An elementary introduction to the Grassmann manifolds XX ...cduran/Ensino/Semestres... · Algebraic...

An elementary introductionto the Grassmann manifolds

XX Brazilian TopologyMeeting

UFPR - UTFPR, Curitiba,2016

Carlos Eduardo Duran Fernandez

Universidade Federal do Parana

Contents

Preface ix

Chapter 1. Basic constructions 11.1. Linear Algebra 41.2. Topology 81.3. An Euclidean presentation of Grassmann manifolds 14

Chapter 2. Geometry 172.1. An application to ordinary differential equations 172.2. Homogeneous spaces and Cartan-Klein geometry 212.3. Congruence of curves in the Grassmann manifold 232.4. Jets of curves in Grassmann manifolds 272.5. Notable subsets of the Grassmann manifold 29

Chapter 3. Algebraic Topology 353.1. Combinatorial structure 353.2. Homotopy 42

Bibliography 51

vii

Preface

There is an enormous amount of mathematics that can be treatedand exemplified in the Grassmann manifolds. The purpose of thesenotes is to give an introduction to the Grassmann manifolds that is aselementary as possible, while also showing the way to more advancedconstructions. Thus, the present work can be considerd as a “pre-quel” to more specific treatments, for example the books of Milnor andStasheff [20] for applications to the topology of vector bundles andcarachteristic classes, Piccione and Tausk [23] for the applications toGeometry and Calculus of Variations, and [11] for the algebraic geom-etry viewpoint. We also introduce the reader to the invariant theoryof curves in the Grassmannian [2, 7, 8]) and the Grassmannian ofC∗-algebra [4, 24]

The main leitmotiv is: how to actually get a grasp on a Grassmannmanifolds and compute? In the author’s experience, the jump fromabstract definition to confident computing, an ability needed for qualityresearch, is a difficult one; here we describe how this jump can be madefor Grassmann manifolds.

The work is divided in three chapters: in the first one we give thebasic definitions and, more importantly, models and parametrizationof the Grassmann manifolds. Then in chapter 2 we keep identifying ob-jects in the Grassmannian in terms of quantities living in more familiarspaces; in particular derivatives of curves and give applications to thecongruence problem. Finally chapter 3 is intended to give a flavor tothe finer topology of the Grassmann manifold: we give the standardCW decomposition in terms of Schubert cells (homology) and describethe isoclinic spheres that have been used as geometric representativesof stable homotopy groups.

These notes are the companion of a minicourse given at the XXBrazilian Topology Meeting. I would like to thank Eduardo Hoefeland the organizers for the invitation to teach this minicourse.

Carlos Eduardo Duran Fernandez

ix

CHAPTER 1

Basic constructions

The Grassmann manifold Grk(Rn) is the set whose points are thek-dimensional subspaces of Rn.1 A look at the simplest case gives someflavor of what is to come:

There is only a single 1-dimensional subspace of R1; therefore theGrassmannian Gr1(R1) is just the singleton {R1}; that was too simple.Consider then the next case: the Grassmannian Gr1(R2); the inter-esting structure of the Grassmann already appears in this level. Eachpoint of the this Grassmann manifold is a straight line through theorigin in R2. We now describe Gr1(R2) in terms we can manage:

We can first think of describing a point ` in this Grassmannian byits equation,

`(m) = {(x, y) ∈ R : y = mx} ,where m is the slope of the line. Thus we are lead to think that theGr1(R2) 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) approximatesthe vertical line as m → +∞, and also as m → −∞. This suggeststhe following: extend the real line by a single “point at infinity” ∞,and then Gr1(R2) is completely parametrized by this space; in a morepedantic language, we have

Proposition 1.1. The map R 3 m 7→ `, where ` is the uniquestraight line through the origin with slope m, and ∞ 7→ {x = 0} givesa natural bijection between R ∪∞ and the Grassmannian Gr1(R2).

Remark 1.2. The slope m of a non-vertical line ` is the uniquenumber characterized by the relation (1,m) ∈ `. Here the unreachablecharacter of vertical line is patent: a vector (1,m) can never belong inthe 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 Grassmannmanifolds of quite general vector spaces, e.g., [4], where they consider the Grass-mann manifold of certain infinite dimensional vector spaces. We can also considervector spaces over fields other than the real numbers R. However in this introduc-tion 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) closeto the vertical line, we must think of the point at infinity as beingclose to large slopes, and thus we have that the Grassmann manifoldGr1(R2) is a circle: the two points of the drawn semi-circle representthe same line {y = 0} and therefore must be identified. The meaning

Figure 1. Gr1(R2) 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 aswith the other. The numbers α and β are arbitrary, with the exceptionof 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 byfixing an equivalence relation.

Exercise. Show that `(α, β) = `(γ, δ) if and only if there is 0 6= λ ∈ Rsuch that λα = γ, λβ = δ. (The same λ must work simultaneously forboth conditions!)

Thus we have

1. BASIC CONSTRUCTIONS 3

Proposition 1.3. The map [(α, β)] 7→ `(α, β) induces a natural bi-

jection between Gr1(R2) and the set of equivalence classes R2−{~0}/ ∼,where (α, β) ∼ (γ, δ) if and only if there is 0 6= λ ∈ R such thatλα = γ, λβ = δ.

How do we recognize Gr1(R2) as a circle in this presentation? Inorder do accomplish this, we need to give a slightly different -moreprecisely, a dual- interpretation of the bijection above. Two pointsuniquely determine a straight line; since we already now that the origin(0, 0) is a point through which the line passes, we only need anotherdifferent point (a, b). Two such points (a, b) and (c, d) determine thesame line if and only if they satisfy the same kind of relation as inproposition 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 coverall the equivalence classes. Thus proposition 1.3 takes the form

Proposition 1.4. The map (a, b) 7→ `(a, b), where (a, b) ∈ S1,induces a natural bijection between Gr1(R2) and the set of equivalenceclasses S1/ ', where (a, b) ' (c, d) if and only if (a, b) = ±(c, d).

Figure 2. Gr1(R2) is identified with S1/±

Proposition 1.4 tells us that Gr1(R2) is in a natural bijective corre-spondence with the circle S1 with antipodal points identified. This iseasily seen to be a circle again: all equivalence classes are contained inthe right half circle {(a, b) ∈ S1 : b ≥ 0}, and now each (a, b) repre-sents exactly one equivalence class with the exception of (0,±1) whichrepresent the same vertical line. The fact that this is a circle is left asa visualization exercise for now; later in section 1.2 we will constructthe tools necessary to give precise statements on these identificationsof the Grassmann manifolds.


Exercise 1.5. Relate all the presentations of the GrassmannianGr1(R2) above; that is, how is the same line ` represented with thedifferent coordinates given by propositions 1.1, 1.3 and 1.4?

Remark 1.6. The Grassmannians Gr1(Rn+1) are called real pro-jective spaces and are denoted by RP n. In general, this kind of processof “describing an object in manageable terms” is usually called coor-dinatization or parametrization. We have thus “coordinatized the realprojective line RP 1”.

1.1. Linear Algebra

We now begin the study proper of the generic Grassmann mani-folds. The methods used to coordinatize Grk(Rn) are similar in spiritto the introduction on RP 1 above, with the difference that the previoussection was written geometrically (note that we wrote about “lines” in-stead of “one-dimensional subspaces”), whereas for general Grassmannmanifolds we use a more involved linear algebraic approach.

Consider first the construction used in proposition 1.4, but withoutnormalizing. Note that we tagged a line by giving a non-zero vector onthe 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-dimensionalsubspace p of Rn, we can choose an ordered basis (~v1, . . . , ~vk) of p; thismotivates the following definition:

Definition 1.7. The Stiefel manifold Stk(Rn) is the set of k-tuplesof linearly independent vectors in Rn.

Remark 1.8. The Stiefel manifold St1(Rn) is just Rn − {~0}.

Remark 1.9. In geometrical contexts, an element of a Stiefel man-ifold is sometimes called a frame.

There is a tautological function π : Stk(Rn)→ Grk(Rn) 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 samesubspace? The answer is easily found if one writes Stk(Rn) adequately:we organize an element (~v1, . . . , ~vk) ∈ Stk(Rn) as the (n× k)-matrix Awhose columns are the vector vi:

A =(~v1| . . . |~vk

),

1.1. LINEAR ALGEBRA 5

where the vertical bars just denote that we are emphasizing the matrixA written as columns. The condition of the vectors being linearlyindependent translates to the matrix A having maximal rank k. Fromnow on each time we mention the Stiefel manifold it will be organizedin matrix form as described above.

Now reflecting on the way matrices are multiplied, we observe thatif we transform A with right multiplication, that is, if B = AX, whereX is a (k × k) matrix, the columns of B are linear combinations ofthe columns of A, and thus the column space of B is contained inthe column space of A. The matrix X is invertible if and only if thecolumns spaces of A and B are actually the same. Then it is easy toshow that

Proposition 1.10. The map π induces a bijection between Grk(Rn)and the quotient Stk(Rn)/ ∼, where A ∼ B if and only if there is aninvertible (k × k) matrix X such that BX = A.

Note that the multiplication by λ on the previous section is on theleft because there we wrote the vector as row vectors instead of columnvectors.

In the presence of the standard Euclidean structure on Rn, thereduction to S1 is extended as follows:

Definition 1.11. The orthogonal Stiefel manifold Vk(Rn) ⊂ Stk(Rn)is the set of (n×k)-matrices A satisfying A>A = Ik, the (k×k) identitymatrix.

Remark 1.12. The orthogonal Stiefel manifold V1(Rn) ⊂ St1(Rn)

is just the unit sphere Sn−1 ⊂ Rn − {~0}.We now have the exact generalization of proposition 1.4:

Proposition 1.13. The map π induces a bijection between Grk(Rn)and the quotient Vk(Rn)/ ', where A ' B if and only if there is anorthogonal (k × k) matrix X such that BX = A.

Exercise 1.14. Construct the appropriate quotient and prove propo-sition 1.13 for an arbitrary Euclidean inner product on Rn.

Let us now see how the “slope” coordinatization of proposition 1.1generalizes. Recall that in the introductory RP 1-case, finite slopes didnot reach all planes, and the set of reachable planes was identified withR. The following definitions generalize these sets of “reachable” and“unreachable” planes:

Definition 1.15. The top cell UV associated to V is the set

UV = {p ∈ Grk(Rn) : p ∩ V = {~0}} .


The train TV of V is the complement of the top cell, that is,

TV = {p ∈ Grk(Rn) : dim(p ∩ V ) > 0} .

The low dimension of the RP 1 case hides the complexity of thetrain: the train of the vertical line {x = 0} in R2 is the line itself. Forhigher dimension, the train is stratified by the increasing dimensions ofp ∩ V .

Let us fix these ideas with some concrete planes in Grk(Rn): fix theplane H = span(~e1, . . . ~ek) ∈ Grk(Rn), where ~ei denotes the canonicalbasis of Rn; that is,

H = {~v = (x1, . . . , xn) ∈ Rn : 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 similarsituation occurs here, with the difference that the “problematic” set iscombinatorially more involved.

Consider the complementary plane V ∈ Grn−k(Rn), H ⊕ V = Rn,

V = {~v = (x1, . . . , xn) ∈ Rn : xi = 0 for i ≤ k} ,which is the analog of the vertical plane {x = 0} of the RP 1 case (butnote that here the dimensions of the horizontal and vertical planes arepossibly different!)

We now follow the lead of remark 1.2: there are many “slopes” withrespect to H and V which collectively characterize the planes in thetop cell UV . Indeed, we have

Lemma 1.16. Let p ∈ Grk(Rn) be a plane in the top cell of V . Thenfor 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},kerπ|p = {0} and then π|V : p → H is an isomorphism. Let ~vi =(π|p)−1(~ei). By construction, vi ∈ p is uniquely determined by ~ei andhas the form ~ei + ~zi, ~zi ∈ V . �

Lemma 1.16 furnishes a great oportunity to introduce a recurringtheme in computations in Grassmann manifolds: writing conditions interms of the Stiefel manifold. Let p ∈ Grkn, and A = (~v1, . . . ~vk) ∈

1.1. LINEAR ALGEBRA 7

Stk(Rn) 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

A =

v11 v12 · · · v1k...

......

vk1 vk2 · · · vkkv(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 ofthe 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

X =

v11 v12 · · · v1k...

......

vk1 vk2 · · · vkk

is invertible. Note that the condition of the top square part of A beinginvertible only depends on the span of the columns; that is, it is acondition on Stk(Rn) that “descends” to a condition on Grk(Rn). Usingproposition 1.10, we see that the columns B = AX−1 also span p, andthis matrix has the form

B =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1z(k+1)1 z(k+1)2 · · · z(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 thetop cell UT as the class of a matrix in Stk(Rn); we have then proved

Theorem 1.17. Let H ∈ Grk(Rn), V ∈ Grn−k(Rn) be complemen-tary subspaces as above, and let Mn−k,k(R) denote the space of (n−k)×kmatrices. Then the map φH,V : Mn−k,k(R)→ Grk(Rn) given by

φH,V (Z) = column space of

(1k×kZ

)is injective, and its image is the top cell UV .


Note that Z = 0 corresponds toH. This is almost the generalizationof proposition 1.1; there we established that the only missing line,the vertical line, could be safely assigned the slope ∞; whereas inproposition 1.17 the missing planes, which are exactly the members ofthe train TV , have much combinatorial complexity and will be studiedin chapter 3. Note thatH corresponds to Z = 0 in this parametrization.

We finish this section on linear-algebraic parametrizations ofGrk(Rn)with a change of point of view. Let S : H → V by a linear transfor-mation. The graph of S is the set ΓS ⊂ H ⊕ V = Rn given by

ΓS = {x+ S(x) : x ∈ H} .The graph is then a map Γ : Lin(H, V ) → Grk(Rn); note that

Γ0 = H. If we fix the canonical basis of H and V above, then lineartransformations correspond to (n− k)× k matrices; then lemma 1.16,can be interpreted as a linear transformation Sp : H → V defined bythe linear extension of Sp(~ei) = ~zi and extend by linearity.

Theorem 1.18. Let H ∈ Grk(Rn), V ∈ Grn−k(Rn) be comple-mentary subspaces as above. Then the graph map Γ : Lin(H, V ) →Grk(Rn), is injective, and its image is the top cell UV .

Exercise 1.19. Prove the results in this section with the words “asabove” removed, that is, with H ∈ Grk(Rn), V ∈ Grn−k(Rn) arbitrarycomplementary subspaces.

Exercise 1.20. Repeat exercise 1.5 in the general situation givenhere.

1.2. Topology

Up to now we have refrained on purpose to include any topolog-ical concepts; hence the word “bijection” in many of the results inthe previous sections. We now introduce a topology on Grk(Rn); thistopology should reflect our intuitions of k-planes through the originbeing “close”.

Let us give a brief review of the quotient topology, which is essentialin the study of the topology of the Grassmann manifolds. Let X be atopological space, ∼ an equivalence relation on X and Y = X/ ∼ thequotient, that is, the set of all equivalence classes. The projection tothe 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 openin Y if and only if π−1(U) is open in X. It is then, by definition, thelargest topology that makes the projection to the quotient continuous.

The key hint on thinking on quotient spaces is: don’t. Even whenworking 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 ofY are projections of open sets of X; but this can be remedied: an set Cin X is saturated if π−1(π(C)) = C, which translates to the property: ifC contains a point x then it also contains all points x1 with x1 ∼ x. Forthe equivalence relation ~0 6= (x, y) ∼ (λx, λy) of the dual of proposition1.3, 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 itscontinuous functions. For quotient spaces, we have

Proposition 1.21. Let π : X → X/ ∼ be the projection to thequotient. The quotient topology is characterized by: a function f :X/ ∼→ Z is continuous if and only if the composition f ◦ π : X → Zis continuous.

Proposition 1.21 in practice means that the best way of definingfunctions from quotients X/ ∼ is to define them as functions on Xthat are constant on equivalence classes; that is, f(x) = f(x1) wheneverx ∼ x1. We will have plenty of opportunities to use this technique.

We now define the canonical topology of Grassmann manifold: notefirst that the space Mnk(R) is n × k real matrices is isomorphic, as avector space, to Rkn. We endow Mnk(R) with the topology induced bythis identification.

Exercise 1.22. Show that Stk(Rn) is an open subset of the spaceof (n × k) matrices. Hint: the rank of A is the same as the rank ofA>A.

The topology of the Stiefel manifold is the one induced as an opensubset of Mnk(R).


Definition 1.23. The Grassmann Grk(Rn) manifold is the topo-logical quotient of the Stiefel manifold Stk(Rn) under the equivalencerelation ∼ of proposition 1.10.

Let us now establish a few facts about the Grassmann manifoldsas topological spaces; this is very good training on the workings of thequotient topology.

Exercise 1.24. Show that the Stiefel mainfold Stk(Rn) is con-nected if n > 1.

Then, the Grassmann manifold, being a continuous image of a con-nected space, is connected.

Consider now Rn endowed with the standard (or any other) Eu-clidean inner product, so that the presentation of Grk(Rn) of proposi-tion 1.13 makes sense. We have

Proposition 1.25. The quotients Vk(Rn)/ ' and Stk(Rn)/ ∼ arehomeomorphic.

Proof. True to our motto of working upstairs and not in the quo-tient, consider the maps i : Vk(Rn) → Stk(Rn) and GS : Stk(Rn) →Vk(Rn) be the inclusion and the Gram-Schmidt process, respectively.

Exercise 1.26. Show that the Gram-Schmidt process is continu-ous.

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

Denote by π∼ and π' the projections to the quotient of the respec-tive equivalence relations ∼ and '. Exercise 1.27 says that π' ◦ GSand π∼ ◦ i are constant on equivalence classes, thus inducing continu-ous functions i0 : Vk(Rn)/ '→ Stk(Rn)/ ∼ and GS0 : Stk(Rn)/ ∼→Vk(Rn). Since GS ◦ i is the identity of Vk(Rn), GS0 ◦ i0 is the identityof Vk(Rn)/ '. In the other direction, η = i ◦ GS is not the identityof Stk(Rn), but for all x ∈ Stk(Rn), x ∼ η(x). Thus i0 ◦ GS0 is theidentity of Stk(Rn). �

Therefore the quotient Vk(Rn) is also the Grassmann manifold,topologically. However geometrically is a very different space.

Corollary 1.28. The Grassmann manifold Grk(Rn) is compact.

Proof. Grk(Rn) = π'(Vk(Rn)) is the continuous image of the com-pact (prove it!) space Vk(Rn). �

1.2. TOPOLOGY 11

Exercise 1.29. A function f : Rn → Rm is homogeneous of degreek if f(λ~x) = λkf(~x). Show that a homogeneous function induces a

continuous map f : RP n−1 → RPm−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 theGrassmann manifold as a homogeneous manifold, where we can applygeometrical methods.

Let T : Rn → Rn be an invertible linear transformation, representedin the canonical basis by a (n× n) invertible matrix T . Then the mapLT : Stk(Rn) → Stk(Rn), LTA = TA, takes classes to classes andtherefore induces a continuous map `T : Grk(Rn) → Grk(Rn). Theinverse of this map is `T−1 , and therefore `T is a homeomorphism.

We have

Lemma 1.31. Let H, H ∈ Grk(Rn), V, V ∈ Grn−k(Rn), such thatRn = H⊕V = H⊕V . Then there is an invertible linear transformationT : Rn → Rn such that T (H) = H, T (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 invertiblelinear transformation of Rn that satisfies the conclusion. �

The lemma means that, for many applications, one can “adjust”some situations in Grk(Rn) to a simple form via a suitable T . As anexample, we finish this section with another relatively simple topolog-ical property of the Grassmann manifolds:

Theorem 1.32. Let V ∈ Grn−k(Rk). Then the top cell UV is openand dense in Grk(Rn).

Proof. By lemma 1.31, we can assume that V is the span of thelast (n− k)-vectors of the canonical basis of Rn, and let H be the spanof the first k-vectors in Rn. The inverse image of the top cell UV isthen the saturated set in Stk(Rn) given by

UV = {A ∈ Stk(Rn) : the top (k × k) square of A is invertible } ,


as in the discussion after lemma 1.16; therefore the top cell is open onGrk(Rn). As for the density, given A ∈ Stk(Rn), consider the matrix

Aε = A+

(ε1k×k

0(n−k)×k

),

where ε is not an eigenvalue of the top (k × k) square of A. �

Now we can completely understand the top cells of a Grassmannmanifold; again by lemma 1.31 we assume the fixed V = span(~ek+1, . . . , ~en).

Theorem 1.33. The slope parametrization of Theorem 1.17 inducesa homeomorphism between Mn×k(R) ' Rkn and the top cell UV .

Proof. Theorem 1.17 shows that the map φH,V is continuous andinjective. Its inverse ψH,V is induced by the map from the inverse image

UV of the top cell as follows: if A =

(Xk×k

Y(n−k)×k

), then ψH,V (A) =

Y(n−k)×kX−1k×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 home-omorphic to Rm implies m = n), in the connected case the number nis well-defined and is called the dimension of the manifold.

We have

Theorem 1.36. The Grassmann manifold Grk(Rn) is a topologicalmanifold of dimension nk.

Proof. We have already established the compacteness (which im-ply 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. TheHausdorff condition is left as an exercise (be careful: continuous imagesof Hausdorff spaces are not necessarely Hausdorff). �

Definition 1.37. A compactification of a Hausdorff topologicalspace 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 inorder to embedd it into a compact manifold.

1.2. TOPOLOGY 13

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

πarctan(x).

We then have

Theorem 1.39. The Grassmann manifold Grk(Rn) is a compacti-fication of Rnk.

Proof. Follows directly from Theorem 1.32. �

Definition 1.40. The one point compactification of a space X isthe space Y = X ∪ {∞}, the map h : X → Y being the tautologicalinclusion, and the topology of Y given by: a neighborhood of points inX are just the neighborhoods of X, and neighborhoods of∞ are givenby the complments of compact sets in X.

Exercise 1.41. Let N = (0, 1) be the “north pole” of the circleS1. Show that the sterographic projection S described below gives ahomeomorphism between S1 and the one-point compactifiction of R.Generalize to a homeomorphism between the one point compactifica-tion of Rn and the unit sphere Sn.

Figure 4. The stereographic projection

The next two exercises show, in different ways, that Gr1(R2) =RP 1 is homeomorphic to the circle S1; thus we have understood andidentified our first Grassmann manifold.

Exercise 1.42. Show that map φ : RP 1 → one point compactification of Rgiven by

φ(`) =

{slope of ` if ` is not vertical,

∞ if ` is vertical.

induces a homeomorphism between RP 1 and the one point compacti-fication of R, and by transitivity and exercise 1.41, a homeomorphism


between RP 1 and the circle S1. Write out all the formulas and in-verses, and be very careful with the continuity at the vertical line/northpole/point at infinity.

We now identify RP 1 as S1 using the quotient presentation. Definethe function f : R2 − {0} → R2 by

f(x, y) =(x2 − y2, 2xy)

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 functionz 7→ (z/|z|)2.

Exercise 1.42 can be generalized:

Exercise 1.44. Redo the beggining of this chapter substitutingthe reals for the complex field C. Show that CP 1 is homemorphic tothe sphere S2 ⊂ R3 ; in this context we call S2 the Riemann sphere.Relate to your complex analysis course. Do the same for the Hamiltonquaternions H, but remember that they are not commutative. Try to dothe same with the Cayley octonions O, which are neither commutativenor associative. See [3].

We finish this section with the following exercise which will be usefulin chapter 3:

Exercise 1.45. The hereditary properties of Grassmann manifolds:Show that the inclusion Rn → Rn+s, (x1, . . . , xn) 7→ (x1, . . . , xn, 0, . . . 0),induces a continuous, injective map Grk(Rn) → Grk(Rn+s), which isa homeomorphism onto its image. Generalize for arbitrary injectivelinear maps.

1.3. An Euclidean presentation of Grassmann manifolds

We have been careful about using only the vector space structure ofRn, the exception being when we described the Grassmann manifold asa quotient of the orthogonal Stiefel manifold Vk(Rn), in order to provecompactness. If add an Euclidean structure to Rn (that is, a positivedefinite inner product), we can give a more confortable alternate de-

scription as a subset of the space Mn(R) ' Rn2of 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 linearalgebra 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 decompo-sition Rn = V ⊕W , then defining q(~x) = ~x for ~x ∈ V and q(~x) = 0for ~x ∈ W defines a projection q. Thus the set of projections is inbijective correspondence with the set of decomposition of Rn in twocomplementary subspaces.

There is then a map Q → ∪nk=0Grk(Rn), given by q 7→ Im(q) (themap 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 subspaceV , there are many complements W with Rn = V ⊕ W ). Thus Q is“bigger” than the Grassmannian, altough it is an interesting space initself [25].

Exercise 1.47. Show that if T is an invertible linear transforma-tion and p is a projection then Im(TpT−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 orthogo-nal V ⊥. Fix an orthonormal basis of Rn so that we translate lineartransformations 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 orthogonalimage and kernel if and only p2 = p = p>.

The previous exercise leads to the set

Π = {p ∈Mn(R) : p2 = 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, andp, p belong in the same connected component if and only if rank(p) =rank(p).

Consider now the map φ : Π→ ∪nk=0Grk(Rn) given as above send-ing a matrix to its column space (which is the image of the lineartransformation it represents).

Exercise 1.50. Show that φ restricted to each connected compo-nent is a homeomorphism. Find an explicit formula for its inverse.


Exercise 1.51. Show that the action of orthogonal matrices (aslinear trasnformations of Rn) on the Grassmannian is translated toconjugation by orthogonal matrices on the space of orthogonal projec-tions (Hint: use exercise 1.47)

Note that we can, at least for the topological properties, forgetabout the images and just use the algebraic properties (square and thetranspose) of the matrices p. This leads to a far-reaching generalizationof the Grassmann manifolds; in any algebraic structure where we havean square and a “transpose” we can define a Grassmann manifold; ifin addition we have a suitable norm we can also perform analysis andtopology. 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 recog-nition and parametrization “up to order zero”, that is, recognizing thepoints of the Grassmann manifolds. In this chapter we will look at thefirst 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 structurein 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 manifoldsthat has motivated the author through the years. Consider a secondorder, ordinary differential equation:

γ′′(t) = F (t, γ(t), γ′(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 un-derstanding of these equations is the linearization: a canonical processwhere one obtains from the equation above a linear differential equationof the form

(2.1) y′′(t) + p(t)y′(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 impliesthat the solutions are defined for all t ∈ R too).

It is a very important problem in this area to find or estimate theconjugate points: given t0 ∈ R, find t1 > t0 (if it exists) such that thereis 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 thesolutions of (2.1) are given by y(t) = a sinh(λ(t − t0)) , which do nothave conjugate points.

Note that, in the positive case, the bigger the constant λ2, theearlier the first conjugate point appears. Generalizing this, a typicaltheorem is the following consequence of the Sturm comparison theorem(see for example chapter X of [15]):

Theorem 2.2. Let y′′ + R(t)y = 0 be a second order ordinarydifferential 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 usingthe Grassmann manifold. For a single second order differential equa-tion, we actually use RP 1.

Recall that the space of solutions of the equation (2.1) is two dimen-sional, generated for example by two solutions ah(t), av(t) satisfying theinitial conditions

ah(t0) = 1, a′h(t0) = 0 and av(t0) = 0, a′v(t0) = 1 .

Let us draw the evolution of the solutions ah, av simultaneously aa curve in the plane R2; that is, consider the curve in R2 given by

A(t) =

(av(t)ah(t)

).

Note that this curve never touches the origin (because of uniquenessof solutions of ordinary differential equations).

Figure 1. Simultaneous drawing of the linearly inde-pendent solutions

2.1. AN APPLICATION TO ORDINARY DIFFERENTIAL EQUATIONS 19

Now the key insight is that the appearance of a conjugate point t1only 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 formcav for some constant c ∈ R, and therefore we can just use av itself fordetecting 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 associatedto the linear second order differential equation (2.1), and we have seenthat conjugate points only depend on the Jacobi curve.

Exercise 2.3. Find an analytical condition on a frame A(t) thattranslates 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 shapeof the curve A cannot happen if the frame A(t) is obtained with theindependent 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 con-jugate 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 generatesπ1RP 1.

Let us describe the higher dimensional case: consider now a systemof second order homogeneous linear differential equations like equation(2.1),

(2.2) ~y′′(t) + ~y′(t)P (t) + ~y(t)Q(t) = 0 ,

where ~y(t) are row vectors in Rn and P and Q are (n × n)-matrixvalued 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 botht1 and t0. We now generalize the one dimensional projective treatment:consider the frame

A(t) =

a11(t) . . . a1n...

...an1(t) ann(t)

a(n+1)1(t) a(n+1)n(t)...

...a(2n)1(t) a(2n)n(t)

where

• The rows ofA(t) are a system of linearly independent solutionsof (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 Stiefelmanifold Stn(R2n) and its projection to the Grassmann manifold is the

2.2. HOMOGENEOUS SPACES AND CARTAN-KLEIN GEOMETRY 21

Jacobi curve `(t) with values in the “half-Grassmannian” Grn(R2n)Now in contrast with the one-dimensional case, it is not enough to justconsider each individual row solution: a conjugate point will typicallyappear as a linear combination of the first n rows that also vanishesfor some other point t1. In order to deal with this complication, weprojectivize:

The space span(~e(n+1), . . . ~e2n} will be called the vertical subspaceV ; 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 theconjugate point problem in terms of the self-intersection properties ofa 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 equationsand 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 Grassmannmanifolds. Before that, in the next short section we will see what wemean by Geometry.

2.2. Homogeneous spaces and Cartan-Klein geometry

Let X be any set. A group of transformations G on X is a subsetof the space of bijections of X that is closed under compositions andinverses (in particular, the identity transformation is in G). Usually,the subset reflects some kind of structure of the space X: homeomor-phisms for topological spaces, isometries for metric spaces, etc. FelixKlein defined a geometry as the study of properties of a space that areinvariant under a group of transformations.

Example 2.4. Plane Euclidean geometry: (in modern terms) thespace is the plane R2 and the group of transformations is the Euclideangroup of distance-preserving maps of R2 (endowed with its canonicalnorm); this group is generated by reflections and includes translationsand rotations. An example of invariant property is the angle betweentwo lines.

Example 2.5. Plane affine geometry: the space is also the planeR2 , but the group is the affine group of transformations that preservestraight lines. This group is generated by invertible linear transforma-tions and translations. The angle between two lines is not invariantunder an affine transformation. An example of affine invariant is the

22 2. GEOMETRY

ratio abac

between the lengths of segmentes defined by three colinearpoints a, b, c.

A central problem is the congruence problem: given two objects ina class in the given geometry, when does there exist a transformationtaking one to the other?

Example 2.6. Two triangles in Euclidean geometry are congruentif 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 lengthof the sides is complete invariant: a set of computable quantities thatdetermine weather or not two objects are congruent.

The specific subset of the bijections of a space can be abstractedaway; given a group G a left action (resp. right action) of G on a spaceX is a group homomorphism (resp. antihomomorphism) of G into thebijections of X. Again, typically the action falls into a rather restrictedsubset of bijections (homeomorphisms, linear maps, etc.) A left actioncan be transformed into a right action and vice-versa by precomposingwith the group inverse map, whih is an antihomomorphism, but werather not do it in our context: left and right actions come concretelyby left and right multiplication of matrices. It is customary to denotea group action by a binary operation: given say a left action which isa homomorphism φ : G → bij(X), we write for example g · x insteadof φ(g)(x). The quotient of an action is the set of classes by the equiv-alence 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 everythingin 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 manifoldStk(Rn) as in chapter 1. The quotient by the right action gives theGrassmann manifold Grk(Rn), and the left action defines Grk(Rn) asa Klein geometry.

Remark 2.7. It is equally valid to consider the Grassmann mani-fold as a geometry using only orthogonal matrices, but this gives a com-pletely different geometry. This illustrates the principle of the same setwith different symmetry groups must be considered a different space.

The congruence problem we study in the next section is aboutcurves in the Grassmann manifold, motivated by section 2.1. Wewant to construct a complete invariant that decides: given two curves`1, `2 : [a, b] → Grk(Rn), 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 objectsand the methods are differentiable the study of invariants is usuallycalled 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 parametrizedcurves in the Grassmannian. The main references we use are

• For RP 1, the work of Flanders ([10]),• For the half-Grassmannian Grn(R2n), [2],• For the “divisible” Grassmannian Grn(Rkn), [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 equivalencefor general maps into Grassmann manifolds is a wide open area of re-search.

In line with the spirit of these notes, the main objective is to learnhow to effectively compute in a Grassmann manifold. The construc-tions described in this section illustrates a main theme: compute inthe Stiefel manifold quantities that only depend on the span of thecolumns.

The first order of business is to do a rough classification of how“twisted” a given curve is: let A : R → Stk(Rn) be a differentiablecurve. The linear flag associated to A is the set of inclusions

{0} ⊂ cs(A(t)) ⊂ cs(A(t)|A′(t)) ⊂ cs(A(t)|A′(t)|A′′(t)) ⊂ · · · ⊂ Rn ,

where “cs” stands for “column space of” and the vertical bar denotesjuxtaposition. The following exercise is then one fundamental instanceof 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 thecurve `(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 ofintegers 0 < k ≤ k1 ≤ · · · ≤ kr ≤ n which is called the Young diagramof `(t) and measures how free the derivatives ` are ([17]).

As is usual in equivalence problems, the nicest invariants occur in ageneric set of maximally twisted curves, since each successive derivativefurnishes more and more information.

24 2. GEOMETRY

Definition 2.9. A curve `(t) in the divisible GrassmannianGrn(Rkn)is said to be fanning if its Young diagram is the maximal sequence0 < n < 2n < · · · < (k − 1)n < kn.

This means that each derivative contributes as much as possibleto 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 ana-lytic condition of exercise 2.3 is that the curve is not fanning.

We now turn to the equivalence problem. We take the route thatmakes the presentation more elementary and compact, by means ofdoing the RP 1-case. The known higher order cases have similar intu-itions 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 inthe 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 thevectors A(t) and A′(t) are linearly independent and therefore a basisof R2. Any vector ~v ∈ R2 can be expressed in a unique way as a linearcombination of A(t) and A′(t), in particular the acceleration A′′(t).Then we have structural equation

(2.3) A′′(t) + 2A′(t)p(t) +A(t)q(t) = 0

for some real-valued functions p(t), q(t) uniquely defined by A(t). Thefactor of 2 in the middle one is convenient for some computations andis consistent with most of the literature. We write p and q on the rightso that they can be considered (1 × 1) matrices which will generalizeto higher dimensions, but each (1× 1) row of A satisfies a the secondorder differential equation of the form

a′′(t) + 2p(t)a′(t) + q(t)a(t) = 0 .

Note that if T : R2 → R2 is an invertible matrix, then if A = TA,then by multiplying both sides of (2.3) we see that the coefficients ofthe 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 istempting to declare then that these coefficients are the invariants weare seeking, and indeed, if our equivalence problem was of fanningcurves of frames in R2 endowed with the action of GL(2,R) then thiswould be right.

Exercise 2.11. Prove the assertion of the last phrase of the previ-ous paragraph.

2.3. CONGRUENCE OF CURVES IN THE GRASSMANN MANIFOLD 25

However, the coefficients p and q change when we use two differentframes to describe the same curve in RP 1. Let us see what happens ifwe 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)

2A′(t)p(t) = 2B′(t)X(t)p(t) + 2B(t)X ′(t)p(t)

A′′(t) = B′′(t)X(t) + 2B′(t)X ′(t) + B(t)X ′′(t)

adding these equations, we have that B satisfies the differential equa-tion with new coefficients

B′′(t)+2B′(t)(X ′(t)

X(t)+ p(t)

)+B(t)

(q(t) + 2

X ′(t)

X(t)p(t) +

X ′′(t)

X(t)

)= 0 .

Or, introducing new notation, B′′(t) + B′(t)p(t) + B(t)q(t) = 0, where

p(t) =X ′(t)

X(t)+ p(t)(2.4)

q(t) = q(t) + 2X ′(t)

X(t)p(t) +

X ′′(t)

X(t)(2.5)

We have

Theorem 2.12. Two fanning frames A(t) and B(t) represent thesame curve in RP 1 if and only if the coefficients of their respectiveequations (2.3) are related by the equations (2.4) and (2.5) for somenon-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 equationX(t)(p(t)− p(t)) = X ′(t) satisfies A(t) = B(t)X(t). �

We want to find a quantity that is invariant under the transforma-tions (p, q) 7→ (p, q). A standard way of doing this is to find a normalform for the coefficients p and q: what is the “nicest” form we can findfor equation (2.3) by transforming the coefficients by the rules givenby 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 thesimpler-looking differential equation

(2.6) B′′(t) + B(t)Q(t) = 0

A fanning frame B is said to be in normal form if it satisfies adifferential 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 con-stant.

Proof. For (1), let A(t) be an arbitrary frame spanning `(t). LetB(t) = A(t)X−1(t) as above. For (2), let C(t),D(t) be two normalframes spanning `(t). Since they both span the same curve, there is anon-vanishing real function X(t) such that C(t) = D(t)X(t). Since theyare both normal the coefficients of the first derivative in their respectivestructure equations vanish. Then by equation (2.4), X ′(t) = 0. �

Exercise 2.14. There is a white lie in the argument above: howdoes one know that there exists a frame spanning `? Use the methodsof 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 re-lated to the original coefficients coming from A by Q(t) = q(t) +p2(t) − p′(t). Given a frame A satisfying (2.3), the function SA(t) :=q(t) + p2(t) − p′(t) is called the Schwarzian of A. Lemma 2.13 thenshows that the Schwarzian depends only on `(t) and not on the specificframe: indeed, the association ` 7→ normal frame spanning ` is almostcanonical: by lemma 2.13, any two normal frames differ by a constant;and then their respective structure equations (2.6) are the same andtherefore gives the same Schwarzian.

Exercise 2.15. Use equations (2.4) and (2.6) to show by directcomputation that the Schwarzian only depends on the curve `(t).

We are now ready for the solution of the equivalence problem:

Theorem 2.16. Two curves `1(t), `2(t) : [a, b] → RP 1 are congru-ent if and only if their Schwarzians coincide.

Proof. The “only if” part follows from the discussion after equa-tion (2.3) and from the Schwarzian being well-defined for curves inRP 1. In order to do the “if” part, let A1(t),A2(t) be normal framesspanning `1(t) and `2(t), respectively. We first find T ∈ GL(2,R) suchthat at the initial point a, TA1(a) = A2(a), TA′1(a) = A′2(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, TA1(t)and A2(t) satisfy the same structure equation, since the Schwarziansare equal by hypothesis, and they have the same initial conditions att = a; by uniqueness of solution of ordinary differential equations,TA1(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 curveswith notable invariants. For example, fanning curves in RP 1 withvanishing Schwarzian are the quotients of normal frames satisfyingA(t)′′ = 0, that is, there are linearly independent vectors ~v0, ~v1 ∈ R2

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

such that `(a) is the x-axis. Let A(t) =

(a1(t)a2(t)

)be a (not necessarily

normal) frame spanning `(t). We have that a1(a) 6= 0 and by continuitya1(t) 6= 0 in some neighbourhood of t = a. Then the frame

M(t) = A(t)1

a1(t)=

(1

m(t)

),

where m(t) = a2(t)a1(t)

is defined around t = a and spans the same line

`(t). Note that this is the slope parametrization of the very beginningof this book. Taking the necessary derivatives, we see that the structureequation of M is

M′′(t) + 2M′′(t)p(t) = 0 ,

where p(t) = −m′′(t)2m′(t)

.

Exercise 2.18. Show that the fanning condition translates tom′(t) 6=0 in this presentation and therefore dividing by m′(t) is allowed.

Therefore our Schwarzian is given by

Q(t) =1

4

{−(m′′(t)

m′(t)

)′+

(m′′(t)

m′(t)

)2},

which is, modulo constants due to our choice of normalizations, theSchwarzian of classical complex analysis. Note that, in the slope parametriza-tion, curves with zero Scwharzian are exactly the Mobius transforma-tions

t 7→ a+ tb

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 deriva-tives 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 theGrassmann manifold itself.

The presentation is necessarily informal, since a complete treatmentneeds the theory of differentiable manifolds and spaces of jets. Theinterested reader can find an unified exposition in the book [16]. LetX and M be differentiable manifolds (objects where it makes sense totalk about derivatives). The space of Jr(X;M) of r-jets of maps ofX onto M is roughly described as follows: each element of Jr(X;M)contains the information of the value of a function f : X → M at agiven point x ∈ X plus the information of its first r derivatives.

Example 2.19. The space Jk(R;Rn) is simply (Rn)k+1. Given acurve γ : R → Rn, its k-jet at a point, say at t = 0, is the evaluatedvector of derivatives (γ(0), γ′(0), . . . γ(k)(0)). This also applies to anyopen subset of Rn.

Let now `(t) : R→ Grk(Rn) be a differentiable curve in the Grass-mann manifold. How to take the derivative of a curve of subspacesis not immediately obvious. We try to follow the method we havebeen using: let A(t) : R → Stk(Rn) be a frame spanning `(t). Weare tempted now to define the derivative of ` as A′. However, thereis the problem is that many frames span the same curve; indeed, fromchapter 1 we know exactly all other frames spanning `(t): they are ofthe 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 Grassmannmanifold, and then they must have “the same” derivative. Taking anactual derivative, and remembering that Leibniz’s rule applies to ma-trix multiplication, we have

(2.7)

B′(t) = A′(t)X(t) +A(t)X ′(t) ,

B′′(t) = A′′X(t) + 2A′(t)X ′(t) +A(t)X ′′(t) ,

...

B(r)(t) =r∑s=0

(r

s

)A(r−s)(t)X(s)(t) .

We now deal with two simple spaces of jets: first the space of jets inthe Stiefel manifolds Jr(R;Stk(Rn) and Jr(R;GL(k,R)) of invertible(k×k) matrices. The reason we called them “simple” is that the target

spaces are open subsets of Rkn and Rk2 , respectively, and then example

2.5. NOTABLE SUBSETS OF THE GRASSMANN MANIFOLD 29

2.19 applies:

Jr(R;Stk(Rn) = {(A0,A1, . . . ,Ar) , A0 ∈ Stk(Rn)} ,Jr(R;GL(k,R)) = {(X0, X1, . . . , Xr) , det(X0) 6= 0} ,

and the important insight is that Jr(R;GL(k,R)) is a group, withmultiplication induced by Leibniz’s rule:(2.8)

(X0, X1, . . . , Xr)(Y0, Y1, . . . , Yr) = (X0Y0, X0Y1+X1Y0, . . . ,r∑s=0

(r

s

)Xr−sYs) .

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

(A0,A1, . . . ,Ar) • (X0, X1, . . . , Xr) = (B0,A1, . . . ,Br) ,

where Bj =∑j

s=0

(rj

)Aj−sXs, that is, defined by Leibniz rule again (but

be careful: formula (2.7) relates functions of t, whereas this last equa-tions relate matrices that are functions and its derivatives evaluated ata point.

Exercise 2.20. Show that the association • defines a group actionof th group Jr(R;GL(k,R)) on Jr(R;Stk(Rn).

Then we can recognize the space of derivatives up to order r ofcurves in the Grassmann manifold also as a quotient:

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

2.5. Notable subsets of the Grassmann manifold

Up to now we have been considering the Grassmann manifoldGrk(Rn),where there is no other structure in Rn other than the vector spacestructure. The only exception being the standard inner product forconsidering orthogonal Stiefel manifolds and orthogonal projections,which was used only to give alternative presentations.

In this section we impose additional structure in the Grassmannmanifold arising from special vector spaces. It is psychologically con-venient now to change the notation slightly and consider Grk(V ), theGrassmannian of a vector space V , with the same definition: it is theset of k-dimensional subspaces in V ; if V is a real, n-dimensional vectorspace, after fixing a base we recover Grk(Rn).

30 2. GEOMETRY

2.5.1. Other fields. The first change, which we mention only inpassing, is to consider vector spaces over other fields; in geometry andtopology we consider mainly the complex field (C) and the Hamiltonquaternions H (which do not form a field but a division ring, havingnon-commutative multiplication; however most of linear algebra can berecovered, with the notable exception of the determinant, []). Then,fixing a basis, we have the spaces Grk(Cn) and Grk(Hn).

Exercise 2.21. Redo this book replacing the reals with the com-plex numbers everywhere. In particular, exercise 1.41 is transformed toa homeomorphism between CP 1 and the Riemann sphere S2. In partic-ular, the results at the end of section 2.3 translate to classical Mobiustransformations and the “extended complex plane” acquires a meaning:it is extended the “missing line” of a given slope parametrization whichis 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 ex-tremely rich in properties. The Hopf fibrations S3 → S2 and S7 → S4

can be seen as V1(C2) → CP 1 and V1(H2) → HP 1, respectively. Thislast fibration has contribuited many constructions in geometry, anddifferential topology; we just mention a topic dear to the author, thegeometry of exotic spheres [5, 6, 9, 12, 13, 18, 28]. The Hopf fi-brations are also the substrate of many construction in Physics; see[31].

Another construction is to try to extend the concepts of this bookusing the Cayley octonions Ca, an 8-dimensional real division algebrawhich is neither commutative nor associative. In contrast with thequaternion case, the lack of associativity is a heavy blow: it meansthat the relation ~v ∼ ~w if and only if there is λ 6= 0 with λ~v = ~w is notan equivalence relation. However, with some care one can recover theCayley projective line CaP 1 and the Cayley projective plane CaP 2, see[3].

Now this section is called “some geometrically defined subsets ofthe Grassmann manifold” and all we have done up to now is extendthe concept to other fields. However, by applying the forgetful functorwe can always consider these Grassmannians as subsets of some realGrassmann manifold: since a complex (resp. quaternionic) vector spaceof dimension n is (canonically) a real vector space of dimension 2n (resp4n), we have

Grk(Cn) ⊂ Gr2k(R2n) , Grk(Hn) ⊂ Gr4k(R4n) .


We can “half forget” and consider a complex vector space as a realvector 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 mapJ : V → V such that J2 = −1. Such a vector space always has evendimension 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 candefine 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 isconsidered a complex vector space.

For the quaternions, a quaternionic structure on a real vector spaceis a pair (I, J) of linear transformations of V satisfying I2 = J2 = −1,IJ = −JI (then IJ = K as a third element and {1, I, J,K} span analgebra isomorphic to the quaternion algebra). A similar discussion asthe complex case applies.

2.5.2. The Lagrangian Grassmannian. The most notable sub-set of the Grassmann manifold is the Lagrangian Grassmannian. Herewe give a brief introduction, limiting ourselves to the study of the slopeparametrization, 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 abilinear 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

ω0((~x, ~y), (~v, ~w)) =n∑i=1

xiwi − yivi .

It can be shown that every symplectic vector space is equivalent tothe canonical example above (and, in particular, a symplectic vectorspace is always even-dimensional).

Given a symplectic vector space (W,ω), we can define “orthogonal-ity” 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 vectorspace, then dimX + dimX⊥ = dimV , and (X⊥)⊥ = X.

32 2. GEOMETRY

However, the antisymmetry of the symplectic form goes againstEuclidean intuition:

Definition 2.25. Let (W,ω) be a symplectic vector space. A sub-space ` ⊂ 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 Lagrangiansubspaces of a symplectic vector space (W,ω). By lemma 2.26, Λn(W,ω)is contained in the half-Grassmannian Grn(W ) where dimW = 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 variationalprinciple, as in Classical Mechanics and Differential Geometry (see e.g.[23]). Generalizations to higher order variational principles are foundin [7].

Let us study the Lagrangian Grassmanian Λn of the canonical sym-plectic vector space (Rn × Rn, ω0). Note that each of the subspaces

H = Rn × {~0} and V = {~0} × Rn are Lagrangian subspaces. Thecanonical basis of Rn 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, . . . ~fnform 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-rem 1.17, a dense open neighbourhood ofH inGrn(R2n) is parametrizedby n× n matrices, the map being given by

φH,V (Z) = column space of

(1k×kZ

),

with H corresponding to Z = 0. Now the question is, which of thoseplanes are Lagrangian? We have

Theorem 2.27. The plane φH,V (Z) is Lagrangian if and only if thematrix Z is symmetric.


Proof. Let ~ci be the ith column of the matrix above. Then

~ci =

(~ei∑r zir

~fr

),

where we have written the vectors ~ei, ~fr as column vectors. Now com-puting the symplectic form on a pair ~ci,~cj of such columns, we have

ω0(~ci,~cj) =∑r,s

ω0(~ei + zir ~fr, ~ej + zjs ~fs) = zji + zij .

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 `⊥ ⊂ `, butthen since dim ` = n we have that ` = `⊥. �

In accordance with the spirit of this notes, we have stated both thestatement and the proof of Theorem 2.27 in a fixed basis of fixed com-plementary subspaces. In reality, this characterization/parametrizationon works for any pair of complementary Lagrangian subspaces on asymplectic vector space (W,ω), and is invariantly defined: an openand dense set of Lagrangian planes is parametrized by the set of sym-metric bilinear forms on one of them. Concretely, in the symplecticsituation we can use the symplectic form to reparametrize the graphmap of theorem 1.18, T : H → V 7→ ΓT , as follows: we send T to thebilinear 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 thegraph ΓT is a Lagrangian plane. This characterization is extremelyuseful: 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 hastwo 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 ideasin Homotopy Theory. We do not have the space in these notes for adescription of Homology and Homotopy and we keep the discussionof these general theories informal; a complete study can be found forexample in the textbook [14]. What we will do is to describe two con-structions that exemplify the Homology and Homotopy of the Grass-mann manifold: for homology, its CW decomposition, a way of com-binatorially constructing the Grassmannian from simpler pieces thatallows the computation of its homology. In relation to homotopy, wewill describe specific generators of some homotopy groups of the Grass-mann manifolds, which have been crucial in Differential Geometry andcurvature. In this section, it is much more convenient to consider anauxiliary Euclidean strcuture of Rn, say the standard one, as opposedto 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 thespace

Dn = {~x ∈ Rn : ||~x|| ≤ 1}with boundary the unit sphere

Sn−1 = {~x ∈ Rn : ||~x|| = 1}

and interior the ball

Bn = {~x ∈ Rn : ||~x|| < 1} .

The author vividly recalls his Algebraic Topology teacher stating:“I am a very humble man. My only ambition in Topology is to under-stand 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 bound-aries, one gets essentially all spaces of interest in Algebraic Topology[19]. This is the idea behind CW complexes. We give some informaldiscussion of CW complexes first and then proceed with the precisedefinitions.

Roughly speaking, a CW complex is built inductively adding disksof 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 complexis 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:


Note that the 1-disk that is “hanging” form the third point wasattached 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 send-

ing all the boundary to the fourth point, whereas the one at the leftwas attached by sending the boundary to the circle formed by previous1-disks.and so on. We only deal with finite CW complexes: at each stage weonly glue a finite amount of disks, and we stop at a certain dimension

3.1. COMBINATORIAL STRUCTURE 37


n which is called the dimension of the CW complex. For infinite CWcomplexes, one has to be careful with the topology one puts in it; seefor example appendix A of [14]. After definition 3.2 it will be implicitthat our CW complexes are always finite.

Let us express the 2-sphere S2 as a CW complexes in two differentways:

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 successiveskeleton 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 hasthe advantage of considering each hemisphere separately. We will seethat this motivates the CW structure on the projective spaces. Note


Figure 5. Another CW-decomposition of the 2-sphere

that the same space can have different decompositions as CW com-plexes.

Let us finish this informal discussion with a CW decomposition ofthe 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 topologicalspace X and a finite collection of maps φα,n : Dn → X, called thecharacteristic maps, satisfying:

(1) Each φα,n restricts to a homeomorphism from the interior of(Dn) onto its image, called a cell enα ⊂ X , and these cells areall disjoint and their union is X.

(2) For each cell enα, φα,n(S(n−1)) is contained in the union of cellsof dimension less than n.


(3) X is covered by the images of the φα,n.

Given a CW complex X, the union of all cells of dimension less thanor equal to k is called the k-skeleton Xk. The maximum dimension nof a disk appearing in the characteristic maps is called the dimensionof the CW complex. The restrictions φα,n|Sn−1 are called the attachingmaps. The definition then also expresses the idea of CW complex canbe tough as gluing n-disks at their boundary to the (n − 1) skeletonthrough the attaching maps.

Exercise 3.3. Find explicit formulas for the characteristic and at-taching maps in the examples.

Exercise 3.4. Show that in a CW complex, the difference of setsXk −Xk−1 is homeomorphic to the disjoint union of open balls Bk.

3.1.2. Schubert symbols and Schubert cells. We now proceedto construct a standard CW structure on the Grassmann manifolds. Asusual, it is instructive to begin with Gr1(Rn+1) = RP n.

Proposition 3.5. The projective spaces RP n admit a CW decom-position with exactly one cell in each dimension 0, 1, . . . n.

Proof. The proof is by induction on n. Recall that by exercise1.45, the inclusions

RP 0 ⊂ RP 1 ⊂ · · · ⊂ RP n

are induced by the chain of inclusions

R1 ⊂ R2 ⊂ · · · ⊂ Rn+1 .

The space RP 0 is a point, that is, a 0-cell. Suppose now RP n−1 admits adecomposition satisfying the conclusion of the theorem. All we have todo now is to construct a characteristic map φn : Dn → RP n satisfyingconditions (1) and (2) of definition 3.2, and condition (3) is translatedto RP n = RP n−1 ∪ φ(Dn). A natural candidate is the top cell ofdefinition 1.15 (hence the name): note that the top cell in the case ofRP n is exactly RP n−RP n−1, and it is homeomorphic to Rn by the slopeparametrization. All that is needed then is to find an adequate mapφn : Dn → RP n such that φn maps (int(Dn)) homeomorphically ontothe 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 Stiefelmanifold St1(Rn+1) defined by

Un+1 = {~v = (v1, . . . , vn+1) ∈ St1(Rn+1) : vn+1 6= 0} .


Consider the map Φn : Dn → St1(Rn+1), given by

Φn(x1, . . . , xn) = (x1, . . . , xn,√

1−∑

x2i ) .

Note that the image of Φ actually falls inside the orthogonal Stiefelmanifold V1(Rn+1), that is, the unit sphere, and the last coordinate isin fact positive. Also, for ~x in the interior of Dn, Φn(~x) ∈ Un+1. Wecan now define φn to be composition of Φn with the projection fromStiefel to Grassmann; this is the analytic content of figure 1 in chapter1.

Exercise 3.6. Show that the functions φn inductively defined abovesatisfy the definition of carachteristic maps of a CW decomposition.Prove that the attaching maps are given by the projection V1(Rk+1)→RP k, that is, Sk → RP k.

�

We now proceed to the general Grassmann manifolds; our discussionis based on chapter 6 of [20] . The inclusions

R0 ⊂ R1 ⊂ · · · ⊂ Rn

still plays a lead role: given a plane ` ∈ Grk(Rn), we consider thesequence of integers

0 ≤ dim(R1∩`) ≤ dim(R2∩`) ≤ · · · ≤ dim(Rn−1∩`) ≤ dim(Rn∩`) = k

Two consecutive numbers in this sequence differ at most by one: in-deed, consider the map p : ` ∪ Ri+1 → R to be the projection onto thelast coordinate of Ri+1. By the rank-nullity theorem,

dim(Im(p)) + dim(ker(p)) = dim(Ri+1 ∩ `),

but ker(p) = Ri ∩ ` and dim(Im(p)) is either zero or one. Thereforethe above sequence must “jump” exactly k times. We codify the in-formation of the places where the jump occurs: a Schubert symbol isan ordered sequence of k integers 1 ≤ σ1 < σ2 < · · · < σk ≤ n. Nowwe associate a Schubert symbol to an element ` ∈ Grk(Rn) by notingthe jump places, that is, σi(`) is characterized by dim(Rσi(`) ∩ `) = iand dim(Rσi(`)+1) ∩ ` = i + 1. Observe that in the case k = 1 ofRP n = Gr1(Rn+1), the sequence has exactly one element σ1. Thengiven a line ` ∈ RP n, σ1(`) measures the Ri such the ` ∈ Ri; forany line in the top cell, σ1(`) = n + 1; and then the Schubert symbolparametrizes the cell decomposition of the projective spaces.

The same phenomenom occurs for the general Grassmannian: givena Schubert symbol σ = 1 ≤ σ1 < σ2 < · · · < σk ≤ n, the Schubert cell


associated to σ is the set

e(σ) = {` ∈ Grk(Rn) : σi(`) = σi} .Obviously, and plane ` belongs to exactly one such Schubert cell.

Define the open half-space Hi ⊂ Ri ⊂ Rn by

H i = {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 suchthat ~vi ∈ Hσi . Therefore we have, as in the RP n case, successfully liftedthe situation to the orthogonal Stiefel manifold Vk(Rn). Consider thenE(σ) to be the lifted cell:

E(σ) = {(~v1, . . . , ~vk) ∈ Vk(Rn) : ~vi ∈ Hσi} .Let E(σ) be the closure of E(σ).

Lemma 3.7. Each E(σ) is homeomorphic to a closed disk Ds, where

s =∑k

i=1 σi − i.

Proof. We proceed by induction on k. For k = 1, there is only oneelement 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,and σ = 1 ≤ σ1 < · · · < σr ≤ n1 < n be the “truncated” symbolobtained by deleting the last element. By the induction hypothesis,E(σ) is a closed disk of dimension

∑ri=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 ofthe 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 ofbeing in Hσr . The set D is clearly homeomorphic to a closed disk ofdimension 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 toa closed disk of dimension σr − r − 1, where D~v1,...,~vr is constructed asabove the first r elements of the basis are fixed but arbitrary elementsof E(σ). Hint: use the orthogonal group to transform (~v1, . . . , ~vr) into

(~b1, . . . ,~br) while fixing (Rσr)⊥.

The previous exercise plus the induction hypothesis suggests thatE(σ) is the product Dα × Dβ ' Dα+β, where α =

∑ri=1 σi − i and


β = σr − r − 1; this would finish the proof of the lemma. However wehave to be careful: given a space Z and a projection π : Z → X suchthat, 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,...,~vru)

where T~v1,...,~vr is a orthogonal map depending continuously on ~v1, . . . , ~vrand transforms ~b1, . . . ,~br into ~v1, . . . , ~vr. This map can be constructedas follows: given unit vectors ~u,~v ∈ RN , let R(~u,~v) be the the orthog-onal 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 thatthe 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. �

We have then covered Grk(Rn) with images of closed disks. It iseasy to see that

(1) The proof of the previous lemma furnishes a homeomorphismbetween 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)in a CW decomposition of Grk(Rn), with r =

∑i σi − i.

Exercise 3.11. Identify the top cell of chapter 1 as one of the cellse(σ).

Exercise 3.12. How many cells of a given dimension does theGrassmann manifold has?

3.2. Homotopy

The idea of homotopy theory is to study deformation classes ofmaps instead of the maps themselves in order to get manageable com-plexity.

3.2. HOMOTOPY 43

3.2.1. Basic Definitions. Let us make precise what we mean bydeformation classes:

Definition 3.13. Let X, Y be two topological spaces. Two con-tinuous maps f0, f1 : X → Y are said to be homotopic if there existsa continuous map H : X × [0, 1] → Y such that f0(x) = H(x, 0) andf1(x) = H(x, 1).

The intuition is that f0 and f1 can be deformed into each otherthrough the deformation parameter t ∈ [0, 1]. The map H is said to bea 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 thepair (Dn, Sn−1)). In this category, we naturally require that mapsf : (X,A) → (Y,B), and homotopies H must satisfy H(a, t) ∈ B forall a ∈ A. A more restrictive but still quite frequent notion is that thehomotopy does not move the set A at all: we said that a homotopyH : X × [0, 1] → Y is said to be relative to A (written “rel A”) ifH(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 categoryof pointed spaces has objects (X, x0), where X is a topological space,x0 ∈ X, and continuous maps f : (X, x0) → (Y, y0) are required tosatisfy f(x0) = y0. The homotopies are then required to be rel {x0}.

Exercise 3.14. Show that in all the cases described above, “ishomotopic to” is an equivalence relation.

The idea now is to study the rough topology of a space by classifyingits “holes”. Each of these holes is detected at dimension n by homotopyclasses of maps (Sn, 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 homo-topy group πn(X, x0) is the set of homotopy classes (rel {N}) of mapsf : (Sn, N)→ (X, x0).

Note that the words in the definition are homotopy groups. Indeed,these sets have rich algebraic structures, the most elementary of thembeing a group structure which is Abelian for n > 1. We will not delveinto this rich theory here (again, see [14]), but it is natural what theneutral element should be:

Definition 3.16. A map f : (Sn, N)→ (X, x0) is said to be null-homtopic if there is a pointed homotopy between f and the constantmap c(θ) ≡ x0.


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

Figure 7. π1 detects the hole in R2 − {~0} but not in R3 − {~0}

3.2.2. Geometric generators of Homotopy groups. In purelytopological applications, a map from a sphere into a space representinga homotopy class can arbitrarily deform the sphere (in a continuousway, of course); after all this is the idea of homotopy.

For a geometer, however it is not only aesthetically pleasing, but anecessity, to obtain representatives of homotopy classes that preservethe “roundness” of the sphere. For example, such geometric generatorsin the works of Rigas [26] lead to the construction of non-negativelycurved metrics on many spaces.

Thus we prefer to generate an element of the first homotopy groupof the torus with something like figure 8 below insted of figure 9.

In this section we will construct special subsets of the Grassmannmanifold that later turn out to be generators of homotopy groups.The proof of these facts is beyond the scope of these notes; howeverthe construction also illustrates that, by focusing of sets explicitly de-fined by elementary methods, we can construct representatives of moreadvanced phenomena which naturally carry geometric structure andinformation.

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 elementof π1(T )

Definition 3.17. A pair of subspaces `, r of Rn are said to beisoclinic if there are real numbers α2, β2 such that, for all ~x1, ~x2 ∈ `and ~y1, ~y2 ∈ r,

〈π`(~y1), π`(~y2)〉 = α2〈~y1, ~y2〉 and 〈πr(~x1), πr(~x2)〉 = β2〈~x1, ~x2〉 ,

where π`, πr are the orthogonal projections onto ` and r, 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 suchcoefficients exist they cannot be negative. Also, we write the projectionon both sides just for aesthetic symmetry: since orthogonal projectionsare self-adjoint,

〈π`(~y1), π`(~y2)〉 = 〈π`(~y1), ~y2〉 = 〈~y1, π`(~y2)〉 .


In principle, ` and r have different dimensions. However this is onlypossible in the degenerate (orthogonal) case: ` and r are orthogonal ifand only if they are isoclinic, with one of (and therefore both) α and βequal to zero. Let us deal with the non-orthogonal case, that is αβ 6= 0:If ` and r satisfy the first equation above, by setting y1 = y2 we findthat kerπ`|r must be trivial, which means that dim r ≤ dim `. Thenthe second condition implies that dim r = dim ` and each projectionrestricted to the other subspace is an isomorphism.

Again by setting y1 = y2 above, we find that in the non-orthogonalcase we get also that α2 ≤ 1, β2 ≤ 1 and the upper bound is onlyattained when ` = r, since the same α, β must work for all the vectorsin 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 onlyif either they are orthogonal and α = β = 0 or dim ` = dim r, `and r belong to the same Grassmann manifold ` ∩ r = {~0}, and theysatisfy one of the equations above (the other being automatic by usinginverses) with a non-zero coefficient that is also strictly less than one.

Let us begin by studying n-dimensional isoclinic spaces in the half-Grassmannian Grn(R2n). Consider the base space `0 = Rn × {~0} ⊂Rn × Rn, and `⊥ its natural complement. Let us find, using the slopeparametrization, all non-orthogonal n-planes r isoclinic to `. Note thatby the discussion after the definition, the projection π`|r : r→ ` mustbe an isomorphism; therefore, any non-orthogonal space isoclinic to `is contained in the top cell centered at ` and the slope parametrizationcovers it. Expressing any r in the slope parametrization of Theorem1.17,

(3.1) r = column space of

(1n×nZ

),

and let us discover conditions on the (n×n) matrix Z that characterizethe isoclinic property. Any vector ~y ∈ r can be written as

~y =

(~x

Z~x

)where ~x ∈ `. Taking the inner product of two such vectors,

〈~y1, ~y2〉 = 〈~x1, ~x2〉+ 〈Z~x1, Z~x2〉 = α2〈~x1, ~x2〉 ,where the first equality is given by the orthogonality of the complemen-tary subspaces chosen and the second is the isoclinic condition. Sinceα2 < 1, it will be useful to write α = cos(θ) with 0 < θ < π/2. Then ris isoclinic to ` if and only if there is a fixed θ such that

〈Z~x1, Z~x2〉 = sin(θ)2〈~x1, ~x2〉 .

3.2. HOMOTOPY 47

This justifies the word isoclinic: ` and r are at “constant angle” θ toeach other. Observing that Z preserves inner products up to a constantmultiple, we have

Proposition 3.18. A n-plane r described in the slope parametriza-tion as in (3.1) is isoclinic to ` if and only if Z = sin(θ)Z is an or-thogonal matrix, Z> = Z−1.

There is a little contraband in the previous proposition: by fix-ing respective basis, we have implicitly identified the complementarysubspaces ` and `⊥.

An alternate viewpoint, from the orthogonal Stiefel manifold, is toconsider now an orthonormal basis of r. If (~x1, . . . ~xn) is an orthonormalbasis of `, then cos(θ)~xi + sin(θ)Z~xi is an orthonormal basis of r. Thismotivates the following construction: consider the transformation ofR2n described by the block matrix form in the canonical basis of Rn ⊕Rn = `⊕ `⊥,

Jr =

(0n −Z>Z 0n

).

The matrix Jr satisfies:

(1) J2r = −I,

(2) J>r = −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 ofr. 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`0 ∈ Grk(RN), there is a projection p : O(N) → Grk(RN) given byp(T ) = T`0. A section is a map that goes in the other direction s :Grk(RN) → O(N) such that p ◦ s(`) = `. In general, a continuoussection does not exists. However, when restricted to isoclinic subspacesto a given one ` as is our case, s(r) = (cos(θ)I + sin(θ)Jr) is such thatp(s(r)) = r.

We want to consider subsets of Grn(R2n) such that all elements aremutually isoclinic. This let us consider a third plane q, that is isoclinic

to ` and r. Being isoclinic to ` means that there is a complex structureJq satisfying (1) - (3) above, and q = (cos(φ)I + sin(φ)Jq)(`) for some

angle 0 < φ < π/2, and

Jq =

(0n −W>

W 0n

),


where W is an orthogonal matrix, W> = W−1. We want to determinethe conditions on Jq that translate q and r being isoclinic.

Exercise 3.19. Use exercise 1.51 to show that the orthogonal pro-jection πr with image r is given by

πr = (cos(θ)I + sin(θ)Jr)

(I 00 0

)(cos(θ)I − sin(θ)Jr)

=

(cos2(θ) cos(θ) sin(θ)Z>

cos(θ) sin(θ)Z sin2(θ)

).

After some computation with the previous exercise as starting point,we find that the isoclinic condition 〈πr~u1, πr~u2〉 = σ2〈~u1, ~u2〉 translatesto

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 ma-trices that produce isoclinic supbspaces:

Definition 3.21. A set of matrices (2n × 2n) complex structuresJ1, . . . , 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 theJi being complex structures, and if i 6= k, they mean that Ji and Jkanticommute: JiJk = −JkJi.

Note that having a system of matrices satisfy the Hurwicz condi-tions, the planes `0, . . . , `r given by `0 = `, `i = (cos(θi)I + sin(θi)Ji)`for certain angles θi are a system of mutually isoclinic planes. Whatwe do now is “filling” this kind of set, finally defining isoclinic spheres:

Definition 3.22. Let J = {J1, . . . , Jk} be a set of complex struc-tures in R2n satisfying the Hurwicz conditions for some n-plane ` ⊂ R2n.The set

SJ = {(a0I + a1J1 + · · ·+ akJk)`, a20 + . . . a2k = 1} ,

is called the isoclinic sphere spanned by J .The results of Rigas [26] (for generators) and Wang [32] (for arbi-

trary elements) imply

Theorem 3.23. Given k ∈ N, isoclinic spheres give representativesof all elements πk(Grr(R2r)) for sufficiently large r.

3.2. HOMOTOPY 49

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

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