# Connections on foliated manifolds 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. The Tanaka-Webster connection on Sasakian manifolds,
5. Connections on 3-Sasakian manifolds,
6. Relevant Background material.

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 will be forthcoming in post 6, 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.

1. Marco

Awesome!