
§ 1.1 Volume expanding endomorphisms without invariant splitting 21
Let R
m
i
= {x ∈ R
n
: k
−
i
+ m < x
i
< k
+
i
+ m} with m ∈ Z, 1 ≤ i ≤ n and x
i
is the
i−th coordinate of x. Thus, U
0
⊂
m∈Z,1≤i≤n
R
m
i
. Denote by L
+
i
= {x ∈ R
n
: x
i
= k
+
i
} and
L
−
i
= {x ∈ R
n
: x
i
= k
−
i
}. Let
˜
f be the lift of f.
The next claim answer the third question stated at the beginning of the section.
Claim 1.3 Let m > 2
√
n be fixed. Given any arc γ in R
n
with diam(γ) > m, t here
exist an arc γ
′
⊂ γ, 1 ≤ i ≤ n and j ∈ Z such that ∂γ
′
∩ (L
+
i
+ j), ∂γ
′
∩ (L
−
i
+ j + 1)
and P
j
i
(γ
′
) ⊂ [k
+
i
+ j, k
−
i
+ j + 1]. Moreover, γ
′
admits a nice cylinder, γ
∗
= π(γ
′
) is in U
c
2
,
diameter of γ
∗
is larger than δ
0
and γ
∗
also admits a nice cylinder contained in U
c
1
.
Proof. Take γ an arc with diameter larger than m, then t he projection of γ in the
i−th coordinate contains an interval of t he kind formed by k
+
i
and 1 + k
−
i
for some i
(or formed by k
+
i
+ j and k
−
i
+ j + 1 for some j ∈ Z). If it is not true, γ would be in
a n−dimensional cube with sides smaller than k
+
i
− k
−
i
< 1 and this cube has diameter
smaller than
√
n, but t his cont r adict the fact that diam(γ) > m. Hence, we may pick an
arc γ
′
in γ such that ∂γ
′
∩ L
+
i
+ j, ∂γ
′
∩ L
−
i
+ j + 1 and P
j
i
(γ
′
) ⊂ [k
+
i
+ j, k
−
i
+ j + 1]
for some 1 ≤ i ≤ n and some j ∈ Z. Therefore, diameter of γ
′
is greater than δ
0
and in
consequence its projection in M also has diameter greater than δ
0
.
Moreover, since the property of the arc γ
′
to have a projection in between k
+
i
+ j and
k
−
i
+1 +j, we may construct a cylinder centered at γ
′
and radius
d
1
2
such that this cylinder
is “f ar” away fro m
U, so this cylinder could b e simply connected or, if it is not simply
connected cylinder, it has ho les that are different from
U. In the case that the cylinder
is not simply connect ed, we consider the convex hull of the cylinder, since the original
cylinder is bounded by L
+
i
+ j a nd L
−
i
+ j + 1, then the convex hull stay in between these
two hyperplanes and therefore it does not intersect
U. By abuse of notation, let us denote
this set by C(γ
′
,
d
1
2
), it is a simply connected cylinder. Observe tha t by construction, this
cylinder will have top and bottom sides, thus C(γ
′
,
d
1
2
) is a nice cylinder.