Let $\langle\cdot,\cdot\rangle$ be the induced inner product on $\Lambda^k(T^*M)$ induced by $g$, scaled in such a way that for an orthonormal frame $\alpha^i$ of one forms on an open set $U\subset M$, we have that: $$ \{\alpha^{i_1}\wedge\cdots \wedge \alpha^{i_k}:i_1<\cdots<i_k\} $$ is an orthonormal frame for $\Lambda^k(T^*M)$.
I am trying to show that: $$ \langle\omega,\eta \rangle_{L^2}=\int_M\langle\omega,\eta\rangle\text{dvol}_g $$ is a non-degenerate inner product on the space of differential $k$ forms with compact support. My proof is as follows:
Blockquote We proceed by contradiction; suppose the $L^2$ inner product is degenerate, then there exists a $k$ form $\omega$ with compact support on $M$ such that: $$\begin{align*} \langle \omega,\eta\rangle_{L^2}=0 \end{align*}$$ for all $\eta\in \Omega^k(M)$ with compact support. Let $\text{supp }\omega=K\subset M$ for some compact set $K$, let $x\in \text{int }K$, and $U$ be an open neighborhood of $x$. Since $K$ is the closure of an open set in $M$, it follows that $U\subset K$. The open neighborhood $U$ then admits an orthornormal frame of $k$ forms: $$ \begin{align*} \{\alpha^{i_1}\wedge \cdots \wedge \alpha^{i_k}:i_1<\cdots <i_k\} \end{align*}$$ In this frame let: $$\begin{align*} \omega=\omega_{i_1\cdots i_k}\alpha^{i_1}\wedge\cdots \wedge \alpha^{i_k} \end{align*}$$ where $\omega_{i_1\cdots i_k}$ are smooth functions on $U$. There exists a positively oriented coordinate chart $\phi$ such that $\phi(x)=0$ and $\phi(U)$ is an open ball of radius $r$ in $\mathbb{R}^n$ centered at $0$. The closed ball $B^{r_0}$ of radius $r_0<r$ is then a nonempty compact subset of $\phi(U)$, and by continuity of $\phi^{-1}$, $L=\phi^{-1}(B^{r_0})\subset\ K$ is then a compact set in $M$. We construct a smooth bump function on $U$ by first defining the smooth function $f$ on $\phi(U)$ by: $$\begin{align*} f(x)=\begin{cases} \exp\left(\frac{r_0}{r_0-(x^1)^2-\cdots -(x^n)^2}\right)\text{ for }&(x^1)^2+\cdots+(x^n)^2<r_0\\ 0 &\text{otherwise} \end{cases} \end{align*}$$ $\phi^{*}f$ is then a smooth function on $U$, satisfying $\text{supp }\phi^{*}f=L$. This function can be smoothly extending to all of $M$, by defining: $$ \begin{align*} h(p)=\begin{cases} \phi^{*}f(p)\text{ for }&p\in U\\ 0&\text{otherwise} \end{cases} \end{align*} $$ Clearly, $\text{supp }h=L$ as well, hence we construct global $k$ forms with compact support equal to $L$ by: $$\begin{align*} \eta^{i_1\cdots i_k}=h\cdot\omega_{i_1\cdots i_k}\alpha^{i_1}\wedge\cdots \wedge \alpha^{i_k} \end{align*}$$ where there are is no implied summation in the line above. We then see that for all $i_1<\cdots<i_k$: $$\begin{align*} \langle \omega,\eta^{i_1\cdots i_k}\rangle_{L^2}=&\int_M\langle \omega,\eta^{i_1\cdots i_k} \rangle \text{dvol}_g\\ =&\int_{\phi(U)}\phi^{-1*}\left(h\cdot\omega_{j_1\cdots j_k}\omega_{i_1\cdots i_k}\langle \alpha^{j_1}\wedge \cdots \wedge \alpha^{j_k}, \alpha^{i_1}\cdots \alpha^{i_k}\rangle\text{dvol}_g\right)\\ =&\pm\int_{\phi(U)} \phi^{-1*}\left(h\cdot \omega_{i_1\cdots i_k}^2\text{dvol}_g\right) \end{align*}$$ where the sign depends on $\langle\cdot,\cdot \rangle$. We have that $h>0$ on $L$ by construction, and clearly $\omega_{i_1\cdots i_k}^2\geq 0$, for all $i_1<\cdots<i_k$. Furthermore, $\omega_{i_1\cdots i_k}$ can't be identically zero on $L$ for all $i_1<\cdots<i_k$ as $L\subset U\subset K$, thus for some ordered multi index $i_1,\cdots,i_k$: $$\begin{align*} \int_M\langle\omega,\eta^{i_1\cdots i_k} \rangle\text{dvol}_g=\int_{\phi(U)} \phi^{-1*}\left(h\cdot \omega_{i_1\cdots i_k}^2\text{dvol}_g\right)>0 \end{align*}$$ a contradiction, so $\langle \cdot,\cdot \rangle_{L^2}$ is a non degenerate inner product on $k$ forms on $M$ with compact support, as desired.
Is there an easier way to do this? Does this even work? I can't exploit positive definiteness of $\langle\cdot,\cdot\rangle$, as we can in the usual Riemannian case, so I didn't really see a clear path from the get go, I just started doing stuff until I got somewhere that made sense.