6
$\begingroup$

I would like to know how to show that the Lie derivative on a differentiable manifold satisfies

\begin{equation*} \mathcal{L}_{[X, Y]} = \mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X \end{equation*}

for any tensor field on which the derivative is applied, where $[X, Y] = XY - YX$ is the commutator of $X$ and $Y$, which are arbitrary vector fields.

Edit:

In component notation, $[X, Y]^\mu = X^\lambda \partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu = X^\lambda \nabla_\lambda Y^\mu - Y^\lambda \nabla_\lambda X^\mu$, where $\partial_\mu$ and $\nabla_\mu$ are respectively the partial and the covariant derivatives with respect to the variable $x^\mu$.

$\endgroup$
6
$\begingroup$

One way to prove this result is to first show it for functions on the manifold, and use this to further prove it for vector fields, co-vector fields and finally general tensor fields. The advantage of this approach as it requires the minimum of formulas.

We will need the following assumptions:

(a) The Lie derivative follows the Leibnitz rule when acting on a product of objects.

(b) $ \mathcal{L}_{X} (f) = X(f) $

Part 1 - Show for a function, f: Using (b) we have $$ \mathcal{L}_X \mathcal{L}_Y f = \mathcal{L}_X ( Y(f)) = X(Y(f)) $$

In a coordinate basis this is:

$$ X^{\rho} \partial_{\rho} (Y^{\sigma} \partial_{\sigma} f) = X^{\rho} ( \partial_{\rho} Y^{\sigma}) (\partial_{\sigma} f) + X^{\rho} Y^{\sigma} (\partial_{\rho} \partial_{\sigma} f) $$

Because $ \partial_{\rho} \partial_{\sigma} f $ is symmetric in $\rho \leftrightarrow \sigma $ then the second term on the RHS is symmetric in $ X \leftrightarrow Y $. Therefore:

$$ \mathcal{L}_X \mathcal{L}_Y f - \mathcal{L}_Y \mathcal{L}_X f = X^{\rho} ( \partial_{\rho} Y^{\sigma}) (\partial_{\sigma} f) - Y^{\rho} ( \partial_{\rho} X^{\sigma}) (\partial_{\sigma} f) = (X^{\rho} ( \partial_{\rho} Y^{\sigma}) - X^{\rho} ( \partial_{\rho} Y^{\sigma}))(\partial_{\sigma} f) $$

$$ = [X,Y]^{\sigma} \partial_{\sigma} f = [X,Y](f) = \mathcal{L}_{[X,Y]} f $$

Part 2: Show for a vector field, $ T = V \in TM $

Using the result from Part 1 we have:

$$ \mathcal{L}_{[X,Y]} (V(f)) = \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) - \mathcal{L}_{Y} \mathcal{L}_{X} (V(f)) $$

Using the Leibnitz rule:

$$ \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) = \mathcal{L}_{X} (( \mathcal{L}_{Y} V)f) + \mathcal{L}_{X}( V(\mathcal{L}_{Y} f) ) = (\mathcal{L}_{X} \mathcal{L}_{Y} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y}f) + (\mathcal{L}_{X} V)(\mathcal{L}_{Y} f) + (\mathcal{L}_{Y} V)(\mathcal{L}_{X} f) $$

The last 2 terms taken together are symmetric in $ X \leftrightarrow Y $. We therefore have that

$$ \mathcal{L}_{[X,Y]} (V(f)) = \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) - \mathcal{L}_{Y} \mathcal{L}_{X} (V(f)) = (\mathcal{L}_{X} \mathcal{L}_{Y} V - \mathcal{L}_{Y} \mathcal{L}_{X} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y} f - \mathcal{L}_{Y} \mathcal{L}_{X}f) $$

But we can also apply the Leibnitz rule to $\mathcal{L}_{[X,Y]} (V(f))$ to give:

$$ \mathcal{L}_{[X,Y]} (V(f)) = (\mathcal{L}_{[X,Y]} V)(f) + V(\mathcal{L}_{[X,Y]} f)) = (\mathcal{L}_{[X,Y]} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y} f - \mathcal{L}_{Y} \mathcal{L}_{X}f))$$

Comparing these 2 expressions we get that:

$$ (\mathcal{L}_{[X,Y]} V)(f) = (\mathcal{L}_{X} \mathcal{L}_{Y} V - \mathcal{L}_{Y} \mathcal{L}_{X} V)(f) $$

This is true for any function hence we have shown the identity for a vector field.

Part 3 - Show for a covector field, $ T = \eta \in T^{*}M $

We consider the action of $ \mathcal{L}_{[X,Y]} $ on the function $ \eta (V) $ where $ \eta $ is a covector field and V is a vector field. This works exactly the same as in part 2 giving us:

$$(\mathcal{L}_{[X,Y]} \eta)(V) = (\mathcal{L}_{X} \mathcal{L}_{Y} \eta - \mathcal{L}_{Y} \mathcal{L}_{X} \eta)(V) $$

This is true for any vector field hence the result is shown for covector fields.

Part 4 - General Tensor field, T

We will now consider the action of $ \mathcal{L}_{[X,Y]} $ on the function $T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s)$, where T is an (r,s) rank tensor, $ X_i $ is a vector field, and $ \eta_i $ is a covector field. This step proceeds similarly to before.

By the result from Part 1:

$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = (\mathcal{L}_{X} \mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $$

Similarly to Part 2 $ \mathcal{L}_{X} \mathcal{L}_{Y} $ acting on $ T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $ gives terms symmetric in $ X \leftrightarrow Y $ when the Lie derivatives hit seperate terms. These terms will cancel with terms contained in $ - \mathcal{L}_{Y} \mathcal{L}_{X} T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $. Therefore we are left with:

$$ (\mathcal{L}_{X} \mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = (\mathcal{L}_{X} \mathcal{L}_{Y}T - \mathcal{L}_{Y} \mathcal{L}_{X}T)(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T((\mathcal{L}_{X}\mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,(\mathcal{L}_{X}\mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$

Using the results from Parts 2 and 3 gives:

$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) =(\mathcal{L}_{X} \mathcal{L}_{Y}T - \mathcal{L}_{Y} \mathcal{L}_{X}T)(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(\mathcal{L}_{[X,Y]}X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,\mathcal{L}_{[X,Y]}X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$

Lastly, the use of the Leibnitz rule gives

$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(\mathcal{L}_{[X,Y]}X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,\mathcal{L}_{[X,Y]}X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$

Comparing the 2 equalities and noting that they are true for any vector fields $ X_1,X_2,...,X_r $ and any covector fields $ \eta_1,\eta_2,...,\eta_s $ proves that the desired result holds for any general tensor field T.

$\endgroup$
2
  • $\begingroup$ I wonder if you could clarify the first step: $L_X L_Y f = X(Y(f))$? By definition $L_Y f = Y(f)$, but shouldn't $L_X (Y(f)) = [X , Y(f)]$ since $Y(f)$ is a vector field? $\endgroup$ Sep 29 '20 at 2:07
  • $\begingroup$ @AlvisNordkovich $Y(f)$ is not a vector field. It's a function. $\endgroup$
    – ChoMedit
    May 8 '21 at 15:42
3
$\begingroup$

To show this for vector fields, we use the relation $$ \mathcal L_XY=[X,Y], $$ and hence the statement follows from the properties of the Lie bracket of vector fields.

For forms, first show that the interior product fulfills $$ i_{[X,Y]}=[\mathcal L_X,i_y], $$ (easy if you use Cartan's formula) and then a quick calculation using Cartan's formula and the fact that the Lie and exterior derivative commute gives $$ \mathcal L_{[X,Y]} \alpha = i_{[X,Y]}d\alpha + di_{[X,Y]}\alpha = \mathcal L_X(i_Y\alpha+di_y\alpha) - (i_y d - di_y)\mathcal L_X\alpha = [\mathcal L_X,\mathcal L_Y]\alpha. $$ and by the Leibniz rule this generalises to arbitrary tensors.

$\endgroup$
2
  • $\begingroup$ Since I'm learning about this material in a physics course, not in a math course, I haven't seen the interior product and Cartan's formula. So I don't really understand your solution. However, I found a way to do it, as explained in my answer. $\endgroup$
    – OpticAl
    Mar 16 '16 at 1:20
  • $\begingroup$ I'm also a physicist, so I can understand your struggle :) These concepts are probably not so new to you - interior product is just contraction. Thus it could be worth the effort to read a bit about exterior derivative, wedge product, interior product (...). At least to me it is often easier to deal with those instead of tensor components with zillions of indices. Your solution seems also fine, though! $\endgroup$ Mar 16 '16 at 8:26
0
$\begingroup$

What I did in the end is something that goes along the lines of a proof by induction. In this sense, it is similar to Markus Heinrich's answer. First, the Lie derivative of an arbitrary $(k, l)$ tensor is

\begin{align*} \mathcal{L}_V {T^{\mu_1 \mu_2 \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} &= V^\sigma \partial_\sigma{T^{\mu_1 \mu_2 \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} \\ &- (\partial_\lambda V^{\mu_1}){T^{\lambda \mu_2 \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} - (\partial_\lambda V^{\mu_2}){T^{\mu_1 \lambda \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} - \ldots \\ &+ (\partial_{\nu_1} V^\lambda){T^{\mu_1 \mu_2 \ldots \mu_k}}_{\lambda \nu_2 \ldots \nu_l} + (\partial_{\nu_2} V^\lambda){T^{\mu_1 \mu_2 \ldots \mu_k}}_{\nu_1 \lambda \ldots \nu_l} + \ldots \end{align*}

For a function $f: M \rightarrow \mathbb{R}$ on the manifold, this reduces to

\begin{align*} \mathcal{L}_V f = V^\mu \partial_\mu f \end{align*}

while for a one-form field $\omega: M \rightarrow T^*_p M$, this reduces to

\begin{align*} \mathcal{L}_V \omega_\mu = V^\nu \partial_\nu f\omega_\mu + (\partial_\mu V^\nu) \omega_\nu \end{align*}

For a vector field, the formula $\mathcal{L}_V U^\mu = [V, U]^\mu$ is simpler than the above formula.

First, I showed that $\mathcal{L}_{[X, Y]} = \mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X$ is true when applied to a function, to a vector field, and to a covector field. It is straightforward using the above formulae. This step stood as the base case for my "proof by induction". Then I used the fact that any tensor can be expressed as the tensor product of vectors and covectors and that the Lie derivative obeys the Leibniz rule: $\mathcal{L}_V (S \otimes T) = (\mathcal{L}_V S) \otimes T + S \otimes (\mathcal{L}_V T)$. As my induction hypothesis, I claimed that $\mathcal{L}_{[X, Y]} T = \mathcal{L}_X \mathcal{L}_Y T - \mathcal{L}_Y \mathcal{L}_X T$ is true for any tensor T. Then, taking two tensors $S$ and $T$ obeying this rule, I proved that a more general tensor $S \otimes T$ also obeyed it:

\begin{align*} \mathcal{L}_{[X, Y]} (S \otimes T) = \mathcal{L}_X \mathcal{L}_Y (S \otimes T) - \mathcal{L}_Y \mathcal{L}_X (S \otimes T) \end{align*} Therefore, by induction, I had what I wanted. $\blacksquare$

NB: The first formula is also true if the partial derivatives are replaced by covariant derivatives.

$\endgroup$
3
  • $\begingroup$ Just one comment: You should (and wanted to) do induction over the rank of the tensor. As you wrote it, your conclusion is wrong. You assume that the statement is true for any tensor, which is not the point of induction. What I also didn't write in my answer is that you can assume tensors to be of the form $S\otimes T$ because of linearity. $\endgroup$ Mar 16 '16 at 8:32
  • $\begingroup$ In this case, should I say that the formula is true for some tensor $T$, instead of for any tensor $T$? $\endgroup$
    – OpticAl
    Mar 16 '16 at 20:21
  • $\begingroup$ You should assume that it is correct for any tensor with co- and contravariant rank $\leq n$ and then show it for an arbitrary tensor $\mathcal T$ with rank $n+1$. Since we can assume that $\mathcal T$ is simple (because of linearity), we can write $\mathcal T= T\otimes S$ for some tensors $S$ and $T$ with ranks $\leq n$ and apply the equation above. $\endgroup$ Mar 17 '16 at 8:35

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.