GAUSS’S LAW AND RESIDUE CALCULUS

IN THE FRAMEWORK OF COHOMOLOGY THEORY

P. MERTIKOPOULOS

A. In this paper, our aim is to incorporate Gauss’s famous law in electrody-
namics and some of Cauchy’s results in residue calculus within the general setting
of deRham cohomology, that will be shown to provide a most natural environment
for their development. Namely, it will be demonstrated that forms defined and
closed on punctured spaces can be assigned a residue similar to the one defined
in complex analysis. This can be achieved in any number of dimensions and is, in
essence, an extension of Gauss’s three dimensional law: monopole fields generate
the deRham cohomology groups of those spaces and with their help we will outline
a method for evaluating certain kinds of multiple improper integrals.

1. I
Gauss’s Law in classical electrodynamics is an extremely elegant and deep
result relating the flow of an electric field out of a given closed surface to the
total electric charge within the surface. In a similar fashion, Cauchy’s residue
theorems connect the integral of a meromorphic function over a closed circuit with
the function’s residuesat the poles contained within the circuit. In this paper, we
will try to combine these two fundamental results and extend them to spaces of
higher dimension, with the help of cohomology theory which provides beautiful
insights into the matter.
The purpose of this introductory section is two-fold: firstly, in an effort to be
self-contained, we will provide clear definitions for most of the notions - especially
the topological ones - that are to be used throughout this paper. Of course, we
will not be original here, but we do urge the reader to be cautious before skipping
the first part of this section as a number of concepts introduced are new. Finally,
towards the end of the section, we will perform a few calculations, necessary for
the main body of the paper to proceed smoothly.
To begin with, let A be a subset of Rn and consider k ∈ N with k ≤ n. Recall that
a proper k-cube1 of A is a smooth C∞ diffeomorphism γ : Ik 7→ A where Ik is just the
k-dimensional unit cube: Ik = [0, 1]k = [0, 1] × · · · × [0, 1]. When we use the term
proper k-cube, we will refer either to the diffeomorphism γ itself or its image in
A, depending on the context; however, when a distinction between the two must
be made, we will denote the image of γ in A by hγi. Now, it is clear that a proper
k-cube can be oriented in precisely two different ways 2, unless k = 0 - in which

Date: November 21, 2001.

Key words and phrases. singular terms, form residues.
1As opposed to a singular k-cube - see [1] - that may possess self-intersections.
2Indeed, any proper k-cube is a compact topological space and, hence, admits a triangulation.
Therefore, since a proper k-cube retains the homotopy type of a k-dimensional cube - see [2] - it is clear
1
2 P. MERTIKOPOULOS

case it admits but a single orientation. A proper k-cube, along with a choice of
orientation will be simply called an oriented k-cell. Finally, if σ and τ are oriented
k-cells that refer to the same proper k-cube taken with opposite orientations, we
will - formally - write: σ = −τ.
Now, if SA is the collection of all oriented k-cells of A and G is a group, we may
proceed to define a k-chain of A with coefficients in G as follows:
Definition 1.1. A k-chain of A with coefficients in G is a mapping f : SA 7→ G
which vanishes almost everywhere3 on SA and which satisfies f (−γ) = − f (γ) for
any oriented k-cell γ.
In the rest of this paper we will be concerned - almost exclusively - with k-chains
whose coefficients are integers; if this is not the case, it will be stated explicitly.
Also, in order to avoid cumbersome notation, if γ is an oriented k-cell, we will
denote by γ the k-chain φ for which φ(±γ) = ±1 and φ(ξ) = 0 for any oriented k-
cell ξ other than ±γ. One can simply define addition of k-chains through addition
of the respective mappings and it can easily be seen that - under addition - the set
of k-chains of A forms a group, denoted by Ck (A) and referred to as the kth chain
group of A. The kth chain group of A can be thought of as a free group generated by
the oriented k-cells of A and subject to the relations: γ + (−γ) = 0 for all γ ∈ SA .
A very important homomorphism that maps Ck (A) 7→ Ck−1 (A) is the boundary
homomorphism ∂ : Ck (A) 7→ Ck−1 (A) that assigns to an arbitrary k-chain C a (k-
1)-chain ∂C that is made up of the “faces” of C 4. We do not wish to provide a
formal definition of ∂, as such a discussion would take us too far afield; as far
as intuition goes, we will say that ∂ maps an oriented k-cell to its faces taken
with the induced orientation and to see how ∂ acts on an arbitrary k-chain, one
simply has to extend linearly - recall that the oriented k-cells are the generators
of Ck (A). Now, as can be seen in [1] and [2] the kernel of ∂ is a subgroup of
Ck (A) denoted by Zk (A) = ker ∂ = {C ∈ Ck (A) : ∂C = 0} and referred to as the
kth cycle group of A; if Z is an element of Zk (A) then Z will be called a k-cycle
of A. In a similar fashion, the image of ∂ is a subgroup of Ck−1 (A) denoted by
Bk−1 (A) = Im∂ = {C ∈ Ck−1 (A) : ∃F ∈ Ck (A) with ∂F = C} and referred to as the
(k − 1)th group of boundaries; if B ∈ Bk−1 (A), then B will be called a (k-1)-boundary
or a bounding (k-1)-cycle. Finally, we define the kth homology group of A to be the
quotient group Hk (A) = Zk (A)/Bk (A) whose elements are the equivalence classes of
homologous k-cycles 5.
Let us consider a k-cycle Z of Rk+1 . Since Hk (Rk+1 ) is trivial 6 we can see that
Z must also be a k-boundary. In other words, there exists a (k+1)-chain F of Rk+1
such that: Z = ∂F. In addition to that, F is unique; for, if G is another (k+1)-chain
of Rk+1 with ∂G = Z, it follows that ∂G = ∂F or, equivalently: ∂(G − F) = 0 which
shows G − F to be a (k+1)-cycle. However, Zk+1 (Rk+1 ) is also trivial, implying that

that we can choose a triangulation that maps any proper k-cube to a k-simplex, and it is a well known
fact that a k-simplex can be oriented in precisely two different ways - except, of course, when k = 0.
3That is, everywhere on S with the possible exception of a finite number of proper k-cubes.
A
4A formal definition of the boundary operator and of the faces of a k-chain can again be found in
[1]. Also, [2] provides great intuition into the the matter.
5That is, k-cycles whose difference is a k-boundary.
6Again, see [2] or practically any other book on algebraic topology.
 GAUSS’S LAW IN COHOMOLOGY THEORY 3

there are no (k+1)-cycles other than 0, the trivial one; hence, G − F = 0 or G = F and
we have shown that F is unique. Thus, we have arrived at the following definition:
Definition 1.2. Let C be a k-cycle of Rk+1 . The unique (k+1)-chain of Rk+1 whose
boundary is C will be called the interior of C and denoted by Int(C). More succinctly:
F = Int(C) ⇔ ∂F = C.
Now, let us consider the punctured space: T = Rk+1 \ {P} where P is an arbitrary
point of Rk+1 . Clearly, this space is homotopy equivalent to the k-sphere: T ' Sk .
Hence, the kth homology group of T will be: Hk (T) ' Hk (Sk ) ' Z - see [2] for
a complete discussion on the matter. Moreover, we can give the unit k-sphere
centered at P the outward orientation 7 to obtain a k-cycle S. Since the homology
class of S: Σ = [S] is non-trivial, we can choose Σ to be the generator for the infinite
cyclic group: Hk (T) = {Σ|∅}. Therefore, if Z ∈ Zk (T) is a k-cycle of T, it follows that
the homology class [Z] 8 can be expressed in terms of the generator Σ as: [Z] = nΣ
where n ∈ Z - recall that Hk (T) ' Z. This expression is obviously unique and leads
us to:
Definition 1.3. Let Z be a k-cycle of Rk+1 \ {P}. The unique integer n ∈ Z for which:
[Z] = nΣ with Σ defined as above, will be referred to as the index or winding number
of Z with respect to the point P and we will write:
n = Ind(Z : O) (1.1)
The notions of a k-cycle, its interior and its index with respect to a given point
P ∈ Rk+1 will play a principal part in this paper, since integration over k-cycles is
what the form residues help us with. However, for the time being, we will leave
topology aside and turn to the study of arrays which are also of great importance.
So, let A be a set and consider k1 , . . . km ∈ N:
Definition 1.4. A collection T of k1 × · · · × km elements of A will be called an array
of type (k1 , . . . km ) with elements in A. The order of T will be the number m while
k j will be referred to as the jth dimension of T. The notation used is T j1 ... jn for the
element hof A corresponding
i to (j1 , . . . jm ) while the array T itself will be denoted
by: T = T j1 ... jn .
A point of the above definition that deserves special mention, is the case of order
zero: m = 0. In that case, the array T will consist of a single element q ∈ A and we
will write: T = [q]. Furthermore, in the rest of this paper, we will be using arrays
of a more special form, namely those whose elements are real numbers and for
which: k1 = . . . = km = k. Thus, henceforward and unless otherwise specified, an
array T of type (k, . . . , k) with elements in R will be simply called a - real - array of
| {z }
m times
order m and dimension k.
As we shall see, one property of square matrices - that is, arrays of order 2 - that
needs to be properly generalized in the case of a real array of order m, is the trace
of the matrix. The exact motivation behind our definition cannot be presented
here but it should become clear later in the paper when we examine whether the
7Actually, the term “outward” is well-defined only for k ≥ 2. When k = 1, we will be giving S1 the
standard - anticlockwise - orientation.
8That is, the equivalence class of k-cycles of T that are homologous to Z
4 P. MERTIKOPOULOS

fundamental forms ΩT are harmonic functions or not. Wihtout further ado, we

have:
Definition 1.5. Let T be a real array of order m ≥ 2 and dimension k. The essence
of T is defined to be the array of order m − 2 and dimension k:
"X k X m Xm #
α β
ess(T) = Ti1 ...i`−1 α...in−1 β...im−2 δi δi . (1.2)
i=1 `=1 n=1
If T is of order m = 0, 1 or if ess(T) is a zero array, T will be called neutral.
Note : in the above definition, we are making use of the summation conven-
tion. To be more precise, a quantity such as: T... j... R... j... is actually shorthand for
...j...
Pk
j=1 T... j... R . Therefore, in relation (1.2), the indices α and β are summed over.
Also, if the above definition is worked out in the case m = 2, it is easy to see that
we will end up with two times the regular trace of the matrix, thus showing that
up to an unimportant multiplicative constant, our definition is a valid extension of
the trace of a matrix to higher order arrays.
The neutral arrays are exactly those arrays which lead to the “useful” singular
terms, that are to be identified with the ordinary multipole fields encountered in
physics. Indeed, we will prove that a singular term is a closed form if and only if
the corresponding array is neutral, and this is one of the main points in this paper.
However, in order to arrive to the definition of fundamental forms and singular
terms, we will make heavy use of spherical symmetry in k-dimensional spaces;
therefore, it is necessary to put aside arrays and introduce a system of “polar”
coordinates in Rk . So, let us consider the space of real numbers Rk and a system of
Euclidean coordinates (x1 , . . . xk ) based at the point O ∈ Rk 9.
q
Definition 1.6. If P(x1 , . . . xk ) is a point of Rk \ {O} and ρ j = x21 + · · · + x2j , the
system of polar coordinates on Rk based at O is defined by the following relations:
q
r = ρk = x21 + · · · + x2k
 xk   xk 
ωk−1 = cos−1 = cos−1
r ρk
..
.
! !
x j+1 x j+1
ωj = cos−1
= cos−1
(1.3)
ρ j+1
q
r2 − x2k − · · · − x2j+2
..
.
 x2   x2 
ω1 = cos−1 q = cos−1
ρ2
x21 + x22

In the above expressions, ω j ∈ [0, π] for 1 < j ≤ k − 1. However, when j = 1,

we let ω j ∈ [0, 2π] and ask in addition that sin ω1 = √ x21 2 ; in this way, all the
x1 +x2

9From now on, and unless otherwise specified, we will assume that k ≥ 3. Things are not funda-
mentally different for the case k = 2 and it will be addressed at the end of the following chapter.
 GAUSS’S LAW IN COHOMOLOGY THEORY 5

ω j ’s are well defined and we have a smooth bijection of Euclidean coordinates

(x1 , . . . xk ) to polar coordinates (r, ω1 , . . . ωk−1 ). However, it should be emphasized
that in the case k = 3 the above definition does not reduce to the usual spherical
coordinates of Rk : the angle ω1 which should correspond to the azimuth φ is not
measured with respect to the x1 axis - we should have used sin in place of cos to
agree with tradition. The definition is presented in this way, so that notation may
be simplified later on when the forms ξi are introduced.
In order to gain intuition, one can express Euclidean coordinates xi in terms of
polar ones simply by inverting the expressions (1.3). By doing just that we obtain:
xk = r cos ωk−1
..
.
xj = r cos ω j−1 sin ω j · · · sin ωk−1 (1.4)
..
.
x1 = r sin ω1 sin ω2 · · · sin ωk−1
Having thus defined the system of polar coordinates on Rk \{O} we may ask what
the Hodge star operator - ? - will yield when applied to the forms dr, dω1 . . . dωk−1 .
To do this, we must supply Rk with a nondegenerate scalar product ( , ) which we
choose to be the standard Euclidean one: (a, b) = kj=1 a j b j . Under this product,
P

one can verify the truth of the following lemma:

Lemma 1.7. Consider the basis ξ of Ω1 (Rk ) defined by:

ξ1 = r sin ω2 · · · sin ωk−1 dω1

..
.
ξj = r sin ω j+1 · · · sin ωk−1 dω j (1.5)
..
.
ξk−1 = r dωk−1
ξk = dr
Then, the basis ξ is orthonormal, that is: (ξi , ξ j ) = δi j and preserves orientation with
respect to the standard basis e = {dx1 , . . . , dxk }.
Proof. To begin with, let us apply the d operator to relations (1.3). For the radial
coordinate r = ρk one obtains the usual expression:
x1 xk
dr = dρk = dx1 + · · · + dxk
r r
with the angular expressions being:
x j+1   ρ2j
sin ω j dω j = x1 dx 1 + · · · + x j dx j − dx j+1 .
ρ3j+1 ρ3j+1
Therefore, if 1 ≤ j ≤ k − 1, we have:
x j+1   ρj
ξj = x1 dx1 + · · · + x j dx j − dx j+1 (1.6)
ρ j ρ j+1 ρ j+1
6 P. MERTIKOPOULOS

ρj
since it can be easily verified that: sin ω j = ρ j+1 and, as a direct consequence:
ρ j+1
sin ω j+1 · · · sin ωk−1 = ρk . Finally, for the case j = k, one has:
x1 xk
ξk = dr = dx1 + · · · + dxk
r r
and we have expressed all elements of ξ in terms of the standard basis elements
k
X
dxi : ξi = Bij dx j where B = [Bi j ] denotes the change of basis matrix.
j=1
Now, proving the orthonormality of the basis ξ is only a straightforward cal-
culation away: simply consider the products (ξi , ξ j ) and use relations (1.6) along
with orthonormality of the basis e to get the desired result: (ξi , ξ j ) = δi j .
In order to show that ξ preserves orientation with respect to e one need only
show that det(B) = 1 > 0 and this can be achieved by performing a simple induction
trick. Indeed, let Dk denote the determinant:
x2 x1 ρ
ρ2 ρ1 − ρ12 0 ... 0
x3 x1 x3 x2 ρ
ρ3 ρ2 ρ3 ρ2 − ρ23 ... 0
Dk = .. .. .. (1.7)
. . .
xk x1 xk x2 xk x3 ρ
ρk−1 ρk ρk−1 ρk ρk−1 ρk ... − ρk−1
k
x1 x2 x3 xk
ρk ρk ρk ... ρk

It should be clear that Dk is just det(B) - recall that locally B maps Λ1 (Rk ) 7→ Λ1 (Rk ).
Moreover, if we expand Dk+1 in minors w.r.t. the (k + 1)th column, we obtain:
ρk  ρk  xk+1 xk+1
Dk+1 = − − Dk + Dk = Dk
ρk+1 ρk+1 ρk+1 ρk+1
since ρ2k+1 is just ρ2k + x2k+1 . Therefore, taking into account that D2 = 1 as can be seen
by an easy direct computation, the induction is complete and det(B) = 1 > 0 for
any number of dimensions. This shows that ξ preserves orientation w.r.t. e and
completes our proof. 

Lemma 1.7 can also be rephrased in terms of the change of basis transformation
B by stating that B is an orientation preserving isometry. Therefore, since B leaves
the volume form µ = dx1 ∧ . . . ∧ dxk intact - that is: µ = ξ1 ∧ . . . ∧ ξk as well - we can
easily deduce how the Hodge star operator will transform the forms ξ j . Indeed,
from the definition of ? one has:
?ξ j ∧ ξ j = (ξ j , ξ j )µ = ξ1 ∧ . . . ∧ ξk = (−1)k−j (µ \ ξ j ) ∧ ξ j
where µ \ ξ j is shorthand for the form ξ1 ∧ . . . ∧ ξ j−1 ∧ ξ j+1 ∧ . . . ∧ ξk . Therefore,
since the ? operator acting on 1-forms is an isomorphism - and, hence, an injection
- mapping Λ1 (Rk ) to Λk−1 (Rk ), we have proven the following lemma:

Lemma 1.8. For the members of the basis ξ under the usual Euclidean product ( , ) one
has:
?ξ j = (−1)k−j µ \ ξ j (1.8)
where µ \ ξ j stands for µ with ξ j deleted, that is: µ \ ξ j = ξ1 ∧ . . . ∧ ξ j−1 ∧ ξ j+1 ∧ . . . ∧ ξk .
 GAUSS’S LAW IN COHOMOLOGY THEORY 7

The two results presented above will be of great use later on, when we evaluate
the integral of a singular term over the (k-1)-dimensional sphere.
8 P. MERTIKOPOULOS

2. S T  F R

In the theory of classical electrodynamics, it is a well known fact that any electric
field with point singularities can be decomposed as the sum of an electric field with
no singularities and a series - finite or infinite - of multipole fields - see for example
[5]. In mathematical terminology, this means that a 1-form which is defined and
closed on 3-space except for a subset A of R3 which is not dense in the space
can be decomposed into a 1-form which is defined and closed on the whole of
R3 and a series of 1-forms which correspond to the multipole fields. This result
is exceedingly similar to the Laurent decomposition of meromorphic functions in
complex analysis and is readily adaptable to higher dimensional spaces once the
multipole fields have been worked out. These “fields” will be referred to as singular
terms of the corresponding order and for the purposes of this paper we may begin
as follows:

Definition 2.1. Consider a real array T = [T j1 ... jm ] of order m and dimension k and
let ΩT denote the 0-form of Rk \ {O} defined by:

T j1 ... jm x j1 · · · x jm
ΩT = (2.1)
r2m+k−2
The 0-form ΩT will be referred to as the fundamental form corresponding to the array
T.

Note : in the above definition, we are making use of the summation conven-
tion. To be more precise, a quantity such as: T... j... R... j... is actually shorthand for
... j...
Pk
j=1 T... j... R .
The fundamental forms are the first step towards introducing the singular terms
in which a given closed (k-1)-form will be decomposed. Hopefully, the motivation
behind our definition will be clear to the reader before the end of this section but,
at this point, no justification can be given without a long and thorough discussion
on the matter.

Definition 2.2. Consider a real array T = [T j1 ... jm ] of order m and dimension k. The
(k − 1)-form defined on Rk \ {O} by:

PT = − ? dΩT

will be referred to as the singular term corresponding to the array T. The order of PT is
defined equal to the order of the array T and PT will be frequently referred to as a
2m -pole.

Before proceeding any further, we must deduce whether the fundamental forms
are harmonic functions on Rk \ {O} a question which is clearly equivalent to asking
whether the singular terms are closed on Rk \ {O}. To this end, we will prove the
following proposition which shows that the answer depends solely on the essence
of the term’s corresponding array.
 GAUSS’S LAW IN COHOMOLOGY THEORY 9

Proposition 2.3. Let T = [T j1 ... jm ] be a real array of order m and dimension k. Then, PT
is closed on Rk \ {O} iff T is neutral, that is: dPT = 0 iff ess(T) = 0.
Proof. To begin with, let us set: εi = ∂Ω ∂xi ; then, the closedness condition: dΩT = 0
T

k
X ∂εi
is equivalent to asking that: = 0. Having said that, the proof is but a few
∂xi
i=1
direct calculations away, and we will merely present an outline to them.
Indeed, if we let T = T j1 ... jm x j1 · · · x jm and set: Ri = ∂T∂xi , we can easily see that:
T xi Ri
εi = −(2m + k − 2)
2m+k
+ 2m+k−2
r r
Therefore, a second differentiation w.r.t. xi will yield:
2
∂εi 2m + k − 2  x  1 ∂Ri
= (2m + k)T 2i − 2Ri xi − T + 2m+k−2 . (2.2)
∂xi r2m+k r r ∂xi
However, the very definition of Ri infers:
Xm
Ri = T j1 ... j`−1 α...jm x j1 · · · x j`−1 δαi · · · x jm
`=1
and, if we multiply by xi and sum over i, we will obtain:
k
X m
X
Ri xi = T = mT .
i=1 `=1
k
X
Moreover, x2i = r2 ; thus, relation 2.2 will yield:
i=1
k k
X ∂ε i 1 X ∂R i
= . (2.3)
∂xi r2m+k−2 ∂xi
i=1 i=1
Now, a direct calculation will reveal that:
k k X
m X
m
X ∂R X
β
T j1 ... j`−1 α...jn−1 β...jm x j1 · · · x j`−1 δαi · · · x jn−1 δi · · · x jm
i
=
∂xi
i=1 i=1 `=1 n=1

and, for PT to be closed, this expression must vanish for all the xi ’s. Therefore, we
must ask that the array
k X
m X
m
β
X
Ti1 ...i`−1 α...in−1 β...im−2 δαi δi
i=1 `=1 n=1
be equal to zero. But, since this array is just ess(T) - compare with definition 1.5 -
and vanishes if and only if T is neutral, we can see that PT is closed if and only if T is
neutral. 
This result will allow us to show that when a singular term corresponding to a
neutral array is integrated over a (k − 1)-cycle of Rk \ {O}, the result will be zero
unless the array is of order 0. To be more precise, we will prove the following
proposition which is the proper extension of Gauss’s law to higher dimensional
spaces:
10 P. MERTIKOPOULOS

Proposition 2.4 (Gauss’s Law). Let Z be a (k − 1)-cycle of Rk \ {O} and T a neutral

array of dimension k and order m. Then:

q(k − 2)Ak−1 Ind(Z : O), if m = 0 and T = [q]
Z 
PT = 

(2.4)
Z 0,
 if m ≥ 1
√
π k
where Ak−1 = 2   is the area spanned by the unit (k-1)-sphere.
Γ k
2

Proof. Let [Z] be the homology class of Z and let Σ denote the homology class of
the unit (k-1)-sphere Sk−1 , taken with the outward orientation and centered at O.
From the definition of Ind(Z : O) we have:
[Z] = Ind(Z : O)Σ ⇒ [Z − Ind(Z : O)Sk−1 ] = 0
Obviously, this implies the existence of a k-chain C of Rk \ {O} with the property:
∂C = Z − Ind(Z : O)Sk−1 10. Therefore, an application of Stokes’ theorem - see for
example [4], [1] or [6] - easily yields:
Z Z Z Z
PT − Ind(Z : O) PT = PT = dPT = 0
Z Sk−1 ∂C C
k
(recall that PT is closed on R \ {O} since the array T is neutral). Hence, we obtain:
Z Z
PT = Ind(Z : O) PT (2.5)
Z Sk−1

and it suffices to evaluate the integral of PT over the (k-1) unit sphere Sk−1 , centered
at O and taken with the outward orientation.
Bearing this in mind, consider the sphere Sε - which is the same as Sk−1 except
that it is of radius ε - and the set:
Jε = {ε} × [0, 2π] × [0, π] × · · · × [0, π]
| {z }
k−2 times
One can readily see that Jε is exactly where Sε is pulled back to under φ - the
transformation
R from polar to euclidean coordinates of Rk - and we are left to
evaluate φ(J ) PT . This task can be accomplished by switching to polar coordinates,
ε
where we get: Z Z Z
PT = φ∗ PT = − φ∗ (?dΩT )
φ(Jε ) Jε Jε
In order to proceed, we will have to express ΩT in terms of the forms ξ j intro-
duced in section 1. Now, to simplify notation, consider the functions: αi = xri where
xi is expressed in polar coordinates via relations (1.4); as a direct consequence, one
T j1 ... jm α j1 ···α jm
has: ΩT = rm+k−2
. Therefore, taking d of ΩT , will yield:
T j1 ... jm α j1 · · · α jm
dΩT = −(m + k − 2) dr + terms containing dωi .
rm+k−1
The differential above can be expressed in terms of the basis ξ simply by considering
that ξ j is a multiple of dω j . Thus, by applying the star operator to dΩT , the terms
10It is not hard to show that this k-chain is, in fact, unique and equal to Int(Z) − Ind(Z : O)B where
k
Bk is the ball bounded by Sk−1 .
 GAUSS’S LAW IN COHOMOLOGY THEORY 11

containing dωi will become wedge products containing the form ξk = dr and these
terms will clearly vanish when pulled back by φ and integrated over Jε - where r
is kept fixed to ε. As a consequence of the above, we may write:

∂ΩT
Z Z Z
PT = − φ (?dΩT ) = −
∗
ξ1 ∧ . . . ∧ ξk−1
Sε Jε Jε ∂r

T j1 ... jm α j1 · · · α jm k−1
Z
= (m + k − 2) r τ
Jε rm+k−1
where τ is defined by:
1
τ= ξ1 ∧ · · · ∧ ξk−1 = sin ω2 . . . sink−2 ωk−1 dω1 ∧ . . . ∧ dωk−1
rk−1
and corresponds to the “solid angle” form of Rk . So, if we let
J = [0, 2π] × [0, π] × · · · × [0, π]
| {z }
k−2 times

and set r = ε in the above integrals, we will obtain the crucial result:

m+k−2
Z Z
PT = T j1 ... jm α j1 · · · α jm τ (2.6)
Sε εm J
R R
For m = 0 and T = [q], the above result reduces to: Sε
PT = q(k−2) J
τ. However,
R
τ is just the solid angle bounded by the sphere S with respect to O; in other
J
k−1
√
( π)k
R
words J τ = Ak−1 where Ak−1 = 2   is the area spanned by the unit sphere Sk−1 .
Γ k
2

Therefore, in conjunction with relation (2.5), we finally get11:

Z
PT = q(k − 2)Ak−1Ind(Z : O)
Z

On the other hand, if m ≥ 1, consider δ , ε; then, by applying (2.6) we get

m+k−2
Z
PT = IT
Sδ δm
R
where IT = J T j1 ...jm α j1 · · · α jm τ. But, since PT is closed on Rk \ {O}, we can use
R R
Stokes’ theorem to show that: S PT = S PT which, combined with our previous
ε δ
results, leads to:
m+k−2 m+k−2
IT = IT
δm εm
Z
Of course, this cannot hold for m ≥ 1, unless IT = 0. This shows that PT must
Z
vanish for m ≥ 1 and completes our proof. 

11In order to obtain the integral over the unit sphere Sk−1 , it suffices to set ε = 1 in (2.6).
12 P. MERTIKOPOULOS

Along with definitions 2.1 and 2.2, the two results presented above clearly
demonstrate how fundamental forms correspond to the potential of an ideal mul-
tipole in k-dimensional spaces, with singular terms corresponding to the fields
themselves. Indeed, if we ignore the presence of the Hodge star operator which
has been included for purposes of integration, we can see that the vector fields
on Rk \ O associated to the singular terms have a single point singularity, both
zero curl and zero divergence and, most important, they also respect Gauss’s Law.
Hence, they are proper extensions of multipoles to Rk with k ≥ 3 and the reader
is actively encouraged to treat them as such; henceforward, intuitive discussion
based on the matter will be based freely upon this fact.
Now, before proceeding any further, we will have to present a few elements
of deRham cohomology theory which provides a natural setting for the study of
closed forms. Of course, we will not venture into detailed discussions as this
would take us beyond the scope of this paper; we will only go as far as to define
the kth - deRham - cohomology space of a space A.
To be more precise, let us consider a submanifold A of Rn and a differential
k-form ω ∈ Ωk (A). Under addition of k-forms and multiplication by a scalar λ ∈ R,
the set of closed k-forms: Zk (A) = {ω ∈ Ωk (A) : dω = 0} attains the structure of
a vector space over the real field R and will be referred to as the kth cocycle space
of A. Moreover, if there exists a (k+1)-form σ ∈ Ωk+1 (A) such that: ω = dσ, we
will say that ω is exact on A. The set of exact forms: Bk (A) = {ω ∈ Ωk (A) :
∃σ ∈ Ωk+1 (A)with dσ = ω} also has the structure of a real vector space and will
be referred to as the kth coboundary space of A. Finally, if ω ∈ Ωk (A), we define the
- deRham - cohomology class of ω to be the set: [ω] = {ψ ∈ Zk (A) : ω − ψ ∈ Bk (A)}.
Bearing this in mind, the kth - deRham - cohomology space of A is defined to be the
ker d
quotient space: Hk (A) = Zk (A)/Bk (A) = = {[ω] : ω ∈ Ωk (A)} - see [6] and [3].
Imd
A very deep and greatly celebrated theorem by deRham - see [6] - states that
for a given smooth manifold M, the spaces Hk (M) and Hk (M) are isomorphic12,
the isomorphism being given by integration. Therefore, if A is a punctured open
ball of of Rk+1 with the interior point Q removed, we will have: Hk (A) ' Hk (A) '
R. Furthermore, from proposition 2.4 we may deduce that the singular term P
corresponding to the array of order 0: T = [1] and based at Q is a closed k-form
which is not exact13. Therefore, since P is not exact, the cohomology class π = [P]
of P will be non-trivial: π , 0 and, consequently, we may choose π as a basis for
Hk (A). So, if ω ∈ Zk (A) is a closed k-form of A, we may write: [ω] = q · π where q is
a unique real number, and we have arrived at the following definition:
Definition 2.5. Let ω be a closed k-form on a punctured open (k+1)-ball A with
an interior point Q removed. The - unique - q ∈ R for which: [ω] = q · π with π
defined as above, will be called the residue of ω at Q and we will write:
Res(ω : Q) = q (2.7)

12In this paper, we have not given the structure of a vector space to the homology groups H (M).
k
However, the homology groups can easily attain this structure if we agree to let k-chains take on real
coefficients - see [2] and [6] for a first discussion on the matter.
13Indeed, if it were exact, its integral over any k-cycle Z would have to be zero, which clearly does
not hold if Ind(Z : Q) , 0.
 GAUSS’S LAW IN COHOMOLOGY THEORY 13

Note: There is a striking resemblance between the residue of a given closed form
at a point Q and the index of a k-cycle with respect to the same point. In a sense,
these notions are dual to one another, much as the notion of a k-form is dual to
that of a k-chain. With this in mind, as the k-spheres centered at the “holes” of a
manifold generate the kth homology group of the manifold, the monopoles based
at the same “holes” provide a basis for the kth - deRham - cohomology space of the
manifold. We will have more to say on this subject later on in this chapter.
One final point that we should stress, is that definition 2.5 allows us to define
the residue of a closed k-form ω on A, even when A is an open (k+1)-ball with a
- finite - collection of points Q = {Q j , j = 1 . . . n} removed. To obtain the residue
of ω at a given point Q j one simply has to consider a neighbourhood of Q j which
contains no other “singular” points14 and apply definition 2.515.

We will see that definition 2.5 is indeed a proper generalization of the traditional
Cauchy residue found in Complex Analysis. To be more precise, we will be able
to derive the following proposition from our previous discussion:
Proposition 2.6. Let A be an open ball of Rk with a point Q removed and consider a
closed (k-1)-form ω ∈ Ωk−1 (A). Then, if Z is a (k-1)-cycle of A, the integral of ω over Z is:
Z
ω = (k − 2)Ak−1 · Ind(Z : Q) · Res(ω : Q) (2.8)
Z

Proof. Let T denote the array of order 0: T = [1]. Then, from the definition of the
residue of ω at Q, we have: [ω] = Res(ω : Q) · π where π = [PT ] and PT is the -
(k-1)-dimensional - singular term associated to T and based at Q. Therefore, there
exists an exact (k-1)-form σ such that: ω = σ + Res(ω : Q)PT and, by integrating
over Z, we obtain: Z Z Z
ω= σ + Res(ω : Q) PT .
Z Z Z
R
But, σ clearly vanishes since σ is exact, while an application of proposition 2.4
ZZ
will yield: PT = (k − 2)Ak−1 Ind(Z : Q). Thus, by combining these results, we
Z
finally get: Z
ω = (k − 2)Ak−1 · Ind(Z : Q) · Res(ω : Q),
Z
as was to be shown. 
As a direct consequence of the above, we may also obtain the following corollary
- which is, in essence, a more intuitive way of rephrasing proposition 2.6:
Corollary 2.7. Let A be a punctured open ball of Rk with an interior point Q removed,
and consider a closed (k-1)-form ω of A. If S is a (k-1)-sphere contained in A, centered at
Q and given the standard - outward - orientation, then we have:
Z
1
Res(ω : Q) = ω (2.9)
(k − 2)Ak−1 S

14Recall that this is possible because the subset of singularities of ω is not dense in A.
15
Alternatively, one could argue that Hk (A) is an n-dimensional vector space and choose the set of
monopoles based at the points Q j as basis vectors for Hk (A).
14 P. MERTIKOPOULOS

Now, we are ready to state and prove the central result of this paper which can
be thought of as a direct generalization to higher dimensional spaces of Cauchy’s
main residue theorem. Without further ado, we have:

Proposition 2.8 (Cauchy’s Residue Theorem). Consider an open ball A of Rk and a

differential (k-1)-form ω which is defined and closed on A except for a nondense collection
of - interior - points Q = {Q j ∈ A, j = 1 . . . n}, that is: ω ∈ Zk−1 (A \ Q). Then, if Z is a
(k-1)-cycle of A \ Q, the following formula holds:
Z n
X
ω = (k − 2)Ak−1 Ind(Z : Q j )Res(ω : Q j ) (2.10)
Z j=1

Proof. Consider the array of order 0: T = [1] and denote the - (k-1)-dimensional -
singular term associated to T and based at Q j by P j . Obviously, Hk−1 (A \ Q) ' Rn
and this allows us to choose the cohomology classes: π j = [P j ] as basis vectors for
Hk−1 (A \ Q). Therefore, since ω is closed on A \ Q, we will have: [ω] = nj=1 π j or,
P

equivalently:
n
X
ω=σ+ Res(ω : Q j )P j
j=1

where σ is an exact (k-1)-form defined on A \ Q.

P Moreover, it can be readily seen
that the homology class of Z is just: [Z] = ni=1 Ind(Z : Qi )Σi where Σi is the
homology class of a sphere Si centered at Qi , taken with the outward orientation
and contained in A - but containing no other elements of Q. Consequently, there
exists a k-chain C of A \ Q such that:
n
X
Z = ∂C + Ind(Z : Qi )Si .
i=1

- compare with proposition

R 2.4 as well. It should be clear that since ω is closed on
A \ Q, the integral ∂C ω will vanish; in addition to that, when σ is integrated over
a sphere Si , it will also yield zero since σ is exact. Bearing this in mind, we obtain:
Z Z n
X Z
ω= ω + Ind(Z : Qi ) ω
Z ∂C i=1 Si
n
X Z  n
X 
= Ind(Z : Qi ) σ+ Res(ω : Q j )P j
i=1 Si j=1
n X
X n Z
= Ind(Z : Qi )Res(ω : Q j ) Pj
i=1 j=1 Si

However, given
R that Res(P j : Q j ) = 1, if we apply propositions 2.4 and 2.6 in order
to evaluate S P j we will get:
i

Z
P j = (k − 2)Ak−1 Ind(Si : Q j ) = (k − 2)Ak−1 δi j
Si
 GAUSS’S LAW IN COHOMOLOGY THEORY 15

as Si contains no other “singular” points of Q except for Qi . Thus, we obtain:

Z X n X n
ω= Ind(Z : Qi )Res(ω : Q j )(k − 2)Ak−1 δi j ⇒
Z i=1 j=1
Z n
X
ω = (k − 2)Ak−1 Ind(Z : Q j )Res(ω : Q j )
Z j=1

and our proof is complete. 

With the aid of this proposition, we may prove an extension to Poincaré’s

lemma16 which will provide us with a criterion of whether a given closed form is
exact or not. To be more precise, we will prove the following proposition:
Proposition 2.9. Let M be a k-submanifold of Rk which is diffeomorphic to an open
k-ball of Rk , possibly with a - nondense - collection of points removed.
Z Suppose further
that ω ∈ Ωq (M) is a differential q-form of M which satisfies: ω = 0 for all q-cycles
Z
Z ∈ Zq (M). Then, ω is exact on M.
Proof. First, we will attack the case q = k − 1, which is of the most interest to us. So,
let usn assume that M iso diffeomorphic to A = B \ Q where B is an open k-ball and
Q = Q j ∈ B, j = 1 . . . n is a - finite - collection of points of B. Moreover, denote by
φ the diffeomorphism that maps A 7→ M : M = φ(A). Then, if W is a (k-1)-cycle of
A, the image of W under φ will be a (k-1)-cycle Z = φ(W) of M and we may write:
Z Z Z
φω=
∗
ω= ω=0
W φ(W) Z

a result which holds for all (k-1)-cycles of A.

Consequently, if we set ψ = φ∗ ω and let P j denote the singular term of A which
corresponds to the array T = [1] and is based at Q j , we will have:
n
X
ψ=σ+ Res(ψ : Q j )P j
j=1

where σ is exact on A. However, if we integrate this expression over an arbitrary

(k-1)-cycle W, we will obtain:
Z n
X
ψ=0⇒ Res(ψ : Q j )Ind(W : Q j ) = 0
W j=1

Clearly, this can hold for all (k-1)-cycles W of A if and only if Res(ψ : Q j ) = 0 for all
j = 1 . . . n. Thus, ψ = σ and we have shown that ψ is exact. Therefore, if τ ∈ Ωk (A)
is a k-form of A with: dτ = ψ, by pushing forward w.r.t. φ, we obtain:
d(φ∗ τ) = φ∗ dτ = φ∗ ψ = φ∗ (φ∗ ω) = ω
and we have shown that ω is exact on M.

16Poincaré’s lemma states that if ω is a closed form on a contractible k-manifold M, then ω has to be
exact as well - see [6], [4], [1] and [3] for a discussion on the matter.
16 P. MERTIKOPOULOS

On the other hand, let ω ∈ Ωq (M) with q < k − 1 and consider q-chains A1 , A2 ∈
Cq (M) which share the same boundary: ∂A1 = ∂A2 = C ∈ Cq−1 (M). It is not hard to
show that: Z Z
ω= ω.
A1 A2
Indeed, if we set Z = A2 − A1 , it should be clear
R that Z is a q-cycle: ∂Z = ∂A2 −
∂A1 = C − C = 0. However, by assumption, Z ω = 0, and this leads us to our
previous result. With this in mind, we may proceed to Z define a linear function
σ̃ : Bq−1 (M) 7→ R which maps C ∈ Bq−1 (M)17 to σ̃[C] = ω where F is a q-chain
F
satisfying: ∂F = C - note that σ̃ isRwell defined
R since, if we choose another q-chain
G with ∂G = C we would have: G ω = F ω = σ̃[C]. Moreover, since σ̃ is defined
via integration of q-forms, we can see that it is at least C1 and we can use the smooth
version of Tietze’s extension theorem - see [6] and [2] - to extend σ̃ to a C1 linear
function σ̄ on Cq−1 (M) which agrees with σ̃ on Bq−1 (M) - of course, this extension is
not and need not be unique. Thus, if d denotes the coboundary operator, mapping
d : Cq−1 (M) 7→ Cq (M)18 we know for a fact that the following diagram commutes -
see for example [4]:
d
Cq−1 (M) −−−−−→ Cq (M)
x
 x


 

 
d
Ωq−1 (M) −−−−−→ Ωq (M)
Denote by ω̄ the linear function on Cq (M) that corresponds to ω. Then, if F is
an arbitrary q-chain with ∂F = C, we can apply the coboundary operator d to σ̄ in
order to obtain: Z
dσ̄[F] = σ̄[∂F] = ω = ω̄[F].
F
This shows that ω̄ = dσ̄. Therefore, if σ ∈ Ωq−1 (A) denotes the form that
corresponds 19 to σ̄ ∈ Cq−1 (M), the commutativity of the diagram above shows that:
dσ = ω and ω is exact on M as was to be shown. 
Now, from the proof of this exactness criterion, we can readily obtain the fol-
lowing very useful corollary:
Corollary 2.10. Let M be a k-submanifold of Rk which is diffeomorphic to an open k-ball
of Rk , possibly with a - finite - number of points removed. Then, a (k-1)-form ω will be
exact on M iff:
Res(φ∗ ω : Q j ) = 0
for all Q j ∈ Q - with φ and Q defined as above20.

In order to see how Poincaré’s lemma can be derived from proposition 2.9,
consider a contractible manifold M and a closed k-form ω on M: dω = 0. Then, if
17Here, we are assuming that the (q-1)-chains take on real coefficients
18Cq (M) refers to the space of q-cochains of M - see [4] for a full account.
19Again, we refer the reader to [4] for a clear account of how a q-cochain is assigned to a differential
q-form.
20See proof of proposition 2.9
 GAUSS’S LAW IN COHOMOLOGY THEORY 17

Z is an arbitrary k-cycle of M, Z will be a bounding k-cycle as well, i.e. there exists

a (k+1)-chain C of M with: ∂C = Z. So, integrating ω over Z will yield:
Z Z Z
ω= ω= dω = 0
Z ∂C C

as ω is closed on M. Therefore, since this results holds for any k-cycle Z, we can
conclude from proposition 2.9 that ω is exact on M, as was to be shown.
So, let us consider a k-form ω defined and closed on a deleted neighbourhood A
of a point Q ∈ Rk+1 and assume that it decomposes into singular terms as follows21:
m
X
ω = ω0 + PT j
j=0

with the singular terms PT j being based at Q and m - possibly equal to ∞ - the order
of the pole at Q. From definition 2.5 we may deduce the following lemma:
Lemma 2.11. If T0 = [q], then the residue of ω at Q is just:
Res(ω : Q) = q. (2.11)
In other words, the residue of a closed form at a given point, is equal to the “coefficient”
of the singular term of order 0, based at that point.
Proof. By assumption, the singular decomposition of ω at P is:
m
X
ω = ω0 + PT j ,
j=0

where T0 = [q] and we are asked to show that Res(ω : Q) = q. Now, if Z is a k-cycle
of A, its interior Int(Z) will be a k-chain of A ∪ Q and, integrating ω0 over Z infers:
Z Z
ω0 = dω0 = 0
Z Int(Z)

since ω0 is 22
Zclosed on A∪Q . Furthermore, if j , 0, we can deduce from proposition
2.4 that: PT j = 0 as well, and these results hold for any k-cycle of A. Therefore,
Z
proposition 2.9 tells us that both ω0 and all of the PT j singular terms - with j , 0 -
are exact on A, and we may write:
ω = σ + PT0
where σ is an exact k-form of A. Thus, for the - deRham - cohomology class of ω,
we obtain: [ω] = [PT0 ].
On the other hand, from definition 2.5 we know that: [ω] = Res(ω : Q)[PQ ]
where PQ is the singular term of A that corresponds to the array T = [1] and which
is based at Q. But, it can be easily seen that PT0 is just q · PQ and, by comparing the
two expressions for [ω] we finally get: Res(ω : Q) = q. 

21Clearly, any such decomposition is unique.

22Recall that Z = ∂Int(Z).
18 P. MERTIKOPOULOS

One point in the above proof that we should stress is the exactness of singular
terms of higher order. This result is of great intuitive value since it exposes the true
nature of monopole terms, and deserves special mention, eventhough it is easily
implied by previous results:
Corollary 2.12. Let Q ∈ Rk+1 . Then, all singular terms based at Q are exact, except for
the monopoles.
This is the same kind of duality that we encountered earlier when dealing with
the index of a k-cycle and the residue of a k-form. We know that all k-cycles of
Rk+1 \ O are also k-boundaries except for those that are homologous to the k-sphere
centered at the origin. Similarly, all k-forms that are closed on Rk+1 \ O are exact
as well, except for those which are cohomologous to the k-monopole based at the
origin. The k-spheres generate the kth homology group of punctured spaces while
the k-monopoles generate the corresponding cohomology space; with this kind of
duality in mind, one can say that monopoles play the same part in function spaces
as spheres play in ordinary geometrical ones.

Now, we are left to actually evaluate the residue of a k-form at one of its “sin-
gular” points. A priori, this need not be easy, as the form’s expression in an
arbitrary coordinate system might be quite complicated, and a singular decom-
position would not be readily apparent. Therefore, we will present a method for
computing the residue of a closed form that closely mimics the “shortcuts” used
in Complex Analysis to compute the residue of an meromorphic function. The
method is described in the following proposition:
Proposition 2.13. Let A be an open ball of Rk and consider a (k-1)-form ω closed on A,
with a pole of order m at a point P ∈ A. Then, the residue of ω at P is equal to:
1 ∂m 
lim m rm+k−2~Lω · ~r

Res(ω : P) = (2.12)
m!(k − 2) r→0 ∂r
where ~Lω = (?−1 ω)] and r is the usual polar coordinate, based at P23.
Proof. Consider a polar coordinate system {r, ω1 , . . . , ωk−1 } based at P; since ω
presents a polePof order m at P, it will decompose into singular terms as fol-
lows: ω = ω0 + mj=0 PT j where ω0 is a (k-1)-form defined and closed on A. Also, set
~L = (?−1 ω0 )] and ~L j = (?−1 PT )] so that: ~Lω = ~L + Pm ~L j . Then, from the definition
j j=0
of singular terms, it follows that:
] ∂Ω
~ T · ~r = −r T j .
~L j · ~r = − dΩT · ~r = −∇Ω

j j
∂r
h j i
Now, if we denote the array T j by: T j = Ts1 ...s j and, in addition, set R j =
j
Ts1 ...s j αs1 · · · αs j , we obtain:
m m
rm+k−2~L j · ~r =
X X
rm−j ( j + k − 2)R j
j=0 j=0

23For the definition of the index raising operator ] see [6]. Roughly speaking, if ω is a 1-form:

ω = f1 dx1 + · · · + fk dxk then ω] = f1~e1 + · · · + fk~ek with ~e j being the standard vector basis for Rk .
 GAUSS’S LAW IN COHOMOLOGY THEORY 19

since R j depends only on the angular coordinates ωi and is independent of r. Now,

by differentiating our previous result m times w.r.t. r, we get:
m m
∂m X m+k−2~ X
r L j · ~
r = p j r−j ( j + k − 2)R j
∂rm
j=0 j=0

where p j = m−1
`=0 (m − j − `). However, p j vanishes for all j, unless j = 0 in which
Q
Qm−1
case: p0 = `=0 (m − `) = m!. Thus, we get:
m
∂m X m+k−2~
r L j · ~r = m!(k − 2)R0
∂rm
j=0

But, R0 is the sole element of the array T0 , and this is exactly the residue of ω at P.
In other words, we have shown that:
m
1 ∂m X m+k−2~
Res(ω : P) = r L j · ~r.
m!(k − 2) ∂rm
j=0

On the other hand, ~L and any of its derivatives are clearly bounded at P. Then, by
applying Leibniz’s rule in order to differentiate the scalar product: rm+k−2~L·~r m times
w.r.t. r, we will obtain terms where r appears with an exponent of at least k − 1.
These terms will clearly vanish as we let r → 0 and we will have:
∂m 
lim m rm+k−2~L · ~r = 0.

r→0 ∂r
Thus, by combining the two results presented above, we finally obtain:
1 ∂m 
lim m rm+k−2~Lω · ~r

Res(ω : P) =
m!(k − 2) r→0 ∂r
which concludes our proof. 

In closing this section, we will demonstrate how our approach can be adapted
in R2 and we will exhibit the connection between ordinary residue calculus and
our construction of form residues. Beffore proceeding, there is only one important
remark that needs to be made and which concerns the special nature of the plane:
namely, if ω ∈ Ω1 (R2 ) then ?ω is also an element of Ω1 (R2 ). This means that
ordinary path integrals and “surface” integrals are essentially the same thing and,
eventhough this appears to be a rather subtle point, it leads to many important
ramifications that will not be addressed in this paper.
For now, the construction of singular terms in the plane is essentially the same as
before, with only one minor change in the definition of fundamental forms which
must be modified as follows:
Definition 2.14. Consider a real array T = [T j1 ... jm ] of order m and dimension 2.
The fundamental form assigned to T will be the 0-form of R2 \ {O}:

−q ln r
 if m = 0 and T = [q],
ΩT = 

j jm (2.13)
 T j1 ... jm x2m1 ···x

if m ≥ 1.
r
In a similar fashion, the singular term associated to T will be the 1-form defined on
R2 \ {O} by:
PT = − ? dΩT (2.14)
20 P. MERTIKOPOULOS

The presence of the logarithm function in the above definition should be familiar
to physicists and is due to the fact that d ln r = r12 (x1 dx1 + x2 dx2 ), thus leading to
the correct singular term of order 0. the rest of our results in this section hold in
the planar case exactly as they hold in higher dimensional spaces, with the notable
exception of the calculating proposition 2.13 which should read as follows:
Proposition 2.15. Let A be an open ball of R2 and consider a 1-form ω defined and closed
on A except for a point P ∈ A where it presents a pole of order m. Then, the residue of ω at
P is:
1 ∂m 
lim m rm~Lω · ~r

Res(ω : P) = (2.15)
m! r→0 ∂r
where ~Lω = (?−1 ω)] and r is the usual polar coordinate, based on P.
Note: the above proposition differs from the higher dimensional case in the sense
that it cannot be derived by 2.13 by merely setting k = 2 as that is not possible due
to the k − 2 factor appearing in the denominator. However, the rest of our results
may be trivially extended to the plane, simply by setting k = 2.
To see the connection between meromorphic functions and closed forms, let us
consider a complex function f (z) = u(z) + iv(z) which is analytic on an open ball
A of the complex plane C except, possibly, for a finite number of poles. If γ is a
locally smooth curve of A - a 1-chain, to be exact - not passing through any of the
poles of f , we will have:
Z Z Z Z Z
f (z)dz = udx − vdy + i vdx + udy = ξ+i η
γ γ γ γ γ

where z = x + iy, ξ = udx − vdy and η = vdx + udy. However, since f is analytic on
A - except for a finite number of poles - the Cauchy-Riemann equations infer that:
ux = v y and vx = u y . Then, we shall have:
dξ = (u y + vx )dy ∧ dx = 0 and
dη = (v y − ux )dy ∧ dx = 0,
revealing ξ and η to be closed. Then, the integral of f over γ is reduced to the study
of the integrals of ξ and η, which clearly fall within the domain of our previous
results. Indeed, if Q is a pole of f , it is quite easy to show that:
h i
< Res( f : Q) = Res(η : Q) and
h i
= Res( f : Q) = −Res(ξ : Q).
The relations above clearly demonstrate how ordinary complex residues can be
identified with the form residues of 1-forms in the plane, and establish our main
claim that Gauss’s law and Cauchy’s residue theorems can both be unified and
given a new interpretation within the framework of deRham cohomology. Then,
we can see that two distinct notions, such as the residue of a meromorphic function
at one of its poles and the total electric charge accumulated at a singular point of
an electric field are, in essence, the same thing, differing only in the setting where
they are encountered. Ultimately, the multipole decomposition is much the same
1 k−1
as Laurent decomposition with the (z−a) k term being interpreted as a 2 -pole:
monopoles correspond to the 1/(z − a) term and these terms generate the first
deRham cohomology group of C \ {a}.
 GAUSS’S LAW IN COHOMOLOGY THEORY 21

However, we must stress that there is some fine print involved and we have only
xdx+ydy
scratched the surface in the planar case. Indeed, 1-forms such as: ω = r2
-
coming from applying d to fundamental forms and not ?d - are perfectly “singular”
(with an intuitive interpretation of the term) and yet, they are not singular terms,
neither can they be expressed as a series of singular terms. Fortunately, they
have zero form residue and they do not interfere with proposition 2.15 but their
very existence implies some very important ramifications. Their existence and
nature seem to stem from the fact that the plane is unique as 1-forms and their
? counterparts, (2-1)-forms, are the same. Still, we can only hint at their impact
on our approach and their study may well form the core of a future paper on the
matter.
22 P. MERTIKOPOULOS

3. E  A

This section is devoted in its entirety to examples and applications illustrating

the use of the concepts we introduced. Bearing this in mind, our approach will
not be very rigorous; instead, we will place emphasis on how the form residues
actually work.
To begin with, we will perform a number of integrations in R2 . Therefore, let
us consider the following differential forms, which are defined on all of R2 \ {O}:
h i
(1) ξ = (x2 +y
1 2 2
2 )2 (x − y )dy − 2xydx
x
h i
(2) ω = x2e+y2 (x sin y − y cos y)dx + (x cos y + y sin y)dy

Furthermore, let C be the unit circle in R2 centered at the origin O(0, 0) and taken
with
R Rthe standard
R - anticlockwise - orientation24. We wish to evaluate the integrals
C
ρ, C ξ and C ω.
Now, by a direct calculation, it can be shown that the forms above are closed in
R2 \ {O}, that is: dξ = dω = 0. In addition to that, one can easily see that they are
closed with ω having a pole of order 0 at O while ξ has an order-1 pole at O; hence,
we can apply propositions 2.8 and 2.15 in order to evaluate the desired integrals.
Indeed, by proposition 2.8 we get:
Z
− = 2π Ind(C : O)Res(− : O) = 2π Res(− : O)
C
since Ind(C : O) is clearly +1. Consequently, we only need to evaluate the form
residues at the origin, and this can be accomplished by means of proposition 2.15:
(1) Evaluation of Res(ξ : O)
By applying proposition 2.15 to ξ - m = 1 - we get:

i]
~Lξ = ?−1 ξ)] = 1 (x2 − y2 )dx + 2xydy = 1 x2 − y2 , 2xy ⇒
 h  
r4 r4
1 x
r~Lξ · ~r = 3 (x3 − xy2 + 2xy2 ) = = α1 25
r r
In polar coordinates, the result above is clearly independent of the radial
∂ ~
 
coordinate r, that is: ∂r rLξ · ~r = 0 and this leads to: Res(ξ : O) = 0 26.
Z
Therefore, we have shown that: ξ = 0.
C
(2) Evaluation of Res(ω : O):
Bearing in mind that ω has a pole of order 0 at O we obtain:

~Lω = ?−1 ω)] =

 ex  
x cos y + y sin y, y cos y − x sin y ⇒
x2 + y2
~Lω · ~r = ex cos y

24In order to be absolutely clear on this point, a parametrization for the cycle we have in mind is:
x = cos θ, y = sin θ with θ ∈ [0, 2π] being the usual angular coordinate.
25Recall that α = x j - see proofs of propositions 2.4 and 2.13.
j r
26This should come as no surprise: ξ is just the singular term associated to the array T = [−1 0].
 GAUSS’S LAW IN COHOMOLOGY THEORY 23

Clearly, r → 0 is equivalent to ~r = (x, y) → (0, 0) = ~0. Therefore:

lim ~Lω · ~r = lim ex cos y = 1

r→0 ~r→~0
Z
Hence, Res(ω : O) = 1 which means that: ω = 2π.
C

In the examples presented above, we have reduced integration to a straightfor-

ward algebraic calculation. We will further illustrate the process by a purposedly
complex example in R4 .
Indeed, let ψ be the differential 3-form defined on R4 \ {O}:

"
1  4
ψ = 8 2xr − (w − 2x + y + z)r2 −
r
 
− 6x w(w − x + y + z) + x2 + (y − 2z)z − x(y + z) dz ∧ dy ∧ dw

+ 2yr4 + (w − x + z)r2 −
 
− 6y w(w − x + y + z) + x2 + (y − 2z)z − x(y + z) dx ∧ dz ∧ dw

+ 2zr4 + (w − x + y − 4z)r2 +
 
− 6z w(w − x + y + z) + x2 + (y − 2z)z − x(y + z) dy ∧ dx ∧ dw

+ 2wr4 + (2w − x + y + z)r2 −
#
 
− 6w w(w − x + y + z) + x2 + (y − 2z)z − x(y + z) dx ∧ dy ∧ dz

Also, let Z denote the unit sphere S3 , centered at O and taken with the standard
Z {dx, dy, dz, dw}. As in the previous examples,
orientation with respect to the basis
we wish to evaluate the integral ψ. A direct calculation will reveal that ψ is
Z
closed in R4 \ {O} and has a pole of order 2 at O. Therefore:
Z
ψ = (4 − 2) · 2π2 Res(ψ : O)Ind(Z : O) = 4π2 Res(ψ : O) 27
Z

and we simply have to evaluate the residue of ψ at O by means of proposition 2.13.

So, by performing a few easy calculations, we obtain:

r4~Lψ · ~r = r4 (?−1 ψ)] · ~r =

2
  
= 2 r2 − 2 2w2 + (w − x + y + z)w + 3x2 − x(y + z) + y(2y + z) + 4
r

27Clearly, Ind(Z : O) = +1.

24 P. MERTIKOPOULOS

∂2
r4~Lψ ·~r = 4 28 which
h i
Thus, by expressing this result in polar coordinates we get: ∂r2
Z
yields: Res(ψ : O) = 1. Hence: ψ = 4π2 .
Z

Note: Having computed the form residue at the origin for ξ, ω and ψ we can also
deduce whether the forms are exact or not. As we saw, only ξ leaves zero residue
at the origin and this means that ξ is the only exact form in the previous examples.
This would be a rather hard problem to solve, but we have seen that it is readily
reduced to a mere algebraic manipulation.

Now, a very important application of complex residue calculus lies in evaluating

real improper integrals. In a similar fashion, we will see that the form residues
which we introduced allow us to handle a large class of multiple - iterated -
improper integrals29. Without further ado, we will proceed to a final series of
examples illustrating the process.
Z ∞ Z ∞
dx1 . . . dxk
Consider the k-fold integral: Ik = ...   k+1 with a > 0.
−∞ −∞
x21 + · · · + x2k + a2
2

First, observe that in Rk+1 the denominator - except for the (k+1) exponent - merely
expresses the distance between a point (x1 , . . . , xk , 0) belonging to the k-dimensional
hyperplane: πk+1 = hx1 , . . . , xk i and the point P = (0, . . . , 0, a). Furthermore, the
spherical symmetry of the integrand and the presence of the (k+1) exponent -
which equals the dimension of Rk+1 - suggest that a singular term of order 0 and
based at P lurks just around the corner. Thus, let us consider a monopole Φ based
at P; its expression in Cartesian coordinates based at O = (0, . . . , 0, 0) will be:
xk+1 − a
Φ= dx1 ∧ . . . ∧ dxk + terms containing dxk+1
rk+1
a
q
where ra = x21 + · · · + x2k + (xk+1 − a)2 denotes the distance between any given
point (x1 , . . . , xk+1 ) and P.
We will integrate Φ over the k-cycle ZR which consists of:
(1) The section CR of the k-sphere S(P, R) - of radius R and centered at P - which
lies “above” πk+1 - that is: xk+1 ≥ 0 - and
(2) the disk DR on πk+1 which is bounded by the intersection of S(P, R) with
πk+1
The orientations given to CR and DR are such to ensure
Z that Z
ZR is taken
Z with the
outward orientation. Bearing this in mind, we have: Φ= Φ+ Φ.
ZR DR CR
Obviously, when we pull back and integrate Φ over DR R ⊆ πk+1 where xk+1 is kept
fixed to 0, the terms containing dxk+1 will vanish and D Φ will be reduced to the
R

28Arguments of dimensional analysis can be used here: it is an easy tas to show that say wx or any
2 r
other dimensionless combination w.r.t. r will not depend on r.
29We will not go so far as to introduce concrete methods like the ones used in Complex Analysis.
However, we hope that the number of examples provided will give the reader a fair notion of the
potential of form residues.
 GAUSS’S LAW IN COHOMOLOGY THEORY 25

following k-fold integral:

dx1 . . . dxk
Z Z
Φ=a   k+1
DR DR
x21 + · · · + x2k + a2
2

Z
Let R → ∞. Clearly, lim Φ = aIk since DR takes up all of πk+1 as R → ∞.
R→∞ DR
Moreover, by computing the residue of Φ by means of proposition 2.13 we get:
√
( π)k−1
Z
Res(Φ : O) = k−1 which yields - proposition 2.8:
1
Φ = Ak = 2π   with Ak
ZR Γ k+1
2
k
denoting the area spanned by the unit k-sphere
Z S . This result is not modified by
R
letting R → ∞ and we obtain: aIk = Ak − lim Φ. But, C Φ is the “solid angle”
R→∞ CR
R

30
spanned by CR and it can be easily seen that as R → ∞ this solid angle is just 21 Ak
since, for R → ∞, CR assumes the shape of a k-hemisphere - which clearly bounds
half the surface spanned by a k-sphere of the same radius. Combining all of the
1
above in a single relation, we obtain: aIk = Ak which leads to our final result:
2
√ k−1
π ( π)
Ik = ·  
a Γ k+1
2

By considering singular terms of higher order we may evaluate other classes

of k-fold integrals. Indeed, set T = [0 . . . 0 1] and let Ψ be the singular term
associated to T and based at (0, . . . , 0, a) ∈ Rk+1 with a > 0. A direct computation
will reveal that in Cartesian coordinates, Ψ can be expressed as follows:

h (xk+1 − a)2 1 i
Ψ = (k + 1) − dx1 ∧ . . . ∧ dxk + terms containing dxk+1
rk+3
a rak+1

Z If ZR , CR , DR are defined as in the previous example, it should be clear that:

Ψ = 0 since Res(Ψ : O) = 0 - recall that Ψ is a singular term of order 1: it leaves
ZR
no residue at P. In addition to that, when Rpulling back and integrating over DR ,
the terms containing dxk+1 will vanish and D Ψ will be reduced to:
R

x21 + ··· + x2k

− ka2
Z Z
Ψ=   k+3 dx1 . . . dxk
DR DR
x21 + · · · + x2k + a2
2

Z
However, if we let R → ∞, it can be shown that the integral Ψ vanishes.
CR
Indeed, from a dimensional standpoint, we can see that the exponent of the xi ’s in
the denominator of Ψ exceeds that of the numerator by k+1 R and we are integrating
over a k-cell: this will implement a scale factor of R1 in C Ψ and, as R → ∞, this
R

30Recall the proof of proposition 2.4.

26 P. MERTIKOPOULOS

R
will cause the integral to vanish31. This shows that for R → ∞, D Ψ must vanish
R R
as well and, if we apply this result to the explicit expression of D Ψ, we obtain:
R
Z ∞ Z ∞
dx1 . . . dxk
Ik − (k + 1)a2 . . .   k+3 =0
−∞ −∞
x21 + · · · + x2k + a2
2

which, by substituting Ik , yields:

Z ∞ Z ∞ √
dx1 . . . dxk π ( π)k−1
... = ·   .
  k+3 (k + 1)a3 Γ k+1
−∞ −∞
x21 + · · · + x2k + a2
2
2

Naturally, the examples presented above are the simplest conceivable ones and,
by merely playing around with the position of the multipole, we would be able to
evaluate integrals with denominators far more complex. Furthermore, as should
be obvious, the method of form residues can be applied to various trigonometric
integrals as well, but, at this point, a simple method in the form of an “algorithm” -
as in the case of Complex Analysis - cannot be presented, but could well constitute
part of a future paper.

R
[1] M. Spivak, A Comprehensive Introduction to Differential Geometry, Publish or Perish, Houston, 1979
[2] M.A. Armstrong, Basic Topology, Springer-Verlag, New York, 1983
[3] V. Guillemin and A. Pollack, Differential Topology, Prentice-Hall, Englewood Cliffs, 1974
[4] P. Bamberg and S. Sternberg, A Course in Mathematics for Students of Physics, vol. I and II, Cambridge
University Press, Cambridge, 1990
[5] J. D. Jackson, Classical Electrodynamics, John Wiley & Sons, New York, 1999
[6] R. Abraham, J.E. Marsden and T. Ratiu, Manifolds, Tensor Analysis, and Applications, Springer-Verlag,
New York, 1988

D  P, U  A

E-mail address: aragorn@ath.forthnet.gr

31Of course, this “proof” lacks rigour. A rigorous argument along these lines is not hard to construct
but, since we are interested in the overall method, and, in keeping with the “applied” style of this section,
we will not venture into details.