9
$\begingroup$

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)?

$\endgroup$

2 Answers 2

11
$\begingroup$

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 $$ T = B_1\otimes 2\alpha_1{\circ}\alpha_n +\cdots+ B_{n-1}\otimes2\alpha_{n-1}{\circ}\alpha_n + B_n\otimes\bigl({\alpha_1}^2+\cdots+{\alpha_{n-1}}^2+2{\alpha_n}^2\bigr). $$

$\endgroup$
6
  • 1
    $\begingroup$ I gather $G$ and $\rho(G)$ must be the stabilizers of points at infinity in $\mathrm{Isom}(\mathbb{CH}^2))$ and $\mathrm{Isom}(\mathbb{H}^2)$, respectively. I think the former is a semidirect product of a Heisenberg group with $\mathbb R$, where $\mathbb R$ permutes horospheres. Now I have trouble visualising the kernel of $\rho$. The center of the Heisenber group must lie in the kernel, while the above mentioned copy of $\mathbb R$ should not. I do not think such $\rho $ exist. What am I missing? Could you describe $\rho$? $\endgroup$ Aug 6, 2017 at 2:44
  • $\begingroup$ @IgorBelegradek: I hope that the revised and generalized answer is helpful. Let me know if you still have questions. $\endgroup$ Aug 9, 2017 at 0:55
  • $\begingroup$ While I understand the idea of your construction, I still do not see how to answer the question in my first comment, i.e., what is the kernel of $\rho$? $\endgroup$ Aug 9, 2017 at 1:27
  • 1
    $\begingroup$ @IgorBelegradek: In the new version, which provides an example for all $n$ (and simplifies the $n=2$ example), the kernel of $\rho_n$ is simply an abelian group of dimension $n$. (The structure equations for the kernel are got by setting all of the $\beta_i$ to zero in the structure equations for $G_n$ itself. This leaves $\mathrm{d}\alpha_i=0$ for $1\le i\le n$, which are the structure equations of an abelian group.) The fibers of $\pi_n$ are copies of $\mathbb{R}^n$ endowed with a flat, complete metric (but, of course, they are not totally geodesic in $\mathbb{CH}^n$). $G_n$ has no center. $\endgroup$ Aug 9, 2017 at 8:55
  • $\begingroup$ Thank you, this helped me understand the situation better. I was not familiar with this way of thinking about $\mathbb{CH}^n$. $\endgroup$ Aug 9, 2017 at 12:14
1
$\begingroup$

This is not exactly an answer (but maybe too long for a comment). Essentially, the only semi-Riemannian submersion from a complex pseudo-hyperbolic space onto any manifold whose fibers are complex, connected, and totally geodesic is the semi-Riemannian submersion from $\mathbb{CH}_1^{2n+1}$ to the quaternionic hyperbolic space $\mathbb{H}H^n$. (This is the twistor space construction over $\mathbb{H}H^n$). This is proved by G. Baditoiu and S. Ianus in Theorem 2.8 of this paper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.