# Bott 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.

## Biquard and Hladky Connections (Connections 4)

I’ve been looking at codimension 3 sub-Riemannian manifolds, so I’m posting out of order. 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.

#### Definition 4.1

Let $$(\mathbb{M},g,\mathcal{H})$$ be a $$4n+3$$-dimensional sub-Riemannian manifold with codimension $$3$$ distribution $$\mathcal{H}$$ such that

• $$\mathcal{H}$$ has a $$Sp(n)Sp(1)$$-structure, that is there exists a rank 3 bundle $$\mathcal{Q}$$ consisting of $$(1,1)$$-tensors on $$\mathcal{H}$$ locally generated by three almost-complex structures $$I_1,I_2,I_3$$ on $$\mathcal{H}$$ satisfying the quaternion relations $$I_1I_2I_3 = -id$$ which are hermitian compatible with the metric, that is
$g(I_j \cdot, I_j \cdot) = g(\cdot, \cdot)$
for $$j \in \{1,2,3\}$$.
• $$\mathcal{H}$$ is locally given as the kernel of a $$1$$-form $$\eta = (\eta_1,\eta_2,\eta_3)$$ with values in $$\mathbb{R}^3$$ such that
$2g(I_jX,Y) = d\eta_j(X,Y)$
for $$j \in \{1,2,3\}$$.

We then call $$(\mathbb{M},g,\mathcal{H},\mathcal{Q})$$ a quaternionic contact manifold or qc manifold.

Remark. These are interesting because they are an example of sub-Riemannian manifolds where
$(\mathcal{L}_{X_\mathcal{V}}g)(Y_\mathcal{H},Z_\mathcal{H}) \neq 0$

In this setting, we have two reference connections, the Hladky connection and Biquard connection.

#### Theorem 4.2 (Biquard)

Let $$(\mathbb{M},g,\mathcal{H},\mathcal{Q})$$ be a quaternionic contact manifold of dimension $$4n+3 > 7$$. Then there exists a unique connection $$\nabla^{Bi}$$ with torsion $$T^{Bi}$$ on $$\mathbb{M}$$ and a unique supplementary distibution $$\mathcal{V}$$ to $$\mathcal{H}$$ such that

• $$\mathcal{H}, \mathcal{V},$$ and $$g$$ are parallel for $$\nabla^{Bi}$$;
• $$T^{Bi}(\mathcal{H},\mathcal{H}) \subseteq \mathcal{V}, T^{Bi}(\mathcal{H},\mathcal{V}) \subseteq \mathcal{H}$$;
• for $$X \in \mathcal{V},$$ the operator $$T^{Bi}(X, \cdot) \colon \mathcal{H} \rightarrow \mathcal{H}$$ is in $$(\mathfrak{sp}(n)\oplus \mathfrak{sp}(1))^\perp \subset \mathfrak{gl}(4n)$$.

The connection $$\nabla^{Bi}$$ is called the Biquard connection on $$(\mathbb{M},g,\mathcal{H},\mathcal{Q})$$.

Biquard also described the vertical space $$\mathcal{V}$$ as being locally generated by vector fields $$\{\xi_1,\xi_2,\xi_3\}$$ such that
$\begin{split} \eta_j(\xi_k) &= \delta_{jk}, \\ (\iota_{\xi_j}d\eta_j)_\mathcal{H} &= 0, \\ (\iota_{\xi_j}d\eta_k)_\mathcal{H} &= -(\iota_{\xi_k}d\eta_j)_\mathcal{H} \end{split}$
The fields $$\{\xi_1,\xi_2,\xi_3\}$$ are called Reeb vector fields, in keeping with the nomenclature for contact manifolds.

Remark. The condition $$T^{Bi}(\mathcal{H},\mathcal{H}) \subseteq \mathcal{V}$$ is equivalent to $$T^{Bi}(X,Y) = -[X,Y]_\mathcal{V}$$ for all $$X,Y \in \mathcal{H}$$.

Remark. Biquard showed moreover that for a qc manifold $$(\mathbb{M},g,\mathcal{H},\mathcal{Q})$$ of dimension $$7$$ there may not be any such fields. Duchemin has shown that the Biquard connection exists for a $$7$$ dimensional qc manifold if we assume the existence of the Reeb vector fields.

We now introduce the concept of an $$r$$-graded sub-Riemannian manifold in order to define Hladky’s connection.

#### Definition 4.3

We call a sub-Riemannian manifold $$(\mathbb{M},g,\mathcal{H})$$ equipped with a choice of supplementary distribution $$\mathcal{V}$$ (that is $$T\mathbb{M} = \mathcal{H} \oplus \mathcal{V}$$) a sub-Riemannian manifold with complement or sRC manifold.

We say that a sRC manifold $$(\mathbb{M},g,\mathcal{H},\mathcal{V})$$ is r-graded if there are smooth constant rank bundles $$\mathcal{V}^{(j)}, 0 < j \leq r$$, such that
$\mathcal{V} = \mathcal{V}^{(1)} \oplus \cdots \oplus \mathcal{V}^{(r)}$
and
$\mathcal{H} \oplus \mathcal{V}^{(j)} \oplus [\mathcal{H},\mathcal{V}^{(j)}] \subseteq \mathcal{H} \oplus \mathcal{V}^{(j)} \oplus \mathcal{V}^{(j+1)}$
for all $$0 \leq j \leq r$$ with the convention that $$\mathcal{V}^{(0)} = \mathcal{H}$$ and $$\mathcal{V}^{(j)} = 0$$ for $$j > r$$.

A metric extension for an r-graded sRC manifold $$(\mathbb{M},g,\mathcal{V},\mathcal{H})$$ is a Riemannian metric $$\tilde g$$ that agrees with $$g$$ on $$\mathcal{H}$$ and makes the split
$T\mathbb{M} = \mathcal{H} \oplus_{1 \leq g \leq r} \mathcal{V}^{(j)}$
orthogonal.

For convenience, we shall denote by $$X^{(j)}$$ a section of $$\mathcal{V}^{(j)}$$ and set
$\hat{\mathcal{V}}^{(j)} = \bigoplus_{k \neq j} \mathcal{V}^{(k)}$

Lie derivatives are not tensorial in general, but we can define on an sRC manifold with metric extension the symmetric tensor $$B^{(j)}$$ by
$B^{(j)}(X,Y,Z) = (\mathcal{L}_Zg)(X,Y)$
for $$X,Y \in \mathcal{V}^{(j)}, Z \in \hat{\mathcal{V}}^{(j)}$$ and setting $$B^{(j)} = 0$$ on the orthogonal complement of $$\mathcal{V}^{(j)} \times \mathcal{V}^{(j)} \times \hat{\mathcal{V}}^{(j)}$$.

We contract these to tensors $$C^{(j)} \colon T\mathbb{M} \times T\mathbb{M} \rightarrow \mathcal{V}^{(j)}$$ defined by
$g(C^{(j)} (X,Y), Z^{(j)}) = B^{(j)}(X,Z^{(j)},Y)$

Remark. The tensors $$B^{(0)}$$ and $$C^{(0)}$$ rely only on the sRC structure, and are independent of the grading and metric extension

#### Definition 4.4

If $$g$$ is a metric extension of a r-graded sRC manifold then there exists a unique connection $${\nabla^{Hl}}^{(r)}$$ with torsion $${T^{Hl}}^{(r)}$$ such that

• $${\nabla^{Hl}}^{(r)}$$ is metric, that is $${\nabla^{Hl}}^{(r)} g = 0$$;
• $$\mathcal{V}^{(j)}$$ is parallel for all $$j$$;
• $${T^{Hl}}^{(r)}(\mathcal{V}^{(j)},\mathcal{V}^{(j)}) \subseteq \hat{\mathcal{V}}^{(j)}$$ for all $$j$$;
• $$g({T^{Hl}}^{(r)}(X^{(j)}, Y^{(k)}),Z^{(j)}) = g({T^{Hl}}^{(r)}(Z^{(j)}, Y^{(k)}),X^{(j)})$$ for all $$j,k$$.

Furthermore, if $$X,Y \in \mathcal{H}$$ then $${\nabla^{Hl}}^{(r)}(X)$$ and $${T^{Hl}}^{(r)}(X,Y)$$ are independent of the choice of grading and metric extension.

The Hladky connection can be expressed explicitly for vector fields $$X,Y,Z \in V^{(j)}, T \in \hat{\mathcal{V}}^{(j)}$$ by
$\begin{split} g({\nabla^{Hl}}^{(r)}_XY, T) &= 0 \\ g({\nabla^{Hl}}^{(r)}_XY, Z) &= g(\nabla^g_XY,Z) \\ {\nabla^{Hl}}^{(r)}_TY &= [T,Y]_j + \frac{1}{2}C^{(j)}(Y,T) \\ \end{split}$
where $$\nabla^g$$ is the Levi-Civita connection.

Remark. An r-graded sRC manifold also admits a k-grading (for all $$1 \leq k < r$$) given by
$\tilde{\mathcal{V}}^{(j)} = \mathcal{V}^{(j)}, 0 \leq j < k, \qquad \tilde{\mathcal{V}}^{(k)} = \bigoplus_{j \geq k} \mathcal{V}^{(j)}$
and then associated to each k-grading there is a connection $${\nabla^{Hl}}^{(k)}$$. For this entire family of connections, $${\nabla^{Hl}}^{(j)} X^{(k)} = {\nabla^{Hl}}^{(r)}X^{(k)}$$ whenever $$0 \leq k < j$$, so in particular for a horizontal vector field $$X$$ it holds that
${\nabla^{Hl}}^{(1)}X = {\nabla^{Hl}}^{(2)}X = \cdots {\nabla^{Hl}}^{(r)}X$
and so the differences between the connections $${\nabla^{Hl}}^{(k)}X$$ can be viewed as a choice of how to differentiate vertical vector fields.

All we need is the trivial 1-grading, but I wonder if the connections associated to higher gradings may be interesting.

#### Theorem 4.5 (Hladky)

Let $$(\mathbb{M},g,\mathcal{H},\mathcal{V})$$ be an r-graded sRC manifold with extended metric. We will call $${\nabla^{Hl}} = {\nabla^{Hl}}^{(1)}$$ the Hladky connection on $$\mathbb{M}$$.

#### Corollary 4.6

The Hladky connection is uniquely determined on a 1-graded sRC manifold with metric extension by the properties

• $$\mathcal{H}, \mathcal{V},$$ and $$g$$ are parallel for $${\nabla^{Hl}}$$;
• $${T^{Hl}}(\mathcal{H},\mathcal{H}) \subseteq \mathcal{V}, {T^{Hl}}(\mathcal{V},\mathcal{V}) \subseteq \mathcal{H},$$;
• $$g({T^{Hl}}(X,Z),Y) = g({T^{Hl}}(Y,Z),X)$$ for $$X,Y \in \mathcal{V}, Z \in \mathcal{H}$$ or $$X,Y \in \mathcal{H}, Z \in \mathcal{V}$$.

Remark. If $$(\mathcal{L}_{X_\mathcal{V}}g)(Y_\mathcal{H}, Z_\mathcal{H}) = 0$$ then $$B^{(j)} = C^{(j)} = 0$$ and the Hladky connection is equivalent to the Bott and Tanno connections. This occurs, for example, in the K-contact case.

Let $$(\mathbb{M},g,\mathcal{H},\mathcal{Q})$$ be a qc manfold (assuming the existence of the Reeb fields in dimension 7.) By the defining theorem for Biquard’s connection there is a unique distribution $$\mathcal{V}$$ such that the Biquard connection is well defined. Then given an orthogonal extension $$\tilde{g}$$ of the metric to $$\mathcal{V}$$, $$(\mathbb{M},\tilde{g},\mathcal{H},\mathcal{V})$$ will be a 1-graded sRC manifold with metric extension and thus have an Hladky connection by defining theorem for Hladky’s connection.

If we extend $$g$$ to $$\mathcal{V} = span(\xi_1,\xi_2,\xi_3)$$ by requiring $$g(\xi_j, \xi_k) = \delta_{jk}$$, it is known that $$\nabla^{Bi} g = 0$$, in agreement with $${\nabla^{Hl}}$$.

QUESTION: Do the Hladky and Biquard connections agree for this extension? Do they even agree on $$\mathcal{H}$$?

Using the explicit expression for the Hladky connection, we see that for $$X \in \mathcal{V},Y \in \mathcal{H}$$,

$\begin{split} {T^{Hl}}(X,Y) &= {\nabla^{Hl}}_XY – {\nabla^{Hl}}_YX – [X,Y] \\ &= [X,Y]_\mathcal{H} + \frac{1}{2} C^\mathcal{H}(Y,X) – [Y,X]_\mathcal{V} – \frac{1}{2}C^\mathcal{V}(X,Y) – [X,Y] \\ &= \frac{1}{2} \left( C^\mathcal{H}(Y,X) – C^\mathcal{V}(X,Y) \right) \\ \end{split}$

using this, we can get expressions for the horizontal and vertical components. For $$Z \in \mathcal{V}$$,
$\begin{split} g({T^{Hl}}(X,Y), Z) &= \frac{1}{2} g \left( C^\mathcal{H}(Y,X) – C^\mathcal{V}(X,Y), Z \right) \\ &= -\frac{1}{2} B^\mathcal{V}(X,Z,Y) \\ &= -\frac{1}{2} (\mathcal{L}_Y g)(X,Z) \\ &= 0 \\ \end{split}$
as desired, so for $$X \in \mathcal{V}$$, we have that $${T^{Hl}}(X, \cdot) \colon \mathcal{H} \rightarrow \mathcal{H}$$ in agreement with $$T^{Bi}$$. Moreover for $$Z \in \mathcal{H}$$,

$\begin{split} g({T^{Hl}}(X,Y), Z) &= \frac{1}{2} g \left( C^\mathcal{H}(Y,X) – C^\mathcal{V}(X,Y), Z \right) \\ &= \frac{1}{2} B^\mathcal{H}(Y,Z,X) \\ &= \frac{1}{2} (\mathcal{L}_X g)(Y,Z) \\ \end{split}$

This is sufficient to show that if $$\frac{1}{2} (\mathcal{L}_X g)(Y,Z) = 0$$ then $${\nabla^{Hl}} = \nabla^{Bi}$$. Otherwise, we need to determine if $${T^{Hl}}(X, \cdot) \colon \mathcal{H} \rightarrow \mathcal{H}$$ is in $$(\mathfrak{sp}(n)\oplus \mathfrak{sp}(1))^\perp \subset \mathfrak{gl}(4n)$$, which doesn’t seem to be the case.

## 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.