Construction of maps S^3 -> S^2 with arbitrary Hopf invariant? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:32:48Z http://mathoverflow.net/feeds/question/3107 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3107/construction-of-maps-s3-s2-with-arbitrary-hopf-invariant Construction of maps S^3 -> S^2 with arbitrary Hopf invariant? jc 2009-10-28T18:32:41Z 2009-10-28T20:03:57Z <p>The well-known Hopf fibration S^1 -> S^3 -> 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 -> S^2 with Hopf invariant not equal to one. Are there any simple expressions for those maps?</p> http://mathoverflow.net/questions/3107/construction-of-maps-s3-s2-with-arbitrary-hopf-invariant/3126#3126 Answer by Eric Wofsey for Construction of maps S^3 -> S^2 with arbitrary Hopf invariant? Eric Wofsey 2009-10-28T19:49:02Z 2009-10-28T19:49:02Z <p>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.</p> http://mathoverflow.net/questions/3107/construction-of-maps-s3-s2-with-arbitrary-hopf-invariant/3128#3128 Answer by Ilya Nikokoshev for Construction of maps S^3 -> S^2 with arbitrary Hopf invariant? Ilya Nikokoshev 2009-10-28T19:57:50Z 2009-10-28T19:57:50Z <p>Actually, yes, there is a construction involving complex projective line. </p> <p>Consider all points (x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>) on a 3-sphere in the 4-dimensional space. Our goal is to map them to <code>S</code><sup>2</sup> which is the same as <code>CP</code><sup>1</sup>. </p> <p>To do this, take a quaternion </p> <blockquote> <p><code>x</code><sub>1</sub> + <code>x</code><sub>2</sub><code>i</code> + <code>x</code><sub>3</sub><code>j</code> + <code>x</code><sub>4</sub><code>k</code></p> </blockquote> <p>raise it to the <code>n</code>-th power (this is that group law on a 3-sphere) and decompose back into two complex numbers z<sub>1</sub> + z<sub>2</sub><code>j</code>. Now <code>z</code><sub>i</sub><code>:z</code><sub>i</sub> is a point of a complex projective line, that's it!</p>