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 $[p_{0},p]$.

  • $\begingroup$ I do not see why this should be the case. Did you at least check the case of the hyperbolic plane, where everything can be computed easily? $\endgroup$ – Benoît Kloeckner Jul 20 '13 at 11:22
  • $\begingroup$ This is clearly true in real-hyperbolic case; you should check complex-hyperbolic case, where bisectors are explicitly computed (see e.g. Goldman's book). $\endgroup$ – Misha Jul 20 '13 at 13:31
  • $\begingroup$ In the complex projective case this follows from the theorem of cosines that can be found in my paper here. Later I noticed that the formula had already been discovered by Shirokov in the 1950s. There is a dual formula in the complex hyperbolic case. $\endgroup$ – Mikhail Katz Jul 24 '13 at 10:20
  • $\begingroup$ @Misha, katz: I think we did not understand the question in the same way. Do you talk about exponentiating a linear subspace from $m$, or an affine subspace from $p_0$? $\endgroup$ – Benoît Kloeckner Jul 24 '13 at 12:25
  • $\begingroup$ As Benoît Kloeckner pointed out, I misunderstood the question. The OP is apparently not exponentiating from the midpoint. My answer is related to showing that the equidistant hypersurface is the image of the exponential map from the midpoint of the segment. $\endgroup$ – Mikhail Katz Jul 24 '13 at 12:29

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.

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  • $\begingroup$ You are right, I misunderstood the question (didn't pay enough attention to the "affine" part). $\endgroup$ – Mikhail Katz Jul 24 '13 at 12:27
  • $\begingroup$ Thanks. Could you expand on the argument in the 2nd paragraph, or provide a reference? I'm just a novice with Riemannian geometry and although I find this intuitive, I don't see the argument. I would also like to know if the following works: extend the geodesic $[p_{0},p]$ to $[-p,p]$ with midpoint $p_{0}$. Compute the bisector at $p_{0}$ in the way you described and then translate it along the geodesic $[-p,p]$ until you get to the midpoint of $[p_{0},p]$. Is that the bisector between $[p_{0},p]$? By translate, I think I mean using a 1 parameter subgroup gotten by exponentiating the... $\endgroup$ – A. Pascal Jul 24 '13 at 13:07
  • $\begingroup$ line through the tangent space at $p_{0}$ in the direction of the geodesic. This is the intuition I had for my original question, but I guess that translation in the tangent space giving an affine hyperplane is simply not compatible with translation in the symmetric space after exponentiating from the tangent space. $\endgroup$ – A. Pascal Jul 24 '13 at 13:10
  • $\begingroup$ @A.Pascal: I have probably been quite optimistic, so I tempered my claim a bit. I will try to think, but in Hadamard manifolds this seems too bold as the bissector is a global object and should not in general be constructed in such a local way. $\endgroup$ – Benoît Kloeckner Jul 24 '13 at 14:10
  • $\begingroup$ (continued) The claim in the second paragraph should hold in rank 1 symmetric spaces, but I am not sure for higher rank ones except in very specific cases. You should try the case of $\mathbb{H}^2\times \mathbb{R}$: it is the simplest higher-rank case, and appears anyway inside every other ones. $\endgroup$ – Benoît Kloeckner Jul 24 '13 at 14:11

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.

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  • $\begingroup$ For those interested in exponentiating from the midpoint, I would be interested in seeing a proof less involved than the one using Leuzinger's detailed results. $\endgroup$ – Mikhail Katz Jul 24 '13 at 12:30
  • $\begingroup$ Thanks. One would like to argue just like this. Leuzinger's result applies to geodesic triangles whose geodesic sides are regular. Is that clearly the case here? What if the geodesic $[p_{0},p]$ is singular? $\endgroup$ – A. Pascal Jul 24 '13 at 13:24
  • $\begingroup$ A generic one is regular, so for the purposes of merely showing that the exponential map gives the equidistant surface that's enough by a suitable passage to the limit. But perhaps there is a simpler argument for showing this that does not require trigonometry on a symmetric space. $\endgroup$ – Mikhail Katz Jul 24 '13 at 13:32
  • $\begingroup$ OK. Thanks. Although the trigonometry is hard to set up, it is intuitive to use, no? $\endgroup$ – A. Pascal Jul 24 '13 at 13:37
  • $\begingroup$ Yes. For example, in the complex projective case, one gets a law of cosines that's identical with the spherical case except for one additional term involving the angle in the tangent space between the complex lines spanned by the two directions. $\endgroup$ – Mikhail Katz Jul 24 '13 at 13:45

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.

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