dim(R ∩`) = k Two consecutive numbers in this sequence differ at most by one: indeed, consider the map p : ` ∪ Ri+1 → R to be the projection onto the last coordinate of Ri+1. By the rank-nullity theorem, i+1 dim(Im(p)) + dim(ker(p)) = dim(R ∩ `), but ker(p) = Ri ∩ ` and dim(Im(p)) is either zero or one. Therefore the above sequence must “jump” exactly k times. We codify the information of the places where the jump occurs: a Schubert symbol is an ordered sequence of k integers 1 ≤ σ1 < σ2 < ··· < σk ≤ n. Now n we associate a Schubert symbol to an element ` ∈ Grk(R ) by noting σi(`) the jump places, that is, σi(`) is characterized by dim(R ∩ `) = i and dim(Rσi(`)+1) ∩ ` = i + 1. Observe that in the case k = 1 of n n+1 RP = Gr1(R ), the sequence has exactly one element σ1. Then n i i given a line ` ∈ RP , σ1(`) measures the R such the ` ∈ R ; for any line in the top cell, σ1(`) = n + 1; and then the Schubert symbol parametrizes the cell decomposition of the projective spaces. The same phenomenom occurs for the general Grassmannian: given a Schubert symbol σ = 1 ≤ σ1 < σ2 < ··· < σk ≤ n, the Schubert cell 3.1. COMBINATORIAL STRUCTURE 41 associated to σ is the set n e(σ) = {` ∈ Grk(R ): σi(`) = σi} . Obviously, and plane ` belongs to exactly one such Schubert cell. i n Define the open half-space Hi ⊂ R ⊂ R by i H = {x1, . . . , xi, 0,... 0) : xi > 0}

Let σ = 1 ≤ σ1 < ··· < σk ≤ n) be a Schubert symbol. A plane ` ∈ e(σ) admits an unique ordered orthonormal basis ~v1, . . . ,~vk such σi n that ~vi ∈ H . Therefore we have, as in the RP case, successfully lifted n the situation to the orthogonal Stiefel manifold Vk(R ). Consider then E(σ) to be the lifted cell:

n σi E(σ) = {(~v1, . . . ,~vk) ∈ Vk(R ): ~vi ∈ H } . Let E¯(σ) be the closure of E(σ). Lemma 3.7. Each E¯(σ) is homeomorphic to a closed disk Ds, where Pk s = i=1 σi − i. Proof. We proceed by induction on k. For k = 1, there is only one element in the Schubert symbol and the proof is the same as for the RP n case. Let σ = 1 ≤ σ1 < ··· < σr < σr+1 ≤ n be a Schubert symbol, andσ ˜ = 1 ≤ σ1 < ··· < σr ≤ n1 < n be the “truncated” symbol obtained by deleting the last element. By the induction hypothesis, ¯ Pr ¯ E(˜σ) is a closed disk of dimension i=1 σi − i . Let us now build E(σ) form E¯(˜σ) in order to apply the induction step. In order to get an idea of the structure of this set, let us begin by fixing the first r elements of such a base as the simplest ones: for i ≤ r ~ let bi be the unit vector with all coordinates zero with the exception of ¯ the σi coordinate, which is taken to be one. Let D ⊂ E(σ) be t ~ ~ D = {(b1,..., br,~vr+1)} ~ where bi is fixed as above and ~vr+1 is free except for the restriction of being in Hσr . The set D is clearly homeomorphic to a closed disk of dimension σr − r − 1, by a reasoning similar to the case r = 1. ¯ Exercise 3.8. Show that D~v1,...,~vr ⊂ E(σ) is a homeomorphic to a closed disk of dimension σr − r − 1, where D~v1,...,~vr is constructed as above the first r elements of the basis are fixed but arbitrary elements ¯ of E(˜σ). Hint: use the orthogonal group to transform (~v1, . . . ,~vr) into ~ ~ σr ⊥ (b1,..., br) while fixing (R ) . The previous exercise plus the induction hypothesis suggests that ¯ α β α+β Pr E(σ) is the product D × D ' D , where α = i=1 σi − i and 42 3. ALGEBRAIC TOPOLOGY

β = σr − r − 1; this would finish the proof of the lemma. However we have to be careful: given a space Z and a projection π : Z → X such that, for any x ∈ X, π−1(x) is homeomorphic to some fixed space Y , does not mean that Z is the product X × Y (e.g., fiber bundles, [30]). But in this case, it works: an explicit homeomorphism φ : E¯(˜σ) × D → E¯(σ) is given by

φ((~v1, . . . ,~vr), ~u) = (~v1, . . . ,~vr,T~v1,...,~vr u) where T~v1,...,~vr is a orthogonal map depending continuously on ~v1, . . . ,~vr ~ ~ and transforms b1,..., br into ~v1, . . . ,~vr. This map can be constructed as follows: given unit vectors ~u,~v ∈ RN , let R(~u,~v) be the the orthogonal map that takes ~u to ~v and is the identity in span(~u,~v)⊥. Exercise 3.9. Write an explicit formula for such R and show that the map ~u,~v, ~x 7→ R(~u,~v)(~x) is continuous.

Then T~v1,...,~vr can be deﬁned as the composition ~ ~ ~ T~v1,...,~vr = R(br,~vr) ◦ R(br−1,~vr−1) ◦ · · · ◦ R(b1,~v1) , which ﬁnishes the proof of the lemma. n We have then covered Grk(R ) with images of closed disks. It is easy to see that (1) The proof of the previous lemma furnishes a homeomorphism between the open disk and E(σ). (2) E(σ) projects homeomorphically to e(σ). (3) The image of the boundary of the disk in E(σ) projects to the union of cells of lower dimension. Then we have Theorem 3.10. Each set e(σ) is the image of an open cell int(Dr) n P in a CW decomposition of Grk(R ), with r = i σi − i. Exercise 3.11. Identify the top cell of chapter 1 as one of the cells e(σ). Exercise 3.12. How many cells of a given dimension does the Grassmann manifold has?

3.2. Homotopy The idea of homotopy theory is to study deformation classes of maps instead of the maps themselves in order to get manageable complexity. 3.2. HOMOTOPY 43

3.2.1. Basic Deﬁnitions. Let us make precise what we mean by deformation classes: Definition 3.13. Let X,Y be two topological spaces. Two continuous maps f0, f1 : X → Y are said to be homotopic if there exists a continuous map H : X × [0, 1] → Y such that f0(x) = H(x, 0) and f1(x) = H(x, 1).

The intuition is that f0 and f1 can be deformed into each other through the deformation parameter t ∈ [0, 1]. The map H is said to be a homotopy between f1 and f2. It is quite useful in Algebraic Topology to consider not only spaces, but pairs of spaces (X,A) where A ⊂ X (think for example of the pair (Dn,Sn−1)). In this category, we naturally require that maps f :(X,A) → (Y,B), and homotopies H must satisfy H(a, t) ∈ B for all a ∈ A. A more restrictive but still quite frequent notion is that the homotopy does not move the set A at all: we said that a homotopy H : X × [0, 1] → Y is said to be relative to A (written “rel A”) if H(a, t) = H(a, 0) for all a ∈ A. In the case A and B are points, the two concepts above merge (since there is no space to move inside a single point!): the category of pointed spaces has objects (X, x0), where X is a topological space, x0 ∈ X, and continuous maps f :(X, x0) → (Y, y0) are required to satisfy f(x0) = y0. The homotopies are then required to be rel {x0}. Exercise 3.14. Show that in all the cases described above, “is homotopic to” is an equivalence relation. The idea now is to study the rough topology of a space by classifying its “holes”. Each of these holes is detected at dimension n by homotopy n classes of maps (S ,N) → (X, x0), where N is, say, the north pole (1, 0,... 0) of Sn.

Definition 3.15. Let (X, x0) be a pointed space. The n-th homotopy group πn(X, x0) is the set of homotopy classes (rel {N}) of maps n f :(S ,N) → (X, x0). Note that the words in the deﬁnition are homotopy groups. Indeed, these sets have rich algebraic structures, the most elementary of them being a group structure which is Abelian for n > 1. We will not delve into this rich theory here (again, see [14]), but it is natural what the neutral element should be: n Definition 3.16. A map f :(S ,N) → (X, x0) is said to be null- homtopic if there is a pointed homotopy between f and the constant map c(θ) ≡ x0. 44 3. ALGEBRAIC TOPOLOGY

Then the class of nullhomotopic maps is the neutral element of the group structure. A nullhomotopic map detects nothing; a n-dimensional “hole” is detected by a map if it is not nullhomotopic. The classical picture is how the ”hole” in R2 −{~0} is detected by a map from S1 that surrounds it, however this same map does not detect the 2-dimensional hole in R3 − {~0}

2 ~ 3 ~ Figure 7. π1 detects the hole in R − {0} but not in R − {0}

3.2.2. Geometric generators of Homotopy groups. In purely topological applications, a map from a sphere into a space representing a homotopy class can arbitrarily deform the sphere (in a continuous way, of course); after all this is the idea of homotopy. For a geometer, however it is not only aesthetically pleasing, but a necessity, to obtain representatives of homotopy classes that preserve the “roundness” of the sphere. For example, such geometric generators in the works of Rigas [26] lead to the construction of non-negatively curved metrics on many spaces. Thus we prefer to generate an element of the first homotopy group of the torus with something like figure 8 below insted of figure 9. In this section we will construct special subsets of the Grassmann manifold that later turn out to be generators of homotopy groups. The proof of these facts is beyond the scope of these notes; however the construction also illustrates that, by focusing of sets explicitly defined by elementary methods, we can construct representatives of more advanced phenomena which naturally carry geometric structure and information. We take the definition from [34]. The reader interested in the in- terplay between these spaces and topology can consult [26, 32] 3.2. HOMOTOPY 45

Figure 8. A good representative of an element of π1(T )

Figure 9. A not so good representative of an element of π1(T )

Definition 3.17. A pair of subspaces `, r of Rn are said to be 2 2 isoclinic if therer are real numbers α , β such that, for all ~x1, ~x2 ∈ ` and ~y1, ~y2 ∈ , 2 r r 2 hπ`(~y1), π`(~y2)i = α h~y1, ~y2i and hπ (~x1), π (~x2)i =rβ h~x1, ~x2i , where π`, πr are the orthogonal projections onto ` and , respectively. Some comments on the definition are in order. Note that setting ~x1 = ~x2 and ~y1 = ~y2 above justifies the squaring of α and β; if such coefficients exist they cannot be negative. Also, we write the projection on both sides just for aesthetic symmetry: since orthogonal projections are self-adjoint,

hπ`(~y1), π`(~y2)i = hπ`(~y1), ~y2i = h~y1, π`(~y2)i . 46 3. ALGEBRAIC TOPOLOGY

In principle, ` and r have different dimensions. However this is only possible in the degenerate (orthogonal) case: ` and r are orthogonal if and only if they are isoclinic, with one of (and therefore both) α and β equal to zero.r Let us deal with the non-orthogonal case, that is αβ 6= 0: If ` and satisfy the first equation above, by settingr y1 = y2 we find that ker π`|r must be trivial, which means that dim ≤ dim `. Then the second condition implies that dim r = dim ` and each projection restricted to the other subspace is an isomorphism. Again by setting y1 = y2 above, we find that in the non-orthogonal case we get also that α2 ≤ 1, β2 ≤ 1 and the upper bound is only attained when ` = r, since the same α, β must work for all the vectors in the equations above. Then if ` 6= r and they are isoclinic, `∩r = {~0}. Thus we find that two subspaces ` 6= r are isoclinic if and only if either they are orthogonal and α = β = 0 or dim ` = dim r, ` and r belong to the same Grassmann manifold ` ∩ r = {~0}, and they satisfy one of the equations above (the other being automatic by using inverses) with a non-zero coefficient that is also strictly less than one. Let us begin by studying n-dimensional isoclinic spaces in the half- 2n n ~ Grassmannian Grn(R ). Consider the base space `0 = R × {0} ⊂ Rn × Rn, and `⊥ its natural complement. Letr us find, using the slope parametrization, all non-orthogonal n-planes isoclinic tor `r. Note that by the discussion after the definition, the projection π`| : → ` must be an isomorphism; therefore, any non-orthogonal space isoclinic to ` is contained in the top cell centered at ` and the slope parametrization covers it. Expressing any r in the slope parametrization of Theorem 1.17, 1 (3.1) r = column space of n×n , Z and let us discover conditions on the (n×n) matrix Z that characterize the isoclinic property. Any vector ~y ∈ r can be written as ~x ~y = Z~x˜ where ~x ∈ `. Taking the inner product of two such vectors, ˜ ˜ 2 h~y1, ~y2i = h~x1, ~x2i + hZ~x1, Z~x2i = α h~x1, ~x2i , where the first equality is given by the orthogonality of the complementary subspaces chosen and the second is the isoclinic condition. Since α2 < 1, it will be useful to write α = cos(θ) with 0 < θ < π/2. Then r is isoclinic to ` if and only if there is a fixed θ such that ˜ ˜ 2 hZ~x1, Z~x2i = sin(θ) h~x1, ~x2i . 3.2. HOMOTOPY 47

This justifies the word isoclinic: ` and r are at “constant angle” θ to each other. Observing that Z˜ preserves inner products up to a constant multiple, we have Proposition 3.18. A n-plane r described in the slope parametrization as in (3.1) is isoclinic to ` if and only if Z = sin(θ)Z˜ is an orthogonal matrix, Z> = Z−1. There is a little contraband in the previous proposition: by fixing respective basis, we have implicitly identified the complementary subspaces ` and `⊥. An alternate viewpoint, from ther orthogonal Stiefel manifold, is to consider now an orthonormal basis of . If (~x1, . . . ~xn) is an orthonormalr basis of `, then cos(θ)~xi + sin(θ)Z~xi is an orthonormal basis of . This motivates the following construction: consider the transformation of R2n described by the block matrix form in the canonical basis of Rn ⊕ Rn = ` ⊕ `⊥, 0 −Z> Jr = n . Z 0n The matrix Jr satisfies: (1) Jr2 = −I, (2) Jr> = −Jr, (3) Jr(`) = `⊥, (4) r = (cos(θ)I + sin(θ)Jr)(`). Items (1) and (2) above mean that Jr is a complex structure on R2n, and together with (3) and (4) determines Jr uniquely as a function of r. Also (cos(θ)I + sin(θ)Jr) belongs to the orthogonal group O(2n) This furnishes a partil section of r to the orthogonal group O(2n); let us clarify the meaning of “partial section”: choosing a base plane N N `0 ∈ Grk(R ), there is a projection p : O(N) → Grk(R ) given by p(T ) = T `0.A section is a map that goes in the other direction s : N Grk(R ) → O(N) such that p ◦ s(`) = `. In general, a continuous section does not exists. However, when restricted to isoclinic subspaces to a given one ` as is our case, s(r) = (cos(θ)I + sin(θ)Jr) is such that p(s(r)) = r. 2n We want to consider subsets of Grn(R ) such that all elements are mutually isoclinic. This let us consider a third plane q, that is isoclinic to ` and r. Being isoclinic to ` means that there is a complex structure Jq satisfying (1) - (3) above, and q = (cos(φ)I + sin(φ)Jq)(`) for some angle 0 < φ < π/2, and 0 −W > Jq = n , W 0n 48 3. ALGEBRAIC TOPOLOGY where W is an orthogonal matrix, W > = W −1. We want to determine the conditions on Jq that translate q and r being isoclinic. Exercise 3.19. Use exercise 1.51 to show that the orthogonal projection πr with image r is given by I 0 πr = (cos(θ)I + sin(θ)Jr) (cos(θ)I − sin(θ)Jr) 0 0 cos2(θ) cos(θ) sin(θ)Z> = . cos(θ) sin(θ)Z sin2(θ) After some computation with the previous exercise as starting point, 2 we find that the isoclinic condition hπr~u1, πr~u2i = σ h~u1, ~u2i translates to JrJq + JqJr = −2λI , where λ is a constant. This λ serves as a kind of “inner product” between the slopes parametrizing isoclinic planes. Exercise 3.20. Give a formula relating λ, φ and θ. We define then an analog of an orthonormal basis for a set of matrices that produce isoclinic supbspaces: Definition 3.21. A set of matrices (2n × 2n) complex structures J1,...,Js is said to satisfy the Hurwicz conditions for a given n-plane ⊥ ` if Ji(`) = ` and JiJk + JkJi = −2δjI. Note that, if i = j, the Hurwicz condition is a consequence of the Ji being complex structures, and if i 6= k, they mean that Ji and Jk anticommute: JiJk = −JkJi. Note that having a system of matrices satisfy the Hurwicz conditions, the planes `0, . . . , `r given by `0 = `, ì = (cos(θi)I + sin(θi)Ji)` for certain angles θi are a system of mutually isoclinic planes. What we do now is “filling” this kind of set, finally defining isoclinic spheres:

Definition 3.22. Let J = {J1,...,Jk} be a set of complex structures in R2n satisfying the Hurwicz conditions for some n-plane ` ⊂ R2n. The set 2 2 SJ = {(a0I + a1J1 + ··· + akJk)`, a0 + . . . ak = 1} , is called the isoclinic sphere spanned by J . The results of Rigas [26] (for generators) and Wang [32] (for arbitrary elements) imply Theorem 3.23. Given k ∈ N, isoclinic spheres give representatives 2r of all elements πk(Grr(R )) for suﬃciently large r. 3.2. HOMOTOPY 49

In reality Rigas’ and Wong’s results concern the so-called stable homotopy groups; see e.g. chapter 4 of [14]

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