Scalar product on manifold. Let $M$ be a closed Riemannian manifold and $\omega$ and $\eta$ two differential forms of the same degree. Then one can consider $\int_M \omega \wedge *\eta$, where $*$ denotes the Hodge star operator. Can you tell me, why this defines a scalar product or at least where I can find a proof of this fact? In particular I would be very interested why this expression is symmetric in $\omega$ and $\eta$ and why it is positive definite.
 A: The Riemannian metric extends to a metric on all tensors of rank $(k,l)$, and in particular it extends to $p$-forms. The very definition of the Hodge star operator is that
$$
\omega \wedge \ast \eta = \langle \omega, \eta \rangle d\mathrm{vol}
$$
where $\langle \omega, \eta \rangle$ is the pairing on $p$-forms induced by the metric, and $d\mathrm{vol}$ is the volume form (or density) induced by the metric. So
$$
\int_M \omega \wedge \ast \eta = \int_M \langle \omega, \eta \rangle d\mathrm{vol}
$$
which is very obviously symmetric and positive definite.
If you are using some other definition of the Hodge star (for example using te $\epsilon$ symbol), then all you have to do is to check that it is equivalent to the condition $\omega \wedge \ast \eta = \langle \omega, \eta \rangle d\mathrm{vol}$.
A: Note: Here I use the following pointwise definition of the Hodge star $\ast$: On an orthonormal basis $\{e^1, \dots, e^n\}$ of $T^\ast_x M$, $$\ast(e^{i_1} \wedge \cdots \wedge e^{i_p}) = e^{j_1} \wedge \cdots \wedge e^{j_{n-p}}$$ for $j_1, \dots, j_{n-p}$ such that $(i_1, \dots, i_p, j_1, \dots, j_{n-p})$ is a positive permutation of $(1, 2, \dots, n)$.
At each point $x \in M$, $\Lambda^p T^\ast_x M$ has an inner product $\langle \cdot, \cdot \rangle$ induced by the Riemannian metric. I claim the following:

Lemma. For any $x \in M$ and $\omega, \eta \in \Lambda^p T^\ast_x M$, $$\omega \wedge \ast \eta = \langle \omega, \eta \rangle ~d\mathrm{vol}.$$

Proof. It suffices to verify this for an orthonormal basis $\{e^1, \dots, e^n\}$ of $T^\ast_x M$. We have that $$e^{i_1} \wedge \cdots \wedge e^{i_p} \wedge \ast(e^{i_1} \wedge \cdots \wedge e^{i_p}) = e^1 \wedge \cdots \wedge e^m = d\mathrm{vol}$$ and $$e^{i_1} \wedge \cdots \wedge e^{i_p} \wedge \ast(e^{j_1} \wedge \cdots \wedge e^{j_p} )= 0$$ if $i_k \neq j_k$ for some $1 \leq k \leq p$. Our claim follows. $~\Box$
From the above, we see that $$(\omega, \eta) = \int_M \omega \wedge \ast \eta = \int_M \langle \omega, \eta \rangle ~d\mathrm{vol},$$ from which the fact that $(\cdot, \cdot)$ is an inner product easily follows from the fact that the pointwise inner products $\langle \cdot, \cdot \rangle$ are inner products.
The Hodge star can be characterized as the unique linear map $$\ast: \Lambda^p T^\ast M \longrightarrow \Lambda^{n-p} T^\ast M$$ such that pointwise we have $$\omega \wedge \ast \eta = \langle \omega, \eta \rangle ~d\mathrm{vol}.$$ 
A: Low level tidbits from Frank Warner, Foundations of Differentiable Manifolds and Lie Groups, exercise 13 on page 79. One may also talk about the inner product on forms this way: homogeneous forms of different degrees have inner product zero. Then let
$$ \langle  w^1 \wedge  \cdots \wedge w^p,  v^1 \wedge  \cdots \wedge v^p   \rangle  = \det \langle w^i, v^j \rangle   $$ and then extend bilinearly to all of $\Lambda^p(T^\ast M).$ Showing that this is well defined is a nice little exercise. 
You have not exactly mentioned this, but the Hodge star is also an isometry between $\Lambda^p(T^\ast M)$ and $\Lambda^{n-p}(T^\ast M).$ This leads us to this, not widely known:
Given a square matrix $M \in SO_n$ decomposed as illustrated  with square blocks $A,D$ and rectangular blocks $B,C,$
$$M  = \left( \begin{array}{cc} 
A & B \\  
 C & D 
 \end{array} \right) ,$$
then   $\det A = \det D.$  It is worth working out how this is describing an isometry. Meanwhile, as a fact about matrices, it has a quick proof: 
$$ \left( \begin{array}{cc}  A & B \\  0 & I  \end{array} \right) 
\left( \begin{array}{cc}  A^t & C^t \\ B^t & D^t  \end{array} \right)   = 
\left( \begin{array}{cc}  I & 0 \\  B^t & D^t  \end{array} \right). $$
