## 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.

## The Bott Connection (Connections 1)

I’d like to begin this blog by discussing some ideas relevant to my current research; to that end, this will be the first in a series of posts about connections on foliated manifolds. The planned sequence is

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.

For this post, I assume that the reader is familiar with Riemannian manifolds, (Koszul) connections, the Levi-Civita connection, foliated manifolds, basic vector fields, and quite a few other things.

Throughout this post, all manifolds will be smooth, oriented, connected, Riemannian, and complete with respect to their metric.

## 1. The Bott Connection on Foliated Manifolds

Let $$(\mathbb{M}, g, \mathcal{F})$$ be a Riemannian manifold of dimension $$n+m$$, equipped with a foliation $$\mathcal{F}$$ which has totally geodesic, $$m$$-dimensional leaves and a bundle-like metric $$g$$. The sub-bundle $$\mathcal{V}$$ of $$T\mathbb{M}$$ formed by vectors tangent to the leaves is referred to as the vertical distribution, and the sub-bundle $$\mathcal{H}$$ of $$T\mathbb{M}$$ which is normal (under $$g$$) to $$\mathcal{V}$$ is referred to as the horizontal distribution.

Our first task will be to define the Bott connection on foliated manifolds. Heuristically, this connection is interesting because it is well adapted to the foliation, making both the vertical and horizontal distributions parallel while also being metric.

#### Theorem 1.1 (Bott Connection).

For $$(\mathbb{M}, g, \mathcal{F})$$ as before, there exists a unique connection $$\nabla^B$$ over $$T\mathbb{M}$$ satisfying the following:

1. $$\nabla^B$$ is metric. That is, $$\nabla^B g = 0$$.
2. If $$Y$$ is an horizontal vector field, $$\nabla^B_XY$$ is horizontal for all vector fields $$X$$.
3. If $$Z$$ is a vertical vector field, $$\nabla^B_XZ$$ is vertical for all vector fields $$X$$.
4. For horizontal vector fields $$X_1,X_2$$ and vertical vector fields $$Z_1,Z_2$$, it holds that $$T^B(X_1,X_2)$$ is vertical and that $$T^B(X_1,Z_1) = T^B(Z_1,Z_2) = 0$$, where $$T^B(X,Y) = \nabla^B_XY – \nabla^B_YX – [X,Y]$$ is the torsion tensor associated to $$\nabla^B$$.

This connection is referred to as the Bott connection on $$(\mathbb{M}, g, \mathcal{F})$$. The proof will proceed in two parts.

##### Part 1. (Uniqueness)

We begin by showing that the Bott connection is necessarily unique. Let $$X,Y,Z$$ be vector fields. Because $$\nabla^B$$ is metric, we have the relations

\begin{align} g(\nabla^B_XY,Z) + g(Y, \nabla^B_XZ) &= X \cdot g(Y,Z) \\ g(\nabla^B_YZ,X) + g(Z, \nabla^B_YX) &= Y \cdot g(Z,X) \\ g(\nabla^B_ZX,Y) + g(X, \nabla^B_ZY) &= Z \cdot g(X,Y) \end{align}

as well as the torsion relations

\begin{align} \nabla^B_XY – \nabla^B_YX &= [X,Y] – \pi_\mathcal{V}[\pi_\mathcal{H}X, \pi_\mathcal{H}Y] \\ \nabla^B_ZX – \nabla^B_XZ &= [Z,X] – \pi_\mathcal{V}[\pi_\mathcal{H}Z, \pi_\mathcal{H}X] \\ \nabla^B_ZY – \nabla^B_YZ &= [Z,Y] – \pi_\mathcal{V}[\pi_\mathcal{H}Z, \pi_\mathcal{H}Y] \end{align}

\begin{align} T^B(X,Y) &= T^B(\pi_\mathcal{V}X + \pi_\mathcal{H}X, \pi_\mathcal{V}Y + \pi_\mathcal{H}Y) \\ &= T^B(\pi_\mathcal{V}X, \pi_\mathcal{V}Y) + T^B(\pi_\mathcal{H}X, \pi_\mathcal{V}Y) + T^B(\pi_\mathcal{V}X, \pi_\mathcal{H}Y) + T^B(\pi_\mathcal{H}X, \pi_\mathcal{H}Y) \\ &= T^B(\pi_\mathcal{H}X, \pi_\mathcal{H}Y) \\ &= \pi_\mathcal{V}\left(\nabla^B_{\pi_\mathcal{H}X}\pi_\mathcal{H}Y – \nabla^B_{\pi_\mathcal{H}Y}\pi_\mathcal{H}X – [\pi_\mathcal{H}X,\pi_\mathcal{H}Y]\right) \\ &= -\pi_\mathcal{V}[\pi_\mathcal{H}X, \pi_\mathcal{H}Y] \end{align}

Alternately summing the metric relations, we find

$2g(\nabla^B_XY, Z) = g(\nabla^B_XY – \nabla^B_YX, Z) + g(\nabla^B_ZX – \nabla^B_XZ , Y) + g(\nabla^B_ZY – \nabla^B_YZ, X) \\ + X \cdot(Y,Z) + Y \cdot g(Z,X) – Z \cdot g(X,Y)$

which, applying the torsion relations, reduces to

\begin{align} 2g(\nabla^B_XY, Z) &= g([X,Y] – \pi_\mathcal{V}[\pi_\mathcal{H}X, \pi_\mathcal{H}Y], Z) \\ &\quad + g([Z,X] – \pi_\mathcal{V}[\pi_\mathcal{H}Z, \pi_\mathcal{H}X], Y) \\ &\quad + g([Z,Y] – \pi_\mathcal{V}[\pi_\mathcal{H}Z, \pi_\mathcal{H}Y], X) \\ &\quad + X \cdot(Y,Z) + Y \cdot g(Z,X) – Z \cdot g(X,Y)\end{align}

The right side of this expression, while a bit messy, is independant of the connection and thus determines the Bott connection uniquely. (Notice, this is the same proceedure that is carried out for the Levi-Civita connection, but isn’t quite as clean thanks to the torsion.)

##### Part 2. (Existence)

To see that the Bott connection exists, we construct it explicitly in terms of $$\nabla^g$$, the Levi-Cevita connection on $$\mathbb{M}$$ associated to the metric $$g$$. Recall that $$\nabla^g$$ is the unique connection on $$\mathbb{M}$$ that is both metric and torsion free (i.e. $$T^g(X,Y) = 0$$.) We define a connection $$\nabla$$ on $$T\mathbb{M}$$ by

$\nabla_XY = \begin{cases} \pi_\mathcal{H}\nabla^g_XY & X,Y \in \Gamma^\infty(\mathcal{H}) \\ \pi_\mathcal{H}[X,Y] & X \in \Gamma^\infty(\mathcal{V}), Y \in \Gamma^\infty(\mathcal{H}) \\ \pi_\mathcal{V}[X,Y] & X \in \Gamma^\infty(\mathcal{H}), Y \in \Gamma^\infty(\mathcal{V}) \\ \pi_\mathcal{V}\nabla^g_XY & X,Y \in \Gamma^\infty(\mathcal{V}) \end{cases}$

That $$\nabla$$ is a connection is clear, verifying the Leibniz property directly. We claim that $$\nabla$$ satisfies the conditions of the Bott connection. Conditions 2 and 3 are immediate, by definition. The rest of the proof will follow by cases, decomposing vector fields as $$X = \pi_\mathcal{V}X + \pi_\mathcal{H}X$$ and using the additive properties of connections.

To show that condition 4 holds, let $$X_i \in \Gamma^\infty(\mathcal{H})$$ and $$Z_i \in \Gamma^\infty(\mathcal{V})$$. Then

\begin{align} T(X_1,X_2) &= \nabla_{X_1}X_2 – \nabla_{X_2}X_1 – [X_1,X_2] \\ &= \pi_\mathcal{H}\nabla_{X_1}X_2 – \pi_\mathcal{H}\nabla^g_{X_2}X_1 – (\nabla^g_{X_1}X_2 – \nabla^g_{X_2}X_1) \\ &= -\pi_\mathcal{V}\nabla^g_{X_1}X_2 + \pi_\mathcal{V}\nabla^g_{X_2}X_1 \\ &= -\pi_\mathcal{V} [X_1,X_2]\end{align}

using the fact that the Levi-Civita connection is torsion free. Similarly,

\begin{align} T(Z_1,Z_2) &= \nabla_{Z_1}Z_2 – \nabla_{Z_2}Z_1 – [Z_1,Z_2] \\ &= \pi_\mathcal{V}\nabla_{Z_1}Z_2 – \pi_\mathcal{V}\nabla^g_{Z_2}Z_1 – (\nabla^g_{Z_1}Z_2 – \nabla^g_{Z_2}Z_1) \\ &= -\pi_\mathcal{H}\nabla^g_{Z_1}Z_2 + \pi_\mathcal{H}\nabla^g_{Z_2}Z_1 \\ &= 0 \end{align}

where the last step follows since the vertical distribution being totally geodesic implies that $$\nabla^g_{Z_i}Z_j$$ is vertical whenever both $$Z_i$$ and $$Z_j$$ are both vertical. Finally,

\begin{align} T(X_1,Z_1) &= \nabla_{X_1}Z_1 – \nabla_{Z_1}X_1 – [X_1,Z_1] \\ &= \pi_\mathcal{V}[X_1,Z_1] – \pi_\mathcal{H}[Z_1,X_1] – [X_1,Z_1] \\ &= 0\end{align}

which shows that $$\nabla$$ satisfies condition 4.

It remains to be shown that $$\nabla$$ is metric. We have that $$\nabla g$$ is given by

$(\nabla g)(X,Y,Z) = X \cdot (g(Y,Z)) – g(\nabla_XY,Z) – g(Y,\nabla_XZ)$

for any vector fields $$X,Y,Z$$.

First, if $$Y \in \Gamma^\infty(\mathcal{H}), Z \in \Gamma^\infty(\mathcal{V})$$ we have by the definition of $$\nabla$$ that $$\nabla_XY \in \Gamma^\infty(\mathcal{H}), \nabla_XZ \in \Gamma^\infty(\mathcal{V})$$ and since the metric splits orthogonally as $$g = g_\mathcal{V} \oplus g_\mathcal{H}$$ each of the terms on the right side vanish, and similarly for $$Y \in \Gamma^\infty(\mathcal{V}), Z \in \Gamma^\infty(\mathcal{H})$$. Thus we only need to consider the cases where $$Y,Z$$ are both vertical or both horizonal.

Now, if $$X,Y,Z \in \Gamma^\infty(\mathcal{H})$$, we see that

\begin{align} (\nabla g)(X,Y,Z) &= X \cdot (g(Y,Z)) – g(\nabla_XY,Z) – g(Y,\nabla_XZ) \\ &= X \cdot (g(Y,Z)) – g(\pi_\mathcal{H}\nabla^g_XY,Z) – g(Y,\pi_\mathcal{H}\nabla^g_XZ) \\ &= X \cdot (g(Y,Z)) – g(\nabla^g_XY,Z) + g(\pi_\mathcal{V}\nabla^g_XY,Z) \\ &\quad – g(Y,\nabla^g_XZ) + g(Y,\pi_\mathcal{V}\nabla^g_XZ) \\ &= (\nabla^gg)(X,Y,Z) + g(\pi_\mathcal{V}\nabla^g_XY,Z) + g(Y,\pi_\mathcal{V}\nabla^g_XZ) \\ &=0\end{align}

using the fact that the Levi-Cevita connection is metric, and the orthogonality of the horizontal and vertical distributions. A similar computation holds for $$X,Y,Z \in \Gamma^\infty(\mathcal{V})$$.

It is useful here to recall that since $$g$$ is a bundle-like metric, $$(M, g, \mathcal{F})$$ is given locally as a submersion $$\phi \colon (V_{T\mathbb{M}},g\vert_{V_{T\mathbb{M}}}) \rightarrow (U_\mathcal{H},g_\mathcal{H})$$; moreover there exists a basis of the plaque $$U_\mathcal{H}$$ given by basic vector fields, so by the additivity of connections we can always consider the horizontal component of vector fields to be basic.

Then, for $$X \in \Gamma^\infty(\mathcal{V}), Y,Z \in \Gamma^\infty(\mathcal{H})$$,

\begin{align}(\nabla g)(X,Y,Z) &= X \cdot (g(Y,Z)) – g(\nabla_XY,Z) – g(Y,\nabla_XZ) \\ &= X \cdot (g(Y,Z)) – g(\pi_\mathcal{H}[X,Y],Z) – g(Y,\pi_\mathcal{H}[X,Z]) \\ &= 0\end{align}

since the Lie bracket $$[X,Y]$$ of a vertical vector field and a basic vector field is always vertical.

Finally, for $$X \in \Gamma^\infty(\mathcal{H}), Y,Z \in \Gamma^\infty(\mathcal{V})$$,

\begin{align} (\nabla g)(X,Y,Z) &= X \cdot (g(Y,Z)) – g(\nabla_XY,Z) – g(Y,\nabla_XZ) \\ &= X \cdot (g(Y,Z)) – g(\pi_\mathcal{V}[X,Y],Z) – g(Y,\pi_\mathcal{V}[X,Z]) \\ &= X \cdot (g(Y,Z)) – g([X,Y],Z) + g(\pi_\mathcal{H}[X,Y],Z) \\ &\quad – g(Y,[X,Z]) + g(Y,\pi_\mathcal{H}[X,Z]) \\ &= X \cdot (g(Y,Z)) – g([X,Y],Z) – g(Y,[X,Z]) \\ &= X \cdot (g(Y,Z)) – g(\nabla^g_XY,Z) + g(\nabla^g_YX,Z) \\ &\quad – g(Y,\nabla^g_XZ) + g(Y,\nabla^g_ZX) \\ &= (\nabla^gg)(X,Y,Z) + g(\nabla^g_YX,Z) + g(Y,\nabla^g_ZX) \\ &= \mathcal{L}_Xg(Y,Z) \\ &= 0\end{align}

since the vertical distribution is totally geodesic if and only if the flow generated by a basic field is an isometry. From the above, we have that $$\nabla$$ satisfies the conditions, and thus $$\nabla = \nabla^B$$ is the Bott connection, completing the proof.