Equidistant hypersurfaces in symmetric space via exponentiation?

Question

Here's some background and notation:

Let $G/K$ be a symmetric space of non-compact type. For concreteness, assume $G$ is in fact a classical simple real Lie group such as SL,SO, or Sp, and $K$ is a maximal compact subgroup in $G$.

The tangent space to $p_{0}=1 \in G/K$ is identified with $\frak{p}$ where $\frak{p}$ is an orthogonal complement to the Lie algebra of $K$ inside the Lie algebra of $G$, with respect to the trace pairing on the latter. Furthermore, the exponential map $\exp: {\frak p} \rightarrow G$ followed by the quotient map $G \rightarrow G/K$ is a diffeomorphism. Put another way, given a point $p \in G/K$, there is a unique $x \in \frak{p}$ so that $\exp(x)=p$, and the unique length-parametrized geodesic connecting $p_{0}$ to $p$ is $\exp(t x/|x|)$.

I would like to understand the "equidistant hypersurface" $H_{p_{0},p}$ between $p_{0}$ and $p$, consisting of those points $q$ in $G/K$ that are equidistant from $p_{0}$ and $p$.

My question is whether this equidistant hypersurface $H_{p_{0},p}$ can be obtained by exponentiating the affine hyperplane in $\frak{p}$ that passes through the centre of and is orthogonal to the line connecting $0$ and $x$, where $\exp(x)=p$.

This seems intuitively correct, but sometimes intuition is misleading in higher rank symmetric spaces.

EDIT: The answer is no for the question as stated (see answers below), so I would like to revise the question to the following:

1. Explain why the bisector between $p_{0}$ and $p$ is obtained by exponentiating the hyperplane orthogonal to $[p_{0},p]$ in the tangent space at the midpoint $m \in [p_{0},p]$.

2. Explain whether the following is true: we can compute the bisector between $p_{0}$ and $p$ by looking at the whole geodesic line through $p_{0}$ and $p$, computing the bisector between $p$ and $-p$ by exponentiating from the tangent space at $p_{0}$, and then translating this bisector along the geodesic from $p_{0}$ to the midpoint of $[p_{0},p]$.

Answer 1

The answer of katz below makes me think there might be some misunderstanding about what the question asks (possibly on my part). I will therefore start by restating it in my own words, then try to answer it. Please tell if the reformulation is wrong. Note that I cast everything in terms of Riemannian geometry.

What might be true is that the bissector is the set of points whose projection to nearest point on $[p_0,p]$ is the middle point $m$; and that this set is obtained by exponentiating from $m$ the linear hyperplane orthogonal to $[p_0,p]$. Now, as I understand it, the question is whether one gets the bissector by exponentiating from $p_0$ a affine hyperplane.

Take the example of the hyperbolic plane: $p_0$ is any point, $v$ any vector in $T_{p_0} \mathbb{H}^2$ and let $A$ be the affine line of $T_{p_0} \mathbb{H}^2$ that contains $v$ and is orthogonal to it in the metric $g_{p_0}$. Parametrize it as the curve $\sigma_t$ and let $\gamma_t = \exp_p(\sigma_t)$ be the curve you would like to be a bissector (of $p_0$ and $p=\exp_{p_0}(2v)$). Note that the vector $\sigma_t\in T_{p_0} \mathbb{H}^2$ makes an angle with $v$ that goes to $\pm\pi/2$ when $t\to\pm\infty$.

Now, the only geodesic $\eta_t$ such that the segment $[p_0,\eta_t]$ makes an angle going to $\pm\pi/2$ with $v$ when $t\to\pm\infty$ is the geodesic going through $p_0$ orthogonally to $[p_0,p]$. But in the real hyperbolic plane, bissectors are precisely the geodesics, therefore, the answer to your question as I understand it is no.

Answer 2

Consider the midpoint $m$ of the geodesic segment $[p_0,p]$, and consider a point $q$ on a geodesic perpendicular to $mp$. The right-angle triangle $pmq$ satisfies a law of cosines due to Leuzinger (page 273, theorem 1). This says that the length of the third side $pq$ is uniquely determined by the infinitesimal data at the vertex $m$ that can be described in terms of the Weyl group action. This shows that the distance $p_0q$ will be the same.

Answer 3

Here is another argument to convince oneself that the affine plane is not sent to the equidistant hypersurface even in the constant curvature case. By well-known dualities, if this were true in the hyperbolic case it would also be true in the spherical case. But now take a pair of opposite points on the sphere, say north and south poles, and the midpoint of a minimizing segment between them (the segment is of course not unique). The equidistant hypersurface in this case is just the equator. But if one exponentiates an affine line in the tangent plane, one will eventually hit the opposite pole, because the affine line must meet the circle of radius $\pi$ in the tangent plane.