Lie Derivatives of Tensor Fields

Lee Smooth Manifolds

Recall that the Lie derivative of a covariant $k$-tensor field $A$ over a vector field $V$ with flow $\theta$ in a smooth manifold $M$ is defined as

$$({\mathcal{L}}_{V}A{)}_p={\frac{d}{dt}|}_{t=0}({\theta }_t^{\ast }A{)}_p=\underset{t\to 0}{\lim }\frac{d({\theta }_t{)}_p^{\ast }(A_{{\theta }_t(p)})-A_p}{t}$$

where $d({\theta }_t{)}_p^{\ast }$ is the pullback of the tensor $A_{{\theta }_t(p)}$ from $T^k(T_{{\theta }_t(p)}^{\ast }M)$ to $T^k(T_p^{\ast }M)$. The limit is interpreted in the canonical vector space topology of $T^k(T_p^{\ast }M)$. We prove some fundamental results and computations involving the Lie derivative.

Theorem 1) Given $M$, $V$, and $A$ as above, the Lie derivative exists for every $p\in M$ and ${\mathcal{L}}_{V}A$ is a smooth $k$-tensor field on $M$.

Proof) Let $(U,(x^i{)}_{1\le i\le n})$ be a smooth chart centered at $p$. By the fundamental ODE theorem, there exists an open interval $J$ containing $0$ and neighborhood $U_0\subset U$ of $p$ such that $\theta$ maps $J\times U_0$ into $U$. Let $\theta (t,x)=({\theta }^1(t,x),\dots ,{\theta }^n(t,x))$ be the component functions of $\theta$ in the coordinates of $U$.

We first compute the matrix $d({\theta }_t{)}_p$:

$$(d({\theta }_t{)}_p{)}_{ij}=\frac{\partial {\theta }^i}{\partial x^j}(t,p)$$

the $({\theta }_t^{\ast }A{)}_p^{a_1,\dots ,a_k}$ component of the $k$-tensor $({\theta }_t^{\ast }A{)}_p$ is given by

$$({\theta }_t^{\ast }A{)}_p\left(\frac{\partial }{\partial x^{a_1}},\dots ,\frac{\partial }{\partial x^{a_j}}\right)=A_{{\theta }_t(p)}\left(\frac{\partial {\theta }^i}{\partial x^{a_1}}(t,p)\frac{\partial }{\partial x^i},\dots ,\frac{\partial {\theta }^i}{\partial x^{a_k}}(t,p)\frac{\partial }{\partial x^i}\right)=A_{{\theta }_t(p)}^{j_1,j_2,\dots ,j_k}\prod _{w=1}^k\frac{\partial {\theta }^{j_w}}{\partial x^{a_w}}(t,p)$$

which is clearly a smooth function of $t$ that collapses to $A_p^{a_1,\dots ,a_k}$ when $t=0$. Thus, the derivative is well-defined and smooth for each component of $({\theta }_t^{\ast }A{)}_p$, completing the proof.

Theorem 2) Let $M$ be a smooth manifold and let $V\in \mathfrak{X}(M)$. Suppose $f$ is a smooth real-valued function (a $0$-tensor field) on $M$, and $A$ and $B$ are smooth covariant $a$- and $b$-tensor fields on $M$. (a) ${\mathcal{L}}_{V}f=Vf$ (b) ${\mathcal{L}}_V(fA)=({\mathcal{L}}_{V}f)A+f{\mathcal{L}}_{V}A$ (c) ${\mathcal{L}}_V(A\otimes B)=({\mathcal{L}}_{V}A)\otimes B+A\otimes {\mathcal{L}}_{V}B$ (d) If $X_1,\dots ,X_k$ are smooth vector fields and $A$ is a smooth $k$-tensor field,

$${\mathcal{L}}_V(A(X_1,\dots ,X_k))=({\mathcal{L}}_{V}A)(X_1,\dots ,X_k)+A({\mathcal{L}}_V{X}_1,\dots ,X_k)+\dots +A(X_1,\dots ,{\mathcal{L}}_V{X}_k)$$

Proof) Let $\theta$ be the flow of $M$. The pullback $({\theta }_t^{\ast }f{)}_p$ evaluates to $({\theta }_t^{\ast }f{)}_p=f({\theta }_t(p))$, and thus

$$({\mathcal{L}}_{V}f{)}_p={\frac{d}{dt}|}_{t=0}f({\theta }_t(p))=d{f}_p({\theta }^{(p{)}^{\prime }}(0))=d{f}_p(V_p)=V_{p}f\Rightarrow {\mathcal{L}}_{V}f=Vf$$

(b), (c), and (d) all follow from the same logic; we provide a full proof of (d). First, assume the equation is being evaluated at some regular point $p$ of $V$. There must then exist some smooth chart $(U,(x^i{)}_{1\le i\le m})$ containing $p$ such that $V=\partial /\partial x^1$ in $U$. In these coordinates, for sufficiently small $t$, ${\theta }_t$ is simply a translation with trivial Jacobian, and thus ${\mathcal{L}}_{V}A$ is the $x^1$-partial derivative of the tensor $A_p$. Analogously, the vector field Lie derivative ${\mathcal{L}}_{V}X$ is the $x^1$-partial derivative of the vector $X_p$. Thus, writing $X_i=X_i^j\partial /\partial x^j$ and $A=A^{j_1,\dots ,j_k}{\epsilon }_{j_1}\otimes \dots \otimes {\epsilon }_{j_k}$,

$${\mathcal{L}}_V(A(X_1,\dots ,X_k){)}_p={\frac{\partial }{\partial x^1}|}_p(A^{j_1,\dots ,j_k}{X}_i^{j_i})$$

by the product rule, the RHS clearly matches the expression in the problem statement.

We now complete (d). If $p$ is in the boundary of the support of $V$, (d) must be true by continuity, while if $p$ is outside the support there must be some neighborhood $U$ of $p$ on which $V$ vanishes. In this case, all the Lie derivatives vanish, and (d) trivially holds.

Corollary 3) If $f\in C^{\infty }(M)$, ${\mathcal{L}}_V(df)=d({\mathcal{L}}_{V}f)$.

Proof) For any $X\in \mathfrak{X}(M)$,

$$({\mathcal{L}}_{V}df)X=V(dfX)-df({\mathcal{L}}_{V}X)=VXf-[V,X]f=XVf=d(Vf)X=d({\mathcal{L}}_{V}f)X$$

Recall that if $\theta$ is a flow on $M$ and $A$ is a smooth tensor field on $M$, $A$ is invariant under $\theta$ if, for every $p\in M$ and $t\in {\mathcal{D}}^{(p)}$,

$$d({\theta }_t{)}_p^{\ast }(A_{{\theta }_t(p)})=A_p$$

Theorem 4) Let $M$ be a smooth manifold and $V\in \mathfrak{X}(M)$. A smooth variant $k$-tensor field $A$ is invariant under the flow of $V$ iff ${\mathcal{L}}_{V}A=0$.

Proof) For any $(t_0,p)$ in the domain of $\theta$,

$${\frac{d}{dt}|}_{t=t_0}({\theta }_t^{\ast }A{)}_p={\frac{d}{ds}|}_{s=0}d({\theta }_{s+t_0}^{\ast }{)}_p(A_{{\theta }_{s+t_0}(p)})={\frac{d}{ds}|}_{s=0}d({\theta }_{t_0}^{\ast }{)}_{p}d({\theta }_s^{\ast }{)}_{{\theta }_{t_0}(p)}(A_{{\theta }_{s+t_0}(p)})$$ $$=d({\theta }_{t_0}^{\ast }{)}_p\left({\frac{d}{ds}|}_{s=0}d({\theta }_s^{\ast }{)}_{{\theta }_{t_0}(p)}(A_{{\theta }_{s+t_0}(p)})\right)=d({\theta }_{t_0}^{\ast }{)}_p({\mathcal{L}}_{V}A)=({\theta }_{t_0}^{\ast }({\mathcal{L}}_{V}A){)}_p$$

Thus, if ${\mathcal{L}}_{V}A=0$, the $t$-derivative of the pullback $({\theta }_t^{\ast }A{)}_p$ vanishes, forcing it to be constant with $({\theta }_t^{\ast }A{)}_p=A_p$ whenever defined. This implies $A$ is invariant under $\theta$. The converse is trivial.