I'm working on a problem from Lee's Introduction to Smooth Manifolds:

Let M be a smooth manifold and $V\in\mathfrak{X} (M)$ (smooth vector field). Show that the Lie derivative operators on covariant tensor fields, $\mathcal{L}_V:\Gamma(T^k T^*M)\to\Gamma(T^k T^*M)$ for $k\geq0$ (where $\Gamma(T^k T^*M)$ is the space of smooth covariant k-tensor fields), are uniquely characterized by the following properties:

(a) $\mathcal{L}_V$ is linear over $\mathbb{R}$

(b) $\mathcal{L}_V f=Vf$ for $f\in\Gamma(T^0 T^*M)=\mathcal{C}^\infty (M)$

(c) $\mathcal{L}_V(A\otimes B)=\mathcal{L}_V(A)\otimes B+A\otimes\mathcal{L}_V(B)$ for $A\in\Gamma(T^k T^*M)$ and $B\in\Gamma(T^l T^*M)$.

(d) $\mathcal{L}_V(\omega(X))=(\mathcal{L}_V\omega)(X)+\omega([V,X])$ for $\omega\in\Gamma(T^1 T^*M),X\in\mathfrak{X}(M)$.

In the lecture we defined the Lie derivative for covariant k-tensor field as

$\left( \mathcal{L}_VA\right)_p=\frac{d}{dt}\left.\right|_{t=0}\left(\theta^*_tA\right)_p=\lim_{t\to 0}\frac{d(\theta_t)^*_p\left( A_{\theta_t(p)}\right)-A_p}{t}$ where $d(\theta_t)^*_p$ is the pullback of $A$ by $\theta_t$ at $p$ ($\theta$ is the flow of $V$).

So far my approach is to derive this formula for $\left( \mathcal{L}_VA\right)_p$ by using the properties (a)-(d) (we did already show in the lecture that the latter definition of the Lie-derivative satisfies properties (a)-(d), so I'm trying to prove the converse now). E.g. one can derive (if I made no mistake)

$\left(\mathcal{L}_V A\right)\left(X_1,\dots ,X_k\right)= \mathcal{L}_V\left( A(X_1,\dots ,X_k)\right)-A(\mathcal{L}_V X_1,X_2,\dots ,X_k)-\dots -A(X_1,\dots,X_{k-1},\mathcal{L}_V X_k)$

Then after some more work I arrived at

$\mathcal{L}_V \left(A_p\left(X_1\left.\right|_p,\dots ,X_k\left.\right|_p\right)\right)=\lim_{t\to0}\frac{A_{\theta^{(p)}(t)}\left(X_1\left.\right|_{\theta^{(p)}(t)},\dots ,X_k\left.\right|_{\theta^{(p)}(t)}\right)-A_p\left(X_1\left.\right|_p,\dots ,X_k\left.\right|_p\right)}{t}$

which looks somehow promising (at least a bit). However, I can't get the other terms to a form which yields the final form of the Lie derivative.

So I am stuck at that point. Could anyone give me a hint how to continue? Or am I completely wrong and there is a much easier approach to the problem? I would appreciate any help.


Properties a)-c) mean that $\mathcal{L}_X$ is a local derivation of the tensor algebra (well, locality is not postulated but can be checked).

From property d), we have $\mathcal{L}_V(\omega(X))=(\mathcal{L}_V\omega)(X)+\omega([V,X])$, but the LHS is $V(\omega(X)) $ by b), therefore we have $$ \mathcal{L}_V\omega)(X)=V(\omega(X))-\omega([V,X]).$$


Theorem: Let $D$ be any local derivation of the covariant tensor algebra. Then $D$'s action is uniquely characterized by its action on grade 0 and grade 1 elements (scalar fields and covector fields).

Proof: Let $T$ be a grade $k$ covaraint tensor field ($T\in\Gamma(T^kT^*M)$). Let $(U,x)$ be a local chart. Then in $U$, we have $$ T=T_{i_1,...,i_k}dx^{i_1}\otimes...\otimes dx^{i_k}, $$ so we have $$ DT=(DT_{i_1...i_k})\otimes dx^{i_1}\otimes...\otimes dx^{i_k}+T_{i_1...i_k}(Ddx^{i_1})\otimes...\otimes dx^{i_k}+..., $$ but all actions of $D$ can be calculated in the above expression, since they only involve $DT_{i_1...i_k}$ and $Ddx^{i}$, which are grade 0 and grade 1 elements respectively. QED.

You have $\mathcal{L}_V$ as a local derivation that acts on functions by $Vf$ and on 1-forms/covector fields by the formula we found above, so we are done.

Since the Lie derivative defined by the flow formula also knows these properties, and we have shown that this is unique, they most coincide.

If needed, show using bump function that $L_V$ is indeed local.

  • $\begingroup$ Nice! Thank you very much for this well-explained proof $\endgroup$
    – Nukular
    Apr 4 '17 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.