what are maurer-cartan forms (-equations)?
dennis westra
More and more often I encounter the so-called maurer-cartan forms and
their structural equations. In my attempt to look for a basic
explanation of what these objects are and where they are needed for in
high-energy physics (why do we want to use them?) I had little succes.
Does anybody know an easy explanation? (Some terminology about Lie
algebra's may be used ...)
thanx, dennis
John Baez
dennis westra <dennis...@hotmail.com> wrote:
>More and more often I encounter the so-called maurer-cartan forms and
>their structural equations. In my attempt to look for a basic
>explanation of what these objects are and where they are needed for in
>high-energy physics (why do we want to use them?) I had little success.
>Does anybody know an easy explanation? (Some terminology about Lie
>algebras may be used ...)
The Maurer-Cartan form is a 1-form on a Lie group taking
values in its Lie algebra. In other words, it's a beast
that eats a tangent vector anywhere on the Lie group and
spits out an element of the Lie algebra, in a linear way.
How do we define this beast? At the identity element of
our Lie group, it's easy. The tangent space at the
identity element of a Lie group *is* its Lie algebra,
by definition. So, the Maurer-Cartan form just eats
a tangent vector at the identity and spits out the
very same thing - but calling it a Lie algebra element!
It's also not hard to define the Maurer-Cartan form
at any other point of a Lie group. The reason is that
we can map any point of our Lie group to the identity
by left multiplication by a suitable element. This
in turn gives a way to map tangent vectors at any point
to tangent vectors at the identity. So, to get the
Maurer-Cartan form we just do that and then say
"Hey, but now it's a Lie algebra element"!
In short, the Maurer-Cartan form is a completely
tautologous thing. For that reason it must either
be completely boring or very, very interesting.
In fact it's very, very interesting, mainly because
its exterior derivative is not zero. There's a very
pretty formula for its exterior derivative, called
the Maurer-Cartan formula.
(Nota bene: I'm thinking of the Maurer-Cartan form as
a single 1-form taking values in the Lie algebra. People
fond of indices will instead pick a basis for the
Lie algebra and get a list of "Maurer-Cartan forms",
which are ordinary 1-forms, one for each basis vector.
It's just a slightly less elegant way of doing the same
thing, so don't sweat it.)
Alfred Einstead
> In short, the Maurer-Cartan form is a completely
> tautologous thing. For that reason it must either
> be completely boring or very, very interesting.
>
> In fact it's very, very interesting, mainly because
> its exterior derivative is not zero. There's a very
> pretty formula for its exterior derivative, called
> the Maurer-Cartan formula.
>
> (Nota bene: I'm thinking of the Maurer-Cartan form as
> a single 1-form taking values in the Lie algebra. People
> fond of indices will instead pick a basis for the
> Lie algebra and get a list of "Maurer-Cartan forms",
> which are ordinary 1-forms, one for each basis vector.
> It's just a slightly less elegant way of doing the same
> thing, so don't sweat it.)
You can have it both ways by observing the convention that
vector fields may be used in place of lower indices, 1-forms
in place of upper indices. Then indexes are just arguments
of linear functionals and the physicists' and mathematicians'
notations are unified into a single convnetion that supersedes
them both.
The components dw(U,V) of the exterior derivative of the
1-form w, in "ordinary" notation are written as:
dw(U,V) = U w(V) - V w(U) - w([U,V]),
which suffers an additional advantage of hyperconflation.
Rendered by the new convention, they read in a much more
illuminating form as:
dw_{UV} = @_U w_V - @_V w_U - f^w_{UV}
(using @ to stand for the curly partial d symbol)
noting that the structure coefficients are defined by
f^w_{UV} = w_{[U,V]} = w([U,V]) = [U,V]^w
(illustrating the convention, in the process).
Thus,
dw = 1/2 dw_{ab} e^a ^ e^b
= 1/2 (@_a w_b - @_b w_a - f^w_{ab}) e^a ^ e^b.
In the case of w = e^c, one has e^c_b = delta^c_b = constant
and
de^c = 1/2 (@_a delta^c_b - @_b delta^c_a - f^c_{ab}) e^a ^ e^b
= -1/2 f^c_{ab} e^a ^ e^b.
To the other topic at hand, there's a detailed reference that provides
the essentials:
www.cpt.univ-mrs.fr/~coque/book
(by Robert Coquereaux).
Paraphrasing node 92: there are, in fact, two Cartan-Maurer forms:
associated respectively with left and right invariant vector fields.
The former is the one normally referred to as The Cartan-Maurer form.
A gauge group is a Lie group acting on a base manifold. So, all
the geometric machinery pertains directly to gauge field theory.
The basis of the Lie algebra map to the fundamental matrices
(tau_1, tau_2, tau_3 in SU(2), lambda_1 through lambda_8 in SU(3),
Y in U(1)_{hypercharge}), the connection maps to the gauge fields
(W^1, W^2, W^3 in SU(2), G^1 through G^8 in SU(3), B in U(1)),
and the curvature form maps to the field strengths. The 2nd
cartan structure equation relates the curvature to the connection.
The first relates the connection to the basis.
So, if G is an N-dimensional Lie group, e the identity, and A = T_e(G)
the Lie algebra of G (T_g(G) denotes the tangent space of G at the
element g in G), then the Cartan-Maurer form
w_g: T_g(G) -> A
is given by the linear operator
w_g: e_a(g) |-> e_a = e_a(e)
where the fields
(e_1(g), e_2(g), ..., e_N(g))
are left invariant vector fields that comprise a basis for T_g(G)
at each g in G.
Denoting the dual basis in T*_g(G) as e^a(g), the definition reads:
w_g = e^a(g) [x] e_a in T*_g(G) [x] A
where [x], here, means "tensor product".
Since the forms are Lie-algebra valued, it's common to use a
combined commutator-wedge product as a notation. This is
defined for Lie-valued 1 forms h = h^a e_a; k = k^a e_a as:
[h^k] = [h^b e_b ^ k^c e_c]
= (h^b ^ k^c) [x] [e_b, e_c]
= (h^b ^ k^c) [x] f^a_{bc} e_a
where the structure coefficients are used in:
[e_b, e_c] = f^a_{bc} e_a.
The exterior derivative w_g (or w for short) is then just
dw = d(e^a(g) [x] e_a)
= de^a(g) [x] e_a
(since e_a = e_a(e) is independent of g). Or, making use of
the relation derived before in the hybrid notation:
dw = -1/2 f^a_{bc} (e^b ^ e^c) [x] e_a.
= -1/2 (e^b ^ e^c) [x] (f^a_{bc} e_a)
= -1/2 [w^w].
Thus, the first structure equation:
dw + 1/2 [w^w] = 0.
This can be written more directly, returning back to the hybrid
index notation. Let h, k be 1-forms defined on the Lie group G
that take on values in the Lie algebra A = T_e(G). Then for
any vectors U, V in T_g(G), we have
h_U = h^b_U e_b; k_V = k^c_V e_c.
The commutator can then be defined as the quadratic form with
the components:
[h,k]_{UV} = [h_U, k_V].
So, then
[h_U, k_V] = [h^b_U e_b, k^c_V e_c]
= h^b_U k^c_V [e_b, e_c]
= h^b_U k^c_V f^a_{bc} e_a.
This is related to the combined wedge/bracket product since
(h^a ^ k^b f^c_{ab} e_c)_{UV}
= ((h^a [x] k^b - k^b [x] h^a) f^c_{ab} e_c)_{UV}
= (h^a_U k^b_V + k^a_U h^b_V) f^c_{ab} e_c
using the antisymmetry of the structure coefficients
f^c_{ab} = -f^c_{ba}.
Thus,
[h^k] = [h,k] + [k,h].
The structure equation may therefore also be written as:
dw + [w,w] = 0.
For the Cartan-Maurer form associated with right-invariant
vector fields, everything follows through ultimately with a
sign change in the structure equation:
dv - [v,v] = 0,
using v to denote the corresponding Cartan-Maurer form.
dennis westra
I have also been reading a little and my idea is that one can easily
get a basis of tangent vectors around the unity element. These vectors
X_i satisfy a Lie rule. This basis can be translated to another point
giving another basis. One does this in such a way to create
left-invariant vectors (this point is unclear to me: what is precisely
meant with this? I've seen the formulae for this, but it is not clear
what it means.) The maurer-cartan forms are then said to be the dual
basis to the X_i, such that if M_j is a maurer cartan form M_j(X_i) =
delta_i,j . My somewhat refined question is: could someone elaborate
on the idea of left-invariant vectors and what the meaning of this is
for maurer-cartan forms?
regards,
d/w
Joris Vankerschaver
dennis...@hotmail.com says...
There's no such thing as a left invariant *vector*, it's a vector *field*
that is left invariant. Remember how John Baez defined the MC-form by
using left translation. So if we pick a basis of the tangent space at
unity in the Lie group, we can left translate the elements of this basis
to any point. Hence, by left translating a vector we obtain a vector
field. Suppose we obtain n vector fields X_1, ..., X_n this way.
Now, what is a left invariant vector field? Suppose we're given a vector
field X on the Lie group. Consider the point h and the value of X in h,
X(h). Now take another point g in the Lie group. We can left translate
X(h) to g and compare it to X(g). If they are the same, we call X left
invariant.
In the first step we *defined* our vector fields X_1, ..., X_n by left
translating, so it's quite obvious that they are left invariant. On the
other hand, we have the Maurer Cartan form, which is as John Baez
explained, also defined by left translation. Now we write out the MC
form in components, let's call these theta_1, ..., theta_n. It's pretty
obvious that in the unity, theta_i ( X_j ) will be 1 if i = j, and 0
otherwise (just write out the definition of theta in the unity). But, as
both theta_i and X_j are left invariant, this result is also valid in any
other point. So theta_1, ..., theta_n are dual to X_1, ..., X_n.
Hope this helps,
Joris Vankerschaver
Igor Khavkine
As anyone who has studied any general relativity or basic differential
geometry knows, there is no canonical way to compare tangent spaces and
tangent vectors attached at different points of the manifold.
Fortunately, if your manifold G is also a Lie group, then there is a
canonical way to do so. Note that for any g in G, the _left_ action of g
on the manifold is given by L_g : G -> G, L_g: h +-> gh. Since G is a Lie
group, the map L_g is a diffeomorphism. This means that you can push
forward any vector from T(e), e being the identity element of G, to T(g)
using L_g*. If you are not familiar with push forwards, just think of L_g*
as the Jacobian of L_g. The same way any vector from T(g) can be pushed
forward to T(h) using L_{hg^-1}*.
So if you start with a tangent vector attached to any point of G, then we
can naturally move this vector around the whole manifold and define a
vector field. Which is smooth because of the smoothness of Lie group
composition ("multiplication"). Such a vector field is called
_left-invariant_, since it is invariant under push forwards by L_g, for
any g in G.
One way to get the Lie algebra corresponding to some Lie group is to take
the space of all left invariant vector fields on it (which is canonically
isomorphic to the tangent space at the identity element), and endow this
vector space with a Lie bracket which is just the usual Lie bracket
between two vector fields.
Given a basis v_j for T(e) you can find a dual basis a^j, that is a basis
for T^*(e) such that a^i(v_j) = Kronecker delta^i_j. But the v_j naturally
give rise to V_j which is a basis for the left invariant vector fields. We
can naturally extend a^j to one forms A^j which are still dual to V_j.
After this point it gets a little tough for me, I know that these forms are
supposed to satisfy some relations involving the exterior differential,
wedge product, and the structure constants of G's Lie algebra. Which seems
rather natural, since all of these things are related to left invariant
vector fields and their Lie brackets. After all, that is where the
structure constants come from. But I haven't gone through the explicit
calculations yet, so I don't have a definite feeling for how they are
related.
Hope this helps.
Igor
Arkadiusz Jadczyk
>After this point it gets a little tough for me, I know that these forms are
>supposed to satisfy some relations involving the exterior differential,
>wedge product, and the structure constants of G's Lie algebra. Which seems
>rather natural, since all of these things are related to left invariant
>vector fields and their Lie brackets. After all, that is where the
>structure constants come from. But I haven't gone through the explicit
>calculations yet, so I don't have a definite feeling for how they are
>related.
The crucial and useful formula here is:
< d omega , X \wedge Y > = L_X <omega , Y> - L_Y <omega, X> - < omega ,
[X, Y] )
where L_X stands for Lie derative along X
valid for any differential form and any two vector fields.
In our case we take for omega the Maurer-Cartan form
on G, and for X and Y two elements of the basis
of left-invariant vector fields.
The first two terms in the above formula then vanish,
as we have Lie derivatives (here, just a differential of a function)
of constant functions. (exercise: why constant?)
The rest follows by comparing the third term of the above
formula with the other side of Maurer-Cartan structure equation.
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--