We will use the following abbreviations: if Rn
has coordinates xi
, we will use ∂i to denote the
differentiation with respect to xi
, i.e. ∂i = ∂
∂xi . In addition, we will use Einstein’s summation
notation where the sum symbol is not written if two indices appear, one as a lower index and one
as an upper index, e.g. the directional derivative is
dxϕ(A) = ∂iϕ · Ai
.
In such a situation, the summation is implicit (and thus, it is necessary to state explicitly if the
summation is not intended). We believe that this highly improves readability of formulas.
1. Tangent vectors, vector fields and derivations
1.1. Partitions of unity.
Definition 1. A support of a function f : M → R is the closure
supp f = {x ∈ M | f(x) = 0}.
Definition 2. Let U = {Ui | i ∈ I} be an open cover of a manifold M. A (smooth) partition of
unity subordinate to U is a collection of functions λi : M → [0, 1] such that supp λi ⊆ Ui, such that
in a neighbourhood of every point x ∈ M, there is only a finite number of non-zero λi and such
that λi = 1.
For simplicity, we will assume M to be compact, it is however not necessary. Then the local
finiteness of the λi in the definition of the partition of a unity can be replaced by the finiteness,
i.e. only a finite number of the λi is nonzero.
Theorem 3. Let M be compact. There exist a partition of unity subordinate to any open cover
of M.
Proof. We choose a finite subcover U1, . . . , Un. We will construct the functions λi inductively and
denote Vi = {x ∈ M | λi(x) = 0}. It is enough to find the λi in such a way that Vi = M, since
then λ = λi > 0 and we may replace each λi by λi/λ.
Thus, suppose that the λj have been constructed for j < i. Then let λi be a function that is
supported in Ui and is positive on a compact set Ci = M ( j<i Vj ∪ j>i Uj). We may verify
inductively that j<i Vj ∪ j≥i Uj = M so that Ci ⊆ Ui and such a function can be chosen by
the following lemma.
Lemma 4. Let C ⊆ U ⊆ M be subsets of a manifold M such that C is compact and U open.
Then there exists a function λ: M → [0, 1] with supp λ ⊆ U that is positive on C.
Proof. For each x ∈ C, we will find a function λx : M → [0, 1] that is positive at x and with
supp λx ⊆ U. Then choose a finite subcover of C by the open sets Vx = {y ∈ M | λx(y) = 0}, say
C ⊆ Vx1
∪ · · · ∪ Vxn
. Then the required function is λ = λx1
+ · · · + λxn
.
It is enough to find the functions λx. Let ϕ: V → Rn
be a chart centred at x (i.e. ϕ(x) = 0)
with V ⊆ U. Let ρε : R → [0, 1] be the function from the next lemma for ε small enough that
[−ε, ε]n
⊆ ϕ(V ). Then we define
λx(y) =
ρε(y1
) · · · ρε(yn
) y ∈ V
0 y /∈ V
where y1
, . . . , yn
are the coordinates of y in the given chart, i.e. yi
= ϕi
(y). Since λx is smooth
on V and on M ϕ−1
([−ε, ε]n
) (where it is zero), it is smooth on M.
Lemma 5. There exists a function ρε : R → [0, 1] such that ρε(0) > 0 and ρε(x) = 0 for |x| ≥ ε.
Proof. The function
λ(x) =
e−1/x
x > 0
0 x ≤ 0
is smooth (that needs to be verified, see Kol´aˇr’s text). We set ρε(x) = λ(ε + x)λ(ε − x).
1
2
1.2. Germs.
Definition 6. We say that two maps f, g: M → N have the same germ at x if they agree in some
neighbourhood of x. We denote the class of f with respect to this relation as germx f, the germ
of f at x, the resulting decomposition is denoted C∞
x (M, N); in the case of smooth functions on
M, we will abbreviate it to C∞
x M.
We will need that germs of maps defined in a neighbourhood of x extend to germs of globally
defined maps (this clearly does not hold for maps themselves – e.g. 1/x does not extend from
(0, ∞)). We will suffice with functions, thus, let f : U → R be a function defined in a neighbourhood
U x. Let λ: M → R be such that λ equals one in a neighbourhood of x and with support
in U. Then the function λ · f, extended outside of U by the zero function, is smooth both on U
and M supp f (where it equals zero); clearly, it represents the same germ at x as the orginial f.
Thus, the restriction map C∞
M → C∞
U induces a map
C∞
x M
∼=
−→ C∞
x U
which we just showed to be surjective; it is injective almost by definition (if two function f, g have
the same germ in U, i.e. agree in a neighbourhood of x in U then this neighbourhood is also a
neighbourhood in M and, thus, the germs of f and g in M agree).
1.3. Tangent vectors.
Theorem 7. Let D: C∞
M → R be an R-linear map that satisfies the following “Leibniz rule at
x0”
D(f · g) = Df · g(x0) + f(x0) · Dg
(we say that D is a derivation at x0). Then there exists a unique tangent vector A ∈ Tx0 M such
that Df = Af, the derivative of f at x in the direction of A.
Proof. First we prove the statement for M = Rn
and x0 = 0. Let f ∈ C∞
Rn
and write
f(x) − f(0) = [f(t · x)]1
t=0 =
1
0
d
dt f(t · x) dt =
1
0
∂f
∂xi (t · x)xi
dt =
1
0
∂f
∂xi (t · x) dt · xi
.
Denoting gi(x) =
1
0
∂f
∂xi (t · x) dt, a smooth function on Rn
with gi(0) = ∂f
∂xi (0), we get
f(x) = f(0) + gi(x)xi
.
Now we apply D to obtain
Df = D(f(0))
0
+ (Dgi · xi
(0)
0
+gi(0) · Dxi
) = ∂f
∂xi (0) · Dxi
= d0f(Dx1
, . . . , Dxn
)
(it is easy to see from D(1 · 1) = D1 · 1 + 1 · D1 that D1 = 0 and consequently also D(f(0)) =
f(0) · D1 = 0), i.e. Df is the derivative of f at 0 in the direction of the vector (Dx1
, . . . , Dxn
).
Let now M and x0 be general and let f be a function that equals zero near x0. Then there
exists a function λ that equals zero near x0 and equals one on the support of f, i.e. f(x) = 0 ⇒
λ(x) = 1. Then we may write
Df = D(λ · f) = Dλ · f(x0) + λ(x0) · Df = 0.
Thus, by the additivity, D agrees on any two functions with the same germ at x0; in other words,
D gives rise to a map D: C∞
x0
M → R. From the diagram
Tx0 U
∼= //
∼=

Derx0 (C∞
U)
∼=

Tx0
M // Derx0
(C∞
M)
we conclude from the local result that D is indeed a derivative in the direction of a unique vector
A ∈ Tx0 M.
3
Theorem 8. Let D: C∞
M → C∞
M be an R-linear map that satisfies the following “Leibniz
rule”
D(f · g) = Df · g + f · Dg
(we say that D is a derivation). Then there exists a unique vector field X ∈ XM such that
Df = Xf, the derivative of f in the direction of X.
Proof. For any x ∈ M, the composition of D with the evaluation map at x (sending f → f(x)),
Dx : C∞
M
D
−→ C∞
M
evx
−−→ R,
is a derivation at x and, thus, there is a unique vector Xx ∈ TxM such that Df(x) = Dxf = Xxf.
It remains to show that x → Xx is a smooth vector field.
Locally, we have seen that the coordinates of Xx are obtained as (Xx)i
= Dxxi
. Choose a coordinate
chart around x0 ∈ M and let λ: M → R be a function that equals one in a neighbourhood
U x0 and with support inside the coordinate chart. Then, for all x ∈ U, we get
(Xx)i
= Dxxi
= Dx(λ · xi
) = D(λ · xi
)(x)
and, since λ · xi
is smooth, so is D(λ · xi
).
1.4. Duality between tangent and cotangent spaces. We consider the collection Px of all
paths through x, i.e. maps γ : R → M satisfying γ(0) = x and the collection Fx of all functions
vanishing at x, i.e. maps f : M → R satisfying f(x) = 0. Then we have the following mapping
Fx × Px → R
(f, γ) → d
dt t=0
f ◦ γ(t).
Now we consider the relations on both Fx and Px that identify those functions/paths that give
rise to the same pairing. In local coordinates, we have
d
dt t=0
f ◦ γ(t) = ∂f
∂xi (x)dγi
dt (0) = ∂f
∂x1 (x) · · · ∂f
∂xn (x)



dγ1
dt (0)
...
dγn
dt (0)



and thus, we see that this relation means having the same derivatives. We denote the class of f by
dxf, the differential of f at x, and the corresponding decomposition T∗
x M. We denote the class of
γ by ˙γ(0) = d
dt t=0
γ(t), the tangent vector to γ at 0, and the corresponding decomposition TxM.
Thus, we obtain an induced mapping
T∗
x M × TxM → R,
(dxf, ˙γ(0)) → d
dt t=0
f ◦ γ(t) = dxf(˙γ(0)),
or simply (ϕ, A) → ϕ(A). Using the local coordinate expression, we see that TxM and T∗
x M are
dual to each other via this pairing. Even without employing coordinates, T∗
x M has an obvious
vector space structure and, thus, we may endow TxM with a vector space structure using the
pairing:
TxM
∼=
−→ Lin(T∗
x M, R) = (T∗
x M)∗
Concretely, this means that the sum A + B of two vectors (represented by paths) is a vector
that differentiates functions as dxf(A + B) = dxf(A) + dxf(B). Locally, it is easy to add the
representing paths, but there is no obvious coordinate-free addition of paths in M (put differently,
any local chart gives a way of adding the representing paths and, while different, they produce
the same tangent vector).
Later, we will also need the tangent vector ˙γ(t0) of a path γ at a general time t0. This is most
easily defined as ˙γ(t0) = d
dt t=0
γ(t0 +t), i.e. by reparametrization. Let us look at a different point
of view: the path γ has an associated tangent map γ∗ : TR → TM and there is a canonical vector
field d
dt on TR (with value at t0 denoted by d
dt t=t0
) and ˙γ(t0) = γ∗( d
dt t=t0
) (this is obvious
because d
dt t=t0
is the tangent vector at 0 of the path t → t0 + t, the same reparametrization map
as above).
4
We will also need a differential dxf of a function f with f(x) = 0. Again, this is defined via
“reparametrization” as dxf = dx(f − f(x)).
2. Lie bracket
We define the Lie bracket through derivations.
Definition 9. Let X and Y be two vector fields. Then it is easy to see that f → XY f − Y Xf
is a derivation and the corresponding vector field is denoted [X, Y ] and called the Lie bracket of
the vector fields X and Y .
It is easy to derive a coordinate formula
XY f − Y Xf =
i,j
Xj
∂j(Y i
∂if) −
i,j
Y j
∂j(Xi
∂if)
=
i,j
(Xj
∂jY i
∂if + Xj
Y i
∂j∂if) −
i,j
(Xj
∂jY i
∂if + Y j
Xi
∂j∂if)
=
i,j
(Xj
∂jY i
− Y j
∂jXi
)∂if
so that [X, Y ] = i,j(Xj
∂jY i
− Y j
∂jXi
)∂i.
Lemma 10. Vector fields X and Y are f-related if and only if
f(FlX
t (x)) = FlY
t (f(x))
In other words f transfers the flow lines of X into the flow lines of Y . We will use this property
quite often.
Proof. Differentiating the given equality, we get
f∗X(x) = d
dt t=0
f(FlX
t (x)) = d
dt t=0
FlY
t (f(x)) = Y (f(x)),
which is precisely the definition of the f-relatedness.
In the opposite direction, given that X and Y are f-related, we wish to prove the equality
from the statement, i.e. we want to prove that γ(t) = f(FlX
t (x)) is an integral curve of Y through
f(x). Since clearly γ(0) = f(x), we need only check that it satisfies the differential equation of an
integral curve:
˙γ(t0) = d
dt t=t0
f(FlX
t (x)) = f∗X(FlX
t0
(x)) = Y (f(FlX
t0
(x)) = Y (γ(t0))
(the third equality uses the f-relatedness).
Definition 11. Let X, Y be two vector fields on a manifold M. Then we denote
(FlX
t )∗
Y (x) = (FlX
−t)∗Y (FlX
t (x)) ∈ TxM
the “pullback” of Y along the flow FlX
t of X. For each x ∈ M it is defined for t small.
We will need a useful property for the proof of the next proposition. It is based on an observation
that for a function ϕ(s, t) of two variables, with values in a vector space, we have
d
dt t=t0
ϕ(t, t) = d
dt t=t0
ϕ(t, t0) + d
dt t=t0
ϕ(t0, t).
Now let X be a time-dependent vector field, i.e. a map X : R×M → TM such that X(t, x) ∈ TxM.
Let f : R × M → R be a “time-dependent” function on M (just a function on R × M). Write Xt
and ft for the vector field and function obtained by plugging in a specific value of t. Then we may
form the directional derivative Xtft and
d
dt t=t0
Xtft = ( d
dt t=t0
Xt)ft0 + Xt0 ( d
dt t=t0
ft). (∗)
(Locally, we have Xtft(x) = X(t, x)i ∂f
∂xi (t, x) and we apply the previous observation.)
Proposition 12. d
dt t=t0
(FlX
t )∗
Y (x) = (FlX
t0
)∗
[X, Y ](x).
5
Proof. First assume that t0 = 0 and let f : M → R be a smooth function. We differentiate f in
the direction of the left hand side:
d
dt t=0
(FlX
t )∗
Y (x) f
(∗)
= d
dt t=0
(FlX
t )∗
Y (x)f
= d
dt t=0
(FlX
−t)∗Y (FlX
t (x))f
= d
dt t=0
Y (FlX
t (x))(f ◦ FlX
−t)
(∗)
= d
dt t=0
Y (x)(f ◦ FlX
−t) + d
dt t=0
Y (FlX
t (x))(f)
(∗)
= Y (x)( d
dt t=0
f ◦ FlX
−t) + d
dt t=0
(Y f)(FlX
t (x))
= Y (x)(−Xf) + X(x)(Y f)
= −(Y Xf)(x) + (XY f)(x) = [X, Y ](x) f
(the steps labeled by (∗) involve the observation made before the proposition, the first in the
opposite direction).
For a general t0, we have (FlX
t )∗
Y (x) = (FlX
t0
)∗
(FlX
t−t0
)∗
Y (x). Since (FlX
t0
)∗
is a linear map we
can interchange with d
dt .
Corollary 13. The following conditions are equivalent:
• [X, Y ] = 0,
• (FlX
t )∗
Y = Y , i.e. Y is FlX
t -related with itself for all t,
• FlX
t FlY
s (x) = FlY
s FlX
t (x), i.e. the flow lines commute.
In general we have FlY
−s FlX
−t FlY
s FlX
t (x) = x + st[X, Y ](x) + o(s, t)2
.
Proof. The equivalence of the first three conditions follows immediately from the previous proposition
– the second condition states that (FlX
t )∗
Y is a constant function of t, i.e. that the derivative
is zero and this is essentially the first condition. At the same time, the second condition is equivalent
to Y being FlX
t -related to itself and this is equivalent to FlX
t preserving the integral curves
of Y , which is precisely the third condition.
Differentiating the commutator of the flows twice, we get
∂
∂t t=0
∂
∂s s=0
FlY
−s FlX
−t FlY
s FlX
t (x) = ∂
∂t t=0
−Y (x) + (FlX
t )∗
Y (x) = [X, Y ](x)
The remaining derivatives of order at most two are clearly zero.
3. Distributions
Theorem 14. If vector fields X1, . . . , Xn are linearly independent and satisfy [Xi, Xj] = 0 then,
in a neighbourhood of any point x, there exists a coordinate chart in which Xi = ∂i.
Proof. We define a map
ϕ: Rn
→ M, (t1
, . . . , tn
) → FlX1
t1 · · · FlXn
tn (x)
(it is defined in a neighbourhood of 0). Since we may interchange the flows (since [Xi, Xj] = 0),
the partial derivatives equal
∂iϕ(t) = ∂i FlXi
ti FlX1
t1 · · · FlXi
ti · · · FlXn
tn (x) = Xi(FlXi
ti FlX1
t1 · · · FlXi
ti · · · FlXn
tn (x)) = Xi(ϕ(t))
and ϕ is a local diffeomorphism. We may thus use its inverse ϕ−1
as a coordinate chart on M and
the above shows that, in this chart, we have ∂i = Xi.
A distribution S on M is called involutive if for every two vector fields X, Y ∈ S their bracket
[X, Y ] also lies in S.
Theorem 15 (Frobenius theorem). If S is involutive then for every x ∈ M there exists a local
coordinate system t1
, . . . , tn
in a neighbourhood U of x such that the vector fields ∂1, . . . , ∂k form
a basis of the distribution S on U. In particular S is integrable.
6
Proof. Let X1, . . . , Xk be local vector fields which span the distribution S near x and choose vector
fields Xk+1, . . . , Xn so that (X1, . . . , Xn) form a basis near x. We then define a map
ϕ : Rn
⊇ U −→ M
(t1
, . . . , tn
) −→ FlX1
t1 · · · FlXn
tn (x)
The partial derivatives at the origin clearly consist of the vectors Xi(x) and thus ϕ is a local
diffeomorphism – its inverse will form our coordinate system.
Let us compute the partial derivative with respect to ti
for i ≤ k at a general point.
∂iϕ(t1
, . . . , tn
) = FlX1
t1
∗
· · · Fl
Xi−1
ti−1
∗
Xi FlXi
ti · · · FlXn
tn (x)
To conclude the proof it is therefore enough to show that for any Y belonging to S the pullbacks
(FlY
t )∗
Xi also belong to S (then the same will be true for pullbacks (FlY
t )∗
X with X ∈ S by
linearity, and we apply the claim to Xi, Fl
Xi−1
ti−1
∗
Xi, etc.) Denote this pullback by
Yi(t) = (FlY
t )∗
Xi(x) ∈ TxM
and write [Y, Xi] = aj
i Xj. By Lemma 12 the paths Yi(t) satisfy the following system of differential
equations
d
dt Yi(t) = (FlY
t )∗
[Y, Xi] = aj
i (FlY
t (x))Yj(t)
We have Yi(0) = Xi(x) ∈ S(x) and, since the system is linear, we must have Yi(t) ∈ S(x) for all t
(namely, there exists a solution of the system d
dt Zi(t) = aj
i (FlY
t (x))Zj(t) with Zi ∈ S(x) and with
Zi(0) = Xi(x). By uniqueness, we must have Yi(t) = Zi(t) and, thus, Yi(t) ∈ S(x).)
Theorem 16 (Frobenius theorem through 1-forms). Let ω: TM → V be a smooth map that
is linear on each TxM (we say that ω is a V -valued 1-form) and surjective. Then ker ω is a
distribution. It is integrable if and only if ω(X) = 0, ω(Y ) = 0 ⇒ dω(X, Y ) = 0.
This uses the exterior differential of the next section.
Proof. In local coordinates on M and in a basis of V , the 1-form ω is given by a matrix of
maximal rank. We may assume that the left most square block is regular in a neighbourhood
of a given point and use the Gauss elimination to make this matrix (E | A). Then ker ω is
given by (xn−k+1
, . . . , xn
)T
= −A(x1
, . . . , xn−k
), proving that it is a (smooth) distribution. Now
dω(X, Y ) = Xω(Y ) − Y ω(X) − ω([X, Y ]) gives easily the result.
4. Exterior differential
4.1. Invariant differentiation of forms. Let us study the invariance of higher derivatives under
the change of coordinates, i.e. let ω be a k-linear form with components ωi1···ik
,
ω = ωi1···ik
dxi1
⊗ · · · ⊗ dxik
.
Differentiating all components in the direction of a vector X0 gives another k-linear form DX0 ω
with
DX0 ω = ∂i0 ωi1···ik
(X0)i0
dxi1
⊗ · · · ⊗ dxik
which we interpret as a (k + 1)-linear form Dω(X0, X1, . . . , Xk) = DX0
ω(X1, . . . , Xk); its coordinates
are (Dω)i0i1···ik
= ∂i0
ωi1···ik
. Now we apply the change of coordinates f to get
f∗
(Dω)j0j1···jk
= (∂i0
ωi1···ik
◦ f) · ∂j0
fi0
· ∂j1
fi1
· · · · · ∂jk
fik
.
On the other hand D(f∗
ω) equals
D(f∗
ω)j0j1···jk
= ∂j0
((ωi1···ik
◦ f) · ∂j1
fi1
· · · · · ∂jk
fik
)
= ∂j0
((ωi1···ik
◦ f)) · ∂j1
fi1
· · · · · ∂jk
fik
+ (ωi1···ik
◦ f) · ∂j1
fi1
· · · · · ∂2
j0jk
fik
· · · · · ∂jk
fik
The first term equals to f∗
(Dω)j0j1···jk
by the chain rule. Now the point is that to get f∗
(Dω) =
D(f∗
ω) (i.e. to get a differential that does not depend on coordinates), we have to get rid of the
7
second term involving the second derivative ∂2
j0jk
fik
. It clearly disappears after antisymmetrization.
Some representation theory would be required to get that no other part is invariant and we
will not attempt to do this here.
Thus, we get an invariant differentiation operator – the exterior differential – on antisymmetric
forms by antisymmetrizing DX0
ω(X1, . . . , Xk); for technical reasons, we multiply the antisymmetrization
by (k+1)!
1!k! , since the form already was antisymmetric in the variables X1, . . . , Xk and
obtain
dω(X0, . . . , Xk) =
i
(−1)i
DXi ω(X0, . . . , Xi, . . . , Xk)
Denoting by φ ∧ ψ = (|φ|+|ψ|)!
|φ|!|ψ|! Alt(φ ⊗ ψ), we obtain for
ω =
i1<···<ik
ωi1···ik
dxi1
∧ · · · ∧ dxik
a coordinate expression
dω =
i0 i1<···<ik
∂i0
ωi1···ik
dxi0
∧ dxi1
∧ · · · ∧ dxik
.
4.2. Coordinate-free formula for the exterior differential. The differentiation operator DX0
satisfies the Leibniz rule
DX0
(ωi1···ik
(X1)i1
· · · (Xk)ik
) = DX0
ωi1···ik
· (X1)i1
· · · (Xk)ik
+ ωi1···ik
(X1)i1
· · · DX0
(Xj)ij
· · · (Xk)ik
which (after subtracting the sum from the right hand side) translates to
DX0
ω(X1, . . . , Xk) = DX0
(ω(X1, . . . , Xk)) − ω(X1, . . . , Xj−1, DX0
Xj, Xj+1, . . . , Xk).
Here the first DX0
on the right is the directional derivative of the function ω(X1, . . . , Xk). The
second appearance is, however, very different and we have [X, Y ] = DXY − DY X. The exterior
differential then equals
dω(X0, . . . , Xk) =
i
(−1)i
DXi
ω(X0, . . . , Xi, . . . , Xk)
=
i
(−1)i
Xiω(X0, . . . , Xi, . . . , Xk)
−
i<j
(−1)i
ω(X0, . . . , Xi, . . . , Xj−1, DXi Xj, Xj+1, . . . , Xk)
−
i>j
(−1)i
ω(X0, . . . , Xj−1, DXi
Xj, Xj+1, . . . , Xi, . . . , Xk)
where we split the antisymmetrization of the second term according to whether i < j or i > j.
Next we move the term DXi Xj onto the first spot (here the sign differs for the two possibilities):
=
i
(−1)i
Xiω(X0, . . . , Xi, . . . , Xk)
+
i<j
(−1)i+j
ω(DXi
Xj, X0, . . . , Xi, . . . , Xj, . . . , Xk)
−
i>j
(−1)i+j
ω(DXi Xj, X0, . . . , Xj, . . . , Xi, . . . , Xk)
8
and finally swap the indices i, j in the last sum and subtract, using DXi
Xj − DXj
Xi = [Xi, Xj],
to obtain the final formula
=
i
(−1)i
Xiω(X0, . . . , Xi, . . . , Xk)
+
i<j
(−1)i+j
ω([Xi, Xj], X0, . . . , Xi, . . . , Xj, . . . , Xk)
5. Integration of forms
5.1. Integral. Let ω be an n-form on an open subset V ⊆ Rn
, say with a compact support.
Writing
ω(x) = a(x) · dx1
∧ · · · ∧ dxn
,
we define
V
ω = · · ·
V
a(x) dx1
· · · dxn
.
The important property of this integral is that for a diffeomorphism ϕ: W → V with positive
Jacobian, we get
W
ϕ∗
ω =
V
ω.
This follows from the theorem about the transformation of the integral. Clearly, the above defined
integral is additive in ω.
Let M be an oriented manifold of dimension n. We assume for simplicity that M is compact.
Let ω be an n-form on M. We would like to define M
ω. Consider the maximal orientationpreserving
atlas on M and choose a partition of unity λi so that supp λi is a subset of a domain
Ui of a chart ϕi : Ui → Vi. Then we define
M
ω =
i Vi
(ϕ−1
i )∗
(λiω)
We note that λiω has a compact support inside Ui and, thus, the pullback (ϕ−1
i )∗
(λiω) has a
compoact support inside Vi. Thus, the integral exists and is finite. It remains to show that it does
not depend on the choice of the partition λi.
Thus, let µi be another partition. Then we get
i Vi
(ϕ−1
i )∗
(λiω) =
i,j Vi
(ϕ−1
i )∗
(λiµjω)
Denoting θ = λiµjω, a compactly supported n-form inside Ui ∩ Uj, whose images we denote
Vij = ϕi(Ui ∩ Uj) and Vji = ϕj(Ui ∩ Uj), we have
Vi
(ϕ−1
i )∗
θ =
Vij
(ϕ−1
i )∗
θ =
Vij
ϕ∗
ij((ϕ−1
j )∗
θ) =
Vji
(ϕ−1
j )∗
θ =
Vj
(ϕ−1
j )∗
θ
by the invariance of the integral with respect to the diffeomorphism ϕij = ϕiϕ−1
j : Vij → Vji.
5.2. Manifolds with boundary. The main idea here is: in exactly the same manner in which
manifolds are built from the Euclidean space Rn
, manifolds with boundary are built from the
Euclidean halfspace Hn+1
= {(x0
, x1
, . . . , xn
) ∈ Rn+1
| x0
≤ 0}. It is however important that we
allow tangent vectors at the boundary hyperplane to be all vectors from Rn+1
, i.e.
THn+1
= Hn+1
× Rn+1
.
Thus, the geometric definition using paths is inappropriate. Derivations work well if we interpret
∂0f(x) for a boundary point x to be the one-sided partial derivative.
Formally, a map between open subsets of the half-spaces is said to be smooth, if all partial
derivatives exist (one-sided where needed) and are continuous. A diffeomorphism between open
subsets of Hn+1
preserves the boundary points, since at an interior point, any (local) diffeomorphism
has a local inverse and as such maps to an interior point.
9
With this notion, we define a (smooth) manifold with boundary W as a topological space,
Hausdorff and with countable basis of topology, equipped with a maximal atlas consisting of
homeomorphisms ϕ: U → V with V an open subset of Hn+1
and with all change of coordinate
maps smooth in the above sense. We define the boundary of W to be the set ∂W of points that
correspond to the boundary points in a chart.
The standard bases (e0, e1, . . . , en) of Hn+1
and (e1, . . . , en) of Rn
are considered positive. We
say that ∂Hn+1
is oriented via its outward normal: The outward normal is by definition e0 (or
any combination x0
e0 + x1
e1 + · · · xn
en with x0
> 0) and a basis (u1, . . . , un) is then a positive
basis of ∂Hn+1
according to this principle if and only if (e0, u1, . . . , un) is a positive basis of Hn+1
.
This gives a way of orienting a boundary ∂W of any oriented manifold with boundary W. We will
always consider ∂W with this induced orientation.
5.3. Stokes’ theorem.
Theorem 17. For a compact manifold with boundary W of dimension n + 1 and an n-form ω on
W, we have
∂W
ω =
W
dω.
(The left hand side is really the integral of the pullback ι∗
ω along the inclusion ι: ∂W → W.)
Proof. Denoting ∂ϕi the restriction of ϕi to the boundaries, ∂ϕi : ∂Ui → ∂Vi, it is enough to
compare the contributions
∂Vi
(∂ϕ−1
i )∗
ι∗
(λiω) and
Vi
(ϕ−1
i )∗
d(λiω).
Denoting j : ∂Vi → Vi the inclusion, the left hand side equals ∂Vi
j∗
(ϕ−1
i )∗
(λiω), while the right
hand side equals Vi
d(ϕ−1
i )∗
(λiω). Thus, denoting θ = (ϕ−1
i )∗
(λiω), we want to show that
∂Vi
j∗
θ =
Vi
dθ.
Since θ is an n-form on Vi ⊆ Hn+1
, we may write
θ =
i
ai · dx0
∧ · · · ∧ dxi ∧ · · · ∧ dxn
.
Since j(x1
, . . . , xn
) = (0, x1
, . . . , xn
), we get j∗
dx0
= 0 and j∗
dxi
= dxi
, for i > 0. Thus,
j∗
θ = a0j · dx1
∧ · · · ∧ dxn
and the integral on the left is
∂Vi
j∗
θ = · · ·
Vi
a0(0, x1
, . . . , xn
) dx1
· · · dxn
.
Now we simplify the integral on the right, i.e. we compute
dθ =
i
∂iai · dxi
∧ dx0
∧ · · · ∧ dxi ∧ · · · ∧ dxn
=
i
(−1)i
∂iai · dx0
∧ · · · ∧ dxn
.
Now the integral simplifies to
Vi
dω =
i
(−1)i
· · ·
Vi
∂iai dx0
· · · dxn
=
i
(−1)i
· · ·
Vi
∂iai dxi
dx0
· · · dxi · · · dxn
10
For i > 0, we get
∞
−∞
∂iai · dxi
= ai|xi=∞ − ai|xi=−∞ = 0 − 0 = 0, while
0
−∞
∂iai · dx0
=
a0|x0=0 − a0|x0=−∞ = a0(0, x1
, . . . , xn
). Thus, the integral also equals
Vi
dω = · · ·
Vi
a0(0, x1
, . . . , xn
) dx1
· · · dxn
.
Remark. It is interesting to see what we would get if we integrated over a cube instead. Then the
“boundary conditions” ai|xi=±∞ = 0 would be replaced by the non-zero restrictions to the faces
of the cube and the resulting formula would be
In+1
dω =
i
(−1)i
∂+
i In+1
ω −
i
(−1)i
∂−
i In+1
ω
where the ∂ε
i In+1
denotes the subset {(x0
, . . . , xn
) ∈ In+1
| xi
= ε}. The signs reflect the
orientations these faces so that the right hand side actually equals ∂In+1 ω.
5.4. Cohomology in top dimension. In order to distinguish compact manifolds without boundary
from those with boundary, we call then closed.
Theorem 18. For any closed oriented Riemannian manifold M of dimension n, Hn
(M) = 0.
Proof. Since every n-form on M is closed, it is enough to find one that is not exact. We
know that oriented Euclidean spaces admit a canonical volume form specified by the requirement
Vol(e1, . . . , en) = 1 for any positive orthonormal basis (e1, . . . , en). In this way, we obtain a
volume form Vol ∈ Ωn
M. In any chart compatible with the orientation,
Vol = a · dx1
∧ · · · ∧ dxn
with a = Vol(∂1, . . . , ∂n) > 0. Thus, M
Vol is an integral of a positive function and as such must
be also positive. On the other hand, if Vol = dθ, we would get
M
Vol =
M
dθ =
∂M
θ = 0,
since ∂M = ∅.
5.5. Homotopy invariance. We would like to show that Hk
Rn
= 0 for k > 0. This will follow
from the following “homotopy invariance” property.
Theorem 19. Let h: [−1, 1] × M → N be a smooth map and denote ht = h(t, −). Then for any
closed k-form ω, we get [h∗
−1ω] = [h∗
1ω] ∈ Hk
M.
Proof. The idea of the proof is simple. Any k-form is determined by its integrals along kdimensional
cubes embedded in M. This is so because any embedding [−1, 1]k
→ M that
maps ∂i(0) to Ai ∈ TxM restricts to the cube [−ε, ε]k
to an embedding iε such that (iε)∗
ω ∼
(2ε)k
ω(A1, . . . , Ak) (equality holds in limε→0).
Now for an embedding i: [−1, 1]k
→ M, we get an associated embedding id ×i: [−1, 1]k+1
→
[−1, 1]×M. Denote by jt : [−1, 1]k
→ [−1, 1]k+1
the embedding given by jt(t1
, . . . , tk
) = (t, t1
, . . . , tk
).
Then h∗
t ω = j∗
t h∗
ω and both j±1 are embeddings as part of the boundary. Thus, the Stokes’ theorem
relates
[−1,1]×[−1,1]k
d(h∗
ω) =
∂([−1,1]×[−1,1]k)
h∗
ω
=
[−1,1]k
h∗
1ω −
[−1,1]k
h∗
−1ω −
[−1,1]×∂[−1,1]k
h∗
ω (∗)
(the first two terms correspond to ∂[−1, 1] × [−1, 1]k
). Writing
d(h∗
ω) = a · dt ∧ dt1
∧ · · · ∧ dtk
,
the integral on the left can be computed using Fubini’s theorem as
[−1,1]×[−1,1]k
d(h∗
ω) =
[−1,1]k [−1,1]
a(t, t1
, . . . , tk
) dt dt1
· · · dtk
.
11
This can be rephrased in terms of an operator K : Ωk+1
(I × M) → Ωk
M, given by the integral
K(η)(x) = [−1,1]
θ(t, x) dt η = dt ∧ θ
0 η(∂t, −, . . . , −) = 0
as
[−1,1]×[−1,1]k
d(h∗
ω) =
[−1,1]k
Kd(h∗
ω) =
[−1,1]k
K(h∗
dω).
The remaining boundary term in (∗) is then
[−1,1]×∂[−1,1]k
h∗
ω =
∂[−1,1]k
K(h∗
ω) =
[−1,1]k
dK(h∗
ω),
again by the Stokes’ theorem. Thus, we have finally obtained
[−1,1]k
h∗
1ω −
[−1,1]k
h∗
−1ω =
[−1,1]k
(dK(h∗
ω) + K(h∗
dω)))
or, in other words, h∗
1ω−h∗
−1ω = dK(h∗
ω)+K(h∗
dω). This implies rather easily the result, since,
for ω closed, the first term on the right vanishes and, thus, the difference on the left is exact, i.e.
the two terms represent the same cohomology class.
In the situation from the above proof, we say that two chain maps (maps that commute with
differentials, such as pullback maps h∗
t ) are chain homotopic if there exists a collection of maps η
(these are the compositions Kh∗
in the proof) such that
g − f = dη + ηd.
Then, f and g induce the same map in cohomology.
Corollary 20. Hk
Rn
= 0 for k > 0.
Proof. There is a homotopy id ∼ 0 between the identity and the constant map onto the zero.
Then for any closed k-form ω we have [ω] = [id∗
ω] = [0∗
ω] = [0].
6. Riemannian geometry
6.1. Preliminary results.
Lemma 21. For every map
F : XM × · · · × XM → C∞
M
that is C∞
M-linear in each variable there exists a unique tensor field ω of type (0, k) such that
F(X1, . . . , Xk) = ω(X1, . . . , Xk).
Proof. We first prove that Fx = evx F is local; we will assume k = 1 here for simplicity. Thus,
let X ∈ XM be zero in a neighbourhood of x ∈ M. Then there exists a function λ such that λ
is zero near x and X = λ · X. Then Fx(X) = Fx(λ · X) = λ(x) · Fx(X) = 0. This allows one to
define Fx on germs of vector fields and, consequently, a local version Fx : XU → R as in the case
of derivations.
Now, for general k, the map F is C∞
M-linear in each variable and thus local in each variable.
Since locally Xj = Xij
∂ij , we obtain
F(X1, . . . , Xk)(x) = Fx(X1, . . . , Xk) = Fx(∂i1
, . . . , ∂ik
)Xi1
(x) · · · Xik
(x),
i.e. we have ω = F(∂i1 , . . . , ∂ik
)dxi1
⊗ · · · ⊗ dxik
.
A similar result holds for maps F : XM × · · · × XM → XM – such maps are given by tensor
fields of type (1, k); the proof is the same.
A slightly generalization of the first part of the proof of the previous lemma is the following
(for simplicity, we state only unary version):
Lemma 22. Suppose that F : XM → C∞
M is R-linear and satisfies Fx(fX) = 0 for each f that
is zero in a neighbourhood of x. Then there exists a unique map F : XU → C∞
U that passes to
the same map Fx : XxM → C∞
x M.
12
6.2. Covariant derivative in vector spaces. First, we describe the local covariant derivative,
i.e. the covariant derivative in a vector space E, or its open subspace U ⊆ E. We will speak
uniformly about vector fields, tensor fields and other fields as associations F : x → F(x) ∈ Φ(TxU)
– for vector fields, Φ(V ) = V itself; for k-forms, Φ(V ) = (Λk
V )∗
; for functions, Φ(V ) = R. To
simplify the notation, we define Φ(TU) = x∈U Φ(TxU), so that F is a map U → Φ(TU) with the
property F(x) ∈ Φ(TxU). Since we have a canonical identification TxU ∼= E, we may reinterpret
these as maps f : U → Φ(E).1
Let now A ∈ TxU be a tangent vector. Then the directional
derivative Af = dxf(A) ∈ Φ(E) can be translated back to an element of Φ(TxU), denoted by
DAF. For a vector field X ∈ XU, we then get another field DXF, given by DXF(x) = DX(x)F.
{fields F : U → Φ(TU)} oo
∼= //
DX

{maps f : U → Φ(E)}
X

{fields F : U → Φ(TU)} oo
∼= // {maps f : U → Φ(E)}
As in the case of the differential in calculus, we also consider the total derivative of F at a
point x, an element DxF ∈ hom(TxM, Φ(TxM)) ∼= Φ(TxM) ⊗ T∗
x M, given by DxF(A) = DAF.
Finally, denoting again Φ(TM) ⊗ T∗
M = x∈U Φ(TxM) ⊗ T∗
x M, this total derivative defines a
field DF : U → Φ(TM) ⊗ T∗
M.
In the special case of vector fields, we have DXY −DY X = [X, Y ]; this will be important later.
In general, DXDY F − DY DXF = D[X,Y ]F. We will also see later that this derivative is really
connected with the vector space structure on E.
We will now speak briefly about parallel transport and geodesics – these are trivial in a vector
space. First we observe that D˙γ(t)F corresponds to
df(˙γ(t)) = d
dt f(γ(t))
and this depends only on the values of F along γ. Thus, we may consider a field G(t) along γ(t),
i.e. G: R → Φ(TU) such that G(t) ∈ Φ(Tγ(t)U); again, this corresponds to a map g: R → Φ(E).
Assuming that there is a field F such that G(t) = F(γ(t)), we also have g(t) = f(γ(t)). We
may thus define D˙γG
def
= D˙γF since we have already seen that the latter corresponds simply to
df(˙γ(t)) = ˙g(t) and thus depends only on G. As a special case, we obtain D˙γ(t) ˙γ(t) = ¨γ(t). In
terms of this derivative, we may define the concept of a field F transporting parallelly along a
path as D˙γF = 0 – this simply means that f is constant along γ. The covariant derivative may
be defined easily using this transport: in order to define D˙γ(0)F, we transport each F(t) along γ
to γ(0) and thus obtain a path R → Φ(Tγ(0)M) and D˙γ(0)F is the usual derivative of this path.
The advantage of this approach is that it avoids going through Φ(E), of course at the cost of
introducing the parallel transport. We will generalize this to Riemannian manifolds later.
We will now explain some simple rules that hold for computing with the local covariant derivative
D. The first one is that for an actual function f : U → R and its corresponding field
F : U → x∈U R = U × R, the derivative is the usual directinal derivative, i.e. DAF corresponds
to Af (most of the times, these can be considered equal).
The second one is that any linear and natural τ = τV : Φ(V ) → Ψ(V ) defines, for a field
F : U → Φ(TU), a field τF : U → Ψ(TU), as the composition t → τ(F(t)). Since τ is linear, we
obtain easily
DA(τF) = τDAF.
Our last tool will be a general Leibniz rule. Let two fields F : U → Φ(TU) and G: U → Ψ(U)
be given. Then we may form their tensor product F ⊗ G: U → Φ(TU) ⊗ Ψ(TU) (again the union
of the Φ(TxU)⊗Ψ(TxU)). This corresponds to the map f ⊗g: U → Φ(E)⊗Ψ(E), whose derivative
is
dx(f ⊗ g)(A) = dxf(A) ⊗ g(x) + f(x) ⊗ dxg(A)
1Formally, we have a single requirement: Φ should be a functor from the category of vector spaces and their
isomorphisms to itself.
13
(express everything in some bases of the Φ(E) and Ψ(E), observe that the coordinates of f ⊗ g
are actual products of the coordinates of f and g and thus, the standard Leibniz rule applies;
alternatively, avoiding coordinates, tensor multiply the defining relations for the differential f(x+
ξ) = f(x) + dxf(ξ) + o(ξ) and the same for g and observe that dxf(ξ) ⊗ dxg(ξ) ∈ o(ξ)). This
yields immediately
DA(F ⊗ G) = DAF ⊗ G + F ⊗ DAG.
Now suppose that F : U → (T∗
U)⊗k
is a tensor field of type (0, k). Then the evaluation map
ev: (V ∗
)⊗k
⊗ V ⊗k
→ R
is linear and natural and yields
DA(ev(F ⊗ X1 ⊗ · · · ⊗ Xk)) = ev(DA(F ⊗ X1 ⊗ · · · ⊗ Xk))
= ev(DAF ⊗ X1 ⊗ · · · ⊗ Xk)
+
i
ev(F ⊗ X1 ⊗ · · · ⊗ DAXi ⊗ · · · ⊗ Xk)
Expanding the evaluation map, this reads
DA(F(X1, . . . , Xk)) = (DAF)(X1, . . . , Xk) +
i
F(X1, . . . , DAXi, . . . , Xk)
The term on the left is the directional derivative. In particular, we will need shortly a formula for
the derivative of a tensor field g of type (0, 2):
(DAg)(X, Y ) = Ag(X, Y ) − g(DAX, Y ) − g(X, DAY ).
A similar result holds for tensor fields of type (1, k).
6.3. Covariant derivative for submanifolds of Euclidean spaces. We start with the following
situation. Let M ⊆ E be a submanifold. Then we have the following concepts available in M:
parallel transport and covariant derivative. We start with the parallel transport which we find
more intuitive. Let γ : R → M be a path and X : R → TM be a vector field along γ, i.e. we assume
X(t) ∈ Tγ(t)M. We say that X transports parallelly along γ in M if ˙X(t) is perpendicular to
Tγ(t)M. Denoting by P(x) the orthogonal projection TxE → TxM, this means P(γ(t)) ˙X(t) = 0.
Denoting XY = P(DXY ), the condition of the parallel transport is thus ˙γX = 0. Since we
have DXY − DY X = [X, Y ] and [X, Y ] is tangent to M if both X and Y are (so that [X, Y ] is
preserved by P), we obtain
XY − Y X = [X, Y ].
We say that the covariant derivative is symmetric. The second property that we need is that
g = 0 where g is the metric on M (we say that g is covariantly constant). First, we have to
introduce the covariant derivative Xg. This is done by postulating the formula from the previous
section,
( Xg)(Y, Z) = X Y, Z − XY, Z − Y, XZ .
Now we claim that this is zero: this follows from the same rule for D (here g corresponds to
the constant map U → E∗
⊗ E∗
, taking every point to the given scalar product in E, yielding
DXg = 0) with the projection P applied (since Z is tangent to M, its product with DXY is the
same as with XY = PDXY ).
6.4. Riemannian manifolds and linear connections.
Definition 23. A Riemannian metric on a smooth manifold M is a choice of a scalar product
on each TxM that depends smoothly on x ∈ M. In detail, it is a tensor field of type (0, 2), i.e.
a smooth map g: M → (T∗
M)⊗2
, that is symmetric and positive definite at each point (i.e. each
gx ∈ (T∗
M)⊗2
should be symmetric and positive definite).
Definition 24. A Riemannian manifold is a manifold M equipped with a Riemannian metric.
Example 25. The Euclidean space with the constant field g. Any submanifold M ⊆ E of a
Euclidean space E with the restriction of the scalar product on E to M.
14
Definition 26. A linear connection on a manifold M is a mapping
: XM × XM → XM,
denoted (X, Y ) → XY , satisfying the conditions
X(Y1 + Y2) = XY1 + XY2,
X(gY ) = Xg · Y + g XY,
X1+X2
Y = X1
Y + X2
Y,
fXY = f XY.
Example 27. The local covariant derivative D on an open subset of a vector space (all properties
are trivial). The covariant derivative XY = PDXY on a submanifold of a Euclidean space –
the only non-trivial axiom is the second one (apply the projection P to the equality Dx(gY ) =
Xg · Y + gDXY and observe that the second term belongs to TM so that it is preserved by P).
Definition 28. We say that a connection is symmetric is XY − Y X = [X, Y ].
Definition 29. A connection on a Riemannian manifold M is metric if
X Y, Z = XY, Z + Y, XZ .
The conditions of a connection imply that it is local (i.e. XY (x) depends only on the germs
of X and Y at x). Let now DXY again denote the usual local covariant derivative in a given
coordinate system on M. Then XY − DXY is C∞
M-linear in both X and Y and, thus, given
by a tensor field Γ of type (1, 2), i.e.
XY = DXY + Γ(X, Y ).
The symmetry of translates to Γ(X, Y ) = Γ(Y, X), i.e. the symmetry of Γ, and the metricity
of translates to
0 = X Y, Z − DXY + Γ(X, Y ), Z − Y, DXZ + Γ(X, Z)
= (DXg)(Y, Z) − Γ(X, Y ), Z − Y, Γ(X, Z) ,
i.e. Γ(X, Y ), Z + Y, Γ(X, Z) = (DXg)(Y, Z). According to the following lemma, this determines
Γ(X, Y ), Z uniquely and, thus, also Γ(X, Y ), so that we obtain the following theorem.
Theorem 30. There exists a unique symmetric and metric connection on a given Riemannian
manifold – it is called the Levi-Civita connection.
Lemma 31. The map sym23 : (S2
V ⊗V )∗
→ (V ⊗S2
V )∗
, given by sym23 ω(X, Y, Z) = ω(X, Y, Z)+
ω(X, Z, Y ) is an isomorphism.
Proof. The spaces have the same dimensions; thus, it is enough to show that the kernel is zero.
But any ω ∈ ker sym23 is symmetric in the first two and antisymmetric in the last two variables,
hence zero.
Remark. In fact, it is not difficult to show that the inverse is given by
((sym23)−1
θ)(X, Y, Z) = 1
2 (θ(X, Y, Z) + θ(Y, X, Z) − θ(Z, X, Y )).
However, we will not make use of this formula.
6.5. Parallel transport, geodesics. The equation for a parallel transport is
0 = ˙γX = D˙γX + Γ(˙γ, X),
i.e. ˙X = −Γ(˙γ, X). This is a differential equation and, locally, a unique solution exists through
each X(0). However, since the solution exists globally for the zero vector, it must exist for any
small vector and then for any vector since the parallel transport is clearly linear – any linear
combination (with constant coefficients) of parallel vector fields is also parallel.
Another observation is that if both X and Y transport parallelly along γ then
˙γ X, Y = ˙γX, Y + X, ˙γY = 0
15
and the scalar product X, Y is constant along γ – we say that the parallel transport preserves
the scalar product (in fact, this is equivalent to the metricity of ).
We denote by Ptγ
t the map Tγ(t0)M → Tγ(t0+t)M obtained by transporting vectors parallelly
along γ. We have thus proved that each Ptγ
t is orthogonal.
A geodesic path is a path γ such that ˙γ transports parallelly along γ. This reads
¨γ = −Γ(˙γ, ˙γ)
and is a differential equation of second order. Again, locally, a unique solution exists with any
given A = ˙γ(0) ∈ Tγ(0)M. We will temporarily denote it γA. Then it is pretty much clear that
γsA(t) = γA(st).
Thus, denoting exp A = γA(1), we obtain γA(t) = γtA(1) = exp tA. The map exp: TM → M is
not defined globally; however, it is defined in a neighbourhood of the zero section of TM, since
exp 0x = x. Each expx : TxM → M is a local diffeomorphism at 0x (since its derivative is the
identity – expx∗0x
A = d
dt t=0
expx(tA) = A).
We will now show how the covariant derivative can be reconstructed from the parallel transport.
Let ei be an orthonormal frame at x and transport it parallelly along a path γ through x. Then
we get vector fields Ei along γ and they will still be orthonormal since parallel transport preserves
scalar product. Let X be a vector field along γ and express it in this orthonormal frame as
X = fi
Ei. Then
˙γX = ˙γ fi
Ei = ˙γfi
· Ei + fi
· ˙γEi
0
= ˙γfi
· Ei.
In other words, expressing the vector field in a parallel orthonormal frame makes it into a function
f : R → Rn
and then the covariant derivative is simply the usual derivative d
dt f.
We remark that fi
(t)ei = Ptγ
−t X(t) so that ˙γ(0)X = d
dt t=0
Ptγ
−t X(t). In plain words,
transporting the vector field X along γ to γ(0) produces a path of vectors in Tγ(0)M and ˙γ(0)X
is then the usual derivative at zero of this function R → TxM with values in a vector space. This
will be important in the next section.
6.6. Second covariant derivative. Let us compute the second covariant derivative of a field
2
F(X, Y ) = F(X, Y ) = X( F(Y )) − F( XY ) = X Y F − X Y F.
Now define the antisymmetric version of 2
F as
2
altF(X, Y ) = 2
F(X, Y ) − 2
F(X, Y ) = X Y F − Y XF − X Y − Y XF,
where the last term becomes [X,Y ]F for a symmetric connection. There are two important
special cases – that of functions where 2
altf(X, Y ) = [X,Y ]− X Y − Y Xf, i.e. the derivative of
f in the direction of (minus) the torsion (which is zero for symmetric connections – clearly, this
is equivalent to the symmetry of ), and that of vector fields where 2
altZ(X, Y ) is the so-called
curvature.
By definition, 2
altF(X, Y ) is C∞
M-linear in both X and Y . We will now show that 2
altF is
C∞
M-linear in F. Thus, let h be a function and compute
2
(hF)(X, Y ) = X Y (hF) − X Y (hF)
= X( Y h · F + h · Y F) − X Y h · F − h · X Y F
= X Y h · F + Y h · XF + Xh · Y F + h · X Y F − X Y h · F − h · X Y F
= h · 2
F(X, Y ) + 2
h(X, Y ) · F + Y h · XF + Xh · Y F
and both 2
h and the remaining terms disappear after antisymmetrization, yielding
2
alt(hF)(X, Y ) = h 2
altF(X, Y )
or simply 2
alt(hF) = h 2
altF.
16
This means that the alternating second derivative is “algebraic” in the sense that it is a value
of a field on (X, Y, F). In fact, it turns out that TM ⊗ T∗
M has a natural action2
on Φ(TM)
and 2
altF(X, Y ) is simply the action of R(X, Y ) on F (here R(X, Y ) is the curvature tensor
evaluated in two arguments, i.e. still a tensor field of type (1, 1); it could be viewed as the map
Z → R(X, Y )Z).
Since 2
alt is zero on functions (the symmetry on ) and since the local covariant derivative
of any field is obtained by differentiating covariantly the coordinates of the field (and these are
functions!), we see that the curvature of a local covariant derivative is zero.
6.7. Curvature. The curvature is defined to be R(X, Y )Z = X Y Z − Y XZ − [X,Y ]Z. It
is a tensor of type (1, 3) that is clearly antisymmetric in the first two variables. We have seen
that the curvature of the Euclidean covariant derivative is zero. In fact, this characterizes the
Euclidean connection, as the following theorem says.
Before going into the proof, we give a geometric meaning to the curvature. Let X, Y be vector
fields that commute, i.e. such that [X, Y ] = 0. Then X Y Z(x) is obtained as the mixed partial
derivative ∂2
∂s∂t A(0, 0) of the vector valued function A(s, t) ∈ TxM given by transporting parallelly
Z(FlY
t (FlX
s (x))) along the flow line of Y back to FlX
s (x) and then along the flow line of X back to
x. A similar formula holds for the second term. We may however define Z by first transporting
Z(0, 0) ∈ TxM along the flow line of Y and then along the flow lines of X so that the second term
actually becomes zero. Thus, we finally obtain
R(X, Y )Z = ∂2
∂s∂t
(s,t)=(0,0)
PtFlX
−s PtFlY
−t PtFlX
s PtFlY
t Z
(it would look slightly better to change both s, t to their opposites – then this becomes a commu-
tator).
Continuing the notation of the above proof, we will show that for R(X, Y )Z = 0 and [X, Y ] = 0,
the parallel transports along the flow lines of X and Y commute: 0 = R(X, Y )Z = X Y Z, so
that Y Z transports parallelly along the flow lines of X. Since Y Z = 0 for s = 0, it must be
zero everywhere, i.e. Z also transports parallelly along the flow lines of Y . In particular, we obtain
Z(s, t) also by transporting Z(0, 0) first parallelly along the flow line of X to get Z(0, t) and then
parallelly along the flow line of Y (this is what we have just proved).
Theorem 32. The following conditions are equivalent.
(1) The curvature is zero.
(2) The parallel transport does not locally depend on the path.
(3) There is an atlas in which all the Γ are zero.
(4) There is an atlas consisting of isometries.
Proof. We will prove (1)⇒(2)⇒(3)⇒(4)⇒(1).
(1)⇒(2): We use the fact that the parallel transports along vector fields X, Y commute when
[X, Y ] = 0. Start with a vector Z0 ∈ TxM and transport it parallelly along the local vector fields
∂1, . . . , ∂n to obtain a vector field Z with Z(x) = Z0. Since Z was obtained by parallel transport
along ∂i (any one could have been used the last), we have ∂i Z = 0. This holds for any i and,
thus, ∂XZ = 0 for any X. In particular, Z transports parallelly along any path, implying that
the parallel transport of Z(x) along a path from x to y is always Z(y).
(2)⇒(3): Suppose that the parallel transport does not locally depend on the path. Start
with a basis (ei) of TxM and transport it locally to a neighbourhood to obtain vector fields Ei.
Then [Ei, Ej] = Ei Ej − Ej Ei = 0 and there exists a chart with Ei = ∂i. In particular,
Γ(Ei, Ej) = Ei Ej − DEi Ej = 0 and Γ = 0.
(3)⇒(4): Clearly, to obtain a local isometry, it is enough to have DXg = 0 for all X (then g is
constant and we may modify the chart by a linear isomorphism). But
0 = ( Xg)(Y, Z) = (DXg)(Y, Z) − Γ(X, Y ), Z − Y, Γ(X, Z) = (DXg)(Y, Z).
(4)⇒(1) is clear since we have XY = DXY in a Euclidean space and the curvature is zero.
2It is the derivative at the identity of the map GL(TxM) → GL(Φ(TxM)), α → Φ(α), given by the functor Φ.
17
6.8. Remarks about covariant derivative of other fields. The above point of view may
also be applied to other fields. Let Φ be a functor from the category of vector spaces and their
isomorphisms to itself3
. By a field F of type Φ, we will understand an association M x → F(x) ∈
Φ(TxM) that will be smooth in the following sense: a local chart ϕ: U → Rn
gives trivializations
ϕ∗x : TxM
∼=
−→ Rn
of each tangent space TxM, x ∈ U, and applying the functor Φ then gives a
map (the expression of the field in local coordinates)
U → Φ(Rn
)
x → Φ(ϕ∗x)(F(x))
that should be smooth.4
Now we apply this in the same way as above, i.e. in the situation
that γ is a path and F a field along γ. Choose a parallel frame (Ei) that gives trivializations
αt : Tγ(t)M
∼=
−→ Rn
(different from those above) and use them to translate the field F to a function
f : R → Φ(Rn
), t → Φ(αt)(F(t)). Then we take the usual derivative d
dt f : R → Φ(Rn
) and
translate it back to a field ˙γF along γ, i.e.
˙γ(t)F = Φ(αt)−1
( d
dt f(t)).
A different parallel frame differs by a constant (and orthogonal) transformation (Ei) = (Ei)P
which corresponds to an automorphism P : Rn
→ Rn
satisfying αt = Pαt. This implies f = Φ(P)f
and, since Φ(P) is linear, we obtain d
dt f(t) = Φ(P) d
dt f, yielding finally the independence of ˙γF
on the choice of the frame.
There are three simple rules that allow one to compute the covariant derivative of the standard
fields.
The zeroth one is that a function is a field of type R, the constant functor taking every vector
space to R and every isomorphism to the identity. Then, independently of the frame, we obtain
f = F and ˙γF = d
dt F. For a global function (as opposed to a function along γ which then
happens to be just F ◦ γ), we obtain F = dF, the differential of F (i.e. AF = dF(A) = AF,
the derivative of F in the direction of the vector A).
The first one is that linear natural transformations commute with the differentiation, i.e. if
τ = τV : Φ(V ) → Ψ(V ) is linear and natural then we define, for a field F of type Φ, a field τF of
type Ψ, as the composition t → τ(F(t)). We obtain easily
(τF) = τ F.
The second concerns the tensor product of fields. Given functors Φ and Ψ, the tensor products
Φ(V ) ⊗ Ψ(V ) clearly form another functor denoted Ψ ⊗ Ψ. Given fields F and G of types Φ and
Ψ respectively, the association x → F(x) ⊗ G(x) ∈ Φ(TxM) ⊗ Ψ(TxM) is a field of type Φ ⊗ Ψ
and the corresponding function is f ⊗ g: R → Φ(Rn
) ⊗ Ψ(Rn
); its derivative is
dt(f ⊗ g)(A) = dtf(A) ⊗ g(t) + f(t) ⊗ dtg(A)
(by tensoring the defining relations for the differential f(t + τ) = f(t) + dtf(τ) + o(τ) and the
same for g, observing that dtf(τ) ⊗ dtg(τ) ∈ o(τ)). This yields immediately
(F ⊗ G) = F ⊗ G + F ⊗ G.
Now suppose that ω is a field of type hom(Φ⊗k
, Ψ). Then the evaluation map
evV : hom(Φ(V )⊗k
, Ψ(V )) ⊗ Φ(V )⊗k
→ Ψ(V )
3It is useful to think of this as a way of producing out of coordinates V
∼=
−→ Rn on V some sort of coordinates
Φ(V )
∼=
−→ Φ(Rn) on Φ(V ), e.g. ΛkV ∗
∼=
−→ Λk(Rn)∗, ω → i1<···<ik
ωi1···ik
fik ∧ · · · ∧ fi1
4More generally, we may equivalently ask for the expression F in any field of frames (Ei) (simply an n-tuple of
vector fields that provides a basis at each point) to be smooth; such a frame does not necessarily come from a local
chart (e.g. we may require an orthonormal frame and it may be impossible to obtain one from a chart).
18
is linear and natural and yields
X(ev(ω ⊗ X1 ⊗ · · · ⊗ Xk)) = ev X(ω ⊗ X1 ⊗ · · · ⊗ Xk)
= ev( Xω ⊗ X1 ⊗ · · · ⊗ Xk)
+
i
ev(ω ⊗ X1 ⊗ · · · ⊗ XXi ⊗ · · · ⊗ Xk)
When the evaluation map is expanded, this reads
X(ω(X1, . . . , Xk)) = ( Xω)(X1, . . . , Xk) +
i
ω(X1, . . . , XXi, . . . , Xk)
Let us compute as an example the second covariant derivative of any field
2
F(X, Y ) = F(X, Y ) = X( F(Y )) − F( XY ) = X Y F − X Y F.
Now define the antisymmetric version of 2
F as
2
altF(X, Y ) = 2
F(X, Y ) − 2
F(X, Y ) = X Y F − Y XF − X Y − Y XF,
where the last term becomes [X,Y ]F for a symmetric connection. There are two important special
cases – that of functions where 2
altf(X, Y ) = [X,Y ]− X Y − Y Xf, i.e. the derivative of f in the
direction of (minus) the torsion, and that of vector fields where 2
altZ(X, Y ) = R(X, Y )Z, i.e. the
curvature (at least when torsion is zero).
7. Spaces of constant curvature
7.1. Sectional curvature. First we observe that R(X, Y )Z, W is also anti-symmetric in the
variables Z, W. This follows from
0 = ( 2
alt Z, W )(X, Y ) = X Y Z, W − Y X Z, W − X Y − Y X Z, W
= R(X, Y )Z, W + Z, R(X, Y )W
(the terms where each Z, W receives one of the X, Y cancel out).
Let p be a 2-dimensional vector subspace of TxM with orthonormal basis (e1, e2). We define the
sectional curvature K(p) = R(e1, e2)e2, e1 . Denoting R(X, Y, Z, W) = R(X, Y )W, Z , we see
that this depends only on X ∧ Y and Z ∧ W. Thus, when replacing these by a linear combination,
the whole expression gets multiplied by the product of the determinants of the transformation
matrices. In the case of K(p), this means the square of an orthogonal matrix and the value does
not change.
7.2. Sphere. We compute the sectional curvature of a sphere Sn
⊆ Rn+1
. Thus, let e1, e2 ∈ Tx0
Sn
be two orthonormal vectors. We extend them to vector fields on Sn
in the following way: think
of the ei as a constant vector field on Rn+1
and project it orthogonally to obtain a vector field Ei
on TSn
; at a point x, this equals
Ei(x) = ei − x, ei x
In fact, this formula prescribes a vector field on Rn+1
– this is useful since we want to apply the
covariant derivative of Rn+1
:
DAEi = − A, ei x − x, ei A
Projecting to TSn
, the first term becomes zero and the second term remains unchanged (since A
is now assumed tangent to Sn
), i.e.
AEi = − x, ei A
This leads to
[Ei, Ej] = Ei
Ej − Ej
Ei = x, ei Ej − x, ej Ei
and finally
R(Ei, Ej)Ej = Ei Ej
Ej
− x,ej Ej
− Ej Ei
Ej
− x,ej Ei
− [Ei,Ej ]Ej
19
Now the derivative of x, ej in the direction of Ei equals Ei, ej , since the function is linear in
x. Thus,
R(Ei, Ej)Ej = − x, ej · Ei
Ej
− x,ej Ei
− Ei, ej Ej
+ x, ej · Ej
Ei
− x,ei Ej
+ Ej, ej Ei
+ x, ej ( x, ei Ej − x, ej Ei)
(the function x, ej is linear in x) so that
R(Ei, Ej)Ej, Ei (x0) = − ei, ej ej, ei + ei, ei ej, ej = 1 · Vol(ei, ej)2
as x0, ej = 0, since ej ∈ Tx0
Sn
.
7.3. Hyperbolic space. We compute the sectional curvature of a hyperbolic space Hn
⊆ Rn+1
equipped with a metric g = −dx0
⊗ dx0
+ dx1
⊗ dx1
+ · · · + dxn
⊗ dxn
, where
Hn
= {x ∈ Rn+1
| g(x, x) = −1}.
Since each x ∈ Hn
generates a 1-dimensional subspace where the metric is negative definite, it
is easy to see that on its orthogonal complement – and this is again TxHn
– the metric must be
positive definite (the inertia theorem). Thus, let e1, e2 ∈ Tx0 Hn
be two orthonormal vectors. We
extend them to vector fields on Hn
, this times the formula is
Ei(x) = ei + x, ei x
(because x, x = −1). Again, this prescribes a vector field on Rn+1
and:
DAEi = A, ei x + x, ei A
Projecting to TSn
, we get
AEi = x, ei A
This leads to the same formula for R(Ei, Ej)Ej as above, only with different signs. Since the
surviving terms in R(Ei, Ej)Ej, Ei contain exactly one scalar product from R(Ei, Ej)Ej, we get
R(Ei, Ej)Ej, Ei (x0) = ei, ej ej, ei − ei, ei ej, ej = −1 · Vol(ei, ej)2
.