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Define an antipodal map A such that $$A: S^{n} \rightarrow S^{n}$$ by $A(x_{1},...,x_{n+1})=(-x_{1},...,-x_{n+1}).$ There are other posts on here proving that if n is odd, then A is homotopic to the identity on $S^{n}$ without concerning how the map A is defined. Will the logic change if my antipodal map A is defined as above?

I am new to the subject so any input will be greatly helpful.

Thank you!

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No, in fact the logic works because of some properties:

every map $f:S^n \to S^n$ induced on homology $f_*:H_n(S_n) \to H_n(S_n)$, but since these are just the group $\mathbb Z$, all homomorphisms look like $a \mapsto ka$. We can take $k$ to be the definition of degree. From this, some things follow:

  1. $deg(r)$ for any reflection about an axis has degree $-1$
  2. $deg(fg)=deg(f)deg(g)$.

Putting these together, and viewing the antipodal map as $n+1$ reflections, we see that $deg(A)=(-1)^{n+1}$.

This is exactly the map the way you define it, and you can see that the degree is $1$ when $n$ is odd, which is the key ingredient in other answers.


By the way, it would have worked to note that the map is $-I$, where $I$ is the identity matrix. This map is in $SO(n)$, the symmetry group of the sphere when $n$ is odd, and since it is path connected, you can always find a homotopy.

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  • $\begingroup$ thanks @SantanaAfton :) $\endgroup$ – Andres Mejia Dec 4 '17 at 17:28
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Your definition is the definition of the "antipodal map". If someone else refers to a map as the antipodal map, they mean exactly the map you are talking about. They may not explicitly state this definition (because they assume their readers know it), but they will use it at some point of their argument. So, to answer your question, no, nothing changes if you use your definition of $A$, because they already are using it (just without saying so).

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  • $\begingroup$ Yeah I maybe should have emphasized that in my answer as well... $\endgroup$ – Andres Mejia Dec 3 '17 at 5:24
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Any map $f: S^n \to S^n$ has an invariant called degree, and it turns out that two such maps are homotopic if and only if their degrees are equal. The degree of the identity map is 1, and the degree of the antipodal map is $(-1)^{n+1}$. So the antipodal map is homotopic to the identity map if and only if $n$ is odd. See Hatcher's Algebraic Topology, Section 2.2, for example.

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