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MANIFOLDS, VECTOR BUNDLES, AND STIEFEL-WHITNEY CLASSES
THESIS
Presented to the Graduate Council of the University of North Texas in Partial Fulfillment of the Requirements
For the Degree of
MASTER OF ARTS
By
Michael D. Green, Honors B.A. Denton, Texas
August, 1990 fV,:)
Green, Michael D., Manifolds, Vector Bundles, and Stiefel-Whitney Classes, Master of Arts (Mathematics), August, 1990, 90 pp., bibliography, 9 titles.
The problem of embedding a manifold in Euclidean space is considered. Manifolds are introduced in Chapter I along with other basic definitions and examples. Chapter II contains a proof of the Regular Value Theorem along with the "Easy" Whitney Embedding Theorem. In Chapter III, vector bundles are introduced and some of their properties are discussed. Chapter IV introduces the Stiefel-Whitney classes and the four properties that characterize them. Finally, in Chapter V, the Stiefel-Whitney classes are used to produce a lower bound on the dimension of Euclidean space that is needed to embed real projective space. TABLE OF CONTENTS
Page Chapter
I. INTRODUCTION AND BASIC DEFINITIONS ...... 1 II. THE WHITNEY EMBEDDING THEOREM...... 9 III. VECTOR BUNDLES.-.-.-.-.-...... 26 IV. STIEFEL-WHITNEY CLASSES...... 48 V. ADDITIONAL PROPERTIES OF STIEFEL-WHITNEY CLASSES...... 75 BIBLIOGRAPHY ..-.-.-.-.-...... 90
iii CHAPTER I
INTRODUCTION AND BASIC DEFINITIONS
The first part of this thesis is concerned with manifolds. Basically speaking, a manifold is an object that looks like Euclidean space in some neighborhood around every point. For example, a balloon is a two dimensional manifold because a small bug with limited eyesight walking on the balloon's surface would think that the balloon is flat, i.e. around every point of the balloon there is a neighborhood that looks like R2.
We start off by giving some basic definitions, and see what facts we can derive from these. Probably the most interesting result we encounter is the Whitney Embedding
Theorem. The discovery of this theorem leads us to ask a very natural question which then leads us to the topic of vector bundles. As a tool for studying vector bundles we develop the Stiefel-Whitney classes.
The development of vector bundles is based largely on the treatment given in [4] and the notation in the entire thesis is consistant with that in [4]. The discussion of manifolds and the proof of the Whitney Embedding Theorem follow, for the most part, the presentation in [2] although the Local Submersion Theorem and the Mini-Sard Theorem come
1 2 from [1]. The development of Stiefel-Whitney classes given here is suggested by a problem in [4, p.171].
We begin, then, by giving a concrete definition of a manifold.
DEFINITION 1.1. Let U C R' be open. A function f:U -- R' (k > 0) is smooth or differentiable of class C if its partial derivatives of all orders exist and are continuous.
DEFINITION 1.2. A subset M C Rk is a smooth manifold of dimension n > 0 if, for each X E M there exists a smooth function
h:U -- R defined on an open set U C R' such that
1) h maps U homeomorphically onto an open neighborhood
V of x in M. 2) For each u E U the vectors .. Oh evaluated at u, are linearly independent.
The image h(U) = V of such a mapping is called a coordinate neighborhood in M, and the triple (U,Vh) is called a local parametrization of M.
If M is a smooth manifold and A C M is also a smooth manifold then we call A a smooth submanifold of M. 3
Sometimes we denote an n-dimensional smooth manifold M by M" when we wish to emphasize its dimension.
Let M" C Rk be an n-dimensional manifold. Let
(U,V,h) be a local parameterization of M with h(u) = x.
DEFINITION 1.3. A vector V E Rk is tangent to Mn at the point X E M" if v can be expressed as a linear combination of the vectors
Oh(u) .. Oh(u)
DEFINITION 1.4. The set of all vectors tangent to
M at the point x is called the tangent space of M at x and is denoted by DMx.
Thus DMx is an n-dimensional vector space over the real numbers.
It would appear that the tangent space depends on the local parameterization. This, however, is not the case as we now show:
First recall the following Theorem from Analysis:
THEOREM 1.5 (Inverse Function Theorem) Suppose that f:R" --+ R" is a smooth function in an open set containing 4
a, and det(Dfa) # 0. Then there is an open set V containing a and an open set V containing f(a) such that f:V - V has a smooth inverse f-1:V - V. i.e. f is a smooth local diffeomorphism at a.
Let (U,Vh) and (WY,k) be two local parameterizations for M". Let X E V fl Y. Then x = h(u) = k(w) for some u E U, w E V. Define
DMh = {t E M I t =Eai , ai E R} i=1
DM={t E M |J t=Eb , bi E R}
Let t E DM . We want to show that t E DMX.
Since the matrix [y(w)] is non-singular we have, from the inverse function theorem, that k has a smooth inverse 0~' in some neighborhood of x. Then the function
f = k o h : h-'(V f Y) -+ W is a smooth homeomorphism in some neighborhood of u. Thus h = k of and k . By the chain rule this is just a linear combination of the . Explicitly we can
Oh nh"k write , . Thus t E DMX.Mk
So DMh C DMt.
A similar argument shows that DMk C DMh. Therefore
DM= DMk and the tangent space is independent of the local parameterization. 5
DEFINITION 1.6. The tangent manifold of M is defined to be the subspace
DM C M x Rk consisting of all pairs (x,v) with x E M, v E DMN. So DM is a smooth manifold of dimension 2n.
Let M C R", N C R' be smooth manifolds. Let x E M and (U,V,h) be a local parameterization of M with h(u) =x.
DEFINITION 1.7. A function f:M -- N is said to be smooth at x if the composition
f o h:U -- N E Rm is smooth throughout some neighborhood of u. We say f is smooth if it is smooth at every point x E M.
DEFINITION 1.8. f:M -- N is called a diffeomorphism if f is one-to-one, onto, and both f and f-1 are smooth.
Let f:M -- N be smooth at x and let v E DM.. Then we can write
n
V = 'h 6
We now define a map Dfx:DM. -. DNfCX) by
n n Dfx( ai= ai0(f o h)
DEFINITION 1.9. The linear transformation Dfx is
called the derivative, or the Jacobian of f at x.
Just as with the tangent space of a manifold we can
show that the derivative of a smooth map is independent of the local parameterization used.
If f:M - N is smooth we can define
Df:DM -- DN where Df(x,v) = (f(x),Dfx(v)).
We conclude this chapter with some examples of manifolds which will prove useful later on.
DEFINITION 1.10. An n-frame in Rnk is an
n-tuple of linearly independent vectors of Rnk*. The
collection of all n-frames in Rnk forms an open subset of the n-fold cartesian product Rnk x ... x Rn+k called the Stiefel manifold Vn(Rn+k). 7
DEFINITION 1.11. The Grassmann manifold G(Rnk)
is the set of all n-dimensional planes through the origin
of the coordinate space R"*.k This is topologized as follows: Define the function
q:Vn(R"l+k) -- Gn(Rn)
which maps each n-frame to the n-plane which it spans. Now
give Gn(Rn*k) the quotient topology induced by q. That
is, a subset U C Gn(Rnk) is open if and only if its inverse image q-1(U) C V(Rn+k) is open.
DEFINITION 1.12. Let R* denote the vector space consisting of those infinite sequences
X = (X1,X2 ,X3 ,...) of real numbers where xi = 0 for all but finitely many values of i.
For a fixed k the subspace consisting of
{(XiX2 ,,. ., ,0,0,...)} C R
will be identified with Rk. Thus
RI C R2 C R3 C ...
with R = UURk. k= I
DEFINITION 1.13. The infinite Grassmann manifold
Gn = Gn(R*) is the set of all n-planes through the origin of R*. 8
A subset U C Gn is open if and only if its intersection with G.(R"+k) is open as a subset of Gn(Rn+k) for each k. This makes sense since 00 Gn (R) = U G(Rn+k). k =0 CHAPTER II
THE WHITNEY EMBEDDING THEOREM
We now take a closer look at manifolds and eventually we come to the embedding theorems. First we need more definitions.
DEFINITION 2.1. Let f:M -+ N be smooth. Then f
is immersive at x E M if Dfx:DMx -- DNf(X) is injective. If f is immersive at every point of M we call f an immersion. If Df. is surjective then we call f submersive at x E M. If f is submersive at every point of M we call f a submersion.
DEFINITION 2.2. f:M -- N is an embedding if f is an immersion in which f maps M homeomorphically onto its image. We indicate this by writing f:M r+ N.
DEFINITION 2.3. Let f:M -- N be a smooth map. Then y E N is called a regular value for f if
Dfx:DMx -- DNf(X) is surjective at every point x E M such that f(x) = y.
9 10
THEOREM 2.4. (Local Submersion Theorem) If f:M' -+ N" is a submersion at x, and y = f(x) then there exist local parameterizations (U,V,h) and (W,Y,k) around x and y for M' and N" respectively such that
f o h(xi,. .. ,xm) = k(xi,...,n) i.e. such that we get the following commutative diagram:
Mm f0N h"i I N"
U C R"' > V C R" where r:R"' -+ R" is the projection map given by
r(x1,.. .,m) = (x 1,...,xn)
PROOF: Choose local parameterizations (U',V,h') and (W,Y,k) such that h'(0) = x, k(0) = f(x) = y and choose a function g such that the following diagram commutes:
M"m f N"
Uh'jk
U' C Rm Lg v Cc'" 11
We can always do this since we could just let
g = k-1 o f o h'. This definition actually defines a
function g for some open set containing 0 which we call U'.
Then since Df. is surjective we must have that Dgo
is surjective. Since Dgo is linear we can think of it as being a matrix with the usual base. Thus, by a linear
change of coordinates, we can write Dgo = [I J 0 ] where I. is the nxn identity matrix.
Now define C:U -+ R" by C(a) = (g(a),an+.,. .. ,am) where a = (a,,...,am).
Then DCO = Im
Thus, by the inverse function theorem, C-' exists and
maps some open neighborhood U of 0 diffeomophically onto U'.
We now have the following commutative diagram:
M" N"
U, - V
Then by setting h = h' o C-' we get the commutative diagram: 12
hIJ k
U T>V
which is what we wanted. o
THEOREM 2.5. (Regular Value Theorem) If y is a
regular value of f:Mm --+ N" then f-1(y) is a submanif old
of M with dim f-1(y) = dim M' - dim N".
PROOF: From the local submersion theorem we can find
two local parameterizations (U,V,h) and (W,Y,k) for Mm
and N" respectively, with h(o) = x and k(O) = y such that
f o h(xi,. .. ,xm) = k(xl,. . .,)
Then the set f-1(y) l V is an open neighborhood of x. Also, in V we have
f-~(y) = {(0,...,O,Xn+1,. ..,xm) xi E R}.
So the set f-1(y) l V is an open neighborhood of x
in f-'(y) that is homeomorphic to R"-".
Thus f~1(y) is a submanif old of Mm with
dim f-'(y) = dim Mm - dim N". o
DEFINITION 2.6. Let f:M --+ N be smooth. Let
A C N be a smooth submanifold of N. Then we say f is 13
transverse to A if
Image (Df.) + DAY = DNy whenever f(x) = y E A. That is, the vectors in Image(Df.) taken with the vectors in DAY span DNy.
Notice that if the submanifold A consists of a single point then the notion of transverse is equivalent to the notion of regularity. We can use the idea of transversality to extend the regular value theorem.
DEFINITION 2.7. The codimension of a submanifold A C N is codim A = dim N - dim A.
THEOREM 2.8. Let f:M -+*N be smooth and let A C N be a submanifold. If f is transverse to A, then f~1(A) is a submanifold of M. Also the codimension of f-1(A) in
M is the same as the codimension of A in N.
PROOF: Let p = dim A and let q = codim A. Then locally we can write a neighborhood in N around the point
(0,0) as U x V C RP x Rq where U x 0 is a neighborhood around 0 in A. We need to show that f-'(U x 0) is a smooth submanifold of M.
f:M --+ U x V is transverse to U x 0 so we have Image Dfx + D(U x O)o = D(U x V)0 14
Where f(x) = 0. Now consider the map g:M -L XUxV - V. The transversality of f implies that the map Dg. is
surjective. So 0 is a regular value for g. Thus, by the regular value theorem, g 1(0) is a smooth submanifold of 1 M. But g-(x) = f-'(U x 0). So f-1(U x 0) is a smooth
submanifold of M. Also dim f-'(U x 0) = dim g-1(0) =dim M - dim V = dim M - codim A so
dim M - dim f-'(U x 0) = codim A and we have
codim f-'(U x 0) = codim A. o
We have defined a manifold in such a way that it is
already embedded in some Euclidean space, however this is
not necessary. Sometimes a manifold is defined as follows:
DEFINITION 2.9. A 2nd countable Hausdorff space M"
is called an n-dimensional manifold if it is locally
homeomorphic to R". That is, around every point x of M" there is a continuous function
h:U -- M"
that takes an open subset U C R" homeomorphically onto an open neighborhood V of x in M".
Just as before we call the triple (U,V,h) a local parameterization of M". 15
Now suppose that for M" we have a local
parameterizations {(UQIV,hj)} so that the collection
{Va} conver M". We call M" smooth if whenever two of the parameterizations, say (U ,V 1 1,hi) and (U2 ,V2 ,h2 ), are
such that Vi f V2 # 0 the composition
h210o hi:hj'(Vfl V2 ) -+ U2 is smooth.
The fact that the two definitions we have for a manifold are equivalent, follows from the following:
THEOREM 2.10. Let Mm be a compact smooth manifold.
Then there exists a smooth embedding of M into Rq for some q.
PROOF: Let m = dim Mm . Let D'(p) C R" be the
closed disk of radius p and center 0.
Now for each point x of M" there exists a set
Ui E R' and a homeomorphism hi:Ui -+ Vi where Vi is an open neighborhood of x. By changing the hi's if necessary
we can require that 0 E Ui and D'(2) C Ui. Now we cover M" by open sets of the form Int hi(D"'(1)). Thus, since M"
is compact, we can find a finite number of local parameterizations {(Ui,Vi,hi)} with the following two properties: 16
m 1) D (2) C U1 and 2) Mm = U {Int h(D"'(i))}
We will now define a smooth function A:R" -- [0,1]
that is equal to one on the ball B"(1) and equal to zero outside the ball B"(2). We construct the function A by
first defining the smooth function a:R -+ #R by
-1 a( _Je for x> 0 0 for x < 0 The graph of a is shown in Figure 2.1.
a(x)
Figure 2.1
Next define p:R- -+ R by
,(x) = a(x - 1)a(2 - x).
The graph of 8 is shown in Figure 2.2. 17
,8(X)
hl "1_ ----- x t I 2 Figure 2.2
This function is close to the shape of the function we want but we still need to work with it. We define another
function 7:R -- [0,1] by
2
7(X) = see Figure 2.3 for a graph of 7.
7(X) 'I
1 2 X--2 Figure 2.3
Finally we can define A:IR" -- [0,1] by
A(X) = 7(Ix). A graph of A defined on RI is shown in Figure 2.4. 18
\(x) 1
-2 -1i Figure 2.4
So we have a smooth map A:R'" - [0,1] that is equal to 1 m m only on D (1) and equal to 0 on R' - D'"(2). Such a function is called a bump function. Define smooth maps:
S= A o hil on h(Uj) ' ( 0 on M - h(Uj) m Since M = U {Int hi(Dm(1))} we have the sets
Bi = Ai(1) C h(Uj) cover M'. m Def ine maps fi:M -+ R' by if fi(X) = {Ai(x)hi(x) x E hi(Ui) 0 if x E M - hi(Uj) and set Rm+1 = (fi,Ai):M -+R' = and
g = (g,92,. .. ,gn):M - Rm1 x ... x m+1 - n(m+1) Then 9 is smooth since it is composed of smooth functions.
If x E Bi then gi is immersive at x since
fi(x) = Ai(x)h 7(X) = h7 1(x) which is an immersion. Therefore g is an immersion. 19
Now we must show that g is injective. Suppose x 0 y with y E Bi for some i. If x E Bi then g(x) # g(y) since fi|Bi = hi'jBi which is a homeomorphism. If x % Bi
then Ai(y) = 1 # Aj(x). Therefore g(x) # g(y). So g is injective.
Now since g is a continuous one-to-one map from M (a
compact space) onto g(M) (a Hausdorff space) we have that
g is a homeomorphism from M onto g(M). Therefore g is an embedding. o
The preceding theorem says that every compact manifold
embeds in Rq for some q. However, we have no idea of how large q must be. Fortunately we will soon prove a theorem
that does give us a bound for q.
First we need some more definitions.
DEFINITION 2.11. Let M be a metric space and
let A be a subset of M. We say that A is nowhere dense
in M if (M - X) = M.
Note that a closed set A is nowhere dense in M if its complement is dense in M.
DEFINITION 2.12 A subset A of R" has
(n - dimensional) measure zero if for all E > 0 there 20
exists a cover {U 1,U2,...} of A by closed rectangles such
that .Ev(Ui) < E, where v(Ui) is the volume of the i=1
rectangle U1 .
Note that if a closed set A has measure zero in R"
then A must be nowhere dense in R", for suppose not, then
there exists a point x EER such that x 0 (R" - 1). That is, there is an open ball B containing x such that
Bfl ( "-X) =0. So B CAIT= A. Then A does not have
measure zero, which is a contradiction.
More generally suppose that A C R" has measure zero
and is a-compact, that is, A is the union of a countable
collection of compact sets. Each of these is nowhere dense
and so A is nowhere dense by the Baire category theorem (since R" is a complete metric space).
THEOREM 2.13. (Mini-Sard Theorem) Let M,N be
manifolds with dim(M) < dim(N). If f:M -+ N is smooth, then f(M) is nowhere dense in N.
In order to prove this we first need some lemmas.
LEMMA 2.14 Let U be an open subset of R" and let f:U -+ R" be smooth. If A C U has measure zero, then so does f(A). 21
PROOF: We may assume I is compact and contained in U since A may be written as a countable union of subsets
of U with this property. Let E > 0. Let V be an open
neighborhood of A such that V is compact and contained
in U. Since V is compact, there exists a constant I
such that If(x) - f(y)| < Ix - y| for all x,y E V.
Cover A with a sequence {U1,U,...} of closed
rectangles, each of which is contained in W, such that
E v(Uj) < . If v(Uj) = ki then there is a rectangle U containing f(Ui) with v(U ) < n n fE ki < = E 1=.E 1 v(U)< 1=1 So f(A) has measure zero. o LEMMA 2.15 Let U C Rm and let m < n. Then U x 0 has measure zero in U x R"*. PROOF: Let c > 0. We can cover U x 0 by a countable number of closed rectangles {U1,U2 ,.-.-.} such that every side of Uj has length 1 except the side that extends in the direction of the nth coordinate axis which we wM let have length 21+. Then .E v(U) = E < E. So U x 0 has measure zero in U x Rnm. e 22 LEMMA 2.16. If U C R' is open and g:U -- R is smooth with m < n, then g(U) has measure zero. PROOF: Consider the composition of maps: U - U x 0 C U x R"n-m'"-- U -. 9-.Rn Since r and g are smooth and since U x 0 has n-measure zero in U x I", we have (g o r)(U x 0) = g(U) has measure zero in R". o We can define a subset A of a smooth k-dimensional manifold M to be of measure zero provided that, for every local parameterization f:U -+ M, the preimage f1 (A) has measure zero in U C Rk. PROOF: (of Mini-Sard Theorem) We can cover M with a countable number of local parameterizations (,Vi,hi). Then f o hi(Ui) has measure zero. Since a countable collection of sets of measure zero has measure zero we see that f(M) has measure zero. Therefore f(M) is nowhere dense in N since f(M) is a-compact. o We are now finally ready to give a stronger embedding theorem: 23 THEOREM 2.17. (Easy Whitney embedding theorem) Let M" be a compact Hausdorff smooth n-dimensional manifold. Then there is an embedding of M in R2n+1. PROOF: We have already shown that M embeds in some R 0 . So we can replace M" by its image under an embedding. Thus we assume M" is a smooth submanifold of R Ifj. q < 2n + 1 there is nothing to prove. Assume that q > 2n + 1. It suffices to show that M" embeds in R-1 since we could then repeat this process until we have finally embedded M" R2n+'. in We identify R0 -1 with {(x1 , 2 , ... ,xq) E Rq XTq = 0}. 1 For v E R_- R-1 let y:Rq__ - R- be projection parallel to v. We want to find a vector v such that rv|M is an embedding of M into Rq-1. We limit our search to unit vectors. For ir to be injective we must require that v is not parallel to any secant of M. That is, if xy E M then (1)V - Y I -T- Y| The kernel of r is just the line through v. Therefore a tangent vector z E DM. is in the kernel of D(ry). only if z is parallel to v. So rvIM is an immersion if, for all (x,z) E DM with z J 0, 24 (2) z To find a vector that satisfies (1) consider the map (M x M) - A -+ Sq-1 by u(x,y) x -Y where A = {(x,y) E M x MJx = y} and Sq-1 = V E Rq : |lvi = 1}. Then v satisfies (1) if and only if v is not in the image of g. Since g is a smooth map, and dim((M x M) - A) = 2n < dim(Sq-1) we have from the Mini-Sard theorem that the image of o is nowhere dense in Sq-1. So every nonempty open subset of Sq-1 contains a point v which is not in the image of a. To find a vector that satisfies (2) note that we need only consider points (x,z) E DM where Jzj = 1. Define the Unit Tangent Manifold of M to be DIM = {(x,z) E DM :z| = 1} Since DM C M x Rq,7DIM C M x Sq-1. Also DIM is a compact submanifold of DM (since M is compact). Define a smooth map r:DIM -+ Sq-i by projection onto Sq-1 That is r(x,z) = z. Geometrically this is just a parallel translation of unit vectors based at a point of M to unit vectors based at 0. Then a vector v satisfies (2) only if v is not in the image of r. Again note that dim(DIM) = 2n - 1 < dim(S-1) 25 So the Mini-Sard theorem says that the image of r is nowhere dense in Sq-1. We see that we have many vectors that satisfy (1) and many vectors that satisfy (2). We want a vector that satisfies both (1) and (2). Since DIM is compact r(DIM) is compact. Since every compact subset of a Hausdorff space is closed, r(DIM) is closed in Sq-1. So Sq-1 - r(DIM) is an open dense subset of Sq-1. Therefore the set V = (Sq-1 - r(DM))n(Rq - R-1) is a nonempty open subset of Sq-1. Every v E V satisfies (2) and from our previous work we know that there is a vector v E V that satisfies (1). So we have found a vector v E Rq - Rq-1 that satisfies (1) and (2). Therefore rvM:M --+ R-1 is an injective immersion. Furthermore rv|M is a one-to-one continuous mapping from a compact space onto its image (which is Hausdorff), so we have that rIM is an injective immersion that maps M homeomorphically onto its image. Therefore 1 rvM is an embedding of M" into R- . 0 In fact, we can even improve upon this result. Whitney has proven the following [8]: THEOREM 2.18. (Whitney embedding theorem) Every paracompact Hausdorff manifold M" of dimension n embeds in R 2,. CHAPTER III VECTOR BUNDLES Whitney's embedding theorem says any n-dimensional manifold M" can be embedded in R". This theorem is the best possible in the sense that there are n-dimensional manifolds that cannot be embedded in R1. One example is the Klein Bottle which is a 2-dimensional manifold that cannot be embedded in R1. Of course there are 2-dimensional manifolds that embed in R1, such as the torus or the 2-sphere. So Whitney's theorem does not give the smallest dimension of Euclidean space that we need to embed these manifolds. How then, given a specific manifold M", can we know if we can embed it in Euclidean space of dimension less than 2n? To answer this question we must know more than just the dimension of the manifold. We now develop tools that tell us more about the manifold in question. The first tool we need is a generalization of the tangent manifold: DEFINITION 3.1. Let B be a fixed topological space called the base space. A real vector bundle ( over B consists of the following: 26 27 1) a topological space E = E(6) called the total space, 2) a continuous map r:E -+ B called the projection map, and 3) for each b E B, the set r-'(b) has the structure of a vector space over the real numbers. In addition a vector bundle must satisfy the following: Local triviality condition: For each b E B there exists a neighborhood U C B, an integer n > 0, and a homeomorphism h: U x R"'+ -(U) so that, for each b E U, the correspondence x i-+ h(b,x) defines an isomorphism between the vector space R" and the vector space r-'(b). The pair (U,h) is called a local coordinate system for 6 about b. If the set U can be chosen to be the entire base space then we call ( a trivial bundle. The vector space r-'(b) is called the fiber over b and is denoted by Fb or Fb(6). If the dimension of Fb is a constant for all b, say n, then we say that 6 is an n-plane bundle. If n = 1 then we call 6 a line bundle. DEFINITION 3.2. Let 6 and y be two vector bundles over the same base space. We say 6 is isomorphic 28 to n if there exists a homeomorphism f:E(6) -E(n) between the total spaces which maps each vector space Fb( ) isomorphically onto the corresponding vector space Fb(f). If 6 is isomorphic to n we write 6 2 n. Actually we only need the function f to be continuous and map fibers isomorphically onto fibers because of the following: THEOREM 3.3. Let 6 and r be vector bundles over B and let f:E(6) -- E(n) be a continuous function which maps each vector space Fb(6) isomorphically onto the corresponding vector space Fb(n). Then f is necessarily a homeomorphism. Hence 6 is isomorphic to n. PROOF: Given any point bo E B, choose local coordinate systems (U,g) for 6, and (V,h) for i, with bo E U f V. Then we must show that the composition (U f V) x R V0f - -9--+ (Ufl V) xRn is a homeomorphism. We know that h o f o g is continuous, one-to-one, and onto, we must show that it has a continuous inverse. 1 Set h o f o g(b,x) = (b,y). Then since f maps Fb() isomorphically onto Fb(27) there is a non-singular nxn matrix [fij(b)] of real numbers that allows us to express 29 y = (yi,... ,yn) in the form n yi =.E fij(b)xij. Let [Fji(b)] denote the inverse matrix of [fij(b)]. Then g-1 o f-1 o h(b,y) = (b,x) where n Xi =E j i( b) yi. 1 =1 Since the Fji(b) depend continuously on the matrix [fij(b)], they depend continuously on b. Thus g-1 o f1 o h is continuous. o EXAMPLE 3.4. Let B be some topological space. The trivial bundle over B is the bundle with total space B x R' for some n > 0. The projection map is r(b,x) = b and the vector space structure given to the fibers is defined by ti(b,X) + t 2 (b,X 2) = (btixi + t2X2 ) for every tl,t2 E R. We denote the trivial n-plane bundle over B by E. THEOREM 3.5. An n-plane bundle over B is trivial if and only if it is isomorphic to e. PROOF: Let ( be a trivial n-plane bundle. Then there is a homeomorphism h:B x R'- -+ K-'(B) so that the correspondence x i- h(b,x) defines an isomorphism between OR and r-'(b). But r~ 1(B) = E(6) and B x R" = E(E"). So 30 we have a homeomorphism h:E(c") -- E(6) that maps each fiber of c" isomorphically onto the corresponding fiber of 6. Thus 67!. Now suppose Lu n. Then there is a homeomorphism h:E(j") -- E(6) which maps each fiber of e" isomorphically onto the corresponding fiber of 6. But then 1 we have h:B x R -+ -(B) so that for each b E B the correspondence x -- + h(b,x) defines an isomorphism between R" and ir~(b). Thus 6 is trivial. o EXAMPLE 3.6. The tangent bundle rM of a smooth manifold M is defined to be the vector bundle with base space M, total space DM and projection map r:DM -- M defined by r(xv) = x. Again the vector space structure in the fibers r-'(x) is defined by tli(X,vI) + t2 (X, v2 ) = (X, tIvI + t2v2). DEFINITION 3.7. Real Projective Space RP" is the set of all unordered pairs {x,-x} where x ranges over the unit sphere S" C R"'. We topologize RP" as a quotient space of S". DEFINITION 3.8. The canonical line bundle 7A over RP" is obtained as follows: The total space E(-A) C RP" x U"' consists of all pairs ({ x},v) such 31 that the vector v is a multiple of x. The projection map r:E(7A) -+ RP' is defined be r({*x},v) = { x}. Thus each fiber -1({ x}) can be identified with the line through x and -x in R"*I. We give each such line its usual vector space structure. To show that -y is, in fact, a vector bundle we must show that the local triviality condition is satisfied: Let U C S" be any open set which contains no antipodal points, and let U, denote the image of U in RP" under the map X '- {*}. Then we define a homeomorphism h:U1 x R -7-,(Ul) by h({*x},t) = ({x},tx) for each (X, t) E U x R. Then (U 1 ,h) is a local coordinate system for 7,1. Hence 7A is locally trivial. o Since a line through the origin in R"' intersects the unit sphere S" at two antipodal points we can think of a point in RP" as being a line through the origin instead of the pair { x}. But a line through the origin is just a 1-plane, so we see that RP" = GI(R"*i) the Grassmann manifold. DEFINITION 3.9. The canonical vector bundle 7"(R"nk) over G1(R"k) is constructed as follows: Let E = E(7"(Rn+k)) be the set of all pairs 32 (n-plane through 0 in gRnk, vector in that n-plane). This is to be topologized as a subspace of Gn(R"*k) x R"*k. The projection map r:E -+ Gn(Rn*k) is defined by r(X,x) = X, and the vector space structure in the fiber over X is defined by t1(X,XI) + t 2(X,X 2) = (XtlX1 + t2X2)- Note that y 1(R" 1) is the same as the line bundle 7 over RP". Similarly we can construct the canonical bundle 7" over Gn by letting E(7") C Gn x RO be the set of all pairs (n-plane through 0 in RO, vector in that n-plane). The projection map r:E(-") -+ Gn is given by r(X,x) = X. The vector space structure in the fibers is defined as before. The infinite Grassmann manifold is useful to us because of the following property. THEOREM 3.10. Any n-plane bundle ( over a paracompact base space admits a bundle map 6 -+ n. In other words, every n-plane bundle over the paracompact base space B is isomorphic to the induced 33 bundle f*(#yn) for some function f:B -- Gn. For this reason n" is called the universal n-plane bundle. PROOF: Let 6 be an n-plane bundle over the paracompact base space B. We first show that there exists a locally finite covering of B by countably many open sets UIU2,...so that 6IUi is trivial for each i. To do this we use the fact that every paracompact space is normal [7, p.147]. Choose a locally finite open covering {V,} so that each 6|V is trivial, and choose an open covering {WY} with V, C V, for each a. Let AOL: B -- +R be a continuous function which takes the value 1 on V and the value 0 outside of V.- For each non-empty finite subset S of the index set {a}, let U(S) C B denote the set of all b E B for which min A(b) > max A,(b). aES aS Let Uk be the union of those sets U(S) for which S has precisely k elements. Then each Uk is open and B = U1 U U2 U U3 U ... since, given b E B, if precisely k of the numbers A,(b) are positive, then b E Uk- If a is any element of the set S note that U(S) C VOL. 34 Since the covering {Va} is locally finite, it follows that {Uk} is locally finite. Furthermore, since each 61V, is trivial, each 6|U(S) is trivial. But the set Uk is equal to the disjoint union of its open subsets U(S). Therefore 6IUk is also trivial. So we have found a countable number of sets UIU ,U2 3,... that cover B such that 6IUi is trivial for each i. Now to construct a bundle map f:6 -- " we first construct a map f:E(6) -- R' which is linear and injective on each fiber of (. Let U1,U2 ,U3,... be as above. Again using the fact that B is normal, there exist open sets QI,Q2,Q3,.-. covering B with qi C Ui. Similarly construct D 1,D 2 ,D 3 ,--- with Ui C Qi. Let Ai:B -- R denote a continuous function which takes the value 1 on 1i and the value 0 outside of Qi. Since 6IUi is trivial there exists a map hi:r-i(U,) -- Rn which maps each fiber of 6IUi linearly onto R". Define h':E(6) -- R by (e) f or r( e) E Ui ' { Ai(r(e))h1 0 for xr(e) 0 1i 35 Then h' is continuous, and is linear on each fiber. Now define f : E( - R" (DR" ED - -... by f(e) = (h'(e),hl2(e),...). Then f is also continuous and maps each fiber injectively. Now define f:E(6) -+ E(") by f(e) = (f(fiber through e),f(e)) Then f is continuous and maps each fiber of E(6) isomorphically onto a fiber of E(-"). Therefore f is a bundle map from 6 to 7". o DEFINITION 3.11. A cross-section of a vector bundle over B is a continuous function s:B -+ E(6) which maps each b E B into the corresponding fiber Fb(6). A cross-section is nowhere zero if s(b) is a non-zero vector of Fb(6) for each b. Now consider a collection {si,...,s,} of cross-sections of a vector bundle 6. DEFINITION 3.12. The cross-sections s1 ,... ,sn are nowhere dependent if, for each b E B, the vectors s1(b),. . . ,s(b) are linearly independent. 36 THEOREM 3.13. An R"-bundle is trivial if and only if ( admits n cross-sections si,...,sn which are nowhere dependent. PROOF: Let si,...,s be cross-sections which are nowhere dependent. Define f :B x R"'-+E by f(b,x) = xis1(b) + --- + xs(b). Then f is continuous and since the si,.,sn are nowhere dependent, f maps each fiber of the trivial bundle e" B isomorphically onto the corresponding fiber of (. From Theorem 3.3, f is a bundle isomorphism. Thus e.c l So ( is trivial. Now suppose that ( is trivial with coordinate system (B,h). Define si(b) = h(b,(O,...,0,1,0,. ..,0)) E Fb() (with the 1 in the i1h place). Since we know the correspondence x F-. h(b,x) is an isomorphism, ,s(b) are n nowhere dependent cross sections. o We need to consider vector bundles in which each fiber has the structure of a Euclidean Vector Space. DEFINITION 3.14. A real valued function p on a finite dimensional vector space V is quadratic if # can 37 be expressed in the form n #(V) =E 1(v)h(v) 1=1 where li and hi are linear. jp is called positive definite if ,u(v) > 0 for v * 0. DEFINITION 3.15. If ps is quadratic, then we define an inner product on the vector space V to be the map from V x V to R given by (v,w) - v.w where vw =((V + w) - '(v) - p(w)) Notice that vv = {p(v + v) - p(v) -pV) =4Ap(V + v) -/p(V) n =2E li(v+v)hi(v+v)- p(v) 1=1 = 2.Eli(v)hi(v) - p(v) S=I =2p(V) - pkv) DEFINITION 3.16. A Euclidean vector space is a real vector space V together with a positive definite quadratic function pL: V -+ R For each v E V we define the norm of v, denoted jvj, by lvI = v-v = 4Tv). 38 DEFINITION 3.17. A Euclidean vector bundle is a real vector bundle 6 together with a continuous function :E(6) -- R such that the restriction of p to each fiber of 6 is positive definite and quadratic. The function p is called a Euclidean metric on 6. In the case of the tangent bundle rM of a smooth manifold a Euclidean metric p:DM -- R is called a Riemannian metric, and M together with IL is called a Riemannian manifold. Probably the most natural Euclidean metric we can give the trivial bundle e"'is p(b,x) = x' + ... + x. so that we have the usual notion of distance in the fibers. Since the tangent bundle of R" is trivial it follows that the smooth manifold R" posseses a standard Riemannian metric. Now for any smooth manifold M C R" the composition DM C DR" -P4 R makes M into a Riemannian manifold. Thus, since every paracompact manifold M" of dimension n embeds in R2n, M" can be given a Riemannian structure. Note, however, that this structure depends upon the embedding. 39 THEOREM 3.18. Let 6 be a trivial vector bundle of dimension n over B, and let p be any Euclidean metric on 6. Then there exist n cross-sections si,...,sn of ( which are normal and orthogonal in the sense that ifi =y si(b).sj(b) = { for each b E B. PROOF: Let s1, .. ,s' be any n cross-sections which are nowhere dependent. We can apply the Gram-Schmidt process to s'1(b),...,s'(b) and obtain a normal orthogonal basis si(b),...,sI(b) for Fb(6). Since the resulting functions si,*...,s are continuous we have our required cross-sections. o Given a vector bundle 6 with projection r:E -- B we can construct other vector bundles from this, as we now show. DEFINITION 3.19. Let 6 be as above. Let B' be a subset of B. Setting E' = r-'(B') and letting :E' -- B' be the restriction of r to E' we obtain a vector bundle which will be denoted by 6|B' and called the restriction of 6 to B'. Each fiber Fb(61B') is equal to the 40 corresponding fiber Fb(6), and is to be given the same vector space structure. Note that using the above terminology we can restate the local triviality condition by saying that for each b E B there is a neighborhood U of b such that the bundle flU is trivial. DEFINITION 3.20. Let 6 be as defined above and let B be 1 a topological space. Given any map f:B 1 - B we can construct the induced bundle f*6 over B as follows: The total space E, of f * is the subset E1 C B1 x E consisting of all pairs (b,e) with f(b) = r(e). The projection map r1 :E1 --+ B1 is defined by r(b,e) = b. Thus we have the commutative diagram E1 E B1 B where f(b,e) = e. The vector space structure in ri(b) is defined by ti(b, el) + t (b,e 2 2 ) = (b, tie 1 + t2e2 ) . Thus f carries each vector space Fb(f*6) isomorphically onto the vector space Ff(b)(6). 41 To see that f*6 is locally trivial let (U,h) be a 1 local coordinate system for 6, set U1 = f- (U) and define hl:U x R' -p-(U1) by hi(b,x) = (bh(f(b),x)). Then (U1,hj) is a local coordinate system for f*6. So f* ~is locally trivial. More generally we have: DEFINITION 3.21. A Bundle ap from r to ( is a continuous function g:E(1) -+ E (6) which carries each vector space Fb(71) isomorphically onto one of the vector spaces Fd(6). If we set 4(b) = d then the function p:B() -+ B(6) is continuous and we have the following commutative diagram E(71) 9 *E(6) B ( 1) B(6) THEOREM 3.22. If g:E(y) - E(6) is a bundle map, and if 4:B(i) -+ B(6) is the corresponding map of base spaces, then 1 is isomorphic to the induced bundle g*6. 42 PROOF: Define h:E( 7) -+ E(-*6) by h(e) = (r(c),g(e)) where r denotes the projection map of n. Since h is continuous and maps each fiber Fb(n) isomorphically onto the corresponding fiber Fb(4*6), h must be a homeomorphism by Theorem 3.3. Thus q L ~*. o Natural operations on vector spaces can also be extended to vector bundles. DEFINITION 3.23. Let 6, and 62 be vector bundles with projection maps wi:Ei - Bi i = 1,2. The Cartesian Product 61 x 62 is defined to be the bundle with projection map 71 x r2 :E1 x E2 -+ BI x B2 where the vector space structure on each fiber (71 x 72)~(cd) = F,(6 1) x Fd(62) is given by ti(c x x d,ei e3 ) + ti(c x d,e 2 x e4) = (c, tie 1 + t2e 2) x (d, tie3 + t2e4) . DEFINITION 3.24. Let 61 and 62 be two vector bundles over the same base space B. Let d:B -+ B x B be given by d(b) = (b, b). The bundle d*(6i x 62) over B is called the Whitney sum of 61 and 62, and will be denoted by 61 '9 2. 43 Now consider the fiber Fb(d*(61 x 62)) = Fb(61 ED2). We have the following commutative diagram Fb(6i1'S 62) d +Fb(61) 19 Fb(62) 72 1 1 B d B x B and we see that Fb(61 ED2) is isomorphic to Fb(61) @Fb(62) where the isomorphism is given by d. DEFINITION 3.25. Let 6 and y be vector bundles over the same base space with E(6) C E(n). Then 6 is a sub-bundle of n (written 6 C n) if each fiber Fb(6) is a sub-vector-space of the corresponding fiber Fb(77). DEFINITION 3.26. Let n be a vector bundle on which a Euclidean metric has been defined and let ( C n be a sub-bundle. Then the orthogonal complement ( of 6 in n is defined by letting the fibers Fb((1) be the subspace of Fb(77) consisting of all vectors v such that v-w = 0 for all w E Fb(6) and letting the total space be E )= Fb(( ). bEB 44 THEOREM 3.27. E(61) is the total space of a sub-bundle 6 C n-. Furthermore n N6( @ 1. Before we prove this we need the following LEMMA 3.28. Let 61 and 62 be sub-bundles of 17 such that each vector space is equal to the direct sum of the sub-spaces Fb(61) and Fb(62). Then 7 6 1 a 2. PROOF (of Lemma): Define f:E(6E D62) --+ E() by f(b, el, e2 ) = el + e2. Then f is continuous and since Fb(61 62) v Fb(61) @ Fb(62) which equals Fb(1) we have from Theorem 3.3 that f is a homeomorphism and thus SN6 (1@ 2. PROOF (of Theorem): Each vector space Fb(y) = Fb(6) @ Fb(61) so from Lemma 3.26 y ED1 provided that 6 is, in fact, a sub-bundle of q. To show this we must show that 6 is locally trivial. Let bo E B and let U be a neighborhood of b0 which is sufficiently small that both 6jU and yjU are trivial. From Theorem 3.16 there exist orthonormal cross-sections s1,...,sm of 6|U and ri,...,r of yJU where m and n are the respective fiber dimensions. Thus the mxn matrix [si(bo)-rj(bo)] has rank m. Renumbering the rj if 45 necessary we may assume that the first m columns are linearly independent. Let V C U be the set consisting of all points b for which the first m columns of the matrix [si(b)-rj(b)] are linearly independent. Consider the square matrix that consists of the first m rows and m columns. Since the determinant function is continuous det1 ({O}) is closed. Also the complement of V, Vc = det'({O}). Hence V is open. We now show that the sections Si,S2,...,Sm,rm+i,...rn of yJU are not linearly dependent at any point of V. Suppose that the sections were linearly dependent at some point b E V. Without loss of generality we can assume that in n si(b) =.E assi(b) + E cjrj(b) j=2 j=m+1 where at least one cj is not zero. Hence M n E aisi(b) = E cjrj(b). 1=1 j=M+1 So at least one of the si(b),... ,SM(b) is in the subspace spanned by rm.(b),..,r-n(b) and thus is orthogonal to r1(b),...,r(b). That is, for at least one value of i, si(b)-rj(b) = 0 if j < m. Hence at least one of the first m columns of the matrix [si(b)-rj(b)] is zero. This contradicts the fact that the first m columns of the matrix are linearly independent. 46 Therefore s1,s2,...,sm,r.1,...,r are not linearly dependent at any point of V. Applying the Gram-Schmidt process to this sequence of cross-sections, we obtain normal orthogonal cross-sections si,... ,sm,Sm+1,*a.a..,sn of VV. Now a local coordinate system h:V x R",-+ E( ) for ( is given by the formula h(b,x) = xism.1(b) + ... + Xn-ms(b). Since the inverse function h-1(e) = ( ir(e) , (e.sm.(r(e)),...,esn(r(e))) ) is continuous, h is a homeomorphism. Thus 61 is locally trivial and we are done. o DEFINITION 3.29. Let M C N be smooth manifolds, and suppose that N is provided with a Riemannian metric. Then the tangent bundle rM is a sub-bundle of the restriction rNIM. We define the normal bundle v of M in N to be the orthogonal complement TM C 7NIM. From the preceeding theorem we have 47 COROLLARY 3.30. For any smooth submanifold M of a smooth Riemannian manifold N the normal bundle v is defined, and TM (Dv NIM CHAPTER IV STIEFEL-VHITNEY CLASSES In this chapter we use the following facts from Algebraic Topology without proof. To each topological space B there corresponds a sequence of groups Hk(B;1 2 ) called the kth singular cohomology groups with coefficients in 92 defined for all integers k > 0. If we let 0 2 H*(B;1H 2 ) = H (B;21 2 ) x H'(B;112 ) x H (B;112) x - - - then H*(B;i12) has the structure of a ring with the multiplication operation being given by the cup product v. Since, in what follows, we always use cohomology and homology with coefficients in E2 we just write H"(B) instead of H"(B;l 2). Let ( be a real n-plane bundle with total space E = E(6), base space B, and projection r:E -+ B. Furthermore assume that a Euclidean metric has been defined on 6. Let EO denote the set of all non-zero vectors in E. Then we define an (n - 1)-plane bundle 6, over E as follows: A point in E0 is specified by a fiber F of together with a non-zero vector v in that fiber. The fiber of 60 over v is by definition, the orthogonal 48 49 complement of v in the vector space F. Then this is a real vector space of dimension (n - 1). In terms of local coordinates, for each point (bv) E E0 (b E B, V E r~1(b) v # 0) def ine the f iber of (o over (b, v) to be the orthogonal complement of v in r-1(b). Then a point in the total space of 6o looks like (b,v,z) where (bv) E Eo and z E r-1(b) with z-v = 0. The projection map ro of 6, is then given by 7r0(bv,z) = (b,v). Thus we have constructed an (n - 1)-plane bundle 60 over E0 . Note that E0 (and hence 60) is independent of the Euclidean metric used since any Euclidean metric p is positive definite which means that pu(v) > 0 for v J 0. We now have the following important Theorem from Algebraic Topology [4, p.106] or [6, p.144]: THEOREM 4.1. (Thom Isomorphism Theorem) The group H"(E,EQ) contains a unique class u such that for each fiber F = r-'(b) the restriction u|(F,FO) E H"l(F,FO) is non-zero. Furthermore the correspondence x - x - u defines an isomorphism Hk(E) -+ Hkn,(E,EO) for every k. (Here w denotes the cup product operation.) 50 DEFINITION 4.2. Let u be as in Theorem 4.1. Then u is called the fundamental cohomology class or the Thom class. If E is the total space of a vector bundle over B with projection map r, then the zero section embeds B as a deformation retract of E with retraction mapping r. Thus the map r:E -+ B induces an isomorphism: T*: Hk(B) -- +Hk(E). DEFINITION 4.3. The Thom isomorphism l:Hk(B) - Hk(E,Eo) is defined to be the composition of the two isomorphisms: Hk(B) -'*+,Hk(E) :Hk +"(E,Eo) . Let ( be an n-plane bundle with projection map r:E -+ B. Restricting r to the non-zero vectors in E we obtain an associated map iro:E0 -+ B. We now have the following. THEOREM 4.4. To any n-plane bundle ( there is associated an exact sequence of the form e -a-- + Hi(B) qne is (B)d Hi(E) qHi'(B) f- -. The above sequence is called the Gysin sequence of g 51 PROOF: We start with the cohomology exact sequence -- + Hi(E,E0) - H(E) -- + Hi(E0 ) -L H 1 (E,EQ) -+ --- of the pair (E,E). The isomorphism VUs:Hi -"(E) -+4 H(E, EO) allows us to substitute Hi-"(E) for Hj(E,Eo) to get -00-- -+ Hi-"(E) --9+ H(E) -- +H (E0) -0 Hi-"*1(E) - - where g(x) = (xv u)IE = xv (u|E). Now substitute the isomorphic cohomology ring H*(B) in place of H*(E). Since the cohomology class u|E in H"(E) corresponds to the top Stiefel-Whitney class wn E H"(B) we have the required exact sequence S - -p-+ Hi-"(B) "- Hi(B) - Hi(E0 ) -+Hi-" (B -- + - - . o Notice that for j < n - 1 in the proof of the above theorem that the groups HjI(B) and Hi-" 1(B) are zero. So we get Hl(E0)-+ 0. 0 -vw" Hi(B) J Thus, by exactness, r* is both injective and surjective. Therefore 7r is an isomorphism from Hi(B) to Hi(E0) for j < n - 1. We are now ready to define the Stiefel-Whitney classes of a vector bundle. 52 Let 6 be an n-plane bundle over the base space B(6) with total space E(6). Let u E H"(E,Eo) be the Thom class. The inclusion i:(E,#) -+ (E,EO) gives rise to the homomorphism: i*:H*(E,E.) -+H*(E). Also the projection r:E -+ B gives rise to an isomorphism: 7r*:0 H*(B) --+ H*(E) . We now define the nth Stiefel-Whitney class to be: Wn( = (*)(i*(u)) E H"(B(6)) For i < n we define the ith Stiefel-Whitney class inductively by the formula: wi(O) = (7r*)1(wi(6O)) For i > n we define wi(6) = 0. The total Stiefel-Whitney class of an n-plane bundle 6 is the element: w(6) = 1 + wi(6) + W2(6) + - +Wn() in the ring H*(B). To see that the above definition really does define all the Stiefel-Whitney classes , consider the following example: Suppose 6 is a 3-plane bundle and we wish to find w (6). By definition 2 w2 (6) = (rg)~'(w 2 (6o)). Now 60 is a 2-plane bundle, so w2(6o) = (r*)~I(i*(u)), thus we have defined w2( ). If, instead, we wanted to find wi(6) we would first need to know w1(6,0). To find this we would construct a 1-plane bundle (6,)o from 6. in the same way 53 we constructed 6o from 6. Now since (6o)o is a 1-plane bundle, w1((60)0) is defined in terms of the Thom class and thus, by backward substitution, we know wj(6o) and finally Note that at this moment we do not know that this definition makes sense. This is because we do not yet know that wn-.1(60) E Im(7r). We will prove that this is true later (after we have proven Property 1 below). We are interested in proving four main properties of the Stiefel-Whitney classes. These are PROPERTY 1. To each n-plane vector bundle 6 there corresponds a sequence of cohomology classes wi(6) E H(B(6)) i = 0,1,2,... with wo(6) = 1 E Hl(B(6)) and wj(6)=0 for i>n. PROPERTY 2. (Naturality) If f:B(6) - B(i) is covered by a bundle map from 6 to y then snim) = f*wi(7) 54 PROPERTY 3. (The Whitney Product Theorem) If 6 and r are vector bundles over the same base space then k Wk(I @7) = wi () Wk-1(0) i=O Note: The symbol will often be omitted. For example: wi(6 'S7) = wi(6) + wi(17) W2(6 27) = W2(6) + wI(6)wI(V) + W2(n) In terms of the total Stiefel-Whitney class we can write w(6I@ 2) = w()w(n) PROPERTY 4. For the line bundle 71 over the circle RPI, the Stiefel-Whitney class wi(eyl) is non-zero. We first prove Property 2. PROOF (of Property 2): Let 6 and n be n-plane bundles, and let f:B(6) -+ B(n) be covered by a bundle map from 6 to n. That is, we have the following commutative diagram E(6) L4 E(n) 7r() B() B(9) where f carries each vector space Fb(6) isomorphically onto some vector space Fd(2). 55 We proceed by induction by assuming that we know that Property 2 holds for (n - 1)-plane bundles. We are then able to show directly that Property 2 holds for line bundles (Case 2 below), and this is what the induction hinges on. Case 1: (i < n) We can assume that the map f preserves inner products since, if not, we can redefine the metric in E(6) as follows: for v,w E E(6) define vw =f(v)-f(w). Forming the spaces EO(6) and Eo(i) as described above, the map f gives rise to a map fo:Eo(6) -+ Eo() which is covered by a bundle map - yo of (n - 1)-plane bundles. Since 6c and yo are (n - 1)-plane bundles we know wi(6) = f*wi(igo) by the induction hypothesis. The following diagram: EO(1) EO(i) TO (6) 1 17(0) B(6) L B(j) commutes. Also, by definition wi(60) = 7*(M)(wi(O) wi(no) = 70(0) (wi(n)) . 56 So Wi() = T()-(wi(f)) = ()-1(f*wi(no)) = S )'foTh roe(o))(wi( )) = ())*(wi(7)) = f*( (9)) . Therefore: wi(6) = f*(wi(n)) . Case 2: (i = n) We have the following commutative diagram E(6) -- 4 E(6) B (6) L. B(() where f maps each fiber Fb(6) isomorphically onto some Fd(q). Now consider the following diagram for any fiber F (F()jo(I)) (F Fo (E(6),IEO(6)) i(E(17),Eo (7)) where k and 1 are injections. Here we are abusing notation slightly, by f in the bottom half of the diagram we really mean f e f|Eo and in the top half of the diagram we really mean f IF 0 f IFo. In cohomology we get 57 H"(F(6),Fo(f)) 1 H"(F(i7),Fo(7)) k* J 11* H"(E(6), Eo(f)) 1 H"(E(i7),Eo(77)) Let U' E H"(E(),Eo(i)) be the Thom class and let U E H"(E(6),EO()) be the Thom class. We want to show that By the definition of the Thom class, *(u/) is nonzero in H"(F(y),F.(7)) and k*(u) is nonzero in H"(F() ,Fo()). But there are only two elements in H"(F(y),Fo(q)) and H"(F(6),Fo(6)), and since fIF* is an isomorphism we must have f*(t*(u')) = k*(u). That is From the following commutative diagram H"(E(77), EO(7)) iH"(E ff1fj (7) H"(B(7)) H"(E(6),EO(6)) 2** H"(E(6)) 7*CH"(B(6)) we get: wn(6) = (r*(U))= *(6))- ( */ = =wn(7) Thus wn() = fwn(77). 0 58 We are now ready to show that w. 1(6o) E Im(?rg). It suffices to show that wn-1(f0) restricts to zero in each 1 fiber F,. But Sn- is a retract of the fiber F0 so we can think of the vectors in the total space of 60 as being tangent vectors of S"-1 (they are orthogonal to vectors in Sn-1). Therefore it suffices to show that wn.1(r) = 0 where r is the tangent bundle of the sphere S"-1. Let w(M) denote the Stiefel-Whitney class of the tangent bundle of the smooth manifold M. The natural map f:S"-l -- RP- 1 given by f(x) = {+x} is locally a diffeomorphism so the induced map Df:DS" 1 -- D(RP"-1) is a bundle map. Using naturality of the top Stiefel-Whitney class only, we know that wn.1(S"- ) = f*wn.i(RPn-1). We now use the following theorem from Algebraic Topology [5, p.403]. THEOREM 4.5. The mod 2 cohomology of projective space RP" is given by H'(Rpn) = for 0 < i < n 0 for i > n We now compute the homomorphism f*:Hn(RPn) -- H"(S") provided that n > 1. 59 Let a E H1(RPn) be the generator for the cohomology of RP". Then f*an = (f*a)n. But f*a E H1(S") = 0 for n > 1. 1 Thus wn.1(S"- ) = f*(wn-i(RP")) = 0. Therefore wn.1(6o) E Im(7r*) and we have finally proven that our definition of Stiefel-Whitney classes makes sense. PROOF (of Property 1): We only need to show that wo(6) = 1 E HI(B(6)) since the rest follows immediately by definition. Since the Thom class u is a non-zero element of H(E,Eo), and since i* and (r*)-l are isomorphisms it follows that wo() = (*)-(i*(u)) is non-zero in H(B(6)). 0 Now if B(6) is connected H (B(6);l2) has only two elements, and we have wo(6) = 1 E H(B(6)). If B(6) is not connected consider a point x E B(6). We want to show that wo() evaluated at [x], the homology class of x, is equal to 1. We have the following commutative diagram: Fw(a)- uiE() {X} >B(6) where i and ^ are the inclusion maps. 60 0 Let w' E H ({}) be the zero-th Stiefel-Whitney class of the vector bundle f|{x}. Since the single point {x} is certainly connected we know that w' = 1. Also the inclusion i:{x} -+ B(6) induces an inclusion map in homology i,:[x] -+ Ho(B()) . Thus [x] = i*[x] . If we let = So We save the proof of Property 3 until last. PROOF (of Property 4): We want to show that wl('yi) is nonzero. First recall the definition of 71. The total space E = E(-yl) consists of all pairs ({*x},v) where {x} E RP' and v is a multiple of x. We can represent every point of E(-yl) as (*(cos(O),sin(O)) ,t(cos(O),sin(O))) for 9 E [0,7r] and t E R. This representation is unique except that (*(cos(0),sin(o),t(cos(O),sin(O)) = (1(1,0),t(1,0)) = 61 (*L(-,0),-t(-,20) )= ( (cos(T),sin(r),-t(cos(r),sin(r)). That is, E can be obtained from the strip [0,ir] x R in the (0, t)-plane by identifying the points (0, t) with (7,-t). Thus E can be thought of as an open Mobius strip (see Figure 4.1). (0, t) / [O,T] x g Figure 4.1 Now the space of vectors of length < 1 in E is a Mobius strip M bounded by a circle aM. M is a deformation retract of E and aM is a deformation retract of E0 . So H*(M,aM) H*(E,EO). Also, if we embed the disk D2 in RP2 the closure of 2 2 RP - D is homeomorphic to M. To see this consider Figure 4.2 where we have shown the closure of RP2 - D2. 62 a Rp2 - D2 Figure 4.2 Now if we cut along the sides labeled b and c in Figure 4.2 we get the two strips that are shown in Figure 4.3. a - I 1% c ib - I I d Cl a 1 b Figure 4.3 Now we identify the sides labeled a as shown in the left side of Figure 4.4. Then identifying sides b and c again we get the Mobius strip shown in the right side of Figure 4.4. 63 b C b C C a e Figure 4.4 So, by the Excision Theorem: H*(M,M) H*(ROP2,D2 ). Hence we have the isomorphisms: Hi(EEO) -- + Hi(M,8M) + Hi(Rp22,D2) +- Hi(Rp2) for i #0. So 1 2 H(EE) = H (RP ' = 2i7 2o)= H0 if i=0,1,2 if > 2 From the sequence of injective maps (EO,#) d (E,#) et+(E,2EO) we get the exact sequence of the pair (E,EO): HI(EEO) a HI(E) -4 HI(EO) -- where HI(E,EO) l22, HI(E) H'(RPI) = H1(S1) y 112, and 1 1 H (Eo) HI(9M) = HI(S ) 2122.So we have - - 12 0 2 112 --- -- Since OM is a deformation retract of EO and M is a deformation retract of E we see that j* sends the generator a E HI(E) to twice a generator 2a E HI(E0). (We 64 see this by noticing that if we trace a path once around OM this corresponds to tracing a path twice around the center circle of M). Since we are working with coefficients in 12 we have j* 0. so kernel(j*) = H2 and, by exactness, the map i* is onto. Now since the Thom class u E HI(E,EQ) is nonzero, i*(u) 0 0. And since the map x*:H'(B) -+ H'(E) is an isomorphism w, = (r*-(i*(u)) # 0. Thus wi(7y') is nonzero. o We now wish to prove Property 3, but before we can do this we need several lemmas: LEMMA 4.6. Let 6 and y be m and n dimensional vector bundles repectively over the base space B. Then wm+n(E@ 7) = wm(6)wn(i). PROOF: This proof makes use of the cohomology cross product operation [5, p.355]. Ve first wish to compute the Stiefel-Whitney class of the product bundle 6 x 1. Let E = E(6), E' = E(n), and let r x r':E x E' -+ B x B be the projection map. Consider the Thom classes U E Hm(E,Eo) ' E H"(E',E'). From Algebraic Topology the cross product is defined and 65 U x U' E H"''"(E x E' ,(E x E')U(EQ x E')). Then Wn+m( X 7) = (7r*)d(i*(U x u')) = (r*)~1(i*(u) x i*(u')) = (r*)-~i*(u) x (r*)-li*(u/) = wm(6) x wn(17). So wn+m(6 x 1) = wm(6) x wn(17). Lifting both sides of this equation back to B by means of the diagonal map d:B -+ B x B we get wm+n( @ 79) = Wm() wn(7). 0 So we have proven the Whitney Product Theorem for the top Stiefel-Whitney class. LEMMA 4.7. If 6 is an n-plane bundle that possesses a nowhere zero cross-section, then wn(6) = 0. PROOF: Let s:B -+ Eo be a cross-section so that the composition B --+ Eo c-+ E - B is the identity map of B. Then the corresponding composition H"(B) ---* lH"(E) -+ H"(E0) -9* H"(B) is the identity map of H"(B). By definition r* maps wn(6) to the restriction of the Thom class u|E. Hence the first two homomorphisms in the above composition map wn(6) to the restriction (uIE)IEo which is zero since the 66 composition H"(E,Eo) -+ H(E) -+ H"(E,) is zero. Then wn( ) = s*((u|E)IEo) = s*(O) = 0. Therefore w,(6) = 0. o LEMMA 4.8. Let 6 be an n-plane bundle, and let Ck be the trivial k-plane bundle. Then w(6 @ ck) = W(6). PROOF: It suffices to prove the Lemma for the case k = 1 since the Lemma then follows from induction. Let s:B -+ E(6 @ rj) be defined by s(b) = (v,i) where v E Fb(6). Then s is a nowhere zero cross-section. By Lemma 4.6 wn.1(6 @ c') = 0. Hence wn+i(6E D1) = Wn.1(6) = 0. Now since s is nonzero it is covered by a bundle map 6 -+ (( @ c'),. Hence s*wi((( @ c)o) = wi(6) by naturality. If the projection is 7ro : (E @Ec)o - ( Ec), then wi(((@ cl)0 ) = rwi(( @ el) again by naturality. So wi(6) = s*wi((E@ c')) = s*7rgwi(6 E c). But s*r is the identity. So wi(6) = wi(E@ E). o In order to prove the Whitney product theorem we first show that the total Stiefel-Whitney class of a Whitney sum w(6 @ r1) can be expressed as one, and only one, polynomial. We are then able to explicity compute the polynomial 67 and see that it is what we want, i.e. w(6 @E) = w(6)w(n). To make things easier we make use of the "universal" n-plane bundle 7' over the infinite Grassmann manifold G. The reason for this is that we can prove the Whitney Product Theorem for the specific case of the Whitney sum of two universal bundles, then since every n-plane vector bundle over the base space B is isomorphic to the induced bundle f*(7") for some map f:B - G., we can get the Whitney Product Theorem for any n-plane bundle. LEMMA 4.9. H*(G,((R*)) is a polynomial ring over 2 generated by the Stiefel-Whitney classes Wi(7"n),..., - wn(Yn).- PROOF: We proceed by induction on n. From Theorem 4.5 we know the Lemma is true for n = 1. Assume the Lemma is true for n - 1. Consider the Gysin sequence -- +H'(Gn) Z24 Hl*"(Gn) T * H*(o +H'(n +-- We first show that H*(E0) can be identified with H*(Gn.1). Construct a map f:E0 -+ Gn- as follows: A point (X,v) in E0 consists of an n-plane through 0 in R' and a non-zero vector v in X. Let f(X,v) = X fl v' be the orthogonal complement of v in X using the 68 standard metric (VI V2,. .) (WliW2,...) =E VJWj j=1 on . Then f(X,v) is a well defined (n - 1)-plane in R'. Now we wish to show that f induces cohomology isomorphisms. We first consider the subbundle 7n(RN) C Y" consisting of n-planes through 0 in N-space where N is large but finite. Let fN :EO (,,n(RN)) +Gn-1 (RN) be the corresponding restriction of f. For any (n - 1)-plane Y in G. 1(NRN) the inverse image f~1(Y) C Eo( n(RN)) consists of all pairs (X,v) where N. v E R is a non-zero vector perpendicular to Y and where X is determined by Y and v (since every vector in X is a linear combination of a vector in Y and v). Thus fN can be identified with the projection map N-N-n+1Ec,(wN-n+1) -- + Gn..1(RN) where o = Nn+1 is the (N - n + 1)-dimensional vector bundle whose fiber over Y E Gn-(1 RN) is the orthogonal complement of Y in RN The Gysin sequence of w is - Hi(G. 1(RN)) j -z+>H i+N-n+(Gn- 1(RN)) T( Hi+N-n+1 (Eo(w)) -+ Hi*+(G .I(RN)) --- where 7r(w) is an isomorphism when (i + N - n + 1) < (N - n + 1), i.e. when i < 2(N - n + 1). 69 But since fNcan be identified with ro (w) we have that fA induces cohomology isomorphisms in dimensions less than 2(N - n + 1). Letting N go to infinity we see that f induces cohomology isomorphisms in all dimensions. Therefore we can insert G.-I in place of E0 in the original Gysin sequence to get - - - -+ Hi(Gn) -- +H *(Gn) -A H'n"(Gn -1) -- + H'I 1(Gn) - where A = (f*)-1. We must show that A maps wi(-y") to wi(y'-1). For i = n we have Awn(7) = (f*)-1(7r*)(r*)-i(i*(u)) = 0 =wi(7-). For i < n the Stiefel-Whitney classes are defined by the formula wi(y") = (r*)-'(w(7")) so rgw(-n) = wi(-"). But the map f:E0 -+ Gn-i is covered by a bundle map -y" - 7n-. Therefore f*wi(7-,1) = wi('y") by naturality. So Awi(7yn) = (f*)-1(Ir)w("yn) = (f*)-1(7r*)(r*) -1w,(7) = (f*)~1wi(,") = W,(7-) We have shown that Aw,(7yn) = wq(-). Now, we have assumed that H*(Gn.1) is a polynomial ring over 1 1 12 generated by wi(7- ),. . . ,wn(7-'). Therefore A is surjective since it maps to each generator of H*(Gn-1). Our Gysin sequence reduces to --+Hi(Gnd)utHin(G) i aHi(Gn-1) -+0 - n s We now show, using induction on i, that every element 70 X E Hl*"(Gn) can be expressed uniquely as a polynomial in the Stiefel-Whitney classes w1(-"),...,wn(,n). Certainly the image of A(x) can be expressed uniquely 1 as a polynomial P(wi(-yn1),... ,wwn- 1(- )) by our main induction hypothesis that H*(Gn-1) is generated by Wi( 7 "n),... ,w.1("~') . Therefore the element X - P(wi("), .. ,wn-1(-")) belongs to the kernel of A, and hence can be expressed as a product ywn(7") for some uniquely determined y E H'(Gn). Now y can be expressed uniquely as a polynomial Q(wi(7"n),..,w(-n(")) by our induction hypothesis for i. Hence X = P(w1(-7),--- ,wn-1(-")) + wn(7")Q(wl(7"), wn(7n)) . The polynomials on the right are unique since if we also had X = P'(w1(-y"),...,w 1("Y)) + wn(ey)Q'(wi(-Y),...,wn(-")) we could apply A to get P(W,(7"1),..., wn-l(7",n-1)) = P'(wi(-Y ), 1 ,n_ (,n-)). So P = P'. Then we could subtract both polynomials to get W."(7") Q' (w,(7e) -.-.- , Wn (7")) = wn (,,")Q(w,(,,") , . .. ,() which implies that Q = Q'. LEMMA 4.10. There exists one and only one polynomial Pmn(wi, -.. , wm;w'/, ... ,us') with 12 coefficients in m + n indeterminants, so that the identity 71 w(E @ ) = Pmn(wi(),. . . ,wm();wi(7),...,wn()) is valid for every real n-plane bundle 6 and n-plane bundle n over a paracompact base space B. PROOF: Let ir:Gm x Gn - Gm be projection onto the first factor, and let 72:G. x Gn -+ Gn be projection onto the second factor. Let rm("')I= and yn2= ?*(-,) Then the Whitney sum 7m @ 7' can be identified with 7 "nx ". From Algebraic Topology we know that the external cohomology cross product operation a,b - a x b = ra rgb induces an isomorphism H* (Gm) 0 H* (Gn) - H* (Gm x GO) since 112 is a field. Therefore, by Lemma 4.9, H*(Gm x Gn) is a polynomial ring over 112 with the algebraically independent generators riwi(7"') = wi(-m') 1 i < m and rgwj (-") = wj(7yn) 1 < j n hence the total Stiefel-Whitney class of 7, @ 7n can be expressed uniquely as a polynomial W @(7'07') = Pmn(wi(77),.- . ,wm(t);wi(y),. .. , )). Now if 6 is an m-plane bundle over B and 9 is an n-plane bundle over B then we can find maps f:B -Gm 72 and g:B -+ G. such that 6 Uf (7"') and 1y Ng*(n). Def ining the map h:B - G. x Gn by h(b) = (f(b),g(b)) we get the following commutative diagram: B h Gm Gm x Gn _-+Gn it follows that h* ( 7) and h*(oy') 17. Hence w(6 @ 1) = h*w( 7 m @ 7,) = Pmn(wi(6),...,wm(6);wi(),...,wn(y)).0 PROOF (of Property 3): We now wish to compute the polynomials Pmn. We proceed by induction. Suppose that w(7-1E@ )= (1 + w (77- 1) + -.. 1 + wm( 7n,)) (1 + wi(-y) + - + 2n7)). Consider the two vector bundles (D c' and e2 over Gm-i x G., where E' is a trivial line bundle. From Lemma 4.7 we have ~,n@E 2 ) = Pmn(wi(71 @ c'),...,wm(7M-, ) 2i7) ,...,W2(7")). From Lemma 4.6 we can ignore the el summand. Thus @(7~79) = Pmn(wi('y'~'), . . .,m17-)0 ),w2(7)). 73 Substituting wi for wi(,ym- 1) and wj for wi(7y) we get from the induction hypothesis that Pmn(Wi,...,Wm-1,OWj,...,W') = (1+ w +- + W.-1)(1 + W" + + w') Introducing a new indeterminant wm we have the congruence Pmn(wi, ... , wm-iwm~w*',..., W/) (1+ w+ + wm)(1+ wj+ + w) (mod wm). But by a similar induction argument we also have Pmn(wi,...,- wm-1,wiw"),...,) WO (1 + w, + + wm)(1 + w + + w/) (mod w). 1 Since 2[wl,- ,wm,wi,. -.,w'] is a unique f actorization domain, it follows that the above polynomials are congruent modulo the product wmwA. That is Pmn(wi,. ,wm-1,wmw'. ,w) = (1 + w, + + wm)(1 + w" + + w") + twmw where t must be a constant for otherwise the Whitney sum 7m@ 7' would have non-zero Stiefel-Whitney classes in dimensions greater that m + n. But we know that wm+n(6 @ 7) = wm( )wn(77) from Lemma 4.4. Therefore t must be zero. So we have w ( @ 7) = h*w(,f, @n) = Pmn(wi(6),.-.-.,wm(6);wi(7),.. .,wn(7)) = (1 + wI() + ... + wm(f))(1 + wi() + ... + wn(17)) = w(6)w(77). o 74 Ve have now proven that the Stiefel-Whitney classes have Properties 1-4. It turns out that these four properties completely characterize Stiefel-Whitney classes [4]. CHAPTER V ADDITIONAL PROPERTIES OF STIEFEL-VHITNEY CLASSES Now we wish to find out more about the Stiefel-Whitney classes. As noted at the end of the last chapter the Stiefel-Whitney classes are completely characterized by Properties 1 - 4. Hence, in what follows, we shall not need to refer back to the original definition of w(6). For the Stiefel-Whitney classes to be of any use we must have the following THEOREM 5.1. If ( is isomorphic to y then wi(6) = wi(n). PROOF: We have the following diagram E(6) > E(i7) B(6) + B(1) where the base spaces are equal and i is the identity map. From Property 2, wi(6) = i*wi(i7). But since i is the 75 76 identity i* must be an isomorphism. Therefore WiM = wi(77). o THEOREM 5.2. If c is a trivial vector bundle, then wi(E) = 0 for i > 0. PROOF: Let E be a trivial n-plane bundle. Then there is a bundle map g:E(E) -- {x} x R" for some point x. If we let be the corresponding map of base spaces we get the commutative diagram E(,E) 9 - {x} x Rn I - I B(E) -- 0 { } By Theorem 3.20 E is isomorphic to the bundle over the point {x}. But H'({x}) = 0 for i > 0. Since wi(bundle over {x}) E H1({x}) = 0 we have wi(e) = 9*wi(bundle over {x}) = j*(0) = 0. Thus wi(E) = 0. o THEOREM 5.3. If e is trivial then wi(C @ 7) = Wi(7). 77 PROOF: From Property 1 we know wo(E) = wo(n) = 1. From Theorem 5.2 we have wi(c) = 0 for i > 0. Finally from Property 3 we get wi(C @ ) = w(E) -wi..j(y) j=O = (1) a wi(y) + 1 0 '- wi.j(7) j=1 = wi(y). o THEOREM 5.4. If ( is an n-plane bundle with a Euclidean metric which possesses a nowhere zero cross-section, then wn(6) = 0. If 6 possesses k cross-sections which are nowhere linearly dependent, then wn-k+1(6) = wn-k+2(6) = ... = wn(6) = 0- That is, the last k Stiefel-Whitney classes are zero. PROOF: From Theorem 3.27 6 2 E De where E is a trivial k-plane bundle and el is an (n - k)-plane bundle. From Theorem 5.3 wi(6) = wi(E @(D ) = wi(c') and since E is an (n - k)-plane bundle, wi(c') = 0 for i > n - k. o THEOREM 5.5. The total Stiefel-Whitney class w(6) = 1 + wi(6) + W2(6) + --- E H*(B(6)) has a multiplicative inverse in H*(B(6)). 78 PROOF: Let w = 1 + w1 + E H*(B). An element W=1+W1 + W 2 + E H*(B) is an inverse of w if wi = 1. Computing this product we see that ww = 1 + (wI + W 1) + (W2 + WiWi + W2) + - + (wn+ w-i+ - - - ++wiwn..I+Wn) + So w has an inverse if there are Stiefel-Whitney classes W1,...,wn that solve the following equations: Wi + Wi = 0 W2 + WIWI + W2 = 0 wn + wn- 1 + -- * + wIn-1+ in = 0 We can solve the first equation for -w to get w1 = w 1 E HI(B) (Remember we are working with coefficients in E2 so that WI= -wI). Plugging this into the second equation we are able to solve for W2: W2 = W1 + w2 E H 2(B). We may continue in this manner so that we are able to compute n for any n by f irst computing wi for i < n. The formula for Wn is 79 n = W1Wn- + W2Wn-2 + + wn-ii E H"(B). Thus w=1+ W1 + W 2 + is an inverse for w in H*(B). o COROLLARY 5.6. Let ( and 7 be vector bundles over the same base space. The equation w(6'@ 97) = w(w(1) can be solved as W(7) = -W()w(E@ 7). In particular, when E y is trivial w(77) = W() PROOF: When E D is trivial w( @ )= 1 from Theorem 5.2. The rest follows immediately from Theorem 5.5. 0 We will make use of a special case of the above corollary. COROLLARY 5.7. (Whitney Duality Theorem) If rTM is the tangent space of a manifold in Euclidean space and v is the normal bundle, then wi(V) = wi(rM). 80 PROOF: From Corollary 3.28 v @ r is trivial. Now by Corollary 5.6 wi(v) = wi(rm). 0 The following theorem illustrates a weakness of the Stiefel-Whitney classes. It states that the tangent bundle of the unit sphere S" has the same Stiefel-Whitney class as the trivial bundle over S". NOTATION: We write w(M) for the total Stiefel-Whitney class of a tangent bundle rM. THEOREM 5.8. For the tangent bundle 7 of the unit sphere S", the class w(S") = 1. PROOF: For the standard embedding S" c R 1, the normal (line) bundle v is trivial since the function h: E(S1) -- + S"I x R given by h(x,tx) = (x,t) is a homeomorphism. Now from the Whitney Product Theorem 1 = w(r @ v) = w(r)w(v). But w(v) = 1 since v is trivial. Thus w(r) = 1. o In order to compute the Stiefel-Whitney classes for 7 over RP" we need the following fact from Algebraic Topology [5, p.4 03]. 81 THEOREM 5.9. The mod 2 cohomology of projective space RP" is given by Hi(Rpn)NfE2 for 0 < i < n -1 0 for i > n Furthermore, if a denotes the nonzero element of H1(RP) then each H'(RP") is generated by the i-fold cup product a. THEOREM 5.10. The total Stiefel-Whitney class of the canonical line bundle 7' over RP" is given by w(7y) = 1 + a. PROOF: From Property 1, wo(yl) = 1 and wi(-yn) = 0 for i > 1. The inclusion j:RP' -- RP" is covered by a bundle map 71 -- 7. By Property 2 j*wl(71) = wl(7) # 0. Therefore w1(-y.) cannot be zero and hence wl(-y2) = a. So w(7-y) = 1 + a. o The following Theorem shows that all of the n Stiefel-Whitney classes of an n-plane bundle may be non-zero. 82 THEOREM 5.11. By definition, the line bundle -n' over RP" is contained, as a subbundle, in the trivial bundle E"*i. Let 7 denote the orthogonal complement of y in E" *. Then w(y)= 1 + a + a2 + + a". PROOF: Since 7E@ -y y E"+1 which is trivial, we have w(-y') = W(y)w( -t1E ) = i-(-yn). Thus w(7y') = (1 + a)-'. To compute (1 + a)' consider the product of (1 + a) with some element (1 + b, + b + ''') E 2 H*(RPn). We wish to find b1,b2,... so that the product is 1. From the formula introduced in the proof of Theorem 2 5.5 we have b, = a, b2 = a , .. ,bn = a 1 2 bn.1 = a" = 0, bn+2 = an = 0, . . . Thus 1 2 (1 + a) = 1 + a + a + --- + a" + 0 + 0 + . So w(-y') = 1 + a + a2 + - - - + a". o We now wish to compute the total Stiefel-Whitney class of the tangent bundle of RP". To do this we first have the following: DEFINITION 5.12. Given two vector bundles ( and r over the same base space B we can construct the vector bundle Hom(6,i!) in the following way: A fiber Fb of Hom(6,y) is just 83 Fb = Hom(Fb(6),Fb(17)) the space consisting of all linear transformations from Fb(6) to Fb(1). The total space E is defined to be the disjoint union E(Hom(6,y)) = Fb- bEB We will be interested in the vector bundle Hom(7y,71). In this case the fibers are Fb = Hom(Fb(7-),Fb(7')) = Hom(L,L') where L is the line through the origin in R" that passes through b. LEMMA 5.13. Let r be the tangent bundle of RP'. Then r Hom(7y,71). PROOF: Let L be a line through the origin in R"'+, intersecting S" in the points kx, and let L' be the complementary n-plane. Let f:S" -- RP' be the map f(x) = { x}. The two tangent vectors (x,v) and (-x,-v) both have the same image under the map Df:DS" -- D(Rp"). Thus the tangent manifold D(RP") can be identified with the set of all pairs {(x,v),(-x,-v)} satisfying x= 1 and x-v = 0. (Remember that X E Sn which is the unit n-sphere, so 84 X-x = 1, and if v is tangent to S" at x it is orthogonal to x, hence x-v = 0). Then we can define a linear map l:L - L' by 1(x) = v. Also note that any map 1:L -- L' determines a unique pair {(x,v),(-x,-v)} satisfying the above conditions. Thus the tangent space to RP" at { x} is isomorphic to the vector space Hom(L,Ll). So we have a continuous function that maps each fiber of r isomorphically onto the corresponding fiber in Hom(7-y,7'). By Theorem 3.3 T Lv Hom(7.1,7). 11 THEOREM 5.14. Let E' be a trivial line bundle over RP". The Whitney sum Tr E cl is isomorphic to the (n + 1)-fold Whitney sum 7.@ 7nD--- -7..@ Hence the total Stiefel-Whitney class is given by w(RP") = (1 + a)"+' = 1 + n + 1 a + (n + 1 )a2 + +n + 1 an. PROOF: A fiber F over the bundle Hom(7y,71) is the collection of all linear functions that map the line L, that passes through the point { x} in RP", to itself. So each linear function f E F is a stretching by a factor of t for some t E R. That is, each function in F can be 85 written as ft where ft(v) = tv. Then the correspondence ft -- t shows that Hom(71,7) is a line bundle. Furthermore Hom(7ny,7) has a nowhere zero cross-section defined by {+x} a-- fi E Hom(L,L) where fi is the identity function that maps the line L to itself by fi(v) = v. Therefore Hom(-y7,7 ) is a trivial line bundle. Now r @(c ' Hom(-y ,7') @ Hom(-y,,7y) Hom(-y ,7y' ED y1) NHom(7n, en"i) L Hom(7,y , f I ( C I ( -... E) C ,) (n+1) -times Hom(7 ,e1) (DHom(-t , EI) 19 --- (D Hom(-n,,El). Each fiber F = r-({x}) of Hom(-y7,ce) has the form F = Hom(L,R) where L is the line through {zx}. Since 7 has a Euclidean metric, the correspondence w - , E Hom(L,R) where 9, is defined by ,w(v) = w-v gives an isomorphism L -- Hom(LR). Thus, by Theorem 3.3 7' 2Hom(7 ,el) . So we have r (D c A. 7@-@2,. (n+1) -times Now from the Whitney Product Theorem w(r) = w(-r @ e ) = w(-Y7)w(-Yn)-.w(Y) = (1 + a)". Using the binomial theorem (and the fact that a"*i = 0) we have 86 w(r) = (1 +a) = 1+ + + 'ja + (n +ja2 + --- + (n + 1)jan Remember that we are working with coefficients in 12 so we are only interested in the binomial coefficients modulo 2. Table 5.1., reproduced from [4, p.46], makes it easy to f ind the total Stief el-Whitney class w(RP") f or n = 1, .. .,10. 1 1 1 RP' 1 0 1 Rp21 1 1 1 P3 1 0 0 0 1 RP 4 1 1 0 0 1 1 5 RP 1 0 1 0 1 0 1 RP6 1 1 1 1 1 1 1 1 RP1 1 0 0 0 0 0 0 0 1 Rp8 1 1 0 0 0 0 0 0 1 1 9 RP 1 0 1 0 0 0 0 0 1 0 1 RPIO 1 1 1 1 0 0 0 0 1 1 1 Table 5.1 The table lists the binomial coefficients modulo 2. Since 87 a"*1 = 0 we ignore the ones on the right hand side of the triangle. As an example, to find w(RP 4) we look at the line on the table: RP4 1 1 0 0 1 1 This means that 4 2 w(RP ) = (1) + (1) a + (0) a + (0) a' + (1) a4 = 1 + a + a4 Remember that we ignore the last 1 in the row since a5 = 0. Now suppose that a manifold M" of dimension n can be immersed in Euclidean space Rnk. Recall that a function f:M" n-+ R"k is an immersion if at every point x E Mn Dfx:DMx R*kI-+ is injective. This means that rM is an n-dimensional vector bundle, and the normal bundle v is a k-dimensional vector bundle. From the Whitney Duality Theorem wi(v) = Wi(M") So the dual Stiefel-Whitney classes Wi(M") = 0 for i > k. We have proven the following THEOREM 5.15. If an n-dimensional manifold M" can be immersed in Rn+k then Wi(M") = 0 for i > k. 88 EXAMPLE: For RP5 we have w(RP5) = 1 + a2 + a then w(Rp5)= 1 + a2 So if we want to immerse RP5 in R5*k then k must be at 5 least 2 (since W2 (OP) # 0). EXAMPLE: For RP9 w(RP 9) = 1 + a2 + a i(RP9) = 1 + a2 + a + a6 . 9 So if RP can be immersed in R9.k then k must be at least 6. COROLLARY 5.16. Let n = 2r for some integer r. If RP" can be immersed in Rn+k, then k must be at least n - 1. PROOF: w(RP") = (1 + a)"+' = (1 + a) (1 + a)". Since we are working modulo 2 and n = 2 we have (1 + a)" = 1 + a". So w(RP") = (1+ a) (1 + a)" + a + a + a" 1 + a + a". Thus w(RP") = 1 + a + a". Then W(RP") = 1 + a + a2 + -- +a"1. Therefore k must be at least n - 1. o 89 Whitney has proven [8] that any smooth compact manifold of dimension n > 1 can be immersed in R2n-. We have shown that for RP", where n is a power of 2, that this is the best we can do. So we have accomplished, at least in part, what we set out to do in Chapter 3. That is, we now have a way to find a lower bound on the dimension of Euclidean space that we need to embed RP". BIBLIOGRAPHY 1. Guillemin, Victor and Alan Pollack. Differential Topology. Englewood Cliffs: Prentice-Hall, 1974. 2. Hirsch, Morris W. Differential Topology. New York: Springer-Verlag, 1976. 4. Milnor, John V. and James D. Stasheff. Characteristic Classes. Princeton: Princeton University Press, 1974. 5. Munkres, James R. Elements of Algebraic Topology. Redwood City: Addison-Wesley, 1984. 6. Vick, James W. Homology Theory. New York: Academic Press, 1973. 7. Willard, Stephen. General Topology. Reading: Addison-Wesley, 1970. 8. Whitney, Hassler. The self-intersections of a smooth n-manifold in 2n-space. Annals of Math. 45 (1944), 220-246. 9. Whitney, Hassler. The singularities of a smooth n-manifold in (2n - 1)-space. Annals of Math. 45 (1944), 247-293. 90