In Riemannian geometry, Gauss's lemma asserts that any sufficiently small sphere centered at a point in a Riemannian manifold is perpendicular to every geodesic through the point. More formally, let M be a Riemannian manifold, equipped with its Levi-Civita connection, and p a point of M. The exponential map is a mapping from the tangent space at p to M:
e
x
p
:
T
p
M
→
M
which is a diffeomorphism in a neighborhood of zero. Gauss' lemma asserts that the image of a sphere of sufficiently small radius in TpM under the exponential map is perpendicular to all geodesics originating at p. The lemma allows the exponential map to be understood as a radial isometry, and is of fundamental importance in the study of geodesic convexity and normal coordinates.
We define the exponential map at
p
∈
M
by
exp
p
:
T
p
M
⊃
B
ϵ
(
0
)
⟶
M
,
v
⟼
γ
p
,
v
(
1
)
,
where
γ
p
,
v
is the unique geodesic with
γ
(
0
)
=
p
and tangent
γ
p
,
v
′
(
0
)
=
v
∈
T
p
M
and
ϵ
0
is chosen small enough so that for every
v
∈
B
ϵ
(
0
)
⊂
T
p
M
the geodesic
γ
p
,
v
is defined in 1. So, if
M
is complete, then, by the Hopf–Rinow theorem,
exp
p
is defined on the whole tangent space.
Let
α
:
I
→
T
p
M
be a curve differentiable in
T
p
M
such that
α
(
0
)
:=
0
and
α
′
(
0
)
:=
v
. Since
T
p
M
≅
R
n
, it is clear that we can choose
α
(
t
)
:=
v
t
. In this case, by the definition of the differential of the exponential in
0
applied over
v
, we obtain:
T
0
exp
p
(
v
)
=
d
d
t
(
exp
p
∘
α
(
t
)
)
|
t
=
0
=
d
d
t
(
exp
p
(
v
t
)
)
|
t
=
0
=
d
d
t
(
γ
(
1
,
p
,
v
t
)
)
|
t
=
0
=
γ
′
(
t
,
p
,
v
)
|
t
=
0
=
v
.
So (with the right identification
T
0
T
p
M
≅
T
p
M
) the differential of
exp
p
is the identity. By the implicit function theorem,
exp
p
is a diffeomorphism on a neighborhood of
0
∈
T
p
M
. The Gauss Lemma now tells that
exp
p
is also a radial isometry.
Let
p
∈
M
. In what follows, we make the identification
T
v
T
p
M
≅
T
p
M
≅
R
n
.
Gauss's Lemma states: Let
v
,
w
∈
B
ϵ
(
0
)
⊂
T
v
T
p
M
≅
T
p
M
and
M
∋
q
:=
exp
p
(
v
)
. Then,
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
)
⟩
q
=
⟨
v
,
w
⟩
p
.
For
p
∈
M
, this lemma means that
exp
p
is a radial isometry in the following sense: let
v
∈
B
ϵ
(
0
)
, i.e. such that
exp
p
is well defined. And let
q
:=
exp
p
(
v
)
∈
M
. Then the exponential
exp
p
remains an isometry in
q
, and, more generally, all along the geodesic
γ
(in so far as
γ
(
1
,
p
,
v
)
=
exp
p
(
v
)
is well defined)! Then, radially, in all the directions permitted by the domain of definition of
exp
p
, it remains an isometry.
Recall that
T
v
exp
p
:
T
p
M
≅
T
v
T
p
M
⊃
T
v
B
ϵ
(
0
)
⟶
T
exp
p
(
v
)
M
.
We proceed in three steps:
T
v
exp
p
(
v
)
=
v
: let us construct a curve
α
:
R
⊃
I
→
T
p
M
such that
α
(
0
)
:=
v
∈
T
p
M
and
α
′
(
0
)
:=
v
∈
T
v
T
p
M
≅
T
p
M
. Since
T
v
T
p
M
≅
T
p
M
≅
R
n
, we can put
α
(
t
)
:=
v
(
t
+
1
)
. We find that, thanks to the identification we have made, and since we are only taking equivalence classes of curves, it is possible to choose
α
(
t
)
=
v
t
(these are exactly the same curves, but shifted because of the domain of definition
I
; however, the identification allows us to gather them around
0
. Hence,
T
v
exp
p
(
v
)
=
d
d
t
(
exp
p
∘
α
(
t
)
)
|
t
=
0
=
d
d
t
γ
(
t
,
p
,
v
)
|
t
=
0
=
v
.
Now let us calculate the scalar product
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
)
⟩
.
We separate
w
into a component
w
T
parallel to
v
and a component
w
N
normal to
v
. In particular, we put
w
T
:=
a
v
,
a
∈
R
.
The preceding step implies directly:
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
)
⟩
=
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
T
)
⟩
+
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
N
)
⟩
We must therefore show that the second term is null, because, according to Gauss's Lemma, we must have:
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
N
)
⟩
=
⟨
v
,
w
N
⟩
=
0.
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
N
)
⟩
=
0
:
Let us define the curve
α
:
[
−
ϵ
,
ϵ
]
×
[
0
,
1
]
⟶
T
p
M
,
(
s
,
t
)
⟼
t
v
+
t
s
w
N
.
Note that
α
(
0
,
1
)
=
v
,
∂
α
∂
t
(
s
,
t
)
=
v
+
s
w
N
,
∂
α
∂
s
(
0
,
t
)
=
t
w
N
.
Let us put:
f
:
[
−
ϵ
,
ϵ
]
×
[
0
,
1
]
⟶
M
,
(
s
,
t
)
⟼
exp
p
(
t
v
+
t
s
w
N
)
,
and we calculate:
T
v
exp
p
(
v
)
=
T
α
(
0
,
1
)
exp
p
(
∂
α
∂
t
(
0
,
1
)
)
=
∂
∂
t
(
exp
p
∘
α
(
s
,
t
)
)
|
t
=
1
,
s
=
0
=
∂
f
∂
t
(
0
,
1
)
and
T
v
exp
p
(
w
N
)
=
T
α
(
0
,
1
)
exp
p
(
∂
α
∂
s
(
0
,
1
)
)
=
∂
∂
s
(
exp
p
∘
α
(
s
,
t
)
)
|
t
=
1
,
s
=
0
=
∂
f
∂
s
(
0
,
1
)
.
Hence
⟨
T
v
exp
p
(
v
)
,
T
v
exp
p
(
w
N
)
⟩
=
⟨
∂
f
∂
t
,
∂
f
∂
s
⟩
(
0
,
1
)
.
We can now verify that this scalar product is actually independent of the variable
t
, and therefore that, for example:
⟨
∂
f
∂
t
,
∂
f
∂
s
⟩
(
0
,
1
)
=
⟨
∂
f
∂
t
,
∂
f
∂
s
⟩
(
0
,
0
)
=
0
,
because, according to what has been given above:
lim
t
→
0
∂
f
∂
s
(
0
,
t
)
=
lim
t
→
0
T
t
v
exp
p
(
t
w
N
)
=
0
being given that the differential is a linear map. This will therefore prove the lemma.
We verify that
∂
∂
t
⟨
∂
f
∂
t
,
∂
f
∂
s
⟩
=
0
: this is a direct calculation. Since the maps
t
↦
f
(
s
,
t
)
are geodesics,
∂
∂
t
⟨
∂
f
∂
t
,
∂
f
∂
s
⟩
=
⟨
D
∂
t
∂
f
∂
t
⏟
=
0
,
∂
f
∂
s
⟩
+
⟨
∂
f
∂
t
,
D
∂
t
∂
f
∂
s
⟩
=
⟨
∂
f
∂
t
,
D
∂
s
∂
f
∂
t
⟩
=
1
2
∂
∂
s
⟨
∂
f
∂
t
,
∂
f
∂
t
⟩
.
Since the maps
t
↦
f
(
s
,
t
)
are geodesics, the function
t
↦
⟨
∂
f
∂
t
,
∂
f
∂
t
⟩
is constant. Thus,
∂
∂
s
⟨
∂
f
∂
t
,
∂
f
∂
t
⟩
=
∂
∂
s
⟨
v
+
s
w
N
,
v
+
s
w
N
⟩
=
2
⟨
v
,
w
N
⟩
=
0.
