Vector Fields

A vector field is a section of the tangent bundle \(TM\) on a smooth manifold \(M\), that is, a continuous map \(X : M \rightarrow TM\) such that \(\pi \circ X = \mathrm{Id}_M\) (where \(\pi : TM \rightarrow M\) is the projection of the tangent bundle). Thus, it is a mapping \(p \mapsto X_p\) that assigns each point \(p \in M\) to a tangent vector \(X_p \in T_pM\). We are typically only interested in smooth vector fields, which are vector fields \(X : M \rightarrow TM\) that are smooth maps.

Coordinate Functions

Given any smooth chart \((U,x^i)\) on a smooth manifold \(M\), each tangent vector \(X_p\) can be written as a linear combination of the coordinate basis vectors \(\partial/\partial x^i\rvert_p\), and we denote the coordinates as \(X^i(p)\), which implies the existence of coordinate functions \(X^i : U \rightarrow \mathbb{R}\):

\[X_p = X^i(p)\frac{\partial}{\partial x^i}\bigg\rvert_p.\]

Smoothness

Next, let's work out what smoothness means for a vector field \(X : M \rightarrow TM\). Given a smooth chart \((U,\varphi)\) on \(M\) and the induced natural chart \((\pi^{-1}(U),\tilde{\varphi})\) on \(TM\), we have, for each \(x \in \widehat{U}\),

\begin{align}\widehat{X}(x) &= \tilde{\varphi} \circ X \circ \varphi^{-1}(x)\\&= \tilde{\varphi}\left(X_{\varphi^{-1}(x)}\right)\\&= \tilde{\varphi}\left(X^i(\varphi^{-1}(x))\frac{\partial}{\partial x^i}\bigg\rvert_{\varphi^{-1}(x)}\right)\\&= (x^1(\varphi^{-1}(x)),\dots,x^n(\varphi^{-1}(x)),X^1(\varphi^{-1}(x)),\dots,X^n(\varphi^{-1}(x)))\\&= (x^1,\dots,x^n,X^1(\varphi^{-1}(x)),\dots,X^n(\varphi^{-1}(x))).\end{align}

Thus, smoothness of \(X\) means that

\[(x^1,\dots,x^n,X^1(\varphi^{-1}(x)),\dots,X^n(\varphi^{-1}(x)))\]

is smooth (for each chart), and thus \(X\) is smooth if and only if each of its component functions is smooth.

Algebraic Properties

The set of all smooth vector fields on a smooth manifold \(M\) is denoted \(\mathfrak{X}(M)\).

This set has several important algebraic properties.

Vector Space Structure

\(\mathfrak{X}(M)\) is a vector space over \(\mathbb{R}\) under pointwise addition and scalar multiplication:

• \((X + Y)_p = X_p + Y_p\) for all vector fields \(X,Y \in \mathfrak{X}(M)\) and points \(p \in M\).
• \((aX)_p = a(X_p)\) for all vector fields \(X \in \mathfrak{X}(M)\) and scalars \(a \in \mathbb{R}\).

Module Structure

For an smooth function \(f \in C^{\infty}(M)\) and vector field \(X \in \mathfrak{X}(M)\), define the product of \(f\) with \(X\) to be the following vector field:

\[(fX)_p = f(p)X_p.\]

Then, using this operation, \(\mathfrak{X}(M)\) is a module over the ring \(C^{\infty}(M)\).

• \begin{align}(f(X + Y))_p &= f(p)(X + Y)_p\\&= f(p)(X_p + Y_p)\\&= f(p)X_p + f(p)Y_p\\&= (fX)_p + (fY)_p.\end{align}
• \begin{align}((f+g)X)_p &= (f+g)(p)X_p\\&= (f(p)+g(p))X_p\\&= f(p)X_p + g(p)X_p\\&= (fX)_p + (gX)_p.\end{align}
• \begin{align}((fg)X)_p &= (fg)(p)X_p\\&= f(p)g(p)X_p\\&= f(p)(gX)_p\\&= (f(gX))_p.\end{align}
• \begin{align}(1X)_p &= 1(p)X_p\\&= X_p.\end{align}

Given a smooth chart \((U,(x^i))\), the \(i\)-th coordinate vector field

\[\frac{\partial}{\partial x^i}\]

is the mapping

\[p \mapsto \frac{\partial}{\partial x^i}\bigg\rvert_p.\]

Thus, using the module structure of \(\mathfrak{X}(M)\), every vector field can be written as a combination of coordinate vector fields (at least when restricted to the respective coordinate domain):

\[X = X^i \frac{\partial}{\partial x^i}.\]

Frames

The coordinate vector fields are an example of a local frame. An ordered tuple of vector fields \((X_1,\dots,X_k)\) defined on some subset \(A\) of a smooth manifold \(M\) is linearly independent if \((X_1\rvert_p,\dots,X_k\rvert_p)\) is linearly independent in \(T_pM\) for every \(p \in A\), and spans the tangent bundle if \((X_1\rvert_p,\dots,X_k\rvert_p)\) spans \(T_pM\) for every \(p \in A\). For a smooth \(n\)-manifold \(M\), an ordered \(n\)-tuple of vector fields \(E_1,\dots,E_n\) defined on an open subset \(U \subseteq M\) is a local frame for \(M\) if \(E_1,\dots,E_n\) is linearly independent and spans the tangent bundle. A global frame is a frame defined on the entire manifold.

Thus, the coordinate vector fields \((\frac{\partial}{\partial x^i})\) comprise a local frame called the coordinate frame.

Derivations

In the post about tangent spaces, we defined tangent vectors as derivations at a point. If tangent vectors are point-wise derivations, the tangent vector fields should be "point-free" derivations.; that is, maps \(C^{\infty}(M) \rightarrow \mathbb{R}\) get extended to maps \(C^{\infty}(M) \rightarrow C^{\infty}(M)\) satisfying analogous, "point-free" properties. A derivation is a map \(X : C^{\infty}(M) \rightarrow C^{\infty}(M)\) that is linear over \(\mathbb{R}\) and satisfies the following product rule for all \(f,g \in C^{\infty}(M)\):

\[X(fg) = f~Xg + g~Xf.\]

If we define the operation \(Xf\) for a smooth vector field \(X\) and smooth function \(f \in C^{\infty}(M)\) for each \(p \in M\) as

\[(Xf)(p) = X_pf,\]

then each vector field is a derivation, since

• \begin{align}X(af)(p) &= X_p(af)\\&= aX_pf\\&= a(Xf)(p).\end{align}
• \begin{align}X(f+g)(p) \\&= X_p(f+g)\\&= X_pf + X_pg\\&= (Xf)(p) + (Xg)(p).\end{align}
• \begin{align}X(fg)(p) &= X_p(fg)\\&= f(p)X_pg + g(p)X_pf\\&= f(p)(Xg)(p) + g(p)(Xf)(p)\\&= (f~Xg)(p) + (g~Xf)(p).\end{align}

More generally, a derivation is a function \(D : A \rightarrow A\) on an algebra \(A\) over a field \(K\) which is a \(K\)-linear map that satisfies \(D(ab) = aD(b) + bD(a)\) for all \(a,b \in A\). For vector fields, the algebra is the ring \(C^{\infty}(M)\) which is an algebra over the field \(\mathbb{R}\). The notion of derivation plays a central role in differential geometry. Tangent vectors are defined as point-wise derivations, vector fields are derivations, covariant derivatives are defined as a variant of a derivation, and the exterior derivative is defined as a variant of a derivation. Essentially every notion of "derivative" is defined using some variation of the concept of a derivation.

We have demonstrated that every vector field gives rise to a derivation. Now we will show that every derivation gives rise to a vector field. Given a derivation \(D : C^{\infty}(M) \rightarrow C^{\infty}(M)\), we define a vector field \(X\) such that

\[X_pf = (Df)(p).\]

We must verify that \(X_p\) is indeed a tangent vector for each \(p \in M\):

• \begin{align}X_p(af) &= D(af)(p)\\&= aD(f)(p)\\&= aX_pf.\end{align}
• \begin{align}X_p(f+g) &= D(f+g)(p)\\&= (D(f)+D(g))(p)\\&= D(f)(p) + D(g)(p)\\&= X_pf + X_pg.\end{align}
• \begin{align}X_p(fg) &= D(fg)(p)\\&= (f~Dg + g~Df)(p)\\&= (f~Dg)(p) + (g~Df)(p)\\&= f(p)(Dg)(p) + g(p)(Df)(p)\\&= f(p)X_pf + g(p)X_pg.\end{align}

Thus, we may identify derivations of \(C^{\infty}(M)\) with smooth vector fields on \(M\). However, we must verify that the vector field \(X\) so defined is smooth.

First, suppose that, for each \(f \in C^{\infty}(M)\), the function \(Xf\) is smooth on \(M\). Let \(U \subseteq M\) be any open subset, and suppose \(f \in C^{\infty}(U)\). Choose any smooth bump function \(\psi\) that is equal to \(1\) on a neighborhood \(A\) of any point \(p \in U\) and supported in \(U\), and let \(\tilde{f} = \psi f\) extended to \(0\) on \(M \setminus \mathrm{supp}(\psi)\). Note that, since tangent vectors are locally determined, for any subset \(V \subseteq U\), \((Xf)\rvert_V = X(f\rvert_V)\). Thus, in particular, \(X\tilde{f} = Xf\) on the neighborhood \(A\), and so, since \(X\tilde{f}\) is smooth, \(Xf\) is smooth on a neighborhood of every point of \(U\), and is therefore smooth.

Next, suppose that the function \(Xf\) is smooth for every \(f \in C^{\infty}(U)\) on every open subset \(U \subseteq M\). For any smooth local coordinates \((U,(x^i))\), we can compute

\begin{align}Xx^i &= \left(X^j \frac{\partial}{\partial x^j}\right)(x^i)\\&= X^j \frac{\partial x^i}{\partial x^j}\\&= X^i \cdot 1\\&= X^i\end{align}

since \(\partial x^i/\partial x^j\) is the constant function \(0\) when \(i \neq j\), and is the constant function \(1\) when \(i = j\). Since \(x^i \in C^{\infty}(U)\), \(Xx^i\) is smooth and thus so is \(X^i\). As we previously established, if each of the component functions of a vector field is smooth, then the vector field is smooth.

Finally, suppose \(X\) is smooth. For any smooth chart \((U,(x^i))\) on any neighborhood \(U\) of a point \(p\), we can compute, for each point \(p' \in U\) and smooth function \(f \in C^{\infty}(M)\),

\begin{align}Xf(p') &= \left(X^i(p') \frac{\partial}{\partial x^i}\bigg\rvert_{p'}\right)f\\&= X^i(p')\frac{\partial \hat{f}}{\partial x^i}(p').\end{align}

As we previously established, since \(X\) is smooth, each of its component functions is smooth. The function \(f\) is smooth by assumption, and thus each of its partial derivatives is smooth. Thus \(Xf\) is the product of smooth functions, and is therefore smooth.

Thus, \(X\) is smooth if and only if \(Xf\) is smooth for each \(f \in C^{\infty}(M)\), and likewise, \(X\) is smooth if and only if \(Xf\) is smooth for each \(f \in C^{\infty}(U)\) for each open set \(U\).

Returning to the equivalence of vector fields and derivations on \(C^{\infty}(M)\), we see that, for any vector field \(X\) defined such that

\[X_pf = (Df)(p)\]

for a derivation \(D\), since \(Xf = Df\) and \(Df\) is smooth by definition, \(X\) is likewise smooth. Thus, we can identify derivations on \(C^{\infty}(M)\) with smooth vector fields.

Pushforwards

If \(F : M \rightarrow N\) is a smooth map between smooth manifolds \(M\) and \(N\) and \(X : M \rightarrow TM\) is a smooth vector field, then the differential \(dF_p(X_p)\) yields vectors in \(T_{F(p)}N\) for each \(p \in M\). If \(F\) is a diffeomorphism, then we can define a vector field \(F_*X\), called the pushforward of \(X\) by \(F\), as follows for each \(q \in N\):

\[(F_*X)_q = dF_{F^{-1}(q)}(X_{F^{-1}(q)}).\]

\(F\) must be a diffeomorphism (and hence injective and surjective). Since \(F\) is bijective, there is a unique point \(F^{-1}(q) \in M\) corresponding to each point \(q \in N\).

Note that the pushforward is necessarily a smooth map since it is the composition of smooth maps, i.e. \(F_*X = dF \circ X \circ F^{-1}\).

Also note that the pushforward is the unique vector field \(Y\) such that \(dF_p(X_p) = Y_{F(p)}\). Two vector fields \(X\) and \(Y\) that enjoy this relationship (for any smooth map \(F\), not just diffeomorphisms) are called F-related. Thus, when \(F\) is a diffeomorphism, the pushforward is the unique vector field that is F-related to \(X\). To see this, suppose that there is another vector field \(Y\) that is F-related to \(X\) for a diffeomorphism \(F\). Then, since \(F\) is a diffeomorphism, for each point \(q \in N\), there exists a point \(p \in M\) such that \(F(p) = q\), and so

\begin{align}(F_*X)_q &= (F_*X)_{F(p)}\\&= dF_{F^{-1}(F(p))}(X_{F^{-1}(F(p))})\\&= dF_p(X_p)\\&= Y_{F(p)}\\&= Y_q.\end{align}

Lie Brackets

Treating smooth vector fields \(X\) and \(Y\) on a smooth manifold \(M\) as derivations, for any smooth function \(f \in C^{\infty}(M)\), we can apply \(X\) to \(f\) to obtain \(Xf\) and \(Y\) to \(Xf\) to obtain \(YXf\). However, the operation \(f \mapsto YXf\) will not, in general, obey the product rule, and so it is not a derivation and therefore not a vector field in general. However, remarkably, the operator \([X,Y]\) defined as

\[[X,Y]f = XYf - YXf\]

is always a vector field. It is linear over \(\mathbb{R}\), since, for any smooth functions \(f,g \in C^{\infty}(M)\),

\begin{align}[X,Y](f+g) &= XY(f+g) - YX(f+g)\\&= X(Yf + Yg) - Y(Xf + Xg)\\&= XYf + XYg - YXf - YXg\\&= [X,Y]f + [X,Y]g,\end{align}

and, for any scalar \(a \in \mathbb{R}\),

\begin{align}[X,Y](af) &= XY(af) - YX(af)\\&= X(aYf) - Y(aXf)\\&= a(XYf) - a(YXf)\\&= a(XYf - YXf)\\&= a[X,Y]f.\end{align}

It obeys the product rule, since

\begin{align}[X,Y](fg) &= XY(fg) - YX(fg)\\&= X(fYg + gYf) - Y(fXg + gXf)\\&= X(fYg) + X(gYf) - Y(fXg) - Y(gXf)\\&= f~XYg + Yg~Xf + g~XYf + Yf~Xg \\&- f~YXg - Xg~Yf - g~YXf - Xf~Yg\\&= f~XYg - f~YXg + g~XYf - g~YXf\\&= f(XYg - YXg) + g(XYf - YXf)\\&= f[X,Y]g + g[X,Y]f.\end{align}

Thus, the Lie Bracket of two vector fields is always a vector field. Its value at a point \(p \in M\) is given by

\[[X,Y]_pf = X_p(Yf) - Y_p(Xf).\]

The Lie Bracket in Coordinates

Next we consider the coordinate expression for the Lie Bracket. Since, for any smooth vector fields \(X\) and \(Y\) on a smooth manifold \(M\), \([X,Y]\) is a smooth vector field, and since smooth vector fields act locally (since tangent vectors act locally), i.e. \(([X,Y]f)\rvert_U = [X,Y](f\rvert_U)\) on any open set \(U \subseteq M\), we therefore need only consider a single smooth chart \((U,(x^i))\). Since we previously determined that \(X = X^i\partial/\partial x^i\), it follows that

\begin{align}[X,Y]f &= \left[X^i \frac{\partial}{\partial x^i}, Y^j \frac{\partial}{\partial x^j}\right]f\\&= \left(X^i \frac{\partial}{\partial x^i}\right)\left(Y^j \frac{\partial}{\partial x^j}\right)f \\&- \left(Y^j \frac{\partial}{\partial x^j}\right)\left(X^i \frac{\partial}{\partial x^i}\right)f\\&= \left(X^i \frac{\partial}{\partial x^i}\right)\left(Y^j \frac{\partial f}{\partial x^j}\right) \\&- \left(Y^j \frac{\partial}{\partial x^j}\right)\left(X^i \frac{\partial f}{\partial x^i}\right)\\&= Y^j\left(X^i \frac{\partial}{\partial x^i}\right)\frac{\partial f}{\partial x^j} + \frac{\partial f}{\partial x^j}\left(X^i \frac{\partial}{\partial x^i}\right)Y^j\\&- X^i\left(Y^j \frac{\partial}{\partial x^j}\right)\frac{\partial f}{\partial x^i} - \frac{\partial f}{\partial x^i}\left(Y^j \frac{\partial}{\partial x^j}\right)X^i\\&= X^iY^j \frac{\partial^2 f}{\partial x^i\partial x^j} + X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial f}{\partial x^j}\\&- X^iY^j \frac{\partial^2 f}{\partial x^j \partial x^i} - Y^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i}\\&= X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial f}{\partial x^j} - Y^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i}.\end{align}

Exchanging the roles of the dummy indices \(i\) and \(j\) in the latter expression, we finally obtain

\begin{align}[X,Y]f &= X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial f}{\partial x^j} - Y^i\frac{\partial X^j}{\partial x^i}\frac{\partial f}{\partial x^j}\\&= \left(X^i\frac{\partial Y^j}{\partial x^i} - Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial f}{\partial x^j}\\&= \left(\left(X^i\frac{\partial}{\partial x^i}\right)Y^j - \left(Y^i\frac{\partial}{\partial x^i}\right)X^j\right)\frac{\partial f}{\partial x^j} \\&= (XY^j - YX^j)\frac{\partial f}{\partial x^j}.\end{align}

Properties of the Lie Bracket

The Lie Bracket enjoys certain important properties.

Bilinearity

For all smooth vector fields \(X,Y,Z\) on a smooth manifold \(M\) and scalars \(a,b \in \mathbb{R}\), we have that

\begin{align}[aX + bY, Z]f &= (aX + bY)Zf - Z(aX + bY)f\\&= aXZf + bYZf - Z(aXf + bYf)\\&= aXZf + bYZf - ZaXf - ZbYf\\&= aXZf + bYZf - aZXf - bZYf\\&= a(XZf - ZXf) + b(YZf - ZYf)\\&= a[X,Z]f + b[Y,Z]f,\end{align}

and also that

\begin{align}[Z,aX+bY]f &= Z(aX+bY)f - (aX + bY)Zf\\&= Z(aXf + bYf) - aXZf - bYZf\\&= Z(aXf) + Z(bYf) - aXZf - bYZf\\&= aZXf + bZYf - aXZd - bYZf\\&= a(ZXf - XZf) + b(ZYf - YZf)\\&= a[Z,X]f + b[Z,Y]f.\end{align}

Thus, \([aX+bY,Z] = a[X,Z] + b[Y,Z]\).

Antisymmetry

\begin{align}[X,Y]f &= XYf - YXf\\&= -(YXf - XYf)\\&= -[Y,X]f .\end{align}

Thus, \([X,Y] = -[Y,X]\).

Jacobi Identity

\begin{align}[X,[Y,Z]]f + [Y,[Z,X]]f + [Z,[X,Y]]f &= X[Y,Z]f - [Y,Z]Xf + Y[Z,X]f \\&- [Z,X]Yf + Z[X,Y]f - [X,Y]Zf\\&= X(YZf - ZYf) - YZXf + ZYXf \\&+ Y(ZXf - XZf) - ZXYf \\&+ XZYf + Z(XYf - YXf)\\&- XYZf + YXZf\\&= XYZf - XZYf - YZXf + ZYXf \\&+ YZXf - YXZf - ZXYf \\&+ XZYf + ZXYf - ZYXf\\&- XYZf + YXZf \\&= 0.\end{align}

Thus, \([X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0\).

These properties make the real vector space \(\mathfrak{X}(M)\) of smooth vector fields on \(M\) together with the Lie Bracket into a Lie Algebra over \(\mathbb{R}\), which is a real vector space \(\mathfrak{g}\) together with a map \((X,Y) \mapsto [X,Y]\) from \(\mathfrak{g} \times \mathfrak{g}\) to \(\mathfrak{g}\) that enjoys bilinearity, antisymmetry, and the Jacobi identity.