SOLUTIONS TO EXAMPLE SHEET 3
A, B be unbounded geodesic metric spaces. Let P be a geodesic of length n in A,
Q be a geodesic of length n in B . Then we can concatenate (a parallel copy
of ) P with Q to make an `1 geodesic P Q. Doing this nitely many times also works.
2
On the other hand, P × Q ⊂ A × B is an isometrically embedded copy of [0, n] , which
we saw contains a part of a logarithmic spiral, of length unbounded as n → ∞, which
does not stay uniformly close (i.e. independently of n) to any geodesic. Since we can do
this for any n, and since the quasi-isometry constant for the spiral is independent of n,
(1) Let
and let
stability of quasi-geodesics fails.
(2) See hints.
(3) (Closest-point projection).
choose
(4)
y∈Y
so that
Suppose that
d(x, y) ≤ d(x, Y ) + 1
Y ⊂ X
(so y is
is
C quasiconvex.
Given
x ∈ X,
roughly a closest point). Suppose
y 0 ∈ Y has the same property. Consider the triangle [x, y][y, y 0 ][y 0 , x]. If y 00 ∈ [y, y 0 ],
00 lies C close to some z ∈ Y , by quasiconvexity. If d(y 00 , x0 ) ≤ δ for some
then y
0
x ∈ [x, y], then d(x, y 00 ) ≤ d(x, x0 ) + δ , so d(x, z) ≤ d(x, x0 ) + C + δ . On the other hand,
d(x, z) ≥ d(x, y) − 1, so d(x0 , y) ≤ δ + C + 1, so d(y, y 00 ) ≤ 2δ + C + 1. Hence the set
of possible such y has diameter bounded in terms of δ and C , so, up to bounded error,
the closest-point projection to Y is well-dened: just map x to some arbitrarily chosen
y ∈ Y with d(x, Y ) + 1 ≥ d(x, y).
If Y is not quasiconvex, one can't do this. The simplest example I can think of is the
2
following type of thing: let X be R, and let Y = {n : n ∈ N}. Then for each n, let
xn = (n2 + (n + 1)2 )/2. Then n2 and (n + 1)2 are equidistant from xn , but 2n + 1far
from each other, so it's not clear how to projection xn to Y .
0
2
The subgroup generated by a and a is isomorphic to Z , and this can't happen in a
hyperbolic group. Let T be the graph with a vertex for each left coset of A and a vertex
for each left coset of B , with an edge joining the vertex corresponding to gA to that
corresponding to hB when the cosets intersect. First note that T is a tree, because of
the free product decomposition. Next, since the vertices of T correspond to uniformly
bounded sets in Γ, and adjacency corresponds to intersection, T is quasi-isometric to Γ.
Putting the above two claims together (along with the fact that trees are hyperbolic and
Γ is hyperbolic. We thus have a cobounded
G by isometries on a hyperbolic space, namely Γ; neither this space, nor the
action, of G is proper, so we don't violate Milnor-Svarc.
The geodesic ngon A1 A2 · · · An is thin if for all i, we have that Ai is contained in
∪j6=i N (Aj ).
Here's a simple proof that geodesic ngons are (n − 2)δ thin in a δ hyperbolic space.
Indeed, let A1 · · · An be an ngon, n ≥ 3. Join the initial point of A1 to the initial
−1
point of An by a geodesic B , to form a geodesic triangle A1 An B and a (n − 1)gon
−1
A1 · · · An−1 B . By induction, the latter is (n−3)δ thin, and the former is δ thin, from
hyperbolicity is QIinvariant) shows that
action of
(5)
which the claim easily follows.
H(A) is the union of all geodesics that start and end in A. Let γ be a
0
0
0
geodesic joining p, q ∈ H(A). The denition provides ap , ap and aq , aq so that p ∈ [ap , ap ]
0
0
0
0
0
and q ∈ [aq , aq ]. Let σ be a geodesic from ap to aq . Then σ ⊂ H(A), since ap , aq ∈ A.
0
0
On the other hand, γ is contained in the union of the 2δ neighborhoods of [ap , p], [aq , q]
(6) By denition,
1
SOLUTIONS TO EXAMPLE SHEET 3
and
σ
2
2δ γ ⊂ N2δ (H(A)),
(to see this, consider the appropriate geodesic quadrilateral, which must be
thin). Each of those three segments is in
H(A),
and we conclude that
as required. This proves quasiconvexity.
Now let
A = {a1 , . . . , an }.
The claim is clear when
there is a quasi-isometric embedding
q0 : T 0 → X
n = 1.
By induction, assume that
of a nite tree
T0
that sends the leaves
0
to a1 , . . . , an−1 . Stability of quasigeodesics shows that im q has the property that, for
0
all i, j , there is a geodesic from ai to aj that lies uniformly close to im q . Since any two
2δ neighborhood of one another, we see
im q 0 is κ0 quasiconvex, where κ0 depends on δ and the quasi-isometry constant of
q 0 . Let p be a closest point of im q 0 to an , and let γ be a geodesic joining an to p.
0
0
Let T be a tree obtained from T by choosing t ∈ T mapping to p and then attaching
0
0
a γ to T by identifying p with t. Extend q to a map q : T → X by sending γ to γ in the
obvious way. Then T is a tree whose leaves get sent by q to a1 , . . . , an . Using the fact
0
that p is a closest point of im q to an , the usual bound-the-δ -overlap argument shows
that geodesics in T go to quasigeodesics in X , so q is a quasi-isometry. It follows from
the construction that q is a (1, f (n))quasi-isometry, where f depends exponentially on
n (one can do a little better with some more care). As above, any geodesic joining any
ai , aj lies κclose to the image of q , so im q is quasiconvex and indeed coarsely coincides
with H({a1 , . . . , an }).
geodesics with the same endpoints are in the
that
(7) This was done in a lecture, as it turns out.