A Bridge between Algebra and Topology: Swan s Theorem

Transcription

1 A Bridge between Algebra and Topology: Swan s Theorem Daniel Hudson Contents 1 Vector Bundles 1 2 Sections of Vector Bundles 3 3 Projective Modules 4 4 Swan s Theorem 5 Introduction Swan s Theorem is a beautiful theorem which connects algebra and topology by asserting that, for compact Hausdorff spaces X, there is a 1-1 correspondence between vector bundles over X and finitely generated C(X)-modules. Perhaps it is not surprising that vector bundles give rise to algebraic structures, since they are families of vector spaces with additional structure, but it is certainly striking that projective modules, which are defined in a purely algebraic setting, have a rich geometric interpretation. In the first section we develop the necessary background for vector bundles, and in the second section we introduction sections of vector bundles. In the third section we develop projective modules. We conclude with the statement and proof of Swan s Theorem. These (brief) notes are to accompany a short talk I gave at my graduate student seminar at UVic in the fall of Comments are corrections are most welcome at drhh@uvic.ca. 1 Vector Bundles Definition 1.1. Let X be a topological space. A (real) family of vector spaces over X is a topological space E and a continuous surjection π : E X such that, for all x X, each E x := π 1 ({x}) is a real vector space such that addition and scalar multiplication are continuous. We call the map π the projection, and E x the fiber over x X. One can equally well talk about complex families of vector spaces over X. In this case, the fibers are complex vector spaces. For simplicity we will restrict ourselves to the real case, but the main result holds for complex vector bundles too. Example 1.1. If X is a topological space, then, X R n are families of vector spaces over X for all n N. It is called the trivial bundle of rank n. Definition 1.2. If π E : E X and π F : F X are families of vector spaces over a topological space X and ϕ : E F is a continuous map, then we say that ϕ is a homomorphism of families (or just homomorphism, for short) if 1

2 1. The following diagram commutes: E ϕ F π E X π F 2. for all x X, the map ϕ x : E x F x is a linear map of vector spaces. If ϕ is a homeomorphism, then we call ϕ an isomorphism, and we write E = F. Note in particular that if we consider points of E as ordered pairs (x, e), where e E x, then condition 1 says that ϕ fixes x. Thus 2 makes sense. Furthermore, observe that if ϕ is an isomorphism, then each ϕ x is an isomorphism of vector spaces. We are now ready to define a vector bundle. Definition 1.3. If a family π : E X is isomorphic to X R n for some n, the we call E trivial. We call E locally trivial if, for all x X, there exists an open neighbourhood U X containing x such that E U := π 1 (U) = U R n for some n (which depends on U). A locally trivial family π : E X is called a vector bundle. Example 1.2. We will define a canonical line bundle (i.e. all the fibers are one dimensional) on CP n, the complex projective space. Define H := {(L, z) CP n C n+1 : z is a point in the line L}. We give H the topology inherited as a subspace of CP n C n+1 and define the projection π : H CP n to be projection onto the first coordinate. Then the inverse image of each point L CP n is that line as a subspace of C n+1, which carries an obvious vector space structure. In particular, the fibers are one dimensional. Thus, to see that H is locally trivial we have to find a section which is locally non-vanishing. Consider the sets U i := {[z 0,..., z n ] CP n : z i 0}. The U i cover CP n, and on each U i we define the map ( s([z 0,..., z n ]) = [z 0,..., z n ], ( z0 z i,..., z n z i )) H[z 0,...,z. n] ( ) Since the i-th coordinate of z0 zi,..., zn z i is 1, we s is non-vanishing on U i, hence is the appropriate section. Thus, H is a vector bundle. In general, most operations that can be done with vector spaces can be done with vector bundles too. The main example of this, for us at least, is the ability to form the direct sum of vector bundles. Given, two vector bundles π E : E X and π F : F X, we define their direct sum, or Whitney sum, as E F := {(e, f) E F : π E (e) = π F (f).} with the projection map π E F : (e, f) π E (e), and given the subspace topology. Fibrewise we have that (E F ) x = E x F x. 2

3 2 Sections of Vector Bundles A critial notion for the proof of Swan s Theorem will be that of a section, which we shall now discuss. Definition 2.1. A section of a family π : E X is a continuous map s : X E such that π s = id X. The collection of all sections of E is denoted Γ(E). Using sections we can come up with an equivalent characterization of vector bundles. Proposition 2.1. A family π : E X is a vector bundle if and only if for each x X there exists an open neighbourhood U containing x and sections {s 1,..., s n } on U such that {s 1 (y),..., s n (y)} is a basis of E y for each y U. Proof. Suppose that ϕ : E U U R n is an isomorphism and let e 1,..., e n be the standard basis of R n. Then we define sections {s 1,..., s n } on U by s i (y) = ϕ 1 (y, e i ), and these clearly have the desired properties. Conversely, if we have such a collection of sections {s 1,..., s n } on U with the desired properties, then we can define an isomorphism φ : U R n E U by φ : (u, (v 1,..., v n )) n v i s i (u). In particular, a vector bundle E is trivial if and only if we can find such sections defined globally. This observation is critical for the main theorem. Definition 2.2. A Hermitian structure on a vector bundle π : E X is a continuous choice of inner product, x on each E x. Here continuous means that for any two sections s, t : X E the map x s(x), t(x) x is a continuous function on X. Proposition 2.2. Any vector bundle E over a compact Hausdorff space X can be given a Hermitian structure. Proof. First, suppose that E = X R n. If, denotes the standard inner product on R n, then we define (x, v), (x, w) x := v, w. To see that this is continuous, let {s 1,..., s n } be a collection of sections which form a basis of E x for each x X. Then for any i, j, the map x s i (x), s j (x) factors as i=1 X (s i,s j ) E E (pr 2,pr 2 ) R n R n, R, which shows that it is continuous. For an arbitrary vector bundle E over X, let {U i } be a finite collection of trivializations of E. By the previous part, each E U i can be equipped with a Hermitian structure, say,, U i. Let {ρ i } be a partition of unity of X subordinate to {U i }. Define { g U i ρi (u) v, w (v, w) = U i if v, w E u E U i, 0 else. Then defines a Hermitian structure on E., X := g U i (, ) 3

4 3 Projective Modules Modules over a ring R provide a generalization of vector spaces by allowing your scalars to come from a general ring, rather than a field. Definition 3.1. Let R be a ring. An R-module is an abelian group A equipped with a multiplication R A A which distributes over the addition in R and in A and satisfies a certain associativity condition; that is 1. (r + s)a = ra + sa for all r, s R, a A, 2. r(a + b) = ra + rb for all r R, a, b A, 3. r(sa) = (rs)a for all r, s R, a A. If R has an identity then we say that A is unitary if 1a = a for all a A. If A and B are R-modules, then a map ϕ : A B is called R-linear if ϕ(ra + b) = rϕ(a) + ϕ(b) for all r R, a, b A. If ϕ is bijective then we say that A and B are isomorphic (as R-modules) and write A = B. Just like with vector spaces, if {A i } i I is a collection of R-modules we define their direct sum i I A i to be the collection of all i-tuples (a i ) i I where a i A i and a i = 0 for all but finitely many i. Definition 3.2. If R is a ring with unity, then a (unitary) R-module A is free is A = i I R for some index set I. We say that A is projective if there is some R-module B such that A B is free. We say that A is finitely generated if there is a surjective homomorphism R n A. Example 3.1. The most relevant example to us is that the sections of a vector bundle E over X, form a (unitary) module over C(X), where C(X) := {f : X R : f is continuous}. If s Γ(E) and f C(X) then we define the (fs)(x) := f(x)s(x). Proposition 2.1 says that a vector bundle E is trivial if and only if Γ(E) is free and finitely generated C(X)-module. Proposition 3.1. A finitely generated R-module A is projective if and only there exists an idempotent P M n (R) such that A = P (R n ) for some n. Proof. If A is finitely generated and projective then there is a split short exact sequence 0 ker(ϕ) R n ϕ A 0 Hence there is an isomorphism φ : A B = R n, where B = ker(ϕ). We obtain the required idempotent P by φ pr 1 φ 1. Conversely, if A = P (R n ) for some idempotent P M n (R), then A (1 P )(R n ) = R n which shows that A is projective. 4

5 4 Swan s Theorem We are now ready to state and prove Swan s theorem. Theorem 4.1 (Swan). Let X be a compact Hausdorff space. Then there exists a 1-1 correspondence between vector bundles over X and finitely generated projective C(X)-modules. Moreover, the correspondence is given by E Γ(E). We break the proof up over a series of lemmas. Lemma 4.1. If E is a vector bundle over X, then Γ(E) is finitely generated. Proof. Let {U i } be a finite covering of X of trivializations of E, and let {ρ i } be a partition of unity subbordinante to {U i }. By Proposition 2.1, we see that each Γ(E U i ) is free and finitely generated. If s Γ(E U i ) is a generator then we extend s to all of E by s(x) = { ρi (x)s(x) for x U i 0 else. Since each Γ(E U i ) is generated by finitely many sections and there are finitely many such U i, this shows that Γ(E) is finitely generated. Lemma 4.2. If E is any vector bundle over X, then there exists a vector bundle E such that E E = X R m for some m. Proof. We will show that we can embed E in a trivial bundle, then we use a Hermitian structure to define a projection onto E, which will in turn define E. Let {U 1,..., U n } be a finite open cover of X with trivializations {ϕ 1,..., ϕ n } and let {ρ 1,..., ρ n } be a partition of unity subordinate to {U 1,..., U n }. If E has rank k (i.e. dim E x = k), then define Φ : E X R nk (x, e) (x, ρ 1 (x) 1/2 ϕ 1 (x, e),..., ρ n (x) 1/2 ϕ n (x, e)). If, be a Hermitian structure on X R nk, then we observe that Φ(x, e) 2 = n ρ i (x) ϕ i (x, e) 2 0, i=1 with equality if and only if (x, e) was the zero vector in E x, hence Φ is injective and E embeds in X R nk. In particular, E is isomorphic to a sub-bundle of X R nk. Now, thinking of E as a sub-bundle of X R nk, we use the Hermitian structure to define the orthogonal projection P x : {x} R nk E x. Using the continuity of the Hermitian form, we see that P is continuous, hence defining E := (1 P )(X R nk ) yields the desired complementary bundle. 5

6 Theorem 4.2. If E is any vector bundle over X, then Γ(E) is a finitely generated and projective C(X)-module. Proof. Combine Lemma 4.1, and that Γ(E) Γ(F ) = Γ(E F ). This proves the first half of Swan s theorem. Before we prove the second part, we recall Proposition 4.2. In terms of C(X) modules, this says that finitely generated projective modules are in 1-1 correspondence with idempotents P M n (C(X)). Since elements of M n (C(X)) correspond to continuous function X M n (R), we see that finitely generated C(X)-modules are in 1-1 correspondence with idempotent valued functions P : X M n (R). Thus, to conclude the proof of Swan s theorem it is sufficient to prove the following. Lemma 4.3. If P : X M n (R) is an idempotent valued function, then Im(P ) := {(x, v) X R n : v Range(P (x))} is a vector bundle over X whence equipped with the subspace topology, and projection given by mapping onto the first coordinate. Moreover, Γ(Im(P )) = P (C(X)) n. Proof. The fibrewise vector space structure is clear, so we must prove local triviality. Fix x 0 X. Then Range(P (x 0 )) is a k-dimensional subspace of R n. Let v 1,... v k be a basis for Range(P (x 0 )) and extend it to a basis v 1,..., v n of R n. The matrix valued function P (x) = [P (x)v 1 P (x)v 2 P (x)v k v k+1 v n ] M n (R) is such that P (x 0 ) is invertible. Indeed, since v 1,..., v k Range(P (x 0 )), we see that P (x 0 ) is a matrix whose columns are a basis for R n. Since GL n (R) is open in M n (R), there exists some open set U X such that P (x) is invertible for all x U. In particular, this says that, for all x U, the vectors P (x)v 1,..., P (x)v k are linearly independent. Thus, setting s i (x) = P (x)v i, i = 1,..., k, x U, gives sections of E U which are a basis on each fibre, hence a trivialzation of E U. Furthermore, a section of Im(P ) is, by definition, a continuous function s on X such that s(x) Range(P (x)), from which we see that s P (C(X) n ). Thus, Γ(Im(P )) = P (C(X) n ), which completes the proof. References [1] Atiyah, M. F., K-Theory, Benjamin, W. A., [2] Emerson, H. Chapter on K-theory, math465/a01/k_theory_chapter. [3] Rosenberg, J. Algebraic K-theory and its Applications, Graduate Texts in Mathematics, Springer,