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Let $\mathbb{S}^3 \subset \mathbb{R}^4$ be the unit $3$-sphere. Is there a classification available for $3$-homogeneous polynomial, unit norm, vector fields on $\mathbb{S}^3$?

More explicitly, a $3$-homogeneous polynomial vector field on $\mathbb{S}^3$ is a restriction to $\mathbb{S}^3$ of a map $P:x \mapsto (P_1, P_2, P_3, P_4)$, where each $P_i$ is a homogeneous polynomial of degree $3$ in $x = (x_1, x_2, x_3, x_4)$, such that $\sum_{i = 1}^4 x_iP_i = 0$. It is of unit norm if $P$ maps $\mathbb{S}^3$ to $\mathbb{S}^3$ or equivalently if $P_1^2 + P_2^2 + P_3^2 + P_4^2 = 1$ when $|x| = 1$ (equivalently just $P_1^2 + P_2^2 + P_3^2 + P_4^2 = (x_1^2 + x_2^2 + x_3^2 + x_4^2)^3$).


Some obervations:

  • Any unit vector field has topological degree $1$ as a map $\mathbb{S}^3 \to \mathbb{S}^3$.
  • If $J$ is a complex structure on $\mathbb{R}^4$ (i.e. $J^2 = -1$ and $J^T = -J$), then $x \mapsto Jx$ is a unit norm vector field on $\mathbb{S}^3$.
  • There's a way to construct $3$-homogeneous vector fields. Given any unit vector fields $F$ and $G$ on $\mathbb{S}^3$, we may take: $$H:= -F.|x|^2 + 2\langle{F, G}\rangle.G$$ It is easy to see that $H$ is also a unit vector field (this construction is due to Wood); also if $F$ and $G$ are $1$-homogeneous, then $H$ is $3$-homogeneous. Choosing $F = J_Fx$ and $G = J_Gx$ for some complex structures $J_F, J_G$, it easy to see that we may create $H$ to be purely of degree $3$ (i.e. it is not a multiple of $|x|^2$ and another linear unit vector field). Is this the only way to create such vector fields?
  • There are other examples. Consider the Hopf map $\mathbb{S}^3 \to \mathbb{S}^2$ given by $$(w, x, y, z) \mapsto (2(wy + xz), 2(xy - wz), -w^2-x^2+y^2 + z^2)$$ It is easy to see that $\mathcal{H} = 2(wy + xz) V_1 + 2(xy - wz) V_2 + (-w^2 -x^2 + y^2 + z^2) V_3$, where $V_1 = Ix$, $V_2 = Jx$, $V_3 = Kx$ are pointwise orthonormal ($I, J, K$ are the quaternion matrices), satisfies the assumptions and cannot be reduced to a linear vector field. Moreover any quadratic map $\mathbb{S}^3 \to \mathbb{S}^2$ reduces to the Hopf map by rotations (by a book from Eells and Rotta).
  • Solutions of the Cartan-Munzner equations $|\nabla F|^2 = |x|^6$ give maps $\mathbb{S}^3 \to \mathbb{S}^3$.
  • See Homogeneous polynomial vector fields tangent to the unit sphere for the computation of the dimension of the space of homogeneous polynomial vector fields, which are not necessarily unit (in our case it is $45$).
  • I am also interested in the classification of unit $3$-homogeneous vector fields on other odd unit spheres, but $\mathbb{S}^3$ might be the simplest non-trivial case.
  • update: a small computation shows that the degree $1$ part $H_1$ (i.e. writing $H = H_1. |x|^2 + H_3$, where $H_{1, 3}$ are harmonic and $1,3$-homogeneous) of the Wood's vector field $H$ (see third point), necessarily comes from an anti-symmetric matrix; on the other hand the degree $1$ part the Hopf vector field $\mathcal{H}$ is not anti-symmetric. Hence $\mathcal{H}$ does not come from Wood's construction. This indicates that the spheres with Hopf fibrations $\mathbb{S}^3, \mathbb{S}^7, \mathbb{S}^{15}$ could be special, while in all other cases the vector fields should come from Wood's construction.
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