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

- The Bott connection on foliated manifolds,
**Tanno’s connection on contact manifolds,**- The equivalence of Bott and Tanno’s connections on \(K\)-contact manifolds with the Reeb foliation,
- 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 *

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

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

##### Proof.

##### 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.

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