Welcome to MathOverflow

Question

3

1

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?

edited Oct 28 2009 at 20:03

Ilya Nikokoshev
9,019●5●30●73

asked Oct 28 2009 at 18:32

jc
4,152●9●24

Eric Wofsey · Accepted Answer · 2009-10-28 19:49:02Z UTC

10

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

Eric Wofsey
8,887●1●26●46

1

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 2009 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 2009 at 19:58

3

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 2009 at 19:59

Ilya Nikokoshev · Answer 2 · 2009-10-28 19:57:50Z UTC

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 z_i:z_i is a point of a complex projective line, that's it!

Construction of maps S^3 -> S^2 with arbitrary Hopf invariant?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.