antipodal map on Sn is homotopic to the identity if and only if n is odd
Lemma.
If X:Sn→Sn is a unit vector field, then
there is a homotopy between the antipodal map on Sn
and the identity map.
Proof.
Regard Sn as a subspace of Rn+1 and define
H:Sn×[0,1]→Rn+1 by
H(v,t)=(cosπt)v+(sinπt)X(v). Since X is a unit
vector field, X(v)⟂ for any . Hence
, so is into . Finally observe that
and . Thus is a homotopy between
the antipodal map and the identity map.
∎
Proposition.
The antipodal map is homotopic to the identity if and only if is odd.
Proof.
If is even, then the antipodal map is the composition
of an odd of reflections
. It
therefore has degree . Since the degree of the identity
map is , the two maps are not homotopic.
Now suppose is odd, say . Regard has a
subspace of . So each point of has
coordinates with . Define
a map by
,
pairwise swapping coordinates and negating the even coordinates.
By construction, for any , we have that
and . Hence is a unit vector field. Applying the
lemma, we conclude that the antipodal map is homotopic to the identity.
∎
References
- 1 Hatcher, A. Algebraic topology, Cambridge University Press, 2002.
- 2 Munkres, J. Elements of algebraic topology, Addison-Wesley, 1984.
Title | antipodal map on is homotopic to the identity if and only if is odd |
---|---|
Canonical name | AntipodalMapOnSnIsHomotopicToTheIdentityIfAndOnlyIfNIsOdd |
Date of creation | 2013-03-22 15:47:33 |
Last modified on | 2013-03-22 15:47:33 |
Owner | mps (409) |
Last modified by | mps (409) |
Numerical id | 5 |
Author | mps (409) |
Entry type | Derivation |
Classification | msc 51M05 |
Classification | msc 15-00 |