Polynomial function from $S^3$ to $S^3$ and quaternions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:06:09Z http://mathoverflow.net/feeds/question/84877 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84877/polynomial-function-from-s3-to-s3-and-quaternions Polynomial function from $S^3$ to $S^3$ and quaternions francis-jamet 2012-01-04T10:52:24Z 2012-01-05T07:42:06Z <p>I am searching the polynomial functions from $S^3$ to $S^3$.</p> <p>($S^3$ is the set of vectors $x$ in $\mathbb{R}^4$ such that $\|x\|=1$)</p> <p>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 $f(x_1,x_2,x_3,x_4)$</p> <p>We generate the set $F$ of functions from $S^3$ to $S^3$ by:</p> <ul> <li><p>the constant functions belong to $F$,</p></li> <li><p>the identity belongs to $F$,</p></li> <li><p>the isometries belong to $F$,</p></li> <li><p>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.</p></li> </ul> <p>Do all the polynomial functions from $S^3$ to $S^3$ belong to $F$ ?</p> <p>We have identified $S^3$ with the set of quaternions $z$ such that $|z|=1$</p> <p>Thanks in advance.</p>