Does the 4-sphere have a nonzero Poisson structure as a Poisson homogeneous space?

Question

It is known that the 4-sphere does not have a symplectic structure. However, it does admit Poisson structures, for example the zero Poisson structure, which is quite boring. Does it have other, more interesting Poisson structures? For example, are there Poisson structures on $\operatorname{Spin}(5)$ (or $SO(5)$) such that $SO(4)$ is a Poisson subgroup, such that $S^4$ is a Poisson homogeneous space?

edit: A Poisson homogeneous space $G/H$ is a quotient space of a Poisson Lie group $G$ by a Poisson Lie subgroup $H$. In this situation, $G/H$ inherits a Poisson bracket from $G$ by projection. I do not explicitly look for constant rank Poisson structure, as the below answer suggests. A definition is given for example in Chari & Pressley, "Quantum Groups", section 1.2.B.

Answer 1

Yes:

imho geometrically the most interesting one is obtained as a quotient $SO(5)/SO(4)$ where the Poisson stucture on $SO(5)$ is not the so-called standard one, but one determined by an element in the maximal torus (sometimes they are called twisted). This is the Poisson analogue of what is mentioned in Quantum symmetry groups of noncommutative spheres - Varilly, Joseph C. Commun.Math.Phys. 221 (2001) 511-523 (where only the quantum counterpart is developed), ie. the Poisson version of the Connes--Landi noncommutative 4--sphere. As I mentioned in the comment above $SO(4)$ is not a Poisson-Lie subgroup of a $SO(5)$ but only a coisotropic subgroup.

The symplectic foliation is very interesting. In fact you have a level function which is a Casimir, so that leaves are contained in the 3-dimensional spheres $t= const.$ (0-dim leaves when $t=0,1$). Inside such spheres 2-dimensional leaves correspond to the usual description of the 3-sphere as two solid tori glued together. Of course the rank drops down to zero in two copies of $S^1$.

I sort of got the impression this is the only way you obtain a Poisson homogeneous structure on $S^4$ starting from a compact Poisson-Lie group.

There is another interesting Poisson structure on $S^4$ coming from Poisson-Lie groups, not homogeneous one but as a double coset. There the symplectic foliation consists only of two leaves: a 0-dimensional one and a $4$--dimensional symplectomorphic to the standard one on $C^2$ (so a sort of "Poisson compactification"). But since I contributed to this maybe it is interesting only for me...

ADDED

About the notion of Poisson homogeneous spaces $G/H$ of a Poisson-Lie group $G$ one may consider:

1) $H$ is a Poisson-Lie subgroup of $G$ (def. of Chari-Pressley, indeed);

2) the projection $G:\to G/H$ is a Poisson map;

3) the action of $G$ on $G/H$ is both a homogeneous action and a Poisson action.

We have $1)\Rightarrow 2) \Rightarrow 3)$ but none of the arrows is reversible. $2)$ is equivalent to $H$ being a coisotropic subgroup of $G$ and is characterized by the fact that there exists a point in $G/H$ in which the Poisson bivector vanishes. Remark that in general the trivial $\pi=0$ Poisson structure on $G/H$ is not necessarily Poisson homogeneous, as you seem to assume, unless the Poisson-Lie structure on the whole $G$ is trivial.

Answer 2

If I understand your question correctly, the answer is that the zero Poisson structure on $S^4$ is the only homogeneous one. In fact, this is the only Poisson structure on $S^4$ of constant rank. To see this, note that, because the $4$-sphere does not admit an almost complex structure, it follows, in particular, that there cannot be a nondegenerate Poisson structure, i.e., one of maximum rank $4$. If there were one of rank $2$ everywhere, then the tangent bundle of $S^4$ would split as the direct sum of two (orientable) $2$-plane bundles, but this is impossible because the Euler class of the tangent bundle of $S^4$ is nonzero. Thus, the only constant rank Poisson structure on the $4$-sphere must have rank $0$.