A vector field on a manifold assigns one tangent vector to each point, and the assignment varies smoothly from point to point.
A fluid is a good picture to keep in mind. The lecture notes of F. Jean suggest thinking of X as the velocity field of a fluid moving on M [1]: the arrow at each point is the velocity of the fluid there.
Once there is an arrow at every point, particles begin to move along the arrows. In the figure the arrows appear first; then particles are released among them, follow the arrows and approach the unit circle.
Formally, a vector field is a smooth section of the tangent bundle TM [2]. It assigns to each point p an element of T_pM, and the assignment is required to be smooth.
In a chart it is written with components. With the coordinate basis it takes the following form [1].
X(x) = \sum_{i=1}^{n} X^{i}(x) \, \frac{\partial}{\partial x^{i}}
X is smooth exactly when the component functions X^i are smooth [2]. The coordinate basis vectors are vector fields themselves: assigning \partial/\partial x^i to each point gives the coordinate fields [1].
The figure shows a saddle surface with its coordinate curves. At the moving point, \partial_1 and \partial_2 are tangent to the coordinate curves, and X is the diagonal of the parallelogram spanned by X^1 \partial_1 and X^2 \partial_2 in the tangent plane.
Tangent vectors were differential operators, so a vector field becomes an operator that sends functions to functions. The value of X \cdot f at a point is the rate at which f changes in the direction of X there.
This action is linear and satisfies the Leibniz rule [1]. An operator with these two properties is called a derivation, and vector fields can be identified with derivations.
X \cdot (fg) = f \, (X \cdot g) + g \, (X \cdot f)
Seen as operators, vector fields can be added and multiplied by scalars in the obvious way, and the set \mathfrak{X}(M) is an infinite-dimensional vector space.
In the figure the thin curves are level curves of f, and the shading is the new function X \cdot f: colored where f increases along X and grey where it decreases. Where the arrows run along the level curves, X \cdot f is 0, and this boundary moves as X changes.
Giving a vector field is the same as writing an ordinary differential equation on the manifold [1].
\dot{q} = X(q), \qquad q \in M
Its solutions are called integral curves: curves c(t) with \dot{c}(t) = X(c(t)) at every time [1]. Existence and uniqueness hold locally. For every point p there is an \eta > 0 such that exactly one integral curve defined for t \in (-\eta, \eta) starts from c_p(0) = p [1].
The proof is short: in a chart the equation becomes an ordinary differential equation in \mathbb{R}^n [1]. As long as everything stays local, nothing specific to manifolds appears.
The figure uses the pendulum equation \dot{x} = y, \dot{y} = -\sin x. Integral curves grow one at a time from different starting points, always tangent to the short segments of the field, and by uniqueness no two of them ever cross.
The dependence on the initial condition is smooth as well. For a neighborhood U_p of p and a time interval I, the map \varphi(t, q) = c_q(t) is smooth [1].
For each fixed t, the map \varphi_t : q \mapsto \varphi(t, q) is a local diffeomorphism. This family is called the flow of X [1].
In the figure a small square with a grid on it is carried by a flow that turns faster near the center. The square is stretched into a spiral but never torn or folded over, and flowing back by \varphi_{-t} returns it exactly to where it started.
The flow forms a local one-parameter group [1]. Each \varphi_t is a local diffeomorphism, t \mapsto \varphi_t is smooth, \varphi_0 is the identity, and the following holds.
\varphi_t \circ \varphi_s = \varphi_{t+s}
The last property is the essential one. Flowing for s and then for t is the same operation as flowing for t + s: addition of times has turned into composition of maps.
In the figure the flow turns and contracts, so its time-1 map rotates and shrinks. The images \varphi_1(U), \varphi_2(U), \varphi_3(U) line up along a logarithmic spiral, and applying \varphi_1 to \varphi_k(U) always gives \varphi_{k+1}(U).
Flows are carried along by diffeomorphisms. If F is a diffeomorphism, the flow of F_{*}X is F \circ \varphi_t \circ F^{-1}, the conjugate of the original flow [1].
Applying two vector fields one after the other, XY, does not give a vector field. Second derivatives appear, and the Leibniz rule fails [1].
Taking the difference, however, the second-order terms cancel [1].
[X, Y] = XY - YX
Checking the Leibniz rule is short: expand XY \cdot (fg), subtract YX \cdot (fg), and the extra terms cancel [1]. In components, each field differentiates the components of the other, and the difference is taken [1].
[X, Y]^{i} = \sum_j \left( X^{j} \frac{\partial Y^{i}}{\partial x^{j}} - Y^{j} \frac{\partial X^{i}}{\partial x^{j}} \right)
The figure expands XY f and YX f, summing over repeated indices. The terms with second derivatives of f agree, because partial derivatives can be taken in either order, so they cancel in the difference and only the components above remain.
What the bracket measures becomes visible when flows are combined. Flow along X_1 for time t, then along X_2 for time t, then back along X_1 and back along X_2, each for time t.
The path need not return to where it started. The gap is the following [1].
\varphi^{2}_{-t} \circ \varphi^{1}_{-t} \circ \varphi^{2}_{t} \circ \varphi^{1}_{t}(p) - p = t^2 [X_1, X_2](p) + o(t^2)
The gap is of order t^2. The first-order terms cancel, so at first sight the path seems to close.
In the figure X_1 is a translation and X_2 a rotation, so the four legs are two segments and two arcs. As t shrinks the gap shrinks like t^2, and the gap divided by t^2 approaches the vector [X_1, X_2](p).
Take X_1 = \partial_x and X_2 = x \, \partial_y on the plane. Their flows are \varphi^1_t(x, y) = (x + t, y) and \varphi^2_s(x, y) = (x, y + sx).
Starting from the origin, the point moves to (t, 0), (t, t^2), (0, t^2), and finally stays at (0, t^2). It does not return to the origin but ends up t^2 higher.
The bracket is [X_1, X_2] = \partial_y, which is exactly the direction of the gap.
The figure moves the whole plane together with the point. The second leg shears the plane and the fourth leg shears it back, and after the four legs every point of the plane, not only the origin, has moved up by t^2.
The gap is the bracket, and the two flows commute exactly when the bracket is 0 [1]. Coordinate fields commute: [\partial_i, \partial_j] = 0, which says again that partial derivatives can be taken in either order.
How long the flow is defined depends on the point. A vector field whose integral curves are all defined for every t \in \mathbb{R} is called complete.
Incomplete fields are easy to find. On \mathbb{R} take X = x^2 \, \partial_x, so the equation is \dot{x} = x^2. The solution with x(0) = 1 is x(t) = 1/(1 - t), which diverges as t \to 1: it reaches infinity in finite time and cannot be continued beyond it.
The figure draws the whole line in a finite width, with infinity at the two edges, and lets time run upward. The path of a particle starting at x_0 > 0 reaches the right edge at the finite time 1/x_0, so the farther out it starts, the sooner it leaves, while the paths on the negative side only approach 0.
On a compact manifold this cannot happen: there is nowhere to escape to, so every vector field is complete. For a complete field the local group becomes an actual group. Each \varphi_t is a diffeomorphism of the whole manifold, and t \mapsto \varphi_t is a group homomorphism from \mathbb{R}.
Linear vector fields are a good example. On \mathbb{R}^n take X(x) = Ax; the equation is \dot{x} = Ax.
The solution is given by the matrix exponential, and the flow is \varphi_t(x) = e^{tA}x. It is defined for every t, so X is complete. Each \varphi_t is a linear map, so in the figure the circle of points is carried to an ellipse at every moment, while each point follows its own spiral.
The bracket of two linear fields X_A and X_B produces the commutator of the matrices [1]: the bracket of the vector fields corresponds to BA - AB.
This is the simplest appearance of the relation between Lie groups and Lie algebras. The failure of flows to commute becomes the failure of matrices to commute.
The bracket is not destroyed by carrying it along a map. If F : M \to N is a diffeomorphism, the following holds [1].
F_{*}[X, Y] = [F_{*}X,\ F_{*}Y]
So the bracket is a structure that does not depend on coordinates, and with it \mathfrak{X}(M) becomes a Lie algebra.
Carrying a vector field along a map requires a diffeomorphism [1]. If the map is not injective, two values compete at the same image point; if it is not surjective, some points receive no value at all.
In the figure the grid and its arrows are bent little by little. At the marked point the carried bracket and the bracket of the carried fields stay on top of each other. At the end the map develops a fold, and two arrows with different directions arrive at the same point.
The bracket can also be read as the rate of change of Y seen while moving with the flow of X.
Take the value of Y at time t along the flow and pull it back to the starting point. Differentiating the result in t defines the Lie derivative of Y along X.
\mathcal{L}_X Y = \frac{d}{dt}\bigg|_{t=0} \bigl( \varphi_{-t} \bigr)_{*} Y
This quantity equals [X, Y], so the bracket is the derivative that measures change along the flow. In the figure the vectors of Y at the points \varphi_t(p) are pulled back to p one after another. Their tips trace a curve, and its velocity at t = 0 is \mathcal{L}_X Y at p.
Unlike the covariant derivative, no connection is needed. The Lie derivative is determined by the flow alone, so it can be used on manifolds with neither a metric nor a connection.
In exchange, \mathcal{L}_X Y is not determined by X_p alone; it depends on X in a neighborhood of p. Here X does more than point in a direction, and in this it differs from the covariant derivative.
When an arrow is attached to every point, can the zero vector be avoided everywhere? The answer depends on the shape of the manifold.
On odd-dimensional spheres it can. Regard S^{2n-1} as a subset of \mathbb{R}^{2n} with coordinates (x^1, y^1, \ldots, x^n, y^n). Then the following field vanishes nowhere [2].
X = \sum_{i=1}^{n} \left( -y^{i} \frac{\partial}{\partial x^{i}} + x^{i} \frac{\partial}{\partial y^{i}} \right)
It only rotates each coordinate plane. It is orthogonal to the position vector, so it is tangent to the sphere, and its length is 1, so it is never 0. For n = 1 it is the unit tangent field of the circle S^1.
For n = 2 every integral curve is a great circle of S^3. The figure sends S^3 into space by stereographic projection, where these circles become a family of circles, any two of which are linked, and a particle runs along each of them.
On even-dimensional spheres it cannot: every vector field has a zero. This is the statement of the hairy ball theorem [2].
The figure draws the arrows of a field on S^2 densely and changes the field continuously. Its two zeros approach each other, merge into one and separate again, but however the field is changed, a zero never disappears.
A stronger question: can one find as many vector fields as the dimension, linearly independent at every point, so that they form a basis of every tangent space?
A manifold where this is possible is called parallelizable. Among the spheres, only S^1, S^3 and S^7 are parallelizable [1]. The even-dimensional spheres are all excluded: they do not carry even one field without zeros, let alone a basis [1].
S^3 and S^7 are on the list because of the multiplication of quaternions and octonions. The existence of an algebra shows up as a geometric property.
For S^1 the figure turns the tangent lines upright one after another. They line up into a cylinder, so the tangent bundle is the product S^1 \times \mathbb{R}, and the field without zeros becomes a horizontal circle on the cylinder.
If the field is tangent to a submanifold, the flow stays inside it. Let N \subset M be a submanifold with X(q) \in T_qN at every point q of N. Then an integral curve starting at a point of N stays in N, at least for a short time [1].
The reason is uniqueness. The restriction of X to N is a vector field on N, so its solutions are also solutions in M. There is only one solution, so the two coincide [1].
This is also why, in a dynamical system with a conserved quantity, the orbits do not leave a level set. If the energy stays constant, the velocity is tangent to the level surface of the energy.
The figure shows the equation of a freely rotating rigid body. The angular momentum vector m keeps its length, so the field is tangent to every sphere |m| = c, and each orbit stays on the sphere it started from.