# Tanno’s Connection

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## The equivalence of Bott and Tanno’s connections (Connections 3)

This post is the third of a series on connections on foliated manifolds.

1. The Bott connection on foliated manifolds,
2. Tanno’s connection on contact manifolds,
3. The equivalence of Bott and Tanno’s connections on $$K$$-contact manifolds with the Reeb foliation,
4. Connections on codimension 3 sub-Riemannian manifolds.

In the last two posts, we have discussed basic properties of the Bott connection on general foliated manifolds and Tanno’s connection on contact manifolds.  Here we will show that the two notions are equivalent under a certain condition on the contact structure.

Throughout this post, all manifolds will be smooth.

## 3. Bott and Tanno’s connections on $$K$$-contact manifolds

The key property we want on a contact manifold is the following:

#### Definition 3.1

Let $$(\mathbb{M},\theta,g)$$ be a contact manifold with compatible metric $$g$$. We call $$\mathbb{M}$$ a $$K$$-contact manifold if the associated Reeb field $$\xi$$ is a Killing field, that is if

$\mathcal{L}_\xi g = 0$

We are interested in $$K$$-contact manifolds because of the following

#### Proposition 3.2

Let $$(\mathbb{M},\theta,g,\mathcal{F}_\xi)$$ be a contact manifold equipped with Reeb foliation $$\mathcal{F}_\xi$$.  Then the following are equivalent:

1. $$(\mathbb{M},\theta,g)$$ is a $$K$$-contact manifold,
2. $$(\mathbb{M},g,\mathcal{F}_\xi)$$ is a totally-geodesic foliation with bundle-like metric $$g$$.

Remark: Boyer and Galicki indicate that they prefer the name bundle-like contact metric manifold to $$K$$-contact manifold, as it is more descriptive and equivalent by the above. I’m not sure of the history of the name, but this makes sense to me.  I’ll probably use the two interchangeably in future posts.

##### Proof.

The equivalence of the $$K$$-contact condition and $$(\mathbb{M},g,\mathcal{F}_\xi)$$ being having a bundle-like metric $$g$$ is by essentially definition since this is equivalent to

$\mathcal{L}_Zg(X,X) = 0$

for $$X \in \Gamma(\mathcal{H}), Z \in \Gamma(\mathcal{V})$$.  To see that $$K$$ contact manifolds are totally-geodesic foliations, observe that

$\begin{split} \mathcal{L}_Xg(Z,Z) &= X\cdot g(Z,Z) – 2g([X,Z],Z) \\ &= 2\theta(Z) \iota_X d\theta(Z) + 2g([Z,X],Z) \\ &= -\mathcal{L}_Z g(X,Z) + Z \cdot g(X,Z) \\ &= 0 \\ \end{split}$

completing the proof.

Remark: I think there must be a nicer way to show that $$K$$-contact manifolds are totally-geodesic, I may update this.

Now we can state the main claim:

#### Theorem 3.3

Let $$(\mathbb{M}, \theta, g)$$ be a $$K$$-contact manifold with Reeb foliation $$\mathcal{F}_\xi$$.  Then the Bott connection $$\nabla^B$$ on $$(\mathbb{M},g,\mathcal{F}_\xi)$$ and Tanno’s connection $$\nabla^T$$ on $$(\mathbb{M},\theta,g)$$ coincide.

#### Proof.

By Proposition 3.2 the Bott connection is well-defined, and both the Bott and Tanno’s connections are unique by definition.  To see that they are equivalent, we need to show that one satisfies the conditions of the other.  We will proceed by showing that Tanno’s connection satisfies the conditions of Theorem 1.1 defining the Bott connection.

1. ($$\nabla^B$$ is metric)
By definition, Tanno’s connection is metric.
2. (If $$Y \in \Gamma(\mathcal{H})$$ then $$\nabla^B_XY \in \Gamma(\mathcal{H})$$)
We have that
$\begin{split} \nabla^T_XY &= -\nabla^T_X(J^2Y) \\ &= -(\nabla^T_X J)(JY) + J(\nabla^T_X(JY)) \\ &= -Q(JY,X) + J(\nabla^T_X(JY)) \\ &= -\left( (\nabla^g_XJ)(JY) – [(\nabla^g_X\theta)(J^2Y)]\xi +\theta(JY)J(\nabla^g_X\xi) \right) + J(\nabla^T_X(JY)) \\ &= -\left( \nabla^g_X(J^2Y) – J(\nabla^g_X(JY)) – \nabla^g_X(\theta Y) + \theta(\nabla^g_XY)\xi \right) + J(\nabla^T_X(JY)) \\ &= -\left( – \nabla^g_XY – J(\nabla^g_X(JY)) + \theta(\nabla^g_XY)\xi \right) + J(\nabla^T_X(JY)) \\ &= – J(\nabla^g_XY) + J(\nabla^g_X(JY)) + J(\nabla^T_X(JY)) \in \Gamma(\mathcal{H}) \end{split}$
3. (If $$Z \in \Gamma(\mathcal{V})$$ then $$\nabla^B_XZ \in \Gamma(\mathcal{V})$$)
By property 2 of Tanno’s connection,
$\nabla^T_XZ = \nabla^T_X(\theta(Z)\xi) = \nabla^T_X(\theta(Z))\xi \in \Gamma(\mathcal{V})$
4. (For $$X_1,X_2 \in \Gamma(\mathcal{H})$$ and $$Z_1,Z_2 \in \Gamma(\mathcal{V})$$ it holds that $$T^B(X_1,X_2) \in \Gamma(\mathcal{V})$$ and $$T^B(Z_1,X_1) = T^B(Z_1,Z_2) = 0$$)
For the first claim, we see that by property 4 of Tanno’s connection,
$T^T(X_1,X_2) = d\theta(X_1,X_2)\xi \in \Gamma(\mathcal{V}).$For the second,
$\begin{split} T^T(Z_1,X_1) &= -T^T(Z_1,J^2X_1) = JT^T(\xi,JX_1) \\ &= – J^2T^T(Z_1,X_1) \\ \end{split}$
using the fact that $$J^2X_1 = -X_1$$ for horizontal vector fields and property 5 of Tanno’s connection.  This implies that $$T^T(Z_1,X_1)$$ is horizontal.  By the definition of the torsion tensor we see that
$T^T(Z_1,X_1) = \nabla^T_{Z_1}X_1 – \nabla^T_{X_1}Z_1 – [Z_1,X_1] = \nabla^T_{Z_1}X_1$
since $$\nabla^T_{X_1}Z_1$$ is vertical by 3, and the bracket vanishes by assuming $$X_1$$ to be basic.  However, the right hand side of this expression is not tensorial in $$X_1$$, and so we conclude that
$T^T(Z_1,X_1) = 0$

Finally,
$T^T(Z_1,Z_2) = \theta(Z_1) \theta(Z_2) T^T(\xi, \xi) = 0$
completing the proof.

## Tanno’s Connection on Contact Manifolds (Connections 2)

This post is the second of a series on connections on foliated manifolds.

1. The Bott connection on foliated manifolds,
2. Tanno’s connection on contact manifolds,
3. The equivalence of Bott and Tanno’s connections on $$K$$-contact manifolds with the Reeb foliation,
4. Connections on codimension 3 sub-Riemannian manifolds.

We’ll be considering Tanno’s connection, which is well adapted to contact structures and thus appropriate for studying the Reeb foliation. Here I assume the reader is familiar with contact manifolds, (Koszul) connections, and quite a few other things.

Throughout this post, all manifolds will be smooth.

## 2. Tanno’s Connection on Contact Manifolds

We call $$(\mathbb{M}, \theta)$$ a contact manifold if $$\mathbb{M}$$ is a $$2n+1$$ dimensional manifold and $$\theta$$ is a 1-form such that $$\theta \wedge (d\theta)^n$$ is a volume form on $$\mathbb{M}$$.

#### Proposition 2.1

Let $$(\mathbb{M}, \theta)$$ be a contact manifold. There exist on $$\mathbb{M}$$ a unique vector field $$\xi$$, a Riemannian metric $$g$$, and a $$(1,1)$$-tensor field $$J$$ such that

1. $$\theta(\xi) = 1$$, $$\iota_\xi d\theta = 0$$,
2. $$g(X,\xi ) = \theta(X)$$ for all vector fields $$X$$,
3. $$2g(X,JY) = d\theta(X,Y)$$, $$J^2X = -X + \theta(X)\xi$$ for all vector fields $$X,Y$$.

$$\xi$$ is called the Reeb vector field, and such a metric is said to be compatible with the contact structure.

A contact manifold $$(\mathbb{M}, \theta)$$ can be canonically equipped with a codimension 1 foliation $$\mathcal{F}_\xi$$ by choosing the horizontal distribution to be $$\mathcal{H} = \ker \theta$$ and the vertical distribution $$\mathcal{V}$$ to be generated by the Reeb vector field $$\xi$$ . This is known as the Reeb foliation.

Proof of some of the above (well-known) claims are forthcoming, see also [bh17] for an introduction to contact manifolds.

#### Theorem 2.2 (Tanno’s Connection)

Let $$(\mathbb{M}, \theta, \xi, g, J, \mathcal{F}_\xi)$$ as above. There exists a unique connection $$\nabla^T$$ on $$T\mathbb{M}$$ satisfying

1. $$\nabla^T\theta = 0$$,
2. $$\nabla^T\xi = 0$$,
3. $$\nabla^T$$ is metric, i.e. $$\nabla^Tg = 0$$,
4. $$T^T(X,Y) = d\theta(X,Y)\xi$$ for any $$X,Y \in \Gamma^\infty(\mathcal{H})$$,
5. $$T^T(\xi,JY) = -JT^T(\xi,Y)$$ for any $$Y \in \Gamma^\infty(T\mathbb{M})$$,
6. $$(\nabla^T_XJ)(Y) = Q(Y,X)$$ for any $$X,Y \in \Gamma^\infty(T\mathbb{M})$$,

where the Tanno tensor $$Q$$ is the $$(1,2)$$-tensor field determined by

$Q^i_{jk} = \nabla^g_kJ^i_j + \xi^iJ^r_j\nabla^g_k\theta_r + J^i_r\nabla^g_k\xi^r\theta_j$

or equivalently

$Q(X,Y) = (\nabla^g_YJ)X + [(\nabla^g_Y\theta)JX]\xi + \theta(X)J(\nabla^g_Y\xi).$

This connection is known as Tanno’s connection, or sometimes as the generalized Tanaka connection. Just as with Bott’s connection, the proof proceeds in two parts.

##### Part 1. (Uniqueness)

We have the usual metric relations

\begin{align} g(\nabla^T_XY,Z) + g(Y, \nabla^T_XZ) &= X \cdot g(Y,Z) \\ g(\nabla^T_YZ,X) + g(Z, \nabla^T_YX) &= Y \cdot g(Z,X) \\ g(\nabla^T_ZX,Y) + g(X, \nabla^T_ZY) &= Z \cdot g(X,Y) \end{align}

which can be summed to show that

$2g(\nabla^T_XY, Z) = g(\nabla^T_XY – \nabla^T_YX, Z) + g(\nabla^T_ZX – \nabla^T_XZ , Y) + g(\nabla^T_ZY – \nabla^T_YZ, X) \\ + X \cdot(Y,Z) + Y \cdot g(Z,X) – Z \cdot g(X,Y).$

By definition,

$\nabla^T_XY – \nabla^T_YX = [X,Y] + T^T(X,Y)$

so it remains to find an expression for $$T^T$$ independent of the connection.

For vertical vector fields $$X,Y$$,

\begin{aligned} T^T(X,Y) &= \nabla^T_XY – \nabla^T_YX – [X,Y] \\ &= \theta(Y)\nabla^T_X\xi + X \cdot \theta(Y) – \theta(X)\nabla^T_Y\xi – Y \cdot \theta(X) – [X,Y] \\ &= X \cdot \theta(Y) – Y \cdot \theta(X) – [X,Y] \\ \end{aligned}

using the the fact that the Reeb vector field is parallel.

For horizontal fields $$X,Y$$

$T^T(X,Y) = d\theta(X,Y)\xi$

is given as condition 4.

Finally, for $$X$$ vertical and $$Y$$ horizontal we have

\begin{aligned} T^T(X,Y) &= -\theta(X)T^T(\xi,J^2Y) \\ &= \theta(X)JT^T(\xi,JY) \\ &= -\theta(X)J^2T^T(\xi,Y) \\ &= -J^2T^T(X,Y) \\ &=T^T(X,Y) – \theta(T^T(X,Y))\xi \\ \theta(T^T(X,Y))\xi &= 0 \\ \end{aligned}

from which we conclude that $$T^T(X,Y)$$ is horizontal, and also

\begin{aligned} \nabla^T_XY &= -\nabla^T_X(J^2Y) \\ &= -(\nabla^T_XJ)(JY) – J(\nabla^T_X(JY)) \\ &= -Q(JY,X) – J((\nabla^T_XJ)Y – J(\nabla^T_XY)) \\ &= -Q(JY,X) – JQ(Y,X) – J^2(\nabla^T_XY) \\ &= -Q(JY,X) – JQ(Y,X) – \nabla^T_XY + \theta(\nabla^T_XY)\xi \\ 2\nabla^T_XY &= -Q(JY,X) – JQ(Y,X) + \theta(\nabla^T_XY)\xi \\ \end{aligned}

which we can apply to the expression for the torsion giving us that

\begin{aligned} 2T^T(X,Y) &= 2\nabla^T_XY – 2\nabla^T_YX – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + \theta(\nabla^T_XY)\xi \\ &\qquad – (-Q(JX,Y) – JQ(X,Y) + \theta(\nabla^T_YX)\xi ) – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + \theta(\nabla^T_XY – \nabla^T_YX)\xi + JQ(X,Y) – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + JQ(X,Y) + \theta(T^T(X,Y) + [X,Y])\xi – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + JQ(X,Y) – \theta([X,Y])\xi – 2[X,Y]. \\ \end{aligned}

From this, we can write an expression for $$g(\nabla^T_XY,Z)$$ independent of $$\nabla^T$$, so it must be unique.
Remark. Notice that we did not need to use condition 1 (that $$\nabla^T\theta = 0$$) to prove uniqueness.

##### Part 2. (Existence)

Following Tanno’s original paper [tan89], we define a connection $$\nabla$$ by its Christoffel symbols

$\overline{\Gamma^i_{jk}} = \Gamma^i_{jk} + \theta_jJ^i_k – \nabla^g_j\xi^i\theta_k + \xi^i\nabla^g_j\theta_k$

or equivalently in coordinate-free notation,

$\nabla_XY = \nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi$

where the $$\Gamma^i_{jk}$$ denote the Christoffel symbols of the Levi-Civita connection $$\nabla^g$$. We claim that $$\nabla$$ is in fact Tanno’s connection.

To prove this, we will verify the conditions explicitly.

###### Condition 1

We have that

\begin{aligned} (\nabla \theta) (X, Y) &= (\nabla_X\theta)(Y) \\ &= X \cdot \theta(Y) – \theta(\nabla_XY) \\ &= X \cdot \theta(Y) – \theta(\nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi) \\ &= X \cdot \theta(Y) – \theta(\nabla^g_XY) – \theta(X)\theta(JY) + \theta(Y)\theta(\nabla^g_X\xi) – [(\nabla^g_X\theta)Y]\theta(\xi) \\ &= X \cdot \theta(Y) – \theta(\nabla^g_XY) – X \cdot \theta(Y) + \theta(\nabla^g_XY) \\ &= 0 \end{aligned}

using, in particular, that $$\theta(J(Y)) = 0$$ since $$J \colon T\mathbb{M} \rightarrow \mathcal{H} = \ker \theta$$, and also that $$\theta(\nabla^g_X\xi) = 0$$ since $$\nabla^g_X\xi \in \mathcal{H}$$. Thus $$\nabla$$ satisfies condition 1.

###### Condition 2

Similarly,

\begin{aligned} (\nabla \xi)(X) &= \nabla_X\xi \\ &= \nabla^g_X\xi + \theta(X)J\xi – \theta(\xi)\nabla^g_X\xi + [(\nabla^g_X\theta)\xi]\xi \\ &= \nabla^g_X\xi – \nabla^g_X\xi + [X \cdot \theta(\xi) – \theta(\nabla^g_X\xi)]\xi \\ &= 0 \end{aligned}

which proves that $$\nabla$$ satisfies condition 2.

###### Condition 3

Again, we show condition 3 directly,

\begin{aligned} (\nabla g) (X,Y,Z) &= (\nabla_Xg)(Y,Z) \\ &= X \cdot g(Y,Z) – g(\nabla_XY, Z) – g(Y, \nabla_XZ) \\ &= X \cdot g(Y,Z) – g(\nabla^g_XY, Z) – g(Y, \nabla^g_XZ) \\ &\qquad – g(\theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi, Z) \\ &\qquad – g(Y, \theta(X)JZ – \theta(Z)\nabla^g_X\xi + [(\nabla^g_X\theta)Z]\xi) \\ &= (\nabla^gg)(X,Y,Z) \\ &\qquad – g(\theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi, Z) \\ &\qquad – g(Y, \theta(X)JZ – \theta(Z)\nabla^g_X\xi + [(\nabla^g_X\theta)Z]\xi) \\ &= – g([(\nabla^g_X\theta)Y]\xi – \theta(Y)\nabla^g_X\xi, Z) \\ &\qquad – g(Y, [(\nabla^g_X\theta)Z]\xi – \theta(Z)\nabla^g_X\xi) \\ &\qquad – g(\theta(X)JY, Z) – g(Y, \theta(X)JZ) \\ &= – \theta(Z)([(\nabla^g_{X_\mathcal{H}}\theta)Y] – g(Y,\nabla^g_{X_\mathcal{H}}\xi)) \\ &\qquad – \theta(Y)([(\nabla^g_{X_\mathcal{H}}\theta)Z] – g(Z,\nabla^g_{X_\mathcal{H}}\xi)) \\ &\qquad – \theta(X)[d\theta(Z,Y) + d\theta(Y,Z)] \\ &= – \theta(Z)(X_\mathcal{H}\cdot g(Y,\xi) – g(\nabla^g_{X_\mathcal{H}}Y,\xi) – g(Y,\nabla^g_{X_\mathcal{H}}\xi)) \\ &\qquad – \theta(Y)(X_\mathcal{H}\cdot g(Z,\xi) – g(\nabla^g_{X_\mathcal{H}}Z,\xi) – g(Z,\nabla^g_{X_\mathcal{H}}\xi)) \\ &= – \theta(Z)(\nabla^gg)(X_\mathcal{H},Y,\xi) – \theta(Y)(\nabla^gg)(X_\mathcal{H},Z,\xi) \\ &= 0 \end{aligned}

using, in particular, that $$d\theta(Y,Z) + d\theta(Z,Y) = 0$$ and $$g(X,\zeta) = \theta(X)$$.

###### Condition 4

To prove that conditions 4 and 5 hold, we will want an explicit expression for the torsion, which we write as

\begin{aligned} T(X,Y) &= \nabla_XY – \nabla_YX – [X,Y] \\ &= \nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi \\ &\qquad – \nabla^g_YX – \theta(Y)JX + \theta(X)\nabla^g_Y\xi – [(\nabla^g_Y\theta)X]\xi \\ &\qquad – [X,Y] \\ &= \theta(X)(JY + \nabla^g_Y\xi) – \theta(Y)(JX + \nabla^g_X\xi) + ([(\nabla^g_X\theta)Y] – [(\nabla^g_Y\theta)X])\xi \\ &= \theta(X)(JY + \nabla^g_Y\xi) – \theta(Y)(JX + \nabla^g_X\xi) + d\theta(X,Y)\xi \\ \end{aligned}

Then to check condition 4, we assume $$X,Y \in \mathcal{H} = \ker \theta$$ so that

\begin{aligned} T(X,Y) &= \theta(X)(JY + \nabla^g_Y\xi) – \theta(Y)(JX + \nabla^g_X\xi) + d\theta(X,Y)\xi \\ &= d\theta(X,Y)\xi \end{aligned}

using the expansion of the exterior derivative on 1-forms given by a torsion free connection.

###### Condition 5

For condition 5, again let $$Y$$ be any vector field, so that

\begin{aligned} T(\xi,Y) &= \theta(\xi)(JY + \nabla^g_Y\xi) – \theta(Y)(J\xi + \nabla^g_\xi\xi) + d\theta(\xi,Y)\xi \\ &= JY + \nabla^g_Y\xi \\ \end{aligned}

Now, if $$Y$$ is a vertical field the conclusion is clear. For $$Y$$ a horizontal field we claim that $$\nabla^g_{JY}\xi + J\nabla^g_Y\xi = 2Y$$ (which will be shown subsequently) and it holds that

\begin{aligned} -JT(\xi, Y) &= -J^2Y – J\nabla^g_Y \xi \\ &= -J^2Y – (2Y – \nabla^g_{JY}\xi) \\ &= J^2Y + \nabla^g_{JY}\xi \\ &= T(\xi, JY) \\ \end{aligned}

and condition 5 follows from the linearity of $$T$$. We complete the case with the following due to F. Baudoin.

##### Lemma. For horizontal $$X,Y$$ it holds that $$\theta((\nabla^g_XJ)Y) = \theta((\nabla^g_YJ)X)$$.

Proof. Recall that $$\theta(\nabla^g_YJ)X) = g((\nabla^g_YJ)X,\xi)$$. Differentiating $$g(JX,\xi) = 0$$ with respect to $$Y$$ we see that

$g((\nabla^g_YJ)X,\xi) + g(JX,\nabla^g_Y\xi) = 0$

so it is enough to prove that

$g(JX, \nabla^g_Y\xi) = g(JY,\nabla^g_X\xi)$

or equivalently

$d\theta(X,\nabla^g_Y\xi) = d\theta(Y,\nabla^g_X\xi).$

We have that

$d\theta(X, \nabla^g_Y\xi) = d\theta(X,\nabla^g_\xi Y + [Y,\xi]) = d\theta(X,\nabla^g_\xi Y) + d\theta(X,[Y,\xi]).$

Using $$\nabla^g_\xi d\theta = 0$$,

$d\theta(X,\nabla^g_\xi Y) = \xi \cdot d\theta(X,Y) – d\theta(\nabla^g_\xi X,Y)$

and similarly using $$\mathcal{L}_\xi d\theta = d\mathcal{L}_\xi \theta = 0$$,

$-d\theta(X,[Y,\xi]) = \xi \cdot d\theta(X,Y) – d\theta([\xi,X],Y).$

From which we see that

$d\theta(X,\nabla^g_Y\xi) = -d\theta(\nabla^g_\xi X,Y) + d\theta([\xi,X],Y) = -d\theta(\nabla^g_X\xi,Y).$

proving the lemma.

##### Claim. For horizontal $$X$$ it holds that $$\nabla^g_{JX}\xi + J\nabla^g_X\xi = 2X$$.

Proof. Let $$Y$$ be horizonal. It holds that

\begin{aligned} g(\nabla^g_{JX}\xi, Y) &= – g(\xi, \nabla^g_{JX}Y) \\ &= -\theta(\nabla^g_{JX}Y) \\ &= -\theta(\nabla^g_Y(JX)) – \theta([JX,Y]) \\ &= d\theta(JX,Y) – \theta(\nabla^g_Y(JX)) \\ &= 2g(X,Y) – \theta(\nabla^g_Y(JX)). \end{aligned}

On the other hand,

\begin{aligned} g(J\nabla_X\xi,Y) &= – g(\nabla^g_X\xi, JY) \\ &= g(\xi, \nabla^g_X(JY)) \\ &= \theta(\nabla^g_X(JY)) \end{aligned}

thus applying the last lemma, the conclusion follows.

###### Condition 6

For the final condition,

\begin{aligned} (\nabla_XJ)Y &= \nabla_X(JY) – J(\nabla_XY) \\ &= \nabla^g_X(JY) + \theta(X)J(JY) – \theta(JY)\nabla^g_X\xi + [(\nabla^g_X\theta)(JY)]\xi \\ &\qquad – J(\nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi) \\ &= \nabla^g_X(JY) + \theta(X)J^2Y + [(\nabla^g_X\theta)JY]\xi – J(\nabla^g_XY) – \theta(X)J^2Y + \theta(Y)J(\nabla^g_X\xi) \\ &= \nabla^g_X(JY) – J(\nabla^g_XY) + [(\nabla^g_X\theta)JY]\xi + \theta(Y)J(\nabla^g_X\xi) \\ &= (\nabla^g_XJ)Y + [(\nabla^g_X\theta)JY]\xi + \theta(Y)J(\nabla^g_X\xi) \\ &= Q(Y,X) \\ \end{aligned}

completing the proof.

We finish by remarking that the case of interest to us is when $$Q=0$$; this condition is equivalent to $$(M,\theta,J)$$ being a strongly pseudoconvex CR manifold. Moreover, $$\xi$$ will be a Killing field, and the foliation will be totally geodesic with bundle-like metric.

### References

[bh17] A. Banyaga, and D. Houenou. A Brief Introduction to Symplectic and Contact Manifolds. Vol. 15, World Scientific, 2017.

[tan89] S. Tanno. Variational problems on contact Riemannian manifolds. Trans. Amer. Math. Soc., 314(1):349–379, 1989.