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math214:02-12

2020-02-12, Wednesday

Tubular Neighborhood Theorem

Let MRnM \subset \R^n be a embedded submanifold. Let NMNM be the normal bundle of MM, defined by NM={(x,v)TRnxM,vw,wTxM}. NM= \{ (x, v) \in T\R^n \mid x \in M, v \perp w, \forall w \in T_x M \}. Thm: NMNM is an embedded nn-dimensional submanifold of TRnT\R^n. Let M0={(x,v)NMv=0}NMM_0 =\{(x,v) \in NM| v = 0\} \subset NM be the zero section.

There is a shift map: E:NMRn,(x,v)x+v.E: NM \to \R^n, \quad (x,v) \mapsto x+v. One can check that for any xM0x \in M_0, dEx:TxNMTxRndE_x: T_x NM \to T_x \R^n is an isomorphism. Hence EE is a diffeomorphism in a neighborhood of xx. Since xx is arbitrary, we have EE is a diffeomorphism in a neighborhood VM0V \supset M_0. More precisely, we want to constraint the shape of UU. We define

Tubular neighborhood of MM in Rn\R^n is a neighborhood UU of MM, that is the diffeomorphic image under EE of an open subset of the form V={(x,v)NM:v<δ(x)} V = \{(x,v) \in NM : |v| < \delta(x) \} for some positive continuous function δ:MR\delta: M \to \R.

See Thm 6.24 for a proof of the existence of a tubular neighborhood. Mainly, one try to construct the 'radius' function δ\delta and show that it is continuous.

We define a map r:UMr: U \to M, that is identity map when restrict to MM, such a map is called a retraction. r:=πE1r: = \pi \circ E^{-1}, where π:NMM\pi: NM \to M is the projection map.

Transversality

Let VV be a finite dimensional vector space, W1,W2W_1, W_2 be two subspaces, we say W1W_1 and W2W_2 are transversal, if W1+W2=VW_1 + W_2 = V.

Consider two embedded submanifolds N1,N2N_1, N_2 in MM. Suppose dimM=m,dimN1=n1,dimN2=n2\dim M =m, \dim N_1 = n_1, \dim N_2 = n_2. If n1+n2<mn_1 + n_2 < m, then intuition tells us that N1N_1 and N2N_2 'generically' does not intersect(no intersection is a special case of transverse intersection). If n1+n2mn_1 + n_2 \geq m, then 'generically' N1N_1 and N2N_2 should intersects transversally, that is, for any pN1N2p \in N_1 \cap N_2, we should have TpN1+TpN2=TpMT_p N_1 + T_p N_2 = T_p M. But what does 'generically' mean? Let's make it more precise.

Let's consider a more general setting. Let SMS \subset M be an embedded submanifold, and let F:NMF: N \to M be any smooth map. We say that FF is transverse to SS, if for every xF1(S)x \in F^{-1}(S), we have TF(x)S+DFx(TxN)=TF(x)M. T_{F(x)} S + DF_x (T_x N) = T_{F(x)} M.

To understand the following theorem, we need to have

  • Theorem 5.8 (Local Slice Criterion for Embedded Submanifolds) Any embedded submanifold SS has slice coordinate chart.
  • Theorem 5.12 (Constant-Rank Level Set Theorem) : If F:MNF: M \to N has constant rank, then for any xF(M)x \in F(M), F1(x)F^{-1}(x) is a properly embedded submanifold of MM.

Theorem (6.30): (a) If F:NMF: N \to M is a smooth map between smooth manifolds, SMS \subset M embedded manifold, and FF is transverse to SS, then F1(S)F^{-1}(S) is an embedded submanifold of NN with the same codimension as SS in MM
(b) If S,SS', S are two embedded submanifolds, that intersects transversely, then SSS \cap S' is an embedded submanifold of MM, with codimM(SS)=codimMS+codimMSco\dim_M (S \cap S') = co\dim_M S + co\dim_M S'.

Sketch of proof: (b) follows from (a) by taking F=ι:SMF = \iota: S' \into M, and SSSMS' \cap S \into S' \into M, composition of embedding is embedding.

To prove (a), we take a point xSF(N)x \in S \cap F(N), and take slice coordinate chart (U,(x1,,xn))(U, (x_1, \cdots, x_n)) of MM adapted to SS, that is, SUS \cap U can be written as {xk+1=0,xn=0}\{x_{k+1}=0, \cap x_n=0\} where k=dimSk =\dim S. Denote φ=(xk+1,,xn):URnk\varphi=(x_{k+1}, \cdots, x_n): U \to \R^{n-k}. Then, consider (φF):F1(U)Rnk(\varphi \circ F): F^{-1}(U) \to \R^{n-k}, thus (φF)1(0)=F1(US)(\varphi \circ F)^{-1}(0) = F^{-1}(U \cap S). One want to show that (φF)(\varphi \circ F) has 00 a regular value, i.e for any x(φF)1(0)x \in (\varphi \circ F)^{-1}(0), D(φF)x:TxNT0RnkD(\varphi \circ F)_x: T_x N \to T_0 \R^{n-k} is surjective. That is guaranteed by transversality condition. \Box

Next, one consider the parametrix transversality theorem.

Let S,M,NS, M, N be smooth manifolds. A smooth family of maps Fs:NMF_s: N \to M parametrized by sSs \in S is a smooth map F:N×SMF: N \times S \to M, whereas Fs()=F(,s)F_s(-) = F(-, s).

Theorem (6.35) Let XMX \subset M be an embedded submanifold, and let F:N×SMF: N \times S \to M be a smooth map. If FF is transversal to XX, then for almost every sSs \in S, Fs:NMF_s: N \to M is transverse to XX.

Proof: Let W=F1(X)W=F^{-1}(X). Let π:N×SS\pi: N \times S \to S, and πW\pi|_W its restriction to WW. By Sard theorem, the critical value of πW\pi|_W is negligible. Suffice to prove that for any point sSs \in S away from the critical value πW\pi|_W, FsF_s is transversal to XX. (remember that if XFs(N)=X \cap F_s(N) = \emptyset, it is still transverse intersection).

Let's do some simple dimension counting. Suppose codimMX=kcodim_M X = k, then codimN×SW=kcodim_{N \times S} W = k. Let dimN=n\dim N = n, dimS=ds\dim S = d_s. If k>nk > n, then dimW=ds+nk<ds\dim W = d_s + n - k < d_s, then for most ss, Fs(N)X=F_s(N) \cap X = \emptyset and we are done.

Now assume knk \leq n. We can see that sSs \in S is a critical value, if and only if N×{s}N \times \{s\} intersects WW non-transversely (hence intersection is non-empty). \Box.

Theorem 6.36 (Transversality Homotopy Theorem). Suppose MM and NN are smooth manifolds and X􏰌MX \subset􏰌 M is an embedded submanifold. Every smooth map f:NMf: N \to M is homotopic to a smooth map g:NMg: N \to M that is transverse to XX.

Sketch of proof: We are going to let S=BkS=B^k, kk-dimensional unit open ball, and build a smooth map F:N×SMF: N \times S \to M that is a submersion, hence transverse to XX. First, we use Whitney embedding theorem to embed MM to Rk\R^k, ι::MRk\iota: :M \to \R^k. Create tubular neighborhood UU of MM in Rk\R^k. Then, we will wiggle the composition ιf:NRk\iota\circ f: N \to \R^k in all the direction, within UU, then project back to MM.

math214/02-12.txt · Last modified: 2020/02/11 22:50 by pzhou