# The $L_2$ inner product of $k$ forms on pseudo riemannian manifold $(M,g)$ is non-degenerate

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: $$\{\alpha^{i_1}\wedge\cdots \wedge \alpha^{i_k}:i_1<\cdots is an orthonormal frame for $$\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 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 $$r_0 is then a nonempty compact subset of $$\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)$$ by: \begin{align*} f(x)=\begin{cases} \exp\left(\frac{r_0}{r_0-(x^1)^2-\cdots -(x^n)^2}\right)\text{ for }&(x^1)^2+\cdots+(x^n)^2 $$\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: \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 $$i_1<\cdots. Furthermore, $$\omega_{i_1\cdots i_k}$$ can't be identically zero on $$L$$ for all $$i_1<\cdots 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.

• Forget integrals, it is linear algebra problem: Does a lorentzian metric on a vector space define an inner product on the top exterior power? Start by working out the 2-dimensional case. Feb 7 at 1:58
• @MoisheKohan I mean that's clear, the induced inner product on $\otimes^nV$ is non degenerate, so it's restriction to the vector subspace $\Lambda^n(V)$ is non degenerate, right? Feb 7 at 2:10
• I am being loose with what I mean by an inner product. I technically mean a scalar valued bilinear symmetric non degenerate map. Feb 7 at 2:28
• I've seen both in the literature, but that's besides the point. Does my proof work to at least show nondegeneracy? Feb 7 at 2:32
• That’s not a counter example, the integral has to sum to zero for all differential forms, for the inner product to be degenerate. I should add that I believe this claim to be true as otherwise the only time one can obtain the Yang-Mills equation is when $M$ is riemannian, which is certainly not the case Feb 7 at 5:02

I didn’t fully read it, but yes that’s essentially it. Given a $$k$$-form $$\omega$$ such that $$\langle \omega,\cdot\rangle_{L^2}=0$$, you fix a point $$p\in M$$, and a sufficiently small open neighborhood $$U$$ of $$p$$, on which we have a pointwise ‘orthonormal’ frame of vector fields, which gives rise to 1-forms which gives rise to ‘orthonormal’ k-forms $$\{\alpha^{i_1}\wedge\cdots\wedge \alpha^{i_k}\}$$. Note that verifying that the wedges of the 1-forms are ‘orthonormal’ with respect to the induced pseudo-inner product on $$\Lambda^k(T^*M)$$ takes a bit more work in the non-Riemannian case… you can’t just say that oh because it’s a subspace of $$T^0_k(TM)$$ the restriction is non-degenerate (e.g in 2D Minkowski, restricting to the 45 degree lines gives the zero tensor which is as far as you can get from non-degeneracy). In any case, this is a linear algebra fact so do it on one vector space first.
From here, it’s a matter of reducing to the case of the classical ‘fundamental lemma of calculus of variations’. Fix an arbitrary smooth function $$f:M\to\Bbb{R}$$ with support that is compact and contained in $$U$$, and fix an increasing multindex $$I=(i_1,\dots, i_k)$$, and consider the $$k$$-form $$f\alpha^I\equiv f\alpha^{i_1}\wedge\cdots\wedge \alpha^{i_k}$$. Ok, very strictly speaking, I’m abusing notation slightly here, because $$f$$ is defined on all of $$M$$, while $$\alpha^I$$ is only defined on $$U$$, so the product is a-priori only defined on $$U$$; but note that since $$f$$ has compact support, I can define it to be $$0$$ outside $$U$$, and the result is a smooth $$k$$-form on $$M$$. Now, we compute: \begin{align} 0&=\langle\omega,f\alpha^I\rangle_{L^2} =\int_M\pm\omega_If\,dV_g =\int_U\pm\omega_If\,dV_g, \end{align} where the first equality is by assumption, the second by ‘orthonormality’ (the $$\pm$$ sign is given by $$\langle\alpha^I,\alpha^I\rangle_{\bigwedge^k(T^*M)}$$ (a constant function with value $$1$$ or $$-1$$ due to ‘orthonormality’)), and the last equal sign is because $$f$$ has support in $$U$$ . Since $$f$$ is an arbitrary smooth function with compact support in $$U$$, this is back to the classical situation, so we conclude that $$\omega_I=0$$. Since the multindex $$I$$ was arbitrary, it follows $$\omega=0$$ on $$U$$. Finally, since $$p$$ was arbitrary, it follows $$\omega=0$$ on $$M$$, proving the non-degeneracy.
What makes your proof slightly unpleasant to read is that you’re doing too many things simultaneously. For instance, you’re defining bump functions explicitly, and you’re trying to reprove the classical fundamental lemma of calculus of variations (this is a standard result, so if you want, prove that first, and then just invoke it here, but don’t do one proof inside of another inside of another). In fact, you can drop the smoothness assumptions and show that if $$\omega$$ is an $$L^2$$ differential $$k$$-form on $$M$$ such that $$\langle\omega,\cdot\rangle_{L^2}=0$$, then $$\omega=0$$ a.e on $$M$$ (with respect to the measure $$dV_g$$); the proof is virtually unchanged.
• @Chris you essentially did it in your special choice of $\eta^I$, which gave you the term $\omega_I^2$. Simplifying completely, if $\int_a^bfg=0$ for all $g$, then in particular for $g=f$, we get $\int_a^bf^2=0$, and hence $f=0$ a.e (or everywhere if you assume continuity). Next, you have to actually compute and show that the ordered wedges are indeed ‘orthonormal’; maybe you find this easy, but my point is that this is not as trivial as you made it sound in the comments. You can’t just say “by restriction” because you’re not dealing with positive-definite inner products. Feb 7 at 7:04
• Yes agreed, I had already shown this earlier with what I was doing, “by restriction” implies a mildly tedious calculation haha. Ok I see, so with this lemma I wouldn’t need to make that special choice of $\eta^I$? Feb 7 at 7:19