Let $M$ be a closed manifold and $TM$ its tangent vector bundle. Let $F\\subset TM$ be a sub-vector bundle of $TM$. Define
\n\\[
\nF ~\\text{is integrable} ~\\Leftrightarrow ~~\\Big(\\forall X,Y\\in \\Gamma(F) \\Big)~ [X,Y]\\in \\Gamma(F),
\n\\]
\nwhere $[\\cdot,\\cdot]$ 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.<\/p>\n
If such an integrable subbundle $F\\subset TM$ exists on $M$, then we call $M$ a \\emph{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 $TM\/F$.<\/p>\n
The following theorem is called vanishing theorem. It is said that Bott found vanishing theorem all at once during classroom teaching. ii) To obtain this theorem, we need a fit connection $\\nabla^F$ on $F$ such that 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 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 Let $M$ be a closed manifold and $TM$ it …<\/p>\n
\n\\begin{thm}
\nIf $i_1+i_2+\\cdots+i_k>(\\dim M-\\dim F)\/2=\\frac{1}{2} codim~F$, then
\n\\begin{equation}
\n 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).
\n\\end{equation}
\n\\end{thm}
\nNote:<\/strong>
\ni) if $codim~F\\geq \\dim F$, then
\n\\[
\n 4(i_1+i_2+\\cdots+i_k)>2~ codim~F\\geq \\dim F+codim~F=\\dim M.
\n\\]
\nSo $H^{4(i_1+\\cdots+i_k)}_{dR}(M,\\R)={0}$. Thus in case of $\\dim F\\leq \\frac{1}{2}\\dim M$ this theorem is trivial.<\/p>\n
\n\\[
\n p_{i_1}(TM\/F,\\nabla^F)\\cdots p_{i_k}(TM\/F,\\nabla^F)=d\\eta, \\quad \\eta \\in \\Omega^*(M).
\n\\]
\nBott constructed a connection $\\widetilde{\\nabla}^F$ (Bott connection) fitted this condition such that
\n\\[
\n p_{i_1}(TM\/F,\\widetilde{\\nabla}^F)\\cdots p_{i_k}(TM\/F,\\widetilde{\\nabla}^F)=0.
\n\\]
\n\\begin{proof}
\nWe take a Riemannian metric $g^{TM}$ on $TM$. Then $TM$ admits an orthogonal decomposition with respect to $g^{TM}$:
\n\\[
\n TM=F\\oplus F^\\bot.
\n\\]
\nHence $TM\/F$ can be identified with $F^{\\bot}$.<\/p>\n
\n\\[
\n \\nabla^F=p\\nabla^{TM}p,\\qquad \\nabla^{F^{\\bot}}=p^{\\bot}\\nabla^{TM}p^{\\bot}.
\n\\]
\nOne 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 $F^{\\bot}$.
\n\\begin{defn}[\\emph{Bott connection}]\\label{def:1-13}
\nFor any $X\\in \\Gamma(TM)$, $U\\in \\Gamma(F^\\bot)$,
\n\\begin{enumerate}
\n \\item[(i)] If $X\\in \\Gamma(F)$, we define
\n \\[
\n \\widetilde{\\nabla}_X^{F^\\bot}U=p^\\bot[X,U];
\n \\]
\n \\item[(ii)] If $X\\in \\Gamma(F^{\\bot})$, set $\\widetilde{\\nabla}_X^{F^\\bot}U=\\nabla_X^{F^\\bot}U$.
\n\\end{enumerate}
\n\\end{defn}
\n\\begin{excs}
\nTo 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).
\n\\end{excs}
\nLet $\\widetilde{R}^{F^\\bot}$ denote the curvature of $\\widetilde{\\nabla}^{F^\\bot}$. About the Bott connection we have a important lemma.
\n\\begin{lem} \\label{lem:1-14}
\nFor any $X,~Y\\in \\Gamma(F)$, one has
\n\\[
\n \\widetilde{R}^{F^\\bot}(X,Y)=0.
\n\\]
\n\\end{lem}
\nSuppose we have proved this lemma.
\nSince $\\widetilde{R}^{F^\\bot}=(\\widetilde{\\nabla}^{F^\\bot})^2\\in \\Omega^2(m, End(F^\\bot))$, $\\widetilde{R}^{F^\\bot}$ can be written as
\n\\[
\n \\widetilde{R}^{F^\\bot}=\\left(\\begin{array}{ccc}
\n & & \\\\
\n & \\omega_{ij} & \\\\
\n & &
\n \\end{array}\\right)
\n , \\quad \\omega_{ij}\\in \\Omega^2(M).
\n\\]
\nBy above lemma, for any $X,Y\\in \\Gamma(F)$ and any $1\\leq i,j\\leq \\dim F^\\bot$, we have $\\omega_{ij}(X,Y)=0$.<\/p>\n
\n\\[
\n \\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^*.
\n\\]
\nThus $a_{mn}=0$, 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
\n\\[
\np_{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).
\n\\]
\nSince $\\dim F^{\\bot}<2(i_1\\cdots+i_k)$, one sees easily from above formula following formula holds
\n\\[
\n p_{i_1}\\left(F^{\\bot},\\widetilde{\\nabla}^F\\right)\\cdots p_{i_k}\\left(F^{\\bot},\\widetilde{\\nabla}^F\\right)=0.
\n\\]
\ni.e.
\n\\[
\n p_{i_1}\\left(TM\/F,\\widetilde{\\nabla}^F\\right)\\cdots p_{i_k}\\left(TM\/F,\\widetilde{\\nabla}^F\\right)=0.
\n\\]
\nThis completes the proof.
\n\\end{proof}
\nNow, the lemma1.14 has a proof not yet. Using following fact, one can easily obtain the proof,
\n\\begin{gather*}
\n[X,Y]\\in \\Gamma(F), \\\\
\np^{\\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],\\\\
\n\\text{for any } X,Y\\in \\Gamma(F) ,\\quad Z \\in \\Gamma(F^{\\bot}).
\n\\end{gather*}<\/p>\n","protected":false},"excerpt":{"rendered":"