I was looking at the submanifolds in the Poincare Ball model of the real hyperbolic space. It has been mentioned in various places that the totally geodesic hypersurfaces are the parts of the sphere that intersect the Euclidean sphere orthogonally. On the other hand, many authors have also used as a fact that the $k$-dimensional totally-geodesic submanifolds of the Poincare Ball are parts of the spheres that intersect the boundary of the Poincare ball orthogonally (see for instance this paper by Berenstein.)
To prove this, I was trying to use the characterization of totally-geodesic submanifolds in the upper sheet of the two sheeted Hyperboloid (since I am a bit more comfortable in that setting). Particularly, I know that given $\mathbb{H}^n = \left\lbrace \left( \xi, \tau \right) \in \mathbb{R}^{n + 1} | \| \xi \|^2 - \tau^2 = -1, \tau > 0 \right\rbrace$, the $k$-dimensional totally-geodesic submanifolds are precisely the intersections of the $(k + 1)$-dimensional subspaces of $\mathbb{R}^{n + 1}$ with $\mathbb{H}^n$. I also know that if $\mathbb{B}^n$ is the Poincare Ball model of $\mathbb{H}^n$, then the stereographic projection $\pi: \mathbb{H}^n \rightarrow \mathbb{B}^n$, given by $\pi \left( \xi, \tau \right) = \frac{\xi}{1 + \tau}$ is an isometry of the two hyperbolic spaces. Hence, $\pi$ must take the totally-geodesic submanifolds in $\mathbb{H}^n$ into totally-geodesic submanifolds in $\mathbb{B}^n$. However, I cannot see in this formulation as to how these images are parts of spheres that intersect the Euclidean sphere orthogonally?
Any insights or references to get a proof of this fact are appreciable!
PS: I also have an intuition that we can characterize the $k$-dimensional totally-geodesic submanifolds in $\mathbb{B}^n$ as follows:
A $k$-dimensional submanifold $M$ in $\mathbb{B}^n$ is totally geodesic if there is a sphere $S$ that intersects $\mathbb{S}^{n - 1}$ orthogonally, $M \subseteq S$, and as a $k$-dimensional submanifold of $S$, $M$ is totally-geodesic. This characterization seems to be consistent with the claim I want to prove. However, an independent proof of this statement is also appreciable!
EDIT: I have tried some calculations that show a partial result. At least for the case of geodesics and codimension-$1$ totally geodesic submanifolds. For the sake of computations and make the notations precise, I will use the same definition of $\mathbb{H}^n$ as mentioned above. First, we look at the case of $\mathbb{H}^2$. Here, the geodesics and the codimension-$1$ totally geodesic submanifolds are the same.
In the upper sheet of hyperboloid, any geodesic is given by the intersection of $\mathbb{H}^2$, with $2$-dimensional subspaces of $\mathbb{R}^3$. Let $V$ be such a subspace. That is to say, there is a vector $\left( a, b, c \right) \in \mathbb{S}^{2}$ such that every point $\left( x, y, z \right) \in V$ satisfies the equation:
$$ax + by + cz = 0.$$
In the intersection $V \cap \mathbb{H}^2$, the points $\left( x, y, z \right)$ additionally satisfy $x^2 + y^2 - z^2 = -1$. On the other hand, the images of these points under the stereographic projection $\pi: \mathbb{H}^2 \rightarrow \mathbb{B}^2$ is given by $\pi \left( x, y, z \right) = \left( \frac{x}{1 + z}, \frac{y}{1 + z} \right) =: \left( u, v \right)$. Now, let us consider the following two cases:
Case I: If $c = 0$, then we have $ax + by = 0$, and this subspace contains the $Z$-axis as a whole. It is natural to expect that the image of the intersection with $\mathbb{H}^2$ is a line through origin in $\mathbb{B}^2$. It is easy to see that this is indeed the case, since we have $u = \frac{x}{1 + z}$, $v = \frac{y}{1 + z}$, and hence $au + bv = \frac{ax + by}{1 + z} = 0$. That is, any geodesic that passes through $\left( 0, 0, 1 \right) \in \mathbb{H}^2$ gets mapped into a line through origin in $\mathbb{B}^2$.
Case II: Consider $c \neq 0$. Then, we have $z = - \frac{ax + by}{c}$. Upon substituting for $u$ and $v$, we get $u = \frac{cx}{c - ax - by}$, $v = \frac{cy}{c - ax - by}$. Solving for $x$ and $y$ in terms of $u$ and $v$, we get
$$x = \frac{c^2 u}{c^2 + acu + bcv}, y = \frac{c^2y}{c^2 + acu + bcv}.$$
We also have from the equation of the hyperboloid, that
$$z^2 = 1 + x^2 + y^2,$$
which, together with the equation of the plane, gives
$$\left( a^2 - c^2 \right) x^2 + \left( b^2 - c^2 \right) y^2 + 2abxy = c^2.$$
Substituting $x, y$ in terms of $u$ and $v$, we get
$$c^2 u^2 + c^2 v^2 + 2acu + 2bcv = - c^2,$$
which is the equation of a circle in $\mathbb{R}^2$, centered at $\left( - \frac{a}{c}, - \frac{b}{c} \right)$ and of radius $\frac{\sqrt{1 - 2c^2}}{c}$. Also, this circle is orthogonal to the unit circle centered at $0$ in $\mathbb{R}^2$.
Similarly, one can proceed for the $(n - 1)$-dimensional totally-geodesic submanifolds and get the result that these are precisely the ``spherical caps" that intersect the unit sphere orthogonally (I am skipping the exact calculations since they are on the same lines as above).
Now, I wish to do the same thing with $k$-dimensional totally-geodesic submanifolds. For the same, I consider the following:
Let $\xi \subseteq \mathbb{H}^n$ be a $k$-dimensional totally-geodesic submanifold. Then there is a $(k + 1)$-dimensional subspace $V$ of $\mathbb{R}^{n + 1}$ such that $\xi = V \cap \mathbb{H}^n$. However, algebraically, to get a $(k + 1)$-dimensional subspace in $\mathbb{R}^{n + 1}$, we require $(n - k)$-many constraints. That is, a $(k + 1)$-dimensional subspace is the intersection of $( n - k)$-many $(n - 1)$-dimensional subspaces. Let $V_1, \cdots, V_{n - k}$ be these subspaces. Then, $\xi = \bigcap\limits_{j = 1}^{n - k} (V_j \cap \mathbb{H}^n )$. Then, the image of $\xi$ under the stereographic projection would be $$. It is at this point, where we notice that for every $j = 1, \cdots, n - k$, the set $\pi \left( V_j \cap \mathbb{H}^n \right)$ is an $(n - 1)$-dimensional totally-geodesic submanifold of $\mathbb{B}^n$, and hence $\pi \left( \xi \right)$ is the intersection of $(n - k)$-many such spherical caps that intersect the unit sphere orthogonally.
This does complete the proof that every $k$-dimensional totally geodesic submanifold in the Poincare Ball model of the real hyperbolic space is a $k$-dimensional part of spherical caps that intersect the boundary (i.e., the unit sphere) orthogonally. However, one question still remains to be solved
Question: Can we say that this $k$-dimensional part of the spherical cap that intersects the unit sphere orthogonally is a $k$-dimensional totally-geodesic submanifold of the sphere?