As a preliminary, the tensor product is functorial, in the sense that it has a natural extension to linear maps. That is, if if we have linear maps between vector spaces $L_1:U_i\to V_1$ and $L_2:U_2\to V_2$, we may define $L_1\otimes L_2:U_1\otimes U_2\to V_1\otimes V_2$ as the unique linear extension of the expression $(L_1\otimes L_2)(u_1\otimes u_2)=L_1(u_1)\otimes L_2(u_2)$, and this product of maps several important properties, namely that the tensor product of isomorphisms is an isomorphism.
It's more convenient here to think of an inner product $g$ on a finite dimensional vector space $V$ as an isomorphism with the dual space $g:V\to V^*$ with inverse
$\tilde{g}:V^*\to V$. The functorality of tensor product ensures that there is a unique extension to controvariant tensor spaces that is compatible with tensor products, namely $g^{\otimes k}:V^{\otimes{k}}\to(V^*)^{\otimes{k}}$, which is also an isomorphism and thus an inner product. Taking appropriate tensor products of $g$ and $\tilde g$ allows one to construct similar isomorphisms on tensors of any signature. Going back to bilinear forms, this gives the proprty $g(v_1\otimes u_1,v_2\otimes u_2)=g(u_1,u_2)g(v_1,v_2)$.
It seems from the snippets you provide that Moroianu simply states this as a known result (usung $g$ for any such inner product, which I will also do), while Bleeker states the coordinate form, which can be derived without too much fuss. Letting $A$, $B$ be tensors of signature $(0,k)$,
$$\begin{align}
g(A,B)&=g\left(A_{i_1\dots i_k}e^{i_1}\otimes\dots\otimes e^{i_k},B_{j_1\dots j_k}e^{j_1}\otimes\dots\otimes e^{j_k}\right) \\
&=A_{i_1\dots i_k}B_{j_1\dots j_k}g\left(e^{i_1}\otimes\dots\otimes e^{i_k},e^{j_1}\otimes\dots\otimes e^{j_k}\right) \\
&=A_{i_1\dots i_k}B_{j_1\dots j_k}g(e^{i_1},e^{j_1})\dots g(e^{i_k},e^{j_k}) \\
&=g^{i_1 j_1}\dots g^{i_k j_k}A_{i_1\dots i_k}B_{j_1\dots j_k}
\end{align}$$
(the generalization to arbitrary signature is the same computation) The wedge product spaces are isomorphic to the spaces of alternating tensors, which are vector subspaces of the tensor spaces, so they inherit the inner product as a restriction of domain.
Another way of doing things is to construct the wedge product spaces independently. The wedge product is also functorial, so we can carry out exactly the same construction replacing $\otimes$ with $\wedge$, and obtain a similar family of inner products These will correspond up to a choice of normalization; in particular, the authors you cite chosen conventions such that the two differ by a factor of $k!$.
Yes, and it has nothing to do with the metric tensor either.
- Let $T$ be any $(0,k)$ tensor field on a smooth manifold $M$.
- Let $X$ be a vector field on $M$ which at points of $S$ is tangent to $S$; let us write $\tilde{X}=X|_S$ for the restriction to $S$.
- Let $\Phi_t$ denote the flow of $X$, and $\tilde{\Phi}_t$ the flow of $\tilde{X}$.
- Let $\iota:S\to M$ be the inclusion.
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.
Definition. ($L$-related tensors)
Let $V,W$ be finite-dimensional vector spaces over a field $\Bbb{F}$, $L:V\to W$ linear and let $\alpha,\beta$ be $(k,l)$ tensors over $V,W$ respectively. We say $\alpha,\beta$ are $L$-related if $L^{\otimes k}\circ\alpha=\beta\circ L^{\otimes l}$.
Some remarks (mostly obvious):
- Here, $k$ denotes the contravariance, $l$ the covariance. Such a tensor is by definition an element $\alpha\in V^{\otimes k}\otimes (V^{*})^{\otimes l}$. This space is naturally isomorphic to $V^{\otimes k}\otimes (V^{\otimes l})^*$ which itself is naturally isomorphic to $\text{Hom}(V^{\otimes l},V^{\otimes k})$. Hence, a $(k,l)$ tensor on $V$ can be viewed as a linear map $\alpha:V^{\otimes l}\to V^{\otimes k}$. Next, the linear map $L:V\to W$ induces linear maps on the tensor spaces $L^{\otimes j}:V^{\otimes j}\to W^{\otimes j}$ for all $j\geq 0$ (when $j=0$ we define the spaces to be the ground field $\Bbb{F}$ and the map to be the identity). You can of course rephrase all of this in terms of multilinear maps, but the above phrasing seems most efficient.
- Relatedness is ‘transitive’ in the sense that if $\alpha$ is $T$-related to $\beta$ and $\beta$ is $S$-related to $\gamma$, then $\alpha$ is $(S\circ T)$-related to $\gamma$.
- If $L$ is injective, then so is each $L^{\otimes j}$, so for any $\beta$, there is at most one $\alpha$ which is $L$-related to $\beta$.
- Dually, if $L$ is surjective, then so is each $L^{\otimes j}$ and so for each $\beta$, there is at most one $\alpha$ which is $L$-related to $\alpha$.
- If $L$ is an isomorphism, then for every $\alpha$, there exists a unique $\beta$ which is $L$-related (in which case we of course write $\beta=L_*\alpha$ or $\alpha=L^*\beta$).
- For a general $L$, if $l=0$ then for every $\alpha$ there exists an $L$-related $\beta$ (not necessarily unique… but $L^{\otimes k}\circ \alpha$ is one such, and we denote it as $L_*\alpha$). Similarly, if $k=0$ then for every $\beta$, there exists an $L$-related $\alpha$ (not necessarily unique, but $\beta\circ L^{\otimes l}$ is one such, and is denoted $L^*\beta$).
Now that we understand the linear-algebra, we can globalize:
Definition. ($f$-related tensor fields)
Let $M,N$ be smooth manifolds, $f:M\to N$ a smooth map, and let $\xi,\eta$ be $(k,l)$-tensor fields on $M,N$ respectively. We say $\xi,\eta$ are $f$-related if $(Tf)^{\otimes k}\circ\xi=\eta\circ(Tf)^{\otimes l}$.
Some remarks again (many of which should be familiar/obvious):
- In this definition, we could impose smoothness of $\xi,\eta$ if we wanted to, or not. Also, the case $(k,l)=(1,0)$, i.e for vector fields, reproduces the usual definition of $f$-related vector fields.
- A $(k,l)$ tensor field $\xi$ on $M$ can be thought of as a section of the tensor bundle $T^{k}_l(TM)$, or what amounts to the same thing, a vector bundle morphism over the base $M$, of $(TM)^{\otimes l}\to (TM)^{\otimes k}$. A similar story holds for $\eta$. The tangent map $Tf:TM\to TN$ over the base map $f$ induces in a natural way vector bundle maps $(Tf)^{\otimes j}:(TM)^{\otimes j}\to (TN)^{\otimes j}$ for all $j\geq 0$ (again, for $j=0$, this is the trivial line bundle, and the map is identified with $f$). So, essentially, we’re doing the linear algebra fiber-by-fiber, as expected.
- Relatedness is again ‘transitive’: if $\xi$ is $f$-related to $\eta$ and $\eta$ is $g$-related to $\zeta$ then $\xi$ is $(g\circ f)$-related to $\eta$ (due to the chain rule $T(g\circ f)=Tg\circ Tf$, and then by applying the linear algebraic result fiberwise).
- If $f$ is an immersion, then for any rough $(k,l)$ tensor field $\eta$ on $N$, there is at most one rough tensor field $\xi$ on $M$ which is $f$-related to $\eta$. This is by doing linear algebra fiberwise. If now $\eta$ is smooth and such a $\xi$ exists, then $\xi$ is necessarily smooth (this is a little more annoying to prove, but essentially follows by the inverse function theorem, in the form of the local immersion theorem, and some more finessing). In particular (with change of notation), if $\iota:S\to M$ is an immersion (in particular if $S\subset M$ is an embedded submanifold) and $T$ is a smooth $(k,l)$ tensor field on $M$ which at points of $S$ is tangent to $S$, then there is a unique smooth tensor field $\tilde{T}$ on $S$ which is $\iota$-related to $T$. More succinctly, the restriction of a smooth tensor field on $M$ which is tangent to an immersed submanifold is smooth.
- If $f$ is a surjective submersion, then for any rough $(k,l)$-tensor field $\xi$ on $M$, there is at most one rough $(k,l)$-tensor field $\eta$ on $N$ which is $f$-related to $\xi$ (do linear algebra fiberwise). Now, if $\xi$ is smooth and $\eta$ exists, then $\eta$ is necessarily smooth (again, this is by the inverse function theorem, in the guise of the local submersion theorem applied carefully).
- If $f$ is a diffeomorphism, then for every $\xi$, there is a unique $f$-related $\eta$, and vice-versa.
- Once again, if $f$ is a general map, then for any $(0,l)$ tensor field $\eta$ (i.e fully covariant) we can find a (not-necessarily unique) $f$-related $(0,l)$ tensor field $\xi$. One such is $\eta\circ (Tf)^{\otimes l}$, and this is denoted $f^*\eta$.
- In the case $(k,0)$, we can consider the map $(Tf)^{\otimes k}\circ \xi$, but here we have to be careful, as this is a vector bundle map over different base manifolds, and the base map is $f$. So, unless $f$ is invertible, we can’t get an actual $(k,0)$ tensor field on $N$.
Next, we have the following easy lemma:
Lemma. (Flows of $f$-related vector fields are $f$-related)
Let $M,N$ be smooth manifolds and $f:M\to N$ a smooth map, and let $X,Y$ be smooth $f$-related vector fields on $M,N$ respectively. Then, their respective flows $\Phi,\Psi$ are $f$-related in the sense that for all $(t,x)$ in the domain of the flow $\Phi$ of $X$, we have $f(\Phi_t(x))=\Psi_t(f(x))$.
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:
Theorem. ($f$-related fields have $f$-related Lie derivatives)
Let $M,N$ be smooth manifolds, $f:M\to N$ smooth, $X,Y$ smooth $f$-related vector fields, and $\xi,\eta$ smooth $f$-related $(k,l)$ tensor fields on $M,N$ respectively. Then, the Lie-derivatives $\mathcal{L}_X\xi,\mathcal{L}_Y\eta$ are smooth $f$-related $(k,l)$ tensor fields.
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.
Best Answer
By an equivalent definition, inner product (contraction) of two $(0,2)$ tensors $A_{ij}$ and $B_{ij}$ is as follows: $$\langle A,B\rangle= A_{i}{\ }^{j}B_{j}{\ }^{i}.$$ and it is also equal to $A^{i}{\ }_{j}B^{j}{\ }_{i}$. One can show that $$\langle A,B\rangle= A(e_i,e_j)B(e_i,e_j)$$ for arbitrary orthonormal frame $\{e_1,\dots,e_n\}$. By this we have
$$\langle df\otimes df,B\rangle=(df\otimes d f)(e_i,e_j)A(e_i,e_j)=df(e_i)df(e_j)A(e_i,e_j)=(e_if)(e_jf)A(e_i,e_j)=A((e_if)e_i,(e_jf)e_j)=A(\nabla f,\nabla f).$$
Note that in general by the definition of $df$ $$df(X)=X.f=\nabla_Xf$$ and $(e_if)e_i=\nabla f=\mathsf{grad} f=df^\sharp$. Now let $A=\mathcal{L}_Xh$.
Now can you prove
@Jesse Madnickclaim in comment? (Proof is same as above.) $$\langle \alpha \otimes \alpha, \beta \rangle = \beta(\alpha^\sharp, \alpha^\sharp)$$