Chapter 6 - Sard's Theorem Flashcards
Define: measure zero in R^n, measure zero in M
Why is measure zero in M a good definition?
Most important sets of measure zero?
A set A < R^n has MEASURE ZERO if for any delta > 0, A can be covered by a countable collection of open rectangles, the sum of whose volumes is less than delta.
If M is a smooth manifold we say A < M has MEASURE ZERO in M if for every smooth chart (U, phi) for M, the subset phi(A int U) N is a smooth map, and A < M has measure zero. Then F(A) has measure zero in N. R^n smooth, then F(A) has measure zero. Prop 6.5
Also, as with R^n a countable union of sets of measure zero in M has measure zero –> reduce to R^n case using charts.
Graphs of continuous functions
pg 127
What is Sard’s Theorem?
Immediate corollaries?
Sard’s Thm. F:M –> N smooth map, then the set of critical values of F has measure zero in N.
Cor. If dim M < dim N, then F(M) has measure 0. (each point of M is a critical point for F)
Cor. If S < M is an immersed submanifold and dim S < dim M, then S has measure zero in M. (apply above cor. to inclusion map S –> M) — think of R into the torus. Dense but measure 0…
What is Whitney Embedding Theorem? Proof idea?
Lemma 6.13 nice proof strategy. Can reduce dimension of ambient space of submanifold by 1 via projection along some vector v so long as ambient dimension N > 2n +1 where n is dim of submanifold.
Above only works for reducing dimension of immersions and compact embeddings. 6.14 fills the gap.
Lemma 6.14. If M admits a smooth embedding into R^N for some N, then admits a proper smooth embedding into R^2n+1.
Thm. Every smooth n-manifold with or without boundary admits a proper smooth embedding into R^2n+1
Strong Whitney Embedding Theorem. If n > 0, every smooth n-manifold admits a smooth embedding into R^2n
pg 131
Discuss Whitney Approximation Theorems
Tubular neighborhoods?
Whitney Approximation for Functions. Suppose M is a smooth manifold and F: M –> R^k is a continuous function. Given any positive continuous function delta:M –> R, there exists a smooth function F tilda: M –> R^k that is delta close to F.
Pf idea. Let U_i be countable open cover of M s.t. diameter F(U_i) is tiny (all values about c_i) M. By Whitney embedding, can consider M as embedded submanifold of R^n and by above we can approximate F by a smooth map into R^n. However, in general the image of this smooth map will not lie in M. To correct for this , we use a smooth retraction from a tubular neighborhood of M onto M.
Thm. Every embedded submanifold of R^n has a tubular neighborhood.
Prop. If U is any tubular neighborhood of embedded submanifold M in R^n, there exists a smooth map r:U–>M that is both a retraction and a smooth submersion
With this in hand, we can state and prove:
Whitney Approximation Theorem: Say F:N –> M is continuous map between smooth manifolds. Then F is homotopic to a smooth map.
Thm. If maps F,G: N –> M are homotopic, then they are smoothly homotopic
Discuss transversality
Suppose M is a smooth manifold. Two embedded submanifolds S, S’ < M intersect TRANSVERSELY if for each p in S int S’, the tangent spaces T_pS and T_pS’ together span T_pM
If F: N –> M is a smooth map and S < M is an embedded submanifold, we say that F is transverse to S if for every x in F^-1(S), the spaces T_F(x) S and dF_x(T_xN) together span T_F(x)M
The desirability of transversality stems from the following theorem:
Thm. Suppose N, M smooth manifolds and S < M is an embedded submanifold.
(a) If F:N –> M is a smooth map that is transverse to S, then F^-1(S) is an embedded submanifold of N whose codimension is equal to the codimension of S in M
(b) If S’ < M is an embedded submanifold that intersects S transversely, then S int S’ is an embedded submanifold of M whose codimension is equal to the sum of the codimensions of S and S’.
Now it turns out that transversality is in some sense generic.
Transversality Homotopy Theorem. Suppose M, N smooth manifolds and X < M is an embedded submanifold. Every smooth map f : N –> M is homotopic to a smooth map g : N –> M that is transverse to X.
