# Construction of maps $f:S^3 \to S^2$ with arbitrary Hopf invariant?

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?

• 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. – Lucas Seco Aug 24 '16 at 19:24

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 $$S^3$$.

• I'd imagine we could equally well post-compose with a degree-n map from S^2 to S^2. – Aaron Mazel-Gee Oct 28 '09 at 19:55
• No, we couldn't compose with S<sup>2</sup> \to S<sup>2</sup> map, that map isn't the fibration. – Ilya Nikokoshev Oct 28 '09 at 19:58
• 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]. – Eric Wofsey Oct 28 '09 at 19:59

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!

• This is the same as my answer, but writing down the group structure explicitly. – Eric Wofsey Oct 28 '09 at 20:02