Is there a (canonical) Riemannian submersion from the complex hyperbolic space $\mathbb C\mathbb H^n$ into the hyperbolic space $\mathbb H^n$?


In the affirmative case, what can we say about the geometry of the fibers or about the O'Neill tensors (T and A)?



Riemannian submersions from complex hyperbolic space into the hyperbolic space
I don't have a complete answer, but here are a few remarks about this that you may find interesting or useful:


The OP didn't specify exactly what was meant by 'complex hyperbolic space' $\mathbb{CH}^n$ and 'hyperbolic space' $\mathbb{H}^n$, in the sense that the sectional curvatures of the two spaces weren't specified. I will take the meaning of $\mathbb{H}^n$ to be the simply-connected complete Riemannian manifold of constant sectional curvature $-1$ and $\mathbb{CH}^n$ to be the simply-connected, complete Hermitian symmetric space of constant holomorphic sectional curvature that contains $\mathbb{H}^n$ as a totally geodesic real submanifold. (Note that the sectional curvature of tangent $2$-planes in $\mathbb{CH}^n$ that are complex lines is $-4$, not $-1$. The sectional curvature function of $\mathbb{CH}^n$ takes values in the interval $[-4,-1]$. By convention, the sectional curvature of $\mathbb{CH}^1$ is $-4$ (so that the natural complex linear embedding $\mathbb{CH}^1\subset\mathbb{CH}^2$ will be an isometry), 
so $\mathbb{CH}^1$ is not isometric to $\mathbb{H}^2$, but to $\mathbb{H}^2$ with its metric divided by $4$.) 


Now, the obvious thing to try to construct a Riemannian submersion from $\mathbb{CH}^n$ to $\mathbb{H}^n$ doesn't work: The 'nearest point' projection from $\mathbb{CH}^n$ to the submanifold $\mathbb{H}^n\subset \mathbb{CH}^n$ (which *is* a smooth submersion) is *not* a Riemannian submersion.


The next obvious thing to try (especially if one wants a 'canonical' Riemannian submersion) is to look for one that is as homogeneous as possible. Now, a homogeneous example *does* exist for all $n$ (see a construction at the end of this answer), and the O'Neill tensors are easy to compute from the formuale. 


Now, it's not clear (to me) that this homogeneous example is the only one or even the 'most homogeneous' one, so it's not obvious that it should be regarded as 'canonical. However, a little calculation shows that a homogeneous Riemannian submersion is unique (up to a natural notion of equivalence) when $n=2$, the first nontrivial case. More precisely, one has the following results:


**Fact A:** There exists a subgroup $G\_n\subset\mathrm{Isom}(\mathbb{CH}^n)$ that acts simply transitively on $\mathbb{CH}^n$, a homomorphism $\rho\_n:G\_n\to\mathrm{Isom}(\mathbb{H}^n)$ such that $\rho\_n(G\_n)$ acts transitively on $\mathbb{H}^n$, and a Riemannian submersion $\pi\_n:\mathbb{CH}^n\to \mathbb{H}^n$ such that $\pi\bigl(g(m)\bigr) = \rho(g)\bigl(\pi(m)\bigr)$ for all $m\in\mathbb{CH}^n$. 


**Fact B:** When $n=2$, the homogeneous Riemannian submersion $\pi\_2$ is unique up to composition with isometries in $\mathbb{CH}^2$ and $\mathbb{H}^2$: If $\widehat\pi:\mathbb{CH}^2\to\mathbb{H}^2$ is a Riemannian submersion with the property that the group $G\subset \mathrm{Isom}(\mathbb{CH}^2)\times\mathrm{Isom}(\mathbb{H}^2)$ consisting of the pairs $(g,h)$ such that $\widehat\pi\bigl(g(m)\bigr) = h\bigl(\widehat\pi(m)\bigr)$ for all $m\in\mathbb{CH}^2$ acts transitively on $\mathbb{CH}^2$, then $\widehat\pi = h\circ \pi\_2\circ g$ for some $(g,h)\in \mathrm{Isom}(\mathbb{CH}^2)\times \mathrm{Isom}(\mathbb{H}^2)$.


**Remark:** There exist many Riemannian submersions $\pi:\mathbb{CH}^2\to\mathbb{H}^2$ that are not homogeneous. For example, there exist examples whose commuting isometry group $G\subset \mathrm{Isom}(\mathbb{CH}^2)\times \mathrm{Isom}(\mathbb{H}^2)$ (as defined above) acts in cohomogeneity $1$ or $2$ on $\mathbb{CH}^2$.


**Details (Version 2):**


Let $\mathrm{SU}(1,n)\subset\mathrm{GL}(n{+}1,\mathbb{C})$ be the connected subgroup such that its canonical left-invariant form has the expression
$$
\gamma = g^{-1}\mathrm{d}g = 
\pmatrix{
-\mathrm{tr}(\phi) & {}^t\bar\omega \\ 
\omega & \phi}
$$
where $\phi$ is of size $n$-by-$n$ satisfying $\phi + {}^t\bar\phi = 0$, and where $\omega$ is a column of $1$-forms of height $n$, and let $\mathrm{U}(n)\subset \mathrm{SU}(1,n)$ be the connected subgroup on which $\omega$ vanishes. Then $\mathbb{CH}^n = \mathrm{SU}(1,n)/\mathrm{U}(n)$, and the pullback of its metric to $\mathrm{SU}(1,n)$ is ${}^t\omega\circ\bar\omega$.


Now let $\iota:G\_n\hookrightarrow\mathrm{SU}(1,n)$ be the connected subgroup of (real) dimension $2n$ such that 
$$
\iota^\*(\gamma) = 
\pmatrix{
-i\,\alpha\_n & \alpha\_1-i\,\beta\_1 & \cdots & \alpha\_{n-1}-i\,\beta\_{n-1} &\alpha\_n-i\,\beta\_n\\
\alpha\_1+i\,\beta\_1 & 0 & \cdots & 0 & -\beta\_1+i\,\alpha\_1\\
\vdots & \vdots & & \vdots&\vdots\\
\alpha\_{n-1}+i\,\beta\_{n-1} & 0 & \cdots& 0 & -\beta\_{n-1} + i\,\alpha\_{n-1}\\
\alpha\_n-i\,\beta\_n & \beta\_1+i\,\alpha\_1 & \cdots & \beta\_{n-1} + i\,\alpha\_{n-1}
&i\,\alpha\_n
}
$$
where $\alpha\_1,\ldots,\alpha\_n,\beta\_1,\cdots\beta\_n$ are linearly independent left-invariant $1$-forms on $G\_n$ that satisfy
$$
\begin{aligned}
\mathrm{d}\alpha\_i &= \alpha\_i\wedge\beta\_n\,,\quad 1\le i < n\\
\mathrm{d}\alpha\_n &= 2\,\alpha\_1\wedge\beta\_1+2\,\alpha\_2\wedge\beta\_2
 +\cdots + 2\,\alpha\_n\wedge\beta\_n\,,\quad\\
\mathrm{d}\beta\_i &= \beta\_i\wedge\beta\_n\,,\quad 1\le i < n\\
\mathrm{d}\beta\_n &= 0.
\end{aligned}
$$
Note that $\iota^\*({}^t\omega\circ\bar\omega) 
= {\alpha\_1}^2+\cdots+{\alpha\_n}^2+{\beta\_1}^2+\cdots+{\beta\_n}^2$, which implies that the projection $g\mapsto g\mathrm{U}(n)$ from $G\_n$ to $\mathbb{CH}^n$ is a submersion, and completeness and left-invariance imply that it is a surjective covering map, so it is a diffeomorphism. In particular, $G\_n$ is simply-connected.


Meanwhile, the above structure equations imply that there exists a unique Lie group homomorphism $\rho:G\to\mathrm{SO}(1,n)$ such that
$$
\rho^{-1}\mathrm{d}\rho 
= \pmatrix{0 & \beta\_1 & \cdots & \beta\_{n-1} & \beta\_n \\
 \beta\_1 & 0 & \cdots & 0 & -\beta\_1\\
 \vdots & \vdots& & \vdots & \vdots\\
 \beta\_{n-1} & 0 & \cdots & 0 & -\beta\_{n-1}\\
 \beta\_n & \beta\_1 & \cdots &\beta\_{n-1} & 0}.
$$


Let $\mathrm{SO}(n)\subset\mathrm{SO}(1,n)$ be the subgroup that fixes the vector $(1,0,\ldots,0)\in\mathbb{R}^{1,n}$, so that $\mathbb{H}^n = \mathrm{SO}(1,n)/\mathrm{SO}(n)$. Then the above structure equations imply that there exists a unique smooth map $\pi\_n:\mathbb{CH}^n\to\mathbb{H}^n$ such that $\pi\bigl(g\mathrm{U}(n)\bigr) = \rho(g)\mathrm{SO}(n)$ for all $g\in G\_n$. Since $\rho$ pulls back the natural metric on $\mathbb{H}^n$ to be ${\beta\_1}^2+\cdots+{\beta\_n}^2$, it follows that, $\pi\_n$ is a Riemannian submersion of $\mathbb{CH}^n$ onto $\mathbb{H}^n$.


**Remark:** Because the horizontal distribution of $\pi\_n$ is defined by the equations $\alpha\_1 = \cdots = \alpha\_n = 0$, the above structure equations imply that the horizontal $n$-plane field of $\pi\_n$ (i.e., the orthogonal plane field to the fibers of $\pi\_n)$ must be integrable. In particular, O'Neill's $A$ tensor vanishes identically in this example. If we let $A\_i$ and $B\_i$ be the dual vector fields to $\alpha\_i$ and $\beta\_i$, then we find that O'Neill's $T$ tensor is
$$
T = B\_1\otimes 2\alpha\_1{\circ}\alpha\_n +\cdots+ B\_{n-1}\otimes2\alpha\_{n-1}{\circ}\alpha\_n + B\_n\otimes\bigl({\alpha\_1}^2+\cdots+{\alpha\_{n-1}}^2+2{\alpha\_n}^2\bigr).
$$


