Scalar product on manifold.

Question

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.

Answer 1

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 $$\omega \wedge \ast \eta = \langle \omega, \eta \rangle ~d\mathrm{vol}.$$

Answer 2

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 $\omega \wedge \ast \eta = \langle \omega, \eta \rangle d\mathrm{vol}$.

Answer 3

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:

$$ \left( \begin{array}{cc} A & B \\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\ B^t & D^t \end{array} \right). $$