INTRODUCTION TO MANIFOLDS — III

Algebra of vector fields. Lie derivative(s).

1. Notations. The space of all C^{∞}-smooth vector fields on a manifold M is denoted by X(M). If v ∈ X(M) is a vector field, then v(x) ∈ T_{x}M ≃ ℝ^{n} is its value at a point x ∈ M.

The flow of a vector field v is denoted by v^{t}:

∀t ∈ ℝ v^{t}: M → M

is a smooth map (automorphism) of M taking a point x ∈ M into the point v^{t}(x) ∈ M which is the t-endpoint of an integral trajectory for the field v, starting at the point x.

♣ Problem 1. Prove that the flow maps for a field v on a compact manifold M form a one-parameter group:

∀t, s ∈ ℝ v^{t+s} = v^{t} ∘ v^{s} = v^{s} ∘ v^{t},

and all v^{t} are diffeomorphisms of M.

♣ Problem 2. What means the formula

\frac{d}{ds}|_{s=0} v^{s} = v

and is it true?

2. Star conventions. The space of all C^{∞}-smooth functions is denoted by C^{∞}(M). If F: M → M is a smooth map (not necessary a diffeomorphism), then there appears a contravariant map

F^{∗}: C^{∞}(M) → C^{∞}(M), F^{∗}: f ↦ F^{∗}f, F^{∗}(x) = f(F (x)).

If F: M → N is a smooth map between two different manifolds, then

F^{∗}: C^{∞}(N) → C^{∞}(M).

Note that the direction of the arrows is reversed!

♣ Problem 3. Prove that C^{∞}(M) is a commutative associative algebra over ℝ with respect to pointwise addition, subtraction and multiplication of functions. Prove that F^{∗} is a homomorphism of this algebra (preserves all the operations). If F: M → N, then F^{∗}: C^{∞}(N) → C^{∞}(M) is a homomorphism also.

Typeset by AMS-TEX

Another star is associated with differentials: if F: M_{1} → M_{2} is a diffeomor-

phism, then

F_{∗}: X(M_{1}) → X(M_{2}), v ↦ F_{∗}v, (F_{∗}v)(x) = \frac{∂F}{∂x} (x) · v(x),

is a covariant (acts in the same direction) map which is:

(1) additive: F_{∗}(v + w) = F_{∗}v + F_{∗}w;

(2) homogeneous: ∀f ∈ C^{∞}(M) F_{∗}(fv) = (F^{∗})^{−1}f · F_{∗}v. (explain this for-

mula!), Why F_{∗} is in general not defined, if F is just a smooth map and not a diffeomorphism?

3. Vector fields as differential operators.

♡ Definition. If v ∈ X(M), then the Lie derivative L_{v} is

L_{v}: C^{∞}(M) → C^{∞}(M), L_{v}f = lim_{t→0} \frac{1}{t} ((v^{t})^{∗}f − f) .

In coordinates:

L_{v}f(a) = lim_{t→0} \frac{f(a + tv + o(t)) − f(a)}{t} = ∑_{j=1}^{n} \frac{∂f}{∂x_{j}} (a)v_{j}.

Properties of the Lie derivative:

(1) L_{v}: C^{∞}(M) → C^{∞}(M) is a linear operator:

L_{v}(f + g) = L_{v}f + L_{v}g, L_{v}(λf) = λL_{v}f;

   (2) the Leibnitz identity holds:

L_{v}(fg) = L_{v}f · g + f · L_{v}g.

(3) The Lie derivative linearly depends on v:

∀f ∈ C^{∞}(M), v, w ∈ X(M)  L_{fv} = fL_{v},  L_{v+w} = L_{v} + L_{w}.

♣ Problem 4. Prove that the Lie derivative is local: for any function f ∈ C^{∞}(M) and any vector field v the value L_{v}f(a) depends only on v(a), so that for any other field w such that w(a) = v(a), L_{v}f(a) = L_{w}f(a).

Theorem. Any differential operator, that is, a map D: C^{∞}(M) → C^{∞}(M) satisfying

D(f + g) = Df + Dg, D(λf) = λDf, D(fg) = f Dg + Df · g, (DiffOper)

is a Lie derivative along a certain vector field v ∈ X.

Idea of the proof. In local coordinates any function can be written as

f(x) = f(a) + ∑_{k=1}^{n}(x_{k} − a_{k})f_{k}(x), f_{k}(a) = \frac{∂f}{∂x_{k}} (a).

Applying the Leibnitz identity, we conclude that D = L_{v}, where v is the vector field with components v_{k} = D(x_{k} − a_{k}). □

Thus sometimes the notation

v = ∑_{k=1}^{n} v_{k}(x)\frac{∂}{∂x_{k}}

is used: such a notation understood as a differential operator, is a vector field from the geometric point of view.

4. Commutator. If v, w ∈ X(M), then D = L_{v}L_{w} − L_{w}L_{v} is a differential operator. Indeed, the Leibnitz formula is trivially satisfied, therefore D = L_{u}, where u ∈ X(M).

♣ Problem 5. Check it!

♡ Definition. If L_{u} = L_{v}L_{w} − L_{w}L_{v}, then u is a commutator of v and w:

u = [v, w].

In coordinates:

L_{u}f = L_{v} (∑_{k} \frac{∂f}{∂x_{k}} w_{k}) − L_{w}(⋯) =

∑_{k,j} (\frac{∂^{2}f}{∂x_{k}∂x_{j}} w_{k}v_{j} + \frac{∂f}{∂x_{k}} \frac{∂w_{k}}{∂x_{j}} v_{j}) − (⋯) =

∑_{j} (∑_{k} \frac{∂w_{j}}{∂x_{k}} v_{k} − ∑_{k} \frac{∂v_{j}}{∂x_{k}} w_{k}) \frac{∂f}{∂x_{j}} ,

therefore

[v, w] = ∑_{j} (∑_{k} \frac{∂w_{j}}{∂x_{k}} v_{k} − ∑_{k} \frac{∂v_{j}}{∂x_{k}} w_{k}) \frac{∂}{∂x_{j}} .

♣ Problem 6.

\frac{∂^{2}}{∂s∂t}|_{s=0,t=0} (f ∘ v^{t} ∘ w^{s} − f ∘ w^{s} ∘ v^{t}) = L_{[v,w]}f.

♣ Problem 7.

[v, w] = −[w, v].

♣ Problem 8. Prove the Jacobi identity

[[u, v], w] + [[v, w], u] + [[w, u], v] = 0.

5. Lie derivation of vector fields.

♡ Definition. The Lie derivative of a vector field w along another field v is

L_{v}w = lim_{t→0} \frac{1}{t}(v_{∗}^{t}w − w ∘ v^{t}).

♣ Problem 9. Check that the above definition makes sense.

Properties of the Lie derivative: if v, w ∈ X(M), f ∈ C^{∞}(M), then: (1) L_{v}v = 0.

(2) L_{v} is linear map from X(M) to itself.

(3) L_{v}(fw) = (L_{v}f)w + f L_{v}w (the Leibnitz property).

Theorem.

L_{v}w = [v, w] (or [w, v]?)

Proof. Let

a = \frac{∂^{2}}{∂s∂t}|_{s=0,t=0} (f ∘ v^{t} ∘ w^{s} − f ∘ w^{s} ∘ v^{t}).

Then

a = lim_{t→0} \frac{1}{t} ( \frac{∂}{∂s}|_{s=0} (⋯)) ,

but

\frac{∂}{∂s}|_{s=0} v^{t} ∘ w^{s} = v_{∗}^{t}w,

therefore

\frac{∂}{∂s}|_{s=0} f ∘ v^{t} ∘ w^{s} = L_{v_{∗}^{t}w}f,

while

\frac{∂}{∂s}|_{s=0} f ∘ w^{s} ∘ v^{t} = L_{w∘v^{t}}f,

and finally

a = L_{L_{v}w}f. □

♣ Problem 10. Is the Lie derivative of a vector field local in the following sense: if two fields v_{1}, v_{2} ∈ X(M) are coinciding on an open neighborhood of a certain point a ∈ M, then for any other field v ∈ X(M)

(L_{v_{1}}w)(a) = (L_{v_{2}}w)(a).

Is it true that the above value is determined by the (common) value v_{i}(a)?

   /black/users2/yakov/pub, filename 4.ppt

E-mail address: yakov@wisdom.weizmann.ac.il, mtwiener@weizmann.weizmann.ac.il