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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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up vote 13 down vote accepted

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.

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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
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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
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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
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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 S2 which is the same as CP1.

To do this, take a quaternion

x1 + x2i + x3j + x4k

raise it to the n-th power (this is that group law on a 3-sphere) and decompose back into two complex numbers z1 + z2j. Now zi:zi is a point of a complex projective line, that's it!

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