I am searching the polynomial functions from $S^3$ to $S^3$.

($S^3$ is the set of vectors $x$ in $\mathbb{R}^4$ such that $\|x\|=1$)

We say $g$ is a polynomial function from $S^3$ to $S^3$, if there exists $f_1,f_2,f_3,f_4 \in \mathbb{R}[X_1,X_2,X_3,X_4]$ and $f:\mathbb{R}^4 \rightarrow \mathbb{R}^4$, $(x_1,x_2,x_3,x_4) \mapsto (f_1(x_1,x_2,x_3,x_4),...,f_4(x_1,x_2,x_3,x_4))$ such that $g: S^3 \rightarrow S^3$, maps $(x_1,x_2,x_3,x_4)$ to $ f(x_1,x_2,x_3,x_4)$

We generate the set $F$ of functions from $S^3$ to $S^3$ by:

  • the constant functions belong to $F$,

  • the identity belongs to $F$,

  • the isometries belong to $F$,

  • if $f,g \in F$, $\overline{f} \in F$, and $f \circ g\in F$, and $ f \times g\in F$, where $\overline{z}$ is the conjuguate of $z$ in the quaternions, and $\times$ is the quaternions product.

Do all the polynomial functions from $S^3$ to $S^3$ belong to $F$ ?

We have identified $S^3$ with the set of quaternions $z$ such that $|z|=1$

Thanks in advance.

  • 4
    $\begingroup$ You don't need to add the isometries separately to $F$. Every isometry of $S^3$ is of the form $g(z) = pzq$ for $p,q\in S^3$ (if it is orientation preseriving) or $g(z) = p{\bar z}q$ for $p,q\in S^3$ (if it is not orientation preserving), so they are already made from your other allowable constructions, since you have added the identity and the constant functions and have closed it under composition and multiplication. $\endgroup$ – Robert Bryant Jan 4 '12 at 13:17
  • $\begingroup$ So, $F$ is the set of functions that can be written $\Pi_i p_i z^{n_i} q_i \overline{z_i}^{m_i}$ $\endgroup$ – user12806 Jan 4 '12 at 14:36
  • 2
    $\begingroup$ The question can be reformulated in such a way that it applies to any algebraic group $G$ over $k$: Let $G(X)$ be the group of all morphisms $X\to G$ of varieties over $k$. The question is, is $G(G)$ generated by $G(k)$ together with the single element $1:G\to G$? (This latter element is the identity, but in this context that is a problematic name for it!) I don't know the answer when $k=\mathbb R$ and $G$ is the quaternion group $S^3$. A related case would be $k=\mathbb C$ and $G=SL_2$. $\endgroup$ – Tom Goodwillie Jan 4 '12 at 22:46
  • $\begingroup$ Thanks. In the reformulation, how $z \mapsto \overline{z}$ is generated ? $\endgroup$ – user12806 Jan 5 '12 at 8:09
  • $\begingroup$ $\overline z=z^{-1}$. $\endgroup$ – Tom Goodwillie Jan 5 '12 at 11:01

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