Hopf invariant

  • Thread starter eok20
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Can someone give me some hints on how to prove the following statement: if [tex] f: S^3 \to S^2, g: S^3 \to S^3[/tex] then [tex]H(f\circ g) = deg~g H(f)[/tex] where H is the Hopf invariant and deg g is the degree of g. I'm pretty clueless on how to start and I don't see how to get the deg to come in since that has to do with 3-d cohomology but the Hopf invariant has to do with 2-d and 4-d cohomology.

Thanks.
 

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  • #3
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I don't know what definitions you are using, but with the definition of the Hopf invariant given at page 227-228 of Bott-Tu and then using the caracterisation of degree found at http://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping#Differential_topology, it is very simple.
Thanks-- I should have clarified. I'm interested in the algebraic topological definition and proof. Here, the Hopf invariant is the integer h such that [tex] \alpha \smile \alpha = h\beta[/tex] where [tex]\alpha[/tex] generates the 2-d cohomology and [tex]\beta[/tex] generates the 4-d cohomology of the mapping cone of [tex]f:S^3 \to S^2[/tex]. The degree of [tex]g:S^3\to S^3[/tex] is the integer d such that [tex] f_* \gamma = d\gamma[/tex] where [tex]\gamma[/tex] generates the 3-d homology of S^3.

Unfortunately, I have very little intuition/feel for algebraic topology and I much prefer the differential forms and differential topology analogs.
 

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