The Lie derivative $L_X T$ of any tensor $T$ along any vector field $X$ is defined directly and only from the underlying manifold structure. If, moreover, the manifold has a symmetric connection, (symmetric means zero torsion), then it is possible to express $\mathcal{L}_X T$ using this connection. Of course, the result is the same no matter which connection is used, (as long as it is symmetric).
Please note, accordingly, there is no meaning in writing $L_X$ and $\bar{L}_X$. This is wrong. The Lie derivative comes first, and then the expression in terms of a connection, not the other way around.
The Lie derivative has two (equivalent) definitions. A dynamical one and an algebraic one. It is very important to appreciate that these reflect two equally useful intuitions. I will not get into this here, (feel free to ask), but I will consider expressing the Lie derivative in terms of a symmetric connection.
If $\nabla$ is a symmetric connection, then for any vector fields $X,Y$
$$
L_XY = [X,Y] = \nabla_XY-\nabla_YX
$$
where the first equality is by definition, but the second equality means $\nabla$ is symmetric.
For $(0,s)$-tensor $T$, (think of the metric if you would like), $L_XT$ is also a $(0,s)$-tensor, (one says that $L_X$ is type preserving), and by definition
$$
L_XT(Y_1,\ldots,Y_s) = X(T(Y_1,\ldots,Y_s)) - \sum_i T(\ldots,[X,Y_i],\ldots)
$$
The first term is
$$
X(T(Y_1,\ldots,Y_s)) = (\nabla_X T)(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_XY_i,\ldots)
$$
By replacing in the definition of $L_XT$ and using the fact that $\nabla$ is symmetric, this yields the expression
$$
L_XT(Y_1,\ldots,Y_s) = \nabla_X T(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_{Y_i}X,\ldots)
$$
Please note this is not the definition of $L_XT$ but only a formula, using the connection $\nabla$. In other words, $L_XT$ does not depend on $\nabla$ but the righ-hand side does.
If $T$ is parallel, (this means $\nabla_X T = 0$ for any $X$),
$$
L_XT(Y_1,\ldots,Y_s) = \sum_i T(\ldots,\nabla_{Y_i}X,\ldots)
$$
You may apply this last formula to the Lie derivative of the Riemannian metric $T=g$, and use the Levi-Civita connection as $\nabla$, to get the ''elasticity tensor'' $L_Xg$ and understand the definition of Killing vector fields, etc.
-- Salem
Best Answer
Yes, and it has nothing to do with the metric tensor either.
The key point is that the flows have a commutative diagram, i.e $\Phi_t\circ \iota:S\to M$ is equal to $\iota\circ\tilde{\Phi}_t:S\to M$, or said differently, $\Phi_t|_S$ equals $\tilde{\Phi}_t$ (with an enlargement of the target). Ok I’m being a little sloppy because the flows might be incomplete so the domains of these maps are not necessarily all of $S$, but the idea stands: if you start at a point $p\in S$ and your initial velocity is $X_p\in T_pS$, then the integral curves of both vector fields $X$ and its restriction $\tilde{X}$ are the same curve with image lying in $S$ (you can view this as uniqueness of solutions to ODEs). With this out of the way, the naturality of pullback is an easy consequence of the ‘dynamic’ definition of Lie derivatives using flows: \begin{align} \iota^*(\mathcal{L}_XT)&=\iota^*\left(\frac{d}{dt}\bigg|_{t=0}\Phi_t^*T\right)\\ &=\frac{d}{dt}\bigg|_{t=0}\iota^*\Phi_t^*T\\ &= \frac{d}{dt}\bigg|_{t=0}(\Phi_t\circ\iota)^*T\\ &= \frac{d}{dt}\bigg|_{t=0}(\iota\circ\tilde{\Phi}_t)^*T\\ &= \frac{d}{dt}\bigg|_{t=0}\tilde{\Phi_t}^*(\iota^*T)\\ &=\mathcal{L}_{\tilde{X}}(\iota^*T). \end{align} In the second equal sign I was able to commute the derivative with $\iota^*$ since the derivative is a limit of difference quotients, and at each point $p\in S$, $\iota^*$ gives a continuous linear map from $T^0_k(T_pM)\to T^0_k(T_pS)$ (linearity is clear from the definition, while one way to argue continuity is that in finite dimensions, all linear maps are continuous). Due to continuity, we can pull the limit out, and due to linearity, it plays nicely with the difference quotient. All the other steps are either by definition, of basic properties of pullbacks, and the ‘commutativity’ of the flows as I mentioned above.
You can of course work things out in local coordinates if you wish: pick an adapted submanifold chart of $S$ in $M$, and note that $X$ restricting to a tangent vector field on $S$ means that some of its components vanish on $S$. Then, you compute both sides and check you get the same thing. This is more ‘down to earth’ and mechanical, but also more tedious.
Edit: Some generalities.
Since you asked and since I always wonder, and forget the precise details, here’s the generalization to arbitrary rank tensor fields. We start off with the linear algebraic case, and then go on to the manifold case.
Some remarks (mostly obvious):
Now that we understand the linear-algebra, we can globalize:
Some remarks again (many of which should be familiar/obvious):
Next, we have the following easy lemma:
To prove this, fix a point $x\in M$, and consider these two functions of $t$. At $t=0$ they are both equal to $f(x)$. Also, by differentiating with respect to $t$, and using $f$-relatedness of the vector fields, it follows that these are both integral curves of the vector field $Y$ (i.e satisfy the ODE $\psi’(t)=Y(\psi(t))$). So, by uniqueness of integral curves, they are equal.
Finally, we can prove the desired result about Lie-derivatives:
It’s actually essentially the same calculation as above, except that at the crucial step, rather than using the property of pullbacks $(ab)^*=b^*a^*$ (which is really just the chain rule), we simply use $f$-relatedness directly and the chain rule more explicitly. Again, one should really fix a point $x\in M$ and consider everything for small $|t|$ etc, but I don’t want to lug this around in the notation right now. So, here we go; let’s denote the flows of $X,Y$ by $\Phi,\Psi$ respectively. Then, \begin{align} (Tf)^{\otimes k}\circ \mathcal{L}_X\xi&=(Tf)^{\otimes k}\circ\left(\frac{d}{dt}\bigg|_{t=0}\Phi_t^*\xi\right)\\ &= (Tf)^{\otimes k}\circ\left(\frac{d}{dt}\bigg|_{t=0}(T\Phi_t)^{\otimes k}\circ\xi\circ(T\Phi_{-t})^{\otimes l}\right)\\ &=\frac{d}{dt}\bigg|_{t=0} T(f\circ \Phi_t)^{\otimes k}\circ\xi\circ (T\Phi_{-t})^{\otimes l}\\ &=\frac{d}{dt}\bigg|_{t=0}T(\Psi_t\circ f)^{\otimes k}\circ \xi\circ (T\Phi_{-t})^{\otimes l}\tag{$f$-related flows}\\ &=\frac{d}{dt}\bigg|_{t=0}(T\Psi_t)^{\otimes k}\circ (Tf)^{\otimes k}\circ\xi\circ (T\Phi_{-t})^{\otimes l}\\ &=\frac{d}{dt}\bigg|_{t=0}(T\Psi_t)^{\otimes k}\circ \eta\circ (Tf)^{\otimes l}\circ (T\Phi_{-t})^{\otimes l}\tag{$f$-related vector fields}\\ &=\frac{d}{dt}\bigg|_{t=0}(T\Psi_t)^{\otimes k}\circ \eta\circ T(f\circ \Phi_{-t})^{\otimes l} \\ &=\frac{d}{dt}\bigg|_{t=0}(T\Psi_t)^{\otimes k}\circ \eta\circ T(\Psi_{-t}\circ f)^{\otimes l}\tag{$f$-related flows}\\ &=\left(\frac{d}{dt}\bigg|_{t=0}\Psi_t^*\eta\right)\circ (Tf)^{\otimes l}\\ &=(\mathcal{L}_{Y}\eta)\circ(Tf)^{\otimes l}. \end{align} This completes the proof.
Once again, it’s all just chain rule, and using $f$-relatedness. Also, the reason why the time derivative commutes with the linear maps in question is the same as above. As a corollary, if you specialize to $f=\iota:S\hookrightarrow M$ being an immersion, then you get the case of restrictions of tensor fields to immersed submanifolds.