So I have been mulling the following question over in my head for awhile now, and want to see if anyone else might have any ideas.  

Begin with $M$ a manifold and suppose that $M$ has an antipodal map $\alpha:M\rightarrow M$, i.e. $\forall m\in M$ one has: 


$\alpha(m)\neq m$  

$\alpha(\alpha(m)) = m$


Let $A^n$ be the usual antipodal map on $S^n$: $A^n(s) = -s$. What I am wondering is if there is always an embedding $e:M\hookrightarrow S^n$ (for some $n$, not necessarily of minimal dimension), such that $e\circ\alpha = A^n|\_{e(M)}$, that is that the antipodal map on the submanifold extends to the entire sphere.


One thing I suspect may be true is that when $M$ has more than one class of antipodal map, the necessary dimension of the sphere may depend on which class of maps I start with. 


As an example, if I start with $M = \mathbb{T}^2$ with coordinates $(\theta,\phi)$, then I can easily define two classes of antipodal maps: 


$\alpha\_1(\theta,\phi) = (\theta+\pi,\phi)$  

$\alpha\_2(\theta,\phi) = (\theta+\pi,\phi+\pi)$ 


I have been able to convince myself with mental pictures that I can embed $\mathbb{T}^2$ into $S^3$ such that $\alpha\_2$ coincides with $A^3$ (although I could even be mistaken about this), but I cannot convince myself that there is an embedding such that $\alpha\_1$ coincides with $A^3$, so perhaps for this class of antipodal map, one must embed in a higher dimensional sphere.


Anyone have any insights or know of any results in this direction?



When does an antipodal map on a manifold extend to the antipodal map on a spheres
Let me elaborate on my comment above. Suppose $M$ is a manifold equipped with a smooth $\Bbb Z\_2$ action that is also free. Then there is an equivariant smooth embedding $M \to S^j$, for some $j$, where we give the sphere the antipodal action. If we let $N$ be the orbit space of this action on $M$, then by Whithney's "easy" embedding theorem, we can
embed $N$ smoothly in $\Bbb RP^j$, where $j = 2\dim M + 1$. Now pass to double covers
To obtain an equivariant smooth embedding of $M$ in $S^j$.


This argument also works when $M$ is a manifold with free $S^1$-action, where we give the odd spheres the usual free action whose orbits give the complex projective spaces.


More generally, suppose $G$ is a finite group acting freely and smoothly on $M$. Suppose 
$V$ is a free orthogonal $G$-representation. Then the above can be generalized to show that there is an equivariant smooth embedding from $M$ into $S(jV)$ where the latter is the unit sphere of $j$-copies of $V$, $j$ large. 


