5Symmetry methods in PDEs
II Integrable Systems
5.1 Lie groups and Lie algebras
So to begin with, we remind ourselves with what a group is!
Definition (Group). A group is a set G with a binary operation
(g
1
, g
2
) 7→ g
1
g
2
called “group multiplication”, satisfying the axioms
(i) Associativity: (g
1
g
2
)g
3
= g
1
(g
2
g
3
) for all g
1
, g
2
, g
3
(ii)
Existence of identity: there is a (unique) identity element
e ∈ G
such that
ge = eg = g
for all g ∈ G
(iii) Inverses exist: for each g ∈ G, there is g
−1
∈ G such that
gg
−1
= g
−1
g = e.
Example. (Z, +) is a group.
What we are really interested in is how groups act on certain sets.
Definition
(Group action)
.
A group
G
acts on a set
X
if there is a map
G × X → X sending (g, x) 7→ g(x) such that
g(h(x)) = (gh)(x), e(x) = x
for all g, h ∈ G and x ∈ X.
Example. The rotation matrices SO(2) acts on R
2
via matrix multiplication.
We are not going to consider groups in general, but we will only talk about
Lie groups, and coordinate changes born of them. For the sake of simplicity, we
are not going to use the “real” definition of Lie group, but use an easier version
that really looks more like the definition of a local Lie group than a Lie group.
The definition will probably be slightly confusing, but it will become clearer
with examples.
Definition
(Lie group)
.
An
m
-dimensional Lie group is a group such that all
the elements depend continuously on
m
parameters, in such a way that the
maps (
g
1
, g
2
)
7→ g
1
g
2
and
g 7→ g
−1
correspond to a smooth function of those
parameters.
In practice, it suffices to check that the map (g
1
, g
2
) 7→ g
1
g
−1
2
is smooth.
So elements of an (
m
-dimensional) Lie group can be written as
g
(
t
), where
t ∈ R
m
. We make the convention that
g
(0) =
e
. For those who are doing
differential geometry, this is a manifold with a group structure such that the
group operations are smooth maps. For those who are doing category theory,
this is a group object in the category of smooth manifolds.
Example. Any element of G = SO(2) can be written as
g(t) =
cos t −sin t
sin t cos t
for
t ∈ R
. So this is a candidate for a 1-dimensional Lie group that depends on
a single parameter
t
. We now have to check that the map (
g
1
, g
2
)
7→ g
1
g
−1
2
is
smooth. We note that
g(t
1
)
−1
= g(−t
1
).
So we have
g(t
1
)g(t
2
)
−1
= g(t
1
)g(−t
2
) = g(t
1
− t
2
).
So the map
(g
1
, g
2
) 7→ g
1
g
−1
2
corresponds to
(t
1
, t
2
) 7→ t
1
− t
2
.
Since this map is smooth, we conclude that
SO
(2) is a 1-dimensional Lie group.
Example. Consider matrices of the form
g(t) =
1 t
1
t
3
0 1 t
2
0 0 1
, t ∈ R
3
It is easy to see that this is a group under matrix multiplication. This is known
as the Heisenberg group. We now check that it is in fact a Lie group. It has three
obvious parameters
t
1
, t
2
, t
3
, and we have to check the smoothness criterion. We
have
g(a)g(b) =
1 a
1
a
3
0 1 a
2
0 0 1
1 b
1
b
3
0 1 b
2
0 0 1
=
1 a
1
+ b
1
a
3
+ b
3
+ a
1
b
2
0 1 a
2
+ b
2
0 0 1
.
We can then write down the inverse
g(b)
−1
=
1 −b
1
b
1
b
2
− b
3
0 1 −b
2
0 0 1
So we have
g(a)g(b)
−1
=
1 a
1
a
3
0 1 a
2
0 0 1
1 −b
1
b
1
b
2
− b
3
0 1 −b
2
0 0 1
=
1 a
1
− b
1
b
1
b
2
− b
3
− a
1
b
2
+ a
3
0 1 a
2
− b
2
0 0 1
This then corresponds to
(a, b) 7→
a
1
− b
1
a
2
− b
2
b
1
b
2
− b
3
− a
1
b
2
+ a
3
,
which is a smooth map! So we conclude that the Heisenberg group is a three-
dimensional Lie group.
Recall that at the beginning of the course, we had vector fields and flow
maps. Flow maps are hard and complicated, while vector fields are nice and easy.
Thus, we often want to reduce the study of flow maps to the study of vector
fields, which can be thought of as the “infinitesimal flow”. For example, checking
that two flows commute is very hard, but checking that the commutator of two
vector fields vanishes is easy.
Here we are going to do the same. Lie groups are hard. To make life easier,
we look at “infinitesimal” elements of Lie groups, and this is known as the Lie
algebra.
We will only study Lie algebras informally, and we’ll consider only the case
of matrix Lie groups, so that it makes sense to add, subtract, differentiate the
elements of the Lie group (in addition to the group multiplication), and the
presentation becomes much easier.
Suppose we have a curve
x
1
(
ε
) in our parameter space passing through 0 at
time 0. Then we can obtain a curve
A(ε) = g(x
1
(t))
in our Lie group G. We set a = A
0
(0), so that
A(ε) = I + εa + o(ε).
We now define the Lie algebra
g
to be the set of all “leading order terms”
a
arising from such curves. We now proceed to show that
g
is in fact a vector
space.
Suppose we have a second curve B(x), which we expand similarly as
B(ε) = I + εb + o(ε).
We will show that a + b ∈ g. Consider the curve
t 7→ A(t)B(t),
using the multiplication in the Lie group. Then we have
A(ε)B(ε) = (I + εa + o(ε))(I + εb + o(ε)) = I + ε(a + b) + o(ε).
So we know a, b ∈ g implies a + b ∈ g.
For scalar multiplication, given λ ∈ R, we can construct a new curve
t 7→ A(λt).
Then we have
A(λε) = I + ε(λa) + o(ε).
So if a ∈ g, then so is λa ∈ g for any λ ∈ R.
So we get that
g
has the structure of a vector space! This is already a little
interesting. Groups are complicated. They have this weird structure and they
are not necessarily commutative. However, we get a nice, easy vector space
structure form the group structure.
It turns out we can do something more fun. The commutator of any two
elements of g is also in g. To see this, we define a curve C(t) for t > 0 by
t 7→ A(
√
t)B(
√
t)A(
√
t)
−1
B(
√
t)
−1
.
We now notice that
A
(
ε
)
−1
=
I −εa
+
o
(
ε
), since if
A
(
ε
)
−1
=
I
+
ε˜a
+
o
(
ε
), then
I = A(ε)A(ε)
−1
= (I + εa + o(ε))(I + ε˜a + o(ε))
= I + ε(a + ˜a) + o(ε)
So we must have ˜a = −a.
Then we have
C(ε) = (I +
√
εa + ···)(I + εb + ···)(I −
√
εa + ···)(I −
√
εb + ···)
= I + ε(ab − ba) + o(ε).
It is an exercise to show that this is actually true, because we have to keep track
of the second order terms we didn’t write out to make sure they cancel properly.
So if a, b ∈ g, then
[a, b]
L
= ab − ba ∈ g.
Vector spaces with this extra structure is called a Lie algebra. The idea is that
the Lie algebra consists of elements of the group infinitesimally close to the
identity. While the product of two elements
a, b
infinitesimally close to the
identity need not remain infinitesimally close to the identity, the commutator
ab − ba does.
Definition
(Lie algebra)
.
A Lie algebra is a vector space
g
equipped with a
bilinear, anti-symmetric map [
·, ·
]
L
:
g ×g → g
that satisfies the Jacobi identity
[a, [b, c]
L
]
L
+ [b, [c, a]
L
]
L
+ [c, [a, b]
L
]
L
= 0.
This antisymmetric map is called the Lie bracket.
If dim g = m, we say the Lie algebra has dimension m.
The main source of Lie algebras will come from Lie groups, but there are
many other examples.
Example. We can set g = R
3
, and
[a, b]
L
= a × b.
It is a straightforward (and messy) check to see that this is a Lie algebra.
Example. Let M be our phase space, and let
g = {f : M → R smooth}.
Then
[f, g]
K
= {f, g}
is a Lie algebra.
Example. We now find the Lie algebra of the matrix group SO(n). We let
G = SO(n) = {A ∈ Mat
n
(R) : AA
T
= I, det A = 1}.
We let A(ε) be a curve in G with A(0) = I. Then we have
I = A(ε)A(ε)
T
= (I + εa + o(ε))(I + εa
T
+ o(ε))
= I + ε(a + a
T
) + o(ε).
So we must have
a
+
a
T
= 0, i.e.
a
is anti-symmetric. The other condition says
1 = det A(ε) = det(I + εa + o(ε)) = 1 + ε tr(a) + o(ε).
So we need tr(a) = 0, but this is already satisfied since A is antisymmetric.
So it looks like the Lie algebra
g
=
so
(
n
) corresponding to
SO
(
n
) is the vector
space of anti-symmetric matrices:
so(n) = {a ∈ Mat
n
(R) : a + a
T
= 0}.
To see this really is the answer, we have to check that every antisymmetric
matrix comes from some curve. It is an exercise to check that the curve
A(t) = exp(at).
works.
We can manually check that g is closed under the commutator:
[a, b]
L
= ab − ba.
Indeed, we have
[a, b]
T
L
= (ab − ba)
T
= b
T
a
T
− a
T
b
T
= ba − ab = −[a, b]
T
L
.
Note that it is standard that if we have a group whose name is in capital
letters (e.g.
SO
(
n
)), then the corresponding Lie algebra is the same thing in
lower case, fraktur letters (e.g. so(n)).
Note that above all else,
g
is a vector space. So (at least if
g
is finite-
dimensional) we can give
g
a basis
{a
i
}
m
i=1
. Since the Lie bracket maps
g ×g → g
,
it must be the case that
[a
i
, a
j
] =
m
X
k=1
c
k
ij
a
k
for some constants c
k
ij
. These are known as the structure constants.