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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.

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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...


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.

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The article you reference is really fun to read. – Turion Dec 3 '13 at 19:25
I agree on your last remark. In fact, I'm actually looking for pairs of Poisson structures on $G$ and $G/H$ such that $G/H$ is a Poisson homogeneous space in some sense. – Turion Dec 10 '13 at 12:24
If your interest in Poisson homogeneous spaces is raised by Chari-Pressley you should read "Algebras of functions on quantum groups I" by Korogodski-Soibelman, AMS 1998. There you will find a nice introduction to Poisson homogeneous spaces. – Nicola Ciccoli Dec 10 '13 at 14:35

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 $0$.

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So you're implying that a homogeneous space should have a Poisson structure of constant rank? But is this what is happening on other homogeneous spaces? For example for $S^2$, we have a nonzero Poisson bivector on $SU(2)$ that vanishes on the diagonal $U(1)$ subgroup. (See Chari & Pressley, Quantum Groups, Example 1.2.5) – Turion Dec 2 '13 at 13:43
@Turion: Perhaps I don't understand what you mean by 'Poisson homogeneous space'. I took that to mean a Poisson structure that is at least locally homogeneous, i.e., any two points have neighborhoods on which the Poisson structures are equivalent. Such a Poisson structure will have necessarily have constant rank. Obviously, the $2$-sphere has homogeneous Poisson structures of (nonzero) constant rank; what I am arguing is that this cannot happen on the $4$-sphere. If you don't assume constant rank (and, hence, no homogeneity), then there certainly could be others. – Robert Bryant Dec 2 '13 at 13:48
a Poisson homogeneous space is the quotient space of two Poisson Lie groups where one is a Poisson subgroup of the other. See my changes to the question. But thanks anyways! – Turion Dec 2 '13 at 14:00
@Turion A Poisson homogeneous space is something much more general than the quotient of a Poisson--Lie group by a Poisson subgroup; if $\pi_G$ is the Poisson bivector on the group $G$ ans $\pi_M$ the one on the manifold it is characterized by: $\pi_M(g\cdot x)=x\cdot\pi_G(g)+g\cdot\pi_M(x)$ (where hopefully the notation is self-explaining...). In the case in which there exists $x$ s.t. $\pi_M(x)=0$ then $\pi_M$ is projected by $\pi_G$, but the fibre is a coisotropic (not Poisson) subgroup. – Nicola Ciccoli Dec 3 '13 at 18:41
@NicolaCiccoli, what definition are you using? I was following Chari and Pressley. Your case seems to be more general. – Turion Dec 3 '13 at 19:04

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