4Hyperbolic geometry

IB Geometry



4.1 Review of derivatives and chain rule
We start by reviewing some facts about taking derivatives, and make explicit
the notation we will use.
Definition (Smooth function). Let
U R
n
be open, and
f
= (
f
1
, ··· , f
m
) :
U R
m
. We say
f
is smooth (or
C
) if each
f
i
has continuous partial derivatives
of each order. In particular, a
C
map is differentiable, with continuous first-
order partial derivatives.
Definition (Derivative). The derivative for a function
f
:
U R
m
at a point
a U
is a linear map d
f
a
:
R
n
R
m
(also written as D
f
(
a
) or
f
0
(
a
)) such that
lim
h0
kf(a + h) f(a) df
a
· hk
khk
0,
where h R
n
.
If
m
= 1, then d
f
a
is expressed as
f
x
a
(a), ··· ,
f
x
n
(a)
, and the linear map
is given by
(h
1
, ··· , h
n
) 7→
n
X
i=1
f
x
i
(a)h
i
,
i.e. the dot product. For a general
m
, this vector becomes a matrix. The Jacobian
matrix is
J(f)
a
=
f
i
x
j
(a)
,
with the linear map given by matrix multiplication, namely
h 7→ J(f)
a
· h.
Example. Recall that a holomorphic (analytic) function of complex variables
f : U C C has a derivative f
0
, defined by
lim
|w|→0
|f(z + w) f (z) f
0
(z)w|
|w|
0
We let f
0
(z) = a + ib and w = h
1
+ ih
2
. Then we have
f
0
(z)w = ah
1
bh
2
+ i(ah
2
+ bh
1
).
We identify
R
2
=
C
. Then
f
:
U R
2
R
2
has a derivative d
f
z
:
R
2
R
2
given by
a b
b a
.
We’re also going to use the chain rule quite a lot. So we shall write it out
explicitly.
Proposition (Chain rule). Let
U R
n
and
V R
p
. Let
f
:
U R
m
and
g : V U be smooth. Then f g : V R
m
is smooth and has a derivative
d(f g)
p
= (df)
g(p)
(dg)
p
.
In terms of the Jacobian matrices, we get
J(f g)
p
= J(f)
g(p)
J(g)
p
.