Orientation (Informal)
Consider two 2-dimensional beings in the plane. Observe that to each
other, everything appears to be a vertical line. Hence, one can really see
how orientation is important since 2-dimensional beings can’t see their
world from the outside-in, a world of confusion will surmount if a choice of
orientation is made, but changes after some time during a morning stroll.
We will now mathematically quantify the possible choices in the plane. Let
p be a point in the plane. If p chooses to let the direction vector (−1, 0) as
her/his forward = positive orientation, we will call this the counterclockwise
direction. The reason to the same is simply because any particle rotating in
the direction of (−1, 0), by choice of p, is moving forward. Similarly, we let
the opposite direction be the clockwise direction = negative orientation. We
remark that there is no reason why one cannot let the clockwise direction
be the positive orientation because again, it is all a matter of choice.
Observe that now when we transition to defining an orientation in 3-space,
we face the same dilemma as our 2-dimensional beings. Our choice here
now become, what is outside and what is inside? To a 3-dimensional being
it seems pointless to need to decide such a thing since it is clear. However,
this is exactly the same situation we encountered with the lower
dimensional example. Although it is clear to us how to deflate a dispute
between two 2-dimensional beings who have chosen opposite orientations, it
take much more thought for 3-dimensional beings to explain why an
observable is the outside or inside of something. Although, a bit hard to
understand, we see that outside and inside is a choice and now we will
make this precise.
On your right hand (with palm toward eyes), let your index finger denote
the y-axis and middle finger denote the z-axis. We say that one has chosen
the position orientation in 3-space, if when you curl the x-axis finger to the
y-axis finger, your thumb points in the outward direction. Mirroring this
process with the left hand, the thumb will point in its outward direction.
The difference between the two choices is that the right hand’s outward is
the left’s inward and vice-versa. We call the first orientation (pos.) the
clockwise orientation and the latter, the counterclockwise orientation.
1
Orientation on Manifolds
From the above informal discussion, we observe that the orientations in
R2 , R3 can we described by an ordered basis. For example, the
counterclockwise orientation say that when moving to the left in the plane,
that is the positive direction. This is the same as saying if you rotate e1 to
e2 , this is the positive direction.
Since we a defining orientation by an ordered tuple of linearly-independent
vectors, it is very natural that we would like to establish an equivalence
relation on them. The most natural equivalence relation would be on the
determinant of the change of basis matrix from one system with a chosen
orientation to another. We say this is natural because the determinant of
this transition matrix will be in Gl(n, R) i.e either positive or negative.
Hence, we’ve partition the set of orientations into two equivalence classes
which is the best you could aim for.
Let B = {v1 , ..., vn } and B 0 = {u1 , ..., un } be basis for a vector space V and
A denote the change of basis matrix from B to B 0 . We say;
[v1 , ..., vn ] ∼ [u1 , ..., un ] ⇐⇒ det(A) > 0
One can show that ∼ is actually an equivalence relation. We will now show
that for an n-dimension vector space, a non-vanishing n-form determines
the orientation. Let B, B 0 be defined as above and for simplicity let us take
n = 2.
ui =
2
X
aij vj where i = 1, 2
j=1
Let β be a non-vanishing 2-form i.e β is alternating. Then we have;
!
β(u1 , u2 ) = β
X
j1
aj11 vj1 ,
X
aj22 vj2
=
j2
X
aj11 aj12 β(vj1 , vj2 )
j1 ,j2
=
X
aj11 aj22 (sgn(σI ))β(v1 , v2 )
j1 ,j2
= det(A) β(v1 , v2 )
2
Orientation Preserving Maps
From the above we see that if β(u1 , u2 ) and β(v1 , v2 ) have the same sign
⇐⇒ det(A) > 0 is [u1 , u2 ] ∼ [v1 , v2 ]. Hence, we say that β specifies the
orientation (v1 , v2 ) if β(v1 , v2 ) > 0. By the above this is a well-defined
notion i.e if (u1 , u2 ) ∼ (v1 , v2 ) then det(A) > 0 ⇒ β(u1 , u2 ) > 0.
Example: dx1 ∧ dx2 determines the clockwise orientation on R2 since
dx1 ∧ dx2 (e1 , e2 ) = 1 > 0.
Observe that if ω, ω 0 are two nowhere-vanishing smooth n-forms on a
manifold n then ω = f ω 0 where f is some non-vanishing real-values function
on M . This follows from the fact that if we let (U, φ) = (U, x1 , ..., xn ) be a
chart on M we have ω = hdx1 ∧ · · · ∧ dxn and ω = gdx1 ∧ · · · ∧ dxn where
f, g are no-where vanishing smooth functions on U . Hence f = h/g is also a
no-where vanishing smooth function on U . On a connected manifold, f is
either always positive or always negative. Thus we can define an equivalence
relation on no-where vanishing forms for a connected orientable manifold.
ω ∼ ω 0 ⇐⇒ ω = f ω 0 with f > 0
From the above remarks, we know that for each point-wise orientation
µ = [(X1 , ..., Xn )] we associate the equivalence class of a smooth no-where
vanishing n-form ω s.t ω(X1 , ..., Xn ) > 0. It requires proof to show that for
a orientable manifold M , such an ω always exists, but we will just accept
this fact.
We say a diffeomorphism F : (N, [ωN ]) → (M, [ωM ]) of oriented manifolds is
orientation-preserving if [F ∗ ωM ] = [ωN ]. Let U, V ⊂ Rn be open with the
standard orientation inherited from Rn . For sake of clarity, let us take
n = 2 i.e F (x, y) = (u(x, y), v(x, y) = (u, v) where x, y are the coordinates
on U and u, v the coordinates on V .
F ∗ (dx ∧ dy) = d(x ◦ F ) ∧ d(y ◦ F )
= (ux vy − uy vx ) du ∧ dv
= det(Jac(F )) du ∧ dv
> 0 ⇐⇒ det(Jac(F )) > 0
3
The Pull back Map
Let φ : M → N be a smooth map, where M, N are manifolds. Then we
know φ∗,p : Tp M → Tφ(p) is the linear map defined by:
φ∗,p Xp (f ) = Xp (f ◦ φ)
We call the map above, the push-forward map since it takes tangent vectors
to tangent vectors. There is another linear map φ∗ which pull-back
differential forms, define by;
∗
φ ω(v1 , ..., vk ) = ω (φ∗ (v1 ), ..., φ∗ (vk ))
φ(p)
where ω ∈ Ωk (N ) and v1 , ..., vk ∈ Tp M . Hence, the RHS of the equation
above is real-valued i.e φ∗ ω ∈ Tp M . To show φ∗ is linear, for simplicity we
take n = 1.
φ∗ (αω + τ )(v1 ) = (αω + τ )(φ∗ (v1 ))
= αω(φ∗ (v1 )) + τ (φ∗ (v1 ))
= αφ∗ ω + φ∗ τ
For the case where ω ∈ Ω0 (N ) i.e ω : N → R (smooth) we have:
φ∗ ω = ω φ(p)
The pull-back map has some very important properties and we give them
below.
• If ω is a smooth k-form, then φ∗ ω is also smooth.
• φ∗ (α ∧ β) = φ∗ α ∧ φ∗ β where α, β are smooth forms.
• φ∗ (dw) = d(φ∗ ω) where ω is a smooth k-form.
4
We will show these properties, but again for simplicity we take n = 1, 2.
Let ω be a 1-form on N and τ be a two form on N , and φ : M → N be a
smooth map. Let (ψ, U ) = (ψ, x1 , ..., xn ) be a chart on M . Then we have:
X
X
ω=
fk dxk and τ =
gij dxi ∧ dxj
1≤i,j≤n
1≤k≤n
!
X
φ∗ (ω ∧ τ ) = φ∗
fk gij dxk ∧ dxi ∧ dxj
1≤k,j,i≤n
=
X
(fk gij ◦ φ) φ∗ (dxk ∧ dxi ∧ dxj )
1≤k,j,i≤n
=
X
(fk ◦ φ)(gij ◦ φ) d(φ∗ xk ) ∧ d(φ∗ xi ) ∧ d(φ∗ xj )
1≤k,j,i≤n
=
X
(fk ◦ φ)(gij ◦ φ) d(xk ◦ φ) ∧ d(xi ◦ φ) ∧ d(xj ◦ φ)
1≤k,j,i≤n
=
X
X
(fk ◦ φ) d(xk φ) ∧
(gij ◦ φ) d(xi ◦ φ) ∧ d(xj ◦ φ)
1≤i,j≤n
1≤k≤n
= φ∗ ω ∧ φ∗ τ
For the above remark, let ω = fk dxk i.e a simple 1-form on N . Then we
have;
φ∗ (dω)(v1 , v2 ) = φ∗
X ∂fk
j
∂xj
!
dxj ∧ dxk
(v1 , v2 ) =
=
X ∂
(f ◦ φ) φ∗ (dxj ∧ dxk )(v1 , v2 )
j
∂x
j
X ∂
(f ◦ φ) dxj ∧ dxk (φ∗ (v1 ), φ∗ (v2 ))
j
∂x
j
d(φ∗ ω) = d(φ∗ (fk dxk )) = d(fk ◦φ dxk (φ∗ (v1 ), φ∗ (v2 ))) =
5
X ∂
(fk ◦φ) dxj ∧dxk (φ∗ (v1 ), φ∗ (v2 ))
j
∂x
j
Change of Variables Formula
The reason why we define such a map is because the change of variables
formula seems to mimic this process. Let us recall what the change of
variables formula says with a very non-rigorous proof sketch.
Suppose the map G : (a, b) × (c, d) = D ⊂ R2 → G(D) = D0 ⊂ R2 is a C 1
diffeo. Let f : D0 → R be a smooth map. Suppose we take D to be a
sufficiently small rectangle, then we have;
ZZ
f dxdy ≈ f (p) · Area(D0 )
D0
= f (G(p0 ))|det(Jac(Gp0 ))| · Area(D)
ZZ
(f ◦ G)|det(Jac(G))| dudv
=
D
To obtain the result for a general open subset U of R2 , we partition U into
small rectangles (patches) then we have;
ZZ
f dxdy ≈
D0
X
f (pij ) · Area(Rij )
i,j
=
X
f (G(p̃)ij ))|det(Jac(Gp˜ij )| · Area(R̃ij )
i,j
ZZ
(f ◦ G)|det(Jac(G))| dudv
=
D
6
Integration on Manifolds
Before we attempt to define integration on a general manifold, we try to see
how we would want to define integration on manifolds which are already in
Euclidean space. There is no generality lost here because by Whitney, we
can embed any smooth manifold into RN for some N .
If U ⊂ Rn is a open set and f dx1 ∧ · ∧ dxn is a smooth form on U , then we
define:
Z
Z
1
n
f dx ∧ · · · ∧ dx =
f dx1 · · · dxn
U
U
For simplicity, let us take φ(x, y) = (u(x, y), v(x, y) = (u, v). We would like
that if φ : U ⊂ R2 → V ⊂ R2 is a diffeomorphism then:
Z
Z
ω=
φ∗ ω
U
V =φ(U )
However, if ω ∈ Ω2 (Rn ) say ω = f dx ∧ dy then:
Z
∗
Z
φ (f dx ∧ dy) =
U
Z
(f ◦ φ) d(x ◦ φ) ∧ d(y ◦ φ) =
U
(f ◦ φ)(ux vy − uy vx ) du ∧ dv
U
Z
(f ◦ φ) det(Jac(φp )) du ∧ dv
=
U
Hence the above gives us ± the Change of Variables result. Thus, we can
get a well-defined definition if we ensure that the reparametrization φ is
orientation preserving. But from the above we know that this amounts to
saying that the determinant for the jacobian of each transition map is
positive.
7
Homotopic Maps
We end the discussion with a remark about homotopic maps. The reason
for the topic is due to the following fact:
Fact: If f0 , f1 : X → Y are homotopic maps and X is a compact manifold
without boundary then for all closed k-forms ω on Y we have;
Z
f0∗ ω
Z
=
X
f1∗ ω
X
Def : A homotopy H : X × I → Y between two maps f, g is a continuous
map s.t H(x, 0) = f and H(x, 1) = g.
Def : We say that f, g : X → Y are homotopic if there exists a homotopy
H(x, t) : X × I → Y such that H(x, 0) = f and H(x, 1) = g.
Example: Let f : Rn → Rn defined by f (x) = x0 and g : Rn → Rn defined
by g(y) = y. Then we can take H(x, t) = x0 (1 − t) + yt. This particular
homotopy has a special name, it is referred to as the straight-line homotopy.
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