Finding the Degree of a Map on S^3 and its Homotopy with the Identity

[WWGD]

Hi: More on Prelims:

We have a map f: S^3 -->S^3 ; S^3 is the 3-sphere , given by:

(x1,x2,x3,x4)-->(-x2,-x3,-x4,-x1).

We're asked to find its degree, and to determine if f is homotopic to the identity.

I computed that f^4 ( i.e., fofofof ) is the identity, and we have that degree is

multiplicative, so that deg(f)^4=1 , so that we can narrow the choices to degf=+/- 1.

Now, I know we can also compute the induced map on top homology, and see if f is

preserving- or reversing- orientation, but I cannot tell which it is; I am trying to

use a 4-simplex , and see if this map preserves or reverses the orientation, but

I cannot see it clearly.

Another choice I am thinking of using is that f , seen as a map from R^4 to itself,

is a linear map, so that we can calculate Det f , to see if f reverses or preserves

orientation, and then maybe argue that f restricted to the subspace S^3 (unfortunately,

S^3 is not a subspace of R^4 ) has the same effect of preserving/reversing

orientation. Any Ideas/Suggestions/Comments?

see well how to do that

 

[zhentil]

You're on the right track: you calculate the determinant of the matrix in R^4, then you check whether it flips the sign on the radial vector field.

 

[Bacle]

I think you can also use this fact:

Given two maps f,g: S^n --->S^n

such that f(x)=/-g(x), then f,g are homotopic thru:

H(x,t)=[(1-t)f(x)+tg(x)]/|| (1-t)f(x)+tg(x) ||

So, your f, and Id are then homotopic and so degf=1