Defining convex combinations in hyperbolic space

Question

Let $\mathbb H^n$ be the $n$-dimensional hyperbolic space. Given a sequence $u_0,\ldots,u_m$ of points in $\mathbb H^n$ and $t_0,\ldots,t_m$ nonnegative real numbers whose sum is $1$, let us define the convex combination $t_0u_0+\ldots+t_mu_m$ by induction on $m$:

• Case $m=1$. We define $t_0u_0+t_1u_0$ to be the only point $x$ of the geodesic arc $[u_0,u_1]$ such that $d(x,u_0)=t_1d(u_0,u_1)$
• Suppose such combinations have been defined for all sequences of $m+1$ points of $\mathbb H^n$. Let $u_0,\ldots,u_{m+1}$ be a sequence of points of $\mathbb H^n$ and $t_0,\ldots,t_{m+1}$ be nonnegative reals whose sum is $1$. If $t_{m+1}=1$, we define $0u_0+\ldots+0u_m+1u_{m+1}:=u_{m+1}$. Otherwise $1-t_{m+1}\neq0$, and $$\sum_{i=0}^m\frac{t_i}{1-t_{m+1}}=\sum_{i=0}^m\frac{t_i}{t_0+\cdots+t_m}=1,$$ and hence $\frac{t_0}{1-t_{m+1}}u_0+\cdots+\frac{t_m}{1-t_{m+1}}u_m$ is defined, and we define $t_0u_0+\cdots+t_{m+1}u_{m+1}$ to be the convex combination $$(t_0+\cdots+t_m)\left(\frac{t_0}{1-t_{m+1}}u_0+\cdots+\frac{t_m}{1-t_{m+1}}u_m\right)+t_{m+1}u_{m+1},$$ using the previous case.

I want to know whether this construction depends on the order of the sequence $u_0,\ldots,u_m$.

It is easy to prove that $t_0u_0+t_1u_1=t_1u_1+t_0u_0$. Using this I have proved that if $t_0u_0+t_1u_1+t_2u_2=t_0u_0+t_2u_2+t_1u_1$ holds for all convex combinations of three points, then the convex combinations of a finite sequence of points in $\mathbb H^n$ does not depend on the order of the sequence.

However cannot see why $t_0u_0+t_1u_1+t_2u_2=t_0u_0+t_2u_2+t_1u_1$ holds. It is enough to prove this in $\mathbb H^2$. My first question is: is this true?

I have managed to prove this assignment is differentiable, and an smooth embedding in the interior of $\Delta^m$ if $u_0,\ldots u_m$ are in general position in $\mathbb H^n$; this means they do not belong to a hyperbolic subspace of dimension $\leq m-1$.

My second question is: if the first question is false, is there a way to define convex combinations of points of $\mathbb H^n$ which does not depend on the order, it is preserved under isometries, and when the points are in general position the restriction of this function is an embedding?

Answer 1

The best way to do this is to use the Lorentzian model of the hyperbolic space. Let $L^+\subset R^{n,1}$ denote the future cone of time-like vectors $v$, $\langle v, v\rangle <0$ (I am using the negative of the bilinear form from the wikipedia article). The projection of $L^+$ to $RP^n$ is then the Klein model of the hyperbolic space $H^n$. Any finite subset $p_1,...,p_k\in H^n$ corresponds to a set of vectors $v_1,...,v_k\in L^+$. The affine convex combination of these vectors is $$ {\mathbf v}=\sum_{i=1}^k t_i v_i, $$ where $t_i\ge 0, \sum_i t_i=1$. Lastly, project ${\mathbf v}$ to $p\in H^n$. This is your convex combination of points $p_1,...,p_k\in H^n$. The construction is clearly independent of the ordering of the points.

Claim. If the vectors $v_1,...,v_k$ are linearly independent then the map ${\mathbf t}\mapsto p$ is 1-1.

Proof. Since the vectors $v_1,...,v_k$ are linearly independent, the affine $k$-dimensional subspace $A$ which they determine in $R^{n,1}$ does not contain the origin. Hence, the projection $A\to RP^n$ is 1-1. The map $$ {\mathbf t}\mapsto {\mathbf v} $$ is injective (again, by the linear independence of the vectors) and its image is contained in $A$. Hence, the composition $$ {\mathbf t}\mapsto {\mathbf v} \mapsto p $$ is 1-1. qed

The last thing to note is that linear independence of the vectors $v_1,...,v_k$ is equivalent to the property that the corresponding points $p_1,...,p_k\in H^n$ span a $k$-dimensional hyperbolic subspace in $H^n$ (i.e. are not contained in a hyperbolic subspace of dimension $< k$).

Answer 2

The answer to this question is: No and Yes.

No: As you defined it, the convex combination depends on the order in which you do the averages.

Let us define the iteration function $I:H^2\times[0,1]\to H^2$ to be such that $I(x,y,\alpha)$ is the (unique) point in the geodesic between $x$ and $y$ which is at a distance $\alpha d(x,y)$ from $x$ and at a distance $(1 - \alpha) d(x,y)$ from $y$.

The question asked, for example, if whenever $\alpha_1 + \alpha_2 + \alpha_3 = 1$, $\alpha_i \ge 0$, do we have

$$I(I(x_1,x_2,\frac{\alpha_1}{\alpha_1 + \alpha_2}), x_3, \alpha_1 + \alpha_2) = I(I(x_2,x_3,\frac{\alpha_2}{\alpha_2 + \alpha_3}), x_1, \alpha_2 + \alpha_3)?$$

The answer is no.

But also, Yes. If you define it correctly, you can get something like what you are seeking: a center of mass.

See: Saul Stahl, "Mass in the hyperbolic plane". Briefly, given two point-masses $(X, x)$ and $(Y,y)$, where $X,Y$ are in the hyperbolic plane and $x, y \ge 0$ are real numbers (the mass values), one may define the centroid $(C, c) = (X,x) \ast (Y,y)$, so that $C$ is the unique point in the geodesic between $X$ and $Y$ such that

$$x \sinh XC = y \sinh YC $$

and $c$ is given by

$$c = x \cosh XC + y \cosh YC $$

($XC$, $YC$, $XY$ are distances between the points)

It then turns out that $\ast$ is associative and commutative. One may then define the convex combination as the point given by the centroid $(u_1, t_1) \ast \dots \ast (u_m, t_m)$.

Let me attach the following image, for illustration. Let us look at the hyperboloid model $H^2$ in Minkowski space $M^3$ with bilinear form $\langle x, y \rangle = x_1 y_1 + x_2 y_2 - x_3 y_3$. Let $u_1 = (0,0,1)$, $u_2 = (\sinh 1,0,\cosh 1)$ and $u_3 = (0, \sinh 1, \cosh1)$. Let us then draw all the centroids $(u_1, t_1) \ast (u_2, t_2) \ast (u_3, t_3)$ for all $t_1 + t_2 + t_3 = 1$ in steps of $0.03$. We get this: