Foliations and the Bott Vanishing Theorem

Qinchao/ 3月 8, 2012/ Index Theory, 数学笔记/ 0 comments

Let $M$ be a closed manifold and $TM$ its tangent vector bundle. Let $F\subset TM$ be a sub-vector bundle of $TM$. Define
F ~\text{is integrable} ~\Leftrightarrow ~~\Big(\forall X,Y\in \Gamma(F) \Big)~ [X,Y]\in \Gamma(F),
where $[\cdot,\cdot]$ is Lie bracket. Note the operator Lie bracket is only defined on tangent bundle. Because there is local one-parameter transforation group on $M$ for tangent bundle, then one can calculate derivative(Lie derivative) along the integral curve. (see[Bai, Shen. An Introduction Rie. Geo.]) Lie bracket does not necessarily exist for general vector bundle.

If such an integrable subbundle $F\subset TM$ exists on $M$, then we call $M$ a foliation foliated by $F$. Let $TM/F$ be the quotient vector bundle of $TM$ by $F$. Let $p_{i_1}(TM/F)$, $\cdots$, $p_{i_1}(TM/F)$ be $k$ Pontrjagin classes of $TM/F$.

The following theorem is called vanishing theorem. It is said that Bott found vanishing theorem all at once during classroom teaching.

定理 1. If $i_1+i_2+\cdots+i_k>(\dim M-\dim F)/2=\frac{1}{2} codim~F$, then
p_{i_1}(TM/F)\cdot p_{i_2}(TM/F)\cdots p_{i_k}(TM/F)=0 \quad \text{in} \quad H^{4(i_1+\cdots+i_k)}_{dR}(M,\R).

i) if $codim~F\geq \dim F$, then
4(i_1+i_2+\cdots+i_k)>2~ codim~F\geq \dim F+codim~F=\dim M.
So $H^{4(i_1+\cdots+i_k)}_{dR}(M,\R)={0}$. Thus in case of $\dim F\leq \frac{1}{2}\dim M$ this theorem is trivial.

ii) To obtain this theorem, we need a fit connection $\nabla^F$ on $F$ such that
p_{i_1}(TM/F,\nabla^F)\cdots p_{i_k}(TM/F,\nabla^F)=d\eta, \quad \eta \in \Omega^*(M).
Bott constructed a connection $\widetilde{\nabla}^F$ (Bott connection) fitted this condition such that
p_{i_1}(TM/F,\widetilde{\nabla}^F)\cdots p_{i_k}(TM/F,\widetilde{\nabla}^F)=0.

证明 . We take a Riemannian metric $g^{TM}$ on $TM$. Then $TM$ admits an orthogonal decomposition with respect to $g^{TM}$:
TM=F\oplus F^\bot.
Hence $TM/F$ can be identified with $F^{\bot}$.

Let $\nabla^{TM}$ be the Levi-Civita connection on $TM$ associated to $g^{TM}$. Let $p$, $p^\bot$ denote the orthogonal projection form $TM$ to $F$, $F^{\bot}$ respectively. Set
\nabla^F=p\nabla^{TM}p,\qquad \nabla^{F^{\bot}}=p^{\bot}\nabla^{TM}p^{\bot}.
One can verifies that $\nabla^F$, $\nabla^{F^{\bot}}$ are connection on $F$, $F^{\bot}$ respectively. And, they preserve the metric on $F$, $F^{\bot}$ induced from $g^{TM}$ respectively. (If want to know more propositions about $\nabla^F$, $\nabla^{F^{\bot}}$, you can read any book about Riemannian sub-manifold.) By using the connection $\nabla^{F^{\bot}}$, Bott constructed a new connection on $F^{\bot}$.
\begin{defn}[Bott connection]\label{def:1-13}
For any $X\in \Gamma(TM)$, $U\in \Gamma(F^\bot)$,

  1. [(i)] If $X\in \Gamma(F)$, we define
  2. [(ii)] If $X\in \Gamma(F^{\bot})$, set $\widetilde{\nabla}_X^{F^\bot}U=\nabla_X^{F^\bot}U$.
To prove ~$\widetilde{\nabla}^{F^\bot}$ is a connection on $F^\bot$. This is trivial by (ii), but maybe we need think the rationality about (i).
Let $\widetilde{R}^{F^\bot}$ denote the curvature of $\widetilde{\nabla}^{F^\bot}$. About the Bott connection we have a important lemma.
\begin{lem} \label{lem:1-14}
For any $X,~Y\in \Gamma(F)$, one has
Suppose we have proved this lemma.
Since $\widetilde{R}^{F^\bot}=(\widetilde{\nabla}^{F^\bot})^2\in \Omega^2(m, End(F^\bot))$, $\widetilde{R}^{F^\bot}$ can be written as
& & \\
& \omega_{ij} & \\
& &
, \quad \omega_{ij}\in \Omega^2(M).
By above lemma, for any $X,Y\in \Gamma(F)$ and any $1\leq i,j\leq \dim F^\bot$, we have $\omega_{ij}(X,Y)=0$.

Let $F^*$, $\left(F^{\bot}\right)^*$ denote the dual bundle of $F$, $F^{\bot}$ respectively. Hence $T^*M=F^*\oplus \left(F^{\bot}\right)^*$. Let $\left\{e_1^*,\cdots,e_s^*\right\}$, $\left\{ f_1^*,\cdots,f_t^*\right\}$ the local basis of $F^*$, $\left(F^{\bot}\right)^*$ respectively. So there exists $a_{mn}$, $b_{\alpha\beta}$, $c_{xy}$$\in C^{\infty}(M)$ such that
\omega_{ij}=\sum_{m,n}a_{mn}e_m^*\otimes e^*_n+\sum_{\alpha,\beta}b_{\alpha,\beta} e_\alpha^*\otimes f_\beta^*+\sum_{x,y}c_{xy}f_x^*\otimes f_y^*.
Thus $a_{mn}=0$, by $\omega_{ij}(X,Y)=0$ for any $X,Y\in \Gamma(F)$. Then every term in $\omega_{ij}$ almost has a factor which is a 1-form. From the definition of Pontrjagin form and i-th Pontrjagin form associated $\widetilde{\nabla}^{F^\bot}$, we know every term of $p_{i_1}(F^{\bot},\widetilde{\nabla}^{F^\bot})$ at lest has $2i_1$ factors in $\left(F^{\bot}\right)^*$. Then
p_{i_1}\left(F^{\bot}, \widetilde{\nabla}^{F^\bot}\right)\cdots p_{i_k}\left(F^{\bot},\widetilde{\nabla}^{F^\bot}\right)\in \Gamma\left(\Lambda^{2(i_1+\cdots+i_k)}\left( \left(F^{\bot}\right)^*\right) \right)\wedge\Omega^*(M).
Since $\dim F^{\bot}<2(i_1\cdots+i_k)$, one sees easily from above formula following formula holds
p_{i_1}\left(F^{\bot},\widetilde{\nabla}^F\right)\cdots p_{i_k}\left(F^{\bot},\widetilde{\nabla}^F\right)=0.
p_{i_1}\left(TM/F,\widetilde{\nabla}^F\right)\cdots p_{i_k}\left(TM/F,\widetilde{\nabla}^F\right)=0.
This completes the proof.

Now, the lemma1.14 has a proof not yet. Using following fact, one can easily obtain the proof,
[X,Y]\in \Gamma(F), \\
p^{\bot}\left[ X,p^{\bot}[Y,Z]\right]=p^{\bot}\big[ X,[Y,Z]-p[Y,Z]\big]=p^{\bot}\big[X,[Y,Z]\big],\\
\text{for any } X,Y\in \Gamma(F) ,\quad Z \in \Gamma(F^{\bot}).

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