## What spaces have well known horofunctions?

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Following Gromov, take a metric space $(X,d)$ and consider $C(X)/\mathbb{R}$ the set of continuous functions to $\mathbb{R}$ with the topology of uniform convergence on compact sets after taking the quotient by constant functions (i.e. two functions are equivalent if the difference is a constant). Embed $X$ into this space by means of the map:

$x \mapsto f(x) = [d(x,\cdot)]$

A horofunction is an element of the closure of $f(X)$ that is not in $f(X)$.

In $\mathbb{R}^d$ all horofunctions are given by inner product with a unit vector.

In the upper-half plane model of the hyperbolic disc the function $h(z) = \log(\text{Im}(z))$ is a horofunction and all others can be obtained by composing with hyperbolic isometries.

What other spaces have well known horofunctions?

For example, are horofunctions of the model geometries $\text{Nil}$, $\text{Sol}$, and $\widetilde{\text{Sl}(2,R)}$ in dimension $3$ known relatively explicitly?

I know there is a general relationship between horofunctions and geodesics (Busseman functions) in non-positively curved spaces. However I'm interested in spaces where one can compute relatively explicitly (e.g. spaces where one knows what the distance function looks like).

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Just to supplement your posting, the horofunction definition (first?) appears in Gromov's 1980 paper, "Hyperbolic manifolds, groups and actions." ihes.fr/~gromov/PDF/hyperbolic_manifold.pdf – Joseph O'Rourke Sep 24 2010 at 18:49
Hi Pablo! Just to see if I've got something.. If X is compact there are no horofunctions, are there? – Michele Triestino Sep 24 2010 at 18:54

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@Michele: You're correct, no horofunctions if the space is compact (the idea is that adding the horofunction one obtains a nice compactification of $X$ by "directions to infinity"). – Pablo Lessa Sep 24 2010 at 19:16
and yet another silly question: does the same conclusion of my previous comment hold for discrete spaces (I'm thinking of graphs, trees actually)? – Michele Triestino Sep 24 2010 at 19:17

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Understanding Busemann functions involves understanding rays, which I think is a difficult task even for surfaces of revolution in $\mathbb R^3$. – Igor Belegradek Sep 24 2010 at 23:25
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## 3 Answers

Of course, the first example should be non-compact Riemannian symmetric spaces, where the Busemann (horofunctions in your terminology) functions are known in a pretty explicit form. I don't think there are other explicit examples.

More comments.

1) Geodesic rays always converge in this Busemann-Gromov compactification, and it has nothing to do with curvature.

2) I never understood why the definition of this compactification is formulated in terms of uniform convergence on compact sets instead of plain pointwise convergence (anyway, for Lipschitz functions the result is the same). Actually, this is a particular case of the general Constantinescu-Cornea compactification (see the book of Brelot "On Topologies and Boundaries in Potential Theory"), other examples of which are the Martin compactification in potential theory or Thurston compactification in the Teichmuller theory.

ADDED REFERENCES

In what concerns Busemann or horofunctions (I prefer to call them Busemann cocycles, because in invariant language they are cocycles, not functions) for symmetric spaces, there are several sources of various degree of explicitness.

(1) For the group $SL(d,\mathbb R)$ (and the associated symmetric space) the Busemann cocycle essentially appears in Furstenberg's formula for the rate of growth of random matrix products, see

MR0163345 (29 #648) Furstenberg, Harry Noncommuting random products. Trans. Amer. Math. Soc. 108 1963 377--428.

From the geometrical point of view the main idea there is that if you want to find the linear rate of growth of $d(o,g_1 g_2\dots g_n o)$, where $o$ is a reference point, and $g_i$ is a stationary sequence of random isometries, then you can look at the increment $d(o,g_1 g_2\dots g_n o)-d(o,g_2 g_3\dots g_n o)=d(g_1^{-1}o,g_2g_3\dots g_n o)-d(o,g_2 g_3\dots g_n o)$, which converges to the Busemann cocycle $\beta_\gamma(o,g^{-1}o)$ provided $g_2g_3\dots g_n o$ converges to a boundary point $\gamma$ in the Busemann compactification. The cocycle itself looks, roughly speaking, like $\log \|gv\|/\|v\|$, where $g$ is the matrix (or its exterior power) representing a point in the symmetric space, and $v$ is a vector representing the boundary point. There is also a lot about it in later papers by Guivarc'h.

(2) Busemann cocycles naturally appear in various works related to compactifications of symmetric spaces. Historically the first source is the monograph of Karpelevich

MR0231321 (37 #6876) Karpelevic, F. I. The geometry of geodesics and the eigenfunctions of the Beltrami-Laplace operator on symmetric spaces. Trudy Moskov. Mat. Obšc. 14 48--185 (Russian); translated as Trans. Moscow Math. Soc. 1965 1967 pp. 51--199. Amer. Math. Soc., Providence, R.I., 1967.

where he explicitly discusses pencils of convergent geodesics and introduces the associated horospheric coordinates. Later expositions are in two books on compactifications of symmetric spaces:

MR1633171 (2000c:31006) Guivarc'h, Yves(F-RENNB-IM); Ji, Lizhen(1-MI); Taylor, J. C.(3-MGL) Compactifications of symmetric spaces. (English summary) Progress in Mathematics, 156. Birkhäuser Boston, Inc., Boston, MA, 1998. xiv+284 pp. ISBN: 0-8176-3899-7

and

MR2189882 (2007d:22030) Borel, Armand(1-IASP); Ji, Lizhen(1-MI) Compactifications of symmetric and locally symmetric spaces. Mathematics: Theory \& Applications. Birkhäuser Boston, Inc., Boston, MA, 2006. xvi+479 pp. ISBN: 978-0-8176-3247-2; 0-8176-3247-6

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Great! Thank you.  Do you have some reference for Busemann functions of non-compact Riemannian symmetric spaces? – Pablo Lessa Sep 25 2010 at 9:53

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One possible place to look (other than symmetric spaces) is the warped product of hyperbolic metrics, a metric of the form $dr^2+e^{C_1r} dx^2 + e^{C_2r}dy^2$, $C_1, C_2 >0$. Then $r$ is a horofunction. If one could compute the horofunction for another point in the Gromov boundary, then since the isometry group acts transitively on these other points of the Gromov boundary, one would have a computation for all points. However, I imagine computing the other horofunctions could be tricky, since I don't know how to compute the geodesics explicitly.

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some remarks, not an answer. It seems, from [J. Cheeger and D. Gromoll, The splitting theorem for manifolds of nonnegative Ricci curvature, J. Diff. Geom. 6 (1971) 119-128.] it follows that horofunctions in Nil, Sol and \tilde Sl(2,R) are super-harmonic? (Ricci curvature is non-positive?) I know that the spaces of horofunction, Busseman functions coincide with Gromov ideal boundary (spaces of rays) when the sectional curvature is non-positive. (For non-negative curvature these ideal boundaries might be different. For the Heisenberg group Heis^{2n+1} with left-invariant metric the Gromov ideal boundary is the sphere S^{2n-1} with Carnot-Caratheodory metric, and geodesics equations are known - but not horofunctions :( )

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Thanks!  I'll look into this.  Any reference for the facts about the Heisenberg group? – Pablo Lessa Sep 24 2010 at 20:04
valeri probably meant to say "Ricci curvature is nonnegative" else the splitting theorem does not apply, but more to the point the homogeneous metrics on Sol, Nil, $SL(2,\mathbb R)$ do not have nonnegative Ricci curvature. Indeed, compact manifolds of nonnegative Ricci curvature have virtually abelian fundamental group, which is not true for compact manifolds modeled on Sol, Nil, $SL(2,\mathbb R)$.  – Igor Belegradek Sep 24 2010 at 21:05
to Pablo: there are some papers by Kaplan (if I remember correctly) and by Marenich in Geom Dedicata 66:2 with calculations. To Igor: of course, Cheeger Gromoll for non-negative, but what I mean is that for non-positive their arguments will give super-harmonic, is this wrong? – valeri Sep 24 2010 at 21:28
@valeri, I haven't though enough of nonpositive Ricci curvature, but I do not see why what you assert is true. The proof of subhamonicity (of Busemann functions on manifolds of nonnegative Ricci curvature) that I know hinges on Laplacian comparison. To prove the latter one uses Cauchy-Schwarz inequality to get a lower bound bound the Hessian in terms of the Laplacian. Then Weitzenbock formula and lower bound on Ricci finishes the proof. If you have nonpositive Ricci, the Cauchy-Schwarz goes the wrong way and this argument does not work (I think). How do you make it work? – Igor Belegradek Sep 24 2010 at 23:20
@Igor - " haven't though enough of nonpositive Ricci curvature" - me too :). My guess was that in Hadamard manifolds horofunctions are smooth, so Laplacian comparison with flat tangent space (where horofunctions are linear) via exponential map will do. Probably not ... I can not state smth right now.  – valeri Sep 25 2010 at 8:21
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