Differential Topology

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Differential Topology. 1. Smooth maps and manifolds. (a) X * R* is a manifold if 2x . X 3 AU * R* open such that x . U +. AU 5 X )+ R8. (where )+ denotes ...
Di¤erential Topology 1. Smooth maps and manifolds ~ RN open such that x 2 U = (a) X RN is a manifold if 8x 2 X 9 U k ~ \ X = R (where = denotes homeomorphism) for some k: We U say that X has dimension k in this case, denoted dim X = k: (b) A mapping f : U RM ! RN with U open in RM is smooth if it has continuous partial derivatives of all orders. In this case, we also say that f is C 1 : More generally, if f is r times continuously di¤erentiable for r 1; we say that f is C r : If X RM is an N arbitrary subset, f : X ! R is said to be smooth if 9 U RM N open with X U and 9 F : U ! R which is smooth and such that f = F jX : (c) A smooth map f : X ! Y (with X; Y subsets of some Euclidean space) is a di¤ eomorphism if f is a homeomorphism and both f and f 1 are smooth. Cr

i. Di¤eomorphic spaces (which we denote by X = Y ) are topologically equivalent. ii. A smooth manifold X RN is a manifold in which every open neighborhood of x 2 X is di¤eomorphic to Rk for some k: iii. Parameterizations are homeomorphic mappings from linear Euclidean spaces to open subsets of a manifold. Since such subsets are, by de…nition, topologically equivalent to a Euclidean space, this de…nition makes sense. To characterize parameterizations, let V = V~ \ X be an open neighborhood of x 2 X for the smooth k-dimenisonal manifold X RM : Let ' : U = V: The map ' is called a parameterization of the neighborhood V: The parameterization allows us to work with a Euclidean coordinate system on U when analyzing the neighborhood V X: Example: S 1 = (x; y) 2 R2 jx2 + y 2 = 1 i h p '1 (x) = x; 1 x2 parameterizes the upper hemi-sphere of S 1 : iv. Product manifolds. Suppose X RM ; Y RN are smooth manifolds of dimension k and ` respectively. For x 2 V X; ' : U ! V is a local parameterization, and for y 2 W Y; : T ! W is a local parameterization. Theorem: X Y RM RN is a smooth manifold. Proof: De…ne ' : U T ! X Y by ('

) (u; t) = [' (u) ;

We need to verify that ' set U T: 1

(t)] :

is a parameterization of the open

A. Bijectivity is easy. Suppose (u0 ; t0 ) 6= (u; t) : Then either ' (u0 ) 6= ' (u) or (t0 ) 6= (t) so ' is one-to-one. Suppose ' isn’t onto. Then 9 (x; y) 6= ' (u; t) for any (u; t) : But 9 u 3 x = ' (u) and t 3 y = (t) ; so this is a contradiction. It remains to check smoothness. B. The inclusion map i1 : U ! U V de…ned by i (u) = [u; 0] is smooth, since the constant function 0 is smooth. Now, '

= i1 (') + i2 ( ) = ['; 0] + [0; ] :

Since both inclusion maps are smooth, and vector addition is smooth, ' is smooth. 1 The inverse map (' ) (x; y) = ' 1 (x) ; 1 (y) can be shown to be smooth in the same way, once we recognize that because ' and are parameterizations, if X Y is not open, the inverse of can be extended to open neighborhoods of points in X Y since we can do such an extension of each component map separately. It now follows that X Y is a smooth manifold, and dim (X Y ) = dim X + dim Y; since ' is a local di¤eomorphism de…ned on U T which is of dimension k + ` in RM +N : (d) Derivatives and Tangents i. Directional derviatives: given f : Rn ! Rm the derivative of f at x in the direction h 2 Rn is dfx (h) = lim

t!0

f (x + th) t

f (x)

:

The derivative map dfx is linear; if h1 ; h2 2 Rn ; then dfx [ h1 + h2 ] = dfx (h1 ) + dfx (h2 ) : To show this formally, let ph (t) = x + th and de…ne g (t) = f [ph (t)] = f [x + th] : Then lim

t!0

f (x + th) t

f (x)

= g 0 (0) = Df (x) h n X @f dpih = : @xi dt i=1

Hence, for …xed x; dfx : Rn ! Rm is linear and can be represented as dfx (h) = Dx f (x) h 2

where Dx f is the Jacobian matrix Dx f =

@fj @xi

for i = 1; :::n; j = 1; ::; m (m

n):

Now, suppose X is a smooth manifold and let ' : U ! X RN be a paraterization of a neighborhood of x 2 X: The tangent space at x to the manifold X is de…ned as Tx X = Im [d'0 ]

RN :

Note that Tx X is a vector subspace of RN : It represents the best linear approximation to the hypersurface X at x when we translate the subspace to the point x; i.e. x + Tx X: Example with the circle: p ' (x) = x; 1 x2 h i '0 (x) = 1; p1x x2 '0

p

2 2

= [1; 1] then spans T p2 S 1 2

(e) The Inverse Function Theorem. i. We would like to be able to characterize various types of mappings f : X ! Y: Case 1: dim X = dim Y: For this case, the simplest possible behavior f can exhibit locally around y = f (x) is for f to map a neighborhood N (x) di¤eomorphically onto a neighborhood M (y) : The map f is then called a local di¤ eomorphism at x: 3

Cr

Inverse Function Theorem: Suppose f : X ! Y is such that dfx : Tx X ! Ty Y is an isomorphism. The f is a local di¤eomorphism. The Inverse Function Theorem is proved as a corollary of the Implicit Function Theorem: Suppose f : X A ! Y; dim X = dim Y; is such that dfx : T(x;a) X ! Ty Y is an isomorphism at y0 = f (x0 ; a0 ) : Then 9 g : N (a0 ) ! M (x0 ) such that f [g (a) ; a] = y0 8a 2 N (a0 ) (check any text on advanced calculus for a proof of this result). The Inverse Function Theorem is obtained from the Implicit Function Theorem by considering g (x; y) = f (x) y: At y0 = f (x0 ) ; g (x; y) = 0: If Dx g (x0 ; y0 ) = Dx f (x0 ) is nonsingular, the Implicit Function Theorem implies that there exists h : N (y0 ) ! M (x0 ) such that 0 = g [h (y) ; y] = f

h (y)

y)f

h (y) = y:

Hence, h = f 1 : ii. Local di¤eomorphism determins an equivalence relation for mappings; speci…cally, we say that f : X ! Y and f 0 : X ! Y are (locally) equivalent if there exist di¤eomorphis and such that the following diagram commutes:

1 To say that the diagram commutes means that f 0 = f : 0 When the diagram commutes, we say that f and f are the same up to di¤eomorphism. Via the notion of equivalence, we can characterize local di¤eomorphisms as mappings which are locally equivalent to the identity. To see this, consider the diagram

4

which is commutative and f and g di¤eomorphisms. If we let 0 = g we can restructure the diagram as

in which case we now have that f is equivalent to the identity. (f) Immersions Local di¤eomorphisms map between spaces of the same dimension. This raises the question of what is the simplest behavior that can be exhibited by a mapping f : X ! Y when dim X < dim Y ? i. For this case, the best we can ask for in the di¤erential case, is that dfx : Tx X ! Tf (x) Y be injective (i.e. one-to-one and into). In this case, we say that f is an immersion at x: If f is immersive at every point, we say that f is an immersion. The canonical immersion is inclusion of i : Rk ,! Rn ; n > k; where i (x1 ; :::; xk ) = [x1 ; :::; xk ; 0; :::; 0] : ii. The Local Immersion Theorem: Suppose f : X ! Y is an immersion at x and y = f (x): Then f is locally equivalent to the canonical immersion on a neighborhood of x; i.e. there exist local coordinates around x and y such that f (x) = [x; 0] : Proof: Choose parameterizations so that the diagram

5

commutes. Now, since f is an immersion, g is, so dg0 is injective. Since dg0 : Rk ! Rn ; we can make a change of basis in Rn such that dg0 is represented by the matrix Ik 0 where Ik is the k k identity matrix. Now, de…ne a map G : U Rn k ! Rn by G [x; z] = g (x) + [0; z] : Then G maps an open set of Rn into Rn and dG0 = In : By the inverse function theorem, then, G is a local di¤eomorphism at 0: Next, note that g (x) = G (x; 0) = G canonical immersion. Since the maps and G are local di¤eomorphisms, so is G and we can use 0 = G as a local parameterization of Y around y: Then, shrinking U and V if necessary, the following diagram commutes:

This establishes the result. iii. Topological issues The canonical immersion is the niecest possible example of a map whose image is a submanifold. This raises the question of whether the image of every immersion is a submanio‡d. In general, the answer to this question is no. 6

Counter-example:

The mapping implied by twisting the circle into a …gure-8 is locally immersive, but the …gure-8 is not a submanifold since no neighborhood of the self-intersection can be homeomorphic to the line. Note, however, that the mapping here is not globally injective since it isn’t one-to-one. Now consider

This mapping is locally immersive and globally injective, but the image is again not a submanifold. Let G : R1 ! S 1 S 1 be de…ned by g (t) = [cos 2 t; sin 2 t] where we identify S 1 S 1 = T 2 with the two-dimensional torus. Next, de…ne G : R2 ! S 1 S 1 by G [x; y] = [g (x) ; g (y)] : G maps R2 onto the torus. As an aside, we can identify I I with the torus by …rst gluing the top edges of the unit square together to form a cylindar, and then glueing the circular ends of the cylindar together to make a torus. Finally, de…ne a mapping of R1 into the torus by restricting

7

G to a line through the origin with irrational slope, as in the …gure below

In the …gure, we use the identi…cation of T 2 with I I to indicate how the line with irrational slope gets mapped onto the torus. In the next diagram, we show how this mapping winds the real line onto the torus:

8

As long as the slope of the line being mapped to T 2 is irrational, it will never go through the same point twice. If the slope of the line were rational, this would no longer be true and the image of the mapping would be a one-dimensional submanifold of T 2 : The last two examples are both global immersions – locally immersive and injective everywhere –without being submanifolds. The "pathologies" in these examples occur because they map points "near in…nity" into small neighborhoods of the tangent space, i.e. they injectively map non-compact regions into compact ones. If we eliminate this behavior, the resulting class of mappings is quite well-behaved. iv. Properness. A map f : X ! Y is said to be proper is the preimage of any compact set K Y is compact in X: A proper, injective immersion is called an embedding. Theorem: An embedding f : X ! Y maps X di¤eomorphically into a submanifold of Y: 9

Proof: To show that f (X) is a submanifold, we show that if W X is open, f (W ) f (X) is open. Since X is a manifold, there then exists an open set V is some Euclidean space and a parameterization ' such that ' (V ) = W: Then (V ) = f ' (V ) = f (W ) is a parameterization of f (W ) : Now, if f (W ) is not open in f (X); there exists a sequence fyi g 2 f (X) not in f (W ) but lim yi = y 2 f (W ) : Since the sequence fy1 ; y2 ; :::g i!1

is compact, properness of f implies that the pre-image of the sequence by f is is compact in X: The fact that f is injective implies that there is a unique xi such that f (xi ) = yi : Similarly, there is a unique x such that f (x) = y: Since the sequence fx1 ; x2 ; :::g lies in a compact subset of X; we can (passing to a subsequence if necessary) assume that xi ! z 2 X: Then f (xi ) ! f (z) ; so by injectivity, z = x: Now, with y 2 f (W ) ; f 1 (y) = x 2 W: Since W is open, xi ! x implies that there exists N such that for all i N; xi 2 W: Then for all i N; f (xi ) 2 f (W ) ; which is a contradiction. Thus, f (W ) is open and f (X) is a submanifold. To show that f : X ! f (X) is a di¤eomorphism, note that since f is bijective, f 1 is well-de…ned. The inverse mapping f 1 is smooth at every y 2 f (X) since f is an immersion.

(g) Submersions We deal next with the case where f : X ! Y has dim X dim Y: Let y = f (x) : Then the strongest local condition we can impose on dfx : Tx X ! Ty Y is that the mapping be surjective, i.e. onto the tangent space at y: If dfx is surjective, f is said to be a submersion at x: If f is a submersion at every point, we say simply that f is a submersion. The canonical submersion is the standard projection of Rk onto R` for k `; with (x1 ; :::; xk ) = [x1 ; :::; x` ] : As was the case with immersions, every submersion is locally equivalent to the canonical submersion. i. Local submersion theorem: suppose f : X ! Y is a submersion at x; and y = f (x) : Then there exist coordinates around x and y such that f (x1 ; :::; xk ) = [x1 ; :::; x` ] on neighborhoods N (x) and M (y): Proof: Given any parameterization

10

with (0) x and (0) = y; we know that dg0 is surjective. By a linear coordinate change in Rk ; we may assume that dg0 : Rk ! R` is represented by the matrix I` 0 (i.e. the canonical projection). Now, de…ne G : U ! Rk by G (a) = [g (a) ; a`+1 ; :::; ak ] : Then dG0 is the identity Ik ; so G is a local di¤eomorphism at 0: Hence, by the inverse function theorem, G 1 exists on a neighborhood of U 0 (0) and maps U 0 di¤eomorphically onto U: By construction, g =canonical submersion G; so g G is canonical submersion. Hence, the following diagram commutes:

and the proof is complete. ii. These three so-called classi…cation theorems on the local behavior of smooth maps have the following extremely useful corollary on the geometric structure of solutions to equations f (x) = y for smooth maps f : X ! Y . De…nition: For a smooth mapf : X ! Y; a point y 2 Y us called a regular value for f if dfx : Tx X ! Ty Y is surjective at every point x such that f (x) = y:

11

Pre-image Theorem: If y is a regular value of f : X ! Y; then the pre-image f 1 (y) X is a submanifold of X with dim f 1 (y) = dim X dim Y: Proof: Suppose f (x) = y and that f is submersive at x: Choose local coordinates around x; y such that f (x1 ; :::; xk ) = [x1 ; :::; x` ] and y corresponds to 0: Let V be the open neighborhood of x on which the coordinate system is de…ned. Then f 1 (y) \ V is the set of points where x1 = 0; :::; x` = 0: Also, the functions x`+1 ; :::; xk form a coordinate system on f 1 (y) \ V; and this set is relatively open in f 1 (y) :

Now, since f is submersize at every x such that f (x) = y (since y is a regular value of f ), every x 2 f 1 (y) has an open neighborhood di¤eomorphic to Rk ` : Hence, f 1 (y) is a submanifold and dim f 1 (y) = k ` = dim X dim Y: : Comment: Suppose y 2 = f (X): Then, since f 1 (y) = , y is a regular value (vacuously). iii. Classi…cation results for regular values dim X > dim Y : If y is regular, then f is a submersion at every x 2 f 1 (y) : 12

dim X = dim Y : If y is regular, then f is a local di¤eomorphism at x 2 f 1 (y) : dim X < dim Y : If y is regular, then @x 3 f (x) = y: iv. Various corollaries of the pre-image theorem Consider ` functions gi : X ! R where dim X = k `; i = 1; :::; ` Proposition: If the smooth functions gi ; i = 1; :::; ` are independent at each point x such that gi (x) = 0; then the zero-set of the collection of functions is a submanifold of X with dimension equal to k `: Proof: Let g(x) = [g1 (x) ; :::; g` (x)] and Z = fx 2 Xjg (x) = 0g : Then Z = g 1 (0) : Hence, by the pre-image theorem, Z will be a submanifold of X if 0 is a regular value of g: This, in turn, requires that dgx : Tx X ! R` be surjective. In matrix form, dgx is represented by dgx = [Dx g] dx where

2

3 Dx g1 6 7 Dx g = 4 ... 5 Dx g`

is the ` k Jacobian matrix of g: Surjectivity of dgx then requires that the matrix Dx g have maximal rank `; which in turn requires that each vector Dx gi be linearly independent of all the others. This is ensured by the independence requirement. The codimension of an arbitrary submanifold Z of X is de…ned as cod Z = dim X dim Z: The pre-image theorem then says that cod f 1 (y) = dim Y: The proposition above says that ` independent functions "cut out" a cod ` manifold at the zero set. v. Example 2 Let M (n) be the set of all n n matrices. M (n) = Rn : Let O(n) be the set of all orthogonal n n matrices (i.e. matrices all of who vectors are mutually orthogonal). Proposition: O(n) is a manifold in M (n): Proof: Let S(n) be the set of symmetric n n matrices. S(n) is a submanifold of M (n) of dimensions n(n + 1)=2 (obtained by

13

requiring that aij = aji ). Hence, we can show that O(n) is a manifold if we can show that f : M (n) ! S(n) given by f (A) = AAT has I as a regular value. (The map f is smooth since linear operators are smooth.) So, assume that AAT = I; and calculate dfA [B] as follows: dfA [B]

f (A + sB) f (A) s T (A + sB) (A + sB) AAT = lim s!0 s T T AA + sBA + sAB T + s2 BB T = lim s!0 s = BAT + AB T =

lim

s!0

AAT

To show that I is a regular value, we need dfA : TA M (n) ! Tf (A) S (n) to be surjective. Since both M (n) and S(n) are Euclidean, they are their own tangent spaces, so we need to show that for any C 2 S (n) there exists B 2 M (n) such that dfA (B) = C; i.e. BAT + AB T = C: Now, since C is symmetric, we can write C = 21 C + 12 C T in which case we solve for B as 1 B = CA 2 (since AAT = I): Given this, then, dfA [B]

1 1 CA AT + A CA 2 2 1 1 = C + A AT C T 2 2 = C

T

=

so the derivative mapping is surjective, and, by the pre-image theorem, f 1 (I) = O (n) is a submanifold of M (n) : Finally, the codimension calculation tells us that dim O(n)

=

dim M (n) dim S(n) n (n + 1) = n2 2 n (n 1) = 2 14

vi. Proposition: Let Z = f ping f : X ! Y: Then

1

(y) for a regular value y of the map-

ker [dfx : Tx X ! Ty Y ] = Tx Z at any point x 2 Z: We illustrate this geometrically below.

Proof of Proposition: Note that f is constant on Z; so dfz = 0 on Tx Z; which implies that Tx Z ker dfx : Since y is a regular value of f; dfx is surjective, so the pre-image theorem implies that dim ker dfx

= dim Tx X = dim Z

dim Ty Y = dim X

dim Y

Hence, Tx Z and ker dfx (which are both linear subspaces) have the same dimension and must therefore coincide. vii. A structural result for local di¤ eomorphisms: Suppose f : X ! Y is smooth, X is compact, dim X = dim Y; and let y be a regular value of f: (Stack of records) Theorem: Under the assumptions above, f 1 (y) is a …nite set fx1 ; :::; xN g ; and there exists a neighbor15

hood U (y) such that f with Vi \ Vj = more,

1

(U ) = V1 [

[ VN

if i 6= j; xi 2 Vi and Vi open for all i: Furtherf jVi : Vi = U:

Proof: By the pre-image theorem, dim X = dim Y and y regular implies dim f 1 (y) = 0; so f 1 (y) is a collection of discrete points. Furthermore, each pre-image point has an open neighborhood not containing other pre-image points, since otherwise, there would exist a sequence xj 2 f 1 (y) with xj ! x0 2 f 1 (y) and no neighborhood of x0 can be mapped di¤eomorphically onto 0 2 Rk and hence, f 1 (y) isn’t a zero dimensional manifold. This establishes that f 1 (y) is countable. Now, since X is compact, every sequence in X clusters, and we can …nd a subsequence which converges in X: Without loss of generality, assume that if #f 1 (y) = 1; xi 2 f 1 (y) converges to some point x: Then, since f is continuous, x 2 f 1 (y) and jDx f (x)j 6= 0; and hence, x must be an element of the countable set of pre-image points, i.e. x = xj for some j: This then implies that the set of pre-image points is …nite. Since f is a local di¤eomorphism at each xj ; there exist neighborhoods Wj (xj ) such that f jWj : Wj = Uj (y) for some open set Uj (y) follows that

Y: Taking the Wj to be open, it X

[j Wj

is a compact set, so f [X [j Wj ] is compact in Y: Since y 2 = f [X [j Wj ] and Y is a manifold, there exists an open set U (y) such that f 1 (U ) Wj for all j: Let Vj = f

1

(U ) \ Wj :

Then f jVj : Vj = U for all j; and the Vj are the open sets we are seeking. illustrate this below.

16

We

(h) Transversality i. We can extend our analysis of regular values to the case where a smooth map f : X ! Y has many "solutions" in the sense of asking when fx 2 Xjf (x) 2 Zg

for a smooth manifold Z Y is a reasonable geometric object – speci…cally, a manifold. The question of whether f 1 (z) is a manifold is, as usual, a local concern; speci…cally, f 1 (z) is a manifold if 8x 2 f 1 (z) 9 U (x) open in X such that f 1 (z) \ U (x) is a manifold. This reduces the study to a question of the regularity of the single point y = f (x): So, suppose y = f (x) 2 Z: Since Z is a manifold, we can "cut out" Z on a neighborhood of y using independent functions via a partial converse to the corollary to the pre-image theorem discussed above. Given this result, pick a neighborhood W (y) . We then have Z \ W (y) = fy 0 2 Y jg1 (y 0 ) = 17

= g` (y 0 ) = 0g

where ` = cod Z: Then, near x; f

1

(z) = fx 2 Xjg1 f (x) =

g` f (x) = 0g : `

Now, let g = [g1 ; :::; g` ] : then g : Y ! R is a submersion (since the ` functions are independent). Now, for U (x) su¢ ciently small, consider g f : U ! R` : By the pre-image theorem, (g f ) 1 (0) will be a manifold if 0 is a regular value of g f: Note that as formulated here, the condition for f 1 (z) to be well-behaved depends on the mapping g which is basically arbitrary. We can eliminate this arbitrariness by reformulating the condition in terms which depend only on the mapping f and the mani…ld Z; as follows. ii. Consider the regularity condition on g f: This requires that d (g f )x : Tx X ! R` be surjective. Since d (g f )x = dgy dfx ; the composite mapping will be surjective if and only if dgjIm dfx maps onto R` : But dgy : Ty Y ! R` is a surjective linear transformation with ker dgy = Ty Z: Since this means that dgy jTy Z ? is an isomorphism, it follows that dgy maps any subspace of Ty Y onto R` precisely when that subspace and Ty Z together span all of Ty Y Applying this observation to the subspace Im dfx ; we have the condition Im dfx + Ty Z = Ty Y When this condition holds (and hence the implied regularity condition holds), we say that f intersects Z transversely, denoted f t Z; for x 2 f 1 (z) : iii. Example Consider f : R ! R3 de…ned by f (t) = t; t; t2 1 : Then dft = [1; 1; 2t] : Let Z = (x; y; z) 2 R3 jz = 0 :

Then f

1

(Z) = t 2 Rjt2

1 = 0 = f1; 1g :

At t = 1; df1 = [1; 1; 1] and Im df1 = diag R3 : Also, T(1;1;0) Z = R2

f0g :

Clearly, then Im df1 + T(1;1;0) Z = R3 : We can show a similar result for t = the following picture: 18

1: Geometrically, we have

iv. To summarize, we have shown the following: Theorem: If the smooth map f : X ! Y has f t Z Y for a submanifold Z; then f 1 (Z) is a submanifold and cod f 1 (Z) = cod Z: The codimension calculation follows from the fact that 0 is a regular value of the mapping g f introduced above. Regularity implies that 1 cod [g f ] (0) = ` = cod Z since Z = g 1 (0) : Note next that if f : X ! Y is the inclusion map of a submanifold X into Y and Z Y is a submanifold, then the condition that i t Z implies that for all x 2 X \ Z Tx X + Tx Z = Tx Y: In this case, we say that the submanifolds X and Z are transverse, or have transverse intersection, or that X and Z are in general position, and we write X t Z: For this case, the previous theorem implies the following

19

Corollary: If X t Z then X \ Z is a submanifold and codX (X \ Z) = codY Z: This implies that codY (X \ Z)

= dim Y dim (X \ Z) = dim Y dim X + dim X dim (X \ Z) = codY X + codX (X \ Z) = codY X + codY Z

i.e. in the ambient space Y; the codimension of the intersections is the sum of the codimensions of the submanifolds. (i) Stability Intuitively, stability refers to properties of an object which persist under small perturbations. Example: Transverse intersection is stable; non-transverse intersections are not. We illustrate this below.

We formalize the notion of stability via the concept of homotopy. De…nition: Two maps f0 : X ! Y , and f1 : X ! Y are smoothly homotopic if there exists a smooth map F :X 20

I!Y

such that F (x; 0) = f0 (x) and F (x; 1) = f1 (x): Theorem: The following classes of smooth maps of a compact manifold X into a manifold Y are stable classes: local di¤eomorphisms immersions submersions f t Z Y for Z a manifold embeddings di¤eomorphisms (See Guillemin and Pollack for proof.) (j) Sard’s theorem One of the most important results in di¤erential topology, though its proof is decidedly analytic. i. Recall that the regular value theorem assets that f 1 (y) is a smooth manifold if y is regular. Sard’s theorem tells us that regular values are common. Theorem (Sard): If f : X ! Y is smooth, then almost every point in Y is a regular value of f: Note: The phrase "almost every" here means every point except for a subset Y0 Y having Lebesgue measure zero. This gives the equivalent statement of the theorem: The critical values of a smooth map f : X ! Y have measure zero (i.e. they constitute a Baire category I suset of Y ). Corollary: The set of regular values of a map f : X ! Y are dense. This follows from the fact that no set of measure zero can contain an open set. Hence, any critical value y0 has regular values in any neighborhood of it, and hence, the regular values are dense because the critical values are nowhere dense. ii. If x is such that f (x) = y is regular, we call x a regular point of f ; if x is such that f (x) is critical, we call x a critical point of f: (Note that values are in the codomain of f; while points are in the domain.) iii. Suppose dim X < dim Y and f : X ! Y is smooth. We saw in de…ning the concept of regularity that in this case, y is regular i¤ f 1 (y) = : Since dim [Im f ] dim X < dim Y; Sard’s theorem implies the same results, since measureY [Im f ] = 0: The notion of stability also applies here, since at any point y = f (x); we can perturb f homotopically to f1 (x) 6= y: Hence, the regular values are only those not hit. (k) Manifolds with boundary The manifolds one typically encounters in economic theory are generally compact (due to scarcity of resources) but also have boundaries, 21

imposed by sign restrictions (for example, non-negativity of consumption). A typical manifold with boundary in economics is the space of normalized prices, for example S ` = p 2 R`+ j kpk = 1 : The boundary of this set occurs where the sphere intersects the boundary of the non-negative orthant of R` (and, in fact, in this case, we actually have a manifold with corners). De…nintion: A subset X RN is a k-dimensional manifold with boundary if every point x has a neighborhood di¤eomorphic to an open set in H k = x 2 Rk jxk 0 : The boundary of X; denoted @X; is the image under some parameterization ' : H k ! X of the boundary of H k = x 2 Rk jxk = 0 : A manifold with corners is de…ned similarly using not a half-space but an orthant. Various results on manifolds with boundaries i. Proposition: The product of a @-manifold X and a boundaryless manifold Y is a @-manifold. Furthermore, @ (X

Y ) = @X

Y

and dim (X

Y ) = dim X + dim Y:

ii. Proposition: If X is a k-dimensional @-manifold, then @X is a (k 1)-dimensional manifold without boundary. iii. Tx (@X) is a linear codimension one subspace of Tx X (for x 2 @X). The linear mapping that de…nes the tangent space is – in terms of a parameterization ' : H k ! X – d'x : Rk ! Tx X restricted to Rk 1 Rk d'x j@H k : Rk

1

! Tx (@X) :

iv. Extension of the pre-image theorem to @-manifold Theorem: Suppose X is a @-manifold, Y a boundaryless manifold, and f : X ! Y is smooth. Let @f : @X ! Y be f j@X : Suppose f t Z Y for Z a submanifold of Y; and @f t Z: Then f 1 (Z) is a @-manifold and @f

1

(Z) = f

1

(Z) \ @X:

v. We can also extend Sard’s theorem: For any smooth map f : X ! Y with X a @-manifold and Y boundaryless, almost every y 2 Y is a regular value of f and @f: 22

(l) Transversal Density The result we develop here is key. We saw previously that t-intersection is a stable property which is preserved under perturbations of mappings. We show here (via Sard’s theorem) that t-intersection is in fact a generic property; any map f which is not transverse to a submanifold Z Y can be perturbed an arbitrarily small amount to a map f 0 which is transverse to Z: The perturbation is accomplished by varying a set of parameters which de…ne the map (for example, the coe¢ cients of a polynomial). To formalize these ideas, we …rst de…ne a family of mappings indexed by some parameter s 2 S (possibly a vector) and denote a typical element of the family by fs : X ! Y: Now, let F :X

S!Y

be de…ned by F (x; s) = fs (x) : We need S to be a manifold (without boundary) and F to be smooth. Theorem: Suppose F : X S ! Y is smooth and only @X 6= : Let Z Y be any boundaryless submanifold of Y: Then, if F t Z and @F t Z; then for almost every s 2 S; fs t Z and @fs t Z: Proof: By the pre-image theorem, W =F

1

(Z)

X

S

is a submanifold with @W = W \ @ (X S) : Let : X S ! S be the natural projection. We show that whenever s 2 S is a regular value of jW : W ! S; then fs t Z and whenever s is a regular value of @ jW : @W ! S; then @fs t Z: Since Sard’s theorem implies that almost every s is a regular value of jW and @ jW ; the theorem will then follow. We illustrate the basic geometric argument below.

23

To show that fs t Z; suppose fs (x) = z 2 Z: By assumption, F (x; s) = z and F t Z; so we know that dF(x;s) T(x;s) (X

S) + Tz Z = Tz Y:

This implies that given any vector a 2 Tz Y; there exists b 2 T(x;s) (X such that dF(x;s) b a 2 Tz Z:

We wish to …nd a vector v 2 Tx X such that dfs (v) this will imply that fs t Z: Now T(x;s) (X

a 2 Tz Z; since

S) = Tx X + Ts S

so b = [w; e] for w 2 Tx X and e 2 Ts S: Now, if e = 0; then we are done, since fs = F jX fsg and dfs (w) = dF(x;s) [w; 0] : 24

S)

In general, e 6= 0, but we can use the projection operator to eliminate it. Since d (x;s) : Tx X Ts S ! Ts S the assumption that s is a regular value of jW (which implies that d maps T(x;s) W onto Ts S) tells us that there is a vector [u; e] 2 T(x;s) W with e 6= 0: But F : W ! Z so dF(x;s) [u; e] 2 Tz Z: Hence, v = w u is the vector we are seeking, since v = w u 2 Tx X and dfs (v)

a = dF(x;s) [[w; e] [u; e]] a = dF(x;s) [w; e] a dF(x;s) [u; e]

and dF(x;s) [w; e]

a 2 Tz Z

and dF(x;s) [u; e] 2 Tz Z:

Note, …nally, that the same argument shows that @fs t Z when s is a regular value of @ : Example –F : R R ! R – F (x; s) = x2 + s – Claim:p F t f0g : To show this, consider F (x; s) = 0; then x= s: Calculate p 2 2x s = DF = : 1 F =0 1 Since DF has rank 1, we have F t f0g : The transverse density theorem then implies that fs (x) t f0g for an open and dense set S^ = fs 2 Rjs 6= 0g : Geometrically, we have

25

2. Mod 2 Degree Theory (a) The mod 2 degree is a topological invariant (like compactness or connectedness) which turns out to be tremendously useful in economic theory establishing existence of equilibrium. Intuitively, the mod 2 degree of the intersection of two manifolds is simply the number of intersections points counted mod 2 (i.e whether the number of points is odd or even), as illustrated below.

26

Note that the parity of the number of intersections will remain invariant as long as the intersections of the manifolds remain transverse under the perturbation. (b) This property allows us to de…ne a global intersection invariant, the mod 2 degree, for all manifolds which are homotopic to one having all transverse intersections with another given manifold. Indeed, if M0 and M1 are manifolds and there exists x 2 M0 \ M1 at which the two manifolds are not in general position, then a small perturbation (i.e. a homotopy) will restore the transverse intersection and we can de…ne the mod 2 degree based on the perturbation. To see the applicability of this concept for general equilibrium, consider our example GE economy, with the equilibrium manifold illustrated below.

27

For the two manifolds, take one to be the equilibrium manifold E and the other the two lines over ! 0 and ! 1 : Since the line over ! 0 intersects E transversely, and the equilibrium associated with ! 0 is unique (a no-trade equilibrium, for example), the mod 2 intersection number is 1. Since the line over ! 0 is homotopic to the one over ! 1 ; the mod 2 intersection number will be the same. If we had chosen a di¤erent ! at which the line over this ! was tangent to one of the critical folds points, this intersection is not transverse, but a small perturbation in one direction or the other will make it so, so we de…ne the mod 2 intersection number for this intersection as 1. Since this number (which we call the mod 2 degree) is now a global invariant, it imples that there must exist at least one equilibrium. (c) Formal Developments i. Given manifolds X; Y; Z with X; Z have complementary dimension if

Y; X and Z are said to

dim X + dim Z = dim Y: If X t Z; then X \ Z is a zero-dimensional manifold (i.e. a collections of points). 28

If, in addition, X and Z are closed and one of them is compact, then the intersection is a …nite set of points. Let # (X \ Z) be the number of intersections points, and call this the intersection number. ii. As noted previously, the intersection number may not be wellde…ned if X 6t Z: Even when it is well-de…ned, it can change if X or Z is perturbed. What is true, however, that that # (X \ Z) mod 2 is invariant, as long as X t Z: We will demonstrate below that the mod 2 intersection number is a homotopy invariant. To set notation, let I2 (X; Z) = # (X \ Z) mod 2 when X t Z: iii. Homotopy considerations Homotopy gives us a mathematically precise way of considering perturbations of a manifold. Given a submanifold X Y , we can view the inclusion of X in Y; { : X ,! Y as simply embedding X in Y: Since inclusion is a mapping, it can be varied homotopically, and, since embeddings are a stable class of mapping, homotopies of { near the identity will still be embeddings. Hence, we can use homotopy to deform X di¤eomorphically. This notion generalizes. Let f : X ! Y be transverse to a submanifold Z Y; where dim X + dim Z = dim Y; with X compact. Then f 1 (Z) is (by the pre-image theorem) a closed zero-dimensional submaifold of X: As before, the mod 2 intersection number of the mapping f with Z is I2 (f; Z) = #f

1

(Z) mod 2:

For an arbitrary smooth map g : X ! Y; de…ne I2 (g; Z) = I2 (f; Z) for any smooth f t Z which is homotopic to g: That this de…nition makes sense is show by the following result. Theorem: If f0 ; f1 : X ! Y and fi t Z for i = 0; 1; then I2 (f0 ; Z) = I2 (f1 ; Z) : Proof: Let F : X I ! Y be a homotopy of f0 ; f1 : By the extension theorem, we can assume that F t Z: Aside: While we don’t prove this result in full generality (see Guillemin and Pollack for this), you can show this result easily for the case where F (x; t) = (1

t)f0 (x) + tf1 (x) :

Then DF =

29

(1

t)Dx f0 + tDx f1 f0 f1

This matrix will have full rank by virtue of the transversality assumptions on f0 and f1 ; and the fact that the two functions are not the same. (If they are the same at some point, perturbing f1 will eliminate this.) With @ (X I) = X f0g [ X f1g and @F

= f0 on X = f1 on X

f0g f1g

it follows that @F t Z: Hence, by the pre-image theorem, F 1 (Z) is a one dimensional manifold with boundary in X: Note that @F

1

(Z)

= F 1 (Z) \ @ (X I) = f0 1 (Z) f0g [ f1 1 (Z)

f1g :

From the claissi…cation theorem for one-manifolds, @F must have an even number of points. Hence,

1

(Z)

#f0 1 (Z) = #f1 1 (Z) mod 2: iv. As a corollary, we have Corollary: If g0 ; g1 : X ! Y are homotopic, then I2 (g0 ; Z) = I2 (g1 ; Z) : Proof: Homotopy is an equivalence relation, so there exists f : h h X ! Y with f t Z and g0 f g1 : v. An important special case of these ideas occurs when X is compact and Y is connected, and dim X = dim Y: Theorem: If F : X ! Y is a smooth map of a compact manifold X into a connected manifold Y and dim X = dim Y; then I2 (f; fyg) is the same for all y 2 Y: In this case, we call I2 (f; fyg) the mod 2 degree of f; denoted deg2 f: Proof: Given y 2 Y; if f 6t fyg ; perturb it to something that is; WLOG, then, we assume f t fyg : By the stack of records theorem, there exists a neighborhood U of y such that f

1

(y) = V1 [

[ Vn

for some n < 1; Vi \ Vj = ; and Vi = U: Hence, I2 (f; fzg) = n mod 2 for all z 2 U: The function y ! I2 (f; fyg) is, therefore, locall constant. Since Y is connected, this function must be globally constant. This theorem continues to be valid of X is not compact, provided that the mapping f is proper: Cr

De…nition: A mapping f : X ! Y is proper if f is compact for K Y compact. 30

1

(K)

X

Example: f (x) = is not compact.

1 x

is not proper since f

1

([0; 1]) = [1; +1]

3. Jets (a) Consider a function f : Rn ! Rm (or open sets in these two spaces). The r th -order Taylor series expansion of f at x0 is x0 )+D2 f (xo ) (x

f (x) = f (x0 )+Df (x0 ) (x

2

x0 ) +

+Dr f (x0 ) (x

r

Here, Dr f (x0 ) is the rth derivative of f: Dr f (x0 ) 2 L [Rn ; :::; Rn ; Rm ]= space of r-linear maps from Rm : Recall that an r-linear map h is such that

r n i=1 R

!

h [x1 ; :::; xi + yi ; :::; xr ] = h [x1 ; :::; xi ; :::; xr ]+ h [x1 ; :::; yi ; :::; xr ] : r

The notation (x x0 ) should not be viewed as a vector product (either innner or outer), but rather as suggesting the Dr f (x0 ) operates on r-tuples of variables (x x0 ) : For an example with m = 1 and r = 2; we have T

f (x) = f (x0 )+Df (x0 ) (x

x0 )+(x

T

x0 ) D2 f (x0 ) (x

x0 ) :

Here, the second derivative matrix D2 f determines the coe¢ T cients of the bilinear function (x x0 ) D2 f (x0 ) (x x0 ) : For r > 2; the components of the rth derivative mapping would determine the coe¢ cients of the r-linear function appearing in the Taylor series expansion. (b) A fundamental observation: The variables x in the Taylor series convey no information about the function f itself. They simply formalize the transformation of the function f into a polynomial T f: All of the important information about f is contained in the multilinear maps speci…ed by the derivatives. (In many instances, since the space of polynomials is dense in the space of relevant functions, we can construct a sequence of Taylor series approximations to f which converge to f in the limit.) This suggests that an alterantive to writing out the full Taylor series is simply to list the derivative maps, together with the evaluation point x0 and the value of f at x0 ; x0 ; f (x0 ) ; Df (x0 ) ; D2 f (x0 ) ; :::; Dr f (x0 ) : This procedure de…nes a mapping j r f : R n ! Rn

Rm

L (Rn ; Rm )

called the r-jet extension of f: 31

L (Rn ; :::; Rn ; Rm )

r

x0 ) +o (kxk ) :

(c) Euclidean Jet-bundles For C r functions f : Rn ! Rm ; de…ne J r [Rn ; Rm ] = Rn

Rm

L (Rn ; Rm )

L (Rn ; :::; Rn ; Rm ) :

We call J r the r-jet bundle over maps from Rn ! Rm :

i. The structure of the spaces of r-linear maps Lr [Rn ; Rm ] L (Rn ; :::; Rn ; Rm ) A. L [Rn ; Rm ] = space of all m n matrices. L [Rn ; Rm ] is canonically homeomorphic to Rmn via the map A 7! a given by 02

a11 B6 .. H(A) = H @4 . am1

..

.

31 a1 n .. 7C = [a ; :::; a ; a ; :::; a ; :::; a ; :::; a ] : 11 1n 21 2n n1 nm . 5A amn

This map is in fact smooth. To see this, let us calculate the derivative of H in the direction of a matrix B: 1 dB H(A) = lim [H (A + tB) t!0 t Now, H(A + tB)

H(A)] :

H(A) has typical element aij + tbij

aij = tbij :

Hence, dB H(A) = b = [b11 ; :::; bnm ] : 2

n

m

B. L [R ; R ] = L [Rn ; L [Rn ; Rm ]] : Via the identi…cation of L [Rn ; Rm ] with Rnm ; we may identify L2 [Rn ; Rm ] with L [Rn ; Rnm ] which is the space of nm m matrices, which map vectors in Rn to vectors in Rnm : Inverting the homeomorphism H; we obtain an m n matrix, i.e. an element of L [Rn ; Rm ] : This establishes the equality above. With L [Rn ; Rm ] identi…ed 2 to Rnm ; we identify L2 [Rn ; Rm ] to Rn m as before. 2 n m n n Now, consider L [R ; R ] = L [R ; R ; Rm ] : A typical element g 2 L2 [Rn ; Rm ] can be written as g (x; y) = [g1 (x; y) ; :::; gm (x; y)] with gi (x; y) = xT Ai y where the matrix Ai is n n and (x; y) 2 Rn Rn : Hence, to identify g; it su¢ ces to know the collection [A1 ; :::; Am ] of n n matrices, and this collection can be identi…ed in the 2 usual way with Rn m : 32

C. One can show by induction that Lr [Rn ; Rm ] = L Rn ; Lr

1

(Rn ; Rm )

and we can then iden…ty Lr (Rn ; Rm ) with Rn

r

m

:

(d) Topologies on function spaces Let C r [Rn ; Rm ] = f 2 C 0 (Rn ; Rm ) jf is r-times continuously di¤erentiable : We wish to de…ne topologies on the set C r [Rn ; Rm ] in order to be able to talk about when a sequence of functions in the space converges to some other function in the space, and when two functions are "close" in some sense. i. The simplest notion of convergence of functions is known as pointwise convergence. We say that a sequence of functions fn 2 C 0 [Rn ; Rm ] converges pointwise to a function g, denoted pw fn ! g; i¤ lim fn (x) = g (x) n!1

for all x 2 Rn : Pointwise convergence is easy to work with, but unfortunately, it may not preserve important properties of f; such as continuity or di¤erentiability. Example: Consider fn (x) = xn for x 2 [0; 1] ; n = 0; 1; 2; ::: This sequence of functions converges pointwise to the function 0 if 0 x < 1 g(x) = : 1 if x = 1 Clearly, for all …nite n; fn is continuous, but the pointwise limit g is not. ii. A more useful notion of convergence is that of uniform convergence. We say that a sequence fn ! g converges uniformly if sup kfn (x) g (x)k ! 0: x2Rn

It can be shown that if fn is continuous for all n; then the limit function g will be continuous. However, even with uniform convergence, properties such as di¤erentiability may not be preserved. Example: Consider fn (x) =

xn for x 2 [0; 1] ; n = 0; 1; 2; ::: n 33

The limit of this sequence of functions is the function g(x) = 0 for x 2 [0; 1] : This convergence is uniform since sup kfn (x)

g (x)k = sup

x2[0;1]

x2[0;1]

xn 1 = ! 0 as n ! 1: n n

The derivative sequence for fn is fn0 = xn

1

:

As we saw in the previous example, this converges pointwise to the function g which is zero everywhere but at x = 1; where it is 1: For any …nite n; the derivative function is continuous, but the limit function is not. iii. We can strengthen the notion of uniform convergence to allow us to control properties of derivatives by de…ning, for example, C 1 uniform convergence. We say that the sequence of functions fn converges C 1 -uniformly to a function g i¤ sup kfn (x)

x2Rn

g (x)k + max sup kDxi fn (x) i=1;:::;mx2Rn

Dxi gxk ! 0:

iv. C r -uniform convergence is de…ned similarly. v. Finally, since it can be di¢ cult to control the behavior of functions "at in…nity" when the domain of the function is Rn ; we de…ne C r -uniform convergence on compacta by restricting the domains of the functions to be (possibly increasing) compact subsets of Rn : This leads to the most useful topology on spaces of smooth functions, the C r compact open topology. We can use the notion of jets introduced above to de…ne the open sets in this topology. For example, taking r = 2; we let N=

f 2 C 2 (Rn ; Rm ) j8x 2 K Rn with K compact, j 2 f (x) 2 N J 2 [Rn ; Rm ] for N open.

de…ne an open neighborhood in the space of functions. Loosely speaking, all f 2 N are such that their values, and the values of their derivatives on the compact set K are "close." (e) The jet transversality theorem i. Given smooth manifolds M and N; neither of which has a boundary, and a smooth submanifold A N; let trK (M; N ; A) = ff 2 C r (M; N ) jf tK Ag where f tK A denotes transverse intersection when f is restricted to a subset K M: We also let tr (M; N ; A) =trM (M; N ; A) : 34

Jet Transversality Theorem: Let M and N be smooth manifolds without boundary, and let A J r (M; N ) be a smooth submanifold. Suppose 1 r < s 1: Then tr (M; N ; j r ; A) = r r r ff 2 C (M; N ) jj f : M ! J (M; N ) t Ag is a residual set, i.e. a countable intersection of dense open sets in C r (M; N ) : ii. We will not prove this result formally, but to see how it might be applied in general equilibrium theory, consider the case of an exchange economy in which the distribution of endowments is …xed and cannot be changed, but in which preferences can be considered parameters. We can write down the …rst-order conditions and market clearing conditions for this economy as Dui (zi + ! i )

ip

p zi m X zi

= 0 for i = 1; :::; m = 0 for i = 1; :::; m =

0:

i=1

If we consider these FOCs as a mapping g de…ned on prices, net trades and the space of smooth, strictly concave utility functions g:

~

iU

S

R`m ! R(`+1)m+`

then if we can show that 0 is a regular value of g; it will follow that for almost every pro…le of utility functions, the associated system of FOCs will also have zero as a regular value, in which ~ S R`m will be a manifold, and case, the preimage of 0 in i U we can apply the previous analysis to determine things like the degree of the natural projection, etc. To show that g t f0g ; we simply apply the jet transversality theorem (via j 1 ui ). iii. Other applications of the jet transversality theorem can be found in my thesis, where I used the FOCs and market clearing conditions for a two-period lived stochastic OLG economy with multiple goods traded each period, under the assumption that the equilibrium prices depend only on exogenous shocks. These equilibrium conditions can be shown to involve more equations than variables, and one then uses an extension of the jet transversality theorem to infer that for most utility functions, zero is a regular value of the equilibrium equations. With more equations than variables, the only way zero can be a regular value is if it is never hit, i.e. if there is no equilibrium.

35