Let $X$ be a smooth compact Riemannian manifold of even dimension $2n$. Using the Hodge star $*: \Omega^r(X) \to \Omega^{2n-r}(X)$ one can define self-dual and anti-self-dual $n$-forms on $X$,
$$
\omega = * \omega \,, \quad \eta = – * \eta \,.
$$
A property of a non-trivial (anti-)self-dual form $\xi$ is that it must have positive (negative) square-integral $\int_X \xi \wedge \xi$,
\begin{align}
\int_X \omega \wedge \omega &\,=\, ~~\int_X \omega \wedge * \omega \,\equiv ~~(\omega , \omega ) > 0 \\
\int_X \eta \wedge \eta ~&\,=\, – \int_X \eta \wedge * \eta ~\, \equiv – (\eta , \eta) \,< 0 .
\end{align}
My question is whether the converse statement holds:
Can any smooth $n$-form with positive (negative) square-integral be made (anti-)self-dual by a choice of smooth metric?
Best Answer
The answer is no. To see why, note that every self-dual $n$-form actually satisfies a stronger pointwise condition: If $\omega$ is self-dual with respect to some Riemannian metric $g$ and choice of orientation, then $\omega \wedge \omega = \omega \wedge *\omega = \langle \omega,\omega\rangle_g dV_g$ everywhere on $X$, where $dV_g$ is the Riemannian volume form and $\langle \cdot,\cdot\rangle_g$ is the pointwise inner product on $n$-forms induced by the metric. Thus a necessary condition for $\omega$ to be self-dual with respect to some metric is that $\omega\wedge\omega$ must be a nonnegative function times a positively oriented volume form. (An analogous statement holds for anti-self-dual forms.)
To see that your original condition is not sufficient, suppose $X$ is a $2n$-dimensional oriented Riemannian manifold with $n$ even, let $\alpha$ be a nontrivial $n$-form such that $\alpha\wedge\alpha \equiv 0$ -- for example, you could choose $\alpha$ to be a wedge of basis $1$-forms in local coordinates, extended to a global form by multiplying by a bump function. It follows also that $$ *\alpha\wedge *\alpha = \langle *\alpha,\alpha\rangle_g\,dV_g = \langle \alpha,*\alpha\rangle_g\,dV_g = \alpha\wedge\alpha\equiv 0. $$ Let $f$ be an arbitrary real-valued $C^\infty$ function on $X$, and let $\omega = \alpha + f{*\alpha}$. Then $$\int_X\omega\wedge\omega = \int_X 2f\alpha\wedge *\alpha = \int_X 2f \langle\alpha,\alpha\rangle_g dV_g. $$ It is an easy matter to choose $f$ in such a way that it attains both positive and negative values, but such that the integral above is positive. The resulting $\omega$ satisfies your condition but is not self-dual with respect to any metric.
(The requirement that $n$ be even is so that $\alpha \wedge *\alpha = *\alpha\wedge\alpha$. If $n$ is odd, then $\omega\wedge\omega=0$ for every $n$-form $\omega$, and the entire question is moot.)
This suggests a more interesting question: On an oriented $2n$-manifold $X$, if $\omega$ is an $n$-form such that $\omega\wedge\omega$ is a nonnegative function times a positively oriented volume form, is there a Riemannian metric with respect to which $\omega$ is self-dual? I don't know the answer.
P.S. It seems to me that "norm" is an awfully inappropriate term for the quantity $\int_X \xi\wedge\xi$, because it's not positive or even nondegenerate, and there are many nontrivial $n$-forms for which it's zero. (Every decomposable form, for example.)