Lie Derivative

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Directional Derivative of a Function and the Lie Derivative

Consider a function \(f\) defined on a manifold \(M\). By definition, the directional derivative of \(f\) at a point \(p\in M\) in the direction of \(v\in T_pM\) is \(v(f)\). The chief reason we cannot use the usual difference quotient

\[\lim_{t \rightarrow0}\frac{f(p+tv)-f(p)}{t}\]

on \(M\) is that \(p+tv\) is not defined. However, if a suitable \(C^\infty\) curve \(\gamma\) on \(M\) is given with \(\gamma(0)=p\) and \(\gamma'(0)=v\), then we know that the directional derivative can be computed by the formula

\[\lim_{h\rightarrow 0}\frac{f(\gamma(h))-f(p)}{h}\tag{1}\]

(§Examples of Differentials, ⁋Definition 1)

Now suppose a vector field \(X\) on \(M\) is given, and consider the problem of finding the directional derivative \(X_pf\) at each point \(p\) in the direction of \(X_p\). Geometrically, this amounts to choosing, for every point \(p\), a curve \(\gamma_p\) satisfying \(\gamma(0)=p\) and \(\gamma'(0)=X_p\), and applying formula (1) above. But we know that such a curve always exists. (§Vector Fields, ⁋Theorem 6)

Definition 1 Fix a manifold \(M\) and a vector field \(X\) defined on it, and suppose a function \(f:M\rightarrow\mathbb{R}\) is given. Then the Lie derivative \(\mathcal{L}_Xf\) of \(f\) is the function defined by the formula

\[(\mathcal{L}_Xf)(p)=\lim_{t\rightarrow 0}\frac{f(\phi^t(p))-f(\phi^0(p))}{t}=\lim_{t\rightarrow 0}\frac{f(\phi^t(p))-f(p)}{t}\]

Of course, by the preceding argument this is nothing other than the directional derivative \(X_pf\). However, the flow \(\phi^t\) used here furnishes not merely a way to differentiate functions, but a way to differentiate every other object defined on \(M\).

Lie Derivative of Vector Fields

The simplest example is the derivative of a vector field. Since a vector field \(Y\) is a map from \(M\) to \(TM\), one might attempt to differentiate it by a method similar to Definition 1 above, but this is far from straightforward. The obstacle here is more fundamental than for functions: \(Y(\phi^t(p))\) is an element of \(T_{\phi^t(p)}\), whereas \(Y(p)\) is an element of \(T_pM\), so there is no way even to form their difference \(Y_{\phi^t(p)}-Y_p\).

Nevertheless, differentiation is still possible in our setting. Recalling §Vector Fields, ⁋Theorem 6, since \(\phi^t\) is a diffeomorphism, \(d\phi^t\) induces an isomorphism from \(T_pM\) to \(T_{\phi^t(p)}\). Moreover, the same theorem tells us that the inverse of this isomorphism is \(d\phi^{-t}\). Therefore, by pulling \(Y_{\phi^t(p)}\) back to \(T_pM\) via \(d\phi^{-t}\), we can make the following definition.

Definition 2 Fix a manifold \(M\) and a vector field \(X\) defined on it, and suppose another vector field \(Y:M\rightarrow TM\) is given. Then the Lie derivative \(\mathcal{L}_XY\) of \(Y\) is the vector field defined by the formula

\[(\mathcal{L}_XY)_p=\lim_{t\rightarrow 0}\frac{(d\phi^{-t})_{\phi^t(p)}(Y_{\phi^t(p)})-Y_p}{t}\]

Lie Derivative of Differential Forms

Naturally, we can continue defining derivatives in this manner. For example, the following gives the method for differentiating a differential form along a vector field \(X\).

Definition 3 Fix a manifold \(M\) and a vector field \(X\) defined on it, and suppose a differential form \(\omega\in\Omega^\ast(M)\) is given. Then the Lie derivative \(\mathcal{L}_X\omega\) of \(\omega\) is the differential form defined by the formula

\[(\mathcal{L}_X\omega)_p=\frac{d}{dt}\bigg|_{t=0}(\phi^t)^\ast\omega_{\phi^t(p)}=\lim_{t\rightarrow 0}\frac{(\phi^t)^\ast\omega_{\phi^t(p)}-\omega_p}{t}\]

Moreover, it is not difficult to extend this definition to arbitrary tensor fields. Since we will not need this immediately, we omit it.

Some parts of the following proposition have already been proved, and some are new, but we omit all proofs as they would be too lengthy to write out in full.

Proposition 4 For any \(X\in\mathfrak{X}(M)\), the following hold.

1. For any \(f\in C^\infty(M)\), we have \(\mathcal{L}_Xf=X(f)\).
2. For any \(Y\in \mathfrak{X}(M)\), we have \(\mathcal{L}_XY=[X,Y]\).
3. \(\mathcal{L}_X\) is a derivation on \(\Omega^\ast(M)\) and commutes with \(d\).
4. \(\mathcal{L}_X=\iota_X\circ d+d\circ\iota_X\).

5. For any \(\omega\in\Omega^k(M)\) and \(X_0, X_1,\ldots, X_k\in\mathfrak{X}(M)\),

   \[\mathcal{L}_{X_0}(\omega(X_1,\ldots, X_k))=(\mathcal{L}_{X_0}\omega)(X_1,\ldots, X_k)+\sum_{i=1}^k\omega(X_1,\ldots, X_{i-1}, \mathcal{L}_{X_0}X_i,X_{i+1},\ldots, X_k)\]

6. For any \(\omega\in\Omega^k(M)\) and \(X_0, X_1,\ldots, X_k\in\mathfrak{X}(M)\),

   \[\begin{aligned}d\omega(X_0,\ldots, X_k)&=\sum_{i=0}^k(-1)^iX_i\omega(X_0,\ldots, \hat{X}_i,\ldots, X_k)\\&\phantom{==}+\sum_{i<j}(-1)^{i+j}\omega([X_i, X_j], X_0,\ldots, \hat{X}_i,\ldots, \hat{X}_j,\ldots, X_k)\end{aligned}\]

In 5 and 6, the hat indicates that the corresponding entry is omitted.

The first statement of this proposition is something we have already verified. The bracket \([-,-]\) appearing in the second and sixth results is an operation between vector fields called the Lie bracket; we will only touch upon it briefly here and treat it in detail in the next post. At any rate, the key point of the second, fifth, and sixth results is that when computing \(\mathcal{L}_X\), one can obtain the Lie derivative through comparatively simple algebraic manipulations rather than appealing to the original definition. The fourth result is known as Cartan’s formula.

Lie Bracket

\(\mathfrak{X}(M)\) is not a \(C^\infty(M)\)-algebra in general. Computing \((XY)(fg)\) directly, we obtain

\[(XY)(fg)=X(f(Yg)+g(Yf))=(Xf)(Yg)+f(XY)g+(Xg)(Yf)+g(XY)f\tag{2}\]

In other words, the two terms \((Xf)(Yg)\) and \((Xg)(Yf)\) are precisely what prevent \(XY\) from satisfying the Leibniz rule, and so we make the following definition.

Definition 5 For \(X,Y\in\mathfrak{X}(M)\), the element \([X,Y]\in\mathfrak{X}(M)\) is defined by the formula

\[[X,Y]f=X(Yf)-Y(Xf)\]

Of course, for this definition to make sense we must verify that the right-hand side actually yields an element of \(\mathfrak{X}(M)\), but this can be done by interchanging the roles of \(X\) and \(Y\) in formula (2) above to obtain \((YX)(fg)\), and then subtracting. Since this definition was designed from the outset to cancel the two problematic terms, the Leibniz rule will hold automatically.

The vector field \([X,Y]\) defined in this way is called the Lie bracket of \(X\) and \(Y\). Since this definition will be of great importance later, it seems worthwhile to collect a few results in advance.

Suppose a \(C^\infty\) map \(F:M\rightarrow N\) between two manifolds \(M\) and \(N\) is given. Then \(dF_p:T_pM\rightarrow T_{F(p)}N\) is the map that sends a tangent vector \(v\) at a point \(p\) of \(M\) to the tangent vector \(dF_p(v)\) at the point \(F(p)\) of \(N\). In general, however, this cannot be done for vector fields: even if a vector field \(X\) on \(M\) is given, we cannot use \(dF_p\) to produce a vector field on \(N\).

For instance, if \(F\) is not surjective, there is no natural way to assign a tangent vector at a point \(q\in N\) lying outside the image of \(F\). Even if we resolve this by restricting the codomain, should \(F\) fail to be injective and satisfy \(F(p_1)=F(p_2)=q\in N\), we would still have to decide which of \(dF_{p_1} v_1\) and \(dF_{p_2} v_2\) to take as the tangent vector at the common point \(q\).

Therefore, rather than attempting to push \(X\in\mathfrak{X}(M)\) forward through \(F\), it is wiser to examine the case in which a vector field \(Y\in\mathfrak{X}(N)\) already given satisfies the desired property.

Definition 6 Let \(F:M\rightarrow N\) be a \(C^\infty\) map. If \(X\in\mathfrak{X}(M)\) and \(Y\in\mathfrak{X}(N)\) satisfy the equation

\[dF_p(X_p)=Y_{F(p)}\]

for all \(p\in M\), then we say that \(X\) and \(Y\) are \(F\)-related.

In other words, \(X\) and \(Y\) being \(F\)-related means that the following diagram commutes.

As we can see by applying the fact that \(X\) is \(C^\infty\) (from Proposition 2) to each function \(f\), whether \(X\) and \(Y\) are \(F\)-related can also be tested by applying them to each function.

Proposition 7 Let \(F:M\rightarrow N\) be a \(C^\infty\) map, and let \(X\in\mathfrak{X}(M)\) and \(Y\in\mathfrak{X}(N)\). Then \(X\) and \(Y\) are \(F\)-related if and only if for every \(f\),

\[X(f\circ F)=(Yf)\circ F\]

Proof

For any point \(p\in M\),

\[X(f\circ F)(p)=X_p(f\circ F)=dF_p(X_p)f\]

and

\[((Yf)\circ F)(p)=(Yf)(F(p))=Y_{F(p)}f\]

Hence the two given conditions are equivalent.

At first we pointed out that if \(F\) is neither surjective nor injective, it is not natural to seek a \(Y\in\mathfrak{X}(N)\) that is \(F\)-related to a given \(X\in\mathfrak{X}(M)\) through \(F\); however, if \(F\) is a diffeomorphism, then the following natural choice exists.

Proposition 8 If \(F:M\rightarrow N\) is a diffeomorphism, then for every \(X\in\mathfrak{X}(M)\) there exists a unique \(Y\in\mathfrak{X}(N)\) such that \(X\) and \(Y\) are \(F\)-related.

Proof

For each \(q\in N\) there exists a unique \(p\in M\) such that \(F(p)=q\). Thus, for each point \(q\in N\), we define \(Y\) by the formula

\[Y_q=dF_p(X_p)\qquad (F(p)=q)\]

Since the above formula must hold for \(Y\) to be \(F\)-related to \(X\), the uniqueness of such a \(Y\) is obvious. Moreover, \(Y:N\rightarrow TN\) is now the composition of the following \(C^\infty\) maps

\[N\overset{F^{-1}}{\longrightarrow}M\overset{X}{\longrightarrow}TM\overset{dF}{\longrightarrow}TN\]

and therefore \(Y\) is \(C^\infty\).

Proposition 9 Let \(F:M\rightarrow N\) be a \(C^\infty\) map. If for \(i=1,2\) we have \(X_i\in\mathfrak{X}(M)\), \(Y_i\in\mathfrak{X}(N)\), and \(X_i\) and \(Y_i\) are \(F\)-related, then \([X_1,X_2]\) is \(F\)-related to \([Y_1,Y_2]\).

Proof

We must show that the formula

\[dF_p([X_1,X_2]_p)=[Y_1,Y_2]_{F(p)}\]

holds for all \(p\). Now for any function \(f\) defined in a neighborhood of \(F(p)\),

\[\begin{aligned}dF_p([X_1,X_2]_p)f&=[X_1,X_2]_p(f\circ F)\\ &=(X_1)_p(X_2(f\circ F))-(X_2)_p(X_1(f\circ F))\\ &=(X_1)_p((Y_2f)\circ F)-(X_2)_p((Y_1f)\circ F)\\ &=dF_p(X_1)_p(Y_2f)-dF_p(X_2)_p(Y_1f)\\ &=(Y_1)_{F(p)}(Y_2f)-(Y_2)_{F(p)}(Y_1f)\\ &=[Y_1,Y_2]_{F(p)}f\end{aligned}\]

Thus we obtain the desired conclusion.

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References

[War] Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups, Graduate texts in mathematics, Springer, 2013
[Lee] John M. Lee. Introduction to Smooth Manifolds, Graduate texts in mathematics, Springer, 2012

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