Additional study material for the course of global
analysis
Radek Such´anek
Autumn semester 2019 / 2020
Exercise. RPn
is orientable ⇐⇒ n is odd.
Solution. • Consider Rn+1
with its standard orientation. Then the linear
map A: Rn+1
→ Rn+1
given by x → A(x) = −x is orientation preserving
⇐⇒ det A = (−1)n+1
> 0 ⇐⇒ n is odd.
• Now consider Sn
⊂ Rn+1
. Note that ν(x) = x defines a global unit
normal vector field for Sn
⊂ Rn+1
, which in turn determines an orientation
on Sn
as follows: a basis (ζ1, ζ2, . . . , ζn) of Tx Sn
is positively oriented
⇐⇒ (ν(x), ζ1, ζ2, . . . , ζn) is positively oriented basis of Tx Rn+1 ∼= Rn+1
.
Let us fix this orientation on Sn
.
• A|Sn : Sn
→ Sn
is orientation preserving ⇐⇒ n is odd.
• Consider the projection
π: Sn
→ Sn
/∼
∼= RPn
under which antipodal points on Sn
get identified, i.e. x ∼ A(x) = −x.
Then one can check that π is a local diffeomorphism, which implies the
isomorphism of tangent spaces
Tx π: Tx Sn ∼=
−→ Tπ(x) RPn
∀x ∈ Sn
(1)
Moreover, note that π ◦ A = π which implies T(π ◦ A) = Tπ. Thus we can
try to define orientation on RPn
by requiring (1) to be orientation preserving
∀x ∈ Sn
. This will lead to a well-defined orientation ⇐⇒ A is orientation
preserving.
Exercise. Suppose (M, ) is a smooth manifold equipped with an affine connection.
Show that
1. the torsion T and curvature R of are (1, 2) and (1, 3) tensor fields on
M
2. if is torsion-free, then for all ξ, η, ζ ∈ Γ(TM) the Bianchi identity holds:
R(ξ, η)ζ + R(η, ζ)ξ + R(ζ, ξ)η = 0 (2)
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Recall that a connection on a smooth manifold M is a bilinear operator
on the space of vector fields : Γ(TM) × Γ(TM) → Γ(TM), (ξ, η) → ξ η
satisfying
• fξ η = f ξ η for all f ∈ C∞
(M)
• ξ(fη) = ξ(f)η + f ξ η
Solution. 1. Torsion T of is given by
T(ξ, η) := ξ η − η ξ − [ξ, η] , (3)
where [−, −] is the usual bracket on vector fields. Since T(ξ, η) is a vector field,
it is automatically a (1, 0) tensor field on M. Then showing that T is bilinear
with respect to the ring of smooth functions C∞
(M) suffices to show that it is
a (1, 2) tensor field. Consider arbitrary f ∈ C∞
(M) and ξ, η, ζ ∈ Γ(TM) and
compute
T(fξ + η, ζ) = fξ+η ζ − ζ(fξ + η) − [fξ + η, ζ]
= f ξ ζ + η ζ − ζ(fξ) − ζ η + [ζ, fξ] − [η, ζ]
= f ξ ζ + η ζ − ζ(f)ξ − f ζ ξ − ζ η + ζ(f)ξ + f[ζ, ξ] − [η, ζ]
= f( ξ ζ − ζ ξ − [ξ, ζ]) + η ζ − ζ η − [η, ζ]
= fT(ξ, ζ) + T(η, ζ)
Similarly for the second argument.
For arbitrary ξ, η, ζ ∈ Γ(TM), the curvature R of is given by
R(ξ, η)ζ := ξ η ζ − η ξ ζ − [ξ,η] ζ (4)
Observe that R(ξ, η)ζ is a vector field and hence a (1, 0) tensor field. We pick
arbitrary f ∈ C∞
(M) and ξ, η, ζ, φ ∈ Γ(TM). It is easy to check that
R(ξ + φ, η)ζ = R(ξ, η)ζ + R(φ, η)ζ
and similarly for other arguments. Now we want to show that R(fξ, η)ζ =
R(ξ, fη)ζ = R(ξ, η)(fζ) = f R(ξ, η)ζ. We start with the first argument and
then, since the computation for the second argument is similar, we proceed
with the third argument.
R(fξ, η)ζ = fξ η ζ − η fξ ζ − [fξ,η] ζ
= f ξ η ζ − η f ξ ζ − −[η,fξ] ζ
= f ξ η ζ − (η(f) ξ ζ + f η ξ ζ) − −η(f)ξ−f[η,ξ]
= ξ η ζ − η(f) ξ ζ − f η ξ ζ + η(f) ξ ζ − f [ξ,η] ζ
= f( ξ η ζ − η ξ ζ − [ξ,η] ζ)
= f R(ξ, η)ζ
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and the computation for the third argument
R(ξ, η)(fζ) = ξ η(fζ) − η ξ(fζ) − [ξ,η](fζ)
= ξ(η(f)ζ + f η ζ) − η(ξ(f)ζ + f ξ ζ) − [ξ, η](f)ζ − f [ξ,η] ζ
= ξ(η(f))ζ + η(f) ξ ζ + ξ(f) η ζ + f ξ η ζ
− (η(ξ(f))ζ + ξ(f) η ζ + η(f) ξ ζ + f η ξ ζ)
− (ξ(η(f))ζ − η(ξ(f))ζ − f [ξ,η] ζ
= f( ξ η ζ − η ξ ζ − [ξ,η] ζ)
= f R(ξ, η)ζ
Solution. 2. Let us express the left-hand side of (2) using
R(ξ, η)ζ + R(η, ζ)ξ + R(ζ, ξ)η = ( ξ η − η ξ − [ξ,η])ζ
+ ( η ζ − ζ η − [η,ζ])ξ
+ ( ζ ξ − ξ ζ − [ζ,ξ])η
The righ-hand side of the last equality is a sum of the following terms
ξ( η ζ − ζ η) − [η,ζ] ξ (5)
η( ζ ξ − ξ ζ) − [ζ,ξ] η (6)
ζ( ξ η − η ξ) − [ξ,η] ζ (7)
For a torsion free connection we have T(ξ, η) = 0 which is by (3) equivalent to
[ξ, η] = ξ η − η ξ . (8)
If we apply (8) on (5), (6) and (7) and sum up we get
ξ[η, ζ] − [η,ζ] ξ + η[ζ, ξ] − [ζ,ξ] η + ζ[ξ, η] − [ξ,η] ζ
Using (8) again we obtain
R(ξ, η)ζ + R(η, ζ)ξ + R(ζ, ξ)η = [ξ, [η, ζ]] + [η, [ζ, ξ]] + [ζ, [ξ, η]]
Since Lie bracket satisfies Jacobi identity, the right-hand side of the last equality
is zero. We conclude
R(ξ, η)ζ + R(η, ζ)ξ + R(ζ, ξ)η = 0 .
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