Lecture 12 - Grassmann Algebra and de Rham Cohomology (Schuller's Geometric Anatomy of Theoretical Physics)
Lecture 12 - Grassmann Algebra and de Rham Cohomology (Schuller's Geometric Anatomy of Theoretical Physics)
Lecture 12 - Grassmann Algebra and de Rham Cohomology (Schuller's Geometric Anatomy of Theoretical Physics)
b) The electromagnetic field strength F is a differential 2-form built from the electric
and magnetic fields, which are also taken to be forms. We will define these later in
some detail.
() : M T M
p 7 ()(p),
where
()(p)(X1 , . . . , Xn ) := ((p)) (X1 ), . . . , (Xn ) ,
for Xi Tp M .
1
The map : n (N ) n (M ) is R-linear, and its action on 0 (M ) is simply
: 0 (M ) 0 (M )
f 7 (f ) := f .
This works for any smooth map , and it leads to a slight modification of our mantra:
The tensor product does not interact well with forms, since the tensor product of two
forms is not necessarily a form. Hence, we define the following.
Definition. Let M be a smooth manifold. We define the wedge (or exterior ) product of
forms as the map
: n (M ) m (M ) n+m (M )
(, ) 7 ,
where
1 X
( )(X1 , . . . , Xn+m ) := sgn()( )(X(1) , . . . , X(n+m) )
n! m!
Sn+m
f g := f g and f = f = f .
Hence
= .
The wedge product is bilinear over C (M ), that is
(f 1 + 2 ) = f 1 + 2 ,
2
The pull-back distributes over the wedge product.
( ) = () ().
() () (p)(X1 , . . . , Xn+m )
1 X
sgn() () () (p)(X(1) , . . . , X(n+m) )
=
n! m!
Sn+m
1 X
= sgn() ()(p)(X(1) , . . . , X(n) )
n! m!
Sn+m
()(p)(X(n+1) , . . . , X(n+m) )
1 X
= sgn()((p)) (X(1) ), . . . , (X(n) )
n! m!
Sn+m
((p)) (X(n+1) ), . . . , (X(n+m) )
1 X
= sgn()( )((p)) (X(1) ), . . . , (X(n+m) )
n! m!
Sn+m
= ( )((p)) (X1 ), . . . , (Xn+m )
= ( )(p)(X1 , . . . , Xn+m ).
: (M ) (M ) (M )
Recall that the direct sum of modules has the Cartesian product of the modules as
underlying set and module operations defined componentwise. Also, note that by algebra
here we really mean algebra over a module.
Example 12.6. Let = + , where 1 (M ) and 3 (M ). Of course, this + is
neither the addition on 1 (M ) nor the one on 3 (M ), but rather that on (M ) and, in
fact, (M ).
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Let n (M ), for some n. Then
= ( + ) = + ,
= (1)nm .
= = .
= a1 an dxa1 dxan
= b1 bm dxb1 dxbm
with 1 a1 < < an dim M and similarly for the bi . The coefficients a1 an and
b1 bm are smooth functions in C (U ). Since dxai , dxbj 1 (M ), we have
Remark 12.9. We should stress that this is only true when and are pure degree forms,
rather than linear combinations of forms of different degrees. Indeed, if , (M ), a
formula like
=
does not make sense in principle, because the different parts of and can have different
commutation behaviours.
4
12.3 The exterior derivative
Recall the definition of the gradient operator at a point p M . We can extend that
definition to define the (R-linear) operator:
d : C (M )
(T M )
f 7 df
Remark 12.10. Locally on some chart (U, x) on M , the covector field (or 1-form) df can
be expressed as
df = a dxa
for some smooth functions i C (U ). To determine what they are, we simply apply both
sides to the vector fields induced by the chart. We have
df b
= (f ) = b f
x xb
and
a
a dx = a (xa ) = a ba = b .
xb xb
Hence, the local expression of df on (U, x) is
df = a f dxa .
d(f g) = g df + f dg.
We can also understand this as an operator that takes in 0-forms and outputs 1-forms
d : 0 (M )
1 (M ).
This can then be extended to an operator which acts on any n-form. We will need the
following definition.
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Definition. The exterior derivative on M is the R-linear operator
d : n (M )
n+1 (M )
7 d
Remark 12.11. Note that the operator d is only well-defined when it acts on forms. In order
to define a derivative operator on general tensors we will need to add extra structure to our
differentiable manifold.
Example 12.12. In the case n = 1, the form d 2 (M ) is given by
Let us check that this is indeed a 2-form, i.e. an antisymmetric, C (M )-multilinear map
d : (T M ) (T M ) C (M ).
Moreover, thanks to this identity, it suffices to check C (M )-linearity in the first argument
only. Additivity is easily checked
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Therefore
[f X, Y ] = f [X, Y ] Y (f )X.
Hence, we can calculate
= f d(X, Y ),
d( ) = d + (1)n d.
Proof. We will work in local coordinates. Let (U, x) be a chart on M and write
d = dA dxA .
Hence
since we have anticommuted the 1-form dB through the n-form dxA , picking up n minus
signs in the process.
(d) = d( ()).
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Proof (sketch). We first show that this holds for 0-forms (i.e. smooth functions).
Let f C (N ), p M and X Tp M . Then
Remark 12.15. Informally, we can write this result as d = d , and say that the exterior
derivative commutes with the pull-back.
However, you should bear in mind that the two ds appearing in the statement are
two different operators. On the left hand side, it is d : n (N ) n+1 (N ), while it is
d : n (M ) n+1 (M ) on the right hand side.
Remark 12.16. Of course, we could also combine the operators d into a single operator
acting on the Grassmann algebra on M
d : (M ) (M )
by linear continuation.
Example 12.17. In the modern formulation of Maxwells electrodynamics, the electric and
magnetic fields E and B are taken to be a 1-form and a 2-form on R3 , respectively:
E := Ex dx + Ey dy + Ez dz
B := Bx dy dz + By dz dx + Bz dx dy.
F := B + E dt.
dF = 0.
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This is called the homogeneous Maxwells equation and it is, in fact, equivalent to the two
homogeneous Maxwells (vectorial) equations
B=0
B
E+ = 0.
t
In order to cast the remaining Maxwells equations into the language of differential forms,
we need a further operation on forms, called the Hodge star operator.
Recall from the standard theory of electrodynamics that the two equations above imply
the existence of the electric and vector potentials and A = (Ax , Ay , Az ), satisfying
B=A
A
E = .
t
Similarly, the equation dF = 0 on R4 implies the existence of an electromagnetic 4-potential
(or gauge potential) form A 1 (R4 ) such that
F = dA.
( Y (T M ) : (X, Y ) = 0) X = 0.
:= pi dq i
:= d 2 (T Q),
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12.4 de Rham cohomology
The last two examples suggest two possible implications. In the electrodynamics example,
we saw that
(dF = 0) ( A : F = dA),
while in the Hamiltonian mechanics example we saw that
( : = d) (d = 0).
closed if d = 0;
exact if n1 (M ) : = d.
The question of whether every closed form is exact and vice versa, i.e. whether the
implications
(d = 0) ( : = d)
hold in general, belongs to the branch of mathematics called cohomology theory, to which
we will now provide an introduction.
The answer for the direction is affirmative thanks to the following result.
d2 d d : n (M ) n+2 (M )
Definition. Given an object which carries some indices, say Ta1 ,...,an , we define the anti-
symmetrization of Ta1 ,...,an as
1 X
T[a1 an ] := sgn() T(a1 )(an ) .
n!
Sn
1 X
T(a1 an ) := T(a1 )(an ) .
n!
Sn
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Of course, we can (anti)symmetrize only some of the indices
1
T ab[cd]e = (T abcde T abdce ).
2
It is easy to check that in a contraction (i.e. a sum), we have
and
Ta1 (ai aj )an S a1 [ai aj ]an = 0.
Proof. This can be shown directly using the definition of d. Here, we will instead show it
by working in local coordinates.
Recall that, locally on a chart (U, x), we can write any form n (M ) as
= a1 an dxa1 dxan .
Then, we have
and hence
d2 = c b a1 an dxc dxb dxa1 dxan .
Since dxc dxb = dxb dxc , we have
c b a1 an = (c b) a1 an .
Thus
We can extend the action of d to the zero vector space 0 := {0} by mapping the zero
in 0 to the zero function in 0 (M ). In this way, we obtain the chain of R-linear maps
d d d d d d d d
0 0 (M ) 1 (M ) n (M ) n+1 (M ) dim M (M ) 0,
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where we now think of the spaces n (M ) as R-vector spaces. Recall from linear algebra
that, given a linear map : V W , one can define the subspace of V
im() := {(v) | v V },
The traditional notation for the spaces on the right hand side above is
so that Z n is the space of closed n-forms and B n is the space of exact n-forms.
Our original question can be restated as: does Z n = B n for all n? We have already
seen that d2 = 0 implies that B n Z n for all n (B n is, in fact, a vector subspace of Z n ).
Unfortunately the equality does not hold in general, but we do have the following result.
Z n = Bn , n > 0.
In the cases where Z n 6= B n , we would like to quantify by how much the closed n-forms
fail to be exact. The answer is provided by the cohomology group.
You can think of the above quotient as Z n / , where is the equivalence relation
: B n .
The answer to our question as it is addressed in cohomology theory is: every exact n-form
on M is also closed and vice versa if, only if,
H n (M )
=vec 0.
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Of course, rather than an actual answer, this is yet another restatement of the question.
However, if we are able to determine the spaces H n (M ), then we do get an answer.
A crucial theorem by de Rham states (in more technical terms) that H n (M ) only
depends on the global topology of M . In other words, the cohomology groups are topological
invariants. This is remarkable because H n (M ) is defined in terms of exterior derivatives,
which have everything to do with the local differentiable structure of M , and a given
topological space can be equipped with several inequivalent differentiable structures.
Example 12.22. Let M be any smooth manifold. We have
H 0 (M )
=vec R(# of connected components of M )
since the closed 0-forms are just the locally constant smooth functions on M . As an
immediate consequence, we have
H 0 (R)
=vec H 0 (S 1 )
=vec R.
H n (M )
=vec 0
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