1. Whitney Embedding Theorem 1.0.1. Let M N Be a Smooth Manifold

Home , Immersion (mathematics), Submanifold

1. Whitney embedding Theorem 1.0.1. Let M n be a smooth manifold. Then M n admits a smooth 2n+1 proper embedding into R . Proof. We will only give a proof when M is compact. It’s enough to show 2n+1 that there exists a smooth 1−1 immersion f : M → R . Such an f will be automatically a proper embedding due to compactness of M . Lemma 1.0.2. Let M n be a compact n-manifold. There exists a a smooth k 1 − 1 immersion f : M → R for some large k depending on M . 0 Proof of Lemma 1.0.2. For any point p choose a coordinate ball Bp con- 0 0 0 taining p and let xp : Bp → Vp be a local coordinate chart where Vp is n 0 an open ball in R . Let Bp ⊂ Bp be a smaller coordinate ball containing 0 p and such that xp(Bp) = Vp is a concentric ball to Vp of smaller radius. ¯ 0 Then Bp is compact and is contained in Bp . We have that ∪p∈M = M is an open cover of M . By compactness we can choose a ﬁnite subcover by 0 0 coordinate balls B1,...,Bm such that xi : Bi → Vi , i = 1, . . . m are local coordinate charts. Let ϕi : M → R be a smooth function such that ϕi = 1 ¯ 0 0 ¯ on Bi , ϕi = 0 on M\Bi and 0 < ϕi < 1 on Bi\Bi . m(n+1) Let f : M → R be given by

f(p) = (ϕ1(p)x1(p), . . . , ϕm(p)xm(p), ϕ1(p), . . . , ϕm(p)) Claim 1: f is 1 − 1. Let p, q ∈ M be distinct points. If p, q ∈ B¯i for some i then ϕi(p) = 0 ϕi(q) = 1 but xi(p) 6= xi(q) since xi is a diﬀeomoprhism on Bi and in 0 particular it’s 1-1 on Bi . Hence ϕi(p)xi(p) 6= ϕi(q)xi(q) and therefore f(p) 6= f(q). If p, q don’t belong to the same B¯i for any i then there is an i such that p ∈ Bi but q∈ / B¯i . then ϕi(p) = 1 but ϕi(q) < 1 and we can again conclude that f(p) 6= f(q). Claim 2: f is an immersion. Let p ∈ M, v ∈ TpM, v 6= 0. Then p ∈ Bi for some i. Bi is open in M and ϕi = 1 on Bi . Therefore d(ϕi · xi)|p(v) = d(xi)|p(v) 6= 0 since xi is an immersion on Bi . Therefore dfp(v) 6= 0 and hence F is an immersion. This ﬁnishes the proof of Claim 2 and hence of Lemma 1.0.2. N Lemma 1.0.3. Let S ⊂ R be a submanifold of dimension n where N > N N ⊥ ∼ N−1 2n + 1. For any nonzero v ∈ R let πv : R → v = R be the orthogo- N−1 nal projection. Then there exists v 6= 0 such that πv|S : S → R is a 1-1 immersion.

Proof of Lemma 1.0.3. Note that ker πv = Rv for any v 6= 0. Observe that if πv(p) = πv(q) for some distinct p, q ∈ S then p − q = N−1 λv for some nonzero λ i.e. [p − q] = [v] ∈ RP . Also, since πv is n linear, d(πv)p = πv for any p ∈ R . Therefore, if (p, u) ∈ TpS, u 6= 0 and N−1 d(πv)p(u) = 0 then u = λv and [u] = [v] ∈ RP . 1 2

N−1 Let h: S×S\∆S → RP be given by h(p, q) = [p−q] and g : TS\0(S) → N−1 RP be given by g(p, u) = [u]. Since N − 1 > 2n = dim S × S = dim TS , N−1 the only way [v] ∈ RP can be a regular value of h (g) is if [v] ∈/ image of h (g). By Sards theorem the set of singular values of h (g) has measure 0 and since the union of two sets of measure zero has measure zero we can ﬁnd N−1 [v] ∈ RP which is a regular value for both g and h. By above that means that [v] is not in the image of either h or g. Again, by above that means that N−1 πv|S : S → R is a 1-1 immersion. To ﬁnish the proof of Whitney’s theorem we repeatedly apply Lemma 1.0.3 to S = f(M) provided by Lemma 1.0.2.

2. Transversality n Deﬁnition 2.0.1. Let S1,S2 be submanifolds in a smooth manifold M . We say that S1 is transverse to S2 (denoted by S1 t S2 ) if for any p ∈ S1∩S2 it holds that TpS1 + TpS2 = TpM . Deﬁnition 2.0.2. Let S be a submanifold in a smooth manifold N n . Let f : M → N be a smooth map. We say that f is transverse to S (denoted −1 by f t S ) if for any p ∈ f S it holds that Tf(p)S + dfp(TpM) = TpN .

• Let S1,S2 be submanifolds in M . Then S1 t S2 iff i1 t S2 where i1 : S1 → M is the canonical inclusion. • Let f : M → N be smooth and let S = c ∈ N be a point which we view as a zero-dimensional submanifold. Then f t S iff c is a regular value of f . • Let S1,S2 be submanifolds in M such that dim S1 + dim S2 < dim M . Then S1 t S2 iff S1 ∩ S2 = ∅. • If f : M → N is a submersion then f t S for any submanifold S in N . Example 2.0.3.

n k l • Let M = R , S1 = R × 0, S2 = 0 × R .Then S1 t S2 iﬀ k + l > n. 2 2 • Let M = R , S1 = R × 0,S2 = {(x, y): y = x }. Then S1 is not transverse to S2 . n n−1 • Let M = R , S1 = S - the unit sphere centered at 0 and let n S2 = V be a vector subspace of R . Then S1 t V . Theorem 2.0.4. Let f : M → N be smooth and transverse to a submanifold S in N . Then P = f −1(S) is a submanifold of M if the same codimension as S , i.e. dim N − dim S = dim M − dim P . Moreover, if M has a boundary and f|∂M is still transverse to S then P = f −1(S) is a submanifold with boundary in M . 3

Deﬁnition 2.0.5. Two maps f0, f1 : X → Y are called homotopic (f0 ∼ f1 ) if there exists a continuous map F : X × [0, 1] → Y such that F (x, 0) = f0(x) and F (x, 1) = f1(x) for any x ∈ X .

Theorem 2.0.6. Let M,N be smooth manifolds and let f0, f1 : M → N be smooth maps. Then f0 is continuously homotopic to f1 iﬀ f0 is smoothly homotopic to f1 i.e. iﬀ there is a smooth map F : M ×[0, 1] → N such that F (x, 0) = f0(x) and F (x, 1) = f1(x) for any x ∈ M . Theorem 2.0.7 (Whitney approximation theorem). Let M,N be smooth manifolds and let f : M → N be a continuous map. Then f is homotopic to a smooth map g : M → N . Moreover, if f is smooth on a closed subset A ⊂ M then g can be chosen so that g|A = f|A . Theorem 2.0.8 (Transversality Approximation theorem). Let M,N be smooth manifolds and let S ⊂ N be a submanifold. Let f : M → N be a smooth map. Then f is homotopic to a smooth map g : M → N such that g t S . Moreover if T ⊂ M is a submanifold and f|T is transverse to S then g can be chosen so that g|T = fT .