The well-known Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$ has explicit constructions involving the geometry of $C^2$ and intersections of complex lines with the $3$-sphere. They don't seem to generalize easily to "higher" Hopf maps from $S^3 \rightarrow S^2$ with Hopf invariant not equal to one. Are there any simple expressions for those maps?
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1$\begingroup$ The Seifert fibrations of the three-sphere with generic fiber a $(p,q)$ torus knot described in [(mathoverflow.net/questions/248116/… has Hopf invariant $pq$ and also has substantial geometrical meaning, so it might also be of interest to who asked the question. $\endgroup$– Lucas SecoCommented Aug 24, 2016 at 19:24
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2 Answers
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You can get them by precomposing with a degree $n$ map from $S^3$ to itself. In particular, this gives an interpretation in terms of the group structure: if $h:S^3 \to S^2$ is the Hopf map (which is just modding out by the subgroup $S^1=U(1)$ of $S^3=Sp(1)$, then a map of Hopf invariant n is given by $x \mapsto h(x^n)$, where $x^n$ is using the group multiplication on $S^3$.
answered Oct 28, 2009 at 19:49
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1$\begingroup$ I'd imagine we could equally well post-compose with a degree-n map from S^2 to S^2. $\endgroup$ Commented Oct 28, 2009 at 19:55
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1$\begingroup$ No, we couldn't compose with S<sup>2</sup>
\toS<sup>2</sup> map, that map isn't the fibration. $\endgroup$ Commented Oct 28, 2009 at 19:58 -
6$\begingroup$ That's not actually true--it is not true in general that a degree n map on S^k induces multiplication by k on the higher homotopy groups. Indeed, the Hopf element in \pi_3(S^2) can be written as the Whitehead product [i,i] of the identity i \in \pi_2(S^2). A degree n map will send this to [ni,ni]=n^2[i,i], not n[i,i]. $\endgroup$ Commented Oct 28, 2009 at 19:59
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Actually, yes, there is a construction involving complex projective line.
Consider all points (x1, x2, x3, x4) on a 3-sphere in the 4-dimensional space. Our goal is to map them to $S^2$ which is the same as $CP^1$
To do this, take a quaternion
$$x_1+x_{2}i+x_3j+x_4k$$
raise it to the $n$-th power (this is that group law on a 3-sphere) and decompose back into two complex numbers $z_1+z_2j$ . Now $z_i:z_i$ is a point of a complex projective line, that's it!
answered Oct 28, 2009 at 19:57
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$\begingroup$ This is the same as my answer, but writing down the group structure explicitly. $\endgroup$ Commented Oct 28, 2009 at 20:02