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Take a Hermitian manifold $(M,I,g)$ where $I$ is the complex structure and $g$ is the Hermitian metric. The associated fundamental $2$-form $ g(\cdot,I(\cdot)) $ captures a lot of the information about the Hermitian geometry of the manifold.

For quaternionic manifolds an analogous situation giving a fundamental $4$-form.

By analogy there "should be" a fundamental $1$-form for a Riemannian manifold. Is this true? If not then why not?

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    $\begingroup$ For the Hermitian manifold $(M, I, g)$, the $2$-form $g(\cdot, I \cdot)$ defines the symplectic $2$-form $\omega$ on $M$. Some argue that Darboux Theorem proves $\omega$ contains zero local geometric information. Moreover, if a riemannian manifold $(M,g)$ had a canonical $1$-form $\alpha$, then it would equivalently have a canonical vector field $X$ defined by $g(\cdot, X)=\alpha(\cdot)$. Except riemannian manifolds do not have canonical vector fields, e.g. euclidean space has no canonical vector field except the constant zero vector field. $\endgroup$
    – JHM
    Commented Jan 24, 2021 at 14:29
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    $\begingroup$ Moreover you might compare Ricci-Levi Civita's memoir eudml.org/doc/157997, especially pp.162, where the nonexistence of first order differential invariants on Riemannian manifolds is established. This nonexistence is old, long forgotten result, and which plays interesting role in proving the nonexistence of any tensorial expression for Einstein's gravitational energy density $T_{00}$. . $\endgroup$
    – JHM
    Commented Jan 24, 2021 at 14:55
  • $\begingroup$ Compare with mathoverflow.net/questions/235358/… $\endgroup$
    – JHM
    Commented Jan 24, 2021 at 16:01

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