1. Pushforward and Pull Back
Let f : Rn→ Rm be a smooth function. For each p ∈ Rn, we define the push-forward dfp: Tp(Rn) → Tf (p)(Rm)
of f at p by
dfp(vp) = (Df (p)(v))f (p)
where Df (p) is the derivative of f at p. The push-forward dfp of f at p is also called a tangent map.
If g : Rm → R is a smooth function, we define the pull back of g via f by f∗(g) = g ◦ f.
Let ω be a k-form on Rm. We want to define a k-form f∗ω on Rn as follows. For each p ∈ Rn, let (v1)p, · · · , (vk)p be vectors in Tp(Rn). We set
(f∗ω)(p)((v1)p, · · · , (vk)p) = ω(f (p)) (dfp((v1)p), · · · , dfp((vk)p)) .
One can show that (f∗ω)(p) : Tp(Rn)⊕k → R is an alternating k-linear form on Tp(Rn)⊕k for each p ∈ Rn. Hence the function
f∗ω : Rn→ Λr(T∗Rn)
sending p to (f∗ω)(p) defines a k-form on Rn and is called the pull back of ω via f.
Let {x1, · · · , xn} be the coordinate functions on Rnand {y1, · · · , ym} be coordinate func- tions on Rm. If f : Rn → Rm, we define fi : Rn → R by fi = yi ◦ f for 1 ≤ i ≤ m.
Then
f (x1, · · · , xn) = (f1(x1, · · · , xn), · · · , fm(x1, · · · , xm)).
Let us compute f∗dyk for 1 ≤ k ≤ m. Let p ∈ Rn. By linear algebra, in terms of the basis {(dxi)p : 1 ≤ i ≤ n} for Tp∗(Rn), the linear functional (f∗dyk)(p) is given by
(f∗dyk)(p) =
n
X
i=1
(f∗dyk)(p)((ei)p)(dxi)p.
Remark. Let V be a n dimensional vector space over a field F. Let {v1, · · · , vn} be a basis for V and {ϕ1, · · · , ϕn} ⊂ V∗ be the dual basis to {v1, · · · , vn}. For any θ ∈ V∗,
θ =
n
X
i=1
θ(vi)ϕi. By definition,
(f∗dyk)(p)((ei)p) = (dyk)f (p)(dfp((ei)p)) = (dyk)f (p)(Df (p)(ei)f (p))
= (dyk)f (p) ∂f1
∂xi(p), · · · ,∂fm
∂xi (p)
f (p)
= ∂fk
∂xi(p).
We obtain that
(f∗dyk)(p) =
n
X
i=1
∂fk
∂xi(p)(dxi)p for any p ∈ Rn. For each 1 ≤ k ≤ m, one has
f∗dyk=
n
X
i=1
∂fk
∂xidxi. Since dfk=Pn
i=1
∂fk
∂xidxi, we also obtain that
f∗dyk= dfk = d(yk◦ f ) = d(f∗yk).
1
2
Theorem 1.1. Let f : Rn→ Rm be a smooth function. Then (1) f∗dyk =
n
X
i=1
∂fk
∂xidxi = dfk.
(2) f∗(ω1+ ω2) = f∗ω1+ f∗ω2 for any r forms ω1 and ω2 on Rm.
(3) f∗(gω) = f∗(g)f∗ω for any smooth function g : Rm → R and for any smooth r-form ω on Rm.
(4) f∗(ψ1∧ · · · ∧ ψr) = f∗ψ1∧ · · · ∧ f∗ψr for any smooth one form ψ1, · · · , ψr on Rm. (5) f∗(ω ∧ η) = f∗ω ∧ f∗η for any r form ω any s form η on Rm.
(6) f∗(drω) = dr(f∗ω) for any r-form ω.
Proof. We have proved the first identity.
(2) Let p ∈ Rn. For any vectors v1, · · · , vr vectors in Tp(Rn), f∗(ω1+ ω2)(p)(v1, · · · , vr) = (ω1+ ω2)(f (p))(dfp(v1), · · · , dfp(vr))
= ω1(f (p))(dfp(v1), · · · , dfp(vr)) + ω2(f (p))(dfp(v1), · · · , dfp(vr))
= (f∗ω1)(p)(v1, · · · , vr) + (f∗ω2)(p)(v1, · · · , fr)
= (f∗ω1+ f∗ω2) (p)(v1, · · · , vr).
This implies that f∗(ω1+ ω2)(p) = (f∗ω1+ f∗ω2) (p) for any p ∈ Rn which implies the desired identity.
The proof of (3) is left to the reader.
Let p ∈ Rn and (v1)p, · · · , (vr)p be vectors in Tp(Rn). Then
f∗(ψ1∧ · · · ∧ ψr)(p)((v1)p, · · · , (vr)p) = (ψ1∧ · · · ∧ ψr)(f (p))(dfp(v1)p, · · · , dfp(vr)p)
= det(ψi(f (p))(dfp(vj)p))
= det((f∗ψi)(p)(vj)p)
= (f∗ψ1∧ · · · ∧ f∗ψr)(p)((v1)p, · · · , (vr)p).
We find that f∗(ψ1 ∧ · · · ∧ ψr)(p) = (f∗ψ1 ∧ · · · ∧ f∗ψr)(p) for any p ∈ Rn which gives f∗(ψ1∧ · · · ∧ ψr) = f∗ψ1∧ · · · ∧ f∗ψr.
(5) follows from (4). Let ω =P
IωIdxI and η =P
JηJdxJ. Then f∗(ω ∧ η) =X
I,J
f∗(ωIηJ)f∗(dxI∧ dxJ)
=X
I,J
f∗ωIf∗ηJf∗dxI∧ f∗dxJ
= f∗ω ∧ f∗η.
(6) For simplicity, we denote dr by d. Let ω = P
IωI(x)dxI. Hence dω =P
IdωI∧ dxI. By (1), (2), (3), (5),
f∗(dω) =X
I
f∗dωI∧ f∗dxI
=X
I
f∗dωI∧ dfI.
3
If we can show that the statement is true for smooth functions, then f∗dωI = d(f∗ωI). By d2= 0, we have d(f∗ωI∧ dfI) = (df∗ωI) ∧ dfI and hence by (3), we have
f∗(dω) =X
I
d(f∗ωI) ∧ dfI= d X
I
f∗ωI· f∗dxI
!
= d f∗ X
I
ωIdxI
!!
= d(f∗ω).
Let g be a smooth function on Rm. Then dg =
m
X
j=1
∂g
∂yj
dyj. By chain rule,
f∗(dg) =
m
X
j=1
∂g
∂yj
f∗dyj =
m
X
j=1 n
X
k=1
∂g
∂yj
∂fj
∂xk
dxk= d(g ◦ f ) = d(f∗g).
Hence the equation f∗(dg) = d(f∗g) is equivalent to the chain rules of derivatives of func- tions.
Theorem 1.2. Let f : Rn→ Rn be a smooth function. Then
f∗(dx1∧ · · · ∧ dxn) = Jf(x)dx1∧ · · · ∧ dxn where Jf(x) = det(Df (x)) is the Jacobian of f at x.
Proof. This is left to the reader as an exercise.
Department of Mathematics, National Cheng Kung University, Taiwan, fjmliou@mail.ncku.edu.tw NCTS, Mathematics
