用户: Eee/Remark on Thom Isomorphism

Definition 0.1. Orientablity of a vector bundle.

 To each fiber there is assigned a preferred generator $u_F\in H^n(F,F_0;\mathbb{Z})$. The local compatiblity condition implies that $\forall b\in B$, there exists a neighborhood $N$, s.t. $\exists u\in H^n({\pi }^{-1}(N),{\pi }^{-1}(N{)}_0;\mathbb{Z})$ such that $u{|}_{(F,F_0)}=u_F\in H^n(F,F_0;\mathbb{Z})$ for any fiber $F$ over $N$.

 Equivalently, we say a vector bundle is orientable if its structure group may be reduced to $G{L}^{+}(n,\mathbb{R})$.

1Thom Isomorphism Theorem with ${\mathbb{Z}}_2$

Cross Product

 Let ${\mathbb{R}}_0^n$ denote the complement of the origin in ${\mathbb{R}}^n$. For any space $X$, we will prove that $$H^m(X)\cong H^{m+n}(X\times {\mathbb{R}}^n,X\times {\mathbb{R}}_0^n).$$This isomorphism can best be described by introducing the cohomology corss product operation.

Definition 1.1. Cross Product

 Suppose that one is given cohomology classes $$a\in H^m(X,A),b\in H^n(Y,B)$$where $A$ is an open subset of $X$ and $B$ is an open subset of $Y$. (If $B$ is vacuous then $A$ need not to be open, and conversely.) Using the projection maps $$p_1:(X\times Y,A\times Y)\to (X,A)$$$$p_2:(X\times Y,X\times B)\to (Y,B)$$the cross product (or external product) $a\times b$ is defined to be the cohomology class $$(p_1^{\ast }a)(p_2^{\ast }b)\in H^{m+n}\left(X\times Y,(A\times Y)\cup (X\times B)\right).$$ It will be convenient to use the abbreviation $(X,A)\times (Y,B)$ for the pair $\left(X\times Y,(A\times Y)\cup (X\times B)\right)$.

Example 1.2. $({\mathbb{R}}^n,{\mathbb{R}}_0^n)$ can be described as the $n$-fold product $(\mathbb{R},{\mathbb{R}}_0)\times \cdots \times (\mathbb{R},{\mathbb{R}}_0)$.

 Consider the triple $(\mathbb{R},{\mathbb{R}}_0,{\mathbb{R}}_{-})$, $({\mathbb{R}}_0,{\mathbb{R}}_{-})\hookrightarrow (\mathbb{R},{\mathbb{R}}_{-})\hookrightarrow (\mathbb{R},{\mathbb{R}}_0)$, then one have the exact sequence $$0=H^0(\mathbb{R},{\mathbb{R}}_{-})\to H^0({\mathbb{R}}_0,{\mathbb{R}}_{-})\stackrel{\delta }{\to }{H}^1(\mathbb{R},{\mathbb{R}}_0)\to H^1(\mathbb{R},{\mathbb{R}}_{-})=0$$hence we have the coboundary isomorphism $H^0({\mathbb{R}}_0,{\mathbb{R}}_{-})\stackrel{\delta }{\to }{H}^1(\mathbb{R},{\mathbb{R}}_0)$, where $$H^0({\mathbb{R}}_0,{\mathbb{R}}_{-})=H^0({\mathbb{R}}_{-},{\mathbb{R}}_{-})\oplus H^0({\mathbb{R}}_{+})=\mathbb{Z}.$$Thus, $H^0(\mathbb{R},{\mathbb{R}}_0)\cong \mathbb{Z}$ and we define $e^1=\delta (1)$.

 Finally, let $e^n\in H^n({\mathbb{R}}^n,{\mathbb{R}}_0^n)$ denote the $n$-fold cross product $e^1\times \cdots \times e^1$.

Theorem 1.3. For any pair $(X,A)$ with $A$ open in $X$, the correspondence $a\mapsto a\times e^n$ defines an isomorphism $$H^m(X,A)\to H^{m+n}\left((X,A)\times ({\mathbb{R}}^n,{\mathbb{R}}_0^n)\right).$$

The Thom Isomorphism Theorem with Coefficient ${\mathbb{Z}}_2$

 We have the isomorphism $$H^j(B)\to H^{j+n}(B\times \mathbb{R},B\times {\mathbb{R}}_0)$$given by the correspondence $y\mapsto y\times e^n$.  Let $B^{\prime }$ be an open subset of $B$. Then we have isomorphism $$H^j(B,B^{\prime })\to H^{j+n}(B\times \mathbb{R},B\times {\mathbb{R}}_0\cup B^{\prime }\times \mathbb{R})$$given by the corrspondence $y\mapsto y\times e^n$.

Theorem 1.4. Thom Isomorphism with ${\mathbb{Z}}_2$ coefficient

 Let $\xi$ be an $n$-plane bundle with projection $\pi :E\to B$. Let $E_0$ denote the complement of $B$ in $E$.

 There is one and only one cohomology class $u\in H^n(E,E_0;{\mathbb{Z}}_2)$ whose restriction to $H^n(F,F_0;{\mathbb{Z}}_2)$ is non-zero for every fiber $F$.

 Furthermore, the correspondence $y\mapsto y⌣u$ gives $$H^j(E;{\mathbb{Z}}_2)\cong H^{j+n}(E,E_0;{\mathbb{Z}}_2)$$for every integer $j$.

Proof.

Proof. The proof will be divided into four cases.

Case 1.	When $\xi$ is trivial.
Case 2.	When $B=B^{\prime }\cup B^{\prime \prime }$, where both $\xi {|}_{B^{\prime }}$ and $\xi {|}_{B^{\prime \prime }}$ are trivial. We shall prove this case by M-V argument and Five lemma.
Case 3.	Suppose $B$ is covered by finitely many open sets, we prove it by induction.
General case	Let $C$ be an arbitrary compact subset of the base space $B$. Then evidently the bundle $\xi {|}_C$ satisfies the hypothesis of Case 3.  The cohomology group $H^n(E,E_0)$ maps isomorphically to the inverse limit of the groups $H^n({\pi }^{-1}(C),{\pi }^{-1}(C{)}_0)$. But each of the latter groups contains one and only one class $u_C$ whose restriction to each fiber is non-zero. It follows immediately that $H^n(E,E_0)$ contains one and only one class $u$ whose restriction to each fiber is non-zero.  Now consider the homomorphism $⌣u$: $H^j(E)\to H^{j+n}(E,E_0)$. Evidently, for each compact subset $C$ of $B$ there is a commutative diagram Passing to the inverse limit, as $C$ varies over all compact subset, it follows that $⌣u$ is itself an isomorphism. This completes the proof.

$\square$

Lemma 1.5. Given $F$ is a field. Let $C$ be an arbitrary compact subset of the base space $B$. Since the union of any two compact sets is compact, we can form the direct limit $$\underrightarrow{\lim }{H}_j(C;F)$$of homology groups as $C$ varies over all compact subsets of $B$, and the corresponding inverse limit $\underleftarrow{\lim }{H}^j(C;F)$ of cohomology groups.

 The natural homomorphism $$H^j(B;F)\to \underleftarrow{\lim }{H}^j(C;F)$$is an isomorphism, and similarly $H^j(E,E_0;F)$ maps isomorphically to $\underleftarrow{\lim }{H}^j({\pi }^{-1}(C),{\pi }^{-1}(C{)}_0)$.

Proof.

Proof. It is easy to show $\underrightarrow{\lim }{H}_j(C)\cong H_k(B)$ since every singular chain on $B$ is contained in some compact subset of $B$.

 Since $F$ is a field, we have $H^j(B)=\mathrm{Hom}(H_j(B);F)$.

$\square$

2Thom Isomorphism Theorem with an arbitrary ring $\Lambda$

Cap product

Definition 2.1. For any space $X$ and any coefficient domain, there is a bilinear pairing operation $$⌢:C^i(X)\otimes C_n(X)\to C_{n-i}(X)$$which can be characterized as follows. For each cochian $b\in C^i(X)$ and each chain $\xi \in C_n(X)$ the cap product $b⌢\xi$ is the unique element of $C_{n-i}(X)$ such that $$⟨a,b⌢\xi ⟩=⟨ab,\xi ⟩$$(1)for all $a\in C^{n-i}(X)$.

Proposition 2.2. Combining the identity (1) with the standard properties of cup products, one can derive the following rules $$(bc)⌢\xi =b⌢(c⌢\xi ),$$(2)$$1⌢\xi =\xi ,$$(3)$$\partial (b⌢\xi )=(\delta b)⌢\xi +(-1{)}^{\dim b}b⌢\partial \xi .$$(4)The last one is given by formula $⟨\delta c,\alpha ⟩+(-1{)}^{\dim c}⟨c,\partial \alpha ⟩=0$.

 From (4) it follows that there is a corresponding operation $$H^i(X)\otimes H_n(X)\to H_{i-n}(X)$$which will also be denoted by $⌢$.

The Thom Isomorphism Theorem with Coefficient in an arbitrary ring $\Lambda$

 Similarly, the cohomology group $H^n({\mathbb{R}}^n,{\mathbb{R}}_0^n;\Lambda )$ is a free $\Lambda$-module, with a single generator $e^n=e^1\times \cdots \times e^1$.

 Let $\xi$ be an oriented $n$-plane bundle. Then for each fiber $F$ of $\xi$ we are given a preferred generator $$u_F\in H^n(F,F_0;\mathbb{Z}).$$Using the unique ring homomorphism $\mathbb{Z}\to \Lambda$, this gives rise to a corresponding generator for $H^n(F,F_0;\Lambda )$ which will also be denoted by the symbol $u_F$.

Theorem 2.3. The Thom Isomorphism Theorem with Coefficient in an arbitrary ring $\mathbf{\Lambda }$

 Let $\xi$ be an oriented $n$-plane bundle. There is one and only one cohomology class $u\in H^n(E,E_0;\Lambda )$ whose restriction to $(F,F_0)$ is equal to $u_F$ for every fiber $F$.

 Furthermore the correspondence $y\mapsto y⌣u$ gives $$H^j(E;\Lambda )\cong H^{j+n}(E,E_0;\Lambda )$$for every integer $j$.

 If $B$ is compact, then the proof is completely analogous to the previous proof ending at Case 3.

 If the ring $\Lambda$ is a field, then the proof is also completely analogous to the previous proof, since the lemma still holds.

Lemma 2.4. The homology group $H_{n-1}(E,E_0;\mathbb{Z})$ is zero.

Proof.

Proof. If $B$ is compact, the Thom isomorphism and the following corollary is ture, hence $$H_{n-1}(E,E_0;\mathbb{Z})\cong H_{-1}(E;\mathbb{Z})=0.$$ For general case, $H_{n-1}(E,E_0;\mathbb{Z})=\underrightarrow{\lim }{H}_{n-1}({\pi }^{-1}(C),{\pi }^{-1}(C{)}_0;\mathbb{Z})=0$.

$\square$

 With $H_{n-1}(E,E_0;\mathbb{Z})=0$, by Universal Coefficient Theorem, $H^n(E,E_0;\mathbb{Z})=\mathrm{Hom}(H_n(E,E_0;\mathbb{Z}),\mathbb{Z})$, hence the lemma follows.

 It follows that there is a unique fundamental cohomology class $u\in H^n(E,E_0;\mathbb{Z})$.

 To prove that the cup product with $u$ induces cohomology isomorphisms, we will make use of the following constructions.

Definition 2.5. A free chain complex over $\mathbb{Z}$ is a sequence of free $\mathbb{Z}$-modules $K_n$ and homomorphisms $$\cdots \to K_n\stackrel{\partial }{\to }{K}_{n-1}\stackrel{\partial }{\to }{K}_{n-2}\to \cdots$$with $\partial \circ \partial =0$.

 A chain mapping $f:K\to K^{\prime }$ of degree $d$ is a sequence of homomorphisms $K_i\to K_{i+d}^{\prime }$ satisfying ${\partial }^{\prime }\circ f=(-1{)}^d(f\circ \partial )$.

Lemma 2.6. Let $f:K\to K^{\prime }$ be a chain mapping, where $K$ and $K^{\prime }$ are free chain complexes over $\mathbb{Z}$. If $f$ induces a cohomology isomorphism $$f^{\ast }:H^{\ast }(K;\Lambda )\to H^{\ast }(K;\Lambda )$$for every coefficient field $A$, then $f$ induces isomorphisms of homology and cohomology with arbitrary coefficients.

 While proving the Thom Isomorphim, we will simultanuously prove the following. The coefficient ring $\mathbb{Z}$ is to be understood.

Corollary 2.7. The correspondence $\eta \mapsto u⌢\eta$ defines an isomorphism from the integral homology group $H_{n+i}(E,E_0)$ to $H_i(E)$.

Proof.

Proof. Choose a singular cocycle $z$ representing the fundamental cohomology class $u$. Then the correspondence $\gamma \mapsto z⌢\gamma$ from $C_{n+i}(E,E_0)$ to $C_i(E)$ satisfies the identity $$\partial (z⌢\gamma )=(-1{)}^{n}z⌢(\partial \gamma ).$$Therefore $$z⌢:C_{\ast }(E,E_0)\to C_{\ast }(E)$$is a chain mapping of degree $-n$. Using the identity $$⟨c,z⌢\gamma ⟩=(c⌣z,\gamma ⟩$$we see that the induced cochain mapping $$(z⌢{)}^{\#}:C^{\ast }(E;\Lambda )\to C^{\ast }(E,E_0;\Lambda )$$is given by $c\mapsto c⌣z$. Here $\Lambda$ can be any ring.

 If the coefficient ring $\Lambda$ is a field, then this cochain mapping induces a cohomology isomorphism. Thus we can apply the last lemma, and concludes that the homomorphisms $$u⌢:H_{i+n}(E,E_0;\Lambda )\to H_i(E;\Lambda )$$and $$⌣u:H^i(E;\Lambda )\to H^{i+n}(E,E_0;\Lambda )$$are actually isomorphisms for arbitrary $\Lambda$.

 In particular, using the isomorphism $⌣u:H^0(E;\Lambda )\to H^n(E,E_0;\Lambda )$, the uniqueness of the fundamental cohomology class $u$ with coefficients in $\Lambda$ can now be verified.

$\square$

3Thom Isomorphism in Differential Forms

 Since $M\times \{0\}$ is a deformation retract of $E$, it follows that $$H^q(E)\cong H^q(M).$$

 If $E$ and $M$ are orientable manifolds of finite type, then $$H_c^q(E)\cong (H^{m+n-q}(E){)}^{\ast }\cong (H^{m+n-q}(M){)}^{\ast }\cong H_c^{q-n}(M)$$by Poincaré duality.

 Using a Mayer-Vietoris argument as in the proof of the Thom isomorphism, if $\pi :E\to M$ is an orientable rank $n$ bundle over a manifold $M$ of finite type, then $$H_c^q(E)\cong H_c^{q-n}(M).$$Note that the orientability assumption on $M$ is removed.

Definition 3.1. Compact Vertical Cohomology

 We define the complex of forms with compact support in the vertical direction ${\Omega }_{cv}^{\ast }(E)$ as follows:

 a smooth $n$-form $\omega$ on $E$ is in ${\Omega }_{cv}^n(E)$ if and only if for every compact set $K$ in $M$, ${\pi }^{-1}(K)\cap \mathrm{Supp}\omega$ is compact.

 If $\omega \in {\Omega }_{cv}^n(E)$, then since $\mathrm{Supp}(\omega {|}_{E_x})\subset \mathrm{Supp}\omega \cap E_x$ is a closed subset of a compact set, then $\mathrm{Supp}(\omega {|}_{E_x})$ is compact. Note that there may be a difference between $\mathrm{Supp}(\omega {|}_{E_x})$ and $\mathrm{Supp}\omega \cap E_x$.

 Thus, although a form in ${\Omega }_{cv}^{\ast }(E)$ need not have compact support in $E$, its restriction to each fiber has compact support.

 The cohomology of this complex, denoted $H_{cv}^{\ast }(E)$, is called the cohomology of $E$ with compact support in the vertical direction, or compact vertical cohomology.

Definition 3.2. Integration along the fiber

 For any oriented vector bundle $\pi :E\to M$ of rank $n$. Locally, let $t_1,\cdots ,t_n$ be the fiber coordinate on $E_U$, a form in ${\Omega }_{cv}^{\ast }(E)$ is locally of type (I) which do not contain as a factor the $n$-form $d{t}_1\cdots d{t}_n$ and the type (II) which do, the map ${\pi }_{\ast }$ is defined by $$\begin{array}{rl}(I)& ({\pi }^{\ast }\phi )f(x,t_1,\cdots ,t_n)d{t}_{i_1}\cdots t_{i_r}\mapsto 0,r<n,\\ (II)& ({\pi }^{\ast }\phi )f(x,t_1,\cdots ,t_n)d{t}_1\cdots d{t}_n\mapsto \phi \cdot {\int }_{E_x}fd{t}_1\cdots d{t}_n,\end{array}$$where $f$ has compact support for each fixed $x$ in $M$ and $\phi$ is a form on $M$.

 The definition is independent of the choice of the oriented trivialization for $E$, hence we get a global form ${\pi }_{\ast }\omega$ on $M$.

Proposition 3.3. Integration along the fiber ${\pi }_{\ast }$ commutes with exterior differentiation $d$.

Proposition 3.4. Projection Formula

 (a) Let $\pi :E\to M$ be an oriented rank $n$ vector bundle, $\tau$ a form on $M$ and $\omega$ a form on $E$ with compact support along the fiber. Then $${\pi }_{\ast }(({\pi }^{\ast }\tau )\wedge \omega )=\tau \wedge {\pi }_{\ast }\omega .$$ (b) Suppose in addition that $M$ is oriented of dimension $m$, $\omega \in {\Omega }_{cv}^q(E)$, and $\tau \in {\Omega }^{m+n-q}(M)$. Then with the local product orientation on $E$ $${\int }_E({\pi }^{\ast }\tau )\wedge \omega ={\int }_M\tau \wedge {\pi }_{\ast }\omega .$$

Theorem 3.5. Thom Isomorphim

 If the vector bundle $\pi :E\to M$ over manifold $M$ of finite type is orientable, then $$H_{cv}^q(E)\cong H^{q-n}(M)$$where $n$ is the rank of $E$.

 The assumption of "finite type" is not necessary.

Definition 3.6. Thom class  Under the Thom isomorphism $\mathcal{T}:H^{\ast }(M)\to H^{\ast +n}(E)$, the iamge of $1$ in $H^0(M)$ determines a cohomology class $\Phi$ in $H_{cv}^n(E)$, called the Thom class of the oriented vector bundle $E$.

 Because ${\pi }_{\ast }\Phi =1$, by the projection formula $${\pi }_{\ast }({\pi }^{\ast }\omega \wedge \Phi )=\omega \wedge {\pi }_{\ast }\Phi =\omega .$$So the Thom isomorphism, which is inverse to ${\pi }_{\ast }$, is given by $$\mathcal{T}()={\pi }^{\ast }()\wedge \Phi .$$

Proposition 3.7. The Thom class $\Phi$ on a rank $n$ oriented vector bundle $E$ can be uniquely characterized as the cohomology class in $H_{cv}^n(E)$ which restricts to the generator of $H_c^n(F)$ on each fiber $F$.

Proposition 3.8. If $E$ and $F$ are two oriented vector bundles over a manifold $M$, and ${\pi }_1$ ${\pi }_2$ are the projections then the Thom class of $E\oplus F$ is $\Phi (E\oplus F)={\pi }_1^{\ast }\Phi (E)\wedge {\pi }_2^{\ast }\Phi (F)$.