# Expository

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## 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.

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.

## Sasakian and Kähler Manifolds 2

The following is essentially the content of the second talk I gave on Sasakian and Kähler manifolds.

## Sasakian Boothby-Wang Fibrations

Theorem (Newlander-Nirenberg)
An almost complex structure $$J$$ is integrable if and only if the Nijenhuis tensor
$N_J(X,Y) = -J^2[X,Y] + J([JX,Y] + [X,JY]) – [JX,JY]$
vanishes.

The Boothby-Wang fibration gives a canonical circle bundle over a symplectic manifold. Recall, of course, that Kähler manifolds are symplectic; there is the following interesting result:

Theorem (Hatakeyama)
Suppose that on a principle fiber bundle $$\pi \colon \mathbb{M} \rightarrow \mathbb{B}$$ over an almost complex manifold $$\mathbb{B}$$ with group $$S^1$$ we can define an almost contact structure $$(\xi,\eta,\Phi)$$. Then if the almost complex structure $$J = \Phi \vert_\mathbb{B}$$ is integrable and the curvature form $$\omega$$ given by $$\pi^*\omega = d\eta$$ on $$\mathbb{B}$$ associated to the contact form $$\eta$$ of $$\mathbb{M}$$ is of type $$(1,1)$$ with respect to the almost complex structure, then the almost contact structure on $$\mathbb{M}$$ is normal.

Proof:
By a paper of Sasaki and Hatakeyama normality of the contact metric structure is equivalent to the vanishing of the tensor
$N(X,Y) = [X,Y] + \Phi [\Phi X,Y] + \Phi [X,\Phi Y] – [\Phi X,\Phi Y] – \eta([X,Y])\xi – d\eta(X,Y)\xi$
which follows from considering the Nijenhuis tensor on the Riemannian cone over $$\mathbb{M}$$. Projecting onto the horizontal and vertical spaces, it is clear that $$N(X,Y) = 0$$ if and only if $$\pi(N(X,Y)) = 0$$ and $$\eta(N(X,Y)) = 0$$. Directly, it can be seen that
$\pi(N_p(X,Y)) = \bar{N}_{\pi(p)}(\pi X,\pi Y)$
where $$\bar{N}$$ is the Nijenhuis tensor associated to the almost complex structure $$J$$ on $$\mathbb{B}$$. Then the Newlander-Nirenberg theorem implies that $$N$$ will vanish only if $$J$$ is integrable.
Moreover,
$\eta(N(X,Y)) = -\eta([\Phi X,\Phi Y]) – d\eta(X,Y)$
and
$-\eta([\Phi X,\Phi Y]) = d\eta(\Phi X,\Phi Y)$
so that $$\eta(N(X,Y))$$ vanishes if any only if
$d\eta(\Phi X,\Phi Y) = d\eta(X,Y)$
which is equivalent to
$\omega(J(\pi X), J(\pi Y)) = \omega( \pi X, \pi Y)$
and so we are done.

Taking the last two results together gives

Theorem (Hatakeyama)
A necessary and sufficient condition for a compact manifold with a regular contact structure to admit an associated normal contact metric structure (and thus be Sasakian) is that the base manifold of the Boothby-Wang fibration of $$\mathbb{M}$$ is Hodge.

Proof:
One direction is the content of the Boothby-Wang theorem. If $$\mathbb{M}$$ is Hodge, it is Kähler, and thus the almost complex structure is integrable. From the first talk, the almost complex structure is compatible with the symplectic form $$\omega$$, which is precisely the statement
$\omega(JX,JY) = \omega(X,Y)$
and so we are done.

## 3-Sasakian manifolds

We want to introduce a generalization of the Kähler-Sasakian correspondence by allowing for the existence of triples of structures obeying a quaternionic relation. We begin with the following

Definition:
Let $$\mathbb{M}$$ be a $$4n$$-dimensional manifold with 3 integrable almost complex structures $$I_1,I_2,I_3$$ such that
$I_iI_j = -\delta_{ij}Id + 2 \epsilon_{ijk}I_k$
Then we call $$(\mathbb{M}, I,J,K)$$ a hyperkähler manifold.

To develop the corresponding Sasakian’ notion, we begin with extending the definition of a contact’ manifold.

Definition:
Let $$\mathbb{M}$$ be a $$4n+3$$-dimensional manifold such that there exists a family of contact structures $$\mathcal{S} = \{\eta(\tau),\xi(\tau),\Phi(\tau)\}$$ parameterized by $$\tau \in S^2$$ satisfying the relations

• $$\Phi(\tau) \circ \Phi(\tau’) – \eta(\tau) \otimes \eta(\tau’) = – \Phi(\tau \times \tau’) – (\tau \cdot \tau’)Id$$
• $$\Phi(\tau)\xi(\tau’) = – \xi(\tau \times \tau’)$$, and
• $$\eta(\tau) \circ \Phi(\tau’) = – \eta(\tau \times \tau’)$$

for all $$\tau, \tau’ \in S^2$$. We then call $$(\mathbb{M}, \{\eta(\tau),\xi(\tau),\Phi(\tau)\})$$ an almost hypercontact manifold. If moreover there exists a Riemannian metric $$g$$ on $$M$$ such that
$g(\Phi(\tau)X, \Phi(\tau)Y) = g(X,Y) – \eta(\tau)(X) \eta(\tau)(Y)$
for all $$\tau \in S^2$$ we call $$(\mathbb{M}, g, \{\eta(\tau),\xi(\tau),\Phi(\tau)\})$$ an almost hypercontact metric manifold.

Remark: Another standard definition comes from a choice of an orthonormal frame on $$\mathbb{R}^3$$, which we will refer to as an almost contact (metric) 3-structure.

Remark: Every compact, orientable 3-manifold admits an almost contact 3-structure.

Attempting to generalize the idea of Sasakian manifolds gives us the following

Proposition:
There exists a one-to-one correspondence between almost hypercontact structures on $$\mathbb{M}$$ and $$\Psi$$-invariant almost hypercomplex structures $$\mathcal{I}$$ on the cone $$C(\mathbb{M}) = \mathbb{M} \times \mathbb{R}^+$$.

and so we define

Definition:
Let $$(\mathbb{M},\mathcal{S}, g)$$ be an almost hypercontact metric manifold. Then if $$(C(\mathbb{M}), \mathcal{I}, g)$$ is hyperkähler, we call $$(\mathbb{M},\mathcal{S}, g)$$ 3-Sasakian.

Proposition:
If $$\mathcal{S} = \{\eta(\tau),\xi(\tau),\Phi(\tau)\}$$ is a 3-Sasakian structure on $$(\mathbb{M},g)$$ then

• $$g(\xi(\tau),\xi(\tau’)) = \tau \cdot \tau’$$
• $$[\xi(\tau),\xi(\tau’)] = 2\xi(\tau \times \tau’)$$
• $$\Phi(\tau) = -\nabla\xi(\tau)$$

Conversely, if $$\mathcal{S}_1,\mathcal{S}_2, \mathcal{S}_3$$ are Sasakian structures on $$(\mathbb{M},g)$$ with Reeb fields $$\xi_1, \xi_2, \xi_3$$ such that

• $$g(\xi_a,\xi_b) = \delta_{ab}$$
• $$[\xi_a,\xi_b] = 2\epsilon_{abc}\xi_c$$

then $$\mathcal{S} = \{\mathcal{S}_1,\mathcal{S}_2, \mathcal{S}_3\}$$ is a 3-Sasakian structure on $$\mathbb{M}$$.

## 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.