An introduction to de Rham cohomology for
multivariable calculus students
K. Joanidis
April 21, 2014
All functions are assumed to be C ∞ .
1
Derivatives
Let D ⊆ Rn be a region. We use WD = {f : D → R} to denote the set of
functions on D, and VD = {F : D → Rn } the set of vector fields on D.
1 Dimension
∇
/ WD
/ VD
curl
WD
2 Dimensions
WD
∇
/ WD
The curl of a 2D vector field is a scalar function given by:
curl F =
∂Q ∂P
−
∂x
∂y
We saw that curl (F) = 0 when F = ∇f , by Clairaut’s theorem. The
converse, that if curl (F) = 0 then there exists f with F = ∇f is sometimes
true, depending on global properties of D.
3 Dimensions
WD
∇
/ VD
curl
/ VD
div
/ WD
The curl of a 3D vector field is another vector field, defined similarly to the
2D analogue.
∂R ∂Q
∂P
∂R
∂Q ∂P
curl F =
−
i+
−
j+
−
k
∂y
∂z
∂z
∂x
∂x
∂y
1
Again, this version satisfies curl (∇f ) = 0 by the symmetry of partial derivatives.
The divergence of a vector field is a scalar function:
div F =
∂P
∂Q ∂R
+
+
∂x
∂y
∂z
We have that div (curl F) = 0.
2
de Rham numbers
If F = ∇f = ∇g then it isn’t necessarily the case that f = g. In fact we can
take g = f + a constant. This means that if we can find a single f such that
∇f = F we can find infinitely many. Because of this we cannot hope to count
the number of solutions, so we will instead keep track of the degrees of freedom.
By linearity ∇f = ∇g if and only if ∇(f − g) = 0. If we have all the solutions
to ∇h = 0 and a single solution to ∇f = F then we can get all the others in
the form of f + h.
Thus it will suffice to study the latter problem, of counting the degrees of
freedom in the solutions of ∇h = 0.
Starting with the one-dimensional case, consider D = R. ∇h = 0 means
that dh
dx = 0, so h must be constant. Therefore, solutions to ∇h = 0 correspond
precisely to constants (scalars), so we have 1 degree of freedom. A more interesting case is if we set D = (0, 1) ∪ (2, 3), the union of open intervals. In this
case ∇h = 0 doesn’t imply that h is constant everywhere, only that it’s constant
on each interval:
(
c1 if 0 < x < 1
h(x) =
c2 if 2 < x < 3
with possibly distinct constants c1 , c2 , giving two degrees of freedom. Thus
the number of degrees of freedom depend on global properties of the region D.
Locally...
Definition 2.1. The zeroth de Rham number is the dimension of the solutions
to ∇h = 0. It is denoted H 0 (D).
A similar question to consider is whether all vector fields on D arise as ∇f .
For D =
R xR the answer is always yes: for a “vector” field k : R → R define
f (x) = 0 k(t)dt. A similar construction shows that in general if D ⊆ R then
there are no vector fields that don’t come from gradients, that is, we have zero
degrees of freedom.
The circle
Derivatives of periodic functions are periodic, but integrals aren’t
necessarily.
2
Functions and vector fields on the circle are given by periodic functions. Let
k : R → R be a vector field on the circle. When D was the real line, we saw
that all vector fields arose as ∇f . On the circle this is no longer true
Z 2π
∇f = f (2π) − f (0) = 0
0
R 2π
This condition is necessary and sufficient: k = ∇f if and only if 0 k(x)dx = 0.
AnyRvector field k is only off by a constant from one that satisfies this condition:
2π
k − 0 k(x)dx. Thus we get one degree of freedom.
Definition 2.2. The first de Rham number, H 1 (D) is the dimension of the
solutions to curl F = 0 up to ∇f . (If curl isn’t defined, we treat it was the zero
function)
Path connectedness and the zeroth de Rham number
We say D is path connected if for any v, w ∈ D we can find a curve r : [0, 1] → D
such that r(0) = v and r(1) = w.
Theorem 2.3. If D is path connected then H 0 (D) = 1.
Proof. suppose that ∇f = 0, and let v be a fixed point of D and w a generic
point, r : [0, 1] → D a path between them.
Z
Z
0=
0 dr =
∇f dr = f (w) − f (v)
C
C
Hence f (w) = f (v) and f is constant on all of D.
More generally, H 0 (D) is the number of path components of D.
Star-shaped region
If D is star shaped (wlog) around the origin, we can show that its first de Rham
number is 0, (that is all vector fields come with potentials) as follows. Let’s
assume that curl F = 0 and try to cook up a potential function. As D is star
shaped, for all
R v ∈ D the curve tv for 0 ≤ t ≤ 1 lies in D. A potential f for F,
must satisfy C F · dr = f (b) − f (a) where a, b are the endpoints of C. With
this is mind, we define f as:
Z
f (x) =
F · dr
C
where C is the straight line from 0 to v. Writing F(x, y) = P (x, y)i + Q(x, y)j
1
we get
Z 1
f (x, y) =
P (tx, ty)x + Q(tx, ty)ydt.
0
1 this
works just as well in higher dimensions
3
We show that f is potential for F by calculating its partial derivatives:
Z 1
∂f
∂
xP (tx, ty) + yQ(tx, ty)dt
(x, y) =
∂x
∂x 0
Z 1
∂
=
(xP (tx, ty) + yQ(tx, ty))dt
∂x
0
Z 1
∂P
∂Q
tx
=
(tx, ty) + P (tx, ty) + ty
(tx, ty)dt
∂x
∂x
0
∂P
Using that curl F = 0, we can replace ∂Q
∂x by ∂y
Z 1
∂P
∂P
=
tx
(tx, ty) + ty
(tx, ty) + P (tx, ty)dt
∂x
∂y
0
and by a nice application of the chain rule and the product rule, this simplifies
to
Z 1
∂
=
t P (tx, ty) + P (tx, ty)dt
∂t
0
Z 1
∂
(tP (tx, ty))dt
=
∂t
0
= P (x, y)
A parallel argument shows that
∂f
∂y (x, y)
= Q(x, y).
Corollary 2.4. For any vector field F on any region D, we can find a local
potential function.
Punctured plane
x
Consider the vector field x2−y
+y 2 i + x2 +y 2 j. We have seen that its integral over
the unit circle is 2π and hence cannot have a potential function. On the other
hand if we restrict its domain D from R2 \ {0} to {(x, y) : x > 0} we see that
F = ∇(tan−1 (y/x)). (Alternatively, the right half plane is star-shaped). This
again shows us that the de Rham number depends on the global properties of
D.
Going back to D = R2 \ {0}, we see that H 1 (D) cannot be zero. A proof
similar to the case of the circle shows that it is in fact 1. The reason for this is
that D can be “stretched” into a cylinder...
Further reading
• Jänich, Klaus, and Leslie D. Kay. “Vector Analysis, Undergraduate Texts
in Mathematics.”(2001).
• Spivak, Michael. Calculus on Manifolds: A Modern Approach to Classical
Theorems of Advanced Calculus. WA Benjamin, 1965.
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