The lecturer here wants the viewer to derive the components of the Lie derivative of a (1,1) tensor-field. To this end, I want to derive the components of the Lie derivative of a covector field:
let $(U,x)$ be a chart, $\omega, \chi$ covector-fields and $X,Y$ vector fields on the smooth manifold $(M,\mathcal{O},\mathcal{A})$, I get:
I guess I have to use the fact, that one can view a co-vector as a (0,1) tensor. So for the simple case of the Lie derivative of a covector (field) $\omega$ I did:
$$
L_X \omega = L_X \left( T(Y)_i \right) {=}^{*}
\left( L_X T\right) (Y)_i + T(L_X Y)_i
$$
The equality with the star follows from the Leibniz rule from the definition of the Lie derivative.
But this is where I'm stuck, since I have no clue, how I would apply the Lie operator to the abstract tensor for the term $(L_X T) (Y)$.
Can someone give me a hint, if I'm on the right path, and how to deal with this term?
In the chart $\omega = \omega_i \text{d}x^i$ and for a (0,2) tensorfield I then get something like
$$
L_X(T(\omega,\chi)) = \omega_i \chi_j L_X (T(dx^i,dx^j)) =
$$
$$
= \bigg( L_X(T)(\omega_i dx^i, \chi_j dx^j)
+ T(L_X(\omega_i dx^i), \chi_j dx^j)
+ T(\omega_idx^i, L_X( \chi_j dx^j)) \bigg)
$$
again with the strange term in the front and the rest beeing single covector-Lie-derivatives.
Best Answer
(1) If $E_i$ is coordinate vector field and $E^i$ is coframe, and if $\phi_t$ is a flow of $X$, then lie derivative is defined as $$ L_X(S\otimes T) =\frac{d}{dt} \phi_t^\ast S\otimes \phi_t^\ast T = L_XS\otimes T + S\otimes L_X T$$
(If $T$ is a vector, the pull back is $(\phi_{-t})_\ast T_t$)
(2) If $T=T_{ij}E^i\otimes E^j$ we will do Lie derivative : In (2) we will do differentiate componentwise First Lie derivative of function is just differential $L_Xf=Xf$ And if $f_{,i}:=E_i(f)$, then $$ L_XE^j (E_i)=\frac{d}{dt}\bigg|_{t=0} \phi_t^\ast E^j(t)(E_i) = \frac{d}{dt} E^j (\phi_\ast E_i) = \frac{d}{dt} E^j( \phi_{,i}^k E_k(t)) = E_i X^j =X^j_{,i} $$ so that $$ L_XT_{ij}=XT_{ij} + T_{lj} X^l_{,i} + T_{il} X^l_{,j} $$