One way I've learned to understand Poisson geometry is to consider it as symplectic geometry with no open conditions – i.e. no condition of nondegeneracy. This idea can be applied to many other theories, including Kahler geometry, Riemannian geometry, Riemann-flat geometry, and hyperKahler geometry.
Of those, only symplectic geometry seems to lead to a deep and interesting theory. Is there a specific reason for that?
EDIT: I guess this comprises a couple of questions.
Question 1: Is there a specific reason other "degeneracy-generalization" theories don't seem to be studied much? Something that makes symplectic nicer than others?
Question 2: Is there some nice geometric or algebraic reason why symplectic geometry should lead to a dimension-independent Poisson geometry, and most other "nondegenerate" theories don't without losing some information? In particular, why no other integrable theories (at least among those checked) give dimension-independent equations?
Or generally: what is it about symplectic geometry that makes Poisson geometry "nice" that other geometries lack?
I've tried to work out "Poisson-Riemann" and "Poisson-Kahler" algebras, and none of them seem to work out as nicely as symplectic geometry. For example, "Poisson-Kahler algebras" can be made, but the relevant equations depend on the dimension in an obvious way (that is, there's an easy way given the dimension to write the equations). "Poisson-Riemann algebras" can be made in multiple ways; the easiest way has no good generalization of the curvature tensor, and the hardest way doesn't seem to work at all. "Flat Poisson-Riemann algebras" seem to make sense, but the equations seem to depend on dimension in a non-obvious way. Is this one reason that Poisson geometry is studied, and other aren't?
One part of the problem with other theories is that they lack a natural way to do a Lie bracket of vector fields coming from functions. As such, there are tensors that can be made using the Lie bracket that can't be generated by the "Riemann bracket" or "Kahler bracket" (generalizing the Poisson bracket). This contrasts with the fact that in Poisson geometry, the Lie bracket on vector fields from functions effectively is the Poisson bracket. Might this be related?
Some additional detail and calculations to demonstrate a couple of points from the above:
For a "Poisson-Riemann algebra" from a Riemannian metric, the easiest way to do it is to define a bracket $\lfloor f, g\rfloor = \langle \nabla f, \nabla g\rangle$. This gives no relations beyond symmetry and the usual tensor relation $\lfloor fg, h\rfloor = f \lfloor g, h\rfloor + g \lfloor f, h\rfloor$. This, similar to the usual symplectic geometry, is a dimension-independent definition, but doesn't include the curvature tensor.
If $f_1, f_2, f_3, f_4$ are functions and $X_1, X_2, X_3, X_4$ are their gradients, then $R(X_1, X_2, X_3, X_4)$ can be calculated:
$R(X_1, X_2, X_3, X_4) = \lfloor f_1, \lfloor f_3, \lfloor f_2, f_4\rfloor \rfloor \rfloor – \lfloor f_1, \lfloor f_4, \lfloor f_2, f_3\rfloor \rfloor \rfloor – \lfloor f_2, \lfloor f_3, \lfloor f_1, f_4\rfloor \rfloor \rfloor + \lfloor f_2, \lfloor f_4, \lfloor f_1, f_3\rfloor \rfloor \rfloor + \lfloor f_3, \lfloor f_1, \lfloor f_2, f_4\rfloor \rfloor \rfloor – \lfloor f_4, \lfloor f_1, \lfloor f_2, f_3\rfloor \rfloor \rfloor – \lfloor f_3, \lfloor f_2, \lfloor f_1, f_4\rfloor \rfloor \rfloor + \lfloor f_4, \lfloor f_2, \lfloor f_1, f_3\rfloor \rfloor \rfloor – \lfloor \lfloor f_1, f_3\rfloor, \lfloor f_2, f_4\rfloor\rfloor + \lfloor \lfloor f_1, f_4\rfloor, \lfloor f_2, f_3\rfloor\rfloor + 2 \langle [X_1, X_2], [X_3, X_4]\rangle – \langle [X_1, X_3], [X_4, X_2]\rangle – \langle [X_1, X_4], [X_2, X_3]\rangle$
The last three terms are key; they are the only terms that can't be directly expressed in terms of $\lfloor ,\rfloor$.
We could extend the algebra to include the curvature with a tensor $R(f_1, f_2, f_3, f_4)$, but we'd then need some relation so that $R$ is uniquely determined if $\lfloor ,\rfloor$ is nondegenerate. As far as I've calculated, there is no dimension-independent way to do this.
If we instead turn to making "Flat Poisson-Riemann algebras", then I think we shouldn't need any extra tensors (as extra tensors lead to local invariants, and flat Riemann manifolds have no local invariants). This should be the closest to being analogous to Poisson geometry, as both should be "Poisson generalizations" of integrable systems. But the relevant equations, unlike in the Poisson geometry setting, seem to depend on dimension (although I'm still working on the exact equations).
The situation is easier for "Poisson-Kahler algebras". A Kahler form gives rise to a (neither symmetric nor antisymmetric) bracket $\lceil f, g\rceil = \lfloor f, g\rfloor + i \{f, g\}$. The conditions on a $2n$-dimensional Kahler manifold are then:
(Pointwise condition): $\sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} \prod_{i = 1}^n \lceil f_i, g_{\sigma(i)} \rceil = 0$
If the structure is nondegenerate, this is equivalent to the division of the space into $n$-dimensional holomorphic and antiholomorphic parts.
(Integrability of the complex structure): Two equations, which can be gotten by each other by "switching sides" in the bracket.
$\sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} (\lceil f_1, \lceil f'_1, g_{\sigma(1)}\rceil \rceil – \lceil f'_1, \lceil f_1, g_{\sigma(1)}\rceil \rceil) \prod_{i = 2}^n \lceil f_i, g_{\sigma(i)}\rceil = 0$
In the case that the structure is nondegenerate, the two equations say that the bundles are integrable – the above equation says that the Lie bracket of two holomorphic vector fields is holomorphic, and its counterpart says the same for antiholomorphic.
(Integrability of the symplectic structure): $\lceil f, \lceil g, h \rceil \rceil – \lceil \lceil f, g\rceil, h\rceil + \lceil \lceil f, h\rceil, g\rceil – \lceil f, \lceil h, g\rceil + \lceil h, \lceil f, g \rceil \rceil – \lceil \lceil h, f\rceil, g\rceil + \lceil \lceil h, g\rceil, f\rceil – \lceil h, \lceil g, f\rceil + \lceil g, \lceil h, f \rceil \rceil – \lceil \lceil g, h\rceil, f\rceil + \lceil \lceil g, f\rceil, h\rceil – \lceil g, \lceil f, h\rceil = 0$
As far as I can tell, "Poisson-Kahler algebras" still have the problem of having no curvature tensor, but even without it the dimension-dependence seems clear.
Best Answer
Look up sub-Riemannian geometry. This is a quite developed theory now, but here the Riemannian metric is defined only along a subbundle of the tangent bundle which is assumed to be completely non-integrable (so that geodesics reach evewhere). You take the inverse of the Riemannian metric and assume that it may be degenerate.
For your curvature expression, see section 2 of
There is a nice relation between this curvature formula and O'Neill's formula for the behavior of curvatures under a Riemannian submersion: exactly the last 3 terms in your formula have further terms when compared to pullbacks; these are the O'Neill terms.