2.1 $\unicode[STIX]{x1D706}$ -dual vector fields and $\unicode[STIX]{x1D706}$ -dual one-forms

Let $(M,\unicode[STIX]{x1D709})$ be a contact manifold and $\unicode[STIX]{x1D706}$ be a contact form with $\ker \unicode[STIX]{x1D706}=\unicode[STIX]{x1D709}$ . Consider its associated decomposition

(2.1) $$\begin{eqnarray}TM=\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\oplus \unicode[STIX]{x1D709}\end{eqnarray}$$

and denote by $\unicode[STIX]{x1D70B}=\unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}:TM\rightarrow \unicode[STIX]{x1D709}$ the associated projection. This decomposition canonically induces the corresponding dual decomposition

(2.2) $$\begin{eqnarray}T^{\ast }M=\unicode[STIX]{x1D709}^{\bot }\oplus (\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\})^{\bot }\end{eqnarray}$$

where $(\cdot )^{\bot }$ is the annihilator of $(\cdot )$ . This gives rise to a decomposition

(2.3) $$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FC}(X_{\unicode[STIX]{x1D706}})\,\unicode[STIX]{x1D706}+\unicode[STIX]{x1D6FC}\circ \unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}.\end{eqnarray}$$

Then we have the following general lemma whose proof immediately follows from (2.2).

Lemma 2.1. For any given one-form $\unicode[STIX]{x1D6FC}$ , there exists a unique $Y_{\unicode[STIX]{x1D6FC}}\in \unicode[STIX]{x1D709}$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}=Y_{\unicode[STIX]{x1D6FC}}\!\rfloor d\unicode[STIX]{x1D706}+\unicode[STIX]{x1D6FC}(X_{\unicode[STIX]{x1D706}})\unicode[STIX]{x1D706}.\end{eqnarray}$$

Definition 2.2. ( $\unicode[STIX]{x1D706}$ -Dual vector field and one-form)

Let $\unicode[STIX]{x1D706}$ be a given contact form of $(M,\unicode[STIX]{x1D709})$ . We define the $\unicode[STIX]{x1D706}$ -dual vector field of a one-form $\unicode[STIX]{x1D6FC}$ to be

$$\begin{eqnarray}\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC}):=Y_{\unicode[STIX]{x1D6FC}}+\unicode[STIX]{x1D6FC}(X_{\unicode[STIX]{x1D706}})\,X_{\unicode[STIX]{x1D706}}.\end{eqnarray}$$

Conversely for any given vector field $X$ we define its $\unicode[STIX]{x1D706}$ -dual one-form by

$$\begin{eqnarray}\sharp _{\unicode[STIX]{x1D706}}(X)=X\!\rfloor d\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}(X)\,\unicode[STIX]{x1D706}.\end{eqnarray}$$

For the simplicity of notation, we denote $\unicode[STIX]{x1D6FC}_{X}:=\sharp _{\unicode[STIX]{x1D706}}(X)$ . By definition, we have the identity

(2.4) $$\begin{eqnarray}\unicode[STIX]{x1D706}(\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC}))=\unicode[STIX]{x1D6FC}(X_{\unicode[STIX]{x1D706}}).\end{eqnarray}$$

The following proposition is immediate from the definitions of the dual vector field and the dual one-forms.

By definition, we have $\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D706})=X_{\unicode[STIX]{x1D706}}$ which corresponds to the $\unicode[STIX]{x1D706}$ -dual to the contact form $\unicode[STIX]{x1D706}$ itself for which $Y_{\unicode[STIX]{x1D706}}=0$ by definition. Obviously when an exact one-form $\unicode[STIX]{x1D6FC}$ is given, the choice of $h$ with $\unicode[STIX]{x1D6FC}=dh$ is unique modulo addition by a constant (on each connected component of $M$ ).

To equip readers with some feeling on the above decomposition which is not common in the literature, we now provide the coordinate expression of $\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})$ and $\unicode[STIX]{x1D6FC}_{X}$ in the Darboux chart $(q_{1},\ldots ,q_{n},p_{1},\ldots ,p_{n},z)$ with respect to the canonical one-form $\unicode[STIX]{x1D706}_{0}=dz-\sum _{i=1}^{n}p_{i}\,dq_{i}$ on $\mathbb{R}^{2n+1}$ or more generally on the one-jet bundle $J^{1}(N)$ of a smooth $n$ -manifold $N$ . However, this coordinate expression will not be used in the rest of the present paper.

We recall that for this contact form, the associated Reeb vector field is nothing but

$$\begin{eqnarray}X_{\unicode[STIX]{x1D706}_{0}}=\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z}.\end{eqnarray}$$

We start with the expression of $\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})$ for a given one-form

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FC}_{0}\,dz+\mathop{\sum }_{i=1}^{n}a_{i}\,dq_{i}+\mathop{\sum }_{j=1}^{n}b_{j}\,dp_{j}.\end{eqnarray}$$

We denote

$$\begin{eqnarray}\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})=v_{0}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z}+\mathop{\sum }_{i=1}^{n}v_{i;q}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}q_{i}}+\mathop{\sum }_{j=1}^{n}v_{j;p}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}p_{j}}.\end{eqnarray}$$

A direct computation using the defining equation of $\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})$ leads to

Proposition 2.4. Consider the standard contact structure

$$\begin{eqnarray}\unicode[STIX]{x1D706}=dz-\mathop{\sum }_{i=1}^{n}p_{i}\,dq_{i}\quad \text{on}~\mathbb{R}^{2n+1}.\end{eqnarray}$$

Then for the given one-form $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FC}_{0}\,dz+\sum _{i=1}^{n}a_{i}\,dq_{i}+\sum _{j=1}^{n}b_{j}\,dp_{j}$ ,

(2.5) $$\begin{eqnarray}\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})=\left(\unicode[STIX]{x1D6FC}_{0}+\mathop{\sum }_{k=1}^{n}p_{k}\,b_{k}\right)\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z}+\mathop{\sum }_{i=1}^{n}b_{i}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}q_{i}}+\mathop{\sum }_{j=1}^{n}(-a_{j}-p_{j}\,\unicode[STIX]{x1D6FC}_{0})\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}p_{j}}.\end{eqnarray}$$

Conversely, for the given

$$\begin{eqnarray}X=v_{0}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z}+\mathop{\sum }_{i=1}^{n}v_{i;q}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}q_{i}}+\mathop{\sum }_{j=1}^{n}v_{j;p}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}p_{j}},\end{eqnarray}$$

we obtain

(2.6) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FC}_{X} & = & \displaystyle \left(v_{0}-\mathop{\sum }_{j=1}^{n}p_{j}\,v_{j;q}\right)dz\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{i=1}^{n}\left(v_{i;p}+\left((v_{0}-\mathop{\sum }_{j=1}^{n}p_{j}\,v_{j;q})p_{i}\right)\right)dq_{i}+\mathop{\sum }_{j=1}^{n}v_{j;q}\,dp_{j}.\end{eqnarray}$$

Proof. Here we first recall the basic identity (2.4).

By definition, $\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})$ is determined by the equation

(2.7) $$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})\!\rfloor \mathop{\sum }_{i=1}^{n}dq_{i}\wedge dp_{i}+\unicode[STIX]{x1D706}(\flat _{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC}))\left(dz-\mathop{\sum }_{i=1}^{n}p_{i}\,dq_{i}\right)\end{eqnarray}$$

in the current case. A straightforward computation leads to the formula (2.5). Then (2.6) can be derived either by inverting this formula or by using the defining equation of $\unicode[STIX]{x1D6FC}_{X}$ , which is further reduced to

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{X}=X\!\rfloor d\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}(X)\,\unicode[STIX]{x1D706}=X\!\rfloor \mathop{\sum }_{i=1}^{n}\,dq_{i}\wedge dp_{i}+\biggl(dz-\mathop{\sum }_{j=1}^{n}p_{j}\,dq_{j}\biggr)(X)\,\unicode[STIX]{x1D706}.\end{eqnarray}$$

We omit the details of the computation. ◻

Example 2.5. Again consider the canonical one-form $dz-\sum _{i=1}^{n}p_{i}\,dq_{i}$ . Then we obtain the following coordinate expression as a special case of (2.5)

(2.8) $$\begin{eqnarray}\flat _{\unicode[STIX]{x1D706}}(dh)=\left(\frac{\unicode[STIX]{x2202}h}{\unicode[STIX]{x2202}z}+\mathop{\sum }_{i=1}^{n}p_{i}\frac{\unicode[STIX]{x2202}h}{\unicode[STIX]{x2202}p_{i}}\right)\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z}+\mathop{\sum }_{i=1}^{n}\frac{\unicode[STIX]{x2202}h}{\unicode[STIX]{x2202}p_{i}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}q_{i}}+\mathop{\sum }_{i=1}^{n}\left(-\frac{\unicode[STIX]{x2202}h}{\unicode[STIX]{x2202}q_{i}}-p_{i}\frac{\unicode[STIX]{x2202}h}{\unicode[STIX]{x2202}z}\right)\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}p_{i}}.\end{eqnarray}$$

2.2 Perturbation of contact forms of $(M,\unicode[STIX]{x1D709})$

In this section, we exploit the discussion on the $\unicode[STIX]{x1D706}$ -dual vector fields and express the Reeb vector fields $X_{f\unicode[STIX]{x1D706}}$ and the projection $\unicode[STIX]{x1D70B}_{f\unicode[STIX]{x1D706}}$ associated to the contact form $f\unicode[STIX]{x1D706}$ for a positive function $f>0$ , in terms of those associated to the given contact form $\unicode[STIX]{x1D706}$ and the $\unicode[STIX]{x1D706}$ -dual vector fields of  $df$ .

Recalling Lemma 2.1, we can write

$$\begin{eqnarray}dg=Y_{dg}\!\rfloor d\unicode[STIX]{x1D706}+dg(X_{\unicode[STIX]{x1D706}})\unicode[STIX]{x1D706}\end{eqnarray}$$

with $Y_{dg}\in \unicode[STIX]{x1D709}$ in a unique way for any smooth function  $g$ . Then by definition, we have $Y_{dg}=\unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}(\flat _{\unicode[STIX]{x1D706}}(dg))$ .

We first compute the following useful explicit formula for the associated Reeb vector fields $X_{f\unicode[STIX]{x1D706}}$ in terms of $X_{\unicode[STIX]{x1D706}}$ and  $Y_{dg}$ .

Proposition 2.6. (Perturbed Reeb vector field)

Denote

$$\begin{eqnarray}Y_{dg}=\unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}(\flat _{\unicode[STIX]{x1D706}}(dg))\end{eqnarray}$$

as above. Then we have

$$\begin{eqnarray}X_{f\unicode[STIX]{x1D706}}=\frac{1}{f}(X_{\unicode[STIX]{x1D706}}+Y_{dg}),\quad g=\log f.\end{eqnarray}$$

Proof. It turns out to be easier to consider $f\,X_{f\unicode[STIX]{x1D706}}$ which we compute below. First we have

(2.9) $$\begin{eqnarray}f\,X_{f\unicode[STIX]{x1D706}}=c\cdot X_{\unicode[STIX]{x1D706}}+\unicode[STIX]{x1D702}\end{eqnarray}$$

with respect to the splitting $TM=\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\oplus \unicode[STIX]{x1D709}$ for some constant $c$ and $\unicode[STIX]{x1D702}\in \unicode[STIX]{x1D709}$ . We evaluate

$$\begin{eqnarray}c=\unicode[STIX]{x1D706}(fX_{f\unicode[STIX]{x1D706}})=(f\unicode[STIX]{x1D706})(X_{f\unicode[STIX]{x1D706}})=1.\end{eqnarray}$$

It remains to derive the formula for $\unicode[STIX]{x1D702}$ . Using the formula

$$\begin{eqnarray}d(f\unicode[STIX]{x1D706})=f\,d\unicode[STIX]{x1D706}+df\wedge \unicode[STIX]{x1D706}\end{eqnarray}$$

and $\unicode[STIX]{x1D706}(\unicode[STIX]{x1D702})=0$ , we compute

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D702}\!\rfloor d\unicode[STIX]{x1D706} & = & \displaystyle (fX_{f\unicode[STIX]{x1D706}})\!\rfloor d\unicode[STIX]{x1D706}\nonumber\\ \displaystyle & = & \displaystyle X_{f\unicode[STIX]{x1D706}}\!\rfloor d(f\unicode[STIX]{x1D706})-X_{f\unicode[STIX]{x1D706}}\!\rfloor (df\wedge \unicode[STIX]{x1D706})\nonumber\\ \displaystyle & = & \displaystyle -X_{f\unicode[STIX]{x1D706}}\!\rfloor (df\wedge \unicode[STIX]{x1D706})\nonumber\\ \displaystyle & = & \displaystyle -X_{f\unicode[STIX]{x1D706}}(f)\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}(X_{f\unicode[STIX]{x1D706}})\,df\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{f}(X_{\unicode[STIX]{x1D706}}+\unicode[STIX]{x1D702})(f)\unicode[STIX]{x1D706}+\frac{1}{f}\unicode[STIX]{x1D706}(X_{\unicode[STIX]{x1D706}}+\unicode[STIX]{x1D702})\,df\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{f}X_{\unicode[STIX]{x1D706}}(f)\unicode[STIX]{x1D706}-\frac{1}{f}\unicode[STIX]{x1D702}(f)\unicode[STIX]{x1D706}+\frac{1}{f}\,df.\nonumber\end{eqnarray}$$

Take value of $X_{\unicode[STIX]{x1D706}}$ for both sides, we get $\unicode[STIX]{x1D702}(f)=0$ , and hence

$$\begin{eqnarray}\unicode[STIX]{x1D702}\!\rfloor d\unicode[STIX]{x1D706}=-\frac{1}{f}X_{\unicode[STIX]{x1D706}}(f)\unicode[STIX]{x1D706}+\frac{1}{f}\,df.\end{eqnarray}$$

Setting $g=\log f$ , we can rewrite this into

$$\begin{eqnarray}\unicode[STIX]{x1D702}\!\rfloor d\unicode[STIX]{x1D706}=-dg(X_{\unicode[STIX]{x1D706}})\unicode[STIX]{x1D706}+dg.\end{eqnarray}$$

In other words, we obtain

(2.10) $$\begin{eqnarray}dg=\unicode[STIX]{x1D702}\!\rfloor d\unicode[STIX]{x1D706}+dg(X_{\unicode[STIX]{x1D706}})\unicode[STIX]{x1D706}.\end{eqnarray}$$

Therefore, by Lemma 2.1, we have obtained $\unicode[STIX]{x1D702}=Y_{dg}$ . Substituting this into (2.9) and dividing it by  $f$ , we have finished the proof.◻

Next we compare the contact projections $\unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}$ with $\unicode[STIX]{x1D70B}_{f\unicode[STIX]{x1D706}}$ associated to $\unicode[STIX]{x1D706}$ and $f\unicode[STIX]{x1D706}$ , respectively.

Proposition 2.7. (Perturbed contact projection)

Let $(M,\unicode[STIX]{x1D709})$ be a contact manifold and let $\unicode[STIX]{x1D706}$ be a contact form that is, $\ker \unicode[STIX]{x1D706}=\unicode[STIX]{x1D709}$ . Let $f$ be a positive smooth function and $f\,\unicode[STIX]{x1D706}$ be its associated contact form. Denote by $\unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}$ and $\unicode[STIX]{x1D70B}_{f\,\unicode[STIX]{x1D706}}$ be their associated $\unicode[STIX]{x1D709}$ -projection. Then

(2.11) $$\begin{eqnarray}\unicode[STIX]{x1D70B}_{f\,\unicode[STIX]{x1D706}}(Z)=\unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}(Z)-\unicode[STIX]{x1D706}(Z)Y_{dg}\end{eqnarray}$$

for the function $g=\log f$ .

Proof. We compute

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70B}_{f\unicode[STIX]{x1D706}}(Z) & = & \displaystyle Z-f\unicode[STIX]{x1D706}(Z)X_{f\unicode[STIX]{x1D706}}=Z-\unicode[STIX]{x1D706}(Z)(fX_{f\unicode[STIX]{x1D706}})\nonumber\\ \displaystyle & = & \displaystyle Z-\unicode[STIX]{x1D706}(Z)X_{\unicode[STIX]{x1D706}}+(\unicode[STIX]{x1D706}(Z)X_{\unicode[STIX]{x1D706}}-\unicode[STIX]{x1D706}(Z)(fX_{f\unicode[STIX]{x1D706}}))\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}Z+\unicode[STIX]{x1D706}(Z)(X_{\unicode[STIX]{x1D706}}-fX_{f\unicode[STIX]{x1D706}})\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D706}}Z-\unicode[STIX]{x1D706}(Z)Y_{dg}.\nonumber\end{eqnarray}$$

This finishes the proof. ◻

2.3 Linearization formula for the perturbed contact form

We next study the relationship between the linearization of $\unicode[STIX]{x1D6F6}_{\unicode[STIX]{x1D706}}(z)={\dot{z}}-TX_{f\unicode[STIX]{x1D706}}(z)$ which we denote by

$$\begin{eqnarray}D^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6F6}(z)(Z)=\unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-T\unicode[STIX]{x1D6FB}_{Z}X_{f\,\unicode[STIX]{x1D706}}\end{eqnarray}$$

with respect to the triad connection of $(M,\unicode[STIX]{x1D706},J)$ (see [Reference Oh and WangOW2, Proposition 7.6]) for a given function  $f$ . Substituting

$$\begin{eqnarray}X_{f\,\unicode[STIX]{x1D706}}=\frac{1}{f}(X_{\unicode[STIX]{x1D706}}+Y_{dg})\end{eqnarray}$$

into this formula, we derive

Lemma 2.8. (Linearization)

Let $\unicode[STIX]{x1D6FB}$ be the triad connection of $(M,f\unicode[STIX]{x1D706},J)$ . Then for any vector field $Z$ along a Reeb orbit $z=(\unicode[STIX]{x1D6FE}(T\cdot ),o_{\unicode[STIX]{x1D6FE}(T\cdot )})$ ,

(2.12) $$\begin{eqnarray}\displaystyle D^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6F6}(z)(Z) & = & \displaystyle \unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-T\left(\frac{1}{f}\unicode[STIX]{x1D6FB}_{Z}X_{\unicode[STIX]{x1D706}}+Z[1/f]X_{\unicode[STIX]{x1D706}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,T\left(\frac{1}{f}\unicode[STIX]{x1D6FB}_{Z}Y_{dg}+Z[1/f]Y_{dg}\right)\end{eqnarray}$$

Proof. Let $\unicode[STIX]{x1D6FB}$ be the triad connection of $(M,f\unicode[STIX]{x1D706},J)$ . Then by definition its torsion $T$ satisfies the axiom $T(X_{\unicode[STIX]{x1D706}},Z)=0$ for any vector field $Z$ on $M$ (see [Reference Oh and WangOW1, Theorem 1]). Using this property, as in [Reference Oh and WangOW2, Section 7], we compute

$$\begin{eqnarray}\displaystyle D^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6F6}(z)(Z) & = & \displaystyle \unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-T\unicode[STIX]{x1D6FB}_{Z}X_{f\unicode[STIX]{x1D706}}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-T\unicode[STIX]{x1D6FB}_{Z}\left(\frac{1}{f}(X_{\unicode[STIX]{x1D706}}+Y_{dg})\right)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-TZ[1/f](X_{\unicode[STIX]{x1D706}}+Y_{dg})-\frac{T}{f}\unicode[STIX]{x1D6FB}_{Z}(X_{\unicode[STIX]{x1D706}}+Y_{dg})\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-T\left(\frac{1}{f}\unicode[STIX]{x1D6FB}_{Z}X_{\unicode[STIX]{x1D706}}+Z[1/f]X_{\unicode[STIX]{x1D706}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,T\left(\frac{1}{f}\unicode[STIX]{x1D6FB}_{Z}Y_{dg}+Z[1/f]Y_{dg}\right),\nonumber\end{eqnarray}$$

which finishes the proof. ◻

We note that when $f\equiv 1$ , the above formula reduces to the standard formula

$$\begin{eqnarray}D^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6F6}(z)(Z)=\unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-T\unicode[STIX]{x1D6FB}_{Z}X_{\unicode[STIX]{x1D706}}\end{eqnarray}$$

which is further reduced to

$$\begin{eqnarray}D^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6F6}(z)(Z)=\unicode[STIX]{x1D6FB}_{t}^{\unicode[STIX]{x1D70B}}Z-\frac{T}{2}({\mathcal{L}}_{X_{\unicode[STIX]{x1D706}}}J)JZ\end{eqnarray}$$

for any contact triad $(M,\unicode[STIX]{x1D706},J)$ . (See [Reference Oh and WangOW2, Section 7] for some discussion on this formula.)

3.1 Definition of Morse–Bott contact form

Let $(M,\unicode[STIX]{x1D709})$ be a contact manifold and $\unicode[STIX]{x1D706}$ be a contact form of $\unicode[STIX]{x1D709}$ . We would like to study the linearization of the equation ${\dot{x}}=X_{\unicode[STIX]{x1D706}}(x)$ along a closed Reeb orbit. Let $\unicode[STIX]{x1D6FE}$ be a closed Reeb orbit of period $T>0$ . In other words, $\unicode[STIX]{x1D6FE}:\mathbb{R}\rightarrow M$ is a solution of $\dot{\unicode[STIX]{x1D6FE}}=X_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FE})$ satisfying $\unicode[STIX]{x1D6FE}(t+T)=\unicode[STIX]{x1D6FE}(t)$ .

By definition, we can write $\unicode[STIX]{x1D6FE}(t)=\unicode[STIX]{x1D719}_{X_{\unicode[STIX]{x1D706}}}^{t}(\unicode[STIX]{x1D6FE}(0))$ for the Reeb flow $\unicode[STIX]{x1D719}^{t}=\unicode[STIX]{x1D719}_{X_{\unicode[STIX]{x1D706}}}^{t}$ of the Reeb vector field $X_{\unicode[STIX]{x1D706}}$ . In particular $p=\unicode[STIX]{x1D6FE}(0)$ is a fixed point of the diffeomorphism $\unicode[STIX]{x1D719}^{T}$ when $\unicode[STIX]{x1D6FE}$ is a closed Reeb orbit of period  $T$ . Since ${\mathcal{L}}_{X_{\unicode[STIX]{x1D706}}}\unicode[STIX]{x1D706}=0$ , the contact diffeomorphism $\unicode[STIX]{x1D719}^{T}$ canonically induces the isomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D6F9}_{z;p}:=d\unicode[STIX]{x1D719}^{T}(p)|_{\unicode[STIX]{x1D709}_{p}}:\unicode[STIX]{x1D709}_{p}\rightarrow \unicode[STIX]{x1D709}_{p}\end{eqnarray}$$

which is the linearized Poincaré return map $\unicode[STIX]{x1D719}^{T}$ restricted to $\unicode[STIX]{x1D709}_{p}$ via the splitting

$$\begin{eqnarray}T_{p}M=\unicode[STIX]{x1D709}_{p}\oplus \mathbb{R}\cdot \{X_{\unicode[STIX]{x1D706}}(p)\}.\end{eqnarray}$$

Denote $\operatorname{Cont}(M,\unicode[STIX]{x1D709})$ the set of contact one-forms with respect to the contact structure $\unicode[STIX]{x1D709}$ and ${\mathcal{L}}(M)=C^{\infty }(S^{1},M)$ the space of loops $z:S^{1}=\mathbb{R}/\mathbb{Z}\rightarrow M$ . We would like to consider the bundle ${\mathcal{L}}$ over the product $(0,\infty )\times {\mathcal{L}}(M)\times \operatorname{Cont}(M,\unicode[STIX]{x1D709})$ whose fiber at $(T,z,\unicode[STIX]{x1D706})$ is given by the space $C^{\infty }(z^{\ast }TM)$ of sections of the bundle $z^{\ast }TM\rightarrow S^{1}$ . We consider the assignment

$$\begin{eqnarray}\unicode[STIX]{x1D6F6}:(T,z,\unicode[STIX]{x1D706})\mapsto {\dot{z}}-T\,X_{\unicode[STIX]{x1D706}}(z)\end{eqnarray}$$

which is a section. Then $(T,z,\unicode[STIX]{x1D706})\in \unicode[STIX]{x1D6F6}^{-1}(0):=\mathfrak{Reeb}(M,\unicode[STIX]{x1D709})$ if and only if there exists some closed Reeb orbit $\unicode[STIX]{x1D6FE}:\mathbb{R}\rightarrow M$ with period  $T$ , such that $z(\cdot )=\unicode[STIX]{x1D6FE}(T\cdot )$ .

We first start with the standard notion, applied to the set of closed Reeb orbits, of Morse–Bott critical manifolds introduced by Bott in [Reference BottBot]:

The following is an immediate consequence of this definition.

Proof. Let $(T_{0},z_{0})$ and $(T_{1},z_{1})$ be two elements in the same connected component of $\mathfrak{Reeb}(M,\unicode[STIX]{x1D706})$ . We connect them by a smooth one-parameter family $(T_{s},z_{s})$ for $0\leqslant s\leqslant 1$ . Since $\dot{z_{s}}=T_{s}X_{\unicode[STIX]{x1D706}}(z_{s})$ and then $T_{s}=\int _{S^{1}}z_{s}^{\ast }\unicode[STIX]{x1D706}$ , it is enough to prove

$$\begin{eqnarray}\frac{d}{ds}T_{s}=\frac{d}{ds}\int _{S^{1}}z_{s}^{\ast }\unicode[STIX]{x1D706}\equiv 0.\end{eqnarray}$$

We compute

$$\begin{eqnarray}\displaystyle \frac{d}{ds}z_{s}^{\ast }\unicode[STIX]{x1D706} & = & \displaystyle z_{s}^{\ast }(d(z_{s}^{\prime }\!\rfloor \unicode[STIX]{x1D706})+z_{s}^{\prime }\!\rfloor d\unicode[STIX]{x1D706})\nonumber\\ \displaystyle & = & \displaystyle d(z_{s}^{\ast }(z_{s}^{\prime }\!\rfloor \unicode[STIX]{x1D706}))+z_{s}^{\ast }(z_{s}^{\prime }\!\rfloor d\unicode[STIX]{x1D706})\nonumber\\ \displaystyle & = & \displaystyle d(z_{s}^{\ast }(z_{s}^{\prime }\!\rfloor \unicode[STIX]{x1D706})).\nonumber\end{eqnarray}$$

Here we use $z_{s}^{\prime }$ to denote the derivative with respect to $s$ . The last equality comes from the fact that $\dot{z_{s}}$ is parallel to $X_{\unicode[STIX]{x1D706}}$ . Therefore, we obtain by Stokes formula that

$$\begin{eqnarray}\frac{d}{ds}\int _{S^{1}}z_{s}^{\ast }\unicode[STIX]{x1D706}=\int _{S^{1}}\,d(z_{s}^{\ast }(z_{s}^{\prime }\!\rfloor \unicode[STIX]{x1D706}))=0\end{eqnarray}$$

and finish the proof. ◻

Now we prove

However, $Q$ may not be embedded in general along the locus of multiple orbits.

Partially following [Reference BourgeoisBou], from now on in the rest of the paper, we always assume $Q$ is embedded and compact. Denote $\unicode[STIX]{x1D714}_{Q}:=i_{Q}^{\ast }d\unicode[STIX]{x1D706}$ and

$$\begin{eqnarray}\ker \unicode[STIX]{x1D714}_{Q}=\{e\in TQ\mid \unicode[STIX]{x1D714}(e,e^{\prime })=0~\text{for any}~e^{\prime }\in TQ\}.\end{eqnarray}$$

We warn readers that the Morse–Bott condition does not imply that the form $\unicode[STIX]{x1D714}_{Q}$ has constant rank, and hence the dimension of this kernel may vary pointwise on  $Q$ . However, if it does, $\ker \unicode[STIX]{x1D714}_{Q}$ defines an integrable distribution and so defines a foliation, denoted by ${\mathcal{F}}$ , on  $Q$ . Since $Q$ is also foliated by closed Reeb orbits and ${\mathcal{L}}_{X_{\unicode[STIX]{x1D706}}}d\unicode[STIX]{x1D706}=0$ , it follows that ${\mathcal{L}}_{X_{\unicode[STIX]{x1D706}}}\unicode[STIX]{x1D714}_{Q}=0$ when we restrict everything on  $Q$ . Therefore, each leaf of the foliation consists of closed Reeb orbits. Motivated by this, we also impose the condition that the two-form $\unicode[STIX]{x1D714}_{Q}$ has constant rank.

3.2 Structure of the locus of closed Reeb orbits

Let $\unicode[STIX]{x1D706}$ be a Morse–Bott contact form of $(M,\unicode[STIX]{x1D709})$ and $X_{\unicode[STIX]{x1D706}}$ its Reeb vector field. Let $Q$ be as in Definition 3.5. In general, $Q$ carries a natural locally free $S^{1}$ -action induced by the Reeb flow $\unicode[STIX]{x1D719}_{X_{\unicode[STIX]{x1D706}}}^{T}$ (see Lemma 3.4). Then by the general theory of compact Lie group actions (see [Reference HelgasonHe] for example), the action has a finite number of orbit types which have their minimal periods, $T/m$ for some integer $m\geqslant 1$ . The set of orbit spaces $Q/S^{1}$ carries natural orbifold structure at each multiple orbit with its isotropy group $\mathbb{Z}/m$ for some  $m$ .

Now we fix a connected component of $Q$ and just denote it by $Q$ itself. Denote $\unicode[STIX]{x1D703}=i_{Q}^{\ast }\unicode[STIX]{x1D706}$ . We note that the two-form $\unicode[STIX]{x1D714}_{Q}=d\unicode[STIX]{x1D703}$ is assumed to have constant rank on $Q$ by the definition of Morse–Bott contact form in Definition 3.5.

The following is an immediate consequence of the definition but exhibits a particularity of the null foliation of the presymplectic manifold $(Q,\unicode[STIX]{x1D714}_{Q})$ arising from the locus of closed Reeb orbits. We note that $\ker \unicode[STIX]{x1D714}_{Q}$ carries a natural splitting

$$\begin{eqnarray}\ker \unicode[STIX]{x1D714}_{Q}=\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\oplus (\ker \unicode[STIX]{x1D714}_{Q}\cap \unicode[STIX]{x1D709}|_{Q}).\end{eqnarray}$$

Therefore, $\ker \unicode[STIX]{x1D714}_{Q}\cap \unicode[STIX]{x1D709}|_{Q}$ defines another foliation ${\mathcal{N}}$ on $Q$ , and hence

(3.1) $$\begin{eqnarray}T{\mathcal{F}}=\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\oplus T{\mathcal{N}}.\end{eqnarray}$$

Note that this splitting is $S^{1}$ -invariant.

We now recall some basic properties of presymplectic manifold [Reference GotayG] and its canonical neighborhood theorem. Fix an $S^{1}$ -equivariant splitting of $TQ$

(3.2) $$\begin{eqnarray}TQ=T{\mathcal{F}}\oplus G=\mathbb{R}X_{\unicode[STIX]{x1D706}}\oplus T{\mathcal{N}}\oplus G\end{eqnarray}$$

by choosing an $S^{1}$ -invariant complementary subbundle $G\subset TQ$ . This splitting is not unique but its choice will not matter for the coming discussions. The null foliation carries a natural transverse symplectic form in general [Reference GotayG]. Since the distribution $T{\mathcal{F}}\subset TQ$ is preserved by Reeb flow, it generates the $S^{1}$ -action thereon in the current context. We denote by

$$\begin{eqnarray}p_{T{\mathcal{N}};G}:TQ\rightarrow T{\mathcal{N}},\qquad p_{G;G}:TQ\rightarrow G\end{eqnarray}$$

the projection to $T{\mathcal{N}}$ and to $G$ , respectively, with respect to the splitting (3.2). We denote by $T^{\ast }{\mathcal{N}}\rightarrow Q$ the associated foliation cotangent bundle, that is, the dual bundle of $T{\mathcal{N}}$ .

We now consider the isomorphism

(3.3) $$\begin{eqnarray}\widetilde{d\unicode[STIX]{x1D706}|_{\unicode[STIX]{x1D709}}}:\unicode[STIX]{x1D709}\rightarrow \unicode[STIX]{x1D709}^{\ast }\end{eqnarray}$$

and fix a splitting $T_{Q}M=TQ\oplus N_{Q}M$ with $T_{Q}M=TM|_{Q}$ so that $N_{Q}M\subset \unicode[STIX]{x1D709}_{Q}$ : this is possible since $\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\subset TQ$ . We can also choose the splitting so that it is $S^{1}$ -equivariant. (See the proof of Proposition 3.9 below.)

This leads to the further splitting

(3.4) $$\begin{eqnarray}\unicode[STIX]{x1D709}|_{Q}=T{\mathcal{N}}\oplus G\oplus N_{Q}M\end{eqnarray}$$

combined with (3.2), which in turn leads to

(3.5) $$\begin{eqnarray}\unicode[STIX]{x1D709}^{\ast }|_{Q}=(G\oplus N_{Q}M)^{\bot }\oplus (T{\mathcal{N}})^{\bot }\end{eqnarray}$$

where $(\cdot )^{\bot }$ denotes the annihilator of $(\cdot )$ . In particular, it induces an isomorphism

(3.6) $$\begin{eqnarray}T^{\ast }{\mathcal{N}}\cong (G\oplus N_{Q}M)^{\bot }\subset \unicode[STIX]{x1D709}^{\ast }|_{Q}.\end{eqnarray}$$

Now we consider the embedding $T^{\ast }{\mathcal{N}}\rightarrow \unicode[STIX]{x1D709}$ defined by the inverse of (3.3), which we denote by

(3.7) $$\begin{eqnarray}(T{\mathcal{N}})^{\#_{d\unicode[STIX]{x1D706}}}=(\widetilde{d\unicode[STIX]{x1D706}})^{-1}(T^{\ast }{\mathcal{N}}).\end{eqnarray}$$

Next we consider the symplectic normal bundle $(TQ)^{d\unicode[STIX]{x1D706}}\subset T_{Q}M$ defined by

(3.8) $$\begin{eqnarray}(TQ)^{d\unicode[STIX]{x1D706}}=\{v\in T_{q}M\mid d\unicode[STIX]{x1D706}(v,w)=0,\forall w\in T_{q}Q\}.\end{eqnarray}$$

We define a vector bundle

(3.9) $$\begin{eqnarray}E=(TQ)^{d\unicode[STIX]{x1D706}}/T{\mathcal{F}},\end{eqnarray}$$

and then have the natural embedding

(3.10) $$\begin{eqnarray}E=(TQ)^{d\unicode[STIX]{x1D706}}/T{\mathcal{F}}{\hookrightarrow}T_{Q}M/TQ=N_{Q}M\end{eqnarray}$$

induced by the inclusion map $(TQ)^{d\unicode[STIX]{x1D706}}{\hookrightarrow}T_{Q}M$ . The following is straightforward to check.

We now consider the exact sequence

$$\begin{eqnarray}0\rightarrow E\rightarrow N_{Q}M\rightarrow N_{Q}M/E\rightarrow 0\end{eqnarray}$$

induced by (3.10). The sequence is $S^{1}$ -equivariant with respect to the natural $S^{1}$ -action thereon. We have the canonical isomorphism

$$\begin{eqnarray}N_{Q}M/E\cong \frac{T_{Q}M}{TQ+(TQ)^{d\unicode[STIX]{x1D706}}}\end{eqnarray}$$

which is $S^{1}$ -equivariant. Then we have an $S^{1}$ -equivariant splitting

(3.11) $$\begin{eqnarray}T_{Q}M=(TQ+(TQ)^{d\unicode[STIX]{x1D706}})\oplus (T{\mathcal{N}})^{\#_{d\unicode[STIX]{x1D706}}}\end{eqnarray}$$

where $(T{\mathcal{N}})^{\#_{d\unicode[STIX]{x1D706}}}$ is the $d\unicode[STIX]{x1D706}$ -dual (3.7). This also induces an embedding

(3.12) $$\begin{eqnarray}T^{\ast }{\mathcal{N}}\rightarrow (T{\mathcal{N}})^{\#_{d\unicode[STIX]{x1D706}}}{\hookrightarrow}T_{Q}M\rightarrow N_{Q}M\end{eqnarray}$$

which is also $S^{1}$ -equivariant.

We now denote $F:=T^{\ast }{\mathcal{N}}\oplus E\rightarrow Q$ . The following proposition provides a local model of the neighborhood of $Q\subset M$ .

Identifying a neighborhood of $Q\subset M$ with a neighborhood of the zero section of $F$ and pulling back the contact form $\unicode[STIX]{x1D706}$ to  $F$ , we may assume that our contact form $\unicode[STIX]{x1D706}$ is defined on a neighborhood of $o_{F}\subset F$ . We also identify with $T^{\ast }{\mathcal{N}}$ and $E$ as their images in  $N_{Q}M$ .

Proof. The action of $S^{1}$ on $Q$ by $t\cdot q=\unicode[STIX]{x1D719}^{t}(q)$ canonically induces a $S^{1}$ action on $T_{Q}M$ by $t\cdot v=(d\unicode[STIX]{x1D719}^{t})(v)$ , for $v\in T_{Q}M$ . Hence it gives rise to the following identity

(3.13) $$\begin{eqnarray}t^{\ast }d\unicode[STIX]{x1D706}=d\unicode[STIX]{x1D706}\end{eqnarray}$$

since the Reeb flow preserves $\unicode[STIX]{x1D706}$ . We first show it is well-defined on $E\rightarrow Q$ , that is, if $v\in (T_{q}Q)^{d\unicode[STIX]{x1D706}}$ , then $t\cdot v\in (T_{t\cdot q}Q)^{d\unicode[STIX]{x1D706}}$ . In fact, by using (3.13), for $w\in T_{t\cdot q}Q$ ,

$$\begin{eqnarray}d\unicode[STIX]{x1D706}(t\cdot v,w)=((\unicode[STIX]{x1D719}^{t})^{\ast }d\unicode[STIX]{x1D706})(v,(d\unicode[STIX]{x1D719}^{t})^{-1}(w))=d\unicode[STIX]{x1D706}(v,(d\unicode[STIX]{x1D719}^{t})^{-1}(w)).\end{eqnarray}$$

This vanishes, since $Q$ consists of closed Reeb orbits and thus $d\unicode[STIX]{x1D719}^{t}$ preserves  $TQ$ .

Secondly, the same identity (3.13) further indicates that this $S^{1}$ action preserves $\unicode[STIX]{x1D6FA}$ on fibers, that is, $t^{\ast }\unicode[STIX]{x1D6FA}=\unicode[STIX]{x1D6FA}$ , and we are done with the proof of this proposition.◻

Summarizing the above discussion, we have concluded that the base $Q$ is decorated by the one-form $\unicode[STIX]{x1D703}:=i_{Q}^{\ast }\unicode[STIX]{x1D706}$ on the base $Q$ and the bundle $E$ is decorated by the fiberwise symplectic 2-form  $\unicode[STIX]{x1D6FA}$ . They satisfy the following additional properties:

1. (1) $Q=o_{F}$ carries an $S^{1}$ -action which is locally free. In particular $Q/S^{1}$ is a smooth orbifold;

2. (2) the one-form $\unicode[STIX]{x1D703}$ is $S^{1}$ -invariant, and $d\unicode[STIX]{x1D703}$ is a presymplectic form;

3. (3) the bundle $E$ carries an $S^{1}$ -action that preserves the fiberwise 2-form $\unicode[STIX]{x1D6FA}$ and hence induces a $S^{1}$ -invariant symplectic vector bundle structure on  $E$ ;

4. (4) the bundle $F=T^{\ast }{\mathcal{N}}\oplus E\rightarrow Q$ carries the direct sum $S^{1}$ -equivariant vector bundle structure compatible to the $S^{1}$ -action on  $Q$ .

We summarize the above discussions into the following theorem.

We say that $Q$ is of prequantization type if the rank of $d\unicode[STIX]{x1D706}|_{Q}$ is maximal and is of isotropic type if the rank of $d\unicode[STIX]{x1D706}|_{Q}$ is zero. The general case will be a mixture of the two.

4.1 The $S^{1}$ -invariant precontact manifold $(Q,\unicode[STIX]{x1D703})$

First, we consider the pair $(Q,\unicode[STIX]{x1D703})$ such that $Q$ carries a nontrivial $S^{1}$ -action preserving the one-form  $\unicode[STIX]{x1D703}$ . After taking the quotient of $S^{1}$ by some finite subgroup, we may assume that the action is effective. We also assume that the action is locally free. Then by the general theory of group actions of compact Lie group (see [Reference HelgasonHe] for example), the action is free on a dense open subset and has only a finite number of different types of isotropy groups. In particular the quotient $P:=Q/S^{1}$ becomes a presymplectic orbifold with a finite number of different types of orbifold points. We denote by $Y$ the vector field generating the $S^{1}$ -action, that is, the $S^{1}$ -action is generated by its flows.

We require that the circle action preserves $\unicode[STIX]{x1D703}$ , that is, ${\mathcal{L}}_{Y}\unicode[STIX]{x1D703}=0$ . Since the action is locally free and free on a dense open subset of $Q$ , we can normalize the action so that

(4.1) $$\begin{eqnarray}\unicode[STIX]{x1D703}(Y)\equiv 1.\end{eqnarray}$$

We denote this normalized vector field by $X_{\unicode[STIX]{x1D703}}$ . We would like to emphasize that $\unicode[STIX]{x1D703}$ may not be a contact form but can be regarded as the connection form of the circle bundle $S^{1}\rightarrow Q\rightarrow P$ over the orbifold $P$ in general. Although $P$ may carry nonempty set of orbifold points, the connection form $\unicode[STIX]{x1D703}$ is assumed to be smooth on  $Q$ .

Similarly as in Lemma 3.7, we also require the presence of $S^{1}$ -invariant splitting

$$\begin{eqnarray}\ker d\unicode[STIX]{x1D703}=\mathbb{R}\{X_{\unicode[STIX]{x1D703}}\}\oplus H\end{eqnarray}$$

such that the subbundle $H$ is also integrable.

With these terminologies introduced, we can rephrase Theorem 3.11 as follows.

Here we use the fact that there exists a canonical embedding

$$\begin{eqnarray}E=(TQ)^{d\unicode[STIX]{x1D706}}/\!\ker d\unicode[STIX]{x1D703}{\hookrightarrow}T_{Q}M/TQ=N_{Q}M\end{eqnarray}$$

and $d\unicode[STIX]{x1D706}|_{(TQ)^{d\unicode[STIX]{x1D706}}}$ canonically induces a bilinear form $[d\unicode[STIX]{x1D706}]_{E}$ on $E=(TQ)^{d\unicode[STIX]{x1D706}}/\ker di_{Q}^{\ast }\unicode[STIX]{x1D706}$ by symplectic reduction.

As before, we denote by ${\mathcal{F}}$ and ${\mathcal{N}}$ the associated foliations on $Q$ , and decompose

$$\begin{eqnarray}T{\mathcal{F}}=\mathbb{R}\{X_{\unicode[STIX]{x1D703}}\}\oplus T{\mathcal{N}}.\end{eqnarray}$$

We define a one-form $\unicode[STIX]{x1D6E9}_{G}$ on $T^{\ast }{\mathcal{N}}$ as follows. For a tangent $\unicode[STIX]{x1D709}\in T_{\unicode[STIX]{x1D6FC}}(T^{\ast }{\mathcal{N}})$ , define

(4.3) $$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{G}(\unicode[STIX]{x1D709}):=\unicode[STIX]{x1D6FC}(p_{T{\mathcal{N}};G}\,d\unicode[STIX]{x1D70B}_{(T^{\ast }{\mathcal{N}})}(\unicode[STIX]{x1D709}))\end{eqnarray}$$

using the splitting

$$\begin{eqnarray}TQ=\mathbb{R}\{X_{\unicode[STIX]{x1D703}}\}\oplus T{\mathcal{N}}\oplus G.\end{eqnarray}$$

By definition, it follows $\unicode[STIX]{x1D6E9}_{G}|_{VT(T^{\ast }{\mathcal{N}})}\equiv 0$ and $d\unicode[STIX]{x1D6E9}_{G}(\unicode[STIX]{x1D6FC})$ is nondegenerate on

$$\begin{eqnarray}\widetilde{T_{q}{\mathcal{N}}}\oplus VT_{\unicode[STIX]{x1D6FC}}T^{\ast }{\mathcal{N}}\cong T_{q}{\mathcal{N}}\oplus T_{q}^{\ast }{\mathcal{N}}\end{eqnarray}$$

which becomes the canonical pairing defined on $T_{q}{\mathcal{N}}\oplus T_{q}^{\ast }{\mathcal{N}}$ under the identification.

4.2 The bundle $E$

We next examine the structure of the $S^{1}$ -equivariant symplectic vector bundle $(E,\unicode[STIX]{x1D6FA})$ .

We denote by $\vec{R}$ the radial vector field which generates the family of radial multiplication

$$\begin{eqnarray}(c,e)\mapsto c\,e.\end{eqnarray}$$

This vector field is invariant under the given $S^{1}$ -action on $E$ , and vanishes on the zero section. By its definition, $d\unicode[STIX]{x1D70B}(\vec{R})=0$ , that is, $\vec{R}$ is in the vertical distribution, denoted by  $VTE$ , of  $TE$ .

Denote the canonical isomorphism $V_{e}TE\cong E_{\unicode[STIX]{x1D70B}(e)}$ by $I_{e;\unicode[STIX]{x1D70B}(e)}$ . It obviously intertwines the scalar multiplication, that is,

$$\begin{eqnarray}I_{e;\unicode[STIX]{x1D70B}(e)}(\unicode[STIX]{x1D707}\,\unicode[STIX]{x1D709})=\unicode[STIX]{x1D707}I_{e;\unicode[STIX]{x1D70B}(e)}(\unicode[STIX]{x1D709})\end{eqnarray}$$

for a scalar $\unicode[STIX]{x1D707}$ . It also satisfies the following identity (4.4) with respect to the derivative of the fiberwise scalar multiplication map $R_{c}:E\rightarrow E$ .

Lemma 4.4. Let $\unicode[STIX]{x1D709}\in V_{e}TE$ . Then

(4.4) $$\begin{eqnarray}I_{c\,e;\unicode[STIX]{x1D70B}(c\,e)}(dR_{c}(\unicode[STIX]{x1D709}))=c\,I_{e;\unicode[STIX]{x1D70B}(e)}(\unicode[STIX]{x1D709})\end{eqnarray}$$

on $E_{\unicode[STIX]{x1D70B}(c\,e)}=E_{\unicode[STIX]{x1D70B}(e)}$ for any constant $c$ .

Proof. We compute

$$\begin{eqnarray}\displaystyle I_{c\,e;\unicode[STIX]{x1D70B}(c\,e)}(dR_{c}(\unicode[STIX]{x1D709})) & = & \displaystyle \left.I_{c\,e;\unicode[STIX]{x1D70B}(c\,e)}\left(\frac{d}{ds}\right|_{s=0}c(e+s\unicode[STIX]{x1D709})\right)\nonumber\\ \displaystyle & = & \displaystyle I_{c\,e;\unicode[STIX]{x1D70B}(c\,e)}(R_{c}(\unicode[STIX]{x1D709}))=c\,I_{e;\unicode[STIX]{x1D70B}(e)}(\unicode[STIX]{x1D709})\nonumber\end{eqnarray}$$

which finishes the proof. ◻

We then define the fiberwise two-form $\unicode[STIX]{x1D6FA}^{v}$ on $VTE\rightarrow E$ by

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{e}^{v}(\unicode[STIX]{x1D709}_{1},\unicode[STIX]{x1D709}_{2})=\unicode[STIX]{x1D6FA}_{\unicode[STIX]{x1D70B}_{F}(e)}(I_{e;\unicode[STIX]{x1D70B}(e)}(\unicode[STIX]{x1D709}_{1}),I_{e;\unicode[STIX]{x1D70B}(e)}(\unicode[STIX]{x1D709}_{2}))\end{eqnarray}$$

for $\unicode[STIX]{x1D709}_{1},\unicode[STIX]{x1D709}_{2}\in V_{e}TE$ , and one-form $\vec{R}\!\rfloor \unicode[STIX]{x1D6FA}^{v}$ , respectively.

Now we introduce an $S^{1}$ -invariant symplectic connection on $E$ and choose the splitting

$$\begin{eqnarray}TE=HTE\oplus VTE.\end{eqnarray}$$

Existence of such an invariant connection follows for example, by averaging over the compact group  $S^{1}$ . We denote by $\widetilde{\unicode[STIX]{x1D6FA}}$ the resulting two-form on  $E$ . We extend the fiberwise form $\unicode[STIX]{x1D6FA}$ of $E$ into the differential two-form $\widetilde{\unicode[STIX]{x1D6FA}}$ on $E$ by setting

$$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D6FA}}_{e}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D701})=\unicode[STIX]{x1D6FA}_{e}^{v}(\unicode[STIX]{x1D709}^{v},\unicode[STIX]{x1D701}^{v}).\end{eqnarray}$$

Denote by $\vec{R}$ the radial vector field of $E\rightarrow Q$ and consider the one-form

(4.5) $$\begin{eqnarray}\vec{R}\!\rfloor \widetilde{\unicode[STIX]{x1D6FA}}\end{eqnarray}$$

which is invariant under the action of $S^{1}$ on $E$ .

Remark 4.5. Suppose $d\unicode[STIX]{x1D706}_{E}(\cdot ,J_{E}\cdot )=:g_{E;J_{E}}$ defines a Hermitian vector bundle $(\unicode[STIX]{x1D709}_{E},g_{E,J},J_{E})$ . Then we can write the radial vector field considered in the previous section as

$$\begin{eqnarray}\vec{R}(e)=\mathop{\sum }_{i=1}^{k}r_{i}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}r_{i}}\end{eqnarray}$$

where $(r_{1},\ldots ,r_{k})$ is the coordinates of $e$ for a local frame $\{e_{1},\ldots ,e_{k}\}$ of the vector bundle  $E$ . By definition, we have

(4.6) $$\begin{eqnarray}I_{e;\unicode[STIX]{x1D70B}_{E}(e)}(\vec{R}(e))=e.\end{eqnarray}$$

Obviously the right-hand side expression does not depend on the choice of local frames. Let $(E,\unicode[STIX]{x1D6FA},J_{E})$ be a Hermitian vector bundle and define $|e|^{2}=g_{F}(e,e)$ . Motivated by the terminology used in [Reference Bott and TuBT], we call the one-form

(4.7) $$\begin{eqnarray}\unicode[STIX]{x1D713}=\unicode[STIX]{x1D713}_{\unicode[STIX]{x1D6FA}}=\left.\frac{1}{r}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}r}\right\rfloor \unicode[STIX]{x1D6FA}^{v}\end{eqnarray}$$

the global angular form for the Hermitian vector bundle $(E,\unicode[STIX]{x1D6FA},J_{E})$ . Note that $\unicode[STIX]{x1D713}$ is defined only on $E\setminus o_{E}$ although $\unicode[STIX]{x1D6FA}$ is globally defined.

We state the following lemma.

Proof. Notice that $\widetilde{\unicode[STIX]{x1D6FA}}$ is compatible with $\unicode[STIX]{x1D6FA}$ in the sense of symplectic fibration and the symplectic vector bundle connection is nothing but the Ehresmann connection induced by $\widetilde{\unicode[STIX]{x1D6FA}}$ , which is a symplectic connection now. Since $\vec{R}$ is vertical, the statement (1) immediately follows from the fact that the symplectic connection is vertical closed.

It remains to prove statement (2). Let $e\in E$ and $\unicode[STIX]{x1D709}_{1}$ , $\unicode[STIX]{x1D709}_{2}\in T_{e}E$ . By definition, we derive

$$\begin{eqnarray}\displaystyle (R_{c}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}})_{e}(\unicode[STIX]{x1D709}_{1},\unicode[STIX]{x1D709}_{2}) & = & \displaystyle \widetilde{\unicode[STIX]{x1D6FA}}_{c\,e}(dR_{c}(\unicode[STIX]{x1D709}_{1}^{v}),dR_{c}(\unicode[STIX]{x1D709}_{2}^{v}))\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FA}_{c\,e}^{v}(dR_{c}(\unicode[STIX]{x1D709}_{1}^{v}),dR_{c}(\unicode[STIX]{x1D709}_{2}^{v}))=c^{2}\widetilde{\unicode[STIX]{x1D6FA}}_{e}(\unicode[STIX]{x1D709}_{1},\unicode[STIX]{x1D709}_{2})\nonumber\end{eqnarray}$$

where we use the equality (4.4) and $\unicode[STIX]{x1D70B}_{F}(c\,e)=\unicode[STIX]{x1D70B}(e)$ for the fourth equality.

This proves $R_{c}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}=c^{2}\widetilde{\unicode[STIX]{x1D6FA}}$ .◻

It follows from Lemma 4.6(2) that we get ${\mathcal{L}}_{\vec{R}}\widetilde{\unicode[STIX]{x1D6FA}}=2\widetilde{\unicode[STIX]{x1D6FA}}$ . By Cartan’s formula, we get

$$\begin{eqnarray}d(\vec{R}\!\rfloor \widetilde{\unicode[STIX]{x1D6FA}})=2\widetilde{\unicode[STIX]{x1D6FA}}.\end{eqnarray}$$

4.3 Canonical contact form and contact structure on  $F$

Let $(Q,\unicode[STIX]{x1D703},H)$ be a given Morse–Bott contact setup and $(E,\unicode[STIX]{x1D6FA})\rightarrow Q$ be any $S^{1}$ -equivariant symplectic vector bundle equipped with an $S^{1}$ -invariant symplectic connection on it.

Now we equip the bundle $F=T^{\ast }{\mathcal{N}}\oplus E$ with a canonical $S^{1}$ -invariant contact form on $F$ . We denote the bundle projections by $\unicode[STIX]{x1D70B}_{E;F}:F\rightarrow E$ and $\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}:F\rightarrow T^{\ast }{\mathcal{N}}$ of the splitting $F=T^{\ast }{\mathcal{N}}\oplus E$ , respectively, and provide the direct sum connection on $F=T^{\ast }{\mathcal{N}}\oplus E$ .

Proof. We define a differential one-form on $F$ explicitly by

(4.8) $$\begin{eqnarray}\unicode[STIX]{x1D706}_{F}=\unicode[STIX]{x1D70B}_{F}^{\ast }\unicode[STIX]{x1D703}+\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }(\vec{R}\!\rfloor \widetilde{\unicode[STIX]{x1D6FA}}).\end{eqnarray}$$

Using Lemma 4.6, we obtain

(4.9) $$\begin{eqnarray}d\unicode[STIX]{x1D706}_{F}=\unicode[STIX]{x1D70B}_{F}^{\ast }\,d\unicode[STIX]{x1D703}+\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\,d\unicode[STIX]{x1D6E9}_{G}+\unicode[STIX]{x1D70B}_{E;F}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}\end{eqnarray}$$

by taking the differential of (4.8).

A moment of examination of this formula gives rise to the following

Proposition 4.8. There exists some $\unicode[STIX]{x1D6FF}>0$ such that the one-form $\unicode[STIX]{x1D706}_{F}$ is a contact form on the disc bundle $D^{\unicode[STIX]{x1D6FF}}(F)$ , where

$$\begin{eqnarray}D^{\unicode[STIX]{x1D6FF}}(F)=\{(q,v)\in F\mid \Vert v\Vert <\unicode[STIX]{x1D6FF}\}\end{eqnarray}$$

such that $\unicode[STIX]{x1D706}_{F}|_{o_{F}}=\unicode[STIX]{x1D703}$ on $Q\cong o_{F}\subset F$ .

Proof. This immediately follows from the formulas (4.8) and (4.9).◻

This proves the statement (1).

For the proof of the statement (2), we first note that the bundle $E$ itself does not depend on the choice of  $G$ . On the other hand, we put the one-form

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{F;G}=\unicode[STIX]{x1D70B}_{F}^{\ast }\unicode[STIX]{x1D703}+\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }(\vec{R}\!\rfloor \widetilde{\unicode[STIX]{x1D6FA}})\end{eqnarray}$$

on $E$ , which depends on $G$ in general because the one-form $\unicode[STIX]{x1D6E9}_{G}$ does. Furthermore, we recall that the projection map $\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}$ also depends on the canonical splitting

(4.10) $$\begin{eqnarray}T_{Q}F=HT_{Q}F\oplus VT_{Q}F\cong TQ\oplus F|_{Q}.\end{eqnarray}$$

Now we fix this splitting $T_{Q}F$ and let $G$ , $G^{\prime }$ be two splittings of  $TQ$

$$\begin{eqnarray}TQ=T{\mathcal{F}}\oplus G=T{\mathcal{F}}\oplus G^{\prime }.\end{eqnarray}$$

Since both $G$ , $G^{\prime }$ are transversal to $T{\mathcal{F}}$ in  $TQ$ , we can represent $G^{\prime }$ as the graph of the bundle map $A_{G}:G\rightarrow T{\mathcal{N}}$ . Then we consider the bundle isomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D713}_{GG^{\prime }}:TQ/\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\rightarrow TQ/\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\end{eqnarray}$$

defined by

$$\begin{eqnarray}\unicode[STIX]{x1D713}_{GG^{\prime }}=\left(\begin{array}{@{}cc@{}}\operatorname{Id}_{T{\mathcal{N}}} & A_{G}\\ 0 & \operatorname{id}_{G}\end{array}\right)\end{eqnarray}$$

under the splitting $TQ=\mathbb{R}\{X_{\unicode[STIX]{x1D706}}\}\oplus T{\mathcal{N}}\oplus G$ . Then $\unicode[STIX]{x1D713}_{GG^{\prime }}(G)=\text{Graph}\,A_{G}$ and $\unicode[STIX]{x1D713}_{GG^{\prime }}|_{T{\mathcal{N}}}=\operatorname{id}_{T{\mathcal{N}}}$ . Therefore, we have $p_{T{\mathcal{N}};G}=p_{T{\mathcal{N}};G^{\prime }}\circ \unicode[STIX]{x1D713}_{GG^{\prime }}$ .

We compute

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E9}_{G}(\unicode[STIX]{x1D6FC})(\unicode[STIX]{x1D702}) & = & \displaystyle \unicode[STIX]{x1D6FC}(p_{T{\mathcal{N}};G}(d\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}}}(\unicode[STIX]{x1D702})))\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FC}(p_{T{\mathcal{N}};G^{\prime }}\circ \unicode[STIX]{x1D713}_{GG^{\prime }}(\unicode[STIX]{x1D702}))=\unicode[STIX]{x1D6E9}_{G^{\prime }}(\unicode[STIX]{x1D6FC})(\unicode[STIX]{x1D713}_{GG^{\prime }}(\unicode[STIX]{x1D702})).\nonumber\end{eqnarray}$$

This proves $\unicode[STIX]{x1D6E9}_{G}=\unicode[STIX]{x1D6E9}_{G^{\prime }}\circ \unicode[STIX]{x1D713}_{GG^{\prime }}$ .◻

Now we study the contact geometry of $(D^{\unicode[STIX]{x1D6FF}}(F),\unicode[STIX]{x1D706}_{F})$ . We first note that the two-form $d\unicode[STIX]{x1D706}_{F}$ is a presymplectic form with one-dimensional kernel such that

$$\begin{eqnarray}d\unicode[STIX]{x1D706}_{F}|_{VTF}=\widetilde{\unicode[STIX]{x1D6FA}}^{v}|_{VTF}.\end{eqnarray}$$

Denote by $\widetilde{X}:=(d\unicode[STIX]{x1D70B}_{F;H})^{-1}(X)$ the horizontal lifting of the vector field $X$ on $Q$ , where

$$\begin{eqnarray}d\unicode[STIX]{x1D70B}_{F;H}:=d\unicode[STIX]{x1D70B}_{F}|_{H}:HTF\rightarrow TQ\end{eqnarray}$$

is the bijection of the horizontal distribution and $TQ$ .

Lemma 4.9. (Reeb vector field)

The Reeb vector field $X_{F}$ of $\unicode[STIX]{x1D706}_{F}$ is given by

$$\begin{eqnarray}X_{F}=\widetilde{X}_{\unicode[STIX]{x1D703}},\end{eqnarray}$$

where $\widetilde{X}_{\unicode[STIX]{x1D703}}$ denotes the horizontal lifting of $X_{\unicode[STIX]{x1D703}}$ to  $F$ .

Proof. We only have to check the defining property $\widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor \unicode[STIX]{x1D706}_{F}=1$ and $\widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor d\unicode[STIX]{x1D706}_{F}=0$ . We first look at

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D706}_{F}(\widetilde{X}_{\unicode[STIX]{x1D703}}) & = & \displaystyle \unicode[STIX]{x1D70B}_{F}^{\ast }\unicode[STIX]{x1D703}(\widetilde{X}_{\unicode[STIX]{x1D703}})+\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}(\widetilde{X}_{\unicode[STIX]{x1D703}})+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }(\vec{R}\!\rfloor \widetilde{\unicode[STIX]{x1D6FA}})(\widetilde{X}_{\unicode[STIX]{x1D703}})\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D703}(X_{\unicode[STIX]{x1D703}})+0+0=1.\nonumber\end{eqnarray}$$

Here $\widetilde{\unicode[STIX]{x1D6FA}}(\vec{R},\widetilde{X}_{\unicode[STIX]{x1D703}})=0$ by definition of $\widetilde{\unicode[STIX]{x1D6FA}}$ since $\widetilde{X}_{\unicode[STIX]{x1D703}}^{v}=0$ . Then we calculate

$$\begin{eqnarray}\displaystyle \widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor d\unicode[STIX]{x1D706}_{F} & = & \displaystyle \widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor (\unicode[STIX]{x1D70B}_{F}^{\ast }\,d\unicode[STIX]{x1D703}+\widetilde{\unicode[STIX]{x1D6FA}}+\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\,d\unicode[STIX]{x1D6E9}_{G})\nonumber\\ \displaystyle & = & \displaystyle \widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor \unicode[STIX]{x1D70B}_{F}^{\ast }\,d\unicode[STIX]{x1D703}+\widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor \widetilde{\unicode[STIX]{x1D6FA}}+\widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor \unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\,d\unicode[STIX]{x1D6E9}_{G}=0.\nonumber\end{eqnarray}$$

We only need to explain why the last term $\widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor \unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\,d\unicode[STIX]{x1D6E9}_{G}$ vanishes. In fact $pr_{{\mathcal{N}};G}\,d\unicode[STIX]{x1D70B}_{F}(\widetilde{X}_{\unicode[STIX]{x1D703}})=pr_{{\mathcal{N}};G}(X_{\unicode[STIX]{x1D703}})=0$ . Using this, the definition of $\unicode[STIX]{x1D6E9}_{G}$ and the $S^{1}$ -equivariance of the vector bundle $F\rightarrow Q$ and the fact that $\widetilde{X}_{\unicode[STIX]{x1D703}}$ is the vector field generating the $S^{1}$ -action, we derive

$$\begin{eqnarray}\displaystyle \widetilde{X}_{\unicode[STIX]{x1D703}}\!\rfloor \unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\,d\unicode[STIX]{x1D6E9}_{G}=0 & & \displaystyle \nonumber\end{eqnarray}$$

by a straightforward computation. This finishes the proof. ◻

Now we calculate the contact structure $\unicode[STIX]{x1D709}_{F}$ .

Lemma 4.10. (Contact distribution)

At each point $(\unicode[STIX]{x1D6FC},e)\in U_{F}\subset F$ , we define two subspaces of $T_{(\unicode[STIX]{x1D6FC},e)}F$

$$\begin{eqnarray}V:=\{\unicode[STIX]{x1D709}_{V}\in T_{(\unicode[STIX]{x1D6FC},e)}F\mid \unicode[STIX]{x1D709}_{V}=-\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}(\unicode[STIX]{x1D702})X_{F}+\widetilde{\unicode[STIX]{x1D702}},\unicode[STIX]{x1D702}\in \ker \unicode[STIX]{x1D703}\}\end{eqnarray}$$

and

$$\begin{eqnarray}W:=\{\unicode[STIX]{x1D709}_{W}\in T_{(\unicode[STIX]{x1D6FC},e)}F\mid \unicode[STIX]{x1D709}_{W}:=-{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}(e,v)X_{F}+v,v\in VTF\},\end{eqnarray}$$

Then $\unicode[STIX]{x1D709}_{F}=V\oplus W$ .

Proof. By straightforward calculation, both $V$ and $W$ are subspaces of $\unicode[STIX]{x1D709}_{F}=\ker \unicode[STIX]{x1D706}_{F}$ .

For any $\unicode[STIX]{x1D709}\in \unicode[STIX]{x1D709}_{F}$ , we decompose $\unicode[STIX]{x1D709}=\unicode[STIX]{x1D709}^{\text{h}}+\unicode[STIX]{x1D709}^{\text{v}}$ using the decomposition $TF=HTF\oplus VTF$ . Since $\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}(\widetilde{\unicode[STIX]{x1D702}})=\unicode[STIX]{x1D6E9}_{G}(\unicode[STIX]{x1D702})$ and we can write $\unicode[STIX]{x1D709}^{\text{v}}=I_{(e;\unicode[STIX]{x1D70B}(e))}(v)$ for a unique $v\in E_{\unicode[STIX]{x1D70B}(e)}$ . Therefore, we need to find $b\in \mathbb{R}$ , $\unicode[STIX]{x1D702}\in \ker \unicode[STIX]{x1D703}$ so that for the horizontal vector $\unicode[STIX]{x1D709}^{\text{h}}=\widetilde{(\unicode[STIX]{x1D702}+bX_{\unicode[STIX]{x1D703}})}$

$$\begin{eqnarray}\displaystyle & \unicode[STIX]{x1D706}_{F}(\unicode[STIX]{x1D709})=0 & \displaystyle \nonumber\\ \displaystyle & -\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}(\unicode[STIX]{x1D702})X_{F}+\widetilde{\unicode[STIX]{x1D702}}\in V & \displaystyle \nonumber\\ \displaystyle & -{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}(e,v)X_{F}+v\in W. & \displaystyle \nonumber\end{eqnarray}$$

Then

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}^{h} & = & \displaystyle \widetilde{(\unicode[STIX]{x1D702}+bX_{\unicode[STIX]{x1D703}})}\nonumber\\ \displaystyle & = & \displaystyle \widetilde{\unicode[STIX]{x1D702}}+bX_{F}\nonumber\end{eqnarray}$$

which determines $\unicode[STIX]{x1D702}\in T_{\unicode[STIX]{x1D70B}(e)}{\mathcal{N}}\oplus G_{\unicode[STIX]{x1D70B}(e)}$ uniquely. We need to determine  $b$ .

Since

$$\begin{eqnarray}\displaystyle 0=\unicode[STIX]{x1D706}_{F}(\unicode[STIX]{x1D709}) & = & \displaystyle \unicode[STIX]{x1D706}_{F}(\widetilde{\unicode[STIX]{x1D702}}+bX_{F}+\unicode[STIX]{x1D709}^{\text{v}})\nonumber\\ \displaystyle & = & \displaystyle b+\unicode[STIX]{x1D706}_{F}(\widetilde{\unicode[STIX]{x1D702}})+\unicode[STIX]{x1D706}_{F}(\unicode[STIX]{x1D709}^{\text{v}})\nonumber\\ \displaystyle & = & \displaystyle b+\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}(\widetilde{\unicode[STIX]{x1D702}})+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}(e,v).\nonumber\end{eqnarray}$$

Then we set $\unicode[STIX]{x1D709}_{W}=-(1/2)\unicode[STIX]{x1D70B}_{E;F}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}(\vec{R},v)X_{F}+v$ for $v$ such that $I_{e;\unicode[STIX]{x1D70B}(e)}(v)=\unicode[STIX]{x1D709}^{v}$ and then finally choose $b=-\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}(\widetilde{\unicode[STIX]{x1D702}})$ so that $\unicode[STIX]{x1D709}_{V}:=-\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G}(\widetilde{\unicode[STIX]{x1D702}})X_{F}+\widetilde{\unicode[STIX]{x1D702}}$ . Therefore, we have proved $\unicode[STIX]{x1D709}_{F}=V+W$ .

To see it is a direct sum, assume

$$\begin{eqnarray}-(\unicode[STIX]{x1D70B}_{T^{\ast }{\mathcal{N}};F}^{\ast }\unicode[STIX]{x1D6E9}_{G})(\widetilde{\unicode[STIX]{x1D702}})X_{F}+\widetilde{\unicode[STIX]{x1D702}}-{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}(e,v)X_{F}+v=0,\end{eqnarray}$$

for some $\unicode[STIX]{x1D702}\in \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D706}}$ and $v\in VTM$ . Apply $d\unicode[STIX]{x1D70B}$ to both sides, and it follows that

$$\begin{eqnarray}-\!((\unicode[STIX]{x1D70B}_{T^{\ast }|CN;F}^{\ast }\unicode[STIX]{x1D6E9}_{G})(\widetilde{\unicode[STIX]{x1D702}})+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D70B}_{E;F}^{\ast }\widetilde{\unicode[STIX]{x1D6FA}}(e,v)X_{\unicode[STIX]{x1D703}})+\unicode[STIX]{x1D702}=0.\end{eqnarray}$$

Hence $\unicode[STIX]{x1D702}=0$ , and then $v=0$ follows since $X_{F}$ is in horizontal part. This finishes the proof.◻