## Sasakian and Kahler Manifolds 1

I have been caught up in end of semester preparations, but further posts in the series on connections on contact foliated manifolds are forthcoming.  In the meantime, I thought I would post the following notes from a lecture on the relationship between Sasakian and Kahler manifolds that I gave to the UConn Complex Geometry seminar on Friday, 23 March, 2018.  I will be giving a second talk on Friday, 20 April, and the notes will subsequently appear here.  Much of the following is adapted from Banyaga and Houenou, A Brief Introduction to Symplectic and Contact manifolds.

## Introduction

In this post, I am interested in discussing the relationship between the following types of manifolds:

• Symplectic
• Contact
• Kahler
• Sasakian

Essentially by definition, Kahler manifolds are always symplectic (even dimensional), and Sasakian manifolds are always contact (odd dimensional.) We will see that there is a strong relationship between these two sets of structures.

Then, in order to investigate the conditions under which a compact contact manifold is Sasakian, we will introduce the Boothby-Wang fibration.

## Symplectic and Contact manifolds

### Symplectic manifolds

Definition:
A symplectic manifold $$(\mathbb{M},\omega)$$ is a $$2n$$-dimensional smooth manifold with a closed, nondegenerate differential 2-form $$\omega$$ called a symplectic form.

Since the symplectic form is a differential 2-form, it must be skew-symmetric, that is $$\omega(X,Y) = – \omega(Y,X)$$. Since $$\omega$$ is nondegenerate, $$\omega^n$$ is a volume form and thus $$\mathbb{M}$$ is oriented.

Definition:
An almost complex structure $$J$$ is an endomorphism of $$T\mathbb{M}$$ such that $$J^2 = -Id$$.

Definition:
We say that an almost complex structure $$J$$ is compatible with the symplectic manifold $$(\mathbb{M},\omega)$$ if

• $$\omega(X,Y) = \omega(JX,JY)$$ for all $$X,Y \in T\mathbb{M}$$, and
• The bilinear form $$g(X,Y) = \omega(X,JY)$$ is symmetric and positive-definite (and thus a Riemannian metric.)

Claim:
Let $$(\mathbb{M},\omega)$$ be a symplectic manifold. Then there exists a compatible almost complex structure, and moreover the set of all compatible almost complex structures is infinite and contractible.

Proof sketch: (Banyaga, Houenou)
Let $$g$$ be any Riemannian metric on $$\mathbb{M}$$ (which can always be done using the explicit construction on the basis elements) and consider the operator $$A = \tilde{g}^{-1} \circ \tilde{\omega}$$ where $$\tilde{g}(X)(Y) = g(X,Y)$$ and similarly $$\tilde{\omega}(X)(Y) = \omega(X,Y)$$. Then
$g(AX,Y) = \omega(X,Y).$
Set $$A^t$$ to be the adjoint of $$A$$ by $$g$$, that is
$g(A^tX,Y) = g(X,AY).$
We see that $$A$$ is skew-symmetric
$\begin{split} g(A^tX,Y) &= g(X,AY) \\ &= g(AY,X) \\ &= \omega(Y,X) \\ &= – \omega(X,Y) \\ &= – g(AX,Y) \\ \end{split}$
and also that $$A^tA$$ is positive-definite
$g(A^tAX,X) = g(AX,AX) > 0, \quad X \neq 0$
so $$A^tA$$ is diagonalizable with positive eigenvalues $$\{\lambda_1, \dots, \lambda_{2n}\}$$. Thus
$A^tA = B \cdot diag(\lambda_1,\dots, \lambda_{2n}) \cdot B^{-1}$
for some matrix $$B$$. Define $$R = \sqrt{A^tA} = B \cdot diag(\sqrt{\lambda_1},\dots, \sqrt{\lambda_{2n}}) \cdot B^{-1}$$ and also $$J = R^{-1}A$$. Then

• $$g(JX,JY) = g(X,Y)$$,
• $$JR = RJ$$, and
• $$J^t = -J$$ so that $$J^2 = -Id$$

It follows that
$\omega(JX,JY) = g(AJX,JY) = g(AX,Y) = \omega(X,Y)$
and
$\omega(X,JX) = g(AX,JX) = g(-JAX,X) = g(RX,X) > 0$
for all $$X \neq 0$$.
We define a new Riemannian metric
$g_J(X,Y) = \omega(X,JX) = \cdots = g(RX,Y)$
which depends on the original choice of $$g$$, of which there are infinitely many. We can construct an explicit homotopy between $$J_1 = J_{g_1}$$ and $$J_2 = J_{g_2}$$ by
$J_t = J_{(tg_1 + (1-t)g_2)}$

Example:

• $$\mathbb{R}^{2n}$$ with coordinates $$(x_1,\dots,x_n,y_1,\dots,y_n)$$ and 2-form
$\omega = dx_1 \wedge dy_1 + \cdots dx_n \wedge dy_n$
is symplectic, since clearly $$d\omega = 0$$ and $$\omega^n \neq 0$$.
• An even dimensional torus $$T^{2n} = \mathbb{R}^{2n}/\mathbb{Z}^{2n}$$ will be a symplectic manifold with $$\omega$$ descending to the quotient from the first example.

Theorem (Darboux)
Let $$\mathbb{M},\omega)$$ be a symplectic manifold. Each point $$p \in \mathbb{M}$$ has an open neighborhood $$U$$ and a chart $$\phi \colon U \rightarrow \mathbb{R}^{2n}$$ such that $$\phi(p) = 0$$ and
$\phi^*(\omega’) = \omega\vert_U$
where $$\omega’$$ is as in example 1 above.

In other words, all symplectic manifolds look the same, locally.

### Contact manifolds

Definition:
A contact manifold $$(\mathbb{M}, \eta)$$ is a $$2n+1$$-dimensional smooth manifold with a differential 1-form $$\eta$$ such that $$\eta \wedge (d\eta)^n$$ is a volume form. $$\eta$$ is called a contact form.

Remark:
Recall, $$\eta \wedge (d\eta)^n$$ is a volume form if it is a nonvanishing $$2n+1$$-form. A contact form gives an orientation on $$\mathbb{M}$$. Observe that for any smooth, nonvanishing function $$\rho$$ on $$\mathbb{M}$$ the 1-form $$\eta’ = \rho\eta$$ will also be a contact form on $$\mathbb{M}$$.

We have the following
Claim:
Let $$(\mathbb{M},\eta)$$ be a contact manifold. There exists a unique vector field $$\xi$$ called the Reeb vector field such that $$\eta(\xi) = 1$$ and $$\iota_\xi d\eta = 0$$.

Proof:
Since $$\eta \wedge (d\eta)^n$$ is nonvanishing, $$d\eta$$ must have rank $$2n$$. Let $$\xi_p \in \ker d\eta$$, and find $$v_1, \dots, v_{2n}$$ so that $$\{\xi_p, v_1, \dots, v_{2n}\}$$ complete a basis of $$T_p\mathbb{M}$$. Then
$\begin{split} 0 &\neq (\eta \wedge (d\eta)^n)(\xi_p,v_1,\dots,v_{2n}) \\ &= \eta(\xi_p) \wedge (d\eta)^n(v_1,\dots,v_{2n}) \\ &\quad + \sum_{i=1}^{2n} (-1)^i \eta(v_i) (d\eta)^n(v_1,\dots,v_{i-1},\xi_p,v_{i+1},\dots,v_{2n} \\ &= \eta(\xi_p) \wedge (d\eta)^n(v_1,\dots,v_{2n}) \end{split}$
since $$\xi_p \in \ker d\eta$$. But then $$\eta(\xi_p) \neq 0$$ for all $$x$$. Normalizing and denoting the result again by $$\xi$$ we get
$\begin{split} \eta(\xi) &= 1 \\ \iota_\xi d\eta &= 0 \end{split}$
as desired.

Claim:
It is always possible to find a Riemannian metric $$g$$ on $$\mathbb{M}$$ such that $$g(X,\xi) = \eta(X)$$. Such a metric is called compatible with the contact structure.

We can sometimes construct a contact manifold from a symplectic one.

Claim: (Contactization of a symplectic manifold)
Let $$(\mathbb{M}, \omega)$$ be a symplectic manifold such that $$\omega$$ is an exact form, that is there exists a 1-form $$\lambda$$ with $$\omega = d\lambda$$. Then $$\mathbb{M}’ = \mathbb{M} \times \mathbb{R}$$ is a contact manifold with contact form $$\eta = \pi^*\lambda + dt$$ where $$t \colon \mathbb{M} \times \mathbb{R} \rightarrow \mathbb{R}$$ and $$\pi \colon \mathbb{M} \times \mathbb{R} \rightarrow \mathbb{M}$$ are the canonical projections.

Proof:
Notice that $$d\eta = d\pi^*\lambda + d^2t = \pi^*d\lambda = \pi^*\omega$$. Thus $$\eta \wedge (d\eta)^n = \eta \wedge (\pi^*\omega)^n$$ has rank $$2n+1$$ and therefore must be a volume form on $$\mathbb{M}’$$.

From a contact manifold we can also construct a symplectic manifold on its cone $$\mathbb{R}^+ \times \mathbb{M}$$. This process is referred to the symplectization of $$\mathbb{M}$$ (see Boyer-Galicki, pg 203.) We will discuss this in further detail.

Claim: (Symplectization of a contact manifold)
Let $$\eta$$ be a 1-form on a $$2n+1$$-dimensional manifold $$\mathbb{M}$$. Then $$\eta$$ is a contact form on $$\mathbb{M}$$ if and only if the 2-form $$\omega = d(r^2\eta) = 2rdr\wedge\eta + r^2d\eta$$ is a symplectic form over the cone $$C(\mathbb{M})$$.

Proof:
If $$(\mathbb{M}, \eta)$$ is a contact manifold, then taking $$\omega = d(r^2\eta)$$ gives a closed, nondegenerate 2-form on $$C(\mathbb{M})$$.

If $$\omega = d(r^2\eta)$$ is a symplectic form on $$C(\mathbb{M})$$, then since $$\omega$$ is closed we see that $$\tilde\eta = r^2\eta$$ is a 1-form on $$C(\mathbb{M}) = \mathbb{M} \times \mathbb{R}^+$$. Then, restricting $$\tilde\eta \vert_{M \times \{1\}} = \eta$$ we see that $$\eta$$ must be a nondegenerate 1-form on $$\mathbb{M}$$. Since $$\omega^{n+1} = (d(r^2\eta))^{n+1} \neq 0$$, it must be that $$(d\eta)^n \neq 0$$ on $$\mathbb{M}$$, and we can conclude that $$\eta$$ is a contact form on $$\mathbb{M}$$.

We will be interested in this example later.

Example:

• $$\mathbb{R}^{2n+1}$$ with coordinates $$(x_1,\dots,x_n,y_1,\dots,y_n,z)$$ and 1-form
$\eta = \sum_{i=1}^{2n}x_idy_i + dz$
is a contact manifold, and has Reeb field
$\xi = \frac{\partial}{\partial z}$
• $$T^3$$ with 1-form
$\eta = \cos(z) dx + \sin(z) dy$
is a contact manifold with Reeb field
$\xi = \cos(z) \frac{\partial}{\partial x} + \sin(z) \frac{\partial}{\partial y}$
• $$S^{2n+1} \subset \mathbb{R}^{2n+2}$$ with 1-form
$\eta = \frac{1}{2}\left(\sum_{i=1}^{n+1} x_idy_x – y_i dx_i\right)$
is a contact manifold, and has Reeb field
$\xi = \sum_{i=1}^n x_i\frac{\partial}{\partial y_i} – y_i \frac{\partial}{\partial x_i}$

Theorem: (Martinet)
Every orientable 3-manifold admits a contact structure.

There is a well-known theorem describing locally the behavior of all contact forms.
Theorem: (Darboux)
Let $$\eta$$ be a contact form on a $$2n+1$$-dimensional manifold $$\mathbb{M}$$. For each point $$p \in \mathbb{M}$$ there exists an open neighborhood $$U$$ of $$p$$ and a chart $$\phi \colon U \rightarrow \mathbb{R}^{2n+1}$$ with $$\phi(p) = 0$$ and
$\phi^*(\eta’) = \eta\vert_U$
where $$\eta’$$ is the standard contact form
$\eta’ = \sum_{i=1}^n x_idy_i + dz$

## Kahler and Sasakian manifolds

### Kahler manifolds

From the complex point of view, a Kahler manifold is defined as follows.
Definition:
An almost Kahler manifold $$(\mathbb{M},J,h)$$ is a smooth manifold with almost complex structure
$J \in End(T\mathbb{M})$
(that is, $$J^2 = -Id$$) and hermitian scalar product
$h \colon T\mathbb{M} \times T\mathbb{M} \rightarrow \mathbb{C}$
(that is, $$h(X,\bar{Y}) = \overline{h(\bar{X},Y)}$$ and $$h(X,\bar{X}) > 0$$ for all $$X \neq 0$$) such that the associated differential 2-form
$\omega(X,Y) = Re\ h(JX,Y)$
is closed.

This can be strengthened as follows.

Definition:
An almost Kahler manifold $$(\mathbb{M},J,h)$$ such that the almost complex structure $$J$$ is integrable is called a Kahler manifold.

It is easy to show that K\”ahler manifolds are always even dimensional (this is a consequence of the existence of an almost complex structure,) and so
Proposition:
A Kahler manifold $$(\mathbb{M},J,h)$$ is a symplectic manifold $$(\mathbb{M},\omega)$$ when equipped with the 2-form
$\omega(X,Y) = Re\ h(JX,Y)$

In fact, there is an equivalent definition of K\”ahler manifolds from the symplectic perspective.
Definition:
A Kahler manifold $$(\mathbb{M},\omega,J)$$ is a symplectic manifold with symplectic form $$\omega$$ and an integrable almost complex structure $$J \in End(T\mathbb{M})$$ such that $$g(X,Y) = \omega(X,JY)$$ is symmetric and positive definite, and thus a Riemannian metric on $$\mathbb{M}$$.

From this definition, we will recover the hermitian scalar product as $$h = g – i\omega$$.

### Sasakian manifolds

Definition:
A contact metric structure on a contact manifold $$(\mathbb{M},\eta)$$ is a triple $$(\xi, J,g)$$ where $$\xi$$ is the Reeb field associated to $$\eta$$, $$g$$ is a Riemannian metric on $$\mathbb{M}$$ and $$J$$ is a $$(1,1)$$-tensor field satisfying

• $$J(\xi) = 0$$,
• $$J^2(X) = -X + \eta(X)\xi$$,
• $$d\eta(X,Y) = g(X,JY)$$, and
• $$g(X,Y) = g(JX,JY) + \eta(X)\eta(Y)$$.

Notice that $$g$$ is then compatible with the contact structure.

Remark: A triple $$(\xi,J,g)$$ that meet conditions 1 and 2 are referred to as an almost contact structure on a contact manifold $$(\mathbb{M},\eta)$$.

Remark: Notice that if $$(\mathbb{M},\eta)$$ is the contactization of a symplectic manifold $$(\mathbb{B},\omega)$$ then $$J$$ restricted to $$\mathbb{B}$$ is an almost contact structure. In fact, by choosing an almost complex structure $$J$$ on $$(\mathbb{B},d\eta,g)$$ (with compatible Riemannian metric $$g$$) and extending $$J$$ it to $$\mathbb{M}$$ by setting $$J(\xi) = 0$$ and extending the $$g$$ by $$g(X,Y) = g(JX,JY) + \eta(X)\eta(Y)$$ we will recover a contact metric structure on $$\mathbb{M}$$.

Example:
$$\mathbb{R}^3$$ with the form
$\eta = dz – ydx$
is contact, by the above. The Reeb field is
$V_3 = \xi = \frac{\partial}{\partial z}$
and the contact distribution $$\mathbb{B} = \ker \eta$$ is spanned by
$V_1 = \frac{\partial}{\partial y} \text{ and } V_2 = y\frac{\partial}{\partial z} + \frac{\partial}{\partial x}$
the compatible metric $$g$$ must satisfy
$g(V_i,V_j) = \delta_{ij}$
so a computation gives us that
$g = \left(\begin{array}{ccc} 1+y^2 & 0 & -y \\ 0 & 1 & 0 \\ -y & 0 & 1 \end{array}\right)$
and we define the almost contact structure by
$J(V_1) = -V_2, \quad J(V_2) = V_1, \quad J(V_3) = J(\xi) = 0$

Theorem:
Every contact manifold admits infinitely many contact metric structures, all of which are homotopic.

Definition:
Let $$(\mathbb{M},g)$$ be a Riemannian manifold. Its Riemannian cone is the Riemannian manifold $$C(\mathbb{M}) = \mathbb{R}^+ \times \mathbb{M}$$ with cone metric
$g_{C(\mathbb{M})} = dr^2 + r^2g$
where $$r \in \mathbb{R}^+$$.

It is clear that there is a one-to-one correspondence between Riemannian metrics on $$\mathbb{M}$$ and cone metrics on $$C(\mathbb{M})$$. Henceforth, denote $$\Psi = r\frac{\partial}{\partial r}$$. We have the following

Claim:
Let $$(\mathbb{M},\xi,\eta,J)$$ be an almost contact manifold. Then we can define a section $$I$$ of the endomorphism bundle of $$TC(\mathbb{M})$$ by
$IY = JY + \eta(Y)\Psi, \quad I\Psi = -\xi$
for $$Y \in T\mathbb{M}$$ (where we abuse notation by identifying $$T(\mathbb{M})$$ with $$T(\mathbb{M}) \times \{0\} \subset TC(\mathbb{M})$$.) Then $$I$$ is an almost complex structure on $$C(\mathbb{M})$$.

Proof:
We verify directly. First, for $$X = \rho\Psi$$,
$I^2 X = I(-\rho\xi) = -\rho J\xi – \rho\eta(\xi)\Psi = -\rho\Psi = -X$
and for $$Y \in T\mathbb{M}$$,
$I^2Y = I(JY + \eta(Y)\Psi) = J^2Y + \eta(JY)\Psi – \eta(Y)\xi = -Y$

Since for any $$X \in TC(\mathbb{M})$$ it holds that $$X = \rho\Psi + Y$$ with $$\rho$$ a smooth function and $$Y \in T(\mathbb{M})$$, we are done.

Recalling the symplectization of a contact manifold, we have the following.

Corollary:
There is a one-to-one correspondence between the contact metric structures $$(\xi,\eta,J,g)$$ on $$\mathbb{M}$$ and almost K\”ahler structures $$(dr^2 + r^2g, d(r^2\eta),I)$$ on $$C(\mathbb{M})$$.

Definition:
An almost contact structure $$(\xi, \eta, J)$$ is said to be normal if the corresponding almost complex structure $$I$$ on $$C(\mathbb{M})$$ is integrable, or equivalently if $$(C(M), dr^2 + r^2g, d(r^2\eta),I)$$ is Kahler.

Definition:
A manifold $$\mathbb{M}$$ with a normal almost contact metric structure $$(\xi,\eta,J,g)$$ is called a Sasakian manifold.

In some sense, then, Sasakian manifolds are an odd-dimensional counterpart to Kahler manifolds.

Example:
$$S^{2n+1} \hookrightarrow S^{2n+1} \times \mathbb{R} = \mathbb{C}^{n+1}$$.

## Boothby-Wang Fibration

We want to understand the necessary conditions for a contact manifold to be Sasakian. To this end, we strengthen the notion of a contact structure. We hereafter assume our manifolds to be compact.

Definition:
Let $$(\mathbb{M},\eta)$$ be a compact contact manifold. The Reeb field $$\xi$$ generates a dynamical system on $$\mathbb{M}$$; if the orbits of $$\xi$$ are periodic with period 1 we call $$(\mathbb{M},\eta)$$ a regular contact manifold.

Remark: If the orbits are periodic with period $$\lambda(p)$$ (which will be a nonvanishing constant on each orbit of $$\xi$$) then we can define $$\eta’ = \frac{1}{\lambda(p)}\eta$$ which will then make $$(\mathbb{M},\eta’)$$ a regular contact manifold. It is necessary to show that $$\lambda(p)$$ is smooth.

Example:
Any Reeb field on the torus $$T^3$$ generates a noncompact integral curve diffeomorphic to $$\mathbb{R}$$, and thus is not a regular contact form. This holds generally for tori, and is a theorem of Blair.

This then gives rise to the following characterization of regular contact manifolds.
Theorem: (Boothby-Wang)
If $$(\mathbb{M},\eta)$$ is a compact, regular contact manifold then

• $$\mathbb{M}$$ is a principal fiber bundle over the set of orbits $$\mathbb{B}$$ with group and fiber $$S^1$$,
• $$\eta$$ is a connection form in this bundle, and
• the base space $$\mathbb{B}$$ is a symplectic manifold whose symplectic form $$\omega$$ given by $$\pi^*\omega = d\eta$$ determines an integral cocycle on $$\mathbb{B}$$, that is $$\omega$$ is a representative of $$H^2(\mathbb{M},\mathbb{Z})$$.

Proof sketch:

• Since $$\xi$$ is never $$0$$, the integral curves must be closed, compact submanifolds of dimension 1, and thus homeomorphic to $$S^1$$. Then $$\xi$$ generates a periodic global one parameter group of transformations on $$\mathbb{M}$$, i.e. an $$S^1$$-action, that leaves no point fixed. We can conclude that $$\pi \colon \mathbb{M} \rightarrow \mathbb{B}$$ is a principal fiber bundle with group and fiber $$S^1$$.
• Notice that $$\mathcal{L}_\xi\eta = 0$$ and $$\mathcal{L}_\xi d\eta = 0$$. Let $$A = \frac{d}{dt}$$ be a basis for the Lie algebra $$\mathfrak{S}^1$$ of $$S^1$$, and set $$\tilde\eta = \eta A$$. We need to show that for $$B \in \mathfrak{S}^1$$, $$\tilde\eta(B^*) = B$$ (where $$B^*$$ is the vector on $$\mathbb{M}$$ induced by $$B$$) and that $$R^*_t\eta(X) = ad(t^{-1})X$$. The first follows since $$A = \xi$$, and the second follows from the fact that $$R^*_t\eta = \eta$$ and the fact that $$S^1$$ is abelian.
• This is essentially a reversal of the contactization of a symplectic manifold. Since $$d\eta$$ has rank $$2n$$ and $$\iota_\xi d\eta = 0$$, it is clear that $$\omega$$ on $$\mathbb{B}$$ given by $$\pi^*\omega = d\eta$$ will be a volume form on $$\mathbb{B}$$, making it a symplectic manifold. Moreover, $$\omega$$ is necessarily exact, and so determines an element of $$H^2(\mathbb{M},\mathbb{R})$$, what remains to be shown is that it is, in fact, integral. This follows from a theorem of Kobayashi.

Moreover, the converse holds as well
Theorem: (Boothby-Wang, converse)
If $$(\mathbb{B},\omega)$$ is a symplectic manifold such that $$\omega$$ is an integral cocycle, there is a principal $$S^1$$ bundle $$\mathbb{M}$$ over $$\mathbb{B}$$ and a 1-form $$\eta$$ on $$\mathbb{M}$$ such that $$(\mathbb{M},\eta)$$ is a contact manifold and the Reeb field of $$(\mathbb{M},\eta)$$ generates the action of $$S^1$$ on the bundle.

Proof sketch:
The same theorem of Kobayashi gives the existence of a circle bundle $$\pi \colon \mathbb{M} \rightarrow \mathbb{B}$$ with connection $$\tilde\eta$$ and structure equation $$d\tilde\eta = \pi^*\omega$$. It holds that $$(d\tilde\eta)^n = \pi^*\omega^n \neq 0$$, so that $$\tilde\eta \wedge (d\tilde\eta)^n \neq 0$$ is a volume form. Letting $$A$$ be a basis for $$\mathfrak{S}^1$$ and defining $$\eta$$ by $$\tilde\eta = \eta A$$ we have that $$\eta(A) = 1$$ and if $$\iota_Xd\omega = 0$$ then $$\iota_{\pi(X)}\omega = 0$$ so $$\pi(X) = 0$$ which implies that $$X$$ is vertical. Thus $$A = \xi$$, the associated vector field to $$\eta$$.

Recall that a Hodge manifold is a Kahler manifold $$(\mathbb{M},g,\omega,J)$$ such that the symplectic form is an integral cocycle.

Corollary:
If $$\mathbb{B}$$ is a compact Hodge manifold , then it has over it a canonically associated circle bundle which is a regular contact manifold.

Example:
The Hopf Fibration: $$S^1 \hookrightarrow S^3 \rightarrow S^2$$.

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

Theorem: (Hatakeyama)
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. Moreover, if the almost complex structure on $$\mathbb{B}$$ is integrable and the curvature form on $$\mathbb{B}$$ associated to the contact form of $$\mathbb{M}$$ is of type $$(1,1)$$ then the almost contact structure on $$\mathbb{M}$$ is normal.

The proof of the theorem is roughly along these lines: The construction of the contact metric structure is similar to the construction in the converse of Boothby-Wang, and the normality condition then follows from consideration of the Nijenhius tensor on B. The Newlander-Nirenberg theorem implies that J is integrable if and only if N = 0, which here can be shown to imply that the almost contact structure is normal.

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.

## References

• Banyaga, A.; Houenou, D. F. A Brief Introduction to Symplectic and Contact Manifolds; Nankai Tracts in Mathematics, Vol. 15; World Scientific: New Jersey, 2017.
• W. M. Boothby and H. C. Wang, On contact manifolds, Ann. of Math., 68(1958), 721-734.
• Foreman, Brendan. Complex contact manifolds and hyperkähler geometry. Kodai Math. J. 23 (2000), no. 1, 12–26. doi:10.2996/kmj/1138044153. https://projecteuclid.org/euclid.kmj/1138044153
• Hatakeyama, Yoji. Some notes on differentiable manifolds with almost contact structures. Tohoku Math. J. (2) 15 (1963), no. 2, 176–181. doi:10.2748/tmj/1178243844. https://projecteuclid.org/euclid.tmj/1178243844
• Kobayashi, Shoshichi. Principal fibre bundles with the 1-dimensional toroidal group. Tohoku Math. J. (2) 8 (1956), no. 1, 29–45. doi:10.2748/tmj/1178245006. https://projecteuclid.org/euclid.tmj/1178245006
• Morimoto, Akihiko. On normal almost contact structures. J. Math. Soc. Japan 15 (1963), no. 4, 420–436. doi:10.2969/jmsj/01540420. https://projecteuclid.org/euclid.jmsj/1260976537

## Connections on foliated manifolds 2

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

1. The Bott connection on foliated manifolds,
2. Tanno’s connection on contact manifolds,
3. The equivalence of Bott and Tanno’s connections on $$K$$-contact manifolds with the Reeb foliation,
4. The Tanaka-Webster connection on Sasakian manifolds,
5. Connections on 3-Sasakian manifolds,
6. Relevant Background material.

We’ll be considering Tanno’s connection, which is well adapted to contact structures and thus appropriate for studying the Reeb foliation. Here I assume the reader is familiar with contact manifolds, (Koszul) connections, and quite a few other things.

Throughout this post, all manifolds will be smooth.

## 2. Tanno’s Connection on Contact Manifolds

We call $$(\mathbb{M}, \theta)$$ a contact manifold if $$\mathbb{M}$$ is a $$2n+1$$ dimensional manifold and $$\theta$$ is a 1-form such that $$\theta \wedge (d\theta)^n$$ is a volume form on $$\mathbb{M}$$.

#### Proposition 2.1

Let $$(\mathbb{M}, \theta)$$ be a contact manifold. There exist on $$\mathbb{M}$$ a unique vector field $$\xi$$, a Riemannian metric $$g$$, and a $$(1,1)$$-tensor field $$J$$ such that

1. $$\theta(\xi) = 1$$, $$\iota_\xi d\theta = 0$$,
2. $$g(X,\xi ) = \theta(X)$$ for all vector fields $$X$$,
3. $$2g(X,JY) = d\theta(X,Y)$$, $$J^2X = -X + \theta(X)\xi$$ for all vector fields $$X,Y$$.

$$\xi$$ is called the Reeb vector field, and such a metric is said to be compatible with the contact structure.

A contact manifold $$(\mathbb{M}, \theta)$$ can be canonically equipped with a codimension 1 foliation $$\mathcal{F}_\xi$$ by choosing the horizontal distribution to be $$\mathcal{H} = \ker \theta$$ and the vertical distribution $$\mathcal{V}$$ to be generated by the Reeb vector field $$\xi$$ . This is known as the Reeb foliation.

Proof of some of the above (well-known) claims will be forthcoming in post 6, see also [bh17] for an introduction to contact manifolds.

#### Theorem 2.2 (Tanno’s Connection)

Let $$(\mathbb{M}, \theta, \xi, g, J, \mathcal{F}_\xi)$$ as above. There exists a unique connection $$\nabla^T$$ on $$T\mathbb{M}$$ satisfying

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

where the Tanno tensor $$Q$$ is the $$(1,2)$$-tensor field determined by

$Q^i_{jk} = \nabla^g_kJ^i_j + \xi^iJ^r_j\nabla^g_k\theta_r + J^i_r\nabla^g_k\xi^r\theta_j$

or equivalently

$Q(X,Y) = (\nabla^g_YJ)X + [(\nabla^g_Y\theta)JX]\xi + \theta(X)J(\nabla^g_Y\xi).$

This connection is known as Tanno’s connection, or sometimes as the generalized Tanaka connection. Just as with Bott’s connection, the proof proceeds in two parts.

##### Part 1. (Uniqueness)

We have the usual metric relations

\begin{align} g(\nabla^T_XY,Z) + g(Y, \nabla^T_XZ) &= X \cdot g(Y,Z) \\ g(\nabla^T_YZ,X) + g(Z, \nabla^T_YX) &= Y \cdot g(Z,X) \\ g(\nabla^T_ZX,Y) + g(X, \nabla^T_ZY) &= Z \cdot g(X,Y) \end{align}

which can be summed to show that

$2g(\nabla^T_XY, Z) = g(\nabla^T_XY – \nabla^T_YX, Z) + g(\nabla^T_ZX – \nabla^T_XZ , Y) + g(\nabla^T_ZY – \nabla^T_YZ, X) \\ + X \cdot(Y,Z) + Y \cdot g(Z,X) – Z \cdot g(X,Y).$

By definition,

$\nabla^T_XY – \nabla^T_YX = [X,Y] + T^T(X,Y)$

so it remains to find an expression for $$T^T$$ independent of the connection.

For vertical vector fields $$X,Y$$,

\begin{aligned} T^T(X,Y) &= \nabla^T_XY – \nabla^T_YX – [X,Y] \\ &= \theta(Y)\nabla^T_X\xi + X \cdot \theta(Y) – \theta(X)\nabla^T_Y\xi – Y \cdot \theta(X) – [X,Y] \\ &= X \cdot \theta(Y) – Y \cdot \theta(X) – [X,Y] \\ \end{aligned}

using the the fact that the Reeb vector field is parallel.

For horizontal fields $$X,Y$$

$T^T(X,Y) = d\theta(X,Y)\xi$

is given as condition 4.

Finally, for $$X$$ vertical and $$Y$$ horizontal we have

\begin{aligned} T^T(X,Y) &= -\theta(X)T^T(\xi,J^2Y) \\ &= \theta(X)JT^T(\xi,JY) \\ &= -\theta(X)J^2T^T(\xi,Y) \\ &= -J^2T^T(X,Y) \\ &=T^T(X,Y) – \theta(T^T(X,Y))\xi \\ \theta(T^T(X,Y))\xi &= 0 \\ \end{aligned}

from which we conclude that $$T^T(X,Y)$$ is horizontal, and also

\begin{aligned} \nabla^T_XY &= -\nabla^T_X(J^2Y) \\ &= -(\nabla^T_XJ)(JY) – J(\nabla^T_X(JY)) \\ &= -Q(JY,X) – J((\nabla^T_XJ)Y – J(\nabla^T_XY)) \\ &= -Q(JY,X) – JQ(Y,X) – J^2(\nabla^T_XY) \\ &= -Q(JY,X) – JQ(Y,X) – \nabla^T_XY + \theta(\nabla^T_XY)\xi \\ 2\nabla^T_XY &= -Q(JY,X) – JQ(Y,X) + \theta(\nabla^T_XY)\xi \\ \end{aligned}

which we can apply to the expression for the torsion giving us that

\begin{aligned} 2T^T(X,Y) &= 2\nabla^T_XY – 2\nabla^T_YX – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + \theta(\nabla^T_XY)\xi \\ &\qquad – (-Q(JX,Y) – JQ(X,Y) + \theta(\nabla^T_YX)\xi ) – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + \theta(\nabla^T_XY – \nabla^T_YX)\xi + JQ(X,Y) – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + JQ(X,Y) + \theta(T^T(X,Y) + [X,Y])\xi – 2[X,Y] \\ &= -Q(JY,X) – JQ(Y,X) + JQ(X,Y) – \theta([X,Y])\xi – 2[X,Y]. \\ \end{aligned}

From this, we can write an expression for $$g(\nabla^T_XY,Z)$$ independent of $$\nabla^T$$, so it must be unique.
Remark. Notice that we did not need to use condition 1 (that $$\nabla^T\theta = 0$$) to prove uniqueness.

##### Part 2. (Existence)

Following Tanno’s original paper [tan89], we define a connection $$\nabla$$ by its Christoffel symbols

$\overline{\Gamma^i_{jk}} = \Gamma^i_{jk} + \theta_jJ^i_k – \nabla^g_j\xi^i\theta_k + \xi^i\nabla^g_j\theta_k$

or equivalently in coordinate-free notation,

$\nabla_XY = \nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi$

where the $$\Gamma^i_{jk}$$ denote the Christoffel symbols of the Levi-Civita connection $$\nabla^g$$. We claim that $$\nabla$$ is in fact Tanno’s connection.

To prove this, we will verify the conditions explicitly.

###### Condition 1

We have that

\begin{aligned} (\nabla \theta) (X, Y) &= (\nabla_X\theta)(Y) \\ &= X \cdot \theta(Y) – \theta(\nabla_XY) \\ &= X \cdot \theta(Y) – \theta(\nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi) \\ &= X \cdot \theta(Y) – \theta(\nabla^g_XY) – \theta(X)\theta(JY) + \theta(Y)\theta(\nabla^g_X\xi) – [(\nabla^g_X\theta)Y]\theta(\xi) \\ &= X \cdot \theta(Y) – \theta(\nabla^g_XY) – X \cdot \theta(Y) + \theta(\nabla^g_XY) \\ &= 0 \end{aligned}

using, in particular, that $$\theta(J(Y)) = 0$$ since $$J \colon T\mathbb{M} \rightarrow \mathcal{H} = \ker \theta$$, and also that $$\theta(\nabla^g_X\xi) = 0$$ since $$\nabla^g_X\xi \in \mathcal{H}$$. Thus $$\nabla$$ satisfies condition 1.

###### Condition 2

Similarly,

\begin{aligned} (\nabla \xi)(X) &= \nabla_X\xi \\ &= \nabla^g_X\xi + \theta(X)J\xi – \theta(\xi)\nabla^g_X\xi + [(\nabla^g_X\theta)\xi]\xi \\ &= \nabla^g_X\xi – \nabla^g_X\xi + [X \cdot \theta(\xi) – \theta(\nabla^g_X\xi)]\xi \\ &= 0 \end{aligned}

which proves that $$\nabla$$ satisfies condition 2.

###### Condition 3

Again, we show condition 3 directly,

\begin{aligned} (\nabla g) (X,Y,Z) &= (\nabla_Xg)(Y,Z) \\ &= X \cdot g(Y,Z) – g(\nabla_XY, Z) – g(Y, \nabla_XZ) \\ &= X \cdot g(Y,Z) – g(\nabla^g_XY, Z) – g(Y, \nabla^g_XZ) \\ &\qquad – g(\theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi, Z) \\ &\qquad – g(Y, \theta(X)JZ – \theta(Z)\nabla^g_X\xi + [(\nabla^g_X\theta)Z]\xi) \\ &= (\nabla^gg)(X,Y,Z) \\ &\qquad – g(\theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi, Z) \\ &\qquad – g(Y, \theta(X)JZ – \theta(Z)\nabla^g_X\xi + [(\nabla^g_X\theta)Z]\xi) \\ &= – g([(\nabla^g_X\theta)Y]\xi – \theta(Y)\nabla^g_X\xi, Z) \\ &\qquad – g(Y, [(\nabla^g_X\theta)Z]\xi – \theta(Z)\nabla^g_X\xi) \\ &\qquad – g(\theta(X)JY, Z) – g(Y, \theta(X)JZ) \\ &= – \theta(Z)([(\nabla^g_{X_\mathcal{H}}\theta)Y] – g(Y,\nabla^g_{X_\mathcal{H}}\xi)) \\ &\qquad – \theta(Y)([(\nabla^g_{X_\mathcal{H}}\theta)Z] – g(Z,\nabla^g_{X_\mathcal{H}}\xi)) \\ &\qquad – \theta(X)[d\theta(Z,Y) + d\theta(Y,Z)] \\ &= – \theta(Z)(X_\mathcal{H}\cdot g(Y,\xi) – g(\nabla^g_{X_\mathcal{H}}Y,\xi) – g(Y,\nabla^g_{X_\mathcal{H}}\xi)) \\ &\qquad – \theta(Y)(X_\mathcal{H}\cdot g(Z,\xi) – g(\nabla^g_{X_\mathcal{H}}Z,\xi) – g(Z,\nabla^g_{X_\mathcal{H}}\xi)) \\ &= – \theta(Z)(\nabla^gg)(X_\mathcal{H},Y,\xi) – \theta(Y)(\nabla^gg)(X_\mathcal{H},Z,\xi) \\ &= 0 \end{aligned}

using, in particular, that $$d\theta(Y,Z) + d\theta(Z,Y) = 0$$ and $$g(X,\zeta) = \theta(X)$$.

###### Condition 4

To prove that conditions 4 and 5 hold, we will want an explicit expression for the torsion, which we write as

\begin{aligned} T(X,Y) &= \nabla_XY – \nabla_YX – [X,Y] \\ &= \nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi \\ &\qquad – \nabla^g_YX – \theta(Y)JX + \theta(X)\nabla^g_Y\xi – [(\nabla^g_Y\theta)X]\xi \\ &\qquad – [X,Y] \\ &= \theta(X)(JY + \nabla^g_Y\xi) – \theta(Y)(JX + \nabla^g_X\xi) + ([(\nabla^g_X\theta)Y] – [(\nabla^g_Y\theta)X])\xi \\ &= \theta(X)(JY + \nabla^g_Y\xi) – \theta(Y)(JX + \nabla^g_X\xi) + d\theta(X,Y)\xi \\ \end{aligned}

Then to check condition 4, we assume $$X,Y \in \mathcal{H} = \ker \theta$$ so that

\begin{aligned} T(X,Y) &= \theta(X)(JY + \nabla^g_Y\xi) – \theta(Y)(JX + \nabla^g_X\xi) + d\theta(X,Y)\xi \\ &= d\theta(X,Y)\xi \end{aligned}

using the expansion of the exterior derivative on 1-forms given by a torsion free connection.

###### Condition 5

For condition 5, again let $$Y$$ be any vector field, so that

\begin{aligned} T(\xi,Y) &= \theta(\xi)(JY + \nabla^g_Y\xi) – \theta(Y)(J\xi + \nabla^g_\xi\xi) + d\theta(\xi,Y)\xi \\ &= JY + \nabla^g_Y\xi \\ \end{aligned}

Now, if $$Y$$ is a vertical field the conclusion is clear. For $$Y$$ a horizontal field we claim that $$\nabla^g_{JY}\xi + J\nabla^g_Y\xi = 2Y$$ (which will be shown subsequently) and it holds that

\begin{aligned} -JT(\xi, Y) &= -J^2Y – J\nabla^g_Y \xi \\ &= -J^2Y – (2Y – \nabla^g_{JY}\xi) \\ &= J^2Y + \nabla^g_{JY}\xi \\ &= T(\xi, JY) \\ \end{aligned}

and condition 5 follows from the linearity of $$T$$. We complete the case with the following due to F. Baudoin.

##### Lemma. For horizontal $$X,Y$$ it holds that $$\theta((\nabla^g_XJ)Y) = \theta((\nabla^g_YJ)X)$$.

Proof. Recall that $$\theta(\nabla^g_YJ)X) = g((\nabla^g_YJ)X,\xi)$$. Differentiating $$g(JX,\xi) = 0$$ with respect to $$Y$$ we see that

$g((\nabla^g_YJ)X,\xi) + g(JX,\nabla^g_Y\xi) = 0$

so it is enough to prove that

$g(JX, \nabla^g_Y\xi) = g(JY,\nabla^g_X\xi)$

or equivalently

$d\theta(X,\nabla^g_Y\xi) = d\theta(Y,\nabla^g_X\xi).$

We have that

$d\theta(X, \nabla^g_Y\xi) = d\theta(X,\nabla^g_\xi Y + [Y,\xi]) = d\theta(X,\nabla^g_\xi Y) + d\theta(X,[Y,\xi]).$

Using $$\nabla^g_\xi d\theta = 0$$,

$d\theta(X,\nabla^g_\xi Y) = \xi \cdot d\theta(X,Y) – d\theta(\nabla^g_\xi X,Y)$

and similarly using $$\mathcal{L}_\xi d\theta = d\mathcal{L}_\xi \theta = 0$$,

$-d\theta(X,[Y,\xi]) = \xi \cdot d\theta(X,Y) – d\theta([\xi,X],Y).$

From which we see that

$d\theta(X,\nabla^g_Y\xi) = -d\theta(\nabla^g_\xi X,Y) + d\theta([\xi,X],Y) = -d\theta(\nabla^g_X\xi,Y).$

proving the lemma.

##### Claim. For horizontal $$X$$ it holds that $$\nabla^g_{JX}\xi + J\nabla^g_X\xi = 2X$$.

Proof. Let $$Y$$ be horizonal. It holds that

\begin{aligned} g(\nabla^g_{JX}\xi, Y) &= – g(\xi, \nabla^g_{JX}Y) \\ &= -\theta(\nabla^g_{JX}Y) \\ &= -\theta(\nabla^g_Y(JX)) – \theta([JX,Y]) \\ &= d\theta(JX,Y) – \theta(\nabla^g_Y(JX)) \\ &= 2g(X,Y) – \theta(\nabla^g_Y(JX)). \end{aligned}

On the other hand,

\begin{aligned} g(J\nabla_X\xi,Y) &= – g(\nabla^g_X\xi, JY) \\ &= g(\xi, \nabla^g_X(JY)) \\ &= \theta(\nabla^g_X(JY)) \end{aligned}

thus applying the last lemma, the conclusion follows.

###### Condition 6

For the final condition,

\begin{aligned} (\nabla_XJ)Y &= \nabla_X(JY) – J(\nabla_XY) \\ &= \nabla^g_X(JY) + \theta(X)J(JY) – \theta(JY)\nabla^g_X\xi + [(\nabla^g_X\theta)(JY)]\xi \\ &\qquad – J(\nabla^g_XY + \theta(X)JY – \theta(Y)\nabla^g_X\xi + [(\nabla^g_X\theta)Y]\xi) \\ &= \nabla^g_X(JY) + \theta(X)J^2Y + [(\nabla^g_X\theta)JY]\xi – J(\nabla^g_XY) – \theta(X)J^2Y + \theta(Y)J(\nabla^g_X\xi) \\ &= \nabla^g_X(JY) – J(\nabla^g_XY) + [(\nabla^g_X\theta)JY]\xi + \theta(Y)J(\nabla^g_X\xi) \\ &= (\nabla^g_XJ)Y + [(\nabla^g_X\theta)JY]\xi + \theta(Y)J(\nabla^g_X\xi) \\ &= Q(Y,X) \\ \end{aligned}

completing the proof.

We finish by remarking that the case of interest to us is when $$Q=0$$; this condition is equivalent to $$(M,\theta,J)$$ being a strongly pseudoconvex CR manifold. Moreover, $$\xi$$ will be a Killing field, and the foliation will be totally geodesic with bundle-like metric.

### References

[bh17] A. Banyaga, and D. Houenou. A Brief Introduction to Symplectic and Contact Manifolds. Vol. 15, World Scientific, 2017.

[tan89] S. Tanno. Variational problems on contact Riemannian manifolds. Trans. Amer. Math. Soc., 314(1):349–379, 1989.

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

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.