Calculating Lie Derivatives for Tensors & Vectors

[Arman777]

I am writing a code to calculate the Lie Derivatives, and so far, I have defined the Covariant derivative

1) for scalar function;

$$\nabla_a\phi \equiv \partial_a\phi~~(1)$$

2) for vectors;

$$\nabla_bV^a = \partial_bV^a + \Gamma^a_{bc}V^c~~(2)$$
$$\nabla_cV_a = \partial_cV_a - \Gamma^b_{ca}V_b~~(3)$$

3) for rank two tensors;

$$\nabla_cT^{ab} = \partial_cT^{ab} + \Gamma^a_{cd}T^{db} + \Gamma^b_{cd}T^{ad}~~(4)$$
$$\nabla_cT^a_b = \partial_cT^a_b + \Gamma^a_{cd}T^d_b - \Gamma^d_{cb}T^a_d~~(5)$$
$$\nabla_cT_{ab} = \partial_cT_{ab} - \Gamma^d_{ca}T_{db} - \Gamma^d_{cb}T_{ad}~~(6)$$

Similarly, I want to obtain a Lie Derivative of a scalar function, a vector (covariant and contravariant), and a tensor with rank 2.

From some research, I have found that.

1) for scalar function;

$$L_X\phi = X^{a}\partial_a\phi$$

So my questions is how can I write

$$L_XV^a, L_XV_a, L_XT^{ab}, L_XT^a_b, L_XT_{ab}$$ in terms of Eqns. $(2), (3), (4), (5), (6)$, If possible. If it's not possible how can I write them in general.

 

[Orodruin]

The Lie derivative is fundamentally different from the covariant derivative defined by a connection and it does not rely on any underlying structure (such as a connection). The general structure of the Lie derivative of a tensor in terms of components is
$$
\mathcal L_X T^{a_1a_2\ldots}_{b_1b_2\ldots} = X^c \partial_c T^{a_1a_2\ldots}_{b_1b_2\ldots} - T^{ca_2\ldots}_{b_1b_2\ldots} \partial_c X^{a_1} - T^{a_1c\ldots}_{b_1b_2\ldots} \partial_c X^{a_2} - \ldots + T^{a_1a_2\ldots}_{cb_2\ldots} \partial_{b_1} X^c + T^{a_1a_2\ldots}_{b_1c\ldots} \partial_{b_2} X^c + \ldots
$$

 

[Arman777]

I have found these,
$$L_XV^a = X^c\partial_cV^a - V^c\partial_cX^a$$
$$L_XV_b = X^c\partial_cV_b + V_c\partial_bX^c$$
$$L_XT^{ab} = X^c\partial_cT^{ab} - T^{cb}\partial_cX^a - T^{ac}\partial_cX^b$$
$$L_XT^a_b = X^c\partial_cT^a_b - T^c_b\partial_cX^a + T^a_c\partial_bX^c$$
$$L_XT_{ab} = X^c\partial_cT_{ab} + T_{cb}\partial_aX^c + T_{ac}\partial_bX^c$$

Can someone check please

 

[ergospherical]

For sake of an example, consider a $(1,1)$ tensor $\mathbf{T}$. The Lie derivative $L_{\mathbf{X}}$ satisfies Leibniz' rule, hence\begin{align*}
L_{\mathbf{X}} ( \mathbf{T} \otimes \mathbf{e}^a \otimes \mathbf{e}_b ) = L_{\mathbf{X}} \mathbf{T} \otimes \mathbf{e}^a \otimes \mathbf{e}_b + \mathbf{T} \otimes L_{\mathbf{X}} \mathbf{e}^a \otimes \mathbf{e}_b + \mathbf{T} \otimes \mathbf{e}^a \otimes L_{\mathbf{X}} \mathbf{e}_b
\end{align*}for arbitrary basis vectors $\mathbf{e}^a$ and $\mathbf{e}_b$. Now contract over all four positions; recall that $L_{\mathbf{X}}$ preserves contractions.\begin{align*}

(\dagger) \quad L_{\mathbf{X}} {T^a}_b &= {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_j \langle L_{\mathbf{X}} \mathbf{e}^a, \mathbf{e}_i \rangle \langle \mathbf{e}^j, \mathbf{e}_b \rangle + {T^{i}}_j \langle \mathbf{e}^a, \mathbf{e}_i \rangle \langle \mathbf{e}^j, L_{\mathbf{X}} \mathbf{e}_b \rangle \\

&= {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_b \langle L_{\mathbf{X}} \mathbf{e}^a, \mathbf{e}_i \rangle + {T^{a}}_j \langle \mathbf{e}^j, L_{\mathbf{X}} \mathbf{e}_b \rangle

\end{align*}By definition the Lie derivative acts on functions as $L_{\mathbf{X}} f = \mathbf{X}f$ hence in coordinate basis $L_{\mathbf{X}} {T^a}_b = X^i \dfrac{\partial {T^a}_b}{\partial x^i}$. Furthermore, for vectors $\mathbf{Y} \in T_p$ and covectors $\boldsymbol{\eta} \in T^*_p$ we have*\begin{align*}
(L_{\mathbf{X}} \mathbf{Y})^i &= \dfrac{\partial Y^i}{\partial x^j} X^j - \dfrac{\partial X^i}{\partial x^j} Y^j \\

(L_{\mathbf{X}} \boldsymbol{\eta})_i &= \dfrac{\partial \eta_i}{\partial x^j} X^j + \dfrac{\partial X^j}{\partial x^i} \eta_j
\end{align*}which imply that for a coordinate basis $\left(L_{\mathbf{X}} \dfrac{\partial}{\partial x^i} \right)^j = -\dfrac{\partial X^j}{\partial x^i}$ and $(L_{\mathbf{X}} dx^i)_j = \dfrac{\partial X^i}{\partial x^j}$. Returning to equation $(\dagger)$,\begin{align*}
X^i \dfrac{\partial {T^a}_b}{\partial x^i} &= {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_b \dfrac{\partial X^a}{\partial x^i} - {T^{a}}_j \dfrac{\partial X^j}{\partial x^b}
\end{align*}which is better re-written as \begin{align*}
{(L_{\mathbf{X}} \mathbf{T})^a}_b &= X^i \dfrac{\partial {T^a}_b}{\partial x^i} - {T^{i}}_b \dfrac{\partial X^a}{\partial x^i} + {T^{a}}_j \dfrac{\partial X^j}{\partial x^b} \\
\end{align*}The mnemonic is "minus the upper indices & plus the lower indices".

*the latter of these identities is deduced from the former. The former comes from the definition $L_{\mathbf{X}} \mathbf{S} |_p = \lim_{t \rightarrow 0} \frac{1}{t} \left( \mathbf{S} |_p - \phi_{t*} \mathbf{S} |_p \right)$.

 

[Arman777]

Thanks a lot. These were helpful

For sake of an example, consider a $(1,1)$ tensor $\mathbf{T}$. The Lie derivative $L_{\mathbf{X}}$ satisfies Leibniz' rule, hence\begin{align*}
L_{\mathbf{X}} ( \mathbf{T} \otimes \mathbf{e}^a \otimes \mathbf{e}_b ) = L_{\mathbf{X}} \mathbf{T} \otimes \mathbf{e}^a \otimes \mathbf{e}_b + \mathbf{T} \otimes L_{\mathbf{X}} \mathbf{e}^a \otimes \mathbf{e}_b + \mathbf{T} \otimes \mathbf{e}^a \otimes L_{\mathbf{X}} \mathbf{e}_b
\end{align*}for arbitrary basis vectors $\mathbf{e}^a$ and $\mathbf{e}_b$. Now contract over all four positions; recall that $L_{\mathbf{X}}$ preserves contractions.\begin{align*}

(\dagger) \quad L_{\mathbf{X}} {T^a}_b &= {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_j \langle L_{\mathbf{X}} \mathbf{e}^a, \mathbf{e}_i \rangle \langle \mathbf{e}^j, \mathbf{e}_b \rangle + {T^{i}}_j \langle \mathbf{e}^a, \mathbf{e}_i \rangle \langle \mathbf{e}^j, L_{\mathbf{X}} \mathbf{e}_b \rangle \\

&= {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_b \langle L_{\mathbf{X}} \mathbf{e}^a, \mathbf{e}_i \rangle + {T^{a}}_j \langle \mathbf{e}^j, L_{\mathbf{X}} \mathbf{e}_b \rangle

\end{align*}By definition the Lie derivative acts on functions as $L_{\mathbf{X}} f = \mathbf{X}f$ hence in coordinate basis $L_{\mathbf{X}} {T^a}_b = X^i \dfrac{\partial {T^a}_b}{\partial x^i}$. Furthermore, for vectors $\mathbf{Y} \in T_p$ and covectors $\boldsymbol{\eta} \in T^*_p$ we have*\begin{align*}
(L_{\mathbf{X}} \mathbf{Y})^i &= \dfrac{\partial Y^i}{\partial x^j} X^j - \dfrac{\partial X^i}{\partial x^j} Y^j \\

(L_{\mathbf{X}} \boldsymbol{\eta})_i &= \dfrac{\partial \eta_i}{\partial x^j} X^j + \dfrac{\partial X^j}{\partial x^i} \eta_j
\end{align*}which imply that for a coordinate basis $\left(L_{\mathbf{X}} \dfrac{\partial}{\partial x^i} \right)^j = -\dfrac{\partial X^j}{\partial x^i}$ and $(L_{\mathbf{X}} dx^i)_j = \dfrac{\partial X^i}{\partial x^j}$. Returning to equation $(\dagger)$,\begin{align*}
X^i \dfrac{\partial {T^a}_b}{\partial x^i} &= {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_b \dfrac{\partial X^a}{\partial x^i} - {T^{a}}_j \dfrac{\partial X^j}{\partial x^b}
\end{align*}which is better re-written as \begin{align*}
{(L_{\mathbf{X}} \mathbf{T})^a}_b &= X^i \dfrac{\partial {T^a}_b}{\partial x^i} - {T^{i}}_b \dfrac{\partial X^a}{\partial x^i} + {T^{a}}_j \dfrac{\partial X^j}{\partial x^b} \\
\end{align*}The mnemonic is "minus the upper indices & plus the lower indices".

*the latter of these identities is deduced from the former. The former comes from the definition $L_{\mathbf{X}} \mathbf{S} |_p = \lim_{t \rightarrow 0} \frac{1}{t} \left( \mathbf{S} |_p - \phi_{t*} \mathbf{S} |_p \right)$.

 

[Arman777]

In all of these calculations $X$ is just a vector right ?

 

[Ibix]

A vector field, technically, since it needs values in a region for $\partial_i X^j$ to make sense, but yes.

 

[ergospherical]

Yeah, as @Ibix said $\mathbf{X}$ is a vector field on the manifold $M$, and for each $p \in M$ there is a curve $\gamma$ such that $\gamma(0) = p$ and $\gamma'(t) = \mathbf{X} |_{\gamma(t)}$. $\mathbf{X}$ generates a flow $\phi_t$ which takes any point $p$ a parameter distance $t$ along the integral curve through $p$. The difference between a tensor field $\mathbf{T} |_p$ evaluated at $p$ and its push-forward $\phi_{t*} \mathbf{T}|_p$ also evaluated at $p$ $-$ per unit parameter $t$ $-$ is essentially the Lie derivative.

 

[Arman777]

thanks

 

[Arman777]

As a side question, does the covariant derivative defined in 3D space ?

 

[Ibix]

It's defined in any dimension, although since 1d manifolds don't have curvature it's unexciting there.

 

[Arman777]

It's defined in any dimension, although since 1d manifolds don't have curvature it's unexciting there.

Okay thanks