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The resulting curve γ is called a horizontal lift of γ. If γ is merely piecewise C 1, the
horizontal lift can still be de¬ned by piecing together horizontal lifts of the C 1 -segments
in the obvious way. Also, if p = p · g0 , then the horizontal lift of γ with initial condition
p is easily seen to be ρg0 —¦ γ .
Let p ∈ P be chosen and set m = π(p). For every piecewise C 1-loop γ: [0, 1] ’ M
based at m, the horizontal lift γ has the property that γ (1) = p · h(γ) for some unique
˜ ˜
h(γ) ∈ G. The holonomy of A at p, denoted by HA (p) is, by de¬nition, the set of all such
elements h(γ) of G where γ ranges over all of the piecewise C 1 closed loops based at m.
I leave it to the reader to show that HA (p · g) = g ’1 HA (p)g and that, if p and p can
be joined by a horizontal curve in P , then HA (p) = HA (p ). Thus, the conjugacy class of
HA (p) in G is independent of p if M is connected.
A basic theorem due to Borel and Lichnerowitz (see [KN]) asserts that HA (p) is always
a Lie subgroup of G.

L.3.19 56
Exercise Set 3:
Actions of Lie Groups

1. Verify the claim made in the lecture that every right (respectively, left) action of a
Lie group on a manifold can be rewritten as a left (respectively, right) action. Is the
assumption that a left action »: G — M ’ M satisfy »(e, m) = m for all m ∈ M really

2. Show that if f: X ’ Y is a map of smooth manifolds for which the rank of f (x): Tx X ’
Tf (x) Y is independent of x, then f ’1 (y) is a (possibly empty) closed, smooth submanifold
of X for all y ∈ Y . Note that this properly generalizes the usual Implicit Function Theorem,
which requires f (x) to be a surjection everywhere in order to conclude that f ’1 (y) is a
smooth submanifold.
(Hint: Suppose that the rank of f (x) is identically k. You want to show that f ’1 (y)
(if non-empty) is a submanifold of X of codimension k. To do this, let x ∈ f ’1 (y) be given
and construct a map ψ: V ’ Rk on a neighborhood V of y so that ψ —¦ f is a submersion
near x. Then show that (ψ —¦ f)’1 ψ(y) (which, by the Implicit Function Theorem, is a
closed codimension k submanifold of the open set f ’1 (V ) ‚ X) is actually equal to f ’1 (y)
on some neighborhood of x. Where do you need the constant rank hypothesis?)

3. This exercise concerns the automorphism groups of Lie algebras and Lie groups.
(i) Show that, for any Lie algebra g, the group of automorphisms Aut(g) de¬ned by

Aut(g) = {a ∈ End(g) | a(x), a(y) = a [x, y] for all x, y ∈ g}

is a closed Lie subgroup of GL(g). Show that its Lie algebra is

der(g) = {a ∈ End(g) | a [x, y] = a(x), y + x, a(y) for all x, y ∈ g}.

(Hint: Show that Aut(g) is the stabilizer of some point in some representation of the
Lie group GL(g).)
(ii) Show that if G is a connected and simply connected Lie group with Lie algebra g,
then the group of (Lie) automorphisms of G is isomorphic to Aut(g).
Show that ad: g ’ End(g) actually has its image in der(g), and that this image is an
ideal in der(g). What is the interpretation of this fact in terms of “inner” and “outer”
automorphisms of G? (Hint: Use the Jacobi identity.)
(iv) Show that if the Killing form of g is non-degenerate, then [g, g] = g. (Hint: Suppose
that [g, g] lies in a proper subspace of g. Then there exists an element y ∈ g so that
κ [x, z], y = 0 for all x, z ∈ g. Show that this implies that [x, y] = 0 for all x ∈ g,
and hence that ad(y) = 0.)
(v) Show that if the Killing form of g is non-degenerate, then der(g) = ad(g). This shows
that all of the automorphisms of a simple Lie algebra are “inner”. (Hint: Show that

E.3.1 57
the set p = a ∈ der(g) | tr a ad(x) = 0 for all x ∈ g is also an ideal in der(g) and
hence that der(g) = p • ad(g) as algebras. Show that this forces p = 0 by considering
what it means for elements of p (which, after all, are derivations of g) to commute
with elements in ad(g).)

4. Consider the 1-parameter group which is generated by the ¬‚ow of the vector ¬eld X in
the plane
‚ ‚
+ sin2 y
X = cos y .
‚x ‚y
Show that this vector ¬eld is complete and hence yields a free R-action on the plane. Let
Z also act on the plane by the action

m · (x, y) = ((’1)m x, y + mπ) .

Show that these two actions commute, and hence together de¬ne a free action of G = R—Z
on the plane. Sketch the orbits and show that, even though the G-orbits of this action are
closed, and the quotient space is Hausdor¬, the quotient space is not a manifold. (The
point of this problem is to warn the student not to make the common mistake of thinking
that the quotient of a manifold by a free Lie group action is a manifold if it is Hausdor¬.)

5. Show that if ρ: M — G ’ M is a right action, then the induced map ρ— : g ’ X(M)
satis¬es ρ— [x, y] = ρ— (x), ρ— (y) .

6. Prove Proposition 2. (Hint: you are trying to ¬nd an open neighborhood U of {e} — M
in G — M and a smooth map »: U ’ M with the requisite properties. To do this, look for
the graph of » as a submanifold “ ‚ G — M — M which contains all the points (e, m, m)
and is tangent to a certain family of vector ¬elds on G — M — M constructed using the
left invariant vector ¬elds on G and the corresponding vector ¬elds on M determined by
the Lie algebra homomorphism φ: g ’ X(M).)

7. Show that, if A: R ’ g is a curve in the Lie algebra of a Lie group G, then there exists
a unique solution to the ordinary di¬erential equation S (t) = RS(t) A(t) with initial
condition S(0) = e. (It is clear that a solution exists on some interval (’µ, µ) in R. The
problem is to show that the solution exists on all of R.)

8. Show that, under the action of GL(n, R) on the space of symmetric n-by-n matrices
de¬ned in the Lecture, every symmetric n-by-n matrix is in the orbit of an Ip,q .

9. This problem examines the geometry of the classical second order equation for one
(i) Rewrite the second-order ODE
d2 x
= F (t)x
as a system of ¬rst-order ODEs of Lie type for an action of SL(2, R) on R2 .

E.3.2 58
(ii) Suppose in particular that F (t) is of the form f(t) + f (t), where f(0) = 0. Use
the solution
x(t) = exp f(„ ) d„

to write down the fundamental solution for this Lie equation up in SL(2, R).
(iii) Explain why the (more general) second order linear ODE

x = a(t)x + b(t)x

is solvable by quadratures once we know a single solution with either x(0) = 0 or
x (0) = 0. (Hint: all two-dimensional Lie groups are solvable.)

10*. Show that the general equation of the form y (x) = f(x) y(x) is not integrable by
quadratures. Speci¬cally, show that there do not exist “universal” functions F0 and F1
of two and three variables respectively so that the function y de¬ned by taking the most
general solution of
u (x) = F0 x, f(x)
y (x) = F1 x, f(x), u(x)
is the general solution of y (x) = f(x) y(x). Note that this shows that the general solution
cannot be got by two quadratures, which one might expect to need since the general
solution must involve two constants of integration. However, it can be shown that no
matter how many quadratures one uses, one cannot get even a particular solution of
y (x) = f(x) y(x) (other than the trivial solution y ≡ 0) by quadrature. (If one could get
a (non-trivial) particular solution this way, then, by two more quadratures, one could get
the general solution.)

11. The point of this exercise is to prove Lie™s theorem (stated below) on (local) group
actions on R. This theorem “explains” the importance of the Riccati equation, and why
there are so few actions of Lie groups on R. Let g ‚ X(R) be a ¬nite dimensional Lie
algebra of vector ¬elds on R with the property that, at every x ∈ R, there is at least one
X ∈ g so that X(x) = 0. (Thus, the (local) ¬‚ows of the vector ¬elds in g do not have any
common ¬xed point.)
(i) For each x ∈ R, let gk ‚ g denote the subspace of vector ¬elds which vanish to order
at least k + 1 at x. (Thus, g’1 = g for all x.) Let g∞ ‚ g denote the intersection
x x

of all the gx . Show that gx = 0 for all x. (Hint: Fix a ∈ R and choose an X ∈ g

so that X(a) = 0. Make a local change of coordinates near a so that X = ‚/‚x on
a neighborhood of a. Note that [X, g∞ ] ‚ g∞ . Now choose a basis Y1 , . . . , YN of g∞
a a a
and note that, near a, we have Yi = fi ‚/‚x for some functions fi . Show that the fi
must satisfy some di¬erential equations and then apply ODE uniqueness. Now go on
from there.)

* This exercise is somewhat di¬cult, but you should enjoy seeing what is involved in
trying to prove that an equation is not solvable by quadratures.

E.3.3 59
(ii) Show that the dimension of g is at most 3. (Hint: First, show that [gj , gk ] =‚ gj+k .
xx x
Now, by part (i), you know that there is a smallest integer N (which may depend on
x) so that gN +1 = 0. Show that if X ∈ g does not vanish at x and YN ∈ g vanishes to
exactly order N at x, then YN ’1 = [X, YN ] vanishes to order exactly N ’ 1. Conclude
that the vectors X, Y0 , . . . , YN (where Yi’1 = [X, Yi ] for i > 0) form a basis of g. Now,
what do you know about [YN ’1 , YN ]?).
(iii) (Lie™s Theorem) Show that, if dim(g) = 2, then g is isomorphic to the (unique) non-
abelian Lie algebra of that dimension and that there is a local change of coordinates
so that
g = {(a + bx)‚/‚x | a, b ∈ R}.
Show also that, if dim(g) = 3, then g is isomorphic to sl(2, R) and that there exist
local changes of coordinates so that

g = {(a + bx + cx2 )‚/‚x | a, b, c ∈ R}.

(In the second case, after you have shown that the algebra is isomorphic to sl(2, R),
show that, at each point of R, there exists a element X ∈ g which does not vanish at
the point and which satis¬es ad(X) = 0. Now put it in the form X = ‚/‚x for
some local coordinate x and ask what happens to the other elements of g.)
(iv) (This is somewhat harder.) Show that if dim(g) = 3, then there is a di¬eomorphism
of R with an open interval I ‚ R so that g gets mapped to the algebra

a, b, c ∈ R
g= (a + b cos x + c sin x) .

In particular, this shows that every local action of SL(2, R) on R is the restriction
of the M¨bius action on RP1 after “lifting” to its universal cover. Show that two
intervals I1 = (0, a) and I2 = (0, b) are di¬eomorphic in such a way as to preserve
the Lie algebra g if and only if either a = b = 2nπ for some positive integer n or else
2nπ < a, b < (2n + 2)π for some positive integer n. (Hint: Show, by a local analysis,
that any vector ¬eld X ∈ g which vanishes at any point of R must have κ(X, X) ≥ 0.
Now choose an X so that κ(X, X) = ’2 and choose a global coodinate x: R ’ R
so that X = ‚/‚x. You must still examine the e¬ect of your choices on the image
interval x(R) ‚ R.)
Lie and his coworkers attempted to classify all of the ¬nite dimensional Lie subalgebras
of the vector ¬elds on Rk , for k ¤ 5, since (they thought) this would give a classi¬cation
of all of the equations of Lie type for at most 5 unknowns. The classi¬cation became
extremely complex and lengthy by dimension 5 and it was abandoned. On the other hand,
the project of classifying the abstract ¬nite dimensional Lie algebras has enjoyed a great
deal of success. In fact, one of the triumphs of nineteenth century mathematics was the
classi¬cation, by Killing and Cartan, of all of the ¬nite dimensional simple Lie algebras
over C and R.

E.3.4 60
Lecture 4:
Symmetries and Conservation Laws

Variational Problems. In this Lecture, I will introduce a particular set of variational
problems, the so-called “¬rst-order particle Lagrangian problems”, which will serve as a
link to the “symplectic” geometry to be developed in the next Lecture.
De¬nition 1: A Lagrangian on a manifold M is a smooth function L: T M ’ R. For any
smooth curve γ: [a, b] ’ M, de¬ne
FL (γ) = L γ(t) dt.


FL is called the functional associated to L.
(The use of the word “functional” here is classical. The reader is supposed to think
of the set of all smooth curves γ: [a, b] ’ M as a sort of in¬nite dimensional manifold and
of FL as a function on it.)
I have deliberately chosen to avoid the (mild) complications caused by allowing less
smoothness for L and γ, though for some purposes, it is essential to do so. The geometric
points that I want to make, however will be clearest if we do not have to worry about
determining the optimum regularity assumptions.
Also, some sources only require L to be de¬ned on some open set in T M. Others
allow L to “depend on t”, i.e., take L to be a function on R — T M. Though I will not
go into any of these (slight) extensions, the reader should be aware that they exist. For
example, see [A].

Example: Suppose that L: T M ’ R restricts to each Tx M to be a positive de¬nite
quadratic form. Then L de¬nes what is usually called a Riemannian metric on M. For a
curve γ in M, the functional FL (γ) is then twice what is usually called the “action” of γ.
This example is, by far, the most commonly occurring Lagrangian in di¬erential geometry.
We will have more to say about this below.

For a Lagrangian L, one is usually interested in ¬nding the curves γ: [a, b] ’ M with
given “endpoint conditions” γ(a) = p and γ(b) = q for which the functional FL (γ) is a
minimum. For example, in the case where L de¬nes a Riemannian metric on M, the curves
with ¬xed endpoints of minimum “action” turn out also to be the shortest curves joining
those endpoints. From calculus, we know that the way to ¬nd minima of a function on a
manifold is to ¬rst ¬nd the “critical points” of the function and then look among those
for the minima. As mentioned before, the set of curves in M can be thought of as a sort
of “in¬nite dimensional” manifold, but I won™t go into details on this point. What I will
do instead is describe what ought to be the set of “curves” in this space (classically called
“variations”) if it were a manifold.

L.4.1 61
Given a curve γ: [a, b] ’ M, a (smooth) variation of γ with ¬xed endpoints is, by
de¬nition, a smooth map
“ : [a, b] — (’µ, µ) ’ M


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